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WP-42F2 FINAL SUBMITTAL DESIGN DOCUMENTATION REPORT – APPENDIX F - STRUCTURAL Fargo-Moorhead Flood Risk Management Project In-Town Levees – 2nd Street/Downtown Area WP-42F2 Engineering and Design Phase May 6, 2016

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Page 1: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

WP-42F2

FINAL SUBMITTAL

DESIGN DOCUMENTATION REPORT –

APPENDIX F - STRUCTURAL

Fargo-Moorhead Flood Risk Management Project

In-Town Levees – 2nd Street/Downtown Area WP-42F2

Engineering and Design Phase

May 6, 2016

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Fargo-Moorhead Area Diversion Report

TABLE OF CONTENTS

Appendix F Structural

F.1 Floodwall Structures

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Fargo-Moorhead Area Diversion Report

APPENDIX F STRUCTURAL

F.1 Floodwall Structures

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3/18/2016

Design Documentation Report – Appendix F.1 Fargo-Moorhead Flood Risk Management Project

In-Town Levees – 2nd Street/Downtown Area Phase WP-42F.2

Engineering and Design Phase

WP-42F.2 95% SUBMITTAL

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Table of Contents

1. Introduction .......................................................................................................................................... 3

1.1 Purpose ......................................................................................................................................... 3

1.2 Floodwall ....................................................................................................................................... 3

2. Design Guidelines and Performance Objectives ................................................................................... 3

2.1 Design Standards .......................................................................................................................... 3

2.2 Hydraulic Loading Conditions ....................................................................................................... 4

2.3 Performance Objectives ................................................................................................................ 5

3. Parameters of Design ............................................................................................................................ 5

3.1 Material Properties ....................................................................................................................... 5

3.1.1 Soil Properties ....................................................................................................................... 5

3.1.2 Unit Weights ......................................................................................................................... 6

3.1.3 Reinforced Concrete ............................................................................................................. 6

3.2 Load Considerations ...................................................................................................................... 6

3.2.1 Vertical Live Load .................................................................................................................. 6

3.2.2 Live Load Surcharge .............................................................................................................. 6

3.2.3 Earth Load ............................................................................................................................. 7

3.2.4 Hydrostatic Load ................................................................................................................... 7

3.2.5 Uplift ..................................................................................................................................... 7

3.2.6 Bearing .................................................................................................................................. 7

3.2.7 Ice and Debris Load ............................................................................................................... 8

3.2.8 Seismic Load .......................................................................................................................... 8

3.2.9 Wind Load ............................................................................................................................. 8

3.3 Frost Protection ............................................................................................................................ 8

3.4 Safety ............................................................................................................................................ 8

4. Structural Analysis and Design .............................................................................................................. 8

4.1 Design Software ............................................................................................................................ 8

4.2 Stability Criteria ............................................................................................................................. 8

4.3 Load Cases ..................................................................................................................................... 9

4.3.1 Construction (Unusual)(Case 5) .......................................................................................... 10

4.3.2 Normal Low Water (Usual)(Case 4) ..................................................................................... 10

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4.3.3 Normal Low Water + Ice (Unusual)(Case 3) ........................................................................ 10

4.3.4 100 yr Flood + Ice (Unusual)(Case 2) .................................................................................. 10

4.3.5 Water at Top of Structure (See Note - Table 5)(Case 1) ..................................................... 10

4.3.6 Global Stability .................................................................................................................... 10

4.4 Concrete Design .......................................................................................................................... 11

4.4.1 General ................................................................................................................................ 11

4.4.2 Minimum Concrete Cover ................................................................................................... 12

4.4.3 Minimum Shrinkage and Temperature Reinforcing ........................................................... 12

4.4.4 Concrete Wall Thickness ..................................................................................................... 12

4.4.5 Lap Splices and Development Lengths ................................................................................ 12

4.5 Joints & Waterstops .................................................................................................................... 13

Tables

Table 1 - Hydraulic Conditions ...................................................................................................................... 4

Table 2 - Load Categories to Satisfy Performance Requirements (Category 1 Structures) .......................... 4

Table 3 - Required Factors of Safety ............................................................................................................. 5

Table 4 - Revised Table F-2: Concrete Design Load Factors .......................................................................... 9

Table 5 - Revised Table F-5: Floodwall Load Cases ....................................................................................... 9

Table 6 - Global Stability Factors of Safety ................................................................................................. 10

Table 7 - Splice and Development Lengths ................................................................................................. 12

Table 8 - Joint Spacing vs Minimum Reinforcement ................................................................................... 13

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1. Introduction

1.1 Purpose

This section outlines the structural design criteria and methods of analysis for the design of the 2nd St. S.

floodwall being included as part of project WP-42A.3. This project includes one section of floodwall.

1.2 Floodwall

The design includes the final design of one section of floodwall as part of project WP-42A.3 located

directly north of 2nd Street South between Main Avenue and 4th Street South. The geometrical

properties of the flood wall section are shown below:

o Floodwall Section A:

• Stem Height 12 ft

• Stem Thickness 15 in

• Footing Width 17 ft

• Footing Thickness 24 in

• Key Depth 2 ft

2. Design Guidelines and Performance Objectives

2.1 Design Standards

The following manuals and publications were used as reference for the design of the floodwall and

removable floodwall closure structures:

• EC 1110-2-6066 Design of I-walls (April 2011)

• EM 385-1-1 Safety and Health Requirements (September 2008)

• EM 1110-2-1612 Ice Engineering (October 2002)

• EM 1110-2-2100 Stability Analysis of Concrete Structures (December 2005)

• EM 1110-2-2102 Waterstops and Other Preformed Joint Materials for Civil Works Structures

(September 1995)

• EM 1110-2-2104 Strength Design for Reinforced-Concrete Hydraulic Structures (June 1992)

• EM 1110-2-2502 Retaining and Flood Walls (September 1989)

• EM 1110-2-2504 Design of Sheet Pile Walls (March 1994)

• EM 1110-2-2705 Design of Closure Structures for Local Flood Protection Projects (March 1994)

• ACI 318-11, Building Code Requirements for Structural Concrete

• Appendix F – Hydraulic Structures Design Guidelines. Part of the Fargo-Moorhead Metropolitan

Area Flood Risk Management Project Design Guidelines.

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2.2 Hydraulic Loading Conditions

Based on coordination efforts between Houston Engineering and the US Army Corp of Engineers,

clarification of the design requirements for the Fargo-Moorhead Flood Risk Management Project was

developed. As part of that clarification, it was decided that all new structures being built in the FM area

are to be constructed to the Pre-Diversion (Category 1) river levels. The required design river levels for

multiple return period events are shown in Table 1.

Clarification was also provided on the return periods used to define the Usual, Unusual, and Extreme

loading conditions. The results are summarized in Table 2 which replaces Table 3-1 of EM 1110-2-2100

Stability Analysis of Concrete Structures.

Both Table 1 and Table 2 were presented to USACE and agreed upon as an amendment to the Appendix

F – Hydraulic Structure Design Guidelines.

Table 1 - Hydraulic Conditions

Hydraulic Conditions

Return Period FMMetro - Existing

(Pre-Diversion – Category 1)

FMMetro Future Condition

(Post-Diversion – Category 2)

10YR 35.0 35.0

50YR 40.4 35.0

100YR 42.1 35.0

300YR 44.9 38.3

500YR 46.3 40.0

750YR 47.0 41.1

Table 2 - Load Categories to Satisfy Performance Requirements (Category 1 Structures)

Load Condition

Categories

Return Period Annual

Exceedance

Probability

River Stage*

Usual 10 Year Event 0.1 35.0

Unusual 750 Year Event or Top of

Wall (whichever is lower)

0.01 47.0

Extreme Top of Structure Not Applicable**

*River Stage water elevations are determined using the FMMetro Existing Conditions,

with protection (Pre Diversion) return period elevations.

**Top of wall is below 750 year event for this project.

The following flood river stage elevations were developed and provided by Houston Engineering Inc:

Datum: NAVD 88

• Q10 = 897.23

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• Q100 = 904.21

• Q750 = 910.00

2.3 Performance Objectives

For this project, performance objectives were based on EC 1110-2-6066 which was developed from the

lessons learned from the performance of New Orleans flood risk reduction system in Hurricane Katrina.

This document is for I-walls, but the performance requirements are applicable to all USACE flood risk

reduction structures. The goal for performance was that normal load events have very high reliability

and very little or no movement. The structure would be expected to withstand less frequent events with

minimal permanent deformation. And for events that are possible but have very little probability of

occurrence, the structure is expected to survive but may deform and may require rehabilitation after

the event. Only the most extreme events fall in the last category.

Table 3 is adopted from Table 6-1 of EC 1110-2-6066, and provides load categories intended to provide

the performance objectives outlined in the previous paragraph. This table was used in place of Table 3-

2, Table 3-4, and Table 3-5 of EM 1110-2-2100. This table was agreed upon between Houston

Engineering and the USACE. In addition, all hydraulic structures for this project were considered Critical

structures with Ordinary Site Information for selecting criteria requirements from EM 1110-2-2100.

Table 3 - Required Factors of Safety

Usual Unusual Extreme

Sliding 2.0 1.5 1.1

Bearing 3.5 3.0 2.0

Overturning 100% of Base in Comp. 75% of Base in Comp. Resultant Within Base

Flotation 1.3 1.2 1.1

3. Parameters of Design The following are the parameters used in the design of the floodwall. These form the basis of design for

this structure and are included here for information purposes.

3.1 Material Properties

Below is a description of all material properties used during the design of the floodwall.

3.1.1 Soil Properties

The following soil parameters were provided by Braun Intertec in the geotechnical report dated May 8,

2014. Soil properties for the clay foundations and backfills are shown below.

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• Moist Unit Weight (γm) 118 pcf

• Saturated Unit Weight (γs) 55.5 pcf

• Friction Angle (θ) 27 degrees draing

0 degrees undrained

• Cohesion (c) 0 psf drained

1000 psf undrained

Areas around the floodwall that will have a proposed levee will first be pre-consolidated in the previous

phases of the project with a surcharge loading, so settlement around the wall will be minimal. The

geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches.

3.1.2 Unit Weights

For the design, the following unit weights were used:

• Water 62.5 pcf

• Reinforced Concrete 150 pcf

• Steel 490 pcf

• Moist Soil 118 pcf

• Saturated Soil 55.5 pcf

3.1.3 Reinforced Concrete

An ultimate concrete strength (f’c) of 4,500 psi and a reinforcing steel yield stress (Fy) of 60,000 psi were

used for design in accordance with Paragraph F.4.1.1 of Appendix F. The minimum clear cover of

reinforcement is in accordance to paragraph 2-6 of EM 1110-2-2104 and is specified in Section 4.4.2 and

is from the nominal surface of the concrete (excluding architectural surface). Reinforced concrete for

the structures was designed according to ACI 318-11 and the EM 1110-2-2104, Strength Design for

Reinforced-Concrete Hydraulic Structures. The specifics of reinforced concrete design are summarized in

Section 4.4 of this Appendix.

3.2 Load Considerations

This section discuss the various required loading scenarios for the floodwall structures. This section is

not meant to be an in depth discussion of all the loading considerations, but rather a brief synopsis of

the different loads being considered. For a summary of loading for each individual load case see section

4.3 of this Appendix.

3.2.1 Vertical Live Load

Vertical live loads were not considered in the design of floodwalls. Based on direction from the USACE,

vertical live loads directly to the top of the wall should not be used when flooding conditions are

present.

3.2.2 Live Load Surcharge

Section F.8.2.1 of Appendix F – Hydraulic Structures Design Guidelines states that a design surcharge

load of 250 psf shall be applied to the soil next the structures for construction and heavy truck loads.

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The 250 psf surcharge will be applied to the top of the footing in the construction load case to account

for compaction equipment above the structure and placed to produce the most unfavorable event. The

250 psf surcharge load was applied as a vertical load above the footing of the floodwall to produce the

most unfavorable event.

3.2.3 Earth Load

Lateral and vertical soil loads were computed and applied in accordance with EM 1110-2-2502 for

shallow or pile founded concrete structures. Because very little movement or rotation is anticipated, at-

rest pressures were applied to the structures per EM 1110-2-2100. However, in sliding analysis of the

floodwall footing, in accordance with EM 1110-2-2502 and further USACE guidance, active and passive

pressures can be used. The following formula was used to calculate K0:

�� = 1 − sin

Where = friction angle of soil

3.2.4 Hydrostatic Load

Hydrostatic lateral and vertical pressures were applied to the structure based on the assumed water

level for each load case (see Section 4.3) at a magnitude of 62.5 psf per foot depth.

• Saturated Soil = Usual event (the 10yr event is lower than the ground level, therefore,

the usual event will be a heavy rain event in which the soil is saturated, but the water

level has not risen)

• Loading to top of wall = Unusual (750 yr event is higher than the top of wall)

• Per request of the City of Fargo Design top of wall = Constructed top of Wall + 1’-0”

• Extreme = N/A (Top of wall is below 750 year event)

3.2.5 Uplift

This structure will commonly be exposed to uplift forces from the soil water table. These loads are due

to seepage of water (during flood events) under the structure assuming a crack exists in the soil on the

wet side of the structure. The uplift caused by seepage causes a more unfavorable event for the

floodwall and therefore, uplift pressures are used. Uplift forces were used in global stability analysis of

the structure (Overturning, Sliding, and Bearing) as well as concrete and reinforcement design of the

structure. Service loads are used in the global stability analysis of the structure. All structures meet the

minimum factors of safety specified in Table 3.

3.2.6 Bearing

Bearing calculations were completed using USACE methodology in the design spreadsheet. Bearing

calculations were done using the same methodology as EM 1110-2-2502. Braun Intertec provided a

maximum allowable bearing capacity of the soil of 9,000 psf., however, this capacity was only used for a

general capacity check. All applicable dead and live loads were included in calculating the maximum

bearing. Service loads were used when calculating bearing pressure exerted by the footing of the

structures. All structures meet the minimum factors of safety specified in Table 3.

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3.2.7 Ice and Debris Load

Ice and debris forces were considered for the floodwall. Ice forces shall be computed according to EM

1110-2-1612 for all structures subject to moving water flow and were designed for moving ice and

debris forces of 500 lb/ft. All concrete is above the 10 year event, therefore higher ice loads due to

crushing were not included (See Appendix F).

3.2.8 Seismic Load

Fargo/Moorhead does not experience major seismic activity. Lateral seismic forces would be very small

and normal water levels are low so hydrodynamic forces are small and any seismic loading would be

negligible for design. Therefore, seismic design was not considered.

3.2.9 Wind Load

Wind loads were considered and can act anytime in the life of the flood wall. In the Fargo-Moorhead

area 30 lb/sqft was used above the water level in accordance with EM 1110-2-2502.

3.3 Frost Protection

Section F.7.4 of Appendix F states that all hydraulic structures shall be founded below the design frost

depth and that the minimum frost depth of foundations is 6 feet below ground surface. In order to make

sure this requirement is met, Houston Engineering assumed 5’ of fill over the top of the footing with a

minimum footing thickness of 18”.

3.4 Safety

All Structures were designed to provide operator and public safety in conformance with EM 385-1-1 and

applicable codes. The top of the floodwall surface is to be designed to prevent the public from easy

access.

4. Structural Analysis and Design

4.1 Design Software

The following software programs were utilized during the design of the floodwall and removable closure

structures:

• Microsoft Excel

• CTWall

• IES QuickRWall 3.0

4.2 Stability Criteria

Global stability of the structure was designed using USACE requirements noted in Table 3. Design of the

floodwall was completed using an internal design spreadsheet. The design spreadsheet uses USACE

methodology mainly obtained from USACE documents EM’s 1110-2-2100, 1110-2-2104, 1110-2-2502,

and EC 1110-2-6066. The spreadsheet analyzes global stability for the entered geometry and reports

factors of safety for Overturning, Sliding, and Bearing. The geometry of the section was optimized to

provide factors of safety that met USACE standards but were not overly conservative. Once a geometry

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for the floodwall section was obtained, the section was entered in the USACE design program CTWALL

to verify results. Our internal design spreadsheet yielded a slightly more conservative design than

CTWALL. Houston Engineering was informed that the windows 7 version of CTWALL incorrectly applies

entered horizontal loads to the floodwall, so the DOS based version was used on all load cases except

the Unusual water to top of wall case since no other entered horizontal loads are applied (The only

horizontal loads are water and soil). All cases except this case have horizontal loads applied to the wall

(wind or ice). After confirming that the wall geometry passed USACE design criteria, the thickness of the

footing and stem of the wall were verified to surpass shear loading requirements using our design

spreadsheet. If the footing or stem had to change thickness, the global stability was once again checked

using both our internal design spreadsheet and CTWALL.

4.3 Load Cases

Reinforced concrete was designed according to EM 1110-2-2104, Strength Design for Reinforced

Concrete Hydraulic Structures using the single load factor method specified in the EM. Overstress

factors are permitted for Unusual and Extreme cases as shown in Table 4. The overstress factors are

consistent with Appendix F Table F-2. For the Usual case only, the hydraulic factor of 1.3 may be

neglected for shear strength design.

Table 4 - Revised Table F-2: Concrete Design Load Factors

Load

Factor

Hydraulic

Factor

Overstress

Factor

Net

Factor

Usual 1.7 1.3 1 2.21

Unusual 1.7 1.3 0.33 1.7

Extreme 1.7 1.3 0.75 1.3

General Load Cases for design are as shown in Table 5. This table was developed during coordination

efforts with the US Army Corps of Engineers and shall supersede Table F-4 of Appendix F – Hydraulic

Structure Design Guidelines.

Table 5 - Revised Table F-5: Floodwall Load Cases

Load Case Type Category 1

1. Construction* Unusual RS 0

2. Normal Low water (10 yr event) Usual RS 35.0

3. Normal Low water + ice (10 yr

event) (special circumstances only)

Unusual RS 35.0

4. 100 yr Flood + Ice Unusual RS 42.1

5. Water at Top of Structure** See

Note

See Note

*Construction load case will be evaluated without hydrostatic loading.

**Unusual if top of structure is below the 750 year event level. Extreme if top of

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structure is above the 750 year event level. If extreme is above 750 year level, then

unusual at 750 year level will also be analyzed in addition to extreme to top of

structure.

Below is discussion of each load case found in Table 5 and their application to the floodwall structure. As

discussed below, some of the load cases were omitted from analysis.

4.3.1 Construction (Unusual)(Case 1)

A construction load case check for loading per EM 2502 was performed. Loadings assumed moist soils.

4.3.2 Normal Low Water (Usual)(Case 2)

For the design of the Category 1 Structures, the pre-diversion 10 year event (35’) for usual loading

condition will be applied. Since the 10 year event is lower than the soil elevation, the usual loading case

assumes a rain event in which the soils will be saturated, but the water level has not risen.

4.3.3 Normal Low Water + Ice (Unusual)(Case 3)

For the design of the Category 1 Structures, a 500 lb/ft load will be applied at the pre-diversion normal

10 year event (35’) for the unusual loading condition. By observation, this load case will not control as

ice is also applied at the 100 year event. The 100 year event will produce more unfavorable results than

the 10 year event which is below the soil elevation. Therefore, this load case was not included in the

results.

4.3.4 100 yr Flood + Ice (Unusual)(Case 4)

For the design of the Category 1 Structures, a 500 lb/ft load will be applied at the pre-diversion 100 year

event (42.1’) for the unusual loading condition. No ice loading will be applied for unusual or extreme

events higher than the 100 year.

4.3.5 Water at Top of Structure (See Note - Table 5)(Case 5)

Since the 750 year event is above the top of the structure, water at the top of the structure will be an

unusual loading condition. No ice loads will be applied at this level.

4.3.6 Global Stability

The Global Stability results for each load case analyzed is listed below. The factors of safety presented in

each table are from the internal floodwall design spreadsheets. The design spreadsheets, CTWall

models, and QuickRWall models for each load case are provided after this report. When using passive

pressures of soil for sliding, in some cases, the loading from the dry side got large enough where there

was more force applied from the dry side due to passive pressures. In these situations, the design

spreadsheet produced a negative factor of safety in which case the factors of safety from CTWall were

reported. This is the case for the wall section for the unusual construction case, as well as the usual

saturated soil case. In both of these cases, CTWall produced a factor of safety that was over 100. The

factors of safety for both overturning and bearing are still given from the internal design spreadsheet.

Table 6 - Global Stability Factors of Safety

Floodwall Unusual Loading Results (Water to top of wall)

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Required Wall Section A

Sliding FOS 1.5 1.58

Overturning Base in

Compression

75% 96%

Bearing FOS 3 3.32

Floodwall Unusual Loading Results (Water to 100 yr event

plus ice)

Required Wall Section A

Sliding FOS 1.5 4.97

Overturning Base in

Compression

75% 100%

Bearing FOS 3 5.96

Floodwall Unusual Loading Results (Construction)

Required Wall Section A

Sliding FOS 1.5 195.62

Overturning Base in

Compression

75% 100%

Bearing FOS 3 7.55

Floodwall Usual Loading Results (Saturated Soils)

Required Wall Section A

Sliding FOS 2.0 221.81

Overturning Base in

Compression

100% 100%

Bearing FOS 3.5 10.02

4.4 Concrete Design

The design of concrete structures includes all concrete associated with the floodwall. Design of the

structures followed the strength design guidelines as set forth in ACI 318-11 and EM 1110-2-2104. The

requirements for concrete design are described below. Calculations for the floodwall are located

following this report which indicates wall and slab thicknesses, reinforcing steel configurations, and

global stability requirements.

4.4.1 General

For the permanent floodwall stem and footing, design moments and shears for wall and slab structures

were obtained from our internal design spreadsheet and analyzed using the same self-developed

Microsoft Excel Spreadsheet mentioned above. The reinforcement for the floodwall was verified as a

secondary check using QuickRWall 4.0. The global analysis portion of the program was not used as it

does not follow USACE criteria.

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4.4.2 Minimum Concrete Cover

The following minimum concrete clear covers were used for design of all storm sewer structures and are

in accordance with EM 1110-2-2104:

Unformed surfaces in contact with foundation. . . . . . . . . . . . . . . . . . . 4 inches

Formed or screeded surfaces subject to cavitation or

abrasion erosion, such as baffle blocks and stilling

basin slabs. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 inches

Formed and screeded surfaces such as stilling basin

walls, chute spillway slabs, and channel lining slabs

on grade:

Equal to or greater than 24 inches in thickness. . . . . . . . . . . . . . 4 inches

Greater than 12 inches and less than 24 inches in

thickness. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 inches

Equal to or less than 12 inches in thickness. . . . . . . . . . . . . . . . . Per ACI 318

4.4.3 Minimum Shrinkage and Temperature Reinforcing

Minimum shrinkage and temperature reinforcing for hydraulic structures are in accordance with EM

1110-2-2104. It states that the area of reinforcement should be 0.0028 times the gross cross-sectional

area, half in each face, with a maximum area equivalent to No. 9 bars at 12 inches for smaller joint

spacing. As seen in section 4.6, the area of reinforcement will increase with larger joint spacing. As joint

spacing approaches 30’, the values in the table would be used rather than 0.0028 listed above.

4.4.4 Concrete Wall Thickness

The wall thicknesses are sufficient to allow proper placement and consolidation of concrete. Also, the

wall thicknesses were designed in multiples of 3 inches to simplify form tie systems. Minimum wall

thickness of 15 inches with two curtains of steel was used in walls where lift height exceed 8 feet.

4.4.5 Lap Splices and Development Lengths

Lap splices and development lengths in reinforcing bars conform to ACI 318-11. Refer to Table 7 for lap

splice and development lengths. Both tables are based on ACI 318 requirements for fc = 4,500 psi

concrete and Grade 60 reinforcement.

Table 7 - Splice and Development Lengths

SPLICES AND DEVELOPMENT LENGTHS

WALLS AND SLABS

BAR

SIZE

LENGTHS OF LAPPED

SPLICES FOR REINF.

(INCHES)

LENGTH OF END ANCHORAGE FOR

DEVELOPMENT OF REINFORCEMENT

(INCHES)

# TOP BARS* OTHERS TOP BARS* OTHERS 90° HOOKS

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3 23 19 19 14 6

4 31 24 24 18 8

5 38 30 30 23 10

6 46 35 35 27 12

7 67 51 51 40 14

8 76 58 58 45 16

9 86 66 66 51 19

* Top bars are horizontal bars so placed that more than 12" of concrete is cast in

the member below the bar. Horizontal bars in walls need not be provided with lap

lengths as required for top bars.

4.5 Joints & Waterstops

Waterstops are embedded in the monolith joints of the floodwalls to stop the passage of water through

the joint. Where permanent, frequent, and/or long term head differential is expected on construction

joints, water stops shall be installed. Waterstops shall be nonmetallic.

If the wall is to be constructed at higher temperature than 55 degrees, the expansion joint will expand

less and contract more. If the wall is to be constructed at lower temperature than 55 degree then the

expansion joint will expand more and contract less. Therefore the selected joint sealant material was

selected to provide +100% /-50% movement capabilities.

If the expansion joints were constructed wider or narrower than 3/4”, the expansion and contraction

movements shall be computed.

Additional considerations should be addressed when longer monolith lengths are required (road

closures, pump stations, gate monoliths, long walls etc) to provide a practical design. Monolith length

and joint spacing may dictate the requirements for more shrinkage and temperature reinforcement than

the specified minimum. Table 8 below provides minimum shrinkage and temperature reinforcement

ratios for longer joint spacing.

Table 8 - Joint Spacing vs Minimum Reinforcement

Length Between Control Joints, ft Minimum Temperature and Shrinkage Reinforcement

Ratio, Grade 60

Less than 30 ft 0.003

30-40 ft 0.004

Greater than 40 ft 0.005

Page 18: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Designer Initials: __________

Reviewer Initials: __________

Review Date: __________

Floodwall Section A

(12 Ft. High Floodwall)

Page 19: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

12 Ft. Floodwall

(Case 5)

Page 20: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Site: 7438-009 Owner:

Case:

User Data:A 15 Inches

B 17 Feet

C 12 FeetD 24 InchesD1 24 Inches

Hwall 13 Feet

Hwater 13 Feet

H2 5 FeetH3 5 Feet

Unit Weight of Soil (ϒsat):

Below Footing 0.118 k/cu. ft.Wet Side 0.118 k/cu. ft.Dry Side 0.118 k/cu. ft.

Ø= Below Footing 27 Degrees0.471 radians

Wet Side 27 Degrees0.471 radians

Dry Side 27 Degrees0.471 radians

SMF= 0.66667Unit Weight of Conc. (ρc) 0.15 k/cu. ft.

Unit Weight of Water (ρw) 0.0625 k/cu. ft.

Soil Capacity 6.2 k/sq. ft.2 Feet Ko Below Footing 0.55

15 Inches Wet Side 0.55Dry Side 0.55

Kp for sliding = 2.663fysteel 60 ksi Cohesion Factor (CF) 1 ksf

f'c(concrete) 4.5 ksi Frost Depth (Fr): 6 ft

Vertical Surcharge (Dry) (P3) 0 psf

Vertical Surcharge (Wet) (P4) 0 psf

Ice/Debris Load NoType of Loading Unusual

Toe Vertical Fill Width (F)= 3.75 ft

*All Dimensions in Calculations are converted to feet

City of Fargo

13' Floodwall Unusual to top of wall

Keyway Depth (G)Keyway Width (L)

Project Number:Fargo Floodwalls

1 − sin∅

1 − sin∅1 − sin∅

Page 21: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Force CalculationsFactor Unit Equation Description

EHW= 1.063 kips/ft Driving Wet Side Water Pressure

EHS= 0.500 kips/ft Driving Wet Side Soil Pressure

EV2= 13.080 kips Wet Side Vertical Component Weights

EV2d 11.000 ft from toe Moment arm for EV2

EV1= 2.213 kips Dry Side Vertical Component Weights

EV1d 1.875 ft from toe Moment arm for EV1

PehW= 9.031 kips Driving Wet Side Water Force

Pehwd 3.667 ft. above footing Moment arm for Pehw

PehS= 2.248 kips Driving Wet Side Soil Force

Pehsd 1.000 ft. above footing Moment arm for Pehs

PS1= 2.813 kips Vertical Weight of Stem and Key

PS1d 5.975 ft from toe Moment arm for PS1

PS2= 5.100 kips Vertical Weight of Footing

PS2d 8.500 ft from toe Moment arm for PS2

EH2= 1.062 kips Resisting Dry Side Soil Pressure

Peh2= 4.779 kips Resisting Dry Side Forces

Peh2d 1.000 ft from toe Moment arm for Peh2

Peh2s 2.248 kips Resisting dry side soil only

Peh2sd 1.000 ft from toe Moment arm for Peh2s

Pice/debris 0.000 kips .5 k/ft at Q100 Driving ice/debris Force

Pice/debrisd 15.000 ft. above footing Moment arm for Pice/debris

Pwind 0 kips Driving wind Force

Pwindd 15.000 ft. above footing Moment arm for Pwind

Applied Forces

(��� +�� + �)(ρ)

(�3 + �� + �)(ɤ�� − ρ)

��� � ρ +H3(C)(ɤ�� − ρ)

�2(�)(ɤ��)

��

2(��� + �� + �)

���

2(�3 + �� + �)

��� + �� + �

3− �

�3 + �� + �

3− �

��� � ρ� + �( )(ρ�)

! − � �� ρ� + �(�)(ρ�)

(�2 + �� + �)(ɤ��)

��"

2(�2 + �� + �)

! −�

2

2

��� � ρ� ! − � −�

2+ � ρ� ! −

2/$%�

! − � �� ρ�! − �

2+ � � ρ� ! −

2/$%"

�2 + �� + �

3− �

��� + ��

��� − ���

2+ ��� + ��

ρ� − ρ ∗ �2 + �� + � +�2 + �� + �

2

�2 + �� + �

3− �

.03(��� −��� )

Page 22: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Water Uplift

Force Calculation Factor Unit Equation DescriptionLZ-Z1= 17.117 ft Length of Seepage path from heel to toe

Δh= 8.000 ft Head differential between wet and dry side

Ldry side 7.000 ft Seepage path on Dry Side

LS 24.117 ft Total Seepage Path

uZ= 1.063 ksf Water Pressure at bottom of key (Wet side)

uZ1= 0.583 ksf Water pressure at bottom of footing (Dry Side)

HLZ-Z1= 5.678 ft Head Loss along Z-Z1

LSc 19.000 ft Length of concrete surface in sliding surface

HLLK = 0.374 ft Head Loss along key

ubottom key = 1.039 ksf Water pressure at bottom of key (Dry side)

HLUK = 0.971 ft Head Loss up key

utop key = 0.877 ksf Water pressure at top of key (Dry side)

P5 = 12.806 kips Water Uplift for Overturning

P5d 9.222 ft

P5sa 13.984 kips Water uplift for sliding along angle

P5sad -9.326 ft Moment Arm for P5sa

Lsss = 26.000 ft

Uf = 0.736 ksf

P5ss = 15.284 Water uplift of sliding along bottom of key

P5ssd -9.015 Moment Arm for P5ss

(!"+ �")

��� − �2

�2 + ��

)*)� + + ,�.+�

(��� +�� + �)(ρ)

��� + �� − ∆0 )*)� 1

(ρ)

∆0 )*)� �

! + �

%�(� )*)�)

(��� + �� + � − � 23)ρ

+ �

%�(� )*)�)

(��� +�� −� 43)ρ

5) + 567��789�,

2 +

5�7:9�, + 5)�2

(! − )

5) + 5)�2

(!)

5)�! −

2! − +

5�7:9�, − 5)�2

! − 2

3! − + 567��789�, ! −

2+5) − 567��789�,

2 ! −

3/$5

5)� !!

2+

5) − 5)�2

!2!

3/$5�

! + � + �� +�2

��� +�� + � − ��� −�2!

<��ρ

5= + 5>2

(!)

5> !!

2+5) − 5>

2!

2!

3/$5��

Moment Arm

for P5

Water pressure at key depth below

footing on dry side of footing

Page 23: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Distance Moment

to Toe(ft) About Toe (k-ft)

PS1 2.81 5.98 16.8 MPS1

PS2 5.10 8.50 43.4 MPS2

P3 0.00 1.88 0.0 MP3

P4 0.00 11.00 0.0 MP4

P5 -12.81 9.22 -118.1 MP5

-4.89 11.8 -57.9

EV2 - Backfill on Heel EV2 13.08 11.00 143.9 MEV2

EV1 - Fill on Toe EV1 2.21 1.88 4.1 MEV1

15.29 9.7 148.0

LL Live Load 0.0 0.00 0.0 MLL

0.0 0.0

EH Horiz. Earth Load PehS 2.25 -1.00 -1.2 MEH

WS Hydrostatic Pressure PehW 9.03 -3.67 -33.1 MPHW

ID Ice/Debris Force Pice/debris 0.00 -15.00 0.0 MID

W Wind Load Pwind 0.00 -15.00 0.0 MW

11.3 -34.3

Fargo Floodwalls

Ve

rtica

l Lo

ad

s

Applied Force Calculations

Total

P (kips)DescriptorLoad

Vertical Loads

DC

Horizontal Loads

EV

Total

Total

Total

Page 24: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Factor Equation Description

EM 1110-2-2502

Reference

PehW2 = 3.955 kips Water on Dry side for Overturning Calcs Figure 4-5

Pehw2d 0.705 ft Moment Arm for Pehw2 N/A

Peh2sb 1.360 ft Dry Side Soil to bottom of Footing Para. 4-8

Peh2sbd 2.333 kips Moment Arm for Peh2sb N/A

RS = 4.581 kips Resisting soil on Dry side to create equilibrium Para. 4-8

RSd -1 ft Moment Arm for RS N/A

Including Uplift

ΣVo 10.399 kips Total Vertical Load Figure 4-5

ΣMo 56.330 kip-ft 4-1

XR 5.417 ft Resultant Location 4-1

bl 16.25121 ft Length of Base in Compression 4-2

b% 95.595354 Percent of Base in Compression 4-2

Required b% 75 Required Base in Compression Appendix F

OK

Neglecting Uplift

ΣVon 23.205 kips Total Vertical Load Figure 4-5

ΣMon 174.430 kip-ft 4-1

XRn 7.517 ft Resultant Location 4-1

bln 22.550724 ft Length of Base in Compression 4-2

b%n 132.65132 Percent of Base in Compression 4-2

Required b%n 75 Required Base in Compression Appendix 4

OK

Global Stability

Overturning:

Fargo Floodwalls

$%� + $%" + $? + $@ + $A + �B1 + �B2

5)�2

�2 + �� +5�7:9�, + 567��789�,

2(�)

�2 + ��3

5)�2

�2 + ��3

�2 + �� + 5�7:9�,−�2

� +567��789�, − 5�7:9�,

2�

−2�3

/$CD"

−�2

EF1� +EF1" +EF? +EF@ +EFA +EGH� +EGH" +EFIJ +EKL +EJ + M% M%N + $CD"�6 $CD"�6N + $CD"($CD"N)(O7)(%E�)

ΣE7

ΣB7

3QR

S�!(100)

$%� + $%" + $? + $@ + �B1 + �B2

EF1� +EF1" +EF? +EF@ +EGH� +EGH" +EFIJ +EKL +EJ + M% M%N + $CD"�6 $CD"�6N + $CD"($CD"N)(O7)(%E�)

ΣE7T

ΣB7T

3QRT

S�T!

(100)

Total Overturning

Moment about

point o

(�2 + ��)(ɤ�� − ρ)(�2 + ��)/2

$CD − $CD" − $CD"�6(O7)(%E�)

Total Overturning

Moment about

point o

11/17/2015

Page 25: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Factor Equation Description

EM 1110-2-2502

Reference

Sliding Along Angle Between Bottom of Key and Bottom of Footing on Dry Side Figure 4-11

SBFa 1.722 kips Soil Below Footing on Sliding Surface Figure 4-11

Pwda 3.684 kips Water on Dry Side Figure 4-11

ΣVa 10.943 kips Sum of Vertical Forces on Sliding Surface Figure 4-11

αa 0.117 radians Angle of slip plane to horizontal plane Figure 4-11

6.710 degrees

Drained Condition

RFswad 5.142 kips Resisting Force from passive soil and water on dry side Figure 4-11

ΣHad 3.89 kips Sum of Horizontal Forces Figure 4-11

N'ad 11.323 kips Normal Force to Sliding Surface Figure 4-11

Tad 2.584 kips Tangential Force to Sliding Surface Figure 4-11

SSad 5.769 kips Drained Shear Strength 4-12

FSad 2.233 Factor of Safety 4-12

OK

Undrained Condition

RFswau 4.232 kips Resisting Force from passive soil and water on dry side Figure 4-11

ΣHau 4.800 kips Sum of Horizontal Forces Figure 4-11

N'au 11.429 kips Normal Force to Sliding Surface Figure 4-11

Tau 3.488 kips Tangential Force to Sliding Surface Figure 4-11

SSau 11.412 kips Undrained Shear Strength 4-12

FSau 3.272 Factor of Safety 4-12

OK

Sliding:

Fargo Floodwalls

Global Stability

� −�!

2(! − )(ɤ��)

5)2

�2 + �� +5) + 5)�

2(�)

%UV + �B1 + �B2 + $%� + $%" + $? + $@ − $A�

atan�!

$+ + $CD" − $+ Y: (.5)

$CD + $.��/+�6 .� + $.T+ − M��+

ΣB cos \ + Σ�+sin(\)

Σ�+ cos \ − ΣBsin(\)

]′+tan(∅)

%%+_+

$+ + $CD" − $+ (.5)

$CD + $.��/+�6 .� + $.T+ − M��`

ΣB cos \ + Σ�`sin(\)

Σ�` cos \ − ΣBsin(\)

��(%E�)√(!" + �")

%%`_`

11/17/2015

Page 26: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Factor Equation Description

EM 1110-2-2502

Reference

Sliding Along Horizontal Surface along Bottom of Key Figure 4-8

SBFs 3.717 kips Soil Below Footing on Sliding Surface

Pwds 3.310 kips Water on Dry Side Figure 4-11

ΣVs 11.638 kips Sum of Vertical Forces on Sliding Surface Figure 4-11

αs = 0 radians Angle of slip plane to horizontal plane Figure 4-11

0 degrees

Drained Condition

RFswsd 5.266 kips Resisting Force from passive soil and water on dry side Figure 4-11

ΣHsd 3.77 kips Sum of Horizontal Forces Figure 4-11

N'sd 11.638 kips Normal Force to Sliding Surface Figure 4-11

Tsd 3.765 kips Tangential Force to Sliding Surface Figure 4-11

SSsd 5.930 kips Drained Shear Strength 4-12

FSsd 1.575 Factor of Safety 4-12

OK

Undrained Condition

RFswsu 4.045 kips Resisting Force from passive soil and water on dry side Figure 4-11

ΣHsu 4.987 kips Sum of Horizontal Forces Figure 4-11

N'su 11.638 kips Normal Force to Sliding Surface Figure 4-11

Tsu 4.987 kips Tangential Force to Sliding Surface Figure 4-11

SSsu 11.333 kips Undrained Shear Strength 4-12

FSsu 2.273 Factor of Safety 4-12

OK

Global Stabilty

Fargo Floodwalls

�(! − )(ɤ��)

5>2(�2 + �� + �)

%UV� + �B1 + �B2 + $%� + $%" + $? + $@ − $A��

$+� + $CD" − $+� Y: (.5)

$CD + $.��/+�6 .� + $.T+ − M���+

ΣB� cos \� + Σ��+sin(\�)

Σ��+ cos \� − ΣB�sin(\�)

]′�+tan(∅)

%%�+_�+

.

$+� + $CD" − $+� (.5)

$CD + $.��/+�6 .� + $.T+ − M���`

ΣB� cos \� + Σ��`sin(\�)

Σ��` cos \� − ΣB�sin(\�)

�� %E� !

%%�`_�`

11/17/2015

Page 27: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Factor Equation Description

EM 1110-2-2502

Reference

Bearing Along Angle Between Bottom of Key and Bottom of Footing on Dry Side Eqns. From 1110-2-2502

ΣHadb 5.35 kips Sum of Horizontal Forces Figure 4-11

N'adb 11.493 kips Normal Force to Bearing Surface Figure 4-11

Tadb 4.032 kips Tangential Force to Bearing Surface Figure 4-11

SBFad 10.500005 ft Moment Arm for SBFa Figure 4-11

ΣMoaB 65.327363 kip-ft

XraB 5.970 ft Resultant Location Figure 4-11

eaB 2.589 ft Eccentricity of resultant Figure 4-11

B'aB 11.939256 ft Effective Base for Bearing Sec 5-2

Drained Condition

q0 = 0.333 k/ft2 Effective Overburden Pressure

y = 1.6007185

Nq = 13.175528 Bearing Capacity Factor 5-3a

Nc = 23.895844 Bearing Capacity Factor 5-3b

Nϒ = 9.4442971 Bearing Capacity Factor 5-3d

εcd = 1.16 Embedment Factor 5-4a

εqd = 1.08 Embedment Factor 5-4c

δd = 0.34 5-5

19.330524

εqi = 0.62 Inclination Factor 5-5a

εϒi = 0.08 Inclination Factor 5-5b

Q1 = 0.00 5-2

Q2 = 2.9267926 5-2

Q3 = 0.2731585 5-2

Qd = 38.205036 5-2

FS = 5-1

3.3241783 Eq. 5-1 Bearing Criteria Satisfied

Global Stabilty

Fargo Floodwalls

Bearing:

ΣE7U

ΣB )*)�2

− QbU

)*)� − 2(CU)

(C,) tan 45 +∅2

de∅ > 0, (]h−1)ijklm(∅)

]h − 1 tan(1.4∅)

1 + 0.2� !n tan 45 +

∅2

de∅ > 0, 1 + 0.1� !n tan 45 +

∅2

atanΣ�ΣB

1 −o+90

"

1 −o+∅

"

q�+(qh.)(rs)(]h)

qh+(qh.)(rs)(]h)

qh+ qh. ɤ�� − ɤ�� ]t /2

!′(u1 + u2 + u3)

(ɤ�� − ɤ�� )(� )

v(tan ∅)

Normal Component to the base of the structure of the

ultimate bearing capacity

EF1� +EF1" +EF? +EF@ + $A� $A�N + EGH� +EGH" +EGI +EFIJ +EKL +EJ + $CD"�($CD"�N)(O7)(%E�)+$+ $CD"N + %UV %UVN

$CD + $.��/+�6 .� + $.T+ + $CD�(O7)(%E�) − $+ − $CD"�(O7)(%E�)

ΣB cos \ + Σ�+6sin(\)

Σ�+6 cos \ − ΣBsin(\)

2(! − )3

11/17/2015

Page 28: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Factor Equation Description

EM 1110-2-2502

Reference

q0 = 0.333 k/ft2 Effective Overburden Pressure

y = 0

Nq = 1 Bearing Capacity Factor 5-3a

Nc = 5.14 Bearing Capacity Factor 5-3c

Nϒ = 0 Bearing Capacity Factor 5-3d

εcd = 1.10 Embedment Factor 5-4a

εqd = 1.00 Embedment Factor 5-4b

δd = 0.34 5-5

19.330524

εqi = 0.62 Inclination Factor 5-5a

εϒi = 0.08 Inclination Factor 5-5b

Q1 = 3.49 5-2

Q2 = 0.2053161 5-2

Q3 = 0 5-2

Qd = 44.088477 5-2

FS = 5-1

3.8360901

Drained q'max 1.30 Undrained q'max 1.30

Bearing Criteria Satisfied

Undrained Condition

Global Stabilty

Fargo Floodwalls

(ɤ�� − ɤ�� )(� )

v(tan ∅)

de∅ = 0, 5.14

de∅ = 0, 1

(C,)tan(45 +∅2)

]h − 1 tan(1.4∅)

1 + 0.2� !n tan 45 +

∅2

atanΣ�ΣB

1 −o+90

"

1 −o+∅

"

q�+(qh.)(rs)(]h)

qh+(qh.)(rs)(]h)

qh+ qh. ɤ�� − ɤ�� ]t /2

!′(u1 + u2 + u3)Normal Component to the base of the structure of the

ultimate bearing capacity

11/17/2015

Page 29: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Factor Equation Description

Load Factor 1.7

b 16.25121 Effective base in compression

qmin 0

Uws 0.676013 Uplift Pressure Wet side of Stem

Utws 9.659888 Total water uplift on wet side

Utwsd 6.391352 Moment arm for U tws

VLw 7.395112 Vertical load on wet side

Bws 0.897423 Bearing at wet side of stem

Btws 5.048546 Total Bearing uplift on wet side of footing

Btwsd 3.7504 Moment arm for B tws

Total Shear on Heel 3.989161

Total Moment on Heel 36.81552

Total Shear on Stem 7.65 Total Shear on Stem

Total Moment on Stem 36.69128

Uds 0.652666 Uplift Pressure Dry side of Stem

Utds 2.316169 Total water uplift on dry side

Utdsd 1.839562 Moment arm for U tds

VLd 1.021331 Vertical load on dry side

Bds 0.997126 Bearing at dry side of stem

Btds 4.30005 Total Bearing uplift on dry side of footing

Btdsd 1.956515 Moment arm for B tds

Total Shear on Toe -5.57382 Total Shear on Toe

Total Moment on Toe -10.9073 Total Moment on Toe

Utkw 0.9375 Water Pressure at top of key (wet side)

Total Shear on Key -7.65 Total Shear on Key

Total Moment on Key -7.66624 Total Moment on Key

Forces for Reinforcement/Shear

Heel Forces

Stem Forces

Toe Forces

Key Forces

$KL +$J + �

2��� + �3 ɤ�� − ρ

�32

Y7 − ρ�22

�2 − �2 ɤ�� − ρ�22

(Y7) jlN�likjb

$KL ��� + $J $.T+N −�12

+ �

2���

���

3+ �3 ɤ�� − ρ

�32

�33

(Y7) − ρ�22

�2�23

−�2 ɤ�� − ρ�22

�23

(Y7) ( jlN�likjb)

x�7:9�, −x�7:9�, −x)�

! − � −

x�7:9�, −x�

2C − L +

x) +x67��789�,

2( )

x�� − 2

� − +x�7:9�, − x�

22 � −

3� − + x67��789�, � −

2

+x) −x67��789�,

2 � −

3

/x��

�B2+ � � ρ� +� ρ� +$@ − x��

r8z − r8z − r8.T

S(� + �)

(B − !��)( jlN�likjb)

�B2+ � � ρ� +� ρ� +$@�2

− x�� x��N − !�� !��N ( jlN�likjb)

x�7:9�, −x�7:9�, −x)�

! − � + � −

x+� +x)�

2(�)

[x)� ��2

+x+� −x)�

2(�3)(�)]/x�+�

�B1 + $? +� �� ρ� −x�+�

r8z − r8z − r8.T

S(�)

r8z + !+�2

(�)

!+� ��2

+r8z −!+�

22�3

� /!�+�

(B + −!�+�)( jlN�likjb)

�B1+ $? +� �� ρ��2

−x�+� x�+�N − !�+� !�+�N ( jlN�likjb)

}eS% ≥ 100,!, S�

}eS% ≥ 100,r8z

!1−

6C!

, 0

}er8.T = 0,!�

2S − � − � , (

!�+ r8.T

2)(�)

}er8.T = 0,S − � − �

3, r8.T �

�2

+!�− r8.T

2�

�3

/!��

x�9 +x=

2� −

x67��789�, +x�7:9�,

2� − M% jlN�likjb

x�9 ��2

+x= − x�9

2�

2�3

−x�7:9�, ��2

−x67��789�, −x�7:9�,

2�

2�3

− M%�2

( jlN�likjb)

(��� +�)(�)

Total

Moment

on Stem

Page 30: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Input: Values in red need to be entered by user Designer: LJB Date: 11/17/2015

Check: Date:Project Number:

Bridge No./Structure ID:

Owner:

Mu = 36.81552 k-ft 21.656 k-ft concrete wall thickness (t) 24 in.

Vu = 3.989161 k 2.347 k Clear Cover 4 in.

Nu = 0.0 k 0.0 k d dimension 19.625 in.

(factored) (service)150 psi

Bar Size = 6

Enter Reinforcement Spacing = 6 in. Reinforcement yield strength = 60 ksiAs = 0.88 in2

Concrete yield strength = 4.5 ksi

Enter design width = 12 in.Concrete yield strength = 4500 psi

Bar Size = None Ec = 4066.84 ksi

Enter Reinforcement Spacing = 6 in. Es = 29000 ksi

Av = 0.00 in2Lightweight Concrete? NO

Mod Factor ƛ = 1Does ACI 350 Apply? NO Flexure φ = 0.9

Shear φ = 0.75

Flexure Analysis: Mr = Phi(Mn)

For members without axial loads.

a = 1.150

Mn = 1,005.8 k-in

Mn= 83.8 k-ft

φMn = 75.44 k-ft

Minimum Reinforcement: (ACI 318 10.5)

As, min = 0.79 in2 not less than 0.785 in2

ACI 318 - 10.5 Requirement = Pass If Fail ------------>

1.33*Mu = 48.96 k-ft

As,req = 0.51 in2

As, min = 0.79 in2

OK - Section and reinforcement sufficient for Flexural Resistance

*If the requirements of 10.5 are not satisfied then As needs to be at least 1/3 greater than that required by analysis. (10.5.3)

OK

She

ar

7438-009 Section Being Designed:Heel of Footing

Fargo Floodwalls Layer of Steel:

City of Fargo

Modulus of elasticity based on concrete unit weight of

Fle

xure

��,8.T =? >��

>�SNS5kmjk�C<<kDlm200S

+

>�

ET = �� ∗ e, ∗ N −l2

l =�� ∗ e,

0.85 ∗ e′� ∗ S

��, �h =.85en�S

e,N − N" −

2E`

.85e′�S

Page 31: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Designer: LJB Date: 11/17/2015

Check: Date:

ACI 318

Governing Specification: USACE

Min. Steel Requirement:

As/Ag (Includes both faces) ≥ 0.0050 EM 1110-2-2104 2.8

Enter Reinforcing configuration of opposite mat:Bar Size = 5

Enter Reinforcement Spacing = 6 in.As = 0.62 in2

Required Min. Steel (both faces): 1.44 sq. in/ft

Max. Spacing Requirements:

s≤5h or s≤18.0 in. 120 in18 in Structural walls/footings:

or for componenets over 36.0 in Spacing can be ≤ 12 in. on faces > 18" thick

s≤12.0 12 in

Maximum Spacing Allowed: 18 in

ACI 350

Min. Steel Requirement: ACI 350-06 7.12.2.1 Minimum reinforcement ratio for Grade 60 steel

Length Between JointsRatio As/Ag 0.0030 ≤ 20 ft

20-30 ft

Steel in opposite face: 0.62 in230-40 ft≥ 40ft

Required Min. Steel: 0.86 sq. in/ft

Max. Spacing Requirements: ACI 350-065 7.12.2.2

Shrinkage and Temperature Reinforcement shall not be spaced greater than12in. apart

Maximum Spacing Allowed: 12 in

0.0040.005

T&S Steel Requirement Met

Max Spacing Requirement Met

Max Spacing Requirement Met

T&S Steel:

Ratio (As/Ag)0.0030.003

0

T&S Steel:

T&S Steel Requirement Met

Layer of Steel:

Section Being Designed:

Heel of Footing

Page 32: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Designer: LJB Date: 11/17/2015Check: Date:

Shear Analysis:

φVn ≥ Vu (ACI 318 Eq. 11-1)

Vn = Vc + Vs (ACI 318 Eq. 11-2)

Calculate d:

d = 19.625 (Loads placed at 'top' of member.) (ACI 318 11.1.3.1)

dv = 19.625 in

Calc. Vc

For members subject to shear & flexure only.Vc = 2* ƛ * √fc * bv * dv ƛ = 1.0 (ACI 318 Eq. 11-3)

bv = 12 in

Vc = 31.6 kips dv = 19.63 in

For members subject to axial compression.Vc = 2*(1 + Nu/(2000*Ag)) * ƛ * √fc * bv * dv Nu = 0 lb (ACI 318 Eq. 11-4)

Ag = 288 in2

Vc = 758.3 kips

Vc = 31.6 kips

Calc. Vs

Shear Reinforcement perpendicular to axis.

Vs = Av * fyt * d / s Av = 0.00 in2

s = 6 in

Vs = 0.0 kips

Vn = 31.6 kips

θVn = 23.7 kips Vu= 3.989161

θVn = 23.7 kips OK - Section Adequate for Shear Resistance

Section Being Designed:Heel of Footing

Layer of Steel:0

Page 33: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Designer: LJB Date: 11/17/2015Check: Date:

Crack Control: ACI 318

Distribution of flexural reinforcement in beams and one-way slabs.

The spacing of reinforcement closest to the tension face, s, shall not exceed:

cc = 4 in (Clear cover)

fs = Calculated stress from unfactored Moment.

But not greater than Mserv = 259.9 kip-in

n = 7.13

x = 4.038 in NA to extreme comp fiber

y = 15.587 in NA to tension centroid

I = 1787.9 in4

fs = 16155.67 psi

s ≤ 27.1 in

Crack Control: ACI 350

Actual Stress:fs = 16155.67 psi

ACI 350-06 10.6.4.1, for normal environmental exposures as defined in 10.6.4.5

β= 1.28

s= 6 in.db= 0.75 in.

fs,max= 32.65 ksi26781.3 psi > 16155.67 psi

OK - Spacing Adequate

OK - Spacing Adequate

fs,max =320

β*SQRT(s2+4(2+

db∕2)

2)

Section Being Designed:Heel of Footing

Layer of Steel:0

< = 15@s,sss

>�− 2.5 i�

12(40,000

e�)

e� = m ∗E�� � ∗ �

}

� =�� ∗ �� ∗ m + 2 ∗ S ∗ N ∗ m

S−�� ∗ mS

� = N − �

} =13− S ∗ �? + m ∗ �� ∗ �"

Page 34: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Input: Values in red need to be entered by user Designer: LJB Date: 11/17/2015

Check: Date:Project Number:

Bridge No./Structure ID:

Owner:

Mu = 36.69128 k-ft 21.583 k-ft concrete wall thickness (t) 15 in.

Vu = 7.65 k 4.500 k Clear Cover 3 in.

Nu = 0.0 k 0.0 k d dimension 11.625 in.

(factored) (service)150 psi

Bar Size = 6

Enter Reinforcement Spacing = 6 in. Reinforcement yield strength = 60 ksiAs = 0.88 in2

Concrete yield strength = 4.5 ksi

Enter design width = 12 in.Concrete yield strength = 4500 psi

Bar Size = None Ec = 4066.84 ksi

Enter Reinforcement Spacing = 6 in. Es = 29000 ksi

Av = 0.00 in2Lightweight Concrete? NO

Mod Factor ƛ = 1Does ACI 350 Apply? NO Flexure φ = 0.9

Shear φ = 0.75

Flexure Analysis: Mr = Phi(Mn)

For members without axial loads.

a = 1.150

Mn = 583.4 k-in

Mn= 48.6 k-ft

φMn = 43.76 k-ft

Minimum Reinforcement: (ACI 318 10.5)

As, min = 0.47 in2 not less than 0.465 in2

ACI 318 - 10.5 Requirement = Pass If Fail ------------>

1.33*Mu = 48.80 k-ft

As,req = 0.88 in2

As, min = 0.47 in2

OK - Section and reinforcement sufficient for Flexural Resistance

*If the requirements of 10.5 are not satisfied then As needs to be at least 1/3 greater than that required by analysis. (10.5.3)

OK

She

ar

7438-009 Section Being Designed:Stem

Fargo Floodwalls Layer of Steel:

City of Fargo

Modulus of elasticity based on concrete unit weight of

Fle

xure

��,8.T =? >��

>�SNS5kmjk�C<<kDlm200S

+

>�

ET = �� ∗ e, ∗ N −l2

l =�� ∗ e,

0.85 ∗ e′� ∗ S

��, �h =.85en�S

e,N − N" −

2E`

.85e′�S

11/17/20158:37 AM

Page 35: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Designer: LJB Date: 11/17/2015

Check: Date:

ACI 318

Governing Specification: USACE

Min. Steel Requirement:

As/Ag (Includes both faces) ≥ 0.0028 EM 1110-2-2104 2.8

Enter Reinforcing configuration of opposite mat:Bar Size = 5

Enter Reinforcement Spacing = 12 in.As = 0.31 in2

Required Min. Steel (both faces): 0.50 sq. in/ft

Max. Spacing Requirements:

s≤5h or s≤18.0 in. 75 in18 in Structural walls/footings:

or for componenets over 36.0 in Spacing can be ≤ 12 in. on faces > 18" thick

s≤12.0 12 in

Maximum Spacing Allowed: 18 in

ACI 350

Min. Steel Requirement: ACI 350-06 7.12.2.1 Minimum reinforcement ratio for Grade 60 steel

Length Between JointsRatio As/Ag 0.0030 ≤ 20 ft

20-30 ft

Steel in opposite face: 0.31 in230-40 ft≥ 40ft

Required Min. Steel: 0.54 sq. in/ft

Max. Spacing Requirements: ACI 350-065 7.12.2.2

Shrinkage and Temperature Reinforcement shall not be spaced greater than12in. apart

Maximum Spacing Allowed: 12 in

0.0040.005

T&S Steel Requirement Met

Max Spacing Requirement Met

Max Spacing Requirement Met

T&S Steel:

Ratio (As/Ag)0.0030.003

0

T&S Steel:

T&S Steel Requirement Met

Layer of Steel:

Section Being Designed:

Stem

11/17/20158:37 AM

Page 36: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Designer: LJB Date: 11/17/2015Check: Date:

Shear Analysis:

φVn ≥ Vu (ACI 318 Eq. 11-1)

Vn = Vc + Vs (ACI 318 Eq. 11-2)

Calculate d:

d = 11.625 (Loads placed at 'top' of member.) (ACI 318 11.1.3.1)

dv = 11.625 in

Calc. Vc

For members subject to shear & flexure only.Vc = 2* ƛ * √fc * bv * dv ƛ = 1.0 (ACI 318 Eq. 11-3)

bv = 12 in

Vc = 18.7 kips dv = 11.63 in

For members subject to axial compression.Vc = 2*(1 + Nu/(2000*Ag)) * ƛ * √fc * bv * dv Nu = 0 lb (ACI 318 Eq. 11-4)

Ag = 180 in2

Vc = 280.7 kips

Vc = 18.7 kips

Calc. Vs

Shear Reinforcement perpendicular to axis.

Vs = Av * fyt * d / s Av = 0.00 in2

s = 6 in

Vs = 0.0 kips

Vn = 18.7 kips

θVn = 14.0 kips Vu= 7.65

θVn = 14.0 kips OK - Section Adequate for Shear Resistance

Section Being Designed:Stem

Layer of Steel:0

11/17/20158:37 AM

Page 37: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Designer: LJB Date: 11/17/2015Check: Date:

Crack Control: ACI 318

Distribution of flexural reinforcement in beams and one-way slabs.

The spacing of reinforcement closest to the tension face, s, shall not exceed:

cc = 3 in (Clear cover)

fs = Calculated stress from unfactored Moment.

But not greater than Mserv = 259.0 kip-in

n = 7.13

x = 3.003 in NA to extreme comp fiber

y = 8.622 in NA to tension centroid

I = 574.8 in4

fs = 27702.77 psi

s ≤ 14.2 in

Crack Control: ACI 350

Actual Stress:fs = 27702.77 psi

ACI 350-06 10.6.4.1, for normal environmental exposures as defined in 10.6.4.5

β= 1.39

s= 6 in.db= 0.75 in.

fs,max= 30.05 ksi26781.3 psi < 27702.77 psi

ACI 350 Requirements Not Met

OK - Spacing Adequate

fs,max =320

β*SQRT(s2+4(2+

db∕2)

2)

Section Being Designed:Stem

Layer of Steel:0

< = 15@s,sss

>�− 2.5 i�

12(40,000

e�)

e� = m ∗E�� � ∗ �

}

� =�� ∗ �� ∗ m + 2 ∗ S ∗ N ∗ m

S−�� ∗ mS

� = N − �

} =13− S ∗ �? + m ∗ �� ∗ �"

11/17/20158:37 AM

Page 38: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Input: Values in red need to be entered by user Designer: LJB Date: 11/17/2015

Check: Date:Project Number:

Bridge No./Structure ID:

Owner:

Mu = 10.90726 k-ft 6.416 k-ft concrete wall thickness (t) 24 in.

Vu = 5.573823 k 3.279 k Clear Cover 4 in.

Nu = 0.0 k 0.0 k d dimension 19.6875 in.

(factored) (service)150 psi

Bar Size = 5

Enter Reinforcement Spacing = 6 in. Reinforcement yield strength = 60 ksiAs = 0.62 in2

Concrete yield strength = 4.5 ksi

Enter design width = 12 in.Concrete yield strength = 4500 psi

Bar Size = None Ec = 4066.84 ksi

Enter Reinforcement Spacing = 6 in. Es = 29000 ksi

Av = 0.00 in2Lightweight Concrete? NO

Mod Factor ƛ = 1Does ACI 350 Apply? NO Flexure φ = 0.9

Shear φ = 0.75

Flexure Analysis: Mr = Phi(Mn)

For members without axial loads.

a = 0.810

Mn = 717.3 k-in

Mn= 59.8 k-ft

φMn = 53.80 k-ft

Minimum Reinforcement: (ACI 318 10.5)

As, min = 0.79 in2 not less than 0.7875 in2

ACI 318 - 10.5 Requirement = Fail If Fail ------------>

1.33*Mu = 14.51 k-ft

As,req = 0.15 in2

As, min = 0.15 in2

OK - Section and reinforcement sufficient for Flexural Resistance

*If the requirements of 10.5 are not satisfied then As needs to be at least 1/3 greater than that required by analysis. (10.5.3)

OK

She

ar

7438-009 Section Being Designed:Toe of Footing

Fargo Floodwalls Layer of Steel:

City of Fargo

Modulus of elasticity based on concrete unit weight of

Fle

xure

��,8.T =? >��

>�SNS5kmjk�C<<kDlm200S

+

>�

ET = �� ∗ e, ∗ N −l2

l =�� ∗ e,

0.85 ∗ e′� ∗ S

��, �h =.85en�S

e,N − N" −

2E`

.85e′�S

11/17/20158:37 AM

Page 39: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Designer: LJB Date: 11/17/2015

Check: Date:

ACI 318

Governing Specification: USACE

Min. Steel Requirement:

As/Ag (Includes both faces) ≥ 0.0050 EM 1110-2-2104 2.8

Enter Reinforcing configuration of opposite mat:Bar Size = 6

Enter Reinforcement Spacing = 6 in.As = 0.88 in2

Required Min. Steel (both faces): 1.44 sq. in/ft

Max. Spacing Requirements:

s≤5h or s≤18.0 in. 120 in18 in Structural walls/footings:

or for componenets over 36.0 in Spacing can be ≤ 12 in. on faces > 18" thick

s≤12.0 12 in

Maximum Spacing Allowed: 18 in

ACI 350

Min. Steel Requirement: ACI 350-06 7.12.2.1 Minimum reinforcement ratio for Grade 60 steel

Length Between JointsRatio As/Ag 0.0030 ≤ 20 ft

20-30 ft

Steel in opposite face: 0.88 in230-40 ft≥ 40ft

Required Min. Steel: 0.86 sq. in/ft

Max. Spacing Requirements: ACI 350-065 7.12.2.2

Shrinkage and Temperature Reinforcement shall not be spaced greater than12in. apart

Maximum Spacing Allowed: 12 in

0.0040.005

T&S Steel Requirement Met

Max Spacing Requirement Met

Max Spacing Requirement Met

T&S Steel:

Ratio (As/Ag)0.0030.003

0

T&S Steel:

T&S Steel Requirement Met

Layer of Steel:

Section Being Designed:

Toe of Footing

11/17/20158:37 AM

Page 40: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Designer: LJB Date: 11/17/2015Check: Date:

Shear Analysis:

φVn ≥ Vu (ACI 318 Eq. 11-1)

Vn = Vc + Vs (ACI 318 Eq. 11-2)

Calculate d:

d = 19.688 (Loads placed at 'top' of member.) (ACI 318 11.1.3.1)

dv = 19.688 in

Calc. Vc

For members subject to shear & flexure only.Vc = 2* ƛ * √fc * bv * dv ƛ = 1.0 (ACI 318 Eq. 11-3)

bv = 12 in

Vc = 31.7 kips dv = 19.69 in

For members subject to axial compression.Vc = 2*(1 + Nu/(2000*Ag)) * ƛ * √fc * bv * dv Nu = 0 lb (ACI 318 Eq. 11-4)

Ag = 288 in2

Vc = 760.7 kips

Vc = 31.7 kips

Calc. Vs

Shear Reinforcement perpendicular to axis.

Vs = Av * fyt * d / s Av = 0.00 in2

s = 6 in

Vs = 0.0 kips

Vn = 31.7 kips

θVn = 23.8 kips Vu= 5.573823

θVn = 23.8 kips OK - Section Adequate for Shear Resistance

Section Being Designed:Toe of Footing

Layer of Steel:0

11/17/20158:37 AM

Page 41: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Designer: LJB Date: 11/17/2015Check: Date:

Crack Control: ACI 318

Distribution of flexural reinforcement in beams and one-way slabs.

The spacing of reinforcement closest to the tension face, s, shall not exceed:

cc = 4 in (Clear cover)

fs = Calculated stress from unfactored Moment.

But not greater than Mserv = 77.0 kip-in

n = 7.13

x = 3.458 in NA to extreme comp fiber

y = 16.229 in NA to tension centroid

I = 1329.9 in4

fs = 6699.907 psi

s ≤ 79.6 in

Crack Control: ACI 350

Actual Stress:fs = 6699.907 psi

ACI 350-06 10.6.4.1, for normal environmental exposures as defined in 10.6.4.5

β= 1.27

s= 6 in.db= 0.625 in.

fs,max= 33.37 ksi26781.3 psi > 6699.907 psi

OK - Spacing Adequate

OK - Spacing Adequate

fs,max =320

β*SQRT(s2+4(2+

db∕2)

2)

Section Being Designed:Toe of Footing

Layer of Steel:0

< = 15@s,sss

>�− 2.5 i�

12(40,000

e�)

e� = m ∗E�� � ∗ �

}

� =�� ∗ �� ∗ m + 2 ∗ S ∗ N ∗ m

S−�� ∗ mS

� = N − �

} =13− S ∗ �? + m ∗ �� ∗ �"

11/17/20158:37 AM

Page 42: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Input: Values in red need to be entered by user Designer: LJB Date: 11/17/2015

Check: Date:Project Number:

Bridge No./Structure ID:

Owner:

Mu = 7.666244 k-ft 4.510 k-ft concrete wall thickness (t) 15 in.

Vu = 7.645077 k 4.497 k Clear Cover 3 in.

Nu = 0.0 k 0.0 k d dimension 11.5625 in.

(factored) (service)150 psi

Bar Size = 7

Enter Reinforcement Spacing = 12 in. Reinforcement yield strength = 60 ksiAs = 0.60 in2

Concrete yield strength = 4.5 ksi

Enter design width = 12 in.Concrete yield strength = 4500 psi

Bar Size = None Ec = 4066.84 ksi

Enter Reinforcement Spacing = 6 in. Es = 29000 ksi

Av = 0.00 in2Lightweight Concrete? NO

Mod Factor ƛ = 1Does ACI 350 Apply? NO Flexure φ = 0.9

Shear φ = 0.75

Flexure Analysis: Mr = Phi(Mn)

For members without axial loads.

a = 0.784

Mn = 402.1 k-in

Mn= 33.5 k-ft

φMn = 30.16 k-ft

Minimum Reinforcement: (ACI 318 10.5)

As, min = 0.47 in2 not less than 0.4625 in2

ACI 318 - 10.5 Requirement = Pass If Fail ------------>

1.33*Mu = 10.20 k-ft

As,req = 0.18 in2

As, min = 0.47 in2

OK - Section and reinforcement sufficient for Flexural Resistance

*If the requirements of 10.5 are not satisfied then As needs to be at least 1/3 greater than that required by analysis. (10.5.3)

OK

She

ar

7438-009 Section Being Designed:Key

Fargo Floodwalls Layer of Steel:

City of Fargo

Modulus of elasticity based on concrete unit weight of

Fle

xure

��,8.T =? >��

>�SNS5kmjk�C<<kDlm200S

+

>�

ET = �� ∗ e, ∗ N −l2

l =�� ∗ e,

0.85 ∗ e′� ∗ S

��, �h =.85en�S

e,N − N" −

2E`

.85e′�S

11/17/20158:37 AM

Page 43: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Designer: LJB Date: 11/17/2015

Check: Date:

ACI 318

Governing Specification: USACE

Min. Steel Requirement:

As/Ag (Includes both faces) ≥ 0.0050 EM 1110-2-2104 2.8

Enter Reinforcing configuration of opposite mat:Bar Size = 7

Enter Reinforcement Spacing = 12 in.As = 0.60 in2

Required Min. Steel (both faces): 0.90 sq. in/ft

Max. Spacing Requirements:

s≤5h or s≤18.0 in. 75 in18 in Structural walls/footings:

or for componenets over 36.0 in Spacing can be ≤ 12 in. on faces > 18" thick

s≤12.0 12 in

Maximum Spacing Allowed: 18 in

ACI 350

Min. Steel Requirement: ACI 350-06 7.12.2.1 Minimum reinforcement ratio for Grade 60 steel

Length Between JointsRatio As/Ag 0.0030 ≤ 20 ft

20-30 ft

Steel in opposite face: 0.60 in230-40 ft≥ 40ft

Required Min. Steel: 0.54 sq. in/ft

Max. Spacing Requirements: ACI 350-065 7.12.2.2

Shrinkage and Temperature Reinforcement shall not be spaced greater than12in. apart

Maximum Spacing Allowed: 12 in

0.0040.005

T&S Steel Requirement Met

Max Spacing Requirement Met

Max Spacing Requirement Met

T&S Steel:

Ratio (As/Ag)0.0030.003

0

T&S Steel:

T&S Steel Requirement Met

Layer of Steel:

Section Being Designed:

Key

11/17/20158:37 AM

Page 44: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Designer: LJB Date: 11/17/2015Check: Date:

Shear Analysis:

φVn ≥ Vu (ACI 318 Eq. 11-1)

Vn = Vc + Vs (ACI 318 Eq. 11-2)

Calculate d:

d = 11.563 (Loads placed at 'top' of member.) (ACI 318 11.1.3.1)

dv = 11.563 in

Calc. Vc

For members subject to shear & flexure only.Vc = 2* ƛ * √fc * bv * dv ƛ = 1.0 (ACI 318 Eq. 11-3)

bv = 12 in

Vc = 18.6 kips dv = 11.56 in

For members subject to axial compression.Vc = 2*(1 + Nu/(2000*Ag)) * ƛ * √fc * bv * dv Nu = 0 lb (ACI 318 Eq. 11-4)

Ag = 180 in2

Vc = 279.2 kips

Vc = 18.6 kips

Calc. Vs

Shear Reinforcement perpendicular to axis.

Vs = Av * fyt * d / s Av = 0.00 in2

s = 6 in

Vs = 0.0 kips

Vn = 18.6 kips

θVn = 14.0 kips Vu= 7.645077

θVn = 14.0 kips OK - Section Adequate for Shear Resistance

Section Being Designed:Key

Layer of Steel:0

11/17/20158:37 AM

Page 45: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Designer: LJB Date: 11/17/2015Check: Date:

Crack Control: ACI 318

Distribution of flexural reinforcement in beams and one-way slabs.

The spacing of reinforcement closest to the tension face, s, shall not exceed:

cc = 3 in (Clear cover)

fs = Calculated stress from unfactored Moment.

But not greater than Mserv = 54.1 kip-in

n = 7.13

x = 2.537 in NA to extreme comp fiber

y = 9.026 in NA to tension centroid

I = 413.8 in4

fs = 8415.816 psi

s ≤ 63.8 in

Crack Control: ACI 350

Actual Stress:fs = 8415.816 psi

ACI 350-06 10.6.4.1, for normal environmental exposures as defined in 10.6.4.5

β= 1.38

s= 12 in.db= 0.875 in.

fs,max= 17.89 ksi26781.3 psi > 8415.816 psi

OK - Spacing Adequate

OK - Spacing Adequate

fs,max =320

β*SQRT(s2+4(2+

db∕2)

2)

Section Being Designed:Key

Layer of Steel:0

< = 15@s,sss

>�− 2.5 i�

12(40,000

e�)

e� = m ∗E�� � ∗ �

}

� =�� ∗ �� ∗ m + 2 ∗ S ∗ N ∗ m

S−�� ∗ mS

� = N − �

} =13− S ∗ �? + m ∗ �� ∗ �"

11/17/20158:37 AM

Page 46: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Input: Values in red need to be entered by user Designer: LJB Date: 11/17/2015

Check: Date:Project Number:

Bridge No./Structure ID:

Owner:

Mu = 0 k-ft 0.000 k-ft concrete wall thickness (t) 24 in.

Vu = 0 k 0.000 k Clear Cover 5 in.

Nu = 0.0 k 0.0 k d dimension 18.5 in.

(factored) (service)150 psi

Bar Size = 8

Enter Reinforcement Spacing = 12 in. Reinforcement yield strength = 60 ksiAs = 0.79 in2

Concrete yield strength = 4.5 ksi

Enter design width = 12 in.Concrete yield strength = 4500 psi

Bar Size = None Ec = 4066.84 ksi

Enter Reinforcement Spacing = 6 in. Es = 29000 ksi

Av = 0.00 in2Lightweight Concrete? NO

Mod Factor ƛ = 1Does ACI 350 Apply? NO Flexure φ = 0.9

Shear φ = 0.75

ACI 318

Governing Specification: USACE

Min. Steel Requirement:

As/Ag (Includes both faces) ≥ 0.0050 EM 1110-2-2104 2.8

Enter Reinforcing configuration of opposite mat:Bar Size = 8

Enter Reinforcement Spacing = 12 in.As = 0.79 in2

Required Min. Steel (both faces): 1.44 sq. in/ft

Max. Spacing Requirements:

s≤5h or s≤18.0 in. 120 in18 in Structural walls/footings:

or for componenets over 36.0 in Spacing can be ≤ 12 in. on faces > 18" thick

s≤12.0 12 in

Maximum Spacing Allowed: 18 in

She

ar

7438-009 Section Being Designed:Longitudinal Footing Reinforcement

Fargo Floodwalls Layer of Steel:

City of Fargo

Modulus of elasticity based on concrete unit weight of

Fle

xure

T&S Steel:

T&S Steel Requirement Met

Max Spacing Requirement Met

Page 47: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Input: Values in red need to be entered by user Designer: LJB Date: 11/17/2015

Check: Date:Project Number:

Bridge No./Structure ID:

Owner:

Mu = 0 k-ft 0.000 k-ft concrete wall thickness (t) 15 in.

Vu = 0 k 0.000 k Clear Cover 3 in.

Nu = 0.0 k 0.0 k d dimension 11.6875 in.

(factored) (service)150 psi

Bar Size = 5

Enter Reinforcement Spacing = 12 in. Reinforcement yield strength = 60 ksiAs = 0.31 in2

Concrete yield strength = 4.5 ksi

Enter design width = 12 in.Concrete yield strength = 4500 psi

Bar Size = None Ec = 4066.84 ksi

Enter Reinforcement Spacing = 6 in. Es = 29000 ksi

Av = 0.00 in2Lightweight Concrete? NO

Mod Factor ƛ = 1Does ACI 350 Apply? NO Flexure φ = 0.9

Shear φ = 0.75

ACI 318

Governing Specification: USACE

Min. Steel Requirement:

As/Ag (Includes both faces) ≥ 0.0028 EM 1110-2-2104 2.8

Enter Reinforcing configuration of opposite mat:Bar Size = 5

Enter Reinforcement Spacing = 12 in.As = 0.31 in2

Required Min. Steel (both faces): 0.50 sq. in/ft

Max. Spacing Requirements:

s≤5h or s≤18.0 in. 75 in18 in Structural walls/footings:

or for componenets over 36.0 in Spacing can be ≤ 12 in. on faces > 18" thick

s≤12.0 12 in

Maximum Spacing Allowed: 18 in

She

ar

7438-009 Section Being Designed:Longitudinal Stem Reinforcement

Fargo Floodwalls Layer of Steel:

City of Fargo

Modulus of elasticity based on concrete unit weight of

Fle

xure

T&S Steel:

T&S Steel Requirement Met

Max Spacing Requirement Met

11/17/20158:37 AM

Page 48: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

13' unusual input top of wall

****************** Echoprint of Input Data ******************

Date: 2015/11/17 Time: 10.53.38

Structural geometry data: Elevation of top of stem (ELTS) = 13.00 ft Height of stem (HTS) = 13.00 ft Thickness top of stem (TTS) = 1.25 ft Thickness bottom of stem (TBS) = 1.25 ft Dist. of batter at bot. of stem (TBSR)= 0.00 ft Depth of heel (THEEL) = 4.00 ft Distance of batter for heel (BTRH) = 0.00 ft Depth of toe (TTOE) = 2.00 ft Width of toe (TWIDTH) = 3.75 ft Distance of batter for toe (BTRT) = 0.00 ft Width of base (BWIDTH) = 17.00 ft Depth of key (HK) = 2.00 ft Width of bottom of key (TK) = 1.25 ft Dist. of batter at bot. of key (BTRK) = 0.00 ft

Structure coordinates:

x (ft) y (ft) ================== 0.00 -4.00 0.00 0.00 12.00 0.00 12.00 13.00 13.25 13.00 13.25 0.00 17.00 0.00 17.00 -2.00 1.25 -2.00 1.25 -4.00

NOTE: X=0 is located at the left-hand side of the structure. The Y values correspond to the actual elevation used.

Structural property data: Unit weight of concrete = 0.150 kcf

Driving side soil property data:

Moist Saturated Elev. Phi c Unit wt. unit wt. Delta soil (deg) (ksf) (kcf) (kcf) (deg) (ft) ======================================================= 27.00 0.000 0.118 0.118 0.00 5.00

Driving side soil geometry:

Soil Batter Distance point (in:1ft) (ft) ============================= 1 0.00 500.00 2 0.00 0.00 3 0.00 500.00

Page 1

Page 49: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

13' unusual input top of wall

Driving side soil profile:

Soil x y point (ft) (ft) ============================= 1 -1488.00 5.00 2 12.00 5.00

Resisting side soil property data:

Moist Saturated Elev. Phi c Unit wt. unit wt. soil Batter (deg) (ksf) (kcf) (kcf) (ft) (in:1ft) ======================================================== 27.00 0.000 0.118 0.118 5.00 0.00

Resisting side soil profile:

Soil x y point (ft) (ft) ============================= 1 13.25 5.00 2 513.25 5.00

Foundation property data: phi for soil-structure interface = 27.00 (deg) c for soil-structure interface = 0.000 (ksf) phi for soil-soil interface = 27.00 (deg) c for soil-soil interface = 0.000 (ksf)

Water data: Driving side elevation = 13.00 ft Resisting side elevation = 5.00 ft Unit weight of water = 0.0625 kcf Seepage pressures computed by Line of Creep method.

Minimum required factors of safety: Sliding FS = 1.50 Overturning = 100.00% base in compression

Crack options: o Crack *is* down to bottom of heel o Computed cracks *will* be filled with water

Strength mobilization factor = 0.6667

At-rest pressures on the resisting side *are used* in the overturning analysis.

Forces on the resisting side *are used* in the sliding analysis.

*Do* iterate in overturning analysis.

***** Summary of Results *****

*************** *** Not Satisfied *** * Overturning * Required base in comp. = 100.00 % *************** Actual base in comp. = 95.51 % Overturning ratio = 1.40

Page 2

Page 50: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

13' unusual input top of wall

To increase stability try one or a combination of the following: 1. Increase the base width 2. Slope the base of the structure 3. Lower the wall base 4. Add/remove key

Xr (measured from toe) = 5.41 ft Resultant ratio = 0.3184 Stem ratio = 0.2206 Base pressure at x= 16.24 ft from toe = 0.0000 ksf Base pressure at toe = 1.2809 ksf

*********** *** Satisfied *** * Sliding * Min. Required = 1.50 *********** Actual FS = 2.15

********************** Output Results **********************

Date: 2015/11/17 Time: 10.53.38

*************************** ** Overturning Results ** ***************************

Solution converged in 1 iterations.

SMF used to calculate K's = 0.6667 Alpha for the SMF = 0.0000 Calculated earth pressure coefficients: Driving side at rest K = 0.0000 Driving side at rest Kc = 0.0000 Resisting side at rest K = 0.5460 Resisting side at rest Kc = 0.7389 At-rest K's for resisting side calculated.

Depth of cracking = 9.00 ft Crack extends to bottom of base of structure.

** Driving side pressures **

Water pressures: Elevation Pressure (ft) (ksf) ====================== 13.00 0.0000 -4.00 1.0625

** Resisting side pressures **

Water pressures: Elevation Pressure (ft) (ksf) ====================== 5.00 0.0000 -2.00 0.5826 -2.00 1.0392 -4.00 1.0625

Page 3

Page 51: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

13' unusual input top of wall

Earth pressures: Elevation Pressure (ft) (ksf) ====================== 5.00 0.0000 -2.00 0.1329

Balancing earth pressures: Elevation Pressure (ft) (ksf) ====================== -2.00 2.2127 -4.00 2.2127

** Uplift pressures **

Water pressures: x-coord. Pressure (ft) (ksf) ====================== 0.00 1.0625 1.25 1.0392 1.25 0.8768 17.00 0.5826

** Forces and moments **

======================================================================== Part | Force (kips) | Mom. Arm | Moment | | Vert. | Horiz.| (ft) | (ft-k) | ======================================================================== Structure: Structure weight........... 7.913 -7.60 -60.15 Structure, driving side: Moist soil................. 0.000 0.00 0.00 Saturated soil............. 7.080 -11.00 -77.88 Water above structure...... 0.000 0.00 0.00 Water above soil........... 6.000 -11.00 -66.00 External vertical loads.... 0.000 0.00 0.00 Ext. horz. pressure loads.. 0.000 0.00 0.00 Ext. horz. line loads...... 0.000 0.00 0.00 Structure, resisting side: Moist soil................. 0.000 0.00 0.00 Saturated soil............. 2.213 -1.88 -4.15 Water above structure...... 0.000 0.00 0.00 Water above soil........... 0.000 0.00 0.00 Driving side: Effective earth loads...... 0.000 0.00 0.00 Shear (due to delta)....... 0.000 0.00 0.00 Horiz. surcharge effects... 0.000 0.00 0.00 Water loads................ 9.031 3.67 33.11 Resisting side: Effective earth loads...... -0.465 2.33 -1.09 Balancing earth load....... -4.425 -1.00 4.43 Water loads................ -4.141 0.64 -2.65 Foundation: Vertical force on base..... -10.399 -5.41 56.28 Uplift..................... -12.806 -9.22 118.10 ======================================================================== ** Statics Check ** SUMS = 0.000 0.000 0.00

Angle of base = 6.71 degreesPage 4

Page 52: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

13' unusual input top of wall Normal force on base = 10.844 kips Shear force on base = 3.180 kips Max. available shear force = 6.391 kips

Base pressure at x= 16.24 ft from toe = 0.0000 ksf Base pressure at toe = 1.2809 ksf

Xr (measured from toe) = 5.41 ft Resultant ratio = 0.3184 Stem ratio = 0.2206 Base in compression = 95.51 % Overturning ratio = 1.40

Volume of concrete = 1.95 cubic yds/ft of wall

NOTE: The engineer shall verify that the computed bearing pressures below the wall do not exceed the allowable foundation bearing pressure, or, perform a bearing capacity analysis using the program CBEAR. Also, the engineer shall verify that the base pressures do not result in excessive differential settlement of the wall foundation.

*********************** ** Sliding Results ** ***********************

Solution converged. Summation of forces = 0.

Horizontal Vertical Wedge Loads Loads Number (kips) (kips) ================================== 1 0.000 0.000 2 9.031 6.000 3 0.000 0.000

Water pressures on wedges:

Top Bottom Wedge press. press. x-coord. press. number (ksf) (ksf) (ft) (ksf) ================================================ 1 0.0000 0.0000 2 0.0000 1.0625 2 17.0000 0.5826 3 0.0000 0.5826

Points of sliding plane: Point 1 (left), x = 0.00 ft, y = -4.00 ft Point 2 (right), x = 17.00 ft, y = -2.00 ft

Depth of cracking = 9.00 ft Crack extends to bottom of base of structure.

Failure Total Weight Submerged Uplift Wedge angle length of wedge length force number (deg) (ft) (kips) (ft) (kips) ======================================================== 1 0.000 0.000 0.000 0.000 0.000

Page 5

Page 53: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

13' unusual input top of wall 2 6.710 17.117 18.916 17.117 14.080 3 38.366 11.278 3.652 11.278 3.285

Wedge Net force number (kips) =================== 1 0.000 2 -3.401 3 3.401 =================== SUM = 0.000

+-----------------------------+ | Factor of safety = 2.152 | +-----------------------------+

*********************** ** Bearing Results ** ***********************

Base width = 17.117 (ft) Xr = 5.412 (ft) Effective base width = 10.899 (ft) (measured along slope) Base slope = 6.7098 (deg)

Phi = 27.000 (deg) C = 0.000 (ksf) Effective Gamma = 0.0555 (kcf)

Normal load = 10.844 (kips) Load inclination = 16.343 (deg) Load eccentricity = 3.109 (ft)

Surcharge = 0.3885 (kips) Embedment = 7.000 (ft) Ground slope = 0.0000 (deg)

Bearing Capacity Factors ============================================= C Q G ============================================= Bearing 23.9422 13.1992 9.4626 Embedment 1.2096 1.1048 1.1048 Inclination 0.6698 0.6698 0.1558 Base Tilt 0.8747 0.8842 0.8842 Ground Slope 1.0000 1.0000 1.0000

Net ultimate bearing pressure = 3.5366 (ksf)

+-------------------------------+ | Factor of safety = 3.554 | +-------------------------------+

Page 6

Page 54: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,
Page 55: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,
Page 56: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

Backfill Pressure (Water Layer)

γeff γsat γw - 55.5 lb ft³ / 62.5 lb ft³ / - 7 lb ft³ / - = = =

φ 27° φsat =

γ 7 lb ft³ γeff / - =

φ φsat =

γ γeff =

At Rest Earth Pressure Theory -

Ko 1 sin φ - 1 sin 27° - 0.5460 = = =

σh Ko γ H 0.5460 7 lb ft³ / - 15 ft 57.33 psf - = = =

Lateral Earth Pressure (water layer)

σh Ko γ H 0.5460 118 lb ft³ / 13 ft 837.6 psf = = =

Lateral Earth Pressure (water layer, stem only)

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Page 57: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

Passive Pressure

7 ft

γ = 118 lb/ft³φ = 27°

579.9 psf

9 ft

217.4 lb/in

1 ft

At Rest Earth Pressure Theory -

Ko 1 sin φ - 1 sin 27° - 0.5460 = = =

σh Ko γ H 0.5460 118 lb ft³ / 9 ft 579.9 psf = = =

Lateral Earth Pressure

Water Pressure

13

ft

-937.5 psf7 ft 585.9 lb/in

7 ft

-812.5 psf440.1 lb/in

σw γw Hw 62.5 lb ft³ / 15 ft 937.5 psf = = =

Lateral Water Pressure

σw γw Hw 62.5 lb ft³ / 13 ft 812.5 psf = = =

Lateral Water Pressure (stem only)

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Page 58: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

Wall/Soil Weights

425 lb/in

203.1 lb/in

31.25 lb/in

590 lb/in184.4 lb/in

Bearing Pressure

1782 psf

242.5 psf

1434 lb/in

6.35 ft

e = 25.85 in

501.8 lb/in

F µ R 0.350 1434 lb in / 501.8 lb in / = = =

Friction

Bearing Pressure Calculation

Contributing ForcesVert Force ...offset Horz Force ...offset OT Moment

Backfill Pressure 0 lb/in - 35.83 lb/in 5 ft -25798.95 in·lb/ftWater Pressure -0 lb/in - -585.94 lb/in 5 ft 421875 in·lb/ftFooting Weight -425 lb/in 8.5 ft 0 lb/in - -520200 in·lb/ftStem Weight -203.13 lb/in 4.38 ft 0 lb/in - -127968.75 in·lb/ftKey Weight -31.25 lb/in 16.38 ft 0 lb/in - -73687.5 in·lb/ftBackfill Weight -590 lb/in 11 ft 0 lb/in - -934560 in·lb/ftSoil over toe Weight -184.38 lb/in 1.88 ft 0 lb/in - -49781.25 in·lb/ft

-1433.75 lb/in -1310121.45 in·lb/ft1310121.45 in·lb ft / -

1433.75 lb in / - 6.35 ft =

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Page 59: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

Overturning Check

Overturning MomentsForce Distance Moment

Backfill pressure (horz) -21.5 lb/in 5 ft -15479.37 in·lb/ftWater pressure 585.9 lb/in 5 ft 421875 in·lb/ft

Total: 406396 in·lb/ft

Resisting MomentsForce Distance Moment

Passive pressure @ toe 130.5 lb/in 1 ft 18788 in·lb/ftFooting Weight -425 lb/in 8.5 ft 520200 in·lb/ftStem Weight -203.13 lb/in 4.38 ft 127969 in·lb/ftKey Weight -31.25 lb/in 16.38 ft 73688 in·lb/ftBackfill Weight -354 lb/in 11 ft 560736 in·lb/ftSoil over toe Weight -110.63 lb/in 1.88 ft 29869 in·lb/ft

Total: 1331249 in·lb/ft

F.S. RM

OTM

1331249 in·lb ft / 406396 in·lb ft /

3.276 > 2.00 OK = = =

Sliding Check

Sliding Force(s)Backfill pressure -21.5 lb/inWater pressure 585.9 lb/inTotal: 564.4 lb/in

Resisting Force(s)Passive pressure @ toe 130.5 lb/inFriction 393.4 lb/inTotal: 523.9 lb/in

F.S. RFSF

523.9 lb in / 564.4 lb in /

0.928 < 1.50 FAILS = = =

Bearing Capacity Check

Bearing pressure < allowable (1397 psf < 6000 psf) - OKBearing resultant eccentricity < allowable (25.85 in < 51 in) - OK

Wall Top Displacement

(based on unfactored service loads)

Deflection due to stem flexural displacement 0.21 inDeflection due to rotation from settlement 0.082 inTotal deflection at top of wall (positive towards toe) 0.292 in

Stability Checks [1.0D + 1.0F + 0.6H]

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Page 60: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

Overturning Check

Overturning MomentsForce Distance Moment

Backfill pressure (horz) -35.83 lb/in 5 ft -25798.95 in·lb/ftWater pressure 585.9 lb/in 5 ft 421875 in·lb/ft

Total: 396076 in·lb/ft

Resisting MomentsForce Distance Moment

Passive pressure @ toe 217.4 lb/in 1 ft 31313 in·lb/ftFooting Weight -425 lb/in 8.5 ft 520200 in·lb/ftStem Weight -203.13 lb/in 4.38 ft 127969 in·lb/ftKey Weight -31.25 lb/in 16.38 ft 73688 in·lb/ftBackfill Weight -590 lb/in 11 ft 934560 in·lb/ftSoil over toe Weight -184.38 lb/in 1.88 ft 49781 in·lb/ft

Total: 1737510 in·lb/ft

F.S. RM

OTM

1737510 in·lb ft / 396076 in·lb ft /

4.387 > 2.00 OK = = =

Sliding Check

Sliding Force(s)Backfill pressure -35.83 lb/inWater pressure 585.9 lb/inTotal: 550.1 lb/in

Resisting Force(s)Passive pressure @ toe 217.4 lb/inFriction 501.8 lb/inTotal: 719.3 lb/in

F.S. RFSF

719.3 lb in / 550.1 lb in /

1.307 < 1.50 FAILS = = =

Bearing Capacity Check

Bearing pressure < allowable (1782 psf < 6000 psf) - OKBearing resultant eccentricity < allowable (25.85 in < 51 in) - OK

Wall Top Displacement

(based on unfactored service loads)

Deflection due to stem flexural displacement 0.21 inDeflection due to rotation from settlement 0.082 inTotal deflection at top of wall (positive towards toe) 0.292 in

Stability Checks [1.0D + 1.0F + 1.0H]

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Page 61: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

13

11.7

10.4

9.1

7.8

6.5

5.2

3.9

2.6

1.3

0-80 -63.33 -46.67 -30 -13.33 3.33 20

Moment (ft·k/ft)

Offset (ft)

Moment

aAs fy

0.85 F'c

0.07 in² in / 60000 psi 0.85 4500 psi

1.15 in = = =

φMn φ As fy d a 2 / - 0.90 0.07 in² in / 60000 psi 11.63 in 1.15 in 2 / - 43.76 ft·k ft / = = =

Capacity (ACI 318-05 10.2) @ 0 ft from base [Negative bending]

aAs fy

0.85 F'c

0.03 in² in / 60000 psi 0.85 4500 psi

0.41 in = = =

φMn φ As fy d a 2 / - 0.90 0.03 in² in / 60000 psi 11.69 in 0.41 in 2 / - 16.02 ft·k ft / = = =

Capacity (ACI 318-05 10.2) @ 0 ft from base [Positive bending]

aAs fy

0.85 F'c

0.07 in² in / 60000 psi 0.85 4500 psi

1.15 in = = =

φMn φ As fy d a 2 / - 0.90 0.07 in² in / 60000 psi 11.63 in 1.15 in 2 / - 43.76 ft·k ft / = = =

Capacity (ACI 318-05 10.2) @ 11.66 ft from base [Negative bending]

aAs fy

0.85 F'c

0.03 in² in / 60000 psi 0.85 4500 psi

0.41 in = = =

φMn φ As fy d a 2 / - 0.90 0.03 in² in / 60000 psi 11.69 in 0.41 in 2 / - 16.02 ft·k ft / = = =

Capacity (ACI 318-05 10.2) @ 11.88 ft from base [Positive bending]

aAs fy

0.85 F'c

0 in² in / 60000 psi 0.85 4500 psi

0 in = = =

φMn φ As fy d a 2 / - 0.90 0 in² in / 60000 psi 11.63 in 0 in 2 / - 0 ft·k ft / = = =

Capacity (ACI 318-05 10.2) @ 13 ft from base [Negative bending]

aAs fy

0.85 F'c

0 in² in / 60000 psi 0.85 4500 psi

0 in = = =

φMn φ As fy d a 2 / - 0.90 0 in² in / 60000 psi 11.69 in 0 in 2 / - 0 ft·k ft / = = =

Capacity (ACI 318-05 10.2) @ 13 ft from base [Positive bending]

Stem Flexural Capacity

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Page 62: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

13

11.7

10.4

9.1

7.8

6.5

5.2

3.9

2.6

1.3

0-15 -9.17 -3.33 2.5 8.33 14.17 20

Shear (k/ft)

Offset (ft)

Shear

Vc 2 F'c d 2 4500 psi 11.63 in 18.72 k ft / = = =

φVn φ Vc 0.750 18.72 k ft / 14.04 k ft / = = =

Shear Capacity (ACI 318-05 11.1.1, 11.3.1) @ 0 ft from base [Positive shear]

Vc 2 F'c d 2 4500 psi 11.63 in 18.72 k ft / = = =

φVn φ Vc 0.750 18.72 k ft / 14.04 k ft / = = =

Shear Capacity (ACI 318-05 11.1.1, 11.3.1) @ 0 ft from base [Negative shear]

Vc 2 F'c d 2 4500 psi 11.63 in 18.72 k ft / = = =

φVn φ Vc 0.750 18.72 k ft / 14.04 k ft / = = =

Shear Capacity (ACI 318-05 11.1.1, 11.3.1) @ 13 ft from base [Positive shear]

Vc 2 F'c d 2 4500 psi 11.63 in 18.72 k ft / = = =

φVn φ Vc 0.750 18.72 k ft / 14.04 k ft / = = =

Shear Capacity (ACI 318-05 11.1.1, 11.3.1) @ 13 ft from base [Negative shear]

Stem Shear Capacity

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Page 63: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

ψe 1.0 uncoated hooked bars =

λ 1.0 normal weight concrete =

ldh 0.02 ψe λ fy

F'c

db 0.02 1.0 1.060000 psi

4500 psi0.75 in 13.42 in = = =

Factoring ldh by the 0.7 multiplier of 12.5.3 a : ldh 9.39 in =

8 db 8 0.75 in 6.0 minimum limit, does not control = =

Main vertical stem bars (bottom end) - Development Length Calculation (ACI 318-05 12.2.3, 12.5)

ψt 1.0 bars are not horizontal =

ψe 1.0 bar not epoxy coated =

ψs 0.80 bars are #6 or smaller =

λ 1.0 normal weight concrete =

s 2 / 6 in 2 / 3 in = =

cover db 2 / + 3 in 0.75 in 2 / + 3.38 in = =

cb 3 in lesser of half spacing, ctr to surface =

Ktr 0.0 no transverse reinforcement =

cb Ktr + db

3 in 0.0 + 0.75 in

4.0 = =

ld3.40

fy

F'c

ψt ψe ψs λ 2.5

db 3.40

60000 psi

4500 psi

1.0 1.0 0.80 1.0 2.5

0.75 in 16.1 in = = =

Main vertical stem bars (top end) - Development Length Calculation (ACI 318-05 12.2.3, 12.5)

ψt 1.0 bars are not horizontal =

ψe 1.0 bar not epoxy coated =

ψs 0.80 bars are #6 or smaller =

λ 1.0 normal weight concrete =

s 2 / 6 in 2 / 3 in = =

cover db 2 / + 3 in 0.75 in 2 / + 3.38 in = =

cb 3 in lesser of half spacing, ctr to surface =

Ktr 0.0 no transverse reinforcement =

cb Ktr + db

3 in 0.0 + 0.75 in

4.0 = =

ld3.40

fy

F'c

ψt ψe ψs λ 2.5

db 3.40

60000 psi

4500 psi

1.0 1.0 0.80 1.0 2.5

0.75 in 16.1 in = = =

Dowels for vertical stem bars (top end) - Development Length Calculation (ACI 318-05 12.2.3, 12.5)

ψe 1.0 uncoated hooked bars =

λ 1.0 normal weight concrete =

ldh 0.02 ψe λ fy

F'c

db 0.02 1.0 1.060000 psi

4500 psi0.63 in 11.18 in = = =

Factoring ldh by the 0.7 multiplier of 12.5.3 a : ldh 7.83 in =

8 db 8 0.63 in 5.0 minimum limit, does not control = =

2nd curtain vertical bars (bottom end) - Development Length Calculation (ACI 318-05 12.2.3, 12.5)

Stem Development/Lap Length Calculations

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Page 64: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

ψt 1.0 bars are not horizontal =

ψe 1.0 bar not epoxy coated =

ψs 0.80 bars are #6 or smaller =

λ 1.0 normal weight concrete =

s 2 / 12 in 2 / 6 in = =

cover db 2 / + 3 in 0.63 in 2 / + 3.31 in = =

cb 3.31 in lesser of half spacing, ctr to surface =

Ktr 0.0 no transverse reinforcement =

cb Ktr + db

3.31 in 0.0 + 0.63 in

5.30 = =

ld3.40

fy

F'c

ψt ψe ψs λ 2.5

db 3.40

60000 psi

4500 psi

1.0 1.0 0.80 1.0 2.5

0.63 in 13.42 in = = =

2nd curtain vertical bars (top end) - Development Length Calculation (ACI 318-05 12.2.3, 12.5)

Stem Development/Lap Length Calculations (continued)

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Page 65: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

Design moment Mu for toe need not exceed moment at stem base:

Mtoe 9.3 ft·k ft < Mstem / 79.01 ft·k ft / = =

Mu 9.3 ft·k ft stem moment does not control / =

Controlling Moment

aAs fy

0.85 F'c

0.05 in² in / 60000 psi 0.85 4500 psi

0.81 in = = =

φMn φ As fy d a 2 / - 0.90 0.05 in² in / 60000 psi 19.69 in 0.81 in 2 / - 53.8 ft·k ft / = = =

φMn 53.8 ft·k ft ≥ Mu / 9.3 ft·k ft / = = �

Flexure Check (ACI 318-05 10.2)

Vc 2 F'c d 2 4500 psi 19.69 in 31.7 k ft / = = =

φVn φ Vc 0.750 31.7 k ft / 23.77 k ft / = = =

φVn 23.77 k ft ≥ Vu / 2.87 k ft / = = �

Shear Check (ACI 318-05 11.1.1, 11.3.1)

β1 0.85 0.05F'c 4000 -

1000 - 0.85 0.05

4500 psi 4000 - 1000

- 0.8250 = = =

aAs fy

0.85 F'c

0.05 in² in / 60000 psi 0.85 4500 psi

0.81 in = = =

εt 0.003d

a β1 / 1 - 0.003

19.69 in0.81 in 14.0 /

1 - 1.0173 = = =

εt 1.0173 ≥ 0.004 = �

Minimum Strain Check (ACI 318-05 10.3.5)

φMn 53.8 ft·k ft ≥ 4 3 / Mu / 4 3 / 9.3 ft·k ft / 12.41 ft·k ft / = = =

Check is waived per ACI 10.5.3�

Minimum Steel Check (ACI 318-05 10.5.1)

ρST_provAST

t sST

1.58 in² in / 24 in 12 in

0.0055 = = =

ρST_min0.0018 60000

fy

0.0018 60000 60000 psi

0.0018 = = =

ρST_min 0.0018 =

ρST_prov 0.0055 ≥ ρST_min 0.0018 = = �18 inch limit governs

sST_max 18 in =

sST 12 in ≤ sST_max 18 in = = �

Shrinkage Temperature Steel (ACI 318-05 7.12.2)

Mu

φMn

9.3 ft·k ft / 53.8 ft·k ft /

0.1730 ratio to represent excess reinforcement = =

ψt 1.0 12 inches or less cast below 4.00 inches - =

ψe 1.0 bar not epoxy coated =

ψs 0.80 bars are #6 or smaller =

λ 1.0 normal weight concrete =

s 2 / 6 in 2 / 3 in = =

cover db 2 / + 4 in 0.63 in 2 / + 4.31 in = =

cb 3 in lesser of half spacing, ctr to surface =

Ktr 0.0 no transverse reinforcement =

cb Ktr + db

3 in 0.0 + 0.63 in

4.80 = =

ld3.40

fy

F'c

ψt ψe ψs λ 2.5

db 3.40

60000 psi

4500 psi

1.0 1.0 0.80 1.0 2.5

0.63 in 13.42 in = = =

Factoring ld by the excess reinforcement ratio 0.1730 per 12.2.5: ld 2.32 in =

12 inch minimum controls

ld_prov 155 in ≥ ld 12 in = = �

Development Check (ACI 318-05 12.12, 12.2.3)

Toe Unfactored Loads

24

in

#5 @ 6 in

Unfactored Loads

300 psf (Self-wt)

590 psf (Soil)

1782 psf 1442 psf

Toe Factored Loads

24

in

#5 @ 6 in

1.7D + 1.7F + 1.7H

510 psf (Self-wt)

1003 psf (Soil)

3029 psf 2452 psf

3029 psf2452 psf

4.6 k/ft

Toe Checks [1.7D + 1.7F + 1.7H]

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Page 66: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

Design moment Mu for heel need not exceed moment at stem base:

Mheel 34.93 ft·k ft < Mstem / 79.01 ft·k ft / = =

Mu 34.93 ft·k ft stem moment does not control / =

Controlling Moment

aAs fy

0.85 F'c

0.07 in² in / 60000 psi 0.85 4500 psi

1.15 in = = =

φMn φ As fy d a 2 / - 0.90 0.07 in² in / 60000 psi 19.63 in 1.15 in 2 / - 75.44 ft·k ft / = = =

φMn 75.44 ft·k ft ≥ Mu / 34.93 ft·k ft / = = �

Flexure Check (ACI 318-05 10.2)

Vc 2 F'c d 2 4500 psi 19.63 in 31.6 k ft / = = =

φVn φ Vc 0.750 31.6 k ft / 23.7 k ft / = = =

φVn 23.7 k ft ≥ Vu / 2.13 k ft / = = �

Shear Check (ACI 318-05 11.1.1, 11.3.1)

β1 0.85 0.05F'c 4000 -

1000 - 0.85 0.05

4500 psi 4000 - 1000

- 0.8250 = = =

aAs fy

0.85 F'c

0.07 in² in / 60000 psi 0.85 4500 psi

1.15 in = = =

εt 0.003d

a β1 / 1 - 0.003

19.63 in1.15 in 14.0 /

1 - 0.7135 = = =

εt 0.7135 ≥ 0.004 = �

Minimum Strain Check (ACI 318-05 10.3.5)

φMn 75.44 ft·k ft ≥ 4 3 / Mu / 4 3 / 34.93 ft·k ft / 46.57 ft·k ft / = = =

Check is waived per ACI 10.5.3�

Minimum Steel Check (ACI 318-05 10.5.1)

ρST_provAST

t sST

1.58 in² in / 24 in 12 in

0.0055 = = =

ρST_min0.0018 60000

fy

0.0018 60000 60000 psi

0.0018 = = =

ρST_min 0.0018 =

ρST_prov 0.0055 ≥ ρST_min 0.0018 = = �18 inch limit governs

sST_max 18 in =

sST 12 in ≤ sST_max 18 in = = �

Shrinkage Temperature Steel (ACI 318-05 7.12.2)

Mu

φMn

34.93 ft·k ft / 75.44 ft·k ft /

0.4630 ratio to represent excess reinforcement = =

ψt 1.30 more than 12 inches cast below 19.25 inches - =

ψe 1.0 bar not epoxy coated =

ψs 0.80 bars are #6 or smaller =

λ 1.0 normal weight concrete =

s 2 / 6 in 2 / 3 in = =

cover db 2 / + 4 in 0.75 in 2 / + 4.38 in = =

cb 3 in lesser of half spacing, ctr to surface =

Ktr 0.0 no transverse reinforcement =

cb Ktr + db

3 in 0.0 + 0.75 in

4.0 = =

ld3.40

fy

F'c

ψt ψe ψs λ 2.5

db 3.40

60000 psi

4500 psi

1.30 1.0 0.80 1.0 2.5

0.75 in 20.93 in = = =

Factoring ld by the excess reinforcement ratio 0.4630 per 12.2.5: ld 9.69 in =

12 inch minimum controls

ld_prov 56 in ≥ ld 12 in = = �

Development Check (ACI 318-05 12.12, 12.2.3)

Heel Unfactored Loads

24

in#6 @ 6 in

Unfactored Loads

300 psf (Concrete self-wt)

590 psf (Soil weight)

1329 psf

242.5 psf

Heel Factored Loads

24

in#6 @ 6 in

1.7D + 1.7F + 1.7H

510 psf (Concrete self-wt)

1003 psf (Soil weight)

412.3-2259 psf (Bearing pressure)

2259 psf

412.3 psf

2.13 k/ft

Heel Checks [1.7D + 1.7F + 1.7H]

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Page 67: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

Stem Internal Forces

-1423.88 psf-1381.25 psf18.23 k/ft

-79.01 ft·k/ft

Stem Internal Forces

13

11.38

9.75

8.13

6.5

4.88

3.25

1.63

0-80 -60 -40 -20 0

Moment (ft·k/ft)

Moment

Stem Internal Forces

13

11.38

9.75

8.13

6.5

4.88

3.25

1.63

00 5 10 15 20

Shear (k/ft)

Shear

Stem Joint Force Transfer

Location Force@ stem base 18.23 k/ft

Stem Internal Forces

-1423.88 psf -1381.25 psf

Stem Forces [1.7D + 1.7F + 1.7H]

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Page 68: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,
Page 69: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,
Page 70: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

φMn 43.76 ft·k ft < 4 3 / Mu / 4 3 / 79.01 ft·k ft / 105.3 ft·k ft / = = =

As_min3 F'c

fyd

3 4500 psi 60000 psi

11.63 in 0.04 in² in / = = =

200 d fy / 200 11.63 in 60000 psi / 0.04 in² in / = =

As 0.07 in² in ≥ As_min / 0.04 in² in / = = �

Minimum Steel Check (ACI 318-05 10.5.1) @ 0 ft from base [Stem in negative flexure]

φMn 43.76 ft·k ft < 4 3 / Mu / 4 3 / 33 ft·k ft / 44 ft·k ft / = = =

As_min3 F'c

fyd

3 4500 psi 60000 psi

11.63 in 0.04 in² in / = = =

200 d fy / 200 11.63 in 60000 psi / 0.04 in² in / = =

As 0.07 in² in ≥ As_min / 0.04 in² in / = = �

Minimum Steel Check (ACI 318-05 10.5.1) @ 3.33 ft from base [Stem in negative flexure]

φMn 0 ft·k ft ≥ 4 3 / Mu / 4 3 / 0 ft·k ft / 0 ft·k ft / = = =

Check is waived per ACI 10.5.3�

Minimum Steel Check (ACI 318-05 10.5.1) @ 13 ft from base [Stem in negative flexure]

β1 0.85 0.05F'c 4000 -

1000 - 0.85 0.05

4500 psi 4000 - 1000

- 0.8250 = = =

aAs fy

0.85 F'c

0.07 in² in / 60000 psi 0.85 4500 psi

1.15 in = = =

εt 0.003d

a β1 / 1 - 0.003

11.63 in1.15 in 14.0 /

1 - 0.4214 = = =

εt 0.4214 ≥ 0.004 = �

Maximum Steel Check (ACI 318-05 10.3.5) @ 0 ft from base [Stem in negative flexure]

β1 0.85 0.05F'c 4000 -

1000 - 0.85 0.05

4500 psi 4000 - 1000

- 0.8250 = = =

aAs fy

0.85 F'c

0.07 in² in / 60000 psi 0.85 4500 psi

1.15 in = = =

εt 0.003d

a β1 / 1 - 0.003

11.63 in1.15 in 14.0 /

1 - 0.4214 = = =

εt 0.4214 ≥ 0.004 = �

Maximum Steel Check (ACI 318-05 10.3.5) @ 3.33 ft from base [Stem in negative flexure]

β1 0.85 0.05F'c 4000 -

1000 - 0.85 0.05

4500 psi 4000 - 1000

- 0.8250 = = =

aAs fy

0.85 F'c

0.07 in² in / 60000 psi 0.85 4500 psi

1.15 in = = =

εt 0.003d

a β1 / 1 - 0.003

11.63 in1.15 in 14.0 /

1 - 0.4214 = = =

εt 0.4214 ≥ 0.004 = �

Maximum Steel Check (ACI 318-05 10.3.5) @ 13 ft from base [Stem in negative flexure]

Stem Miscellaneous Checks [1.7D + 1.7F + 1.7H]

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Page 71: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

ρhAs_horz shorz /

t

0.62 in² 12 in / 15 in

0.0034 = = =

ρh_min 0.0020 bars No. 5 or less, not less than 60 ksi =

ρh 0.0034 ≥ ρh_min 0.0020 = = �3 twall 3 15 in 45 in = =

18 inch limit governs

smax 18 in =

shorz 12 in ≤ shorz_max 18 in = = �

Wall Horizontal Steel (ACI 318-05 14.3.3, 14.3.5)

Mu

φMn

79.01 ft·k ft / 43.76 ft·k ft /

1.8057 ratio to represent excess reinforcement = =

ψe 1.0 uncoated hooked bars =

λ 1.0 normal weight concrete =

ldh 0.02 ψe λ fy

F'c

db 0.02 1.0 1.060000 psi

4500 psi0.75 in 13.42 in = = =

Factoring ldh by the 0.7 multiplier of 12.5.3 a : ldh 9.39 in =

8 db 8 0.75 in 6.0 minimum limit, does not control = =

ldh_prov 20 in ≥ ldh 9.39 in = = �

Development Check (ACI 318-05 12.12, 12.2.3)

ψt 1.0 bars are not horizontal =

ψe 1.0 bar not epoxy coated =

ψs 0.80 bars are #6 or smaller =

λ 1.0 normal weight concrete =

s 2 / 6 in 2 / 3 in = =

cover db 2 / + 3 in 0.75 in 2 / + 3.38 in = =

cb 3 in lesser of half spacing, ctr to surface =

Ktr 0.0 no transverse reinforcement =

cb Ktr + db

3 in 0.0 + 0.75 in

4.0 = =

ld3.40

fy

F'c

ψt ψe ψs λ 2.5

db 3.40

60000 psi

4500 psi

1.0 1.0 0.80 1.0 2.5

0.75 in 16.1 in = = =

llap 1.3 ld 1.3 16.1 in 20.93 in = = =

llap_prov 40 in ≥ llap 20.93 in = = �1 5 / llap 1 5 / 20.93 in 4.1859 ≤ 6.0 = =

strans 0 in ≤ 1 5 / llap 1 5 / 20.93 in 4.1859 = = = �

Lap Splice Checks (ACI 318-05 12.14.2.3, 12.15.1, 12.15.2) - #6 lap with #6, from 0 ft to 3.33 ft (from stem base)

Stem Miscellaneous Checks [1.7D + 1.7F + 1.7H] (continued)

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Page 72: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

12 Ft. Floodwall

(Case 4)

Page 73: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Site: 7438-009 Owner:

Case:

User Data:A 15 Inches

B 17 Feet

C 12 FeetD 24 InchesD1 24 Inches

Hwall 13 Feet

Hwater 9.31 Feet

H2 5 FeetH3 5 Feet

Unit Weight of Soil (ϒsat):

Below Footing 0.118 k/cu. ft.Wet Side 0.118 k/cu. ft.Dry Side 0.118 k/cu. ft.

Ø= Below Footing 27 Degrees0.471 radians

Wet Side 27 Degrees0.471 radians

Dry Side 27 Degrees0.471 radians

SMF= 0.66667Unit Weight of Conc. (ρc) 0.15 k/cu. ft.

Unit Weight of Water (ρw) 0.0625 k/cu. ft.

Soil Capacity 6.2 k/sq. ft.2 Feet Ko Below Footing 0.55

15 Inches Wet Side 0.55Dry Side 0.55

Kp for sliding = 2.663fysteel 60 ksi Cohesion Factor (CF) 1 ksf

f'c(concrete) 4.5 ksi Frost Depth (Fr): 6 ft

Vertical Surcharge (Dry) (P3) 0 psf

Vertical Surcharge (Wet) (P4) 0 psf

Ice/Debris Load YesType of Loading Unusual

Toe Vertical Fill Width (F)= 3.75 ft

*All Dimensions in Calculations are converted to feet

City of Fargo

13' Floodwall Unusual 100 yr event plus ice

Keyway Depth (G)Keyway Width (L)

Project Number:Fargo Floodwalls

1 − sin∅

1 − sin∅1 − sin∅

Page 74: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Force CalculationsFactor Unit Equation Description

EHW= 0.832 kips/ft Driving Wet Side Water Pressure

EHS= 0.500 kips/ft Driving Wet Side Soil Pressure

EV2= 10.313 kips Wet Side Vertical Component Weights

EV2d 11.000 ft from toe Moment arm for EV2

EV1= 2.213 kips Dry Side Vertical Component Weights

EV1d 1.875 ft from toe Moment arm for EV1

PehW= 5.536 kips Driving Wet Side Water Force

Pehwd 2.437 ft. above footing Moment arm for Pehw

PehS= 2.248 kips Driving Wet Side Soil Force

Pehsd 1.000 ft. above footing Moment arm for Pehs

PS1= 2.813 kips Vertical Weight of Stem and Key

PS1d 5.975 ft from toe Moment arm for PS1

PS2= 5.100 kips Vertical Weight of Footing

PS2d 8.500 ft from toe Moment arm for PS2

EH2= 1.062 kips Resisting Dry Side Soil Pressure

Peh2= 4.779 kips Resisting Dry Side Forces

Peh2d 1.000 ft from toe Moment arm for Peh2

Peh2s 2.248 kips Resisting dry side soil only

Peh2sd 1.000 ft from toe Moment arm for Peh2s

Pice/debris 0.500 kips .5 k/ft at Q100 Driving ice/debris Force

Pice/debrisd 11.310 ft. above footing Moment arm for Pice/debris

Pwind 0.1107 kips Driving wind Force

Pwindd 13.155 ft. above footing Moment arm for Pwind

Applied Forces

(��� +�� + �)(ρ)

(�3 + �� + �)(ɤ�� − ρ)

��� � ρ +H3(C)(ɤ�� − ρ)

�2(�)(ɤ��)

��

2(��� + �� + �)

���

2(�3 + �� + �)

��� + �� + �

3− �

�3 + �� + �

3− �

��� � ρ� + �( )(ρ�)

! − � �� ρ� + �(�)(ρ�)

(�2 + �� + �)(ɤ��)

��"

2(�2 + �� + �)

! −�

2

2

��� � ρ� ! − � −�

2+ � ρ� ! −

2/$%�

! − � �� ρ�! − �

2+ � � ρ� ! −

2/$%"

�2 + �� + �

3− �

��� + ��

��� − ���

2+ ��� + ��

ρ� − ρ ∗ �2 + �� + � +�2 + �� + �

2

�2 + �� + �

3− �

.03(��� −��� )

Page 75: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Water Uplift

Force Calculation Factor Unit Equation DescriptionLZ-Z1= 17.117 ft Length of Seepage path from heel to toe

Δh= 4.310 ft Head differential between wet and dry side

Ldry side 7.000 ft Seepage path on Dry Side

LS 24.117 ft Total Seepage Path

uZ= 0.832 ksf Water Pressure at bottom of key (Wet side)

uZ1= 0.516 ksf Water pressure at bottom of footing (Dry Side)

HLZ-Z1= 3.059 ft Head Loss along Z-Z1

LSc 19.000 ft Length of concrete surface in sliding surface

HLLK = 0.201 ft Head Loss along key

ubottom key = 0.819 ksf Water pressure at bottom of key (Dry side)

HLUK = 0.523 ft Head Loss up key

utop key = 0.674 ksf Water pressure at top of key (Dry side)

P5 = 10.402 kips Water Uplift for Overturning

P5d 9.033 ft

P5sa 11.454 kips Water uplift for sliding along angle

P5sad -9.165 ft Moment Arm for P5sa

Lsss = 26.000 ft

Uf = 0.656 ksf

P5ss = 12.645 Water uplift of sliding along bottom of key

P5ssd -8.835 Moment Arm for P5ss

(!"+ �")

��� − �2

�2 + ��

)*)� + + ,�.+�

(��� +�� + �)(ρ)

��� + �� − ∆0 )*)� 1

(ρ)

∆0 )*)� �

! + �

%�(� )*)�)

(��� + �� + � − � 23)ρ

+ �

%�(� )*)�)

(��� +�� −� 43)ρ

5) + 567��789�,

2 +

5�7:9�, + 5)�2

(! − )

5) + 5)�2

(!)

5)�! −

2! − +

5�7:9�, − 5)�2

! − 2

3! − + 567��789�, ! −

2+5) − 567��789�,

2 ! −

3/$5

5)� !!

2+

5) − 5)�2

!2!

3/$5�

! + � + �� +�2

��� +�� + � − ��� −�2!

<��ρ

5= + 5>2

(!)

5> !!

2+5) − 5>

2!

2!

3/$5��

Moment Arm

for P5

Water pressure at key depth below

footing on dry side of footing

Page 76: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Distance Moment

to Toe(ft) About Toe (k-ft)

PS1 2.81 5.98 16.8 MPS1

PS2 5.10 8.50 43.4 MPS2

P3 0.00 1.88 0.0 MP3

P4 0.00 11.00 0.0 MP4

P5 -10.40 9.03 -94.0 MP5

-2.49 13.6 -33.8

EV2 - Backfill on Heel EV2 10.31 11.00 113.4 MEV2

EV1 - Fill on Toe EV1 2.21 1.88 4.1 MEV1

12.53 9.4 117.6

LL Live Load 0.0 0.00 0.0 MLL

0.0 0.0

EH Horiz. Earth Load PehS 2.25 -1.00 -1.2 MEH

WS Hydrostatic Pressure PehW 5.54 -2.44 -13.5 MPHW

ID Ice/Debris Force Pice/debris 0.50 -11.31 -5.7 MID

W Wind Load Pwind 0.11 -13.16 -1.5 MW

8.4 -21.8

Fargo Floodwalls

Ve

rtica

l Lo

ad

s

Applied Force Calculations

Total

P (kips)DescriptorLoad

Vertical Loads

DC

Horizontal Loads

EV

Total

Total

Total

Page 77: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Factor Equation Description

EM 1110-2-2502

Reference

PehW2 = 3.298 kips Water on Dry side for Overturning Calcs Figure 4-5

Pehw2d 0.809 ft Moment Arm for Pehw2 N/A

Peh2sb 1.360 ft Dry Side Soil to bottom of Footing Para. 4-8

Peh2sbd 2.333 kips Moment Arm for Peh2sb N/A

RS = 1.743 kips Resisting soil on Dry side to create equilibrium Para. 4-8

RSd -1 ft Moment Arm for RS N/A

Including Uplift

ΣVo 10.035 kips Total Vertical Load Figure 4-5

ΣMo 65.255 kip-ft 4-1

XR 6.502 ft Resultant Location 4-1

bl 19.507337 ft Length of Base in Compression 4-2

b% 114.74904 Percent of Base in Compression 4-2

Required b% 75 Required Base in Compression Appendix F

OK

Neglecting Uplift

ΣVon 20.438 kips Total Vertical Load Figure 4-5

ΣMon 159.221 kip-ft 4-1

XRn 7.791 ft Resultant Location 4-1

bln 23.371882 ft Length of Base in Compression 4-2

b%n 137.48166 Percent of Base in Compression 4-2

Required b%n 75 Required Base in Compression Appendix 4

OK

Global Stability

Overturning:

Fargo Floodwalls

$%� + $%" + $? + $@ + $A + �B1 + �B2

5)�2

�2 + �� +5�7:9�, + 567��789�,

2(�)

�2 + ��3

5)�2

�2 + ��3

�2 + �� + 5�7:9�,−�2

� +567��789�, − 5�7:9�,

2�

−2�3

/$CD"

−�2

EF1� +EF1" +EF? +EF@ +EFA +EGH� +EGH" +EFIJ +EKL +EJ + M% M%N + $CD"�6 $CD"�6N + $CD"($CD"N)(O7)(%E�)

ΣE7

ΣB7

3QR

S�!(100)

$%� + $%" + $? + $@ + �B1 + �B2

EF1� +EF1" +EF? +EF@ +EGH� +EGH" +EFIJ +EKL +EJ + M% M%N + $CD"�6 $CD"�6N + $CD"($CD"N)(O7)(%E�)

ΣE7T

ΣB7T

3QRT

S�T!

(100)

Total Overturning

Moment about

point o

(�2 + ��)(ɤ�� − ρ)(�2 + ��)/2

$CD − $CD" − $CD"�6(O7)(%E�)

Total Overturning

Moment about

point o

11/17/2015

Page 78: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Factor Equation Description

EM 1110-2-2502

Reference

Sliding Along Angle Between Bottom of Key and Bottom of Footing on Dry Side Figure 4-11

SBFa 1.722 kips Soil Below Footing on Sliding Surface Figure 4-11

Pwda 3.152 kips Water on Dry Side Figure 4-11

ΣVa 10.705 kips Sum of Vertical Forces on Sliding Surface Figure 4-11

αa 0.117 radians Angle of slip plane to horizontal plane Figure 4-11

6.710 degrees

Drained Condition

RFswad 5.318 kips Resisting Force from passive soil and water on dry side Figure 4-11

ΣHad 0.83 kips Sum of Horizontal Forces Figure 4-11

N'ad 10.729 kips Normal Force to Sliding Surface Figure 4-11

Tad -0.428 kips Tangential Force to Sliding Surface Figure 4-11

SSad 5.466 kips Drained Shear Strength 4-12

FSad -12.778 Factor of Safety 4-12

OK

Undrained Condition

RFswau 3.966 kips Resisting Force from passive soil and water on dry side Figure 4-11

ΣHau 2.181 kips Sum of Horizontal Forces Figure 4-11

N'au 10.887 kips Normal Force to Sliding Surface Figure 4-11

Tau 0.915 kips Tangential Force to Sliding Surface Figure 4-11

SSau 11.412 kips Undrained Shear Strength 4-12

FSau 12.467 Factor of Safety 4-12

OK

Sliding:

Fargo Floodwalls

Global Stability

� −�!

2(! − )(ɤ��)

5)2

�2 + �� +5) + 5)�

2(�)

%UV + �B1 + �B2 + $%� + $%" + $? + $@ − $A�

atan�!

$+ + $CD" − $+ Y: (.5)

$CD + $.��/+�6 .� + $.T+ − M��+

ΣB cos \ + Σ�+sin(\)

Σ�+ cos \ − ΣBsin(\)

]′+tan(∅)

%%+_+

$+ + $CD" − $+ (.5)

$CD + $.��/+�6 .� + $.T+ − M��`

ΣB cos \ + Σ�`sin(\)

Σ�` cos \ − ΣBsin(\)

��(%E�)√(!" + �")

%%`_`

11/17/2015

Page 79: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Factor Equation Description

EM 1110-2-2502

Reference

Sliding Along Horizontal Surface along Bottom of Key Figure 4-8

SBFs 3.717 kips Soil Below Footing on Sliding Surface

Pwds 2.951 kips Water on Dry Side Figure 4-11

ΣVs 11.510 kips Sum of Vertical Forces on Sliding Surface Figure 4-11

αs = 0 radians Angle of slip plane to horizontal plane Figure 4-11

0 degrees

Drained Condition

RFswsd 5.385 kips Resisting Force from passive soil and water on dry side Figure 4-11

ΣHsd 0.76 kips Sum of Horizontal Forces Figure 4-11

N'sd 11.510 kips Normal Force to Sliding Surface Figure 4-11

Tsd 0.762 kips Tangential Force to Sliding Surface Figure 4-11

SSsd 5.864 kips Drained Shear Strength 4-12

FSsd 7.698 Factor of Safety 4-12

OK

Undrained Condition

RFswsu 3.865 kips Resisting Force from passive soil and water on dry side Figure 4-11

ΣHsu 2.282 kips Sum of Horizontal Forces Figure 4-11

N'su 11.510 kips Normal Force to Sliding Surface Figure 4-11

Tsu 2.282 kips Tangential Force to Sliding Surface Figure 4-11

SSsu 11.333 kips Undrained Shear Strength 4-12

FSsu 4.967 Factor of Safety 4-12

OK

Global Stabilty

Fargo Floodwalls

�(! − )(ɤ��)

5>2(�2 + �� + �)

%UV� + �B1 + �B2 + $%� + $%" + $? + $@ − $A��

$+� + $CD" − $+� Y: (.5)

$CD + $.��/+�6 .� + $.T+ − M���+

ΣB� cos \� + Σ��+sin(\�)

Σ��+ cos \� − ΣB�sin(\�)

]′�+tan(∅)

%%�+_�+

.

$+� + $CD" − $+� (.5)

$CD + $.��/+�6 .� + $.T+ − M���`

ΣB� cos \� + Σ��`sin(\�)

Σ��` cos \� − ΣB�sin(\�)

�� %E� !

%%�`_�`

11/17/2015

Page 80: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Factor Equation Description

EM 1110-2-2502

Reference

Bearing Along Angle Between Bottom of Key and Bottom of Footing on Dry Side Eqns. From 1110-2-2502

ΣHadb 2.99 kips Sum of Horizontal Forces Figure 4-11

N'adb 10.982 kips Normal Force to Bearing Surface Figure 4-11

Tadb 1.723 kips Tangential Force to Bearing Surface Figure 4-11

SBFad 10.500005 ft Moment Arm for SBFa Figure 4-11

ΣMoaB 72.793992 kip-ft

XraB 6.800 ft Resultant Location Figure 4-11

eaB 1.759 ft Eccentricity of resultant Figure 4-11

B'aB 13.599898 ft Effective Base for Bearing Sec 5-2

Drained Condition

q0 = 0.333 k/ft2 Effective Overburden Pressure

y = 1.6007185

Nq = 13.175528 Bearing Capacity Factor 5-3a

Nc = 23.895844 Bearing Capacity Factor 5-3b

Nϒ = 9.4442971 Bearing Capacity Factor 5-3d

εcd = 1.14 Embedment Factor 5-4a

εqd = 1.07 Embedment Factor 5-4c

δd = 0.16 5-5

8.917254

εqi = 0.81 Inclination Factor 5-5a

εϒi = 0.45 Inclination Factor 5-5b

Q1 = 0.00 5-2

Q2 = 3.8172486 5-2

Q3 = 1.713704 5-2

Qd = 75.220391 5-2

FS = 5-1

6.8496617 Eq. 5-1 Bearing Criteria Satisfied

Global Stabilty

Fargo Floodwalls

Bearing:

ΣE7U

ΣB )*)�2

− QbU

)*)� − 2(CU)

(C,) tan 45 +∅2

de∅ > 0, (]h−1)ijklm(∅)

]h − 1 tan(1.4∅)

1 + 0.2� !n tan 45 +

∅2

de∅ > 0, 1 + 0.1� !n tan 45 +

∅2

atanΣ�ΣB

1 −o+90

"

1 −o+∅

"

q�+(qh.)(rs)(]h)

qh+(qh.)(rs)(]h)

qh+ qh. ɤ�� − ɤ�� ]t /2

!′(u1 + u2 + u3)

(ɤ�� − ɤ�� )(� )

v(tan ∅)

Normal Component to the base of the structure of the

ultimate bearing capacity

EF1� +EF1" +EF? +EF@ + $A� $A�N + EGH� +EGH" +EGI +EFIJ +EKL +EJ + $CD"�($CD"�N)(O7)(%E�)+$+ $CD"N + %UV %UVN

$CD + $.��/+�6 .� + $.T+ + $CD�(O7)(%E�) − $+ − $CD"�(O7)(%E�)

ΣB cos \ + Σ�+6sin(\)

Σ�+6 cos \ − ΣBsin(\)

2(! − )3

11/17/2015

Page 81: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Factor Equation Description

EM 1110-2-2502

Reference

q0 = 0.333 k/ft2 Effective Overburden Pressure

y = 0

Nq = 1 Bearing Capacity Factor 5-3a

Nc = 5.14 Bearing Capacity Factor 5-3c

Nϒ = 0 Bearing Capacity Factor 5-3d

εcd = 1.09 Embedment Factor 5-4a

εqd = 1.00 Embedment Factor 5-4b

δd = 0.16 5-5

8.917254

εqi = 0.81 Inclination Factor 5-5a

εϒi = 0.45 Inclination Factor 5-5b

Q1 = 4.54 5-2

Q2 = 0.2702814 5-2

Q3 = 0 5-2

Qd = 65.415666 5-2

FS = 5-1

5.9568313

Drained q'max 1.05 Undrained q'max 1.05

Bearing Criteria Satisfied

Undrained Condition

Global Stabilty

Fargo Floodwalls

(ɤ�� − ɤ�� )(� )

v(tan ∅)

de∅ = 0, 5.14

de∅ = 0, 1

(C,)tan(45 +∅2)

]h − 1 tan(1.4∅)

1 + 0.2� !n tan 45 +

∅2

atanΣ�ΣB

1 −o+90

"

1 −o+∅

"

q�+(qh.)(rs)(]h)

qh+(qh.)(rs)(]h)

qh+ qh. ɤ�� − ɤ�� ]t /2

!′(u1 + u2 + u3)Normal Component to the base of the structure of the

ultimate bearing capacity

11/17/2015

Page 82: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Factor Equation Description

Load Factor 1.7

b 17 Effective base in compression

qmin 0.256797

Uws 0.565999 Uplift Pressure Wet side of Stem

Utws 7.697898 Total water uplift on wet side

Utwsd 6.314901 Moment arm for U tws

VLw 6.589602 Vertical load on wet side

Bws 0.814546 Bearing at wet side of stem

Btws 6.428057 Total Bearing uplift on wet side of footing

Btwsd 4.958783 Moment arm for B tws

Total Shear on Heel 0.274627

Total Moment on Heel 8.904932

Total Shear on Stem 4.31 Total Shear on Stem

Total Moment on Stem 22.08889

Uds 0.55342 Uplift Pressure Dry side of Stem

Utds 2.004574 Total water uplift on dry side

Utdsd 1.85294 Moment arm for U tds

VLd 1.332926 Vertical load on dry side

Bds 0.872645 Bearing at dry side of stem

Btds 3.599224 Total Bearing uplift on dry side of footing

Btdsd 1.931749 Moment arm for B tds

Total Shear on Toe -3.85271 Total Shear on Toe

Total Moment on Toe -7.49588 Total Moment on Toe

Utkw 0.706875 Water Pressure at top of key (wet side)

Total Shear on Key -2.89 Total Shear on Key

Total Moment on Key -2.89718 Total Moment on Key

Forces for Reinforcement/Shear

Heel Forces

Stem Forces

Toe Forces

Key Forces

$KL +$J + �

2��� + �3 ɤ�� − ρ

�32

Y7 − ρ�22

�2 − �2 ɤ�� − ρ�22

(Y7) jlN�likjb

$KL ��� + $J $.T+N −�12

+ �

2���

���

3+ �3 ɤ�� − ρ

�32

�33

(Y7) − ρ�22

�2�23

−�2 ɤ�� − ρ�22

�23

(Y7) ( jlN�likjb)

x�7:9�, −x�7:9�, −x)�

! − � −

x�7:9�, −x�

2C − L +

x) +x67��789�,

2( )

x�� − 2

� − +x�7:9�, − x�

22 � −

3� − + x67��789�, � −

2

+x) −x67��789�,

2 � −

3

/x��

�B2+ � � ρ� +� ρ� +$@ − x��

r8z − r8z − r8.T

S(� + �)

(B − !��)( jlN�likjb)

�B2+ � � ρ� +� ρ� +$@�2

− x�� x��N − !�� !��N ( jlN�likjb)

x�7:9�, −x�7:9�, −x)�

! − � + � −

x+� +x)�

2(�)

[x)� ��2

+x+� −x)�

2(�3)(�)]/x�+�

�B1 + $? +� �� ρ� −x�+�

r8z − r8z − r8.T

S(�)

r8z + !+�2

(�)

!+� ��2

+r8z −!+�

22�3

� /!�+�

(B + −!�+�)( jlN�likjb)

�B1+ $? +� �� ρ��2

−x�+� x�+�N − !�+� !�+�N ( jlN�likjb)

}eS% ≥ 100,!, S�

}eS% ≥ 100,r8z

!1−

6C!

, 0

}er8.T = 0,!�

2S − � − � , (

!�+ r8.T

2)(�)

}er8.T = 0,S − � − �

3, r8.T �

�2

+!�− r8.T

2�

�3

/!��

x�9 +x=

2� −

x67��789�, +x�7:9�,

2� − M% jlN�likjb

x�9 ��2

+x= − x�9

2�

2�3

−x�7:9�, ��2

−x67��789�, −x�7:9�,

2�

2�3

− M%�2

( jlN�likjb)

(��� +�)(�)

Total

Moment

on Stem

Page 83: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Input: Values in red need to be entered by user Designer: LJB Date: 11/17/2015

Check: Date:Project Number:

Bridge No./Structure ID:

Owner:

Mu = 8.904932 k-ft 5.238 k-ft concrete wall thickness (t) 24 in.

Vu = 0.274627 k 0.162 k Clear Cover 4 in.

Nu = 0.0 k 0.0 k d dimension 19.625 in.

(factored) (service)150 psi

Bar Size = 6

Enter Reinforcement Spacing = 6 in. Reinforcement yield strength = 60 ksiAs = 0.88 in2

Concrete yield strength = 4.5 ksi

Enter design width = 12 in.Concrete yield strength = 4500 psi

Bar Size = None Ec = 4066.84 ksi

Enter Reinforcement Spacing = 6 in. Es = 29000 ksi

Av = 0.00 in2Lightweight Concrete? NO

Mod Factor ƛ = 1Does ACI 350 Apply? NO Flexure φ = 0.9

Shear φ = 0.75

Flexure Analysis: Mr = Phi(Mn)

For members without axial loads.

a = 1.150

Mn = 1,005.8 k-in

Mn= 83.8 k-ft

φMn = 75.44 k-ft

Minimum Reinforcement: (ACI 318 10.5)

As, min = 0.79 in2 not less than 0.785 in2

ACI 318 - 10.5 Requirement = Pass If Fail ------------>

1.33*Mu = 11.84 k-ft

As,req = 0.12 in2

As, min = 0.79 in2

OK - Section and reinforcement sufficient for Flexural Resistance

*If the requirements of 10.5 are not satisfied then As needs to be at least 1/3 greater than that required by analysis. (10.5.3)

OK

She

ar

7438-009 Section Being Designed:Heel of Footing

Fargo Floodwalls Layer of Steel:

City of Fargo

Modulus of elasticity based on concrete unit weight of

Fle

xure

��,8.T =? >��

>�SNS5kmjk�C<<kDlm200S

+

>�

ET = �� ∗ e, ∗ N −l2

l =�� ∗ e,

0.85 ∗ e′� ∗ S

��, �h =.85en�S

e,N − N" −

2E`

.85e′�S

Page 84: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Designer: LJB Date: 11/17/2015

Check: Date:

ACI 318

Governing Specification: USACE

Min. Steel Requirement:

As/Ag (Includes both faces) ≥ 0.0050 EM 1110-2-2104 2.8

Enter Reinforcing configuration of opposite mat:Bar Size = 5

Enter Reinforcement Spacing = 6 in.As = 0.62 in2

Required Min. Steel (both faces): 1.44 sq. in/ft

Max. Spacing Requirements:

s≤5h or s≤18.0 in. 120 in18 in Structural walls/footings:

or for componenets over 36.0 in Spacing can be ≤ 12 in. on faces > 18" thick

s≤12.0 12 in

Maximum Spacing Allowed: 18 in

ACI 350

Min. Steel Requirement: ACI 350-06 7.12.2.1 Minimum reinforcement ratio for Grade 60 steel

Length Between JointsRatio As/Ag 0.0030 ≤ 20 ft

20-30 ft

Steel in opposite face: 0.62 in230-40 ft≥ 40ft

Required Min. Steel: 0.86 sq. in/ft

Max. Spacing Requirements: ACI 350-065 7.12.2.2

Shrinkage and Temperature Reinforcement shall not be spaced greater than12in. apart

Maximum Spacing Allowed: 12 in

0.0040.005

T&S Steel Requirement Met

Max Spacing Requirement Met

Max Spacing Requirement Met

T&S Steel:

Ratio (As/Ag)0.0030.003

0

T&S Steel:

T&S Steel Requirement Met

Layer of Steel:

Section Being Designed:

Heel of Footing

Page 85: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Designer: LJB Date: 11/17/2015Check: Date:

Shear Analysis:

φVn ≥ Vu (ACI 318 Eq. 11-1)

Vn = Vc + Vs (ACI 318 Eq. 11-2)

Calculate d:

d = 19.625 (Loads placed at 'top' of member.) (ACI 318 11.1.3.1)

dv = 19.625 in

Calc. Vc

For members subject to shear & flexure only.Vc = 2* ƛ * √fc * bv * dv ƛ = 1.0 (ACI 318 Eq. 11-3)

bv = 12 in

Vc = 31.6 kips dv = 19.63 in

For members subject to axial compression.Vc = 2*(1 + Nu/(2000*Ag)) * ƛ * √fc * bv * dv Nu = 0 lb (ACI 318 Eq. 11-4)

Ag = 288 in2

Vc = 758.3 kips

Vc = 31.6 kips

Calc. Vs

Shear Reinforcement perpendicular to axis.

Vs = Av * fyt * d / s Av = 0.00 in2

s = 6 in

Vs = 0.0 kips

Vn = 31.6 kips

θVn = 23.7 kips Vu= 0.274627

θVn = 23.7 kips OK - Section Adequate for Shear Resistance

Section Being Designed:Heel of Footing

Layer of Steel:0

Page 86: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Designer: LJB Date: 11/17/2015Check: Date:

Crack Control: ACI 318

Distribution of flexural reinforcement in beams and one-way slabs.

The spacing of reinforcement closest to the tension face, s, shall not exceed:

cc = 4 in (Clear cover)

fs = Calculated stress from unfactored Moment.

But not greater than Mserv = 62.9 kip-in

n = 7.13

x = 4.038 in NA to extreme comp fiber

y = 15.587 in NA to tension centroid

I = 1787.9 in4

fs = 3907.731 psi

s ≤ 143.5 in

Crack Control: ACI 350

Actual Stress:fs = 3907.731 psi

ACI 350-06 10.6.4.1, for normal environmental exposures as defined in 10.6.4.5

β= 1.28

s= 6 in.db= 0.75 in.

fs,max= 32.65 ksi26781.3 psi > 3907.731 psi

OK - Spacing Adequate

OK - Spacing Adequate

fs,max =320

β*SQRT(s2+4(2+

db∕2)

2)

Section Being Designed:Heel of Footing

Layer of Steel:0

< = 15@s,sss

>�− 2.5 i�

12(40,000

e�)

e� = m ∗E�� � ∗ �

}

� =�� ∗ �� ∗ m + 2 ∗ S ∗ N ∗ m

S−�� ∗ mS

� = N − �

} =13− S ∗ �? + m ∗ �� ∗ �"

Page 87: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Input: Values in red need to be entered by user Designer: LJB Date: 11/17/2015

Check: Date:Project Number:

Bridge No./Structure ID:

Owner:

Mu = 22.08889 k-ft 12.993 k-ft concrete wall thickness (t) 15 in.

Vu = 4.314733 k 2.538 k Clear Cover 3 in.

Nu = 0.0 k 0.0 k d dimension 11.625 in.

(factored) (service)150 psi

Bar Size = 6

Enter Reinforcement Spacing = 6 in. Reinforcement yield strength = 60 ksiAs = 0.88 in2

Concrete yield strength = 4.5 ksi

Enter design width = 12 in.Concrete yield strength = 4500 psi

Bar Size = None Ec = 4066.84 ksi

Enter Reinforcement Spacing = 6 in. Es = 29000 ksi

Av = 0.00 in2Lightweight Concrete? NO

Mod Factor ƛ = 1Does ACI 350 Apply? NO Flexure φ = 0.9

Shear φ = 0.75

Flexure Analysis: Mr = Phi(Mn)

For members without axial loads.

a = 1.150

Mn = 583.4 k-in

Mn= 48.6 k-ft

φMn = 43.76 k-ft

Minimum Reinforcement: (ACI 318 10.5)

As, min = 0.47 in2 not less than 0.465 in2

ACI 318 - 10.5 Requirement = Pass If Fail ------------>

1.33*Mu = 29.38 k-ft

As,req = 0.52 in2

As, min = 0.47 in2

OK - Section and reinforcement sufficient for Flexural Resistance

*If the requirements of 10.5 are not satisfied then As needs to be at least 1/3 greater than that required by analysis. (10.5.3)

OK

She

ar

7438-009 Section Being Designed:Stem

Fargo Floodwalls Layer of Steel:

City of Fargo

Modulus of elasticity based on concrete unit weight of

Fle

xure

��,8.T =? >��

>�SNS5kmjk�C<<kDlm200S

+

>�

ET = �� ∗ e, ∗ N −l2

l =�� ∗ e,

0.85 ∗ e′� ∗ S

��, �h =.85en�S

e,N − N" −

2E`

.85e′�S

11/17/20158:39 AM

Page 88: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Designer: LJB Date: 11/17/2015

Check: Date:

ACI 318

Governing Specification: USACE

Min. Steel Requirement:

As/Ag (Includes both faces) ≥ 0.0028 EM 1110-2-2104 2.8

Enter Reinforcing configuration of opposite mat:Bar Size = 5

Enter Reinforcement Spacing = 12 in.As = 0.31 in2

Required Min. Steel (both faces): 0.50 sq. in/ft

Max. Spacing Requirements:

s≤5h or s≤18.0 in. 75 in18 in Structural walls/footings:

or for componenets over 36.0 in Spacing can be ≤ 12 in. on faces > 18" thick

s≤12.0 12 in

Maximum Spacing Allowed: 18 in

ACI 350

Min. Steel Requirement: ACI 350-06 7.12.2.1 Minimum reinforcement ratio for Grade 60 steel

Length Between JointsRatio As/Ag 0.0030 ≤ 20 ft

20-30 ft

Steel in opposite face: 0.31 in230-40 ft≥ 40ft

Required Min. Steel: 0.54 sq. in/ft

Max. Spacing Requirements: ACI 350-065 7.12.2.2

Shrinkage and Temperature Reinforcement shall not be spaced greater than12in. apart

Maximum Spacing Allowed: 12 in

0.0040.005

T&S Steel Requirement Met

Max Spacing Requirement Met

Max Spacing Requirement Met

T&S Steel:

Ratio (As/Ag)0.0030.003

0

T&S Steel:

T&S Steel Requirement Met

Layer of Steel:

Section Being Designed:

Stem

11/17/20158:39 AM

Page 89: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Designer: LJB Date: 11/17/2015Check: Date:

Shear Analysis:

φVn ≥ Vu (ACI 318 Eq. 11-1)

Vn = Vc + Vs (ACI 318 Eq. 11-2)

Calculate d:

d = 11.625 (Loads placed at 'top' of member.) (ACI 318 11.1.3.1)

dv = 11.625 in

Calc. Vc

For members subject to shear & flexure only.Vc = 2* ƛ * √fc * bv * dv ƛ = 1.0 (ACI 318 Eq. 11-3)

bv = 12 in

Vc = 18.7 kips dv = 11.63 in

For members subject to axial compression.Vc = 2*(1 + Nu/(2000*Ag)) * ƛ * √fc * bv * dv Nu = 0 lb (ACI 318 Eq. 11-4)

Ag = 180 in2

Vc = 280.7 kips

Vc = 18.7 kips

Calc. Vs

Shear Reinforcement perpendicular to axis.

Vs = Av * fyt * d / s Av = 0.00 in2

s = 6 in

Vs = 0.0 kips

Vn = 18.7 kips

θVn = 14.0 kips Vu= 4.314733

θVn = 14.0 kips OK - Section Adequate for Shear Resistance

Section Being Designed:Stem

Layer of Steel:0

11/17/20158:39 AM

Page 90: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Designer: LJB Date: 11/17/2015Check: Date:

Crack Control: ACI 318

Distribution of flexural reinforcement in beams and one-way slabs.

The spacing of reinforcement closest to the tension face, s, shall not exceed:

cc = 3 in (Clear cover)

fs = Calculated stress from unfactored Moment.

But not greater than Mserv = 155.9 kip-in

n = 7.13

x = 3.003 in NA to extreme comp fiber

y = 8.622 in NA to tension centroid

I = 574.8 in4

fs = 16677.63 psi

s ≤ 28.5 in

Crack Control: ACI 350

Actual Stress:fs = 16677.63 psi

ACI 350-06 10.6.4.1, for normal environmental exposures as defined in 10.6.4.5

β= 1.39

s= 6 in.db= 0.75 in.

fs,max= 30.05 ksi26781.3 psi > 16677.63 psi

OK - Spacing Adequate

OK - Spacing Adequate

fs,max =320

β*SQRT(s2+4(2+

db∕2)

2)

Section Being Designed:Stem

Layer of Steel:0

< = 15@s,sss

>�− 2.5 i�

12(40,000

e�)

e� = m ∗E�� � ∗ �

}

� =�� ∗ �� ∗ m + 2 ∗ S ∗ N ∗ m

S−�� ∗ mS

� = N − �

} =13− S ∗ �? + m ∗ �� ∗ �"

11/17/20158:39 AM

Page 91: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Input: Values in red need to be entered by user Designer: LJB Date: 11/17/2015

Check: Date:Project Number:

Bridge No./Structure ID:

Owner:

Mu = 7.495881 k-ft 4.409 k-ft concrete wall thickness (t) 24 in.

Vu = 3.852706 k 2.266 k Clear Cover 4 in.

Nu = 0.0 k 0.0 k d dimension 19.6875 in.

(factored) (service)150 psi

Bar Size = 5

Enter Reinforcement Spacing = 6 in. Reinforcement yield strength = 60 ksiAs = 0.62 in2

Concrete yield strength = 4.5 ksi

Enter design width = 12 in.Concrete yield strength = 4500 psi

Bar Size = None Ec = 4066.84 ksi

Enter Reinforcement Spacing = 6 in. Es = 29000 ksi

Av = 0.00 in2Lightweight Concrete? NO

Mod Factor ƛ = 1Does ACI 350 Apply? NO Flexure φ = 0.9

Shear φ = 0.75

Flexure Analysis: Mr = Phi(Mn)

For members without axial loads.

a = 0.810

Mn = 717.3 k-in

Mn= 59.8 k-ft

φMn = 53.80 k-ft

Minimum Reinforcement: (ACI 318 10.5)

As, min = 0.79 in2 not less than 0.7875 in2

ACI 318 - 10.5 Requirement = Fail If Fail ------------>

1.33*Mu = 9.97 k-ft

As,req = 0.10 in2

As, min = 0.10 in2

OK - Section and reinforcement sufficient for Flexural Resistance

*If the requirements of 10.5 are not satisfied then As needs to be at least 1/3 greater than that required by analysis. (10.5.3)

OK

She

ar

7438-009 Section Being Designed:Toe of Footing

Fargo Floodwalls Layer of Steel:

City of Fargo

Modulus of elasticity based on concrete unit weight of

Fle

xure

��,8.T =? >��

>�SNS5kmjk�C<<kDlm200S

+

>�

ET = �� ∗ e, ∗ N −l2

l =�� ∗ e,

0.85 ∗ e′� ∗ S

��, �h =.85en�S

e,N − N" −

2E`

.85e′�S

11/17/20158:39 AM

Page 92: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Designer: LJB Date: 11/17/2015

Check: Date:

ACI 318

Governing Specification: USACE

Min. Steel Requirement:

As/Ag (Includes both faces) ≥ 0.0050 EM 1110-2-2104 2.8

Enter Reinforcing configuration of opposite mat:Bar Size = 6

Enter Reinforcement Spacing = 6 in.As = 0.88 in2

Required Min. Steel (both faces): 1.44 sq. in/ft

Max. Spacing Requirements:

s≤5h or s≤18.0 in. 120 in18 in Structural walls/footings:

or for componenets over 36.0 in Spacing can be ≤ 12 in. on faces > 18" thick

s≤12.0 12 in

Maximum Spacing Allowed: 18 in

ACI 350

Min. Steel Requirement: ACI 350-06 7.12.2.1 Minimum reinforcement ratio for Grade 60 steel

Length Between JointsRatio As/Ag 0.0030 ≤ 20 ft

20-30 ft

Steel in opposite face: 0.88 in230-40 ft≥ 40ft

Required Min. Steel: 0.86 sq. in/ft

Max. Spacing Requirements: ACI 350-065 7.12.2.2

Shrinkage and Temperature Reinforcement shall not be spaced greater than12in. apart

Maximum Spacing Allowed: 12 in

0.0040.005

T&S Steel Requirement Met

Max Spacing Requirement Met

Max Spacing Requirement Met

T&S Steel:

Ratio (As/Ag)0.0030.003

0

T&S Steel:

T&S Steel Requirement Met

Layer of Steel:

Section Being Designed:

Toe of Footing

11/17/20158:39 AM

Page 93: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Designer: LJB Date: 11/17/2015Check: Date:

Shear Analysis:

φVn ≥ Vu (ACI 318 Eq. 11-1)

Vn = Vc + Vs (ACI 318 Eq. 11-2)

Calculate d:

d = 19.688 (Loads placed at 'top' of member.) (ACI 318 11.1.3.1)

dv = 19.688 in

Calc. Vc

For members subject to shear & flexure only.Vc = 2* ƛ * √fc * bv * dv ƛ = 1.0 (ACI 318 Eq. 11-3)

bv = 12 in

Vc = 31.7 kips dv = 19.69 in

For members subject to axial compression.Vc = 2*(1 + Nu/(2000*Ag)) * ƛ * √fc * bv * dv Nu = 0 lb (ACI 318 Eq. 11-4)

Ag = 288 in2

Vc = 760.7 kips

Vc = 31.7 kips

Calc. Vs

Shear Reinforcement perpendicular to axis.

Vs = Av * fyt * d / s Av = 0.00 in2

s = 6 in

Vs = 0.0 kips

Vn = 31.7 kips

θVn = 23.8 kips Vu= 3.852706

θVn = 23.8 kips OK - Section Adequate for Shear Resistance

Section Being Designed:Toe of Footing

Layer of Steel:0

11/17/20158:39 AM

Page 94: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Designer: LJB Date: 11/17/2015Check: Date:

Crack Control: ACI 318

Distribution of flexural reinforcement in beams and one-way slabs.

The spacing of reinforcement closest to the tension face, s, shall not exceed:

cc = 4 in (Clear cover)

fs = Calculated stress from unfactored Moment.

But not greater than Mserv = 52.9 kip-in

n = 7.13

x = 3.458 in NA to extreme comp fiber

y = 16.229 in NA to tension centroid

I = 1329.9 in4

fs = 4604.428 psi

s ≤ 120.3 in

Crack Control: ACI 350

Actual Stress:fs = 4604.428 psi

ACI 350-06 10.6.4.1, for normal environmental exposures as defined in 10.6.4.5

β= 1.27

s= 6 in.db= 0.625 in.

fs,max= 33.37 ksi26781.3 psi > 4604.428 psi

OK - Spacing Adequate

OK - Spacing Adequate

fs,max =320

β*SQRT(s2+4(2+

db∕2)

2)

Section Being Designed:Toe of Footing

Layer of Steel:0

< = 15@s,sss

>�− 2.5 i�

12(40,000

e�)

e� = m ∗E�� � ∗ �

}

� =�� ∗ �� ∗ m + 2 ∗ S ∗ N ∗ m

S−�� ∗ mS

� = N − �

} =13− S ∗ �? + m ∗ �� ∗ �"

11/17/20158:39 AM

Page 95: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Input: Values in red need to be entered by user Designer: LJB Date: 11/17/2015

Check: Date:Project Number:

Bridge No./Structure ID:

Owner:

Mu = 2.897183 k-ft 1.704 k-ft concrete wall thickness (t) 15 in.

Vu = 2.88578 k 1.698 k Clear Cover 3 in.

Nu = 0.0 k 0.0 k d dimension 11.5625 in.

(factored) (service)150 psi

Bar Size = 7

Enter Reinforcement Spacing = 12 in. Reinforcement yield strength = 60 ksiAs = 0.60 in2

Concrete yield strength = 4.5 ksi

Enter design width = 12 in.Concrete yield strength = 4500 psi

Bar Size = None Ec = 4066.84 ksi

Enter Reinforcement Spacing = 6 in. Es = 29000 ksi

Av = 0.00 in2Lightweight Concrete? NO

Mod Factor ƛ = 1Does ACI 350 Apply? NO Flexure φ = 0.9

Shear φ = 0.75

Flexure Analysis: Mr = Phi(Mn)

For members without axial loads.

a = 0.784

Mn = 402.1 k-in

Mn= 33.5 k-ft

φMn = 30.16 k-ft

Minimum Reinforcement: (ACI 318 10.5)

As, min = 0.47 in2 not less than 0.4625 in2

ACI 318 - 10.5 Requirement = Pass If Fail ------------>

1.33*Mu = 3.85 k-ft

As,req = 0.07 in2

As, min = 0.47 in2

OK - Section and reinforcement sufficient for Flexural Resistance

*If the requirements of 10.5 are not satisfied then As needs to be at least 1/3 greater than that required by analysis. (10.5.3)

OK

She

ar

7438-009 Section Being Designed:Key

Fargo Floodwalls Layer of Steel:

City of Fargo

Modulus of elasticity based on concrete unit weight of

Fle

xure

��,8.T =? >��

>�SNS5kmjk�C<<kDlm200S

+

>�

ET = �� ∗ e, ∗ N −l2

l =�� ∗ e,

0.85 ∗ e′� ∗ S

��, �h =.85en�S

e,N − N" −

2E`

.85e′�S

11/17/20158:39 AM

Page 96: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Designer: LJB Date: 11/17/2015

Check: Date:

ACI 318

Governing Specification: USACE

Min. Steel Requirement:

As/Ag (Includes both faces) ≥ 0.0050 EM 1110-2-2104 2.8

Enter Reinforcing configuration of opposite mat:Bar Size = 7

Enter Reinforcement Spacing = 12 in.As = 0.60 in2

Required Min. Steel (both faces): 0.90 sq. in/ft

Max. Spacing Requirements:

s≤5h or s≤18.0 in. 75 in18 in Structural walls/footings:

or for componenets over 36.0 in Spacing can be ≤ 12 in. on faces > 18" thick

s≤12.0 12 in

Maximum Spacing Allowed: 18 in

ACI 350

Min. Steel Requirement: ACI 350-06 7.12.2.1 Minimum reinforcement ratio for Grade 60 steel

Length Between JointsRatio As/Ag 0.0030 ≤ 20 ft

20-30 ft

Steel in opposite face: 0.60 in230-40 ft≥ 40ft

Required Min. Steel: 0.54 sq. in/ft

Max. Spacing Requirements: ACI 350-065 7.12.2.2

Shrinkage and Temperature Reinforcement shall not be spaced greater than12in. apart

Maximum Spacing Allowed: 12 in

0.0040.005

T&S Steel Requirement Met

Max Spacing Requirement Met

Max Spacing Requirement Met

T&S Steel:

Ratio (As/Ag)0.0030.003

0

T&S Steel:

T&S Steel Requirement Met

Layer of Steel:

Section Being Designed:

Key

11/17/20158:39 AM

Page 97: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Designer: LJB Date: 11/17/2015Check: Date:

Shear Analysis:

φVn ≥ Vu (ACI 318 Eq. 11-1)

Vn = Vc + Vs (ACI 318 Eq. 11-2)

Calculate d:

d = 11.563 (Loads placed at 'top' of member.) (ACI 318 11.1.3.1)

dv = 11.563 in

Calc. Vc

For members subject to shear & flexure only.Vc = 2* ƛ * √fc * bv * dv ƛ = 1.0 (ACI 318 Eq. 11-3)

bv = 12 in

Vc = 18.6 kips dv = 11.56 in

For members subject to axial compression.Vc = 2*(1 + Nu/(2000*Ag)) * ƛ * √fc * bv * dv Nu = 0 lb (ACI 318 Eq. 11-4)

Ag = 180 in2

Vc = 279.2 kips

Vc = 18.6 kips

Calc. Vs

Shear Reinforcement perpendicular to axis.

Vs = Av * fyt * d / s Av = 0.00 in2

s = 6 in

Vs = 0.0 kips

Vn = 18.6 kips

θVn = 14.0 kips Vu= 2.88578

θVn = 14.0 kips OK - Section Adequate for Shear Resistance

Section Being Designed:Key

Layer of Steel:0

11/17/20158:39 AM

Page 98: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Designer: LJB Date: 11/17/2015Check: Date:

Crack Control: ACI 318

Distribution of flexural reinforcement in beams and one-way slabs.

The spacing of reinforcement closest to the tension face, s, shall not exceed:

cc = 3 in (Clear cover)

fs = Calculated stress from unfactored Moment.

But not greater than Mserv = 20.5 kip-in

n = 7.13

x = 2.537 in NA to extreme comp fiber

y = 9.026 in NA to tension centroid

I = 413.8 in4

fs = 3180.457 psi

s ≤ 181.2 in

Crack Control: ACI 350

Actual Stress:fs = 3180.457 psi

ACI 350-06 10.6.4.1, for normal environmental exposures as defined in 10.6.4.5

β= 1.38

s= 12 in.db= 0.875 in.

fs,max= 17.89 ksi26781.3 psi > 3180.457 psi

OK - Spacing Adequate

OK - Spacing Adequate

fs,max =320

β*SQRT(s2+4(2+

db∕2)

2)

Section Being Designed:Key

Layer of Steel:0

< = 15@s,sss

>�− 2.5 i�

12(40,000

e�)

e� = m ∗E�� � ∗ �

}

� =�� ∗ �� ∗ m + 2 ∗ S ∗ N ∗ m

S−�� ∗ mS

� = N − �

} =13− S ∗ �? + m ∗ �� ∗ �"

11/17/20158:39 AM

Page 99: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Input: Values in red need to be entered by user Designer: LJB Date: 11/17/2015

Check: Date:Project Number:

Bridge No./Structure ID:

Owner:

Mu = 0 k-ft 0.000 k-ft concrete wall thickness (t) 24 in.

Vu = 0 k 0.000 k Clear Cover 5 in.

Nu = 0.0 k 0.0 k d dimension 18.5 in.

(factored) (service)150 psi

Bar Size = 8

Enter Reinforcement Spacing = 12 in. Reinforcement yield strength = 60 ksiAs = 0.79 in2

Concrete yield strength = 4.5 ksi

Enter design width = 12 in.Concrete yield strength = 4500 psi

Bar Size = None Ec = 4066.84 ksi

Enter Reinforcement Spacing = 6 in. Es = 29000 ksi

Av = 0.00 in2Lightweight Concrete? NO

Mod Factor ƛ = 1Does ACI 350 Apply? NO Flexure φ = 0.9

Shear φ = 0.75

ACI 318

Governing Specification: USACE

Min. Steel Requirement:

As/Ag (Includes both faces) ≥ 0.0050 EM 1110-2-2104 2.8

Enter Reinforcing configuration of opposite mat:Bar Size = 8

Enter Reinforcement Spacing = 12 in.As = 0.79 in2

Required Min. Steel (both faces): 1.44 sq. in/ft

Max. Spacing Requirements:

s≤5h or s≤18.0 in. 120 in18 in Structural walls/footings:

or for componenets over 36.0 in Spacing can be ≤ 12 in. on faces > 18" thick

s≤12.0 12 in

Maximum Spacing Allowed: 18 in

She

ar

7438-009 Section Being Designed:Longitudinal Footing Reinforcement

Fargo Floodwalls Layer of Steel:

City of Fargo

Modulus of elasticity based on concrete unit weight of

Fle

xure

T&S Steel:

T&S Steel Requirement Met

Max Spacing Requirement Met

Page 100: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Input: Values in red need to be entered by user Designer: LJB Date: 11/17/2015

Check: Date:Project Number:

Bridge No./Structure ID:

Owner:

Mu = 0 k-ft 0.000 k-ft concrete wall thickness (t) 15 in.

Vu = 0 k 0.000 k Clear Cover 3 in.

Nu = 0.0 k 0.0 k d dimension 11.6875 in.

(factored) (service)150 psi

Bar Size = 5

Enter Reinforcement Spacing = 12 in. Reinforcement yield strength = 60 ksiAs = 0.31 in2

Concrete yield strength = 4.5 ksi

Enter design width = 12 in.Concrete yield strength = 4500 psi

Bar Size = None Ec = 4066.84 ksi

Enter Reinforcement Spacing = 6 in. Es = 29000 ksi

Av = 0.00 in2Lightweight Concrete? NO

Mod Factor ƛ = 1Does ACI 350 Apply? NO Flexure φ = 0.9

Shear φ = 0.75

ACI 318

Governing Specification: USACE

Min. Steel Requirement:

As/Ag (Includes both faces) ≥ 0.0028 EM 1110-2-2104 2.8

Enter Reinforcing configuration of opposite mat:Bar Size = 5

Enter Reinforcement Spacing = 12 in.As = 0.31 in2

Required Min. Steel (both faces): 0.50 sq. in/ft

Max. Spacing Requirements:

s≤5h or s≤18.0 in. 75 in18 in Structural walls/footings:

or for componenets over 36.0 in Spacing can be ≤ 12 in. on faces > 18" thick

s≤12.0 12 in

Maximum Spacing Allowed: 18 in

She

ar

7438-009 Section Being Designed:Longitudinal Stem Reinforcement

Fargo Floodwalls Layer of Steel:

City of Fargo

Modulus of elasticity based on concrete unit weight of

Fle

xure

T&S Steel:

T&S Steel Requirement Met

Max Spacing Requirement Met

11/17/20158:39 AM

Page 101: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

13'ICE

****************** Echoprint of Input Data ******************

Date: **/11/16 Time: 15.22.53

Company name: HEI Project name: 2nd St. S Project location: Wall location: Computed by: ljb

Structural geometry data: Elevation of top of stem (ELTS) = 13.00 ft Height of stem (HTS) = 13.00 ft Thickness top of stem (TTS) = 1.25 ft Thickness bottom of stem (TBS) = 1.25 ft Dist. of batter at bot. of stem (TBSR)= .00 ft Depth of heel (THEEL) = 4.00 ft Distance of batter for heel (BTRH) = .00 ft Depth of toe (TTOE) = 2.00 ft Width of toe (TWIDTH) = 3.75 ft Distance of batter for toe (BTRT) = .00 ft Width of base (BWIDTH) = 17.00 ft Depth of key (HK) = 2.00 ft Width of bottom of key (TK) = 1.25 ft Dist. of batter at bot. of key (BTRK) = .00 ft

Structure coordinates:

x (ft) y (ft) ================== .00 -4.00 .00 .00 12.00 .00 12.00 13.00 13.25 13.00 13.25 .00 17.00 .00 17.00 -2.00 1.25 -2.00 1.25 -4.00

NOTE: X=0 is located at the left-hand side of the structure. The Y values correspond to the actual elevation used.

Structural property data: Unit weight of concrete = .150 kcf

Driving side soil property data:

Moist Saturated Elev. Phi c Unit wt. unit wt. Delta soil (deg) (ksf) (kcf) (kcf) (deg) (ft) =======================================================

Page 1

Page 102: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

13'ICE 27.00 .000 .118 .118 .00 5.00

Driving side soil geometry:

Soil Batter Distance point (in:1ft) (ft) ============================= 1 .00 500.00 2 .00 .00 3 .00 500.00

Driving side soil profile:

Soil x y point (ft) (ft) ============================= 1 -1488.00 5.00 2 12.00 5.00

Resisting side soil property data:

Moist Saturated Elev. Phi c Unit wt. unit wt. soil Batter (deg) (ksf) (kcf) (kcf) (ft) (in:1ft) ======================================================== 27.00 .000 .118 .118 5.00 .00

Resisting side soil profile:

Soil x y point (ft) (ft) ============================= 1 13.25 5.00 2 513.25 5.00

Foundation property data: phi for soil-structure interface = 27.00 (deg) c for soil-structure interface = .000 (ksf) phi for soil-soil interface = 27.00 (deg) c for soil-soil interface = .000 (ksf)

Water data: Driving side elevation = 9.31 ft Resisting side elevation = 5.00 ft Unit weight of water = .0624 kcf Seepage pressures computed by Line of Creep method.

Horizontal pressure data:

Elevation Pressure (ft) (ksf) ========================== 13.00 .03 9.31 .03

Horizontal line load data:

Elevation Force (ft) (kips) ======================== 9.31 .50

Minimum required factors of safety:Page 2

Page 103: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

13'ICE Sliding FS = 1.50 Overturning = 75.00% base in compression

Crack options: o Crack *is* down to bottom of heel o Computed cracks *will* be filled with water

Strength mobilization factor = .6667

At-rest pressures on the resisting side *are used* in the overturning analysis.

Forces on the resisting side *are used* in the sliding analysis.

*Do* iterate in overturning analysis.

***** Summary of Results *****

Project name: 2nd St. S

*************** *** Satisfied *** * Overturning * Required base in comp. = 75.00 % *************** Actual base in comp. = 100.00 % Overturning ratio = 1.59

Xr (measured from toe) = 6.48 ft Resultant ratio = .3814 Stem ratio = .2206 Base pressure at heel = .1703 ksf Base pressure at toe = 1.0117 ksf

*********** *** Satisfied *** * Sliding * Min. Required = 1.50 *********** Actual FS = 10.47

********************** Output Results **********************

Date: **/11/16 Time: 15.22.53

Company name: HEI Project name: 2nd St. S Project location: Wall location: Computed by: ljb

*************************** ** Overturning Results ** ***************************

Solution converged in 1 iterations.

SMF used to calculate K's = .6667Page 3

Page 104: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

13'ICE Alpha for the SMF = .0000 Calculated earth pressure coefficients: Driving side at rest K = .0000 Driving side at rest Kc = .0000 Resisting side at rest K = .5460 Resisting side at rest Kc = .7389 At-rest K's for resisting side calculated.

Depth of cracking = 9.00 ft Crack extends to bottom of base of structure.

** Driving side pressures **

Water pressures: Elevation Pressure (ft) (ksf) ====================== 9.31 .0000 -4.00 .8305

** Resisting side pressures **

Water pressures: Elevation Pressure (ft) (ksf) ====================== 5.00 .0000 -2.00 .5149 -2.00 .8180 -4.00 .8305

Earth pressures: Elevation Pressure (ft) (ksf) ====================== 5.00 .0000 -2.00 .1699

Balancing earth pressures: Elevation Pressure (ft) (ksf) ====================== -2.00 1.0464 -4.00 1.0464

** Uplift pressures **

Water pressures: x-coord. Pressure (ft) (ksf) ====================== .00 .8305 1.25 .8180 1.25 .6731 17.00 .5149

** Forces and moments **

======================================================================== Part | Force (kips) | Mom. Arm | Moment | | Vert. | Horiz.| (ft) | (ft-k) | ======================================================================== Structure:

Page 4

Page 105: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

13'ICE Structure weight........... 7.913 -7.60 -60.15 Structure, driving side: Moist soil................. .000 .00 .00 Saturated soil............. 7.080 -11.00 -77.88 Water above structure...... .000 .00 .00 Water above soil........... 3.227 -11.00 -35.50 External vertical loads.... .000 .00 .00 Ext. horz. pressure loads.. .111 13.15 1.46 Ext. horz. line loads...... .500 11.31 5.66 Structure, resisting side: Moist soil................. .000 .00 .00 Saturated soil............. 2.213 -1.88 -4.15 Water above structure...... .000 .00 .00 Water above soil........... .000 .00 .00 Driving side: Effective earth loads...... .000 .00 .00 Shear (due to delta)....... .000 .00 .00 Horiz. surcharge effects... .000 .00 .00 Water loads................ 5.527 2.44 13.47 Resisting side: Effective earth loads...... -.595 2.33 -1.39 Balancing earth load....... -2.093 -1.00 2.09 Water loads................ -3.451 .74 -2.55 Foundation: Vertical force on base..... -10.047 -6.48 65.13 Uplift..................... -10.385 -9.03 93.82 ======================================================================== ** Statics Check ** SUMS = .000 .000 .00

Angle of base = 6.71 degrees Normal force on base = 10.223 kips Shear force on base = .905 kips Max. available shear force = 6.074 kips

Base pressure at heel = .1703 ksf Base pressure at toe = 1.0117 ksf

Xr (measured from toe) = 6.48 ft Resultant ratio = .3814 Stem ratio = .2206 Base in compression = 100.00 % Overturning ratio = 1.59

Volume of concrete = 1.95 cubic yds/ft of wall

NOTE: The engineer shall verify that the computed bearing pressures below the wall do not exceed the allowable foundation bearing pressure, or, perform a bearing capacity analysis using the program CBEAR. Also, the engineer shall verify that the base pressures do not result in excessive differential settlement of the wall foundation.

*********************** ** Sliding Results ** ***********************

Solution converged. Summation of forces = 0.

Horizontal VerticalPage 5

Page 106: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

13'ICE Wedge Loads Loads Number (kips) (kips) ================================== 1 .000 .000 2 6.138 3.227 3 .000 .000

Water pressures on wedges:

Top Bottom Wedge press. press. x-coord. press. number (ksf) (ksf) (ft) (ksf) ================================================ 1 .0000 .0000 2 .0000 .8305 2 17.0000 .5149 3 .0000 .5149

Points of sliding plane: Point 1 (left), x = .00 ft, y = -4.00 ft Point 2 (right), x = 17.00 ft, y = -2.00 ft

Depth of cracking = 9.00 ft Crack extends to bottom of base of structure.

Failure Total Weight Submerged Uplift Wedge angle length of wedge length force number (deg) (ft) (kips) (ft) (kips) ======================================================== 1 .000 .000 .000 .000 .000 2 6.710 17.117 18.916 17.117 11.515 3 43.648 10.142 3.031 10.142 2.611

Wedge Net force number (kips) =================== 1 .000 2 -3.001 3 3.002 =================== SUM = .001

+-----------------------------+ | Factor of safety = 10.468 | +-----------------------------+

Page 6

Page 107: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

Concrete f'c = 4500 psiRebar Fy = 60000 psiUnit Weight = 150 lb/ft³15 in

17 ft

3.75 ft 12 ft

24

in

15

ft 13

ft

15.75 ft15 in

2 ft

40

in

7 ft

8 ft

#5 @ 12 in (S&T)#6 @ 6 in#5 @ 12 in (S&T)#5 @ 12 in#6 @ 6 in (dowels)

Heel Bars: #6 @ 6 inToe Bars: #5 @ 6 inFooting S/T Bars: #8 @ 12 in

40

in

#5 @ 12 in (S&T)#6 @ 6 in#5 @ 12 in (S&T)#5 @ 12 in

Design Detail

Check Summary

Ratio Check Provided Required Combination----- Stability Checks -----

0.333 Overturning 6.01 2.00 1.0D + 1.0F + 0.6H + 0.6W0.990 Sliding 1.51 1.50 1.0D + 1.0F + 0.6H + 0.6W0.254 Bearing Pressure 6000 psf 1527 psf 1.0D + 1.0F + 1.0H + 0.6W0.339 Bearing Eccentricity 17.29 in 51 in 1.0D + 1.0F + 1.0H + 0.6W

----- Toe Checks -----0.087 Shear 23.77 k/ft 2.06 k/ft 1.7D + 1.7F + 1.7H + 1.7W0.125 Moment 53.8 ft·k/ft 6.71 ft·k/ft 1.7D + 1.7F + 1.7H + 1.7W0.004 Min Strain 1.0173 0.0040 1.7D + 1.7F + 1.7H + 1.7W0.000 Min Steel 0.05 in² 0 in² 1.7D + 1.7F + 1.7H + 1.7W0.077 Development 155 in 12 in 1.7D + 1.7F + 1.7H + 1.7W0.667 S&T Max Spacing 12 in 18 in 1.7D + 1.7F + 1.7H + 1.7W0.328 S&T Min Rho 0.0055 0.0018 1.7D + 1.7F + 1.7H + 1.7W

----- Heel Checks -----0.025 Shear 23.7 k/ft 0.6 k/ft 1.7D + 1.7F + 1.7H + 1.7W0.244 Moment 75.44 ft·k/ft 18.4 ft·k/ft 1.7D + 1.7F + 1.7H + 1.7W0.006 Min Strain 0.7135 0.0040 1.7D + 1.7F + 1.7H + 1.7W0.000 Min Steel 0.07 in² 0 in² 1.7D + 1.7F + 1.7H + 1.7W0.214 Development 56 in 12 in 1.7D + 1.7F + 1.7H + 1.7W0.667 S&T Max Spacing 12 in 18 in 1.7D + 1.7F + 1.7H + 1.7W0.328 S&T Min Rho 0.0055 0.0018 1.7D + 1.7F + 1.7H + 1.7W

----- Stem Checks -----0.880 Moment 43.76 ft·k/ft 38.5 ft·k/ft 1.7D + 1.7F + 1.7H + 1.7W0.727 Shear 14.04 k/ft 10.2 k/ft 1.7D + 1.7F + 1.7H + 1.7W0.009 Max Steel 0.4214 0.0040 1.7D + 1.7F + 1.7H + 1.7W0.532 Min Steel 0.07 in²/in 0.04 in²/in 1.7D + 1.7F + 1.7H + 1.7W0.413 Base Development 20 in 8.26 in 1.7D + 1.7F + 1.7H + 1.7W0.523 Lap Splice Length 40 in 20.93 in 1.7D + 1.7F + 1.7H + 1.7W0.000 Lap Splice Spacing 0 in 4.19 in 1.7D + 1.7F + 1.7H + 1.7W0.581 Horz Bar Rho 0.0034 0.0020 1.7D + 1.7F + 1.7H + 1.7W0.667 Horz Bar Spacing 12 in 18 in 1.7D + 1.7F + 1.7H + 1.7W

Criteria

Building Code IBC 2006Concrete Load Combs USCoE UnusualMasonry Load Combs MSJC 02/05 (ASD)Stability Load Combs ASCE 7-10 (ASD)Restrained Against Sliding NoNeglect Bearing At Heel NoUse Vert. Comp. for OT NoUse Vert. Comp. for Sliding NoUse Vert. Comp. for Bearing YesUse Surcharge for Sliding & OT NoUse Surcharge for Bearing YesNeglect Soil Over Toe NoNeglect Backfill Wt. for Coulomb NoFactor Soil Weight As Dead NoUse Passive Force for OT YesAssume Pressure To Top YesExtend Backfill Pressure To Key Bottom NoUse Toe Passive Pressure for Bearing NoRequired F.S. for OT 2.00Required F.S. for Sliding 1.50Has Different Safety Factors for Seismic NoAllowable Bearing Pressure 6000 psfReq'd Bearing Location Middle halfWall Friction Angle 25°Friction Coefficent 0.35Soil Reaction Modulus 172800 lb/ft³

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Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

Loads5

ft

7 ft

7 ft

8 ft

γ = 118 lb/ft³φ = 27°7

ft

γ = 118 lb/ft³φ = 27°

9.3

1 ft

-135.5 psf

Loading Options/AssumptionsPassive pressure neglects top 0 ft of soil.

Load Combinations

USCoE Unusual ...1.7D + 1.7F + 1.7H + 1.7W

...1.7D + 1.7F + 1.7H

Backfill Pressure

5 ft

7 ft

7 ft

8 ft

γ = 118 lb/ft³φ = 27°

43.23 psf

7 ft

20.37 lb/in7 ft

3.7

7 ft

-599.84 psf232.7 lb/in

3.1

ft

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Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

Backfill Pressure (Water Layer)

γeff γsat γw - 55.5 lb ft³ / 62.5 lb ft³ / - 7 lb ft³ / - = = =

φ 27° φsat =

γ 7 lb ft³ γeff / - =

φ φsat =

γ γeff =

At Rest Earth Pressure Theory -

Ko 1 sin φ - 1 sin 27° - 0.5460 = = =

σh Ko γ H 0.5460 7 lb ft³ / - 11.31 ft 43.23 psf - = = =

Lateral Earth Pressure (water layer)

σh Ko γ H 0.5460 118 lb ft³ / 9.31 ft 599.8 psf = = =

Lateral Earth Pressure (water layer, stem only)

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Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

Passive Pressure

7 ft

γ = 118 lb/ft³φ = 27°

579.9 psf

9 ft

217.4 lb/in

1 ft

At Rest Earth Pressure Theory -

Ko 1 sin φ - 1 sin 27° - 0.5460 = = =

σh Ko γ H 0.5460 118 lb ft³ / 9 ft 579.9 psf = = =

Lateral Earth Pressure

Water Pressure

9.3

1 ft

-706.88 psf7 ft

333.1 lb/in7 ft

-581.88 psf 225.7 lb/in

σw γw Hw 62.5 lb ft³ / 11.31 ft 706.9 psf = = =

Lateral Water Pressure

σw γw Hw 62.5 lb ft³ / 9.31 ft 581.9 psf = = =

Lateral Water Pressure (stem only)

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Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

Manually Specified Lateral Stem Pressure

-135.5 psf 500 lb/ft

Wall/Soil Weights

425 lb/in

203.1 lb/in

31.25 lb/in

590 lb/in184.4 lb/in

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Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

Bearing Pressure

1527 psf

497.5 psf

1434 lb/in

7.06 ft

e = 17.29 in

501.8 lb/in

F µ R 0.350 1434 lb in / 501.8 lb in / = = =

Friction

Bearing Pressure Calculation

Contributing ForcesVert Force ...offset Horz Force ...offset OT Moment

Backfill Pressure 0 lb/in - 20.37 lb/in 3.77 ft -11059 in·lb/ftWater Pressure -0 lb/in - -333.11 lb/in 3.77 ft 180841 in·lb/ftManual Lateral Pressure -0 lb/in - -41.67 lb/in 13.15 ft 78930 in·lb/ftFooting Weight -425 lb/in 8.5 ft 0 lb/in - -520200 in·lb/ftStem Weight -203.13 lb/in 4.38 ft 0 lb/in - -127968.75 in·lb/ftKey Weight -31.25 lb/in 16.38 ft 0 lb/in - -73687.5 in·lb/ftBackfill Weight -590 lb/in 11 ft 0 lb/in - -934560 in·lb/ftSoil over toe Weight -184.38 lb/in 1.88 ft 0 lb/in - -49781.25 in·lb/ft

-1433.75 lb/in -1457485.33 in·lb/ft1457485.33 in·lb ft / -

1433.75 lb in / - 7.06 ft =

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Page 113: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

Overturning Check

Overturning MomentsForce Distance Moment

Backfill pressure (horz) -20.37 lb/in 3.77 ft -11059 in·lb/ftWater pressure 333.1 lb/in 3.77 ft 180841 in·lb/ftManual lateral pressure 25 lb/in 13.15 ft 47358 in·lb/ft

Total: 217140 in·lb/ft

Resisting MomentsForce Distance Moment

Passive pressure @ toe 217.4 lb/in 1 ft 31313 in·lb/ftFooting Weight -425 lb/in 8.5 ft 520200 in·lb/ftStem Weight -203.13 lb/in 4.38 ft 127969 in·lb/ftKey Weight -31.25 lb/in 16.38 ft 73688 in·lb/ftBackfill Weight -590 lb/in 11 ft 934560 in·lb/ftSoil over toe Weight -184.38 lb/in 1.88 ft 49781 in·lb/ft

Total: 1737510 in·lb/ft

F.S. RM

OTM

1737510 in·lb ft / 217140 in·lb ft /

8.002 > 2.00 OK = = =

Sliding Check

Sliding Force(s)Backfill pressure -20.37 lb/inWater pressure 333.1 lb/inManual lateral pressure 25 lb/inTotal: 337.7 lb/in

Resisting Force(s)Passive pressure @ toe 217.4 lb/inFriction 501.8 lb/inTotal: 719.3 lb/in

F.S. RFSF

719.3 lb in / 337.7 lb in /

2.130 > 1.50 OK = = =

Bearing Capacity Check

Bearing pressure < allowable (1527 psf < 6000 psf) - OKBearing resultant eccentricity < allowable (17.29 in < 51 in) - OK

Wall Top Displacement

(based on unfactored service loads)

Deflection due to stem flexural displacement 0.098 inDeflection due to rotation from settlement 0.055 inTotal deflection at top of wall (positive towards toe) 0.153 in

Stability Checks [1.0D + 1.0F + 1.0H + 0.6W]

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Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

Overturning Check

Overturning MomentsForce Distance Moment

Backfill pressure (horz) -12.22 lb/in 3.77 ft -6635.4 in·lb/ftWater pressure 333.1 lb/in 3.77 ft 180841 in·lb/ftManual lateral pressure 25 lb/in 13.15 ft 47358 in·lb/ft

Total: 221564 in·lb/ft

Resisting MomentsForce Distance Moment

Passive pressure @ toe 130.5 lb/in 1 ft 18788 in·lb/ftFooting Weight -425 lb/in 8.5 ft 520200 in·lb/ftStem Weight -203.13 lb/in 4.38 ft 127969 in·lb/ftKey Weight -31.25 lb/in 16.38 ft 73688 in·lb/ftBackfill Weight -354 lb/in 11 ft 560736 in·lb/ftSoil over toe Weight -110.63 lb/in 1.88 ft 29869 in·lb/ft

Total: 1331249 in·lb/ft

F.S. RM

OTM

1331249 in·lb ft / 221564 in·lb ft /

6.008 > 2.00 OK = = =

Sliding Check

Sliding Force(s)Backfill pressure -12.22 lb/inWater pressure 333.1 lb/inManual lateral pressure 25 lb/inTotal: 345.9 lb/in

Resisting Force(s)Passive pressure @ toe 130.5 lb/inFriction 393.4 lb/inTotal: 523.9 lb/in

F.S. RFSF

523.9 lb in / 345.9 lb in /

1.515 > 1.50 OK = = =

Bearing Capacity Check

Bearing pressure < allowable (1197 psf < 6000 psf) - OKBearing resultant eccentricity < allowable (17.29 in < 51 in) - OK

Wall Top Displacement

(based on unfactored service loads)

Deflection due to stem flexural displacement 0.098 inDeflection due to rotation from settlement 0.055 inTotal deflection at top of wall (positive towards toe) 0.153 in

Stability Checks [1.0D + 1.0F + 0.6H + 0.6W]

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Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

13

11.7

10.4

9.1

7.8

6.5

5.2

3.9

2.6

1.3

0-50 -38.33 -26.67 -15 -3.33 8.33 20

Moment (ft·k/ft)

Offset (ft)

Moment

aAs fy

0.85 F'c

0.07 in² in / 60000 psi 0.85 4500 psi

1.15 in = = =

φMn φ As fy d a 2 / - 0.90 0.07 in² in / 60000 psi 11.63 in 1.15 in 2 / - 43.76 ft·k ft / = = =

Capacity (ACI 318-05 10.2) @ 0 ft from base [Negative bending]

aAs fy

0.85 F'c

0.03 in² in / 60000 psi 0.85 4500 psi

0.41 in = = =

φMn φ As fy d a 2 / - 0.90 0.03 in² in / 60000 psi 11.69 in 0.41 in 2 / - 16.02 ft·k ft / = = =

Capacity (ACI 318-05 10.2) @ 0 ft from base [Positive bending]

aAs fy

0.85 F'c

0.07 in² in / 60000 psi 0.85 4500 psi

1.15 in = = =

φMn φ As fy d a 2 / - 0.90 0.07 in² in / 60000 psi 11.63 in 1.15 in 2 / - 43.76 ft·k ft / = = =

Capacity (ACI 318-05 10.2) @ 11.66 ft from base [Negative bending]

aAs fy

0.85 F'c

0.03 in² in / 60000 psi 0.85 4500 psi

0.41 in = = =

φMn φ As fy d a 2 / - 0.90 0.03 in² in / 60000 psi 11.69 in 0.41 in 2 / - 16.02 ft·k ft / = = =

Capacity (ACI 318-05 10.2) @ 11.88 ft from base [Positive bending]

aAs fy

0.85 F'c

0 in² in / 60000 psi 0.85 4500 psi

0 in = = =

φMn φ As fy d a 2 / - 0.90 0 in² in / 60000 psi 11.63 in 0 in 2 / - 0 ft·k ft / = = =

Capacity (ACI 318-05 10.2) @ 13 ft from base [Negative bending]

aAs fy

0.85 F'c

0 in² in / 60000 psi 0.85 4500 psi

0 in = = =

φMn φ As fy d a 2 / - 0.90 0 in² in / 60000 psi 11.69 in 0 in 2 / - 0 ft·k ft / = = =

Capacity (ACI 318-05 10.2) @ 13 ft from base [Positive bending]

Stem Flexural Capacity

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Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

13

11.7

10.4

9.1

7.8

6.5

5.2

3.9

2.6

1.3

0-15 -10 -5 0 5 10 15

Shear (k/ft)

Offset (ft)

Shear

Vc 2 F'c d 2 4500 psi 11.63 in 18.72 k ft / = = =

φVn φ Vc 0.750 18.72 k ft / 14.04 k ft / = = =

Shear Capacity (ACI 318-05 11.1.1, 11.3.1) @ 0 ft from base [Positive shear]

Vc 2 F'c d 2 4500 psi 11.63 in 18.72 k ft / = = =

φVn φ Vc 0.750 18.72 k ft / 14.04 k ft / = = =

Shear Capacity (ACI 318-05 11.1.1, 11.3.1) @ 0 ft from base [Negative shear]

Vc 2 F'c d 2 4500 psi 11.63 in 18.72 k ft / = = =

φVn φ Vc 0.750 18.72 k ft / 14.04 k ft / = = =

Shear Capacity (ACI 318-05 11.1.1, 11.3.1) @ 13 ft from base [Positive shear]

Vc 2 F'c d 2 4500 psi 11.63 in 18.72 k ft / = = =

φVn φ Vc 0.750 18.72 k ft / 14.04 k ft / = = =

Shear Capacity (ACI 318-05 11.1.1, 11.3.1) @ 13 ft from base [Negative shear]

Stem Shear Capacity

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Page 117: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

ψe 1.0 uncoated hooked bars =

λ 1.0 normal weight concrete =

ldh 0.02 ψe λ fy

F'c

db 0.02 1.0 1.060000 psi

4500 psi0.75 in 13.42 in = = =

Factoring ldh by the 0.7 multiplier of 12.5.3 a : ldh 9.39 in =

8 db 8 0.75 in 6.0 minimum limit, does not control = =

Main vertical stem bars (bottom end) - Development Length Calculation (ACI 318-05 12.2.3, 12.5)

ψt 1.0 bars are not horizontal =

ψe 1.0 bar not epoxy coated =

ψs 0.80 bars are #6 or smaller =

λ 1.0 normal weight concrete =

s 2 / 6 in 2 / 3 in = =

cover db 2 / + 3 in 0.75 in 2 / + 3.38 in = =

cb 3 in lesser of half spacing, ctr to surface =

Ktr 0.0 no transverse reinforcement =

cb Ktr + db

3 in 0.0 + 0.75 in

4.0 = =

ld3.40

fy

F'c

ψt ψe ψs λ 2.5

db 3.40

60000 psi

4500 psi

1.0 1.0 0.80 1.0 2.5

0.75 in 16.1 in = = =

Main vertical stem bars (top end) - Development Length Calculation (ACI 318-05 12.2.3, 12.5)

ψt 1.0 bars are not horizontal =

ψe 1.0 bar not epoxy coated =

ψs 0.80 bars are #6 or smaller =

λ 1.0 normal weight concrete =

s 2 / 6 in 2 / 3 in = =

cover db 2 / + 3 in 0.75 in 2 / + 3.38 in = =

cb 3 in lesser of half spacing, ctr to surface =

Ktr 0.0 no transverse reinforcement =

cb Ktr + db

3 in 0.0 + 0.75 in

4.0 = =

ld3.40

fy

F'c

ψt ψe ψs λ 2.5

db 3.40

60000 psi

4500 psi

1.0 1.0 0.80 1.0 2.5

0.75 in 16.1 in = = =

Dowels for vertical stem bars (top end) - Development Length Calculation (ACI 318-05 12.2.3, 12.5)

ψe 1.0 uncoated hooked bars =

λ 1.0 normal weight concrete =

ldh 0.02 ψe λ fy

F'c

db 0.02 1.0 1.060000 psi

4500 psi0.63 in 11.18 in = = =

Factoring ldh by the 0.7 multiplier of 12.5.3 a : ldh 7.83 in =

8 db 8 0.63 in 5.0 minimum limit, does not control = =

2nd curtain vertical bars (bottom end) - Development Length Calculation (ACI 318-05 12.2.3, 12.5)

Stem Development/Lap Length Calculations

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Page 118: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

ψt 1.0 bars are not horizontal =

ψe 1.0 bar not epoxy coated =

ψs 0.80 bars are #6 or smaller =

λ 1.0 normal weight concrete =

s 2 / 12 in 2 / 6 in = =

cover db 2 / + 3 in 0.63 in 2 / + 3.31 in = =

cb 3.31 in lesser of half spacing, ctr to surface =

Ktr 0.0 no transverse reinforcement =

cb Ktr + db

3.31 in 0.0 + 0.63 in

5.30 = =

ld3.40

fy

F'c

ψt ψe ψs λ 2.5

db 3.40

60000 psi

4500 psi

1.0 1.0 0.80 1.0 2.5

0.63 in 13.42 in = = =

2nd curtain vertical bars (top end) - Development Length Calculation (ACI 318-05 12.2.3, 12.5)

Stem Development/Lap Length Calculations (continued)

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Page 119: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

Design moment Mu for toe need not exceed moment at stem base:

Mtoe 6.71 ft·k ft < Mstem / 38.5 ft·k ft / = =

Mu 6.71 ft·k ft stem moment does not control / =

Controlling Moment

aAs fy

0.85 F'c

0.05 in² in / 60000 psi 0.85 4500 psi

0.81 in = = =

φMn φ As fy d a 2 / - 0.90 0.05 in² in / 60000 psi 19.69 in 0.81 in 2 / - 53.8 ft·k ft / = = =

φMn 53.8 ft·k ft ≥ Mu / 6.71 ft·k ft / = = �

Flexure Check (ACI 318-05 10.2)

Vc 2 F'c d 2 4500 psi 19.69 in 31.7 k ft / = = =

φVn φ Vc 0.750 31.7 k ft / 23.77 k ft / = = =

φVn 23.77 k ft ≥ Vu / 2.06 k ft / = = �

Shear Check (ACI 318-05 11.1.1, 11.3.1)

β1 0.85 0.05F'c 4000 -

1000 - 0.85 0.05

4500 psi 4000 - 1000

- 0.8250 = = =

aAs fy

0.85 F'c

0.05 in² in / 60000 psi 0.85 4500 psi

0.81 in = = =

εt 0.003d

a β1 / 1 - 0.003

19.69 in0.81 in 14.0 /

1 - 1.0173 = = =

εt 1.0173 ≥ 0.004 = �

Minimum Strain Check (ACI 318-05 10.3.5)

φMn 53.8 ft·k ft ≥ 4 3 / Mu / 4 3 / 6.71 ft·k ft / 8.94 ft·k ft / = = =

Check is waived per ACI 10.5.3�

Minimum Steel Check (ACI 318-05 10.5.1)

ρST_provAST

t sST

1.58 in² in / 24 in 12 in

0.0055 = = =

ρST_min0.0018 60000

fy

0.0018 60000 60000 psi

0.0018 = = =

ρST_min 0.0018 =

ρST_prov 0.0055 ≥ ρST_min 0.0018 = = �18 inch limit governs

sST_max 18 in =

sST 12 in ≤ sST_max 18 in = = �

Shrinkage Temperature Steel (ACI 318-05 7.12.2)

Mu

φMn

6.71 ft·k ft / 53.8 ft·k ft /

0.1246 ratio to represent excess reinforcement = =

ψt 1.0 12 inches or less cast below 4.00 inches - =

ψe 1.0 bar not epoxy coated =

ψs 0.80 bars are #6 or smaller =

λ 1.0 normal weight concrete =

s 2 / 6 in 2 / 3 in = =

cover db 2 / + 4 in 0.63 in 2 / + 4.31 in = =

cb 3 in lesser of half spacing, ctr to surface =

Ktr 0.0 no transverse reinforcement =

cb Ktr + db

3 in 0.0 + 0.63 in

4.80 = =

ld3.40

fy

F'c

ψt ψe ψs λ 2.5

db 3.40

60000 psi

4500 psi

1.0 1.0 0.80 1.0 2.5

0.63 in 13.42 in = = =

Factoring ld by the excess reinforcement ratio 0.1246 per 12.2.5: ld 1.67 in =

12 inch minimum controls

ld_prov 155 in ≥ ld 12 in = = �

Development Check (ACI 318-05 12.12, 12.2.3)

Toe Unfactored Loads

24

in

#5 @ 6 in

Unfactored Loads

300 psf (Self-wt)

590 psf (Soil)

1527 psf 1300 psf

Toe Factored Loads

24

in

#5 @ 6 in

1.7D + 1.7F + 1.7H + 1.7W

510 psf (Self-wt)

1003 psf (Soil)

2595 psf 2209 psf

6.71 ft·k/ft

3.33 k/ft

Toe Checks [1.7D + 1.7F + 1.7H + 1.7W]

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Page 120: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

Design moment Mu for heel need not exceed moment at stem base:

Mheel 18.4 ft·k ft < Mstem / 38.5 ft·k ft / = =

Mu 18.4 ft·k ft stem moment does not control / =

Controlling Moment

aAs fy

0.85 F'c

0.07 in² in / 60000 psi 0.85 4500 psi

1.15 in = = =

φMn φ As fy d a 2 / - 0.90 0.07 in² in / 60000 psi 19.63 in 1.15 in 2 / - 75.44 ft·k ft / = = =

φMn 75.44 ft·k ft ≥ Mu / 18.4 ft·k ft / = = �

Flexure Check (ACI 318-05 10.2)

Vc 2 F'c d 2 4500 psi 19.63 in 31.6 k ft / = = =

φVn φ Vc 0.750 31.6 k ft / 23.7 k ft / = = =

φVn 23.7 k ft ≥ Vu / 0.6 k ft / = = �

Shear Check (ACI 318-05 11.1.1, 11.3.1)

β1 0.85 0.05F'c 4000 -

1000 - 0.85 0.05

4500 psi 4000 - 1000

- 0.8250 = = =

aAs fy

0.85 F'c

0.07 in² in / 60000 psi 0.85 4500 psi

1.15 in = = =

εt 0.003d

a β1 / 1 - 0.003

19.63 in1.15 in 14.0 /

1 - 0.7135 = = =

εt 0.7135 ≥ 0.004 = �

Minimum Strain Check (ACI 318-05 10.3.5)

φMn 75.44 ft·k ft ≥ 4 3 / Mu / 4 3 / 18.4 ft·k ft / 24.54 ft·k ft / = = =

Check is waived per ACI 10.5.3�

Minimum Steel Check (ACI 318-05 10.5.1)

ρST_provAST

t sST

1.58 in² in / 24 in 12 in

0.0055 = = =

ρST_min0.0018 60000

fy

0.0018 60000 60000 psi

0.0018 = = =

ρST_min 0.0018 =

ρST_prov 0.0055 ≥ ρST_min 0.0018 = = �18 inch limit governs

sST_max 18 in =

sST 12 in ≤ sST_max 18 in = = �

Shrinkage Temperature Steel (ACI 318-05 7.12.2)

Mu

φMn

18.4 ft·k ft / 75.44 ft·k ft /

0.2440 ratio to represent excess reinforcement = =

ψt 1.30 more than 12 inches cast below 19.25 inches - =

ψe 1.0 bar not epoxy coated =

ψs 0.80 bars are #6 or smaller =

λ 1.0 normal weight concrete =

s 2 / 6 in 2 / 3 in = =

cover db 2 / + 4 in 0.75 in 2 / + 4.38 in = =

cb 3 in lesser of half spacing, ctr to surface =

Ktr 0.0 no transverse reinforcement =

cb Ktr + db

3 in 0.0 + 0.75 in

4.0 = =

ld3.40

fy

F'c

ψt ψe ψs λ 2.5

db 3.40

60000 psi

4500 psi

1.30 1.0 0.80 1.0 2.5

0.75 in 20.93 in = = =

Factoring ld by the excess reinforcement ratio 0.2440 per 12.2.5: ld 5.11 in =

12 inch minimum controls

ld_prov 56 in ≥ ld 12 in = = �

Development Check (ACI 318-05 12.12, 12.2.3)

Heel Unfactored Loads

24

in#6 @ 6 in

Unfactored Loads

300 psf (Concrete self-wt)

590 psf (Soil weight)

1224 psf497.5 psf

Heel Factored Loads

24

in#6 @ 6 in

1.7D + 1.7F + 1.7H + 1.7W

510 psf (Concrete self-wt)

1003 psf (Soil weight)

845.7-2081 psf (Bearing pressure)

18.4 ft·k/ft

0.6 k/ft

Heel Checks [1.7D + 1.7F + 1.7H + 1.7W]

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Page 121: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

Stem Internal Forces

-1019.72 psf-989.19 psf

-230.35 psf

10.2 k/ft

-38.5 ft·k/ft

Stem Internal Forces

13

11.38

9.75

8.13

6.5

4.88

3.25

1.63

0-40 -30 -20 -10 0

Moment (ft·k/ft)

Moment

Stem Internal Forces

13

11.38

9.75

8.13

6.5

4.88

3.25

1.63

00 3 6 9 12

Shear (k/ft)

Shear

Stem Joint Force Transfer

Location Force@ stem base 10.2 k/ft

Stem Internal Forces

-1019.72 psf -989.19 psf

-230.35 psf

Stem Forces [1.7D + 1.7F + 1.7H + 1.7W]

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Page 122: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

13

11.7

10.4

9.1

7.8

6.5

5.2

3.9

2.6

1.3

0-50 -38.33 -26.67 -15 -3.33 8.33 20

Moment (ft·k/ft)

Offset (ft)

Moment

φMn 43.76 ft·k ft ≥ Mu / 38.5 ft·k ft / = = �

Check (ACI 318-05 Ch 10) @ 0 ft from base

φMn 43.76 ft·k ft ≥ Mu / 0.2 ft·k ft / = = �

Check (ACI 318-05 Ch 10) @ 11.66 ft from base

φMn 42.83 ft·k ft ≥ Mu / 0.2 ft·k ft / = = �

Check (ACI 318-05 Ch 10) @ 11.69 ft from base

Stem Moment Checks [1.7D + 1.7F + 1.7H + 1.7W]

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Page 123: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

13

11.7

10.4

9.1

7.8

6.5

5.2

3.9

2.6

1.3

0-15 -10 -5 0 5 10 15

Shear (k/ft)

Offset (ft)

Shear

φVn 14.04 k ft ≥ Vu / 10.2 k ft / = = �

Shear Check (ACI 318-05 Ch 11.1.1) @ 0 ft from base

Stem Shear Checks [1.7D + 1.7F + 1.7H + 1.7W]

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Page 124: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

φMn 43.76 ft·k ft < 4 3 / Mu / 4 3 / 38.5 ft·k ft / 51.34 ft·k ft / = = =

As_min3 F'c

fyd

3 4500 psi 60000 psi

11.63 in 0.04 in² in / = = =

200 d fy / 200 11.63 in 60000 psi / 0.04 in² in / = =

As 0.07 in² in ≥ As_min / 0.04 in² in / = = �

Minimum Steel Check (ACI 318-05 10.5.1) @ 0 ft from base [Stem in negative flexure]

φMn 43.76 ft·k ft ≥ 4 3 / Mu / 4 3 / 14.57 ft·k ft / 19.42 ft·k ft / = = =

Check is waived per ACI 10.5.3�

Minimum Steel Check (ACI 318-05 10.5.1) @ 3.33 ft from base [Stem in negative flexure]

φMn 0 ft·k ft ≥ 4 3 / Mu / 4 3 / 0 ft·k ft / 0 ft·k ft / = = =

Check is waived per ACI 10.5.3�

Minimum Steel Check (ACI 318-05 10.5.1) @ 13 ft from base [Stem in negative flexure]

β1 0.85 0.05F'c 4000 -

1000 - 0.85 0.05

4500 psi 4000 - 1000

- 0.8250 = = =

aAs fy

0.85 F'c

0.07 in² in / 60000 psi 0.85 4500 psi

1.15 in = = =

εt 0.003d

a β1 / 1 - 0.003

11.63 in1.15 in 14.0 /

1 - 0.4214 = = =

εt 0.4214 ≥ 0.004 = �

Maximum Steel Check (ACI 318-05 10.3.5) @ 0 ft from base [Stem in negative flexure]

β1 0.85 0.05F'c 4000 -

1000 - 0.85 0.05

4500 psi 4000 - 1000

- 0.8250 = = =

aAs fy

0.85 F'c

0.07 in² in / 60000 psi 0.85 4500 psi

1.15 in = = =

εt 0.003d

a β1 / 1 - 0.003

11.63 in1.15 in 14.0 /

1 - 0.4214 = = =

εt 0.4214 ≥ 0.004 = �

Maximum Steel Check (ACI 318-05 10.3.5) @ 3.33 ft from base [Stem in negative flexure]

β1 0.85 0.05F'c 4000 -

1000 - 0.85 0.05

4500 psi 4000 - 1000

- 0.8250 = = =

aAs fy

0.85 F'c

0.07 in² in / 60000 psi 0.85 4500 psi

1.15 in = = =

εt 0.003d

a β1 / 1 - 0.003

11.63 in1.15 in 14.0 /

1 - 0.4214 = = =

εt 0.4214 ≥ 0.004 = �

Maximum Steel Check (ACI 318-05 10.3.5) @ 13 ft from base [Stem in negative flexure]

ρhAs_horz shorz /

t

0.62 in² 12 in / 15 in

0.0034 = = =

ρh_min 0.0020 bars No. 5 or less, not less than 60 ksi =

ρh 0.0034 ≥ ρh_min 0.0020 = = �3 twall 3 15 in 45 in = =

18 inch limit governs

smax 18 in =

shorz 12 in ≤ shorz_max 18 in = = �

Wall Horizontal Steel (ACI 318-05 14.3.3, 14.3.5)

Stem Miscellaneous Checks [1.7D + 1.7F + 1.7H + 1.7W]

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Page 125: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

Mu

φMn

38.5 ft·k ft / 43.76 ft·k ft /

0.8799 ratio to represent excess reinforcement = =

ψe 1.0 uncoated hooked bars =

λ 1.0 normal weight concrete =

ldh 0.02 ψe λ fy

F'c

db 0.02 1.0 1.060000 psi

4500 psi0.75 in 13.42 in = = =

Factoring ldh by the 0.7 multiplier of 12.5.3 a : ldh 9.39 in =

Factoring ldh by the excess reinforcement ratio 0.8799 per 12.5.3 d : ldh 8.26 in =

8 db 8 0.75 in 6.0 minimum limit, does not control = =

ldh_prov 20 in ≥ ldh 8.26 in = = �

Development Check (ACI 318-05 12.12, 12.2.3)

ψt 1.0 bars are not horizontal =

ψe 1.0 bar not epoxy coated =

ψs 0.80 bars are #6 or smaller =

λ 1.0 normal weight concrete =

s 2 / 6 in 2 / 3 in = =

cover db 2 / + 3 in 0.75 in 2 / + 3.38 in = =

cb 3 in lesser of half spacing, ctr to surface =

Ktr 0.0 no transverse reinforcement =

cb Ktr + db

3 in 0.0 + 0.75 in

4.0 = =

ld3.40

fy

F'c

ψt ψe ψs λ 2.5

db 3.40

60000 psi

4500 psi

1.0 1.0 0.80 1.0 2.5

0.75 in 16.1 in = = =

llap 1.3 ld 1.3 16.1 in 20.93 in = = =

llap_prov 40 in ≥ llap 20.93 in = = �1 5 / llap 1 5 / 20.93 in 4.1859 ≤ 6.0 = =

strans 0 in ≤ 1 5 / llap 1 5 / 20.93 in 4.1859 = = = �

Lap Splice Checks (ACI 318-05 12.14.2.3, 12.15.1, 12.15.2) - #6 lap with #6, from 0 ft to 3.33 ft (from stem base)

Stem Miscellaneous Checks [1.7D + 1.7F + 1.7H + 1.7W] (continued)

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Page 126: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

12 Ft. Floodwall

(Case 1)

Page 127: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Site: 7438-009 Owner:

Case:

User Data:A 15 Inches

B 17 Feet

C 12 FeetD 24 InchesD1 24 Inches

Hwall 13 Feet

Hwater 5 Feet

H2 5 FeetH3 5 Feet

Unit Weight of Soil (ϒsat):

Below Footing 0.118 k/cu. ft.Wet Side 0.118 k/cu. ft.Dry Side 0.118 k/cu. ft.

Ø= Below Footing 27 Degrees0.471 radians

Wet Side 27 Degrees0.471 radians

Dry Side 27 Degrees0.471 radians

SMF= 0.66667Unit Weight of Conc. (ρc) 0.15 k/cu. ft.

Unit Weight of Water (ρw) 0.0625 k/cu. ft.

Soil Capacity 6.2 k/sq. ft.2 Feet Ko Below Footing 0.55

15 Inches Wet Side 0.55Dry Side 0.55

Kp for sliding = 2.663fysteel 60 ksi Cohesion Factor (CF) 1 ksf

f'c(concrete) 4.5 ksi Frost Depth (Fr): 6 ft

Vertical Surcharge (Dry) (P3) 250 psf

Vertical Surcharge (Wet) (P4) 250 psf

Ice/Debris Load NoType of Loading Unusual

Toe Vertical Fill Width (F)= 3.75 ft

*All Dimensions in Calculations are converted to feet

City of Fargo

13' Floodwall Unusual construction

Keyway Depth (G)Keyway Width (L)

Project Number:Fargo Floodwalls

1 − sin∅

1 − sin∅1 − sin∅

Page 128: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Force CalculationsFactor Unit Equation Description

EHW= 0.563 kips/ft Driving Wet Side Water Pressure

EHS= 0.500 kips/ft Driving Wet Side Soil Pressure

EV2= 7.080 kips Wet Side Vertical Component Weights

EV2d 11.000 ft from toe Moment arm for EV2

EV1= 2.213 kips Dry Side Vertical Component Weights

EV1d 1.875 ft from toe Moment arm for EV1

PehW= 2.531 kips Driving Wet Side Water Force

Pehwd 1.000 ft. above footing Moment arm for Pehw

PehS= 2.248 kips Driving Wet Side Soil Force

Pehsd 1.000 ft. above footing Moment arm for Pehs

PS1= 2.813 kips Vertical Weight of Stem and Key

PS1d 5.975 ft from toe Moment arm for PS1

PS2= 5.100 kips Vertical Weight of Footing

PS2d 8.500 ft from toe Moment arm for PS2

EH2= 1.062 kips Resisting Dry Side Soil Pressure

Peh2= 4.779 kips Resisting Dry Side Forces

Peh2d 1.000 ft from toe Moment arm for Peh2

Peh2s 2.248 kips Resisting dry side soil only

Peh2sd 1.000 ft from toe Moment arm for Peh2s

Pice/debris 0.000 kips .5 k/ft at Q100 Driving ice/debris Force

Pice/debrisd 7.000 ft. above footing Moment arm for Pice/debris

Pwind 0.24 kips Driving wind Force

Pwindd 11.000 ft. above footing Moment arm for Pwind

Applied Forces

(��� +�� + �)(ρ)

(�3 + �� + �)(ɤ�� − ρ)

��� � ρ +H3(C)(ɤ�� − ρ)

�2(�)(ɤ��)

��

2(��� + �� + �)

���

2(�3 + �� + �)

��� + �� + �

3− �

�3 + �� + �

3− �

��� � ρ� + �( )(ρ�)

! − � �� ρ� + �(�)(ρ�)

(�2 + �� + �)(ɤ��)

��"

2(�2 + �� + �)

! −�

2

2

��� � ρ� ! − � −�

2+ � ρ� ! −

2/$%�

! − � �� ρ�! − �

2+ � � ρ� ! −

2/$%"

�2 + �� + �

3− �

��� + ��

��� − ���

2+ ��� + ��

ρ� − ρ ∗ �2 + �� + � +�2 + �� + �

2

�2 + �� + �

3− �

.03(��� −��� )

Page 129: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Water Uplift

Force Calculation Factor Unit Equation DescriptionLZ-Z1= 17.117 ft Length of Seepage path from heel to toe

Δh= 0.000 ft Head differential between wet and dry side

Ldry side 7.000 ft Seepage path on Dry Side

LS 24.117 ft Total Seepage Path

uZ= 0.563 ksf Water Pressure at bottom of key (Wet side)

uZ1= 0.438 ksf Water pressure at bottom of footing (Dry Side)

HLZ-Z1= 0.000 ft Head Loss along Z-Z1

LSc 19.000 ft Length of concrete surface in sliding surface

HLLK = 0.000 ft Head Loss along key

ubottom key = 0.563 ksf Water pressure at bottom of key (Dry side)

HLUK = 0.000 ft Head Loss up key

utop key = 0.438 ksf Water pressure at top of key (Dry side)

P5 = 7.594 kips Water Uplift for Overturning

P5d 8.662 ft

P5sa 8.500 kips Water uplift for sliding along angle

P5sad -8.854 ft Moment Arm for P5sa

Lsss = 26.000 ft

Uf = 0.563 ksf

P5ss = 9.563 Water uplift of sliding along bottom of key

P5ssd -8.500 Moment Arm for P5ss

(!"+ �")

��� − �2

�2 + ��

)*)� + + ,�.+�

(��� +�� + �)(ρ)

��� + �� − ∆0 )*)� 1

(ρ)

∆0 )*)� �

! + �

%�(� )*)�)

(��� + �� + � − � 23)ρ

+ �

%�(� )*)�)

(��� +�� −� 43)ρ

5) + 567��789�,

2 +

5�7:9�, + 5)�2

(! − )

5) + 5)�2

(!)

5)�! −

2! − +

5�7:9�, − 5)�2

! − 2

3! − + 567��789�, ! −

2+5) − 567��789�,

2 ! −

3/$5

5)� !!

2+

5) − 5)�2

!2!

3/$5�

! + � + �� +�2

��� +�� + � − ��� −�2!

<��ρ

5= + 5>2

(!)

5> !!

2+5) − 5>

2!

2!

3/$5��

Moment Arm

for P5

Water pressure at key depth below

footing on dry side of footing

Page 130: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Distance Moment

to Toe(ft) About Toe (k-ft)

PS1 2.81 5.98 16.8 MPS1

PS2 5.10 8.50 43.4 MPS2

P3 0.94 1.88 1.8 MP3

P4 3.00 11.00 33.0 MP4

P5 -7.59 8.66 -65.8 MP5

4.26 6.8 29.1

EV2 - Backfill on Heel EV2 7.08 11.00 77.9 MEV2

EV1 - Fill on Toe EV1 2.21 1.88 4.1 MEV1

9.29 8.8 82.0

LL Live Load 0.0 0.00 0.0 MLL

0.0 0.0

EH Horiz. Earth Load PehS 2.25 -1.00 -1.2 MEH

WS Hydrostatic Pressure PehW 2.53 -1.00 -2.5 MPHW

ID Ice/Debris Force Pice/debris 0.00 -7.00 0.0 MID

W Wind Load Pwind 0.24 -11.00 -2.6 MW

5.0 -6.4

Fargo Floodwalls

Ve

rtica

l Lo

ad

s

Applied Force Calculations

Total

P (kips)DescriptorLoad

Vertical Loads

DC

Horizontal Loads

EV

Total

Total

Total

Page 131: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Factor Equation Description

EM 1110-2-2502

Reference

PehW2 = 2.531 kips Water on Dry side for Overturning Calcs Figure 4-5

Pehw2d 1.000 ft Moment Arm for Pehw2 N/A

Peh2sb 1.360 ft Dry Side Soil to bottom of Footing Para. 4-8

Peh2sbd 2.333 kips Moment Arm for Peh2sb N/A

RS = -0.495 kips Resisting soil on Dry side to create equilibrium Para. 4-8

RSd -1 ft Moment Arm for RS N/A

Including Uplift

ΣVo 13.549 kips Total Vertical Load Figure 4-5

ΣMo 110.173 kip-ft 4-1

XR 8.132 ft Resultant Location 4-1

bl 24.39482 ft Length of Base in Compression 4-2

b% 143.49894 Percent of Base in Compression 4-2

Required b% 75 Required Base in Compression Appendix F

OK

Neglecting Uplift

ΣVon 21.143 kips Total Vertical Load Figure 4-5

ΣMon 175.950 kip-ft 4-1

XRn 8.322 ft Resultant Location 4-1

bln 24.966364 ft Length of Base in Compression 4-2

b%n 146.86096 Percent of Base in Compression 4-2

Required b%n 75 Required Base in Compression Appendix 4

OK

Global Stability

Overturning:

Fargo Floodwalls

$%� + $%" + $? + $@ + $A + �B1 + �B2

5)�2

�2 + �� +5�7:9�, + 567��789�,

2(�)

�2 + ��3

5)�2

�2 + ��3

�2 + �� + 5�7:9�,−�2

� +567��789�, − 5�7:9�,

2�

−2�3

/$CD"

−�2

EF1� +EF1" +EF? +EF@ +EFA +EGH� +EGH" +EFIJ +EKL +EJ + M% M%N + $CD"�6 $CD"�6N + $CD"($CD"N)(O7)(%E�)

ΣE7

ΣB7

3QR

S�!(100)

$%� + $%" + $? + $@ + �B1 + �B2

EF1� +EF1" +EF? +EF@ +EGH� +EGH" +EFIJ +EKL +EJ + M% M%N + $CD"�6 $CD"�6N + $CD"($CD"N)(O7)(%E�)

ΣE7T

ΣB7T

3QRT

S�T!

(100)

Total Overturning

Moment about

point o

(�2 + ��)(ɤ�� − ρ)(�2 + ��)/2

$CD − $CD" − $CD"�6(O7)(%E�)

Total Overturning

Moment about

point o

11/17/2015

Page 132: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Factor Equation Description

EM 1110-2-2502

Reference

Sliding Along Angle Between Bottom of Key and Bottom of Footing on Dry Side Figure 4-11

SBFa 1.722 kips Soil Below Footing on Sliding Surface Figure 4-11

Pwda 2.531 kips Water on Dry Side Figure 4-11

ΣVa 14.364 kips Sum of Vertical Forces on Sliding Surface Figure 4-11

αa 0.117 radians Angle of slip plane to horizontal plane Figure 4-11

6.710 degrees

Drained Condition

RFswad 5.524 kips Resisting Force from passive soil and water on dry side Figure 4-11

ΣHad -2.75 kips Sum of Horizontal Forces Figure 4-11

N'ad 13.944 kips Normal Force to Sliding Surface Figure 4-11

Tad -4.412 kips Tangential Force to Sliding Surface Figure 4-11

SSad 7.105 kips Drained Shear Strength 4-12

FSad -1.610 Factor of Safety 4-12

OK

Undrained Condition

RFswau 3.655 kips Resisting Force from passive soil and water on dry side Figure 4-11

ΣHau -0.884 kips Sum of Horizontal Forces Figure 4-11

N'au 14.163 kips Normal Force to Sliding Surface Figure 4-11

Tau -2.556 kips Tangential Force to Sliding Surface Figure 4-11

SSau 11.412 kips Undrained Shear Strength 4-12

FSau -4.464 Factor of Safety 4-12

OK

Sliding:

Fargo Floodwalls

Global Stability

� −�!

2(! − )(ɤ��)

5)2

�2 + �� +5) + 5)�

2(�)

%UV + �B1 + �B2 + $%� + $%" + $? + $@ − $A�

atan�!

$+ + $CD" − $+ Y: (.5)

$CD + $.��/+�6 .� + $.T+ − M��+

ΣB cos \ + Σ�+sin(\)

Σ�+ cos \ − ΣBsin(\)

]′+tan(∅)

%%+_+

$+ + $CD" − $+ (.5)

$CD + $.��/+�6 .� + $.T+ − M��`

ΣB cos \ + Σ�`sin(\)

Σ�` cos \ − ΣBsin(\)

��(%E�)√(!" + �")

%%`_`

11/17/2015

Page 133: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Factor Equation Description

EM 1110-2-2502

Reference

Sliding Along Horizontal Surface along Bottom of Key Figure 4-8

SBFs 3.717 kips Soil Below Footing on Sliding Surface

Pwds 2.531 kips Water on Dry Side Figure 4-11

ΣVs 15.297 kips Sum of Vertical Forces on Sliding Surface Figure 4-11

αs = 0 radians Angle of slip plane to horizontal plane Figure 4-11

0 degrees

Drained Condition

RFswsd 5.524 kips Resisting Force from passive soil and water on dry side Figure 4-11

ΣHsd -2.75 kips Sum of Horizontal Forces Figure 4-11

N'sd 15.297 kips Normal Force to Sliding Surface Figure 4-11

Tsd -2.753 kips Tangential Force to Sliding Surface Figure 4-11

SSsd 7.794 kips Drained Shear Strength 4-12

FSsd -2.831 Factor of Safety 4-12

OK

Undrained Condition

RFswsu 3.655 kips Resisting Force from passive soil and water on dry side Figure 4-11

ΣHsu -0.884 kips Sum of Horizontal Forces Figure 4-11

N'su 15.297 kips Normal Force to Sliding Surface Figure 4-11

Tsu -0.884 kips Tangential Force to Sliding Surface Figure 4-11

SSsu 11.333 kips Undrained Shear Strength 4-12

FSsu -12.822 Factor of Safety 4-12

OK

Global Stabilty

Fargo Floodwalls

�(! − )(ɤ��)

5>2(�2 + �� + �)

%UV� + �B1 + �B2 + $%� + $%" + $? + $@ − $A��

$+� + $CD" − $+� Y: (.5)

$CD + $.��/+�6 .� + $.T+ − M���+

ΣB� cos \� + Σ��+sin(\�)

Σ��+ cos \� − ΣB�sin(\�)

]′�+tan(∅)

%%�+_�+

.

$+� + $CD" − $+� (.5)

$CD + $.��/+�6 .� + $.T+ − M���`

ΣB� cos \� + Σ��`sin(\�)

Σ��` cos \� − ΣB�sin(\�)

�� %E� !

%%�`_�`

11/17/2015

Page 134: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Factor Equation Description

EM 1110-2-2502

Reference

Bearing Along Angle Between Bottom of Key and Bottom of Footing on Dry Side Eqns. From 1110-2-2502

ΣHadb 0.24 kips Sum of Horizontal Forces Figure 4-11

N'adb 14.294 kips Normal Force to Bearing Surface Figure 4-11

Tadb -1.440 kips Tangential Force to Bearing Surface Figure 4-11

SBFad 10.500005 ft Moment Arm for SBFa Figure 4-11

ΣMoaB 117.11955 kip-ft

XraB 8.153 ft Resultant Location Figure 4-11

eaB 0.405 ft Eccentricity of resultant Figure 4-11

B'aB 16.30698 ft Effective Base for Bearing Sec 5-2

Drained Condition

q0 = 0.333 k/ft2 Effective Overburden Pressure

y = 1.6007185

Nq = 13.175528 Bearing Capacity Factor 5-3a

Nc = 23.895844 Bearing Capacity Factor 5-3b

Nϒ = 9.4442971 Bearing Capacity Factor 5-3d

εcd = 1.12 Embedment Factor 5-4a

εqd = 1.06 Embedment Factor 5-4c

δd = -0.10 5-5

-5.752631

εqi = 1.13 Inclination Factor 5-5a

εϒi = 1.47 Inclination Factor 5-5b

Q1 = 0.00 5-2

Q2 = 5.2641707 5-2

Q3 = 6.6661073 5-2

Qd = 194.54681 5-2

FS = 5-1

13.610382 Eq. 5-1 Bearing Criteria Satisfied

Global Stabilty

Fargo Floodwalls

Bearing:

ΣE7U

ΣB )*)�2

− QbU

)*)� − 2(CU)

(C,) tan 45 +∅2

de∅ > 0, (]h−1)ijklm(∅)

]h − 1 tan(1.4∅)

1 + 0.2� !n tan 45 +

∅2

de∅ > 0, 1 + 0.1� !n tan 45 +

∅2

atanΣ�ΣB

1 −o+90

"

1 −o+∅

"

q�+(qh.)(rs)(]h)

qh+(qh.)(rs)(]h)

qh+ qh. ɤ�� − ɤ�� ]t /2

!′(u1 + u2 + u3)

(ɤ�� − ɤ�� )(� )

v(tan ∅)

Normal Component to the base of the structure of the

ultimate bearing capacity

EF1� +EF1" +EF? +EF@ + $A� $A�N + EGH� +EGH" +EGI +EFIJ +EKL +EJ + $CD"�($CD"�N)(O7)(%E�)+$+ $CD"N + %UV %UVN

$CD + $.��/+�6 .� + $.T+ + $CD�(O7)(%E�) − $+ − $CD"�(O7)(%E�)

ΣB cos \ + Σ�+6sin(\)

Σ�+6 cos \ − ΣBsin(\)

2(! − )3

11/17/2015

Page 135: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Factor Equation Description

EM 1110-2-2502

Reference

q0 = 0.333 k/ft2 Effective Overburden Pressure

y = 0

Nq = 1 Bearing Capacity Factor 5-3a

Nc = 5.14 Bearing Capacity Factor 5-3c

Nϒ = 0 Bearing Capacity Factor 5-3d

εcd = 1.07 Embedment Factor 5-4a

εqd = 1.00 Embedment Factor 5-4b

δd = -0.10 5-5

-5.752631

εqi = 1.13 Inclination Factor 5-5a

εϒi = 1.47 Inclination Factor 5-5b

Q1 = 6.25 5-2

Q2 = 0.3769299 5-2

Q3 = 0 5-2

Qd = 107.99801 5-2

FS = 5-1

7.555478

Drained q'max 0.96 Undrained q'max 0.96

Bearing Criteria Satisfied

Undrained Condition

Global Stabilty

Fargo Floodwalls

(ɤ�� − ɤ�� )(� )

v(tan ∅)

de∅ = 0, 5.14

de∅ = 0, 1

(C,)tan(45 +∅2)

]h − 1 tan(1.4∅)

1 + 0.2� !n tan 45 +

∅2

atanΣ�ΣB

1 −o+90

"

1 −o+∅

"

q�+(qh.)(rs)(]h)

qh+(qh.)(rs)(]h)

qh+ qh. ɤ�� − ɤ�� ]t /2

!′(u1 + u2 + u3)Normal Component to the base of the structure of the

ultimate bearing capacity

11/17/2015

Page 136: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Factor Equation Description

Load Factor 1.7

b 17 Effective base in compression

qmin 0.77116

Uws 0.4375 Uplift Pressure Wet side of Stem

Utws 5.40625 Total water uplift on wet side

Utwsd 6.155347 Moment arm for U tws

VLw 8.64875 Vertical load on wet side

Bws 0.905201 Bearing at wet side of stem

Btws 10.05816 Total Bearing uplift on wet side of footing

Btwsd 5.840081 Moment arm for B tws

Total Shear on Heel -2.396

Total Moment on Heel -13.0693

Total Shear on Stem 0.41 Total Shear on Stem

Total Moment on Stem 3.671978

Uds 0.4375 Uplift Pressure Dry side of Stem

Utds 1.640625 Total water uplift on dry side

Utdsd 1.875 Moment arm for U tds

VLd 2.634375 Vertical load on dry side

Bds 0.919163 Bearing at dry side of stem

Btds 3.525401 Total Bearing uplift on dry side of footing

Btdsd 1.888924 Moment arm for B tds

Total Shear on Toe -1.51474 Total Shear on Toe

Total Moment on Toe -2.92359 Total Moment on Toe

Utkw 0.4375 Water Pressure at top of key (wet side)

Total Shear on Key 0.84 Total Shear on Key

Total Moment on Key 0.841434 Total Moment on Key

Forces for Reinforcement/Shear

Heel Forces

Stem Forces

Toe Forces

Key Forces

$KL +$J + �

2��� + �3 ɤ�� − ρ

�32

Y7 − ρ�22

�2 − �2 ɤ�� − ρ�22

(Y7) jlN�likjb

$KL ��� + $J $.T+N −�12

+ �

2���

���

3+ �3 ɤ�� − ρ

�32

�33

(Y7) − ρ�22

�2�23

−�2 ɤ�� − ρ�22

�23

(Y7) ( jlN�likjb)

x�7:9�, −x�7:9�, −x)�

! − � −

x�7:9�, −x�

2C − L +

x) +x67��789�,

2( )

x�� − 2

� − +x�7:9�, − x�

22 � −

3� − + x67��789�, � −

2

+x) −x67��789�,

2 � −

3

/x��

�B2+ � � ρ� +� ρ� +$@ − x��

r8z − r8z − r8.T

S(� + �)

(B − !��)( jlN�likjb)

�B2+ � � ρ� +� ρ� +$@�2

− x�� x��N − !�� !��N ( jlN�likjb)

x�7:9�, −x�7:9�, −x)�

! − � + � −

x+� +x)�

2(�)

[x)� ��2

+x+� −x)�

2(�3)(�)]/x�+�

�B1 + $? +� �� ρ� −x�+�

r8z − r8z − r8.T

S(�)

r8z + !+�2

(�)

!+� ��2

+r8z −!+�

22�3

� /!�+�

(B + −!�+�)( jlN�likjb)

�B1+ $? +� �� ρ��2

−x�+� x�+�N − !�+� !�+�N ( jlN�likjb)

}eS% ≥ 100,!, S�

}eS% ≥ 100,r8z

!1−

6C!

, 0

}er8.T = 0,!�

2S − � − � , (

!�+ r8.T

2)(�)

}er8.T = 0,S − � − �

3, r8.T �

�2

+!�− r8.T

2�

�3

/!��

x�9 +x=

2� −

x67��789�, +x�7:9�,

2� − M% jlN�likjb

x�9 ��2

+x= − x�9

2�

2�3

−x�7:9�, ��2

−x67��789�, −x�7:9�,

2�

2�3

− M%�2

( jlN�likjb)

(��� +�)(�)

Total

Moment

on Stem

Page 137: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Input: Values in red need to be entered by user Designer: LJB Date: 11/17/2015

Check: Date:Project Number:

Bridge No./Structure ID:

Owner:

Mu = 13.06932 k-ft 7.688 k-ft concrete wall thickness (t) 24 in.

Vu = 2.396003 k 1.409 k Clear Cover 4 in.

Nu = 0.0 k 0.0 k d dimension 19.625 in.

(factored) (service)150 psi

Bar Size = 6

Enter Reinforcement Spacing = 6 in. Reinforcement yield strength = 60 ksiAs = 0.88 in2

Concrete yield strength = 4.5 ksi

Enter design width = 12 in.Concrete yield strength = 4500 psi

Bar Size = None Ec = 4066.84 ksi

Enter Reinforcement Spacing = 6 in. Es = 29000 ksi

Av = 0.00 in2Lightweight Concrete? NO

Mod Factor ƛ = 1Does ACI 350 Apply? NO Flexure φ = 0.9

Shear φ = 0.75

Flexure Analysis: Mr = Phi(Mn)

For members without axial loads.

a = 1.150

Mn = 1,005.8 k-in

Mn= 83.8 k-ft

φMn = 75.44 k-ft

Minimum Reinforcement: (ACI 318 10.5)

As, min = 0.79 in2 not less than 0.785 in2

ACI 318 - 10.5 Requirement = Pass If Fail ------------>

1.33*Mu = 17.38 k-ft

As,req = 0.18 in2

As, min = 0.79 in2

OK - Section and reinforcement sufficient for Flexural Resistance

*If the requirements of 10.5 are not satisfied then As needs to be at least 1/3 greater than that required by analysis. (10.5.3)

OK

She

ar

7438-009 Section Being Designed:Heel of Footing

Fargo Floodwalls Layer of Steel:

City of Fargo

Modulus of elasticity based on concrete unit weight of

Fle

xure

��,8.T =? >��

>�SNS5kmjk�C<<kDlm200S

+

>�

ET = �� ∗ e, ∗ N −l2

l =�� ∗ e,

0.85 ∗ e′� ∗ S

��, �h =.85en�S

e,N − N" −

2E`

.85e′�S

Page 138: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Designer: LJB Date: 11/17/2015

Check: Date:

ACI 318

Governing Specification: USACE

Min. Steel Requirement:

As/Ag (Includes both faces) ≥ 0.0050 EM 1110-2-2104 2.8

Enter Reinforcing configuration of opposite mat:Bar Size = 5

Enter Reinforcement Spacing = 6 in.As = 0.62 in2

Required Min. Steel (both faces): 1.44 sq. in/ft

Max. Spacing Requirements:

s≤5h or s≤18.0 in. 120 in18 in Structural walls/footings:

or for componenets over 36.0 in Spacing can be ≤ 12 in. on faces > 18" thick

s≤12.0 12 in

Maximum Spacing Allowed: 18 in

ACI 350

Min. Steel Requirement: ACI 350-06 7.12.2.1 Minimum reinforcement ratio for Grade 60 steel

Length Between JointsRatio As/Ag 0.0030 ≤ 20 ft

20-30 ft

Steel in opposite face: 0.62 in230-40 ft≥ 40ft

Required Min. Steel: 0.86 sq. in/ft

Max. Spacing Requirements: ACI 350-065 7.12.2.2

Shrinkage and Temperature Reinforcement shall not be spaced greater than12in. apart

Maximum Spacing Allowed: 12 in

0.0040.005

T&S Steel Requirement Met

Max Spacing Requirement Met

Max Spacing Requirement Met

T&S Steel:

Ratio (As/Ag)0.0030.003

0

T&S Steel:

T&S Steel Requirement Met

Layer of Steel:

Section Being Designed:

Heel of Footing

Page 139: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Designer: LJB Date: 11/17/2015Check: Date:

Shear Analysis:

φVn ≥ Vu (ACI 318 Eq. 11-1)

Vn = Vc + Vs (ACI 318 Eq. 11-2)

Calculate d:

d = 19.625 (Loads placed at 'top' of member.) (ACI 318 11.1.3.1)

dv = 19.625 in

Calc. Vc

For members subject to shear & flexure only.Vc = 2* ƛ * √fc * bv * dv ƛ = 1.0 (ACI 318 Eq. 11-3)

bv = 12 in

Vc = 31.6 kips dv = 19.63 in

For members subject to axial compression.Vc = 2*(1 + Nu/(2000*Ag)) * ƛ * √fc * bv * dv Nu = 0 lb (ACI 318 Eq. 11-4)

Ag = 288 in2

Vc = 758.3 kips

Vc = 31.6 kips

Calc. Vs

Shear Reinforcement perpendicular to axis.

Vs = Av * fyt * d / s Av = 0.00 in2

s = 6 in

Vs = 0.0 kips

Vn = 31.6 kips

θVn = 23.7 kips Vu= 2.396003

θVn = 23.7 kips OK - Section Adequate for Shear Resistance

Section Being Designed:Heel of Footing

Layer of Steel:0

Page 140: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Designer: LJB Date: 11/17/2015Check: Date:

Crack Control: ACI 318

Distribution of flexural reinforcement in beams and one-way slabs.

The spacing of reinforcement closest to the tension face, s, shall not exceed:

cc = 4 in (Clear cover)

fs = Calculated stress from unfactored Moment.

But not greater than Mserv = 92.3 kip-in

n = 7.13

x = 4.038 in NA to extreme comp fiber

y = 15.587 in NA to tension centroid

I = 1787.9 in4

fs = 5735.179 psi

s ≤ 94.6 in

Crack Control: ACI 350

Actual Stress:fs = 5735.179 psi

ACI 350-06 10.6.4.1, for normal environmental exposures as defined in 10.6.4.5

β= 1.28

s= 6 in.db= 0.75 in.

fs,max= 32.65 ksi26781.3 psi > 5735.179 psi

OK - Spacing Adequate

OK - Spacing Adequate

fs,max =320

β*SQRT(s2+4(2+

db∕2)

2)

Section Being Designed:Heel of Footing

Layer of Steel:0

< = 15@s,sss

>�− 2.5 i�

12(40,000

e�)

e� = m ∗E�� � ∗ �

}

� =�� ∗ �� ∗ m + 2 ∗ S ∗ N ∗ m

S−�� ∗ mS

� = N − �

} =13− S ∗ �? + m ∗ �� ∗ �"

Page 141: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Input: Values in red need to be entered by user Designer: LJB Date: 11/17/2015

Check: Date:Project Number:

Bridge No./Structure ID:

Owner:

Mu = 3.671978 k-ft 2.160 k-ft concrete wall thickness (t) 15 in.

Vu = 0.408 k 0.240 k Clear Cover 3 in.

Nu = 0.0 k 0.0 k d dimension 11.625 in.

(factored) (service)150 psi

Bar Size = 6

Enter Reinforcement Spacing = 6 in. Reinforcement yield strength = 60 ksiAs = 0.88 in2

Concrete yield strength = 4.5 ksi

Enter design width = 12 in.Concrete yield strength = 4500 psi

Bar Size = None Ec = 4066.84 ksi

Enter Reinforcement Spacing = 6 in. Es = 29000 ksi

Av = 0.00 in2Lightweight Concrete? NO

Mod Factor ƛ = 1Does ACI 350 Apply? NO Flexure φ = 0.9

Shear φ = 0.75

Flexure Analysis: Mr = Phi(Mn)

For members without axial loads.

a = 1.150

Mn = 583.4 k-in

Mn= 48.6 k-ft

φMn = 43.76 k-ft

Minimum Reinforcement: (ACI 318 10.5)

As, min = 0.47 in2 not less than 0.465 in2

ACI 318 - 10.5 Requirement = Pass If Fail ------------>

1.33*Mu = 4.88 k-ft

As,req = 0.08 in2

As, min = 0.47 in2

OK - Section and reinforcement sufficient for Flexural Resistance

*If the requirements of 10.5 are not satisfied then As needs to be at least 1/3 greater than that required by analysis. (10.5.3)

OK

She

ar

7438-009 Section Being Designed:Stem

Fargo Floodwalls Layer of Steel:

City of Fargo

Modulus of elasticity based on concrete unit weight of

Fle

xure

��,8.T =? >��

>�SNS5kmjk�C<<kDlm200S

+

>�

ET = �� ∗ e, ∗ N −l2

l =�� ∗ e,

0.85 ∗ e′� ∗ S

��, �h =.85en�S

e,N − N" −

2E`

.85e′�S

11/17/20158:41 AM

Page 142: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Designer: LJB Date: 11/17/2015

Check: Date:

ACI 318

Governing Specification: USACE

Min. Steel Requirement:

As/Ag (Includes both faces) ≥ 0.0028 EM 1110-2-2104 2.8

Enter Reinforcing configuration of opposite mat:Bar Size = 5

Enter Reinforcement Spacing = 12 in.As = 0.31 in2

Required Min. Steel (both faces): 0.50 sq. in/ft

Max. Spacing Requirements:

s≤5h or s≤18.0 in. 75 in18 in Structural walls/footings:

or for componenets over 36.0 in Spacing can be ≤ 12 in. on faces > 18" thick

s≤12.0 12 in

Maximum Spacing Allowed: 18 in

ACI 350

Min. Steel Requirement: ACI 350-06 7.12.2.1 Minimum reinforcement ratio for Grade 60 steel

Length Between JointsRatio As/Ag 0.0030 ≤ 20 ft

20-30 ft

Steel in opposite face: 0.31 in230-40 ft≥ 40ft

Required Min. Steel: 0.54 sq. in/ft

Max. Spacing Requirements: ACI 350-065 7.12.2.2

Shrinkage and Temperature Reinforcement shall not be spaced greater than12in. apart

Maximum Spacing Allowed: 12 in

0.0040.005

T&S Steel Requirement Met

Max Spacing Requirement Met

Max Spacing Requirement Met

T&S Steel:

Ratio (As/Ag)0.0030.003

0

T&S Steel:

T&S Steel Requirement Met

Layer of Steel:

Section Being Designed:

Stem

11/17/20158:41 AM

Page 143: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Designer: LJB Date: 11/17/2015Check: Date:

Shear Analysis:

φVn ≥ Vu (ACI 318 Eq. 11-1)

Vn = Vc + Vs (ACI 318 Eq. 11-2)

Calculate d:

d = 11.625 (Loads placed at 'top' of member.) (ACI 318 11.1.3.1)

dv = 11.625 in

Calc. Vc

For members subject to shear & flexure only.Vc = 2* ƛ * √fc * bv * dv ƛ = 1.0 (ACI 318 Eq. 11-3)

bv = 12 in

Vc = 18.7 kips dv = 11.63 in

For members subject to axial compression.Vc = 2*(1 + Nu/(2000*Ag)) * ƛ * √fc * bv * dv Nu = 0 lb (ACI 318 Eq. 11-4)

Ag = 180 in2

Vc = 280.7 kips

Vc = 18.7 kips

Calc. Vs

Shear Reinforcement perpendicular to axis.

Vs = Av * fyt * d / s Av = 0.00 in2

s = 6 in

Vs = 0.0 kips

Vn = 18.7 kips

θVn = 14.0 kips Vu= 0.408

θVn = 14.0 kips OK - Section Adequate for Shear Resistance

Section Being Designed:Stem

Layer of Steel:0

11/17/20158:41 AM

Page 144: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Designer: LJB Date: 11/17/2015Check: Date:

Crack Control: ACI 318

Distribution of flexural reinforcement in beams and one-way slabs.

The spacing of reinforcement closest to the tension face, s, shall not exceed:

cc = 3 in (Clear cover)

fs = Calculated stress from unfactored Moment.

But not greater than Mserv = 25.9 kip-in

n = 7.13

x = 3.003 in NA to extreme comp fiber

y = 8.622 in NA to tension centroid

I = 574.8 in4

fs = 2772.429 psi

s ≤ 208.9 in

Crack Control: ACI 350

Actual Stress:fs = 2772.429 psi

ACI 350-06 10.6.4.1, for normal environmental exposures as defined in 10.6.4.5

β= 1.39

s= 6 in.db= 0.75 in.

fs,max= 30.05 ksi26781.3 psi > 2772.429 psi

OK - Spacing Adequate

OK - Spacing Adequate

fs,max =320

β*SQRT(s2+4(2+

db∕2)

2)

Section Being Designed:Stem

Layer of Steel:0

< = 15@s,sss

>�− 2.5 i�

12(40,000

e�)

e� = m ∗E�� � ∗ �

}

� =�� ∗ �� ∗ m + 2 ∗ S ∗ N ∗ m

S−�� ∗ mS

� = N − �

} =13− S ∗ �? + m ∗ �� ∗ �"

11/17/20158:41 AM

Page 145: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Input: Values in red need to be entered by user Designer: LJB Date: 11/17/2015

Check: Date:Project Number:

Bridge No./Structure ID:

Owner:

Mu = 2.923594 k-ft 1.720 k-ft concrete wall thickness (t) 24 in.

Vu = 1.514744 k 0.891 k Clear Cover 4 in.

Nu = 0.0 k 0.0 k d dimension 19.6875 in.

(factored) (service)150 psi

Bar Size = 5

Enter Reinforcement Spacing = 6 in. Reinforcement yield strength = 60 ksiAs = 0.62 in2

Concrete yield strength = 4.5 ksi

Enter design width = 12 in.Concrete yield strength = 4500 psi

Bar Size = None Ec = 4066.84 ksi

Enter Reinforcement Spacing = 6 in. Es = 29000 ksi

Av = 0.00 in2Lightweight Concrete? NO

Mod Factor ƛ = 1Does ACI 350 Apply? NO Flexure φ = 0.9

Shear φ = 0.75

Flexure Analysis: Mr = Phi(Mn)

For members without axial loads.

a = 0.810

Mn = 717.3 k-in

Mn= 59.8 k-ft

φMn = 53.80 k-ft

Minimum Reinforcement: (ACI 318 10.5)

As, min = 0.79 in2 not less than 0.7875 in2

ACI 318 - 10.5 Requirement = Fail If Fail ------------>

1.33*Mu = 3.89 k-ft

As,req = 0.04 in2

As, min = 0.04 in2

OK - Section and reinforcement sufficient for Flexural Resistance

*If the requirements of 10.5 are not satisfied then As needs to be at least 1/3 greater than that required by analysis. (10.5.3)

OK

She

ar

7438-009 Section Being Designed:Toe of Footing

Fargo Floodwalls Layer of Steel:

City of Fargo

Modulus of elasticity based on concrete unit weight of

Fle

xure

��,8.T =? >��

>�SNS5kmjk�C<<kDlm200S

+

>�

ET = �� ∗ e, ∗ N −l2

l =�� ∗ e,

0.85 ∗ e′� ∗ S

��, �h =.85en�S

e,N − N" −

2E`

.85e′�S

11/17/20158:41 AM

Page 146: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Designer: LJB Date: 11/17/2015

Check: Date:

ACI 318

Governing Specification: USACE

Min. Steel Requirement:

As/Ag (Includes both faces) ≥ 0.0050 EM 1110-2-2104 2.8

Enter Reinforcing configuration of opposite mat:Bar Size = 6

Enter Reinforcement Spacing = 6 in.As = 0.88 in2

Required Min. Steel (both faces): 1.44 sq. in/ft

Max. Spacing Requirements:

s≤5h or s≤18.0 in. 120 in18 in Structural walls/footings:

or for componenets over 36.0 in Spacing can be ≤ 12 in. on faces > 18" thick

s≤12.0 12 in

Maximum Spacing Allowed: 18 in

ACI 350

Min. Steel Requirement: ACI 350-06 7.12.2.1 Minimum reinforcement ratio for Grade 60 steel

Length Between JointsRatio As/Ag 0.0030 ≤ 20 ft

20-30 ft

Steel in opposite face: 0.88 in230-40 ft≥ 40ft

Required Min. Steel: 0.86 sq. in/ft

Max. Spacing Requirements: ACI 350-065 7.12.2.2

Shrinkage and Temperature Reinforcement shall not be spaced greater than12in. apart

Maximum Spacing Allowed: 12 in

0.0040.005

T&S Steel Requirement Met

Max Spacing Requirement Met

Max Spacing Requirement Met

T&S Steel:

Ratio (As/Ag)0.0030.003

0

T&S Steel:

T&S Steel Requirement Met

Layer of Steel:

Section Being Designed:

Toe of Footing

11/17/20158:41 AM

Page 147: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Designer: LJB Date: 11/17/2015Check: Date:

Shear Analysis:

φVn ≥ Vu (ACI 318 Eq. 11-1)

Vn = Vc + Vs (ACI 318 Eq. 11-2)

Calculate d:

d = 19.688 (Loads placed at 'top' of member.) (ACI 318 11.1.3.1)

dv = 19.688 in

Calc. Vc

For members subject to shear & flexure only.Vc = 2* ƛ * √fc * bv * dv ƛ = 1.0 (ACI 318 Eq. 11-3)

bv = 12 in

Vc = 31.7 kips dv = 19.69 in

For members subject to axial compression.Vc = 2*(1 + Nu/(2000*Ag)) * ƛ * √fc * bv * dv Nu = 0 lb (ACI 318 Eq. 11-4)

Ag = 288 in2

Vc = 760.7 kips

Vc = 31.7 kips

Calc. Vs

Shear Reinforcement perpendicular to axis.

Vs = Av * fyt * d / s Av = 0.00 in2

s = 6 in

Vs = 0.0 kips

Vn = 31.7 kips

θVn = 23.8 kips Vu= 1.514744

θVn = 23.8 kips OK - Section Adequate for Shear Resistance

Section Being Designed:Toe of Footing

Layer of Steel:0

11/17/20158:41 AM

Page 148: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Designer: LJB Date: 11/17/2015Check: Date:

Crack Control: ACI 318

Distribution of flexural reinforcement in beams and one-way slabs.

The spacing of reinforcement closest to the tension face, s, shall not exceed:

cc = 4 in (Clear cover)

fs = Calculated stress from unfactored Moment.

But not greater than Mserv = 20.6 kip-in

n = 7.13

x = 3.458 in NA to extreme comp fiber

y = 16.229 in NA to tension centroid

I = 1329.9 in4

fs = 1795.85 psi

s ≤ 324.1 in

Crack Control: ACI 350

Actual Stress:fs = 1795.85 psi

ACI 350-06 10.6.4.1, for normal environmental exposures as defined in 10.6.4.5

β= 1.27

s= 6 in.db= 0.625 in.

fs,max= 33.37 ksi26781.3 psi > 1795.85 psi

OK - Spacing Adequate

OK - Spacing Adequate

fs,max =320

β*SQRT(s2+4(2+

db∕2)

2)

Section Being Designed:Toe of Footing

Layer of Steel:0

< = 15@s,sss

>�− 2.5 i�

12(40,000

e�)

e� = m ∗E�� � ∗ �

}

� =�� ∗ �� ∗ m + 2 ∗ S ∗ N ∗ m

S−�� ∗ mS

� = N − �

} =13− S ∗ �? + m ∗ �� ∗ �"

11/17/20158:41 AM

Page 149: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Input: Values in red need to be entered by user Designer: LJB Date: 11/17/2015

Check: Date:Project Number:

Bridge No./Structure ID:

Owner:

Mu = 0.841434 k-ft 0.495 k-ft concrete wall thickness (t) 15 in.

Vu = 0.841433 k 0.495 k Clear Cover 3 in.

Nu = 0.0 k 0.0 k d dimension 11.5625 in.

(factored) (service)150 psi

Bar Size = 7

Enter Reinforcement Spacing = 12 in. Reinforcement yield strength = 60 ksiAs = 0.60 in2

Concrete yield strength = 4.5 ksi

Enter design width = 12 in.Concrete yield strength = 4500 psi

Bar Size = None Ec = 4066.84 ksi

Enter Reinforcement Spacing = 6 in. Es = 29000 ksi

Av = 0.00 in2Lightweight Concrete? NO

Mod Factor ƛ = 1Does ACI 350 Apply? NO Flexure φ = 0.9

Shear φ = 0.75

Flexure Analysis: Mr = Phi(Mn)

For members without axial loads.

a = 0.784

Mn = 402.1 k-in

Mn= 33.5 k-ft

φMn = 30.16 k-ft

Minimum Reinforcement: (ACI 318 10.5)

As, min = 0.47 in2 not less than 0.4625 in2

ACI 318 - 10.5 Requirement = Pass If Fail ------------>

1.33*Mu = 1.12 k-ft

As,req = 0.02 in2

As, min = 0.47 in2

OK - Section and reinforcement sufficient for Flexural Resistance

*If the requirements of 10.5 are not satisfied then As needs to be at least 1/3 greater than that required by analysis. (10.5.3)

OK

She

ar

7438-009 Section Being Designed:Key

Fargo Floodwalls Layer of Steel:

City of Fargo

Modulus of elasticity based on concrete unit weight of

Fle

xure

��,8.T =? >��

>�SNS5kmjk�C<<kDlm200S

+

>�

ET = �� ∗ e, ∗ N −l2

l =�� ∗ e,

0.85 ∗ e′� ∗ S

��, �h =.85en�S

e,N − N" −

2E`

.85e′�S

11/17/20158:41 AM

Page 150: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Designer: LJB Date: 11/17/2015

Check: Date:

ACI 318

Governing Specification: USACE

Min. Steel Requirement:

As/Ag (Includes both faces) ≥ 0.0050 EM 1110-2-2104 2.8

Enter Reinforcing configuration of opposite mat:Bar Size = 7

Enter Reinforcement Spacing = 12 in.As = 0.60 in2

Required Min. Steel (both faces): 0.90 sq. in/ft

Max. Spacing Requirements:

s≤5h or s≤18.0 in. 75 in18 in Structural walls/footings:

or for componenets over 36.0 in Spacing can be ≤ 12 in. on faces > 18" thick

s≤12.0 12 in

Maximum Spacing Allowed: 18 in

ACI 350

Min. Steel Requirement: ACI 350-06 7.12.2.1 Minimum reinforcement ratio for Grade 60 steel

Length Between JointsRatio As/Ag 0.0030 ≤ 20 ft

20-30 ft

Steel in opposite face: 0.60 in230-40 ft≥ 40ft

Required Min. Steel: 0.54 sq. in/ft

Max. Spacing Requirements: ACI 350-065 7.12.2.2

Shrinkage and Temperature Reinforcement shall not be spaced greater than12in. apart

Maximum Spacing Allowed: 12 in

0.0040.005

T&S Steel Requirement Met

Max Spacing Requirement Met

Max Spacing Requirement Met

T&S Steel:

Ratio (As/Ag)0.0030.003

0

T&S Steel:

T&S Steel Requirement Met

Layer of Steel:

Section Being Designed:

Key

11/17/20158:41 AM

Page 151: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Designer: LJB Date: 11/17/2015Check: Date:

Shear Analysis:

φVn ≥ Vu (ACI 318 Eq. 11-1)

Vn = Vc + Vs (ACI 318 Eq. 11-2)

Calculate d:

d = 11.563 (Loads placed at 'top' of member.) (ACI 318 11.1.3.1)

dv = 11.563 in

Calc. Vc

For members subject to shear & flexure only.Vc = 2* ƛ * √fc * bv * dv ƛ = 1.0 (ACI 318 Eq. 11-3)

bv = 12 in

Vc = 18.6 kips dv = 11.56 in

For members subject to axial compression.Vc = 2*(1 + Nu/(2000*Ag)) * ƛ * √fc * bv * dv Nu = 0 lb (ACI 318 Eq. 11-4)

Ag = 180 in2

Vc = 279.2 kips

Vc = 18.6 kips

Calc. Vs

Shear Reinforcement perpendicular to axis.

Vs = Av * fyt * d / s Av = 0.00 in2

s = 6 in

Vs = 0.0 kips

Vn = 18.6 kips

θVn = 14.0 kips Vu= 0.841433

θVn = 14.0 kips OK - Section Adequate for Shear Resistance

Section Being Designed:Key

Layer of Steel:0

11/17/20158:41 AM

Page 152: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Designer: LJB Date: 11/17/2015Check: Date:

Crack Control: ACI 318

Distribution of flexural reinforcement in beams and one-way slabs.

The spacing of reinforcement closest to the tension face, s, shall not exceed:

cc = 3 in (Clear cover)

fs = Calculated stress from unfactored Moment.

But not greater than Mserv = 5.9 kip-in

n = 7.13

x = 2.537 in NA to extreme comp fiber

y = 9.026 in NA to tension centroid

I = 413.8 in4

fs = 923.7057 psi

s ≤ 642.1 in

Crack Control: ACI 350

Actual Stress:fs = 923.7057 psi

ACI 350-06 10.6.4.1, for normal environmental exposures as defined in 10.6.4.5

β= 1.38

s= 12 in.db= 0.875 in.

fs,max= 17.89 ksi26781.3 psi > 923.7057 psi

OK - Spacing Adequate

OK - Spacing Adequate

fs,max =320

β*SQRT(s2+4(2+

db∕2)

2)

Section Being Designed:Key

Layer of Steel:0

< = 15@s,sss

>�− 2.5 i�

12(40,000

e�)

e� = m ∗E�� � ∗ �

}

� =�� ∗ �� ∗ m + 2 ∗ S ∗ N ∗ m

S−�� ∗ mS

� = N − �

} =13− S ∗ �? + m ∗ �� ∗ �"

11/17/20158:41 AM

Page 153: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Input: Values in red need to be entered by user Designer: LJB Date: 11/17/2015

Check: Date:Project Number:

Bridge No./Structure ID:

Owner:

Mu = 0 k-ft 0.000 k-ft concrete wall thickness (t) 24 in.

Vu = 0 k 0.000 k Clear Cover 5 in.

Nu = 0.0 k 0.0 k d dimension 18.5 in.

(factored) (service)150 psi

Bar Size = 8

Enter Reinforcement Spacing = 12 in. Reinforcement yield strength = 60 ksiAs = 0.79 in2

Concrete yield strength = 4.5 ksi

Enter design width = 12 in.Concrete yield strength = 4500 psi

Bar Size = None Ec = 4066.84 ksi

Enter Reinforcement Spacing = 6 in. Es = 29000 ksi

Av = 0.00 in2Lightweight Concrete? NO

Mod Factor ƛ = 1Does ACI 350 Apply? NO Flexure φ = 0.9

Shear φ = 0.75

ACI 318

Governing Specification: USACE

Min. Steel Requirement:

As/Ag (Includes both faces) ≥ 0.0050 EM 1110-2-2104 2.8

Enter Reinforcing configuration of opposite mat:Bar Size = 8

Enter Reinforcement Spacing = 12 in.As = 0.79 in2

Required Min. Steel (both faces): 1.44 sq. in/ft

Max. Spacing Requirements:

s≤5h or s≤18.0 in. 120 in18 in Structural walls/footings:

or for componenets over 36.0 in Spacing can be ≤ 12 in. on faces > 18" thick

s≤12.0 12 in

Maximum Spacing Allowed: 18 in

She

ar

7438-009 Section Being Designed:Longitudinal Footing Reinforcement

Fargo Floodwalls Layer of Steel:

City of Fargo

Modulus of elasticity based on concrete unit weight of

Fle

xure

T&S Steel:

T&S Steel Requirement Met

Max Spacing Requirement Met

Page 154: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Input: Values in red need to be entered by user Designer: LJB Date: 11/17/2015

Check: Date:Project Number:

Bridge No./Structure ID:

Owner:

Mu = 0 k-ft 0.000 k-ft concrete wall thickness (t) 15 in.

Vu = 0 k 0.000 k Clear Cover 3 in.

Nu = 0.0 k 0.0 k d dimension 11.6875 in.

(factored) (service)150 psi

Bar Size = 5

Enter Reinforcement Spacing = 12 in. Reinforcement yield strength = 60 ksiAs = 0.31 in2

Concrete yield strength = 4.5 ksi

Enter design width = 12 in.Concrete yield strength = 4500 psi

Bar Size = None Ec = 4066.84 ksi

Enter Reinforcement Spacing = 6 in. Es = 29000 ksi

Av = 0.00 in2Lightweight Concrete? NO

Mod Factor ƛ = 1Does ACI 350 Apply? NO Flexure φ = 0.9

Shear φ = 0.75

ACI 318

Governing Specification: USACE

Min. Steel Requirement:

As/Ag (Includes both faces) ≥ 0.0028 EM 1110-2-2104 2.8

Enter Reinforcing configuration of opposite mat:Bar Size = 5

Enter Reinforcement Spacing = 12 in.As = 0.31 in2

Required Min. Steel (both faces): 0.50 sq. in/ft

Max. Spacing Requirements:

s≤5h or s≤18.0 in. 75 in18 in Structural walls/footings:

or for componenets over 36.0 in Spacing can be ≤ 12 in. on faces > 18" thick

s≤12.0 12 in

Maximum Spacing Allowed: 18 in

She

ar

7438-009 Section Being Designed:Longitudinal Stem Reinforcement

Fargo Floodwalls Layer of Steel:

City of Fargo

Modulus of elasticity based on concrete unit weight of

Fle

xure

T&S Steel:

T&S Steel Requirement Met

Max Spacing Requirement Met

11/17/20158:41 AM

Page 155: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

13'CONS

****************** Echoprint of Input Data ******************

Date: **/11/16 Time: 15.01.08

Company name: HEI Project name: 2nd St. S Project location: Wall location: Computed by: ljb

Structural geometry data: Elevation of top of stem (ELTS) = 13.00 ft Height of stem (HTS) = 13.00 ft Thickness top of stem (TTS) = 1.25 ft Thickness bottom of stem (TBS) = 1.25 ft Dist. of batter at bot. of stem (TBSR)= .00 ft Depth of heel (THEEL) = 4.00 ft Distance of batter for heel (BTRH) = .00 ft Depth of toe (TTOE) = 2.00 ft Width of toe (TWIDTH) = 3.75 ft Distance of batter for toe (BTRT) = .00 ft Width of base (BWIDTH) = 17.00 ft Depth of key (HK) = 2.00 ft Width of bottom of key (TK) = 1.25 ft Dist. of batter at bot. of key (BTRK) = .00 ft

Structure coordinates:

x (ft) y (ft) ================== .00 -4.00 .00 .00 12.00 .00 12.00 13.00 13.25 13.00 13.25 .00 17.00 .00 17.00 -2.00 1.25 -2.00 1.25 -4.00

NOTE: X=0 is located at the left-hand side of the structure. The Y values correspond to the actual elevation used.

Structural property data: Unit weight of concrete = .150 kcf

Driving side soil property data:

Moist Saturated Elev. Phi c Unit wt. unit wt. Delta soil (deg) (ksf) (kcf) (kcf) (deg) (ft) =======================================================

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Page 156: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

13'CONS 27.00 .000 .118 .118 .00 5.00

Driving side soil geometry:

Soil Batter Distance point (in:1ft) (ft) ============================= 1 .00 500.00 2 .00 .00 3 .00 500.00

Driving side soil profile:

Soil x y point (ft) (ft) ============================= 1 -1488.00 5.00 2 12.00 5.00

Resisting side soil property data:

Moist Saturated Elev. Phi c Unit wt. unit wt. soil Batter (deg) (ksf) (kcf) (kcf) (ft) (in:1ft) ======================================================== 27.00 .000 .118 .118 5.00 .00

Resisting side soil profile:

Soil x y point (ft) (ft) ============================= 1 13.25 5.00 2 513.25 5.00

Foundation property data: phi for soil-structure interface = 27.00 (deg) c for soil-structure interface = .000 (ksf) phi for soil-soil interface = 27.00 (deg) c for soil-soil interface = .000 (ksf)

Water data: Driving side elevation = 5.00 ft Resisting side elevation = 5.00 ft Unit weight of water = .0624 kcf Seepage pressures computed are hydrostatic.

Uniform load data: Magnitude of load = .25 k/ft

Horizontal pressure data:

Elevation Pressure (ft) (ksf) ========================== 13.00 .03 5.00 .03

Minimum required factors of safety: Sliding FS = 1.50 Overturning = 75.00% base in compression

Crack options:Page 2

Page 157: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

13'CONS o Crack *is* down to bottom of heel o Computed cracks *will* be filled with water

Strength mobilization factor = .6667

At-rest pressures on the resisting side *are used* in the overturning analysis.

Forces on the resisting side *are used* in the sliding analysis.

*Do* iterate in overturning analysis.

***** Summary of Results *****

Project name: 2nd St. S

*************** *** Satisfied *** * Overturning * Required base in comp. = 75.00 % *************** Actual base in comp. = 100.00 % Overturning ratio = 2.53

Xr (measured from toe) = 8.55 ft Resultant ratio = .5028 Stem ratio = .2206 Base pressure at heel = .7549 ksf Base pressure at toe = .7302 ksf

*********** *** Satisfied *** * Sliding * Min. Required = 1.50 *********** Actual FS = 195.62

********************** Output Results **********************

Date: **/11/16 Time: 15.01.08

Company name: HEI Project name: 2nd St. S Project location: Wall location: Computed by: ljb

*************************** ** Overturning Results ** ***************************

Solution converged in 1 iterations.

SMF used to calculate K's = .6667 Alpha for the SMF = .0000 Calculated earth pressure coefficients: Driving side at rest K = .0000 Driving side at rest Kc = .0000

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Page 158: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

13'CONS Resisting side at rest K = .5460 Resisting side at rest Kc = .7389 At-rest K's for resisting side calculated.

Depth of cracking = 9.00 ft Crack extends to bottom of base of structure.

** Driving side pressures **

Water pressures: Elevation Pressure (ft) (ksf) ====================== 5.00 .0000 -4.00 .5616

Surcharge pressures: Elev. Press. (ft) (ksf) ===================

** Resisting side pressures **

Water pressures: Elevation Pressure (ft) (ksf) ====================== 5.00 .0000 -2.00 .4368 -2.00 .5616 -4.00 .5616

Earth pressures: Elevation Pressure (ft) (ksf) ====================== 5.00 .0000 -2.00 .2125

Balancing earth pressures: Elevation Pressure (ft) (ksf) ====================== -2.00 .3143 -4.00 .3143

** Uplift pressures **

Water pressures: x-coord. Pressure (ft) (ksf) ====================== .00 .5616 1.25 .5616 1.25 .4368 17.00 .4368

** Forces and moments **

======================================================================== Part | Force (kips) | Mom. Arm | Moment | | Vert. | Horiz.| (ft) | (ft-k) | ========================================================================

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Page 159: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

13'CONS Structure: Structure weight........... 7.913 -7.60 -60.15 Structure, driving side: Moist soil................. .000 .00 .00 Saturated soil............. 7.080 -11.00 -77.88 Water above structure...... .000 .00 .00 Water above soil........... .000 .00 .00 External vertical loads.... 3.000 -11.00 -33.00 Ext. horz. pressure loads.. .240 11.00 2.64 Ext. horz. line loads...... .000 .00 .00 Structure, resisting side: Moist soil................. .000 .00 .00 Saturated soil............. 2.213 -1.88 -4.15 Water above structure...... .000 .00 .00 Water above soil........... .000 .00 .00 Driving side: Effective earth loads...... .000 .00 .00 Shear (due to delta)....... .000 .00 .00 Horiz. surcharge effects... .000 .00 .00 Water loads................ 2.527 1.00 2.53 Resisting side: Effective earth loads...... -.744 2.33 -1.74 Balancing earth load....... -.629 -1.00 .63 Water loads................ -2.652 .92 -2.44 Foundation: Vertical force on base..... -12.623 -8.55 107.89 Uplift..................... -7.582 -8.66 65.67 ======================================================================== ** Statics Check ** SUMS = .000 -1.257 .00

Angle of base = 6.71 degrees Normal force on base = 12.463 kips Shear force on base = -2.099 kips Max. available shear force = 7.216 kips

Base pressure at heel = .7549 ksf Base pressure at toe = .7302 ksf

Xr (measured from toe) = 8.55 ft Resultant ratio = .5028 Stem ratio = .2206 Base in compression = 100.00 % Overturning ratio = 2.53

Volume of concrete = 1.95 cubic yds/ft of wall

NOTE: The engineer shall verify that the computed bearing pressures below the wall do not exceed the allowable foundation bearing pressure, or, perform a bearing capacity analysis using the program CBEAR. Also, the engineer shall verify that the base pressures do not result in excessive differential settlement of the wall foundation.

*********************** ** Sliding Results ** ***********************

Factor of safety > 100. Last iteration printed.

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Page 160: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

13'CONS Horizontal Vertical Wedge Loads Loads Number (kips) (kips) ================================== 1 .000 .000 2 2.767 3.000 3 .000 .000

Water pressures on wedges:

Top Bottom Wedge press. press. x-coord. press. number (ksf) (ksf) (ft) (ksf) ================================================ 1 .0000 .0000 2 .0000 .5616 2 17.0000 .4368 3 .0000 .4368

Points of sliding plane: Point 1 (left), x = .00 ft, y = -4.00 ft Point 2 (right), x = 17.00 ft, y = -2.00 ft

Depth of cracking = 9.00 ft Crack extends to bottom of base of structure.

Failure Total Weight Submerged Uplift Wedge angle length of wedge length force number (deg) (ft) (kips) (ft) (kips) ======================================================== 1 .000 .000 .000 .000 .000 2 6.710 17.117 18.916 17.117 8.545 3 45.131 9.877 2.878 9.877 2.157

Wedge Net force number (kips) =================== 1 .000 2 -.153 3 2.898 =================== SUM = 2.745

+-----------------------------+ | Factor of safety = 195.625 | +-----------------------------+

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Page 161: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

Concrete f'c = 4500 psiRebar Fy = 60000 psiUnit Weight = 150 lb/ft³15 in

17 ft

3.75 ft 12 ft

24

in

15

ft 13

ft

15.75 ft15 in

2 ft

40

in

7 ft

8 ft

#5 @ 12 in (S&T)#6 @ 6 in#5 @ 12 in (S&T)#5 @ 12 in#6 @ 6 in (dowels)

Heel Bars: #6 @ 6 inToe Bars: #5 @ 6 inFooting S/T Bars: #8 @ 12 in

40

in

#5 @ 12 in (S&T)#6 @ 6 in#5 @ 12 in (S&T)#5 @ 12 in

Design Detail

Check Summary

Ratio Check Provided Required Combination----- Stability Checks -----

0.091 Overturning 22.07 2.00 1.0D + 1.0F + 0.6H + 0.6W0.445 Sliding 3.37 1.50 1.0D + 1.0F + 0.6H + 0.6W0.203 Bearing Pressure 6000 psf 1221 psf 1.0D + 1.0F + 1.0H + 0.6W0.137 Bearing Eccentricity 7.01 in 51 in 1.0D + 1.0F + 1.0L + 0.6H

----- Toe Checks -----0.046 Shear 23.77 k/ft 1.1 k/ft 1.7D + 1.7F + 1.7L + 1.7H + 1....0.067 Moment 53.8 ft·k/ft 3.59 ft·k/ft 1.7D + 1.7F + 1.7L + 1.7H + 1....0.004 Min Strain 1.0173 0.0040 1.7D + 1.7F + 1.7L + 1.7H + 1....0.000 Min Steel 0.05 in² 0 in² 1.7D + 1.7F + 1.7L + 1.7H + 1....0.077 Development 155 in 12 in 1.7D + 1.7F + 1.7L + 1.7H + 1....0.667 S&T Max Spacing 12 in 18 in 1.7D + 1.7F + 1.7L + 1.7H + 1....0.328 S&T Min Rho 0.0055 0.0018 1.7D + 1.7F + 1.7L + 1.7H + 1....

----- Heel Checks -----0.163 Shear 23.7 k/ft 3.86 k/ft 1.7D + 1.7F + 1.7L + 1.7H + 1....0.108 Moment 75.44 ft·k/ft 8.17 ft·k/ft 1.7D + 1.7F + 1.7L + 1.7H + 1....0.006 Min Strain 0.7135 0.0040 1.7D + 1.7F + 1.7L + 1.7H + 1....0.000 Min Steel 0.07 in² 0 in² 1.7D + 1.7F + 1.7L + 1.7H + 1....0.214 Development 56 in 12 in 1.7D + 1.7F + 1.7L + 1.7H + 1....0.667 S&T Max Spacing 12 in 18 in 1.7D + 1.7F + 1.7L + 1.7H + 1....0.328 S&T Min Rho 0.0055 0.0018 1.7D + 1.7F + 1.7L + 1.7H + 1....

----- Stem Checks -----0.187 Moment 43.76 ft·k/ft 8.17 ft·k/ft 1.7D + 1.7F + 1.7L + 1.7H + 1....0.221 Shear 14.04 k/ft 3.11 k/ft 1.7D + 1.7F + 1.7L + 1.7H + 1....0.009 Max Steel 0.4214 0.0040 1.7D + 1.7F + 1.7L + 1.7H + 1....0.000 Min Steel 0.07 in²/in 0 in²/in 1.7D + 1.7F + 1.7L + 1.7H + 1....0.300 Base Development 20 in 6 in 1.7D + 1.7F + 1.7L + 1.7H + 1....0.523 Lap Splice Length 40 in 20.93 in 1.7D + 1.7F + 1.7L + 1.7H + 1....0.000 Lap Splice Spacing 0 in 4.19 in 1.7D + 1.7F + 1.7L + 1.7H + 1....0.581 Horz Bar Rho 0.0034 0.0020 1.7D + 1.7F + 1.7L + 1.7H + 1....0.667 Horz Bar Spacing 12 in 18 in 1.7D + 1.7F + 1.7L + 1.7H + 1....

Criteria

Building Code IBC 2006Concrete Load Combs USCoE UnusualMasonry Load Combs MSJC 02/05 (ASD)Stability Load Combs ASCE 7-10 (ASD)Restrained Against Sliding NoNeglect Bearing At Heel NoUse Vert. Comp. for OT NoUse Vert. Comp. for Sliding NoUse Vert. Comp. for Bearing YesUse Surcharge for Sliding & OT NoUse Surcharge for Bearing YesNeglect Soil Over Toe NoNeglect Backfill Wt. for Coulomb NoFactor Soil Weight As Dead NoUse Passive Force for OT YesAssume Pressure To Top YesExtend Backfill Pressure To Key Bottom NoUse Toe Passive Pressure for Bearing NoRequired F.S. for OT 2.00Required F.S. for Sliding 1.50Has Different Safety Factors for Seismic NoAllowable Bearing Pressure 6000 psfReq'd Bearing Location Middle halfWall Friction Angle 25°Friction Coefficent 0.35Soil Reaction Modulus 172800 lb/ft³

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Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

Loads5

ft

7 ft

7 ft

8 ft

γ = 118 lb/ft³φ = 27°7

ft

γ = 118 lb/ft³φ = 27°

5 ft

250 psf

-30 psf

Loading Options/AssumptionsPassive pressure neglects top 0 ft of soil.

Load Combinations

USCoE Unusual ...1.7D + 1.7F + 1.7L + 1.7H + 1.7W

...1.7D + 1.7F + 1.7L + 1.7H

Backfill Pressure

5 ft

7 ft

7 ft

8 ft

γ = 118 lb/ft³φ = 27°

26.75 psf

7 ft

7.8 lb/in7 ft

2.3

3 ft

-322.15 psf 67.11 lb/in

1.6

7 ft

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Page 163: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

Backfill Pressure (Water Layer)

γeff γsat γw - 55.5 lb ft³ / 62.5 lb ft³ / - 7 lb ft³ / - = = =

φ 27° φsat =

γ 7 lb ft³ γeff / - =

φ φsat =

γ γeff =

At Rest Earth Pressure Theory -

Ko 1 sin φ - 1 sin 27° - 0.5460 = = =

σh Ko γ H 0.5460 7 lb ft³ / - 7 ft 26.75 psf - = = =

Lateral Earth Pressure (water layer)

σh Ko γ H 0.5460 118 lb ft³ / 5 ft 322.1 psf = = =

Lateral Earth Pressure (water layer, stem only)

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Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

Passive Pressure

7 ft

γ = 118 lb/ft³φ = 27°

579.9 psf

9 ft

217.4 lb/in

1 ft

At Rest Earth Pressure Theory -

Ko 1 sin φ - 1 sin 27° - 0.5460 = = =

σh Ko γ H 0.5460 118 lb ft³ / 9 ft 579.9 psf = = =

Lateral Earth Pressure

Water Pressure

5 ft

-437.5 psf7 ft

127.6 lb/in7 ft

-312.5 psf 65.1 lb/in

σw γw Hw 62.5 lb ft³ / 7 ft 437.5 psf = = =

Lateral Water Pressure

σw γw Hw 62.5 lb ft³ / 5 ft 312.5 psf = = =

Lateral Water Pressure (stem only)

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Page 165: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

Uniform Surcharge Pressure

250 psf 7 ft

0 lb/in7 ft

0 lb/in

At Rest Earth Pressure Theory -

σsur Ko q 0.0 250 psf 0 psf = = =

Lateral Surcharge Pressure

Manually Specified Lateral Stem Pressure

-30 psf240 lb/ft

Wall/Soil Weights

425 lb/in

203.1 lb/in

31.25 lb/in

590 lb/in184.4 lb/in

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Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

Bearing Pressure

1221 psf803.3 psf

1434 lb/in

7.92 ft

e = 7.01 in

414.3 lb/in

F µ R 0.350 1184 lb in / 414.3 lb in / = = =

Friction

Bearing Pressure Calculation

Contributing ForcesVert Force ...offset Horz Force ...offset OT Moment

Backfill Pressure 0 lb/in - 7.8 lb/in 2.33 ft -2621.94 in·lb/ftWater Pressure -0 lb/in - -127.6 lb/in 2.33 ft 42875 in·lb/ftManual Lateral Pressure -0 lb/in - -20 lb/in 11 ft 31680 in·lb/ftFooting Weight -425 lb/in 8.5 ft 0 lb/in - -520200 in·lb/ftStem Weight -203.13 lb/in 4.38 ft 0 lb/in - -127968.75 in·lb/ftKey Weight -31.25 lb/in 16.38 ft 0 lb/in - -73687.5 in·lb/ftBackfill Weight -590 lb/in 11 ft 0 lb/in - -934560 in·lb/ftSoil over toe Weight -184.38 lb/in 1.88 ft 0 lb/in - -49781.25 in·lb/ft

-1433.75 lb/in -1634264.44 in·lb/ft1634264.44 in·lb ft / -

1433.75 lb in / - 7.92 ft =

Note: Bearing resultant used for friction calcs is 1184 lb/in - reduced per user options (for sliding check).

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Page 167: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

Overturning Check

Overturning MomentsForce Distance Moment

Backfill pressure (horz) -4.68 lb/in 2.33 ft -1573.16 in·lb/ftWater pressure 127.6 lb/in 2.33 ft 42875 in·lb/ftManual lateral pressure 0 lb/in 11 ft 0 in·lb/ft

Total: 41302 in·lb/ft

Resisting MomentsForce Distance Moment

Passive pressure @ toe 130.5 lb/in 1 ft 18788 in·lb/ftFooting Weight -425 lb/in 8.5 ft 520200 in·lb/ftStem Weight -203.13 lb/in 4.38 ft 127969 in·lb/ftKey Weight -31.25 lb/in 16.38 ft 73688 in·lb/ftBackfill Weight -354 lb/in 11 ft 560736 in·lb/ftSoil over toe Weight -110.63 lb/in 1.88 ft 29869 in·lb/ft

Total: 1331249 in·lb/ft

F.S. RM

OTM

1331249 in·lb ft / 41302 in·lb ft /

32.232 > 2.00 OK = = =

Sliding Check

Sliding Force(s)Backfill pressure -4.68 lb/inWater pressure 127.6 lb/inManual lateral pressure 0 lb/inTotal: 122.9 lb/in

Resisting Force(s)Passive pressure @ toe 130.5 lb/inFriction 324.8 lb/inTotal: 455.3 lb/in

F.S. RFSF

455.3 lb in / 122.9 lb in /

3.704 > 1.50 OK = = =

Bearing Capacity Check

Bearing pressure < allowable (957 psf < 6000 psf) - OKBearing resultant eccentricity < allowable (7.01 in < 51 in) - OK

Wall Top Displacement

(based on unfactored service loads)

Deflection due to stem flexural displacement 0.019 inDeflection due to rotation from settlement 0.022 inTotal deflection at top of wall (positive towards toe) 0.041 in

Stability Checks [1.0D + 1.0F + 1.0L + 0.6H]

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Page 168: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

Overturning Check

Overturning MomentsForce Distance Moment

Backfill pressure (horz) -7.8 lb/in 2.33 ft -2621.94 in·lb/ftWater pressure 127.6 lb/in 2.33 ft 42875 in·lb/ftManual lateral pressure 12 lb/in 11 ft 19008 in·lb/ft

Total: 59261 in·lb/ft

Resisting MomentsForce Distance Moment

Passive pressure @ toe 217.4 lb/in 1 ft 31313 in·lb/ftFooting Weight -425 lb/in 8.5 ft 520200 in·lb/ftStem Weight -203.13 lb/in 4.38 ft 127969 in·lb/ftKey Weight -31.25 lb/in 16.38 ft 73688 in·lb/ftBackfill Weight -590 lb/in 11 ft 934560 in·lb/ftSoil over toe Weight -184.38 lb/in 1.88 ft 49781 in·lb/ft

Total: 1737510 in·lb/ft

F.S. RM

OTM

1737510 in·lb ft / 59261 in·lb ft /

29.320 > 2.00 OK = = =

Sliding Check

Sliding Force(s)Backfill pressure -7.8 lb/inWater pressure 127.6 lb/inManual lateral pressure 12 lb/inTotal: 131.8 lb/in

Resisting Force(s)Passive pressure @ toe 217.4 lb/inFriction 414.3 lb/inTotal: 631.8 lb/in

F.S. RFSF

631.8 lb in / 131.8 lb in /

4.793 > 1.50 OK = = =

Bearing Capacity Check

Bearing pressure < allowable (1221 psf < 6000 psf) - OKBearing resultant eccentricity < allowable (7.01 in < 51 in) - OK

Wall Top Displacement

(based on unfactored service loads)

Deflection due to stem flexural displacement 0.019 inDeflection due to rotation from settlement 0.022 inTotal deflection at top of wall (positive towards toe) 0.041 in

Stability Checks [1.0D + 1.0F + 1.0H + 0.6W]

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Page 169: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

Overturning Check

Overturning MomentsForce Distance Moment

Backfill pressure (horz) -4.68 lb/in 2.33 ft -1573.16 in·lb/ftWater pressure 127.6 lb/in 2.33 ft 42875 in·lb/ftManual lateral pressure 12 lb/in 11 ft 19008 in·lb/ft

Total: 60310 in·lb/ft

Resisting MomentsForce Distance Moment

Passive pressure @ toe 130.5 lb/in 1 ft 18788 in·lb/ftFooting Weight -425 lb/in 8.5 ft 520200 in·lb/ftStem Weight -203.13 lb/in 4.38 ft 127969 in·lb/ftKey Weight -31.25 lb/in 16.38 ft 73688 in·lb/ftBackfill Weight -354 lb/in 11 ft 560736 in·lb/ftSoil over toe Weight -110.63 lb/in 1.88 ft 29869 in·lb/ft

Total: 1331249 in·lb/ft

F.S. RM

OTM

1331249 in·lb ft / 60310 in·lb ft /

22.073 > 2.00 OK = = =

Sliding Check

Sliding Force(s)Backfill pressure -4.68 lb/inWater pressure 127.6 lb/inManual lateral pressure 12 lb/inTotal: 134.9 lb/in

Resisting Force(s)Passive pressure @ toe 130.5 lb/inFriction 324.8 lb/inTotal: 455.3 lb/in

F.S. RFSF

455.3 lb in / 134.9 lb in /

3.374 > 1.50 OK = = =

Bearing Capacity Check

Bearing pressure < allowable (957 psf < 6000 psf) - OKBearing resultant eccentricity < allowable (7.01 in < 51 in) - OK

Wall Top Displacement

(based on unfactored service loads)

Deflection due to stem flexural displacement 0.019 inDeflection due to rotation from settlement 0.022 inTotal deflection at top of wall (positive towards toe) 0.041 in

Stability Checks [1.0D + 1.0F + 0.6H + 0.6W]

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Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

13

11.7

10.4

9.1

7.8

6.5

5.2

3.9

2.6

1.3

0-50 -38.33 -26.67 -15 -3.33 8.33 20

Moment (ft·k/ft)

Offset (ft)

Moment

aAs fy

0.85 F'c

0.07 in² in / 60000 psi 0.85 4500 psi

1.15 in = = =

φMn φ As fy d a 2 / - 0.90 0.07 in² in / 60000 psi 11.63 in 1.15 in 2 / - 43.76 ft·k ft / = = =

Capacity (ACI 318-05 10.2) @ 0 ft from base [Negative bending]

aAs fy

0.85 F'c

0.03 in² in / 60000 psi 0.85 4500 psi

0.41 in = = =

φMn φ As fy d a 2 / - 0.90 0.03 in² in / 60000 psi 11.69 in 0.41 in 2 / - 16.02 ft·k ft / = = =

Capacity (ACI 318-05 10.2) @ 0 ft from base [Positive bending]

aAs fy

0.85 F'c

0.07 in² in / 60000 psi 0.85 4500 psi

1.15 in = = =

φMn φ As fy d a 2 / - 0.90 0.07 in² in / 60000 psi 11.63 in 1.15 in 2 / - 43.76 ft·k ft / = = =

Capacity (ACI 318-05 10.2) @ 11.66 ft from base [Negative bending]

aAs fy

0.85 F'c

0.03 in² in / 60000 psi 0.85 4500 psi

0.41 in = = =

φMn φ As fy d a 2 / - 0.90 0.03 in² in / 60000 psi 11.69 in 0.41 in 2 / - 16.02 ft·k ft / = = =

Capacity (ACI 318-05 10.2) @ 11.88 ft from base [Positive bending]

aAs fy

0.85 F'c

0 in² in / 60000 psi 0.85 4500 psi

0 in = = =

φMn φ As fy d a 2 / - 0.90 0 in² in / 60000 psi 11.63 in 0 in 2 / - 0 ft·k ft / = = =

Capacity (ACI 318-05 10.2) @ 13 ft from base [Negative bending]

aAs fy

0.85 F'c

0 in² in / 60000 psi 0.85 4500 psi

0 in = = =

φMn φ As fy d a 2 / - 0.90 0 in² in / 60000 psi 11.69 in 0 in 2 / - 0 ft·k ft / = = =

Capacity (ACI 318-05 10.2) @ 13 ft from base [Positive bending]

Stem Flexural Capacity

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Page 171: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

13

11.7

10.4

9.1

7.8

6.5

5.2

3.9

2.6

1.3

0-15 -10 -5 0 5 10 15

Shear (k/ft)

Offset (ft)

Shear

Vc 2 F'c d 2 4500 psi 11.63 in 18.72 k ft / = = =

φVn φ Vc 0.750 18.72 k ft / 14.04 k ft / = = =

Shear Capacity (ACI 318-05 11.1.1, 11.3.1) @ 0 ft from base [Positive shear]

Vc 2 F'c d 2 4500 psi 11.63 in 18.72 k ft / = = =

φVn φ Vc 0.750 18.72 k ft / 14.04 k ft / = = =

Shear Capacity (ACI 318-05 11.1.1, 11.3.1) @ 0 ft from base [Negative shear]

Vc 2 F'c d 2 4500 psi 11.63 in 18.72 k ft / = = =

φVn φ Vc 0.750 18.72 k ft / 14.04 k ft / = = =

Shear Capacity (ACI 318-05 11.1.1, 11.3.1) @ 13 ft from base [Positive shear]

Vc 2 F'c d 2 4500 psi 11.63 in 18.72 k ft / = = =

φVn φ Vc 0.750 18.72 k ft / 14.04 k ft / = = =

Shear Capacity (ACI 318-05 11.1.1, 11.3.1) @ 13 ft from base [Negative shear]

Stem Shear Capacity

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Page 172: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

ψe 1.0 uncoated hooked bars =

λ 1.0 normal weight concrete =

ldh 0.02 ψe λ fy

F'c

db 0.02 1.0 1.060000 psi

4500 psi0.75 in 13.42 in = = =

Factoring ldh by the 0.7 multiplier of 12.5.3 a : ldh 9.39 in =

8 db 8 0.75 in 6.0 minimum limit, does not control = =

Main vertical stem bars (bottom end) - Development Length Calculation (ACI 318-05 12.2.3, 12.5)

ψt 1.0 bars are not horizontal =

ψe 1.0 bar not epoxy coated =

ψs 0.80 bars are #6 or smaller =

λ 1.0 normal weight concrete =

s 2 / 6 in 2 / 3 in = =

cover db 2 / + 3 in 0.75 in 2 / + 3.38 in = =

cb 3 in lesser of half spacing, ctr to surface =

Ktr 0.0 no transverse reinforcement =

cb Ktr + db

3 in 0.0 + 0.75 in

4.0 = =

ld3.40

fy

F'c

ψt ψe ψs λ 2.5

db 3.40

60000 psi

4500 psi

1.0 1.0 0.80 1.0 2.5

0.75 in 16.1 in = = =

Main vertical stem bars (top end) - Development Length Calculation (ACI 318-05 12.2.3, 12.5)

ψt 1.0 bars are not horizontal =

ψe 1.0 bar not epoxy coated =

ψs 0.80 bars are #6 or smaller =

λ 1.0 normal weight concrete =

s 2 / 6 in 2 / 3 in = =

cover db 2 / + 3 in 0.75 in 2 / + 3.38 in = =

cb 3 in lesser of half spacing, ctr to surface =

Ktr 0.0 no transverse reinforcement =

cb Ktr + db

3 in 0.0 + 0.75 in

4.0 = =

ld3.40

fy

F'c

ψt ψe ψs λ 2.5

db 3.40

60000 psi

4500 psi

1.0 1.0 0.80 1.0 2.5

0.75 in 16.1 in = = =

Dowels for vertical stem bars (top end) - Development Length Calculation (ACI 318-05 12.2.3, 12.5)

ψe 1.0 uncoated hooked bars =

λ 1.0 normal weight concrete =

ldh 0.02 ψe λ fy

F'c

db 0.02 1.0 1.060000 psi

4500 psi0.63 in 11.18 in = = =

Factoring ldh by the 0.7 multiplier of 12.5.3 a : ldh 7.83 in =

8 db 8 0.63 in 5.0 minimum limit, does not control = =

2nd curtain vertical bars (bottom end) - Development Length Calculation (ACI 318-05 12.2.3, 12.5)

Stem Development/Lap Length Calculations

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Page 173: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

ψt 1.0 bars are not horizontal =

ψe 1.0 bar not epoxy coated =

ψs 0.80 bars are #6 or smaller =

λ 1.0 normal weight concrete =

s 2 / 12 in 2 / 6 in = =

cover db 2 / + 3 in 0.63 in 2 / + 3.31 in = =

cb 3.31 in lesser of half spacing, ctr to surface =

Ktr 0.0 no transverse reinforcement =

cb Ktr + db

3.31 in 0.0 + 0.63 in

5.30 = =

ld3.40

fy

F'c

ψt ψe ψs λ 2.5

db 3.40

60000 psi

4500 psi

1.0 1.0 0.80 1.0 2.5

0.63 in 13.42 in = = =

2nd curtain vertical bars (top end) - Development Length Calculation (ACI 318-05 12.2.3, 12.5)

Stem Development/Lap Length Calculations (continued)

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Page 174: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

Design moment Mu for toe need not exceed moment at stem base:

Mtoe 3.59 ft·k ft < Mstem / 8.17 ft·k ft / = =

Mu 3.59 ft·k ft stem moment does not control / =

Controlling Moment

aAs fy

0.85 F'c

0.05 in² in / 60000 psi 0.85 4500 psi

0.81 in = = =

φMn φ As fy d a 2 / - 0.90 0.05 in² in / 60000 psi 19.69 in 0.81 in 2 / - 53.8 ft·k ft / = = =

φMn 53.8 ft·k ft ≥ Mu / 3.59 ft·k ft / = = �

Flexure Check (ACI 318-05 10.2)

Vc 2 F'c d 2 4500 psi 19.69 in 31.7 k ft / = = =

φVn φ Vc 0.750 31.7 k ft / 23.77 k ft / = = =

φVn 23.77 k ft ≥ Vu / 1.1 k ft / = = �

Shear Check (ACI 318-05 11.1.1, 11.3.1)

β1 0.85 0.05F'c 4000 -

1000 - 0.85 0.05

4500 psi 4000 - 1000

- 0.8250 = = =

aAs fy

0.85 F'c

0.05 in² in / 60000 psi 0.85 4500 psi

0.81 in = = =

εt 0.003d

a β1 / 1 - 0.003

19.69 in0.81 in 14.0 /

1 - 1.0173 = = =

εt 1.0173 ≥ 0.004 = �

Minimum Strain Check (ACI 318-05 10.3.5)

φMn 53.8 ft·k ft ≥ 4 3 / Mu / 4 3 / 3.59 ft·k ft / 4.78 ft·k ft / = = =

Check is waived per ACI 10.5.3�

Minimum Steel Check (ACI 318-05 10.5.1)

ρST_provAST

t sST

1.58 in² in / 24 in 12 in

0.0055 = = =

ρST_min0.0018 60000

fy

0.0018 60000 60000 psi

0.0018 = = =

ρST_min 0.0018 =

ρST_prov 0.0055 ≥ ρST_min 0.0018 = = �18 inch limit governs

sST_max 18 in =

sST 12 in ≤ sST_max 18 in = = �

Shrinkage Temperature Steel (ACI 318-05 7.12.2)

Mu

φMn

3.59 ft·k ft / 53.8 ft·k ft /

0.0667 ratio to represent excess reinforcement = =

ψt 1.0 12 inches or less cast below 4.00 inches - =

ψe 1.0 bar not epoxy coated =

ψs 0.80 bars are #6 or smaller =

λ 1.0 normal weight concrete =

s 2 / 6 in 2 / 3 in = =

cover db 2 / + 4 in 0.63 in 2 / + 4.31 in = =

cb 3 in lesser of half spacing, ctr to surface =

Ktr 0.0 no transverse reinforcement =

cb Ktr + db

3 in 0.0 + 0.63 in

4.80 = =

ld3.40

fy

F'c

ψt ψe ψs λ 2.5

db 3.40

60000 psi

4500 psi

1.0 1.0 0.80 1.0 2.5

0.63 in 13.42 in = = =

Factoring ld by the excess reinforcement ratio 0.0667 per 12.2.5: ld 0.89 in =

12 inch minimum controls

ld_prov 155 in ≥ ld 12 in = = �

Development Check (ACI 318-05 12.12, 12.2.3)

Toe Unfactored Loads

24

in

#5 @ 6 in

Unfactored Loads

300 psf (Self-wt)

590 psf (Soil)

1221 psf 1129 psf

Toe Factored Loads

24

in

#5 @ 6 in

1.7D + 1.7F + 1.7L + 1.7H + 1.7W

510 psf (Self-wt)

1003 psf (Soil)

2075 psf 1919 psf

1003 psf

1.82 k/ft

Toe Checks [1.7D + 1.7F + 1.7L + 1.7H + 1.7W]

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Page 175: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

Design moment Mu for heel need not exceed moment at stem base:

Mheel 29.19 ft·k ft ≥ Mstem / 8.17 ft·k ft / = =

Mu 8.17 ft·k ft stem base moment controls / =

Controlling Moment

aAs fy

0.85 F'c

0.07 in² in / 60000 psi 0.85 4500 psi

1.15 in = = =

φMn φ As fy d a 2 / - 0.90 0.07 in² in / 60000 psi 19.63 in 1.15 in 2 / - 75.44 ft·k ft / = = =

φMn 75.44 ft·k ft ≥ Mu / 8.17 ft·k ft / = = �

Flexure Check (ACI 318-05 10.2)

Vc 2 F'c d 2 4500 psi 19.63 in 31.6 k ft / = = =

φVn φ Vc 0.750 31.6 k ft / 23.7 k ft / = = =

φVn 23.7 k ft ≥ Vu / 3.86 k ft / = = �

Shear Check (ACI 318-05 11.1.1, 11.3.1)

β1 0.85 0.05F'c 4000 -

1000 - 0.85 0.05

4500 psi 4000 - 1000

- 0.8250 = = =

aAs fy

0.85 F'c

0.07 in² in / 60000 psi 0.85 4500 psi

1.15 in = = =

εt 0.003d

a β1 / 1 - 0.003

19.63 in1.15 in 14.0 /

1 - 0.7135 = = =

εt 0.7135 ≥ 0.004 = �

Minimum Strain Check (ACI 318-05 10.3.5)

φMn 75.44 ft·k ft ≥ 4 3 / Mu / 4 3 / 8.17 ft·k ft / 10.89 ft·k ft / = = =

Check is waived per ACI 10.5.3�

Minimum Steel Check (ACI 318-05 10.5.1)

ρST_provAST

t sST

1.58 in² in / 24 in 12 in

0.0055 = = =

ρST_min0.0018 60000

fy

0.0018 60000 60000 psi

0.0018 = = =

ρST_min 0.0018 =

ρST_prov 0.0055 ≥ ρST_min 0.0018 = = �18 inch limit governs

sST_max 18 in =

sST 12 in ≤ sST_max 18 in = = �

Shrinkage Temperature Steel (ACI 318-05 7.12.2)

Mu

φMn

8.17 ft·k ft / 75.44 ft·k ft /

0.1083 ratio to represent excess reinforcement = =

ψt 1.30 more than 12 inches cast below 19.25 inches - =

ψe 1.0 bar not epoxy coated =

ψs 0.80 bars are #6 or smaller =

λ 1.0 normal weight concrete =

s 2 / 6 in 2 / 3 in = =

cover db 2 / + 4 in 0.75 in 2 / + 4.38 in = =

cb 3 in lesser of half spacing, ctr to surface =

Ktr 0.0 no transverse reinforcement =

cb Ktr + db

3 in 0.0 + 0.75 in

4.0 = =

ld3.40

fy

F'c

ψt ψe ψs λ 2.5

db 3.40

60000 psi

4500 psi

1.30 1.0 0.80 1.0 2.5

0.75 in 20.93 in = = =

Factoring ld by the excess reinforcement ratio 0.1083 per 12.2.5: ld 2.27 in =

12 inch minimum controls

ld_prov 56 in ≥ ld 12 in = = �

Development Check (ACI 318-05 12.12, 12.2.3)

Heel Unfactored Loads

24

in#6 @ 6 in

Unfactored Loads

300 psf (Concrete self-wt)

590 psf (Soil weight)

250 psf (Uniform Surcharge)

1098 psf 803.3 psf

Heel Factored Loads

24

in#6 @ 6 in

1.7D + 1.7F + 1.7L + 1.7H + 1.7W

510 psf (Concrete self-wt)

1003 psf (Soil weight)

425 psf (Uniform Surcharge)

1366-1867 psf (Bearing pressure)

3.86 k/ft

Heel Checks [1.7D + 1.7F + 1.7L + 1.7H + 1.7W]

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Page 176: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

Stem Internal Forces

-547.65 psf-531.25 psf

-51 psf

3.11 k/ft

-8.17 ft·k/ft

Stem Internal Forces

13

11.38

9.75

8.13

6.5

4.88

3.25

1.63

0-9 -6.75 -4.5 -2.25 0

Moment (ft·k/ft)

Moment

Stem Internal Forces

13

11.38

9.75

8.13

6.5

4.88

3.25

1.63

00 1 2 3 4

Shear (k/ft)

Shear

Stem Joint Force Transfer

Location Force@ stem base 3.11 k/ft

Stem Internal Forces

-547.65 psf -531.25 psf

-51 psf

Stem Forces [1.7D + 1.7F + 1.7L + 1.7H + 1.7W]

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Page 177: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

13

11.7

10.4

9.1

7.8

6.5

5.2

3.9

2.6

1.3

0-50 -38.33 -26.67 -15 -3.33 8.33 20

Moment (ft·k/ft)

Offset (ft)

Moment

φMn 43.76 ft·k ft ≥ Mu / 8.17 ft·k ft / = = �

Check (ACI 318-05 Ch 10) @ 0 ft from base

φMn 43.76 ft·k ft ≥ Mu / 0.04 ft·k ft / = = �

Check (ACI 318-05 Ch 10) @ 11.66 ft from base

φMn 42.83 ft·k ft ≥ Mu / 0.04 ft·k ft / = = �

Check (ACI 318-05 Ch 10) @ 11.69 ft from base

Stem Moment Checks [1.7D + 1.7F + 1.7L + 1.7H + 1.7W]

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Page 178: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

13

11.7

10.4

9.1

7.8

6.5

5.2

3.9

2.6

1.3

0-15 -10 -5 0 5 10 15

Shear (k/ft)

Offset (ft)

Shear

φVn 14.04 k ft ≥ Vu / 3.11 k ft / = = �

Shear Check (ACI 318-05 Ch 11.1.1) @ 0 ft from base

Stem Shear Checks [1.7D + 1.7F + 1.7L + 1.7H + 1.7W]

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Page 179: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

φMn 43.76 ft·k ft ≥ 4 3 / Mu / 4 3 / 8.17 ft·k ft / 10.89 ft·k ft / = = =

Check is waived per ACI 10.5.3�

Minimum Steel Check (ACI 318-05 10.5.1) @ 0 ft from base [Stem in negative flexure]

φMn 43.76 ft·k ft ≥ 4 3 / Mu / 4 3 / 2.51 ft·k ft / 3.35 ft·k ft / = = =

Check is waived per ACI 10.5.3�

Minimum Steel Check (ACI 318-05 10.5.1) @ 3.33 ft from base [Stem in negative flexure]

φMn 0 ft·k ft ≥ 4 3 / Mu / 4 3 / 0 ft·k ft / 0 ft·k ft / = = =

Check is waived per ACI 10.5.3�

Minimum Steel Check (ACI 318-05 10.5.1) @ 13 ft from base [Stem in negative flexure]

β1 0.85 0.05F'c 4000 -

1000 - 0.85 0.05

4500 psi 4000 - 1000

- 0.8250 = = =

aAs fy

0.85 F'c

0.07 in² in / 60000 psi 0.85 4500 psi

1.15 in = = =

εt 0.003d

a β1 / 1 - 0.003

11.63 in1.15 in 14.0 /

1 - 0.4214 = = =

εt 0.4214 ≥ 0.004 = �

Maximum Steel Check (ACI 318-05 10.3.5) @ 0 ft from base [Stem in negative flexure]

β1 0.85 0.05F'c 4000 -

1000 - 0.85 0.05

4500 psi 4000 - 1000

- 0.8250 = = =

aAs fy

0.85 F'c

0.07 in² in / 60000 psi 0.85 4500 psi

1.15 in = = =

εt 0.003d

a β1 / 1 - 0.003

11.63 in1.15 in 14.0 /

1 - 0.4214 = = =

εt 0.4214 ≥ 0.004 = �

Maximum Steel Check (ACI 318-05 10.3.5) @ 3.33 ft from base [Stem in negative flexure]

β1 0.85 0.05F'c 4000 -

1000 - 0.85 0.05

4500 psi 4000 - 1000

- 0.8250 = = =

aAs fy

0.85 F'c

0.07 in² in / 60000 psi 0.85 4500 psi

1.15 in = = =

εt 0.003d

a β1 / 1 - 0.003

11.63 in1.15 in 14.0 /

1 - 0.4214 = = =

εt 0.4214 ≥ 0.004 = �

Maximum Steel Check (ACI 318-05 10.3.5) @ 13 ft from base [Stem in negative flexure]

ρhAs_horz shorz /

t

0.62 in² 12 in / 15 in

0.0034 = = =

ρh_min 0.0020 bars No. 5 or less, not less than 60 ksi =

ρh 0.0034 ≥ ρh_min 0.0020 = = �3 twall 3 15 in 45 in = =

18 inch limit governs

smax 18 in =

shorz 12 in ≤ shorz_max 18 in = = �

Wall Horizontal Steel (ACI 318-05 14.3.3, 14.3.5)

Stem Miscellaneous Checks [1.7D + 1.7F + 1.7L + 1.7H + 1.7W]

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Page 180: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

Mu

φMn

8.17 ft·k ft / 43.76 ft·k ft /

0.1867 ratio to represent excess reinforcement = =

ψe 1.0 uncoated hooked bars =

λ 1.0 normal weight concrete =

ldh 0.02 ψe λ fy

F'c

db 0.02 1.0 1.060000 psi

4500 psi0.75 in 13.42 in = = =

Factoring ldh by the 0.7 multiplier of 12.5.3 a : ldh 9.39 in =

Factoring ldh by the excess reinforcement ratio 0.1867 per 12.5.3 d : ldh 1.75 in =

8 db 8 0.75 in 6.0 minimum limit, does not control = =

6 inch minimum controls

ldh_prov 20 in ≥ ldh 6 in = = �

Development Check (ACI 318-05 12.12, 12.2.3)

ψt 1.0 bars are not horizontal =

ψe 1.0 bar not epoxy coated =

ψs 0.80 bars are #6 or smaller =

λ 1.0 normal weight concrete =

s 2 / 6 in 2 / 3 in = =

cover db 2 / + 3 in 0.75 in 2 / + 3.38 in = =

cb 3 in lesser of half spacing, ctr to surface =

Ktr 0.0 no transverse reinforcement =

cb Ktr + db

3 in 0.0 + 0.75 in

4.0 = =

ld3.40

fy

F'c

ψt ψe ψs λ 2.5

db 3.40

60000 psi

4500 psi

1.0 1.0 0.80 1.0 2.5

0.75 in 16.1 in = = =

llap 1.3 ld 1.3 16.1 in 20.93 in = = =

llap_prov 40 in ≥ llap 20.93 in = = �1 5 / llap 1 5 / 20.93 in 4.1859 ≤ 6.0 = =

strans 0 in ≤ 1 5 / llap 1 5 / 20.93 in 4.1859 = = = �

Lap Splice Checks (ACI 318-05 12.14.2.3, 12.15.1, 12.15.2) - #6 lap with #6, from 0 ft to 3.33 ft (from stem base)

Stem Miscellaneous Checks [1.7D + 1.7F + 1.7L + 1.7H + 1.7W] (continued)

QuickRWall 4.0 (iesweb.com) H:\Fargo\JBN\7400\7438\12_...\13' Unusual construction.rwd Page 20 of 20 Tuesday 11/17/15 9:13 AM

Page 181: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

12 Ft. Floodwall

(Case 2)

Page 182: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Site: 7438-009 Owner:

Case:

User Data:A 15 Inches

B 17 Feet

C 12 FeetD 24 InchesD1 24 Inches

Hwall 13 Feet

Hwater 5 Feet

H2 5 FeetH3 5 Feet

Unit Weight of Soil (ϒsat):

Below Footing 0.118 k/cu. ft.Wet Side 0.118 k/cu. ft.Dry Side 0.118 k/cu. ft.

Ø= Below Footing 27 Degrees0.471 radians

Wet Side 27 Degrees0.471 radians

Dry Side 27 Degrees0.471 radians

SMF= 0.66667Unit Weight of Conc. (ρc) 0.15 k/cu. ft.

Unit Weight of Water (ρw) 0.0625 k/cu. ft.

Soil Capacity 6.2 k/sq. ft.2 Feet Ko Below Footing 0.55

15 Inches Wet Side 0.55Dry Side 0.55

Kp for sliding = 2.663fysteel 60 ksi Cohesion Factor (CF) 1 ksf

f'c(concrete) 4.5 ksi Frost Depth (Fr): 6 ft

Vertical Surcharge (Dry) (P3) 0 psf

Vertical Surcharge (Wet) (P4) 0 psf

Ice/Debris Load NoType of Loading Usual

Toe Vertical Fill Width (F)= 3.75 ft

*All Dimensions in Calculations are converted to feet

City of Fargo

13' Floodwall Usual

Keyway Depth (G)Keyway Width (L)

Project Number:Fargo Floodwalls

1 − sin∅

1 − sin∅1 − sin∅

Page 183: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Force CalculationsFactor Unit Equation Description

EHW= 0.563 kips/ft Driving Wet Side Water Pressure

EHS= 0.500 kips/ft Driving Wet Side Soil Pressure

EV2= 7.080 kips Wet Side Vertical Component Weights

EV2d 11.000 ft from toe Moment arm for EV2

EV1= 2.213 kips Dry Side Vertical Component Weights

EV1d 1.875 ft from toe Moment arm for EV1

PehW= 2.531 kips Driving Wet Side Water Force

Pehwd 1.000 ft. above footing Moment arm for Pehw

PehS= 2.248 kips Driving Wet Side Soil Force

Pehsd 1.000 ft. above footing Moment arm for Pehs

PS1= 2.813 kips Vertical Weight of Stem and Key

PS1d 5.975 ft from toe Moment arm for PS1

PS2= 5.100 kips Vertical Weight of Footing

PS2d 8.500 ft from toe Moment arm for PS2

EH2= 1.062 kips Resisting Dry Side Soil Pressure

Peh2= 4.779 kips Resisting Dry Side Forces

Peh2d 1.000 ft from toe Moment arm for Peh2

Peh2s 2.248 kips Resisting dry side soil only

Peh2sd 1.000 ft from toe Moment arm for Peh2s

Pice/debris 0.000 kips .5 k/ft at Q100 Driving ice/debris Force

Pice/debrisd 7.000 ft. above footing Moment arm for Pice/debris

Pwind 0.24 kips Driving wind Force

Pwindd 11.000 ft. above footing Moment arm for Pwind

Applied Forces

(��� +�� + �)(ρ)

(�3 + �� + �)(ɤ�� − ρ)

��� � ρ +H3(C)(ɤ�� − ρ)

�2(�)(ɤ��)

��

2(��� + �� + �)

���

2(�3 + �� + �)

��� + �� + �

3− �

�3 + �� + �

3− �

��� � ρ� + �( )(ρ�)

! − � �� ρ� + �(�)(ρ�)

(�2 + �� + �)(ɤ��)

��"

2(�2 + �� + �)

! −�

2

2

��� � ρ� ! − � −�

2+ � ρ� ! −

2/$%�

! − � �� ρ�! − �

2+ � � ρ� ! −

2/$%"

�2 + �� + �

3− �

��� + ��

��� − ���

2+ ��� + ��

ρ� − ρ ∗ �2 + �� + � +�2 + �� + �

2

�2 + �� + �

3− �

.03(��� −��� )

Page 184: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Water Uplift

Force Calculation Factor Unit Equation DescriptionLZ-Z1= 17.117 ft Length of Seepage path from heel to toe

Δh= 0.000 ft Head differential between wet and dry side

Ldry side 7.000 ft Seepage path on Dry Side

LS 24.117 ft Total Seepage Path

uZ= 0.563 ksf Water Pressure at bottom of key (Wet side)

uZ1= 0.438 ksf Water pressure at bottom of footing (Dry Side)

HLZ-Z1= 0.000 ft Head Loss along Z-Z1

LSc 19.000 ft Length of concrete surface in sliding surface

HLLK = 0.000 ft Head Loss along key

ubottom key = 0.563 ksf Water pressure at bottom of key (Dry side)

HLUK = 0.000 ft Head Loss up key

utop key = 0.438 ksf Water pressure at top of key (Dry side)

P5 = 7.594 kips Water Uplift for Overturning

P5d 8.662 ft

P5sa 8.500 kips Water uplift for sliding along angle

P5sad -8.854 ft Moment Arm for P5sa

Lsss = 26.000 ft

Uf = 0.563 ksf

P5ss = 9.563 Water uplift of sliding along bottom of key

P5ssd -8.500 Moment Arm for P5ss

(!"+ �")

��� − �2

�2 + ��

)*)� + + ,�.+�

(��� +�� + �)(ρ)

��� + �� − ∆0 )*)� 1

(ρ)

∆0 )*)� �

! + �

%�(� )*)�)

(��� + �� + � − � 23)ρ

+ �

%�(� )*)�)

(��� +�� −� 43)ρ

5) + 567��789�,

2 +

5�7:9�, + 5)�2

(! − )

5) + 5)�2

(!)

5)�! −

2! − +

5�7:9�, − 5)�2

! − 2

3! − + 567��789�, ! −

2+5) − 567��789�,

2 ! −

3/$5

5)� !!

2+

5) − 5)�2

!2!

3/$5�

! + � + �� +�2

��� +�� + � − ��� −�2!

<��ρ

5= + 5>2

(!)

5> !!

2+5) − 5>

2!

2!

3/$5��

Moment Arm

for P5

Water pressure at key depth below

footing on dry side of footing

Page 185: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Distance Moment

to Toe(ft) About Toe (k-ft)

PS1 2.81 5.98 16.8 MPS1

PS2 5.10 8.50 43.4 MPS2

P3 0.00 1.88 0.0 MP3

P4 0.00 11.00 0.0 MP4

P5 -7.59 8.66 -65.8 MP5

0.32 -17.6 -5.6

EV2 - Backfill on Heel EV2 7.08 11.00 77.9 MEV2

EV1 - Fill on Toe EV1 2.21 1.88 4.1 MEV1

9.29 8.8 82.0

LL Live Load 0.0 0.00 0.0 MLL

0.0 0.0

EH Horiz. Earth Load PehS 2.25 -1.00 -1.2 MEH

WS Hydrostatic Pressure PehW 2.53 -1.00 -2.5 MPHW

ID Ice/Debris Force Pice/debris 0.00 -7.00 0.0 MID

W Wind Load Pwind 0.24 -11.00 -2.6 MW

5.0 -6.4

Fargo Floodwalls

Ve

rtica

l Lo

ad

s

Applied Force Calculations

Total

P (kips)DescriptorLoad

Vertical Loads

DC

Horizontal Loads

EV

Total

Total

Total

Page 186: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Factor Equation Description

EM 1110-2-2502

Reference

PehW2 = 2.531 kips Water on Dry side for Overturning Calcs Figure 4-5

Pehw2d 1.000 ft Moment Arm for Pehw2 N/A

Peh2sb 1.360 ft Dry Side Soil to bottom of Footing Para. 4-8

Peh2sbd 2.333 kips Moment Arm for Peh2sb N/A

RS = -0.495 kips Resisting soil on Dry side to create equilibrium Para. 4-8

RSd -1 ft Moment Arm for RS N/A

Including Uplift

ΣVo 9.611 kips Total Vertical Load Figure 4-5

ΣMo 75.415 kip-ft 4-1

XR 7.847 ft Resultant Location 4-1

bl 23.539693 ft Length of Base in Compression 4-2

b% 138.46878 Percent of Base in Compression 4-2

Required b% 100 Required Base in Compression Appendix F

OK

Neglecting Uplift

ΣVon 17.205 kips Total Vertical Load Figure 4-5

ΣMon 141.193 kip-ft 4-1

XRn 8.206 ft Resultant Location 4-1

bln 24.619466 ft Length of Base in Compression 4-2

b%n 144.82039 Percent of Base in Compression 4-2

Required b%n 100 Required Base in Compression Appendix 4

OK

Global Stability

Overturning:

Fargo Floodwalls

$%� + $%" + $? + $@ + $A + �B1 + �B2

5)�2

�2 + �� +5�7:9�, + 567��789�,

2(�)

�2 + ��3

5)�2

�2 + ��3

�2 + �� + 5�7:9�,−�2

� +567��789�, − 5�7:9�,

2�

−2�3

/$CD"

−�2

EF1� +EF1" +EF? +EF@ +EFA +EGH� +EGH" +EFIJ +EKL +EJ + M% M%N + $CD"�6 $CD"�6N + $CD"($CD"N)(O7)(%E�)

ΣE7

ΣB7

3QR

S�!(100)

$%� + $%" + $? + $@ + �B1 + �B2

EF1� +EF1" +EF? +EF@ +EGH� +EGH" +EFIJ +EKL +EJ + M% M%N + $CD"�6 $CD"�6N + $CD"($CD"N)(O7)(%E�)

ΣE7T

ΣB7T

3QRT

S�T!

(100)

Total Overturning

Moment about

point o

(�2 + ��)(ɤ�� − ρ)(�2 + ��)/2

$CD − $CD" − $CD"�6(O7)(%E�)

Total Overturning

Moment about

point o

11/17/2015

Page 187: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Factor Equation Description

EM 1110-2-2502

Reference

Sliding Along Angle Between Bottom of Key and Bottom of Footing on Dry Side Figure 4-11

SBFa 1.722 kips Soil Below Footing on Sliding Surface Figure 4-11

Pwda 2.531 kips Water on Dry Side Figure 4-11

ΣVa 10.427 kips Sum of Vertical Forces on Sliding Surface Figure 4-11

αa 0.117 radians Angle of slip plane to horizontal plane Figure 4-11

6.710 degrees

Drained Condition

RFswad 5.524 kips Resisting Force from passive soil and water on dry side Figure 4-11

ΣHad -2.75 kips Sum of Horizontal Forces Figure 4-11

N'ad 10.034 kips Normal Force to Sliding Surface Figure 4-11

Tad -3.952 kips Tangential Force to Sliding Surface Figure 4-11

SSad 5.112 kips Drained Shear Strength 4-12

FSad -1.294 Factor of Safety 4-12

OK

Undrained Condition

RFswau 3.655 kips Resisting Force from passive soil and water on dry side Figure 4-11

ΣHau -0.884 kips Sum of Horizontal Forces Figure 4-11

N'au 10.252 kips Normal Force to Sliding Surface Figure 4-11

Tau -2.096 kips Tangential Force to Sliding Surface Figure 4-11

SSau 11.412 kips Undrained Shear Strength 4-12

FSau -5.444 Factor of Safety 4-12

OK

Sliding:

Fargo Floodwalls

Global Stability

� −�!

2(! − )(ɤ��)

5)2

�2 + �� +5) + 5)�

2(�)

%UV + �B1 + �B2 + $%� + $%" + $? + $@ − $A�

atan�!

$+ + $CD" − $+ Y: (.5)

$CD + $.��/+�6 .� + $.T+ − M��+

ΣB cos \ + Σ�+sin(\)

Σ�+ cos \ − ΣBsin(\)

]′+tan(∅)

%%+_+

$+ + $CD" − $+ (.5)

$CD + $.��/+�6 .� + $.T+ − M��`

ΣB cos \ + Σ�`sin(\)

Σ�` cos \ − ΣBsin(\)

��(%E�)√(!" + �")

%%`_`

11/17/2015

Page 188: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Factor Equation Description

EM 1110-2-2502

Reference

Sliding Along Horizontal Surface along Bottom of Key Figure 4-8

SBFs 3.717 kips Soil Below Footing on Sliding Surface

Pwds 2.531 kips Water on Dry Side Figure 4-11

ΣVs 11.360 kips Sum of Vertical Forces on Sliding Surface Figure 4-11

αs = 0 radians Angle of slip plane to horizontal plane Figure 4-11

0 degrees

Drained Condition

RFswsd 5.524 kips Resisting Force from passive soil and water on dry side Figure 4-11

ΣHsd -2.75 kips Sum of Horizontal Forces Figure 4-11

N'sd 11.360 kips Normal Force to Sliding Surface Figure 4-11

Tsd -2.753 kips Tangential Force to Sliding Surface Figure 4-11

SSsd 5.788 kips Drained Shear Strength 4-12

FSsd -2.103 Factor of Safety 4-12

OK

Undrained Condition

RFswsu 3.655 kips Resisting Force from passive soil and water on dry side Figure 4-11

ΣHsu -0.884 kips Sum of Horizontal Forces Figure 4-11

N'su 11.360 kips Normal Force to Sliding Surface Figure 4-11

Tsu -0.884 kips Tangential Force to Sliding Surface Figure 4-11

SSsu 11.333 kips Undrained Shear Strength 4-12

FSsu -12.822 Factor of Safety 4-12

OK

Global Stabilty

Fargo Floodwalls

�(! − )(ɤ��)

5>2(�2 + �� + �)

%UV� + �B1 + �B2 + $%� + $%" + $? + $@ − $A��

$+� + $CD" − $+� Y: (.5)

$CD + $.��/+�6 .� + $.T+ − M���+

ΣB� cos \� + Σ��+sin(\�)

Σ��+ cos \� − ΣB�sin(\�)

]′�+tan(∅)

%%�+_�+

.

$+� + $CD" − $+� (.5)

$CD + $.��/+�6 .� + $.T+ − M���`

ΣB� cos \� + Σ��`sin(\�)

Σ��` cos \� − ΣB�sin(\�)

�� %E� !

%%�`_�`

11/17/2015

Page 189: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Factor Equation Description

EM 1110-2-2502

Reference

Bearing Along Angle Between Bottom of Key and Bottom of Footing on Dry Side Eqns. From 1110-2-2502

ΣHadb 0.24 kips Sum of Horizontal Forces Figure 4-11

N'adb 10.383 kips Normal Force to Bearing Surface Figure 4-11

Tadb -0.980 kips Tangential Force to Bearing Surface Figure 4-11

SBFad 10.500005 ft Moment Arm for SBFa Figure 4-11

ΣMoaB 82.361737 kip-ft

XraB 7.899 ft Resultant Location Figure 4-11

eaB 0.660 ft Eccentricity of resultant Figure 4-11

B'aB 15.798016 ft Effective Base for Bearing Sec 5-2

Drained Condition

q0 = 0.333 k/ft2 Effective Overburden Pressure

y = 1.6007185

Nq = 13.175528 Bearing Capacity Factor 5-3a

Nc = 23.895844 Bearing Capacity Factor 5-3b

Nϒ = 9.4442971 Bearing Capacity Factor 5-3d

εcd = 1.12 Embedment Factor 5-4a

εqd = 1.06 Embedment Factor 5-4c

δd = -0.09 5-5

-5.391268

εqi = 1.12 Inclination Factor 5-5a

εϒi = 1.44 Inclination Factor 5-5b

Q1 = 0.00 5-2

Q2 = 5.2340383 5-2

Q3 = 6.3278467 5-2

Qd = 182.65485 5-2

FS = 5-1

17.590926 Eq. 5-1 Bearing Criteria Satisfied

Global Stabilty

Fargo Floodwalls

Bearing:

ΣE7U

ΣB )*)�2

− QbU

)*)� − 2(CU)

(C,) tan 45 +∅2

de∅ > 0, (]h−1)ijklm(∅)

]h − 1 tan(1.4∅)

1 + 0.2� !n tan 45 +

∅2

de∅ > 0, 1 + 0.1� !n tan 45 +

∅2

atanΣ�ΣB

1 −o+90

"

1 −o+∅

"

q�+(qh.)(rs)(]h)

qh+(qh.)(rs)(]h)

qh+ qh. ɤ�� − ɤ�� ]t /2

!′(u1 + u2 + u3)

(ɤ�� − ɤ�� )(� )

v(tan ∅)

Normal Component to the base of the structure of the

ultimate bearing capacity

EF1� +EF1" +EF? +EF@ + $A� $A�N + EGH� +EGH" +EGI +EFIJ +EKL +EJ + $CD"�($CD"�N)(O7)(%E�)+$+ $CD"N + %UV %UVN

$CD + $.��/+�6 .� + $.T+ + $CD�(O7)(%E�) − $+ − $CD"�(O7)(%E�)

ΣB cos \ + Σ�+6sin(\)

Σ�+6 cos \ − ΣBsin(\)

2(! − )3

11/17/2015

Page 190: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Factor Equation Description

EM 1110-2-2502

Reference

q0 = 0.333 k/ft2 Effective Overburden Pressure

y = 0

Nq = 1 Bearing Capacity Factor 5-3a

Nc = 5.14 Bearing Capacity Factor 5-3c

Nϒ = 0 Bearing Capacity Factor 5-3d

εcd = 1.08 Embedment Factor 5-4a

εqd = 1.00 Embedment Factor 5-4b

δd = -0.09 5-5

-5.391268

εqi = 1.12 Inclination Factor 5-5a

εϒi = 1.44 Inclination Factor 5-5b

Q1 = 6.21 5-2

Q2 = 0.3740903 5-2

Q3 = 0 5-2

Qd = 104.05511 5-2

FS = 5-1

10.021227

Drained q'max 0.75 Undrained q'max 0.75

Bearing Criteria Satisfied

Undrained Condition

Global Stabilty

Fargo Floodwalls

(ɤ�� − ɤ�� )(� )

v(tan ∅)

de∅ = 0, 5.14

de∅ = 0, 1

(C,)tan(45 +∅2)

]h − 1 tan(1.4∅)

1 + 0.2� !n tan 45 +

∅2

atanΣ�ΣB

1 −o+90

"

1 −o+∅

"

q�+(qh.)(rs)(]h)

qh+(qh.)(rs)(]h)

qh+ qh. ɤ�� − ɤ�� ]t /2

!′(u1 + u2 + u3)Normal Component to the base of the structure of the

ultimate bearing capacity

11/17/2015

Page 191: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Factor Equation Description

Load Factor 2.21

b 17 Effective base in compression

qmin 0.512644

Uws 0.4375 Uplift Pressure Wet side of Stem

Utws 5.40625 Total water uplift on wet side

Utwsd 6.155347 Moment arm for U tws

VLw 5.64875 Vertical load on wet side

Bws 0.682298 Bearing at wet side of stem

Btws 7.169657 Total Bearing uplift on wet side of footing

Btwsd 5.716046 Moment arm for B tws

Total Shear on Heel -3.36121

Total Moment on Heel -17.5241

Total Shear on Stem 0.53 Total Shear on Stem

Total Moment on Stem 4.773571

Uds 0.4375 Uplift Pressure Dry side of Stem

Utds 1.640625 Total water uplift on dry side

Utdsd 1.875 Moment arm for U tds

VLd 1.696875 Vertical load on dry side

Bds 0.699971 Bearing at dry side of stem

Btds 2.724297 Total Bearing uplift on dry side of footing

Btdsd 1.897806 Moment arm for B tds

Total Shear on Toe -2.2706 Total Shear on Toe

Total Moment on Toe -4.39469 Total Moment on Toe

Utkw 0.4375 Water Pressure at top of key (wet side)

Total Shear on Key 1.09 Total Shear on Key

Total Moment on Key 1.093864 Total Moment on Key

Forces for Reinforcement/Shear

Heel Forces

Stem Forces

Toe Forces

Key Forces

$KL +$J + �

2��� + �3 ɤ�� − ρ

�32

Y7 − ρ�22

�2 − �2 ɤ�� − ρ�22

(Y7) jlN�likjb

$KL ��� + $J $.T+N −�12

+ �

2���

���

3+ �3 ɤ�� − ρ

�32

�33

(Y7) − ρ�22

�2�23

−�2 ɤ�� − ρ�22

�23

(Y7) ( jlN�likjb)

x�7:9�, −x�7:9�, −x)�

! − � −

x�7:9�, −x�

2C − L +

x) +x67��789�,

2( )

x�� − 2

� − +x�7:9�, − x�

22 � −

3� − + x67��789�, � −

2

+x) −x67��789�,

2 � −

3

/x��

�B2+ � � ρ� +� ρ� +$@ − x��

r8z − r8z − r8.T

S(� + �)

(B − !��)( jlN�likjb)

�B2+ � � ρ� +� ρ� +$@�2

− x�� x��N − !�� !��N ( jlN�likjb)

x�7:9�, −x�7:9�, −x)�

! − � + � −

x+� +x)�

2(�)

[x)� ��2

+x+� −x)�

2(�3)(�)]/x�+�

�B1 + $? +� �� ρ� −x�+�

r8z − r8z − r8.T

S(�)

r8z + !+�2

(�)

!+� ��2

+r8z −!+�

22�3

� /!�+�

(B + −!�+�)( jlN�likjb)

�B1+ $? +� �� ρ��2

−x�+� x�+�N − !�+� !�+�N ( jlN�likjb)

}eS% ≥ 100,!, S�

}eS% ≥ 100,r8z

!1−

6C!

, 0

}er8.T = 0,!�

2S − � − � , (

!�+ r8.T

2)(�)

}er8.T = 0,S − � − �

3, r8.T �

�2

+!�− r8.T

2�

�3

/!��

x�9 +x=

2� −

x67��789�, +x�7:9�,

2� − M% jlN�likjb

x�9 ��2

+x= − x�9

2�

2�3

−x�7:9�, ��2

−x67��789�, −x�7:9�,

2�

2�3

− M%�2

( jlN�likjb)

(��� +�)(�)

Total

Moment

on Stem

Page 192: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Input: Values in red need to be entered by user Designer: LJB Date: 11/17/2015

Check: Date:Project Number:

Bridge No./Structure ID:

Owner:

Mu = 17.52405 k-ft 7.929 k-ft concrete wall thickness (t) 24 in.

Vu = 3.361205 k 1.521 k Clear Cover 4 in.

Nu = 0.0 k 0.0 k d dimension 19.625 in.

(factored) (service)150 psi

Bar Size = 6

Enter Reinforcement Spacing = 6 in. Reinforcement yield strength = 60 ksiAs = 0.88 in2

Concrete yield strength = 4.5 ksi

Enter design width = 12 in.Concrete yield strength = 4500 psi

Bar Size = None Ec = 4066.84 ksi

Enter Reinforcement Spacing = 6 in. Es = 29000 ksi

Av = 0.00 in2Lightweight Concrete? NO

Mod Factor ƛ = 1Does ACI 350 Apply? NO Flexure φ = 0.9

Shear φ = 0.75

Flexure Analysis: Mr = Phi(Mn)

For members without axial loads.

a = 1.150

Mn = 1,005.8 k-in

Mn= 83.8 k-ft

φMn = 75.44 k-ft

Minimum Reinforcement: (ACI 318 10.5)

As, min = 0.79 in2 not less than 0.785 in2

ACI 318 - 10.5 Requirement = Pass If Fail ------------>

1.33*Mu = 23.31 k-ft

As,req = 0.24 in2

As, min = 0.79 in2

OK - Section and reinforcement sufficient for Flexural Resistance

*If the requirements of 10.5 are not satisfied then As needs to be at least 1/3 greater than that required by analysis. (10.5.3)

OK

She

ar

7438-009 Section Being Designed:Heel of Footing

Fargo Floodwalls Layer of Steel:

City of Fargo

Modulus of elasticity based on concrete unit weight of

Fle

xure

��,8.T =? >��

>�SNS5kmjk�C<<kDlm200S

+

>�

ET = �� ∗ e, ∗ N −l2

l =�� ∗ e,

0.85 ∗ e′� ∗ S

��, �h =.85en�S

e,N − N" −

2E`

.85e′�S

Page 193: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Designer: LJB Date: 11/17/2015

Check: Date:

ACI 318

Governing Specification: USACE

Min. Steel Requirement:

As/Ag (Includes both faces) ≥ 0.0050 EM 1110-2-2104 2.8

Enter Reinforcing configuration of opposite mat:Bar Size = 5

Enter Reinforcement Spacing = 6 in.As = 0.62 in2

Required Min. Steel (both faces): 1.44 sq. in/ft

Max. Spacing Requirements:

s≤5h or s≤18.0 in. 120 in18 in Structural walls/footings:

or for componenets over 36.0 in Spacing can be ≤ 12 in. on faces > 18" thick

s≤12.0 12 in

Maximum Spacing Allowed: 18 in

ACI 350

Min. Steel Requirement: ACI 350-06 7.12.2.1 Minimum reinforcement ratio for Grade 60 steel

Length Between JointsRatio As/Ag 0.0030 ≤ 20 ft

20-30 ft

Steel in opposite face: 0.62 in230-40 ft≥ 40ft

Required Min. Steel: 0.86 sq. in/ft

Max. Spacing Requirements: ACI 350-065 7.12.2.2

Shrinkage and Temperature Reinforcement shall not be spaced greater than12in. apart

Maximum Spacing Allowed: 12 in

0.0040.005

T&S Steel Requirement Met

Max Spacing Requirement Met

Max Spacing Requirement Met

T&S Steel:

Ratio (As/Ag)0.0030.003

0

T&S Steel:

T&S Steel Requirement Met

Layer of Steel:

Section Being Designed:

Heel of Footing

Page 194: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Designer: LJB Date: 11/17/2015Check: Date:

Shear Analysis:

φVn ≥ Vu (ACI 318 Eq. 11-1)

Vn = Vc + Vs (ACI 318 Eq. 11-2)

Calculate d:

d = 19.625 (Loads placed at 'top' of member.) (ACI 318 11.1.3.1)

dv = 19.625 in

Calc. Vc

For members subject to shear & flexure only.Vc = 2* ƛ * √fc * bv * dv ƛ = 1.0 (ACI 318 Eq. 11-3)

bv = 12 in

Vc = 31.6 kips dv = 19.63 in

For members subject to axial compression.Vc = 2*(1 + Nu/(2000*Ag)) * ƛ * √fc * bv * dv Nu = 0 lb (ACI 318 Eq. 11-4)

Ag = 288 in2

Vc = 758.3 kips

Vc = 31.6 kips

Calc. Vs

Shear Reinforcement perpendicular to axis.

Vs = Av * fyt * d / s Av = 0.00 in2

s = 6 in

Vs = 0.0 kips

Vn = 31.6 kips

θVn = 23.7 kips Vu= 3.361205

θVn = 23.7 kips OK - Section Adequate for Shear Resistance

Section Being Designed:Heel of Footing

Layer of Steel:0

Page 195: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Designer: LJB Date: 11/17/2015Check: Date:

Crack Control: ACI 318

Distribution of flexural reinforcement in beams and one-way slabs.

The spacing of reinforcement closest to the tension face, s, shall not exceed:

cc = 4 in (Clear cover)

fs = Calculated stress from unfactored Moment.

But not greater than Mserv = 95.2 kip-in

n = 7.13

x = 4.038 in NA to extreme comp fiber

y = 15.587 in NA to tension centroid

I = 1787.9 in4

fs = 5915.415 psi

s ≤ 91.4 in

Crack Control: ACI 350

Actual Stress:fs = 5915.415 psi

ACI 350-06 10.6.4.1, for normal environmental exposures as defined in 10.6.4.5

β= 1.28

s= 6 in.db= 0.75 in.

fs,max= 32.65 ksi26781.3 psi > 5915.415 psi

OK - Spacing Adequate

OK - Spacing Adequate

fs,max =320

β*SQRT(s2+4(2+

db∕2)

2)

Section Being Designed:Heel of Footing

Layer of Steel:0

< = 15@s,sss

>�− 2.5 i�

12(40,000

e�)

e� = m ∗E�� � ∗ �

}

� =�� ∗ �� ∗ m + 2 ∗ S ∗ N ∗ m

S−�� ∗ mS

� = N − �

} =13− S ∗ �? + m ∗ �� ∗ �"

Page 196: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Input: Values in red need to be entered by user Designer: LJB Date: 11/17/2015

Check: Date:Project Number:

Bridge No./Structure ID:

Owner:

Mu = 4.773571 k-ft 2.160 k-ft concrete wall thickness (t) 15 in.

Vu = 0.5304 k 0.240 k Clear Cover 3 in.

Nu = 0.0 k 0.0 k d dimension 11.625 in.

(factored) (service)150 psi

Bar Size = 6

Enter Reinforcement Spacing = 6 in. Reinforcement yield strength = 60 ksiAs = 0.88 in2

Concrete yield strength = 4.5 ksi

Enter design width = 12 in.Concrete yield strength = 4500 psi

Bar Size = None Ec = 4066.84 ksi

Enter Reinforcement Spacing = 6 in. Es = 29000 ksi

Av = 0.00 in2Lightweight Concrete? NO

Mod Factor ƛ = 1Does ACI 350 Apply? NO Flexure φ = 0.9

Shear φ = 0.75

Flexure Analysis: Mr = Phi(Mn)

For members without axial loads.

a = 1.150

Mn = 583.4 k-in

Mn= 48.6 k-ft

φMn = 43.76 k-ft

Minimum Reinforcement: (ACI 318 10.5)

As, min = 0.47 in2 not less than 0.465 in2

ACI 318 - 10.5 Requirement = Pass If Fail ------------>

1.33*Mu = 6.35 k-ft

As,req = 0.11 in2

As, min = 0.47 in2

OK - Section and reinforcement sufficient for Flexural Resistance

*If the requirements of 10.5 are not satisfied then As needs to be at least 1/3 greater than that required by analysis. (10.5.3)

OK

She

ar

7438-009 Section Being Designed:Stem

Fargo Floodwalls Layer of Steel:

City of Fargo

Modulus of elasticity based on concrete unit weight of

Fle

xure

��,8.T =? >��

>�SNS5kmjk�C<<kDlm200S

+

>�

ET = �� ∗ e, ∗ N −l2

l =�� ∗ e,

0.85 ∗ e′� ∗ S

��, �h =.85en�S

e,N − N" −

2E`

.85e′�S

11/17/20158:42 AM

Page 197: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Designer: LJB Date: 11/17/2015

Check: Date:

ACI 318

Governing Specification: USACE

Min. Steel Requirement:

As/Ag (Includes both faces) ≥ 0.0028 EM 1110-2-2104 2.8

Enter Reinforcing configuration of opposite mat:Bar Size = 5

Enter Reinforcement Spacing = 12 in.As = 0.31 in2

Required Min. Steel (both faces): 0.50 sq. in/ft

Max. Spacing Requirements:

s≤5h or s≤18.0 in. 75 in18 in Structural walls/footings:

or for componenets over 36.0 in Spacing can be ≤ 12 in. on faces > 18" thick

s≤12.0 12 in

Maximum Spacing Allowed: 18 in

ACI 350

Min. Steel Requirement: ACI 350-06 7.12.2.1 Minimum reinforcement ratio for Grade 60 steel

Length Between JointsRatio As/Ag 0.0030 ≤ 20 ft

20-30 ft

Steel in opposite face: 0.31 in230-40 ft≥ 40ft

Required Min. Steel: 0.54 sq. in/ft

Max. Spacing Requirements: ACI 350-065 7.12.2.2

Shrinkage and Temperature Reinforcement shall not be spaced greater than12in. apart

Maximum Spacing Allowed: 12 in

0.0040.005

T&S Steel Requirement Met

Max Spacing Requirement Met

Max Spacing Requirement Met

T&S Steel:

Ratio (As/Ag)0.0030.003

0

T&S Steel:

T&S Steel Requirement Met

Layer of Steel:

Section Being Designed:

Stem

11/17/20158:42 AM

Page 198: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Designer: LJB Date: 11/17/2015Check: Date:

Shear Analysis:

φVn ≥ Vu (ACI 318 Eq. 11-1)

Vn = Vc + Vs (ACI 318 Eq. 11-2)

Calculate d:

d = 11.625 (Loads placed at 'top' of member.) (ACI 318 11.1.3.1)

dv = 11.625 in

Calc. Vc

For members subject to shear & flexure only.Vc = 2* ƛ * √fc * bv * dv ƛ = 1.0 (ACI 318 Eq. 11-3)

bv = 12 in

Vc = 18.7 kips dv = 11.63 in

For members subject to axial compression.Vc = 2*(1 + Nu/(2000*Ag)) * ƛ * √fc * bv * dv Nu = 0 lb (ACI 318 Eq. 11-4)

Ag = 180 in2

Vc = 280.7 kips

Vc = 18.7 kips

Calc. Vs

Shear Reinforcement perpendicular to axis.

Vs = Av * fyt * d / s Av = 0.00 in2

s = 6 in

Vs = 0.0 kips

Vn = 18.7 kips

θVn = 14.0 kips Vu= 0.5304

θVn = 14.0 kips OK - Section Adequate for Shear Resistance

Section Being Designed:Stem

Layer of Steel:0

11/17/20158:42 AM

Page 199: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Designer: LJB Date: 11/17/2015Check: Date:

Crack Control: ACI 318

Distribution of flexural reinforcement in beams and one-way slabs.

The spacing of reinforcement closest to the tension face, s, shall not exceed:

cc = 3 in (Clear cover)

fs = Calculated stress from unfactored Moment.

But not greater than Mserv = 25.9 kip-in

n = 7.13

x = 3.003 in NA to extreme comp fiber

y = 8.622 in NA to tension centroid

I = 574.8 in4

fs = 2772.429 psi

s ≤ 208.9 in

Crack Control: ACI 350

Actual Stress:fs = 2772.429 psi

ACI 350-06 10.6.4.1, for normal environmental exposures as defined in 10.6.4.5

β= 1.39

s= 6 in.db= 0.75 in.

fs,max= 30.05 ksi26781.3 psi > 2772.429 psi

OK - Spacing Adequate

OK - Spacing Adequate

fs,max =320

β*SQRT(s2+4(2+

db∕2)

2)

Section Being Designed:Stem

Layer of Steel:0

< = 15@s,sss

>�− 2.5 i�

12(40,000

e�)

e� = m ∗E�� � ∗ �

}

� =�� ∗ �� ∗ m + 2 ∗ S ∗ N ∗ m

S−�� ∗ mS

� = N − �

} =13− S ∗ �? + m ∗ �� ∗ �"

11/17/20158:42 AM

Page 200: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Input: Values in red need to be entered by user Designer: LJB Date: 11/17/2015

Check: Date:Project Number:

Bridge No./Structure ID:

Owner:

Mu = 4.394686 k-ft 1.989 k-ft concrete wall thickness (t) 24 in.

Vu = 2.270603 k 1.027 k Clear Cover 4 in.

Nu = 0.0 k 0.0 k d dimension 19.6875 in.

(factored) (service)150 psi

Bar Size = 5

Enter Reinforcement Spacing = 6 in. Reinforcement yield strength = 60 ksiAs = 0.62 in2

Concrete yield strength = 4.5 ksi

Enter design width = 12 in.Concrete yield strength = 4500 psi

Bar Size = None Ec = 4066.84 ksi

Enter Reinforcement Spacing = 6 in. Es = 29000 ksi

Av = 0.00 in2Lightweight Concrete? NO

Mod Factor ƛ = 1Does ACI 350 Apply? NO Flexure φ = 0.9

Shear φ = 0.75

Flexure Analysis: Mr = Phi(Mn)

For members without axial loads.

a = 0.810

Mn = 717.3 k-in

Mn= 59.8 k-ft

φMn = 53.80 k-ft

Minimum Reinforcement: (ACI 318 10.5)

As, min = 0.79 in2 not less than 0.7875 in2

ACI 318 - 10.5 Requirement = Fail If Fail ------------>

1.33*Mu = 5.84 k-ft

As,req = 0.06 in2

As, min = 0.06 in2

OK - Section and reinforcement sufficient for Flexural Resistance

*If the requirements of 10.5 are not satisfied then As needs to be at least 1/3 greater than that required by analysis. (10.5.3)

OK

She

ar

7438-009 Section Being Designed:Toe of Footing

Fargo Floodwalls Layer of Steel:

City of Fargo

Modulus of elasticity based on concrete unit weight of

Fle

xure

��,8.T =? >��

>�SNS5kmjk�C<<kDlm200S

+

>�

ET = �� ∗ e, ∗ N −l2

l =�� ∗ e,

0.85 ∗ e′� ∗ S

��, �h =.85en�S

e,N − N" −

2E`

.85e′�S

11/17/20158:42 AM

Page 201: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Designer: LJB Date: 11/17/2015

Check: Date:

ACI 318

Governing Specification: USACE

Min. Steel Requirement:

As/Ag (Includes both faces) ≥ 0.0050 EM 1110-2-2104 2.8

Enter Reinforcing configuration of opposite mat:Bar Size = 6

Enter Reinforcement Spacing = 6 in.As = 0.88 in2

Required Min. Steel (both faces): 1.44 sq. in/ft

Max. Spacing Requirements:

s≤5h or s≤18.0 in. 120 in18 in Structural walls/footings:

or for componenets over 36.0 in Spacing can be ≤ 12 in. on faces > 18" thick

s≤12.0 12 in

Maximum Spacing Allowed: 18 in

ACI 350

Min. Steel Requirement: ACI 350-06 7.12.2.1 Minimum reinforcement ratio for Grade 60 steel

Length Between JointsRatio As/Ag 0.0030 ≤ 20 ft

20-30 ft

Steel in opposite face: 0.88 in230-40 ft≥ 40ft

Required Min. Steel: 0.86 sq. in/ft

Max. Spacing Requirements: ACI 350-065 7.12.2.2

Shrinkage and Temperature Reinforcement shall not be spaced greater than12in. apart

Maximum Spacing Allowed: 12 in

0.0040.005

T&S Steel Requirement Met

Max Spacing Requirement Met

Max Spacing Requirement Met

T&S Steel:

Ratio (As/Ag)0.0030.003

0

T&S Steel:

T&S Steel Requirement Met

Layer of Steel:

Section Being Designed:

Toe of Footing

11/17/20158:42 AM

Page 202: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Designer: LJB Date: 11/17/2015Check: Date:

Shear Analysis:

φVn ≥ Vu (ACI 318 Eq. 11-1)

Vn = Vc + Vs (ACI 318 Eq. 11-2)

Calculate d:

d = 19.688 (Loads placed at 'top' of member.) (ACI 318 11.1.3.1)

dv = 19.688 in

Calc. Vc

For members subject to shear & flexure only.Vc = 2* ƛ * √fc * bv * dv ƛ = 1.0 (ACI 318 Eq. 11-3)

bv = 12 in

Vc = 31.7 kips dv = 19.69 in

For members subject to axial compression.Vc = 2*(1 + Nu/(2000*Ag)) * ƛ * √fc * bv * dv Nu = 0 lb (ACI 318 Eq. 11-4)

Ag = 288 in2

Vc = 760.7 kips

Vc = 31.7 kips

Calc. Vs

Shear Reinforcement perpendicular to axis.

Vs = Av * fyt * d / s Av = 0.00 in2

s = 6 in

Vs = 0.0 kips

Vn = 31.7 kips

θVn = 23.8 kips Vu= 2.270603

θVn = 23.8 kips OK - Section Adequate for Shear Resistance

Section Being Designed:Toe of Footing

Layer of Steel:0

11/17/20158:42 AM

Page 203: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Designer: LJB Date: 11/17/2015Check: Date:

Crack Control: ACI 318

Distribution of flexural reinforcement in beams and one-way slabs.

The spacing of reinforcement closest to the tension face, s, shall not exceed:

cc = 4 in (Clear cover)

fs = Calculated stress from unfactored Moment.

But not greater than Mserv = 23.9 kip-in

n = 7.13

x = 3.458 in NA to extreme comp fiber

y = 16.229 in NA to tension centroid

I = 1329.9 in4

fs = 2076.526 psi

s ≤ 278.9 in

Crack Control: ACI 350

Actual Stress:fs = 2076.526 psi

ACI 350-06 10.6.4.1, for normal environmental exposures as defined in 10.6.4.5

β= 1.27

s= 6 in.db= 0.625 in.

fs,max= 33.37 ksi26781.3 psi > 2076.526 psi

OK - Spacing Adequate

OK - Spacing Adequate

fs,max =320

β*SQRT(s2+4(2+

db∕2)

2)

Section Being Designed:Toe of Footing

Layer of Steel:0

< = 15@s,sss

>�− 2.5 i�

12(40,000

e�)

e� = m ∗E�� � ∗ �

}

� =�� ∗ �� ∗ m + 2 ∗ S ∗ N ∗ m

S−�� ∗ mS

� = N − �

} =13− S ∗ �? + m ∗ �� ∗ �"

11/17/20158:42 AM

Page 204: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Input: Values in red need to be entered by user Designer: LJB Date: 11/17/2015

Check: Date:Project Number:

Bridge No./Structure ID:

Owner:

Mu = 1.093864 k-ft 0.495 k-ft concrete wall thickness (t) 15 in.

Vu = 1.093863 k 0.495 k Clear Cover 3 in.

Nu = 0.0 k 0.0 k d dimension 11.5625 in.

(factored) (service)150 psi

Bar Size = 7

Enter Reinforcement Spacing = 12 in. Reinforcement yield strength = 60 ksiAs = 0.60 in2

Concrete yield strength = 4.5 ksi

Enter design width = 12 in.Concrete yield strength = 4500 psi

Bar Size = None Ec = 4066.84 ksi

Enter Reinforcement Spacing = 6 in. Es = 29000 ksi

Av = 0.00 in2Lightweight Concrete? NO

Mod Factor ƛ = 1Does ACI 350 Apply? NO Flexure φ = 0.9

Shear φ = 0.75

Flexure Analysis: Mr = Phi(Mn)

For members without axial loads.

a = 0.784

Mn = 402.1 k-in

Mn= 33.5 k-ft

φMn = 30.16 k-ft

Minimum Reinforcement: (ACI 318 10.5)

As, min = 0.47 in2 not less than 0.4625 in2

ACI 318 - 10.5 Requirement = Pass If Fail ------------>

1.33*Mu = 1.45 k-ft

As,req = 0.03 in2

As, min = 0.47 in2

OK - Section and reinforcement sufficient for Flexural Resistance

*If the requirements of 10.5 are not satisfied then As needs to be at least 1/3 greater than that required by analysis. (10.5.3)

OK

She

ar

7438-009 Section Being Designed:Key

Fargo Floodwalls Layer of Steel:

City of Fargo

Modulus of elasticity based on concrete unit weight of

Fle

xure

��,8.T =? >��

>�SNS5kmjk�C<<kDlm200S

+

>�

ET = �� ∗ e, ∗ N −l2

l =�� ∗ e,

0.85 ∗ e′� ∗ S

��, �h =.85en�S

e,N − N" −

2E`

.85e′�S

11/17/20158:42 AM

Page 205: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Designer: LJB Date: 11/17/2015

Check: Date:

ACI 318

Governing Specification: USACE

Min. Steel Requirement:

As/Ag (Includes both faces) ≥ 0.0050 EM 1110-2-2104 2.8

Enter Reinforcing configuration of opposite mat:Bar Size = 7

Enter Reinforcement Spacing = 12 in.As = 0.60 in2

Required Min. Steel (both faces): 0.90 sq. in/ft

Max. Spacing Requirements:

s≤5h or s≤18.0 in. 75 in18 in Structural walls/footings:

or for componenets over 36.0 in Spacing can be ≤ 12 in. on faces > 18" thick

s≤12.0 12 in

Maximum Spacing Allowed: 18 in

ACI 350

Min. Steel Requirement: ACI 350-06 7.12.2.1 Minimum reinforcement ratio for Grade 60 steel

Length Between JointsRatio As/Ag 0.0030 ≤ 20 ft

20-30 ft

Steel in opposite face: 0.60 in230-40 ft≥ 40ft

Required Min. Steel: 0.54 sq. in/ft

Max. Spacing Requirements: ACI 350-065 7.12.2.2

Shrinkage and Temperature Reinforcement shall not be spaced greater than12in. apart

Maximum Spacing Allowed: 12 in

0.0040.005

T&S Steel Requirement Met

Max Spacing Requirement Met

Max Spacing Requirement Met

T&S Steel:

Ratio (As/Ag)0.0030.003

0

T&S Steel:

T&S Steel Requirement Met

Layer of Steel:

Section Being Designed:

Key

11/17/20158:42 AM

Page 206: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Designer: LJB Date: 11/17/2015Check: Date:

Shear Analysis:

φVn ≥ Vu (ACI 318 Eq. 11-1)

Vn = Vc + Vs (ACI 318 Eq. 11-2)

Calculate d:

d = 11.563 (Loads placed at 'top' of member.) (ACI 318 11.1.3.1)

dv = 11.563 in

Calc. Vc

For members subject to shear & flexure only.Vc = 2* ƛ * √fc * bv * dv ƛ = 1.0 (ACI 318 Eq. 11-3)

bv = 12 in

Vc = 18.6 kips dv = 11.56 in

For members subject to axial compression.Vc = 2*(1 + Nu/(2000*Ag)) * ƛ * √fc * bv * dv Nu = 0 lb (ACI 318 Eq. 11-4)

Ag = 180 in2

Vc = 279.2 kips

Vc = 18.6 kips

Calc. Vs

Shear Reinforcement perpendicular to axis.

Vs = Av * fyt * d / s Av = 0.00 in2

s = 6 in

Vs = 0.0 kips

Vn = 18.6 kips

θVn = 14.0 kips Vu= 1.093863

θVn = 14.0 kips OK - Section Adequate for Shear Resistance

Section Being Designed:Key

Layer of Steel:0

11/17/20158:42 AM

Page 207: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Designer: LJB Date: 11/17/2015Check: Date:

Crack Control: ACI 318

Distribution of flexural reinforcement in beams and one-way slabs.

The spacing of reinforcement closest to the tension face, s, shall not exceed:

cc = 3 in (Clear cover)

fs = Calculated stress from unfactored Moment.

But not greater than Mserv = 5.9 kip-in

n = 7.13

x = 2.537 in NA to extreme comp fiber

y = 9.026 in NA to tension centroid

I = 413.8 in4

fs = 923.7057 psi

s ≤ 642.1 in

Crack Control: ACI 350

Actual Stress:fs = 923.7057 psi

ACI 350-06 10.6.4.1, for normal environmental exposures as defined in 10.6.4.5

β= 1.38

s= 12 in.db= 0.875 in.

fs,max= 17.89 ksi26781.3 psi > 923.7057 psi

OK - Spacing Adequate

OK - Spacing Adequate

fs,max =320

β*SQRT(s2+4(2+

db∕2)

2)

Section Being Designed:Key

Layer of Steel:0

< = 15@s,sss

>�− 2.5 i�

12(40,000

e�)

e� = m ∗E�� � ∗ �

}

� =�� ∗ �� ∗ m + 2 ∗ S ∗ N ∗ m

S−�� ∗ mS

� = N − �

} =13− S ∗ �? + m ∗ �� ∗ �"

11/17/20158:42 AM

Page 208: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Input: Values in red need to be entered by user Designer: LJB Date: 11/17/2015

Check: Date:Project Number:

Bridge No./Structure ID:

Owner:

Mu = 0 k-ft 0.000 k-ft concrete wall thickness (t) 24 in.

Vu = 0 k 0.000 k Clear Cover 5 in.

Nu = 0.0 k 0.0 k d dimension 18.5 in.

(factored) (service)150 psi

Bar Size = 8

Enter Reinforcement Spacing = 12 in. Reinforcement yield strength = 60 ksiAs = 0.79 in2

Concrete yield strength = 4.5 ksi

Enter design width = 12 in.Concrete yield strength = 4500 psi

Bar Size = None Ec = 4066.84 ksi

Enter Reinforcement Spacing = 6 in. Es = 29000 ksi

Av = 0.00 in2Lightweight Concrete? NO

Mod Factor ƛ = 1Does ACI 350 Apply? NO Flexure φ = 0.9

Shear φ = 0.75

ACI 318

Governing Specification: USACE

Min. Steel Requirement:

As/Ag (Includes both faces) ≥ 0.0050 EM 1110-2-2104 2.8

Enter Reinforcing configuration of opposite mat:Bar Size = 8

Enter Reinforcement Spacing = 12 in.As = 0.79 in2

Required Min. Steel (both faces): 1.44 sq. in/ft

Max. Spacing Requirements:

s≤5h or s≤18.0 in. 120 in18 in Structural walls/footings:

or for componenets over 36.0 in Spacing can be ≤ 12 in. on faces > 18" thick

s≤12.0 12 in

Maximum Spacing Allowed: 18 in

She

ar

7438-009 Section Being Designed:Longitudinal Footing Reinforcement

Fargo Floodwalls Layer of Steel:

City of Fargo

Modulus of elasticity based on concrete unit weight of

Fle

xure

T&S Steel:

T&S Steel Requirement Met

Max Spacing Requirement Met

Page 209: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Input: Values in red need to be entered by user Designer: LJB Date: 11/17/2015

Check: Date:Project Number:

Bridge No./Structure ID:

Owner:

Mu = 0 k-ft 0.000 k-ft concrete wall thickness (t) 15 in.

Vu = 0 k 0.000 k Clear Cover 3 in.

Nu = 0.0 k 0.0 k d dimension 11.6875 in.

(factored) (service)150 psi

Bar Size = 5

Enter Reinforcement Spacing = 12 in. Reinforcement yield strength = 60 ksiAs = 0.31 in2

Concrete yield strength = 4.5 ksi

Enter design width = 12 in.Concrete yield strength = 4500 psi

Bar Size = None Ec = 4066.84 ksi

Enter Reinforcement Spacing = 6 in. Es = 29000 ksi

Av = 0.00 in2Lightweight Concrete? NO

Mod Factor ƛ = 1Does ACI 350 Apply? NO Flexure φ = 0.9

Shear φ = 0.75

ACI 318

Governing Specification: USACE

Min. Steel Requirement:

As/Ag (Includes both faces) ≥ 0.0028 EM 1110-2-2104 2.8

Enter Reinforcing configuration of opposite mat:Bar Size = 5

Enter Reinforcement Spacing = 12 in.As = 0.31 in2

Required Min. Steel (both faces): 0.50 sq. in/ft

Max. Spacing Requirements:

s≤5h or s≤18.0 in. 75 in18 in Structural walls/footings:

or for componenets over 36.0 in Spacing can be ≤ 12 in. on faces > 18" thick

s≤12.0 12 in

Maximum Spacing Allowed: 18 in

She

ar

7438-009 Section Being Designed:Longitudinal Stem Reinforcement

Fargo Floodwalls Layer of Steel:

City of Fargo

Modulus of elasticity based on concrete unit weight of

Fle

xure

T&S Steel:

T&S Steel Requirement Met

Max Spacing Requirement Met

11/17/20158:42 AM

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13'USUAL

****************** Echoprint of Input Data ******************

Date: **/11/16 Time: 15.28.00

Company name: HEI Project name: 2nd St. S Project location: Wall location: Computed by: ljb

Structural geometry data: Elevation of top of stem (ELTS) = 13.00 ft Height of stem (HTS) = 13.00 ft Thickness top of stem (TTS) = 1.25 ft Thickness bottom of stem (TBS) = 1.25 ft Dist. of batter at bot. of stem (TBSR)= .00 ft Depth of heel (THEEL) = 4.00 ft Distance of batter for heel (BTRH) = .00 ft Depth of toe (TTOE) = 2.00 ft Width of toe (TWIDTH) = 3.75 ft Distance of batter for toe (BTRT) = .00 ft Width of base (BWIDTH) = 17.00 ft Depth of key (HK) = 2.00 ft Width of bottom of key (TK) = 1.25 ft Dist. of batter at bot. of key (BTRK) = .00 ft

Structure coordinates:

x (ft) y (ft) ================== .00 -4.00 .00 .00 12.00 .00 12.00 13.00 13.25 13.00 13.25 .00 17.00 .00 17.00 -2.00 1.25 -2.00 1.25 -4.00

NOTE: X=0 is located at the left-hand side of the structure. The Y values correspond to the actual elevation used.

Structural property data: Unit weight of concrete = .150 kcf

Driving side soil property data:

Moist Saturated Elev. Phi c Unit wt. unit wt. Delta soil (deg) (ksf) (kcf) (kcf) (deg) (ft) =======================================================

Page 1

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13'USUAL 27.00 .000 .118 .118 .00 5.00

Driving side soil geometry:

Soil Batter Distance point (in:1ft) (ft) ============================= 1 .00 500.00 2 .00 .00 3 .00 500.00

Driving side soil profile:

Soil x y point (ft) (ft) ============================= 1 -1488.00 5.00 2 12.00 5.00

Resisting side soil property data:

Moist Saturated Elev. Phi c Unit wt. unit wt. soil Batter (deg) (ksf) (kcf) (kcf) (ft) (in:1ft) ======================================================== 27.00 .000 .118 .118 5.00 .00

Resisting side soil profile:

Soil x y point (ft) (ft) ============================= 1 13.25 5.00 2 513.25 5.00

Foundation property data: phi for soil-structure interface = 27.00 (deg) c for soil-structure interface = .000 (ksf) phi for soil-soil interface = 27.00 (deg) c for soil-soil interface = .000 (ksf)

Water data: Driving side elevation = 5.00 ft Resisting side elevation = 5.00 ft Unit weight of water = .0624 kcf Seepage pressures computed are hydrostatic.

Horizontal pressure data:

Elevation Pressure (ft) (ksf) ========================== 13.00 .03 5.00 .03

Minimum required factors of safety: Sliding FS = 2.00 Overturning = 100.00% base in compression

Crack options: o Crack *is* down to bottom of heel o Computed cracks *will* be filled with water

Page 2

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13'USUAL Strength mobilization factor = .6667

At-rest pressures on the resisting side *are used* in the overturning analysis.

Forces on the resisting side *are used* in the sliding analysis.

*Do* iterate in overturning analysis.

***** Summary of Results *****

Project name: 2nd St. S

*************** *** Satisfied *** * Overturning * Required base in comp. = 100.00 % *************** Actual base in comp. = 100.00 % Overturning ratio = 2.07

Xr (measured from toe) = 7.78 ft Resultant ratio = .4578 Stem ratio = .2206 Base pressure at heel = .4227 ksf Base pressure at toe = .7094 ksf

*********** *** Satisfied *** * Sliding * Min. Required = 2.00 *********** Actual FS = 221.81

********************** Output Results **********************

Date: **/11/16 Time: 15.28.00

Company name: HEI Project name: 2nd St. S Project location: Wall location: Computed by: ljb

*************************** ** Overturning Results ** ***************************

Solution converged in 1 iterations.

SMF used to calculate K's = .6667 Alpha for the SMF = .0000 Calculated earth pressure coefficients: Driving side at rest K = .0000 Driving side at rest Kc = .0000 Resisting side at rest K = .5460 Resisting side at rest Kc = .7389 At-rest K's for resisting side calculated.

Page 3

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13'USUAL

Depth of cracking = 9.00 ft Crack extends to bottom of base of structure.

** Driving side pressures **

Water pressures: Elevation Pressure (ft) (ksf) ====================== 5.00 .0000 -4.00 .5616

** Resisting side pressures **

Water pressures: Elevation Pressure (ft) (ksf) ====================== 5.00 .0000 -2.00 .4368 -2.00 .5616 -4.00 .5616

Earth pressures: Elevation Pressure (ft) (ksf) ====================== 5.00 .0000 -2.00 .2125

Balancing earth pressures: Elevation Pressure (ft) (ksf) ====================== -2.00 .3143 -4.00 .3143

** Uplift pressures **

Water pressures: x-coord. Pressure (ft) (ksf) ====================== .00 .5616 1.25 .5616 1.25 .4368 17.00 .4368

** Forces and moments **

======================================================================== Part | Force (kips) | Mom. Arm | Moment | | Vert. | Horiz.| (ft) | (ft-k) | ======================================================================== Structure: Structure weight........... 7.913 -7.60 -60.15 Structure, driving side: Moist soil................. .000 .00 .00 Saturated soil............. 7.080 -11.00 -77.88 Water above structure...... .000 .00 .00 Water above soil........... .000 .00 .00 External vertical loads.... .000 .00 .00

Page 4

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13'USUAL Ext. horz. pressure loads.. .240 11.00 2.64 Ext. horz. line loads...... .000 .00 .00 Structure, resisting side: Moist soil................. .000 .00 .00 Saturated soil............. 2.213 -1.88 -4.15 Water above structure...... .000 .00 .00 Water above soil........... .000 .00 .00 Driving side: Effective earth loads...... .000 .00 .00 Shear (due to delta)....... .000 .00 .00 Horiz. surcharge effects... .000 .00 .00 Water loads................ 2.527 1.00 2.53 Resisting side: Effective earth loads...... -.744 2.33 -1.74 Balancing earth load....... -.629 -1.00 .63 Water loads................ -2.652 .92 -2.44 Foundation: Vertical force on base..... -9.623 -7.78 74.89 Uplift..................... -7.582 -8.66 65.67 ======================================================================== ** Statics Check ** SUMS = .000 -1.257 .00

Angle of base = 6.71 degrees Normal force on base = 9.484 kips Shear force on base = -1.749 kips Max. available shear force = 5.698 kips

Base pressure at heel = .4227 ksf Base pressure at toe = .7094 ksf

Xr (measured from toe) = 7.78 ft Resultant ratio = .4578 Stem ratio = .2206 Base in compression = 100.00 % Overturning ratio = 2.07

Volume of concrete = 1.95 cubic yds/ft of wall

NOTE: The engineer shall verify that the computed bearing pressures below the wall do not exceed the allowable foundation bearing pressure, or, perform a bearing capacity analysis using the program CBEAR. Also, the engineer shall verify that the base pressures do not result in excessive differential settlement of the wall foundation.

*********************** ** Sliding Results ** ***********************

Factor of safety > 100. Last iteration printed.

Horizontal Vertical Wedge Loads Loads Number (kips) (kips) ================================== 1 .000 .000 2 2.767 .000 3 .000 .000

Page 5

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13'USUAL Water pressures on wedges:

Top Bottom Wedge press. press. x-coord. press. number (ksf) (ksf) (ft) (ksf) ================================================ 1 .0000 .0000 2 .0000 .5616 2 17.0000 .4368 3 .0000 .4368

Points of sliding plane: Point 1 (left), x = .00 ft, y = -4.00 ft Point 2 (right), x = 17.00 ft, y = -2.00 ft

Depth of cracking = 9.00 ft Crack extends to bottom of base of structure.

Failure Total Weight Submerged Uplift Wedge angle length of wedge length force number (deg) (ft) (kips) (ft) (kips) ======================================================== 1 .000 .000 .000 .000 .000 2 6.710 17.117 18.916 17.117 8.545 3 45.509 9.813 2.840 9.813 2.143

Wedge Net force number (kips) =================== 1 .000 2 -.517 3 2.897 =================== SUM = 2.380

+-----------------------------+ | Factor of safety = 221.807 | +-----------------------------+

Page 6

Page 216: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

Concrete f'c = 4500 psiRebar Fy = 60000 psiUnit Weight = 150 lb/ft³15 in

17 ft

3.75 ft 12 ft

24

in

15

ft 13

ft

15.75 ft15 in

2 ft

40

in

7 ft

8 ft

#5 @ 12 in (S&T)#6 @ 6 in#5 @ 12 in (S&T)#5 @ 12 in#6 @ 6 in (dowels)

Heel Bars: #6 @ 6 inToe Bars: #5 @ 6 inFooting S/T Bars: #8 @ 12 in

40

in

#5 @ 12 in (S&T)#6 @ 6 in#5 @ 12 in (S&T)#5 @ 12 in

Design Detail

Check Summary

Ratio Check Provided Required Combination----- Stability Checks -----

0.091 Overturning 22.07 2.00 1.0D + 1.0F + 0.6H + 0.6W0.515 Sliding 3.88 2.00 1.0D + 1.0F + 0.6H + 0.6W0.203 Bearing Pressure 6000 psf 1221 psf 1.0D + 1.0F + 1.0H + 0.6W0.137 Bearing Eccentricity 7.01 in 51 in 1.0D + 1.0F + 1.0H + 0.6W

----- Toe Checks -----0.060 Shear 23.77 k/ft 1.42 k/ft 2.2D + 2.21F + 2.21H + 2.21W0.086 Moment 53.8 ft·k/ft 4.65 ft·k/ft 2.2D + 2.21F + 2.21H + 2.21W0.004 Min Strain 1.0173 0.0040 2.2D + 2.21F + 2.21H + 2.21W0.000 Min Steel 0.05 in² 0 in² 2.2D + 2.21F + 2.21H + 2.21W0.077 Development 155 in 12 in 2.2D + 2.21F + 2.21H + 2.21W0.667 S&T Max Spacing 12 in 18 in 2.2D + 2.21F + 2.21H + 2.21W0.328 S&T Min Rho 0.0055 0.0018 2.2D + 2.21F + 2.21H + 2.21W

----- Heel Checks -----0.067 Shear 23.7 k/ft 1.59 k/ft 2.2D + 2.21F + 2.21H + 2.21WInfinity Moment 75.44 ft·k/ft Infinite 2.2D + 2.21F + 2.21H + 2.21W0.006 Min Strain 0.7135 0.0040 2.2D + 2.21F + 2.21H + 2.21W0.898 Min Steel 0.07 in² 0.07 in² 2.2D + 2.21F + 2.21H + 2.21W0.374 Development 56 in 20.93 in 2.2D + 2.21F + 2.21H + 2.21W0.667 S&T Max Spacing 12 in 18 in 2.2D + 2.21F + 2.21H + 2.21W0.328 S&T Min Rho 0.0055 0.0018 2.2D + 2.21F + 2.21H + 2.21W

----- Stem Checks -----0.243 Moment 43.76 ft·k/ft 10.62 ft·k/ft 2.2D + 2.21F + 2.21H + 2.21W0.288 Shear 14.04 k/ft 4.04 k/ft 2.2D + 2.21F + 2.21H + 2.21W0.009 Max Steel 0.4214 0.0040 2.2D + 2.21F + 2.21H + 2.21W0.000 Min Steel 0.07 in²/in 0 in²/in 2.2D + 2.21F + 2.21H + 2.21W0.300 Base Development 20 in 6 in 2.2D + 2.21F + 2.21H + 2.21W0.523 Lap Splice Length 40 in 20.93 in 2.2D + 2.21F + 2.21H + 2.21W0.000 Lap Splice Spacing 0 in 4.19 in 2.2D + 2.21F + 2.21H + 2.21W0.581 Horz Bar Rho 0.0034 0.0020 2.2D + 2.21F + 2.21H + 2.21W0.667 Horz Bar Spacing 12 in 18 in 2.2D + 2.21F + 2.21H + 2.21W

Criteria

Building Code IBC 2006Concrete Load Combs USCoE UsualMasonry Load Combs MSJC 02/05 (ASD)Stability Load Combs ASCE 7-10 (ASD)Restrained Against Sliding NoNeglect Bearing At Heel NoUse Vert. Comp. for OT NoUse Vert. Comp. for Sliding NoUse Vert. Comp. for Bearing YesUse Surcharge for Sliding & OT NoUse Surcharge for Bearing YesNeglect Soil Over Toe NoNeglect Backfill Wt. for Coulomb NoFactor Soil Weight As Dead NoUse Passive Force for OT YesAssume Pressure To Top YesExtend Backfill Pressure To Key Bottom NoUse Toe Passive Pressure for Bearing NoRequired F.S. for OT 2.00Required F.S. for Sliding 2.00Has Different Safety Factors for Seismic NoAllowable Bearing Pressure 6000 psfReq'd Bearing Location Middle halfWall Friction Angle 25°Friction Coefficent 0.35Soil Reaction Modulus 172800 lb/ft³

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Page 217: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

Loads5

ft

7 ft

7 ft

8 ft

γ = 118 lb/ft³φ = 27°7

ft

γ = 118 lb/ft³φ = 27°

5 ft

-30 psf

Loading Options/AssumptionsPassive pressure neglects top 0 ft of soil.

Load Combinations

USCoE Usual ...2.2D + 2.21F + 2.21H + 2.21W

...2.2D + 2.21F + 2.21H

Backfill Pressure

5 ft

7 ft

7 ft

8 ft

γ = 118 lb/ft³φ = 27°

26.75 psf

7 ft

7.8 lb/in7 ft

2.3

3 ft

-322.15 psf 67.11 lb/in

1.6

7 ft

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Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

Backfill Pressure (Water Layer)

γeff γsat γw - 55.5 lb ft³ / 62.5 lb ft³ / - 7 lb ft³ / - = = =

φ 27° φsat =

γ 7 lb ft³ γeff / - =

φ φsat =

γ γeff =

At Rest Earth Pressure Theory -

Ko 1 sin φ - 1 sin 27° - 0.5460 = = =

σh Ko γ H 0.5460 7 lb ft³ / - 7 ft 26.75 psf - = = =

Lateral Earth Pressure (water layer)

σh Ko γ H 0.5460 118 lb ft³ / 5 ft 322.1 psf = = =

Lateral Earth Pressure (water layer, stem only)

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Page 219: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

Passive Pressure

7 ft

γ = 118 lb/ft³φ = 27°

579.9 psf

9 ft

217.4 lb/in

1 ft

At Rest Earth Pressure Theory -

Ko 1 sin φ - 1 sin 27° - 0.5460 = = =

σh Ko γ H 0.5460 118 lb ft³ / 9 ft 579.9 psf = = =

Lateral Earth Pressure

Water Pressure

5 ft

-437.5 psf7 ft

127.6 lb/in7 ft

-312.5 psf 65.1 lb/in

σw γw Hw 62.5 lb ft³ / 7 ft 437.5 psf = = =

Lateral Water Pressure

σw γw Hw 62.5 lb ft³ / 5 ft 312.5 psf = = =

Lateral Water Pressure (stem only)

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Page 220: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

Manually Specified Lateral Stem Pressure

-30 psf240 lb/ft

Wall/Soil Weights

425 lb/in

203.1 lb/in

31.25 lb/in

590 lb/in184.4 lb/in

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Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

Bearing Pressure

1221 psf803.3 psf

1434 lb/in

7.92 ft

e = 7.01 in

501.8 lb/in

F µ R 0.350 1434 lb in / 501.8 lb in / = = =

Friction

Bearing Pressure Calculation

Contributing ForcesVert Force ...offset Horz Force ...offset OT Moment

Backfill Pressure 0 lb/in - 7.8 lb/in 2.33 ft -2621.94 in·lb/ftWater Pressure -0 lb/in - -127.6 lb/in 2.33 ft 42875 in·lb/ftManual Lateral Pressure -0 lb/in - -20 lb/in 11 ft 31680 in·lb/ftFooting Weight -425 lb/in 8.5 ft 0 lb/in - -520200 in·lb/ftStem Weight -203.13 lb/in 4.38 ft 0 lb/in - -127968.75 in·lb/ftKey Weight -31.25 lb/in 16.38 ft 0 lb/in - -73687.5 in·lb/ftBackfill Weight -590 lb/in 11 ft 0 lb/in - -934560 in·lb/ftSoil over toe Weight -184.38 lb/in 1.88 ft 0 lb/in - -49781.25 in·lb/ft

-1433.75 lb/in -1634264.44 in·lb/ft1634264.44 in·lb ft / -

1433.75 lb in / - 7.92 ft =

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Page 222: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

Overturning Check

Overturning MomentsForce Distance Moment

Backfill pressure (horz) -7.8 lb/in 2.33 ft -2621.94 in·lb/ftWater pressure 127.6 lb/in 2.33 ft 42875 in·lb/ftManual lateral pressure 12 lb/in 11 ft 19008 in·lb/ft

Total: 59261 in·lb/ft

Resisting MomentsForce Distance Moment

Passive pressure @ toe 217.4 lb/in 1 ft 31313 in·lb/ftFooting Weight -425 lb/in 8.5 ft 520200 in·lb/ftStem Weight -203.13 lb/in 4.38 ft 127969 in·lb/ftKey Weight -31.25 lb/in 16.38 ft 73688 in·lb/ftBackfill Weight -590 lb/in 11 ft 934560 in·lb/ftSoil over toe Weight -184.38 lb/in 1.88 ft 49781 in·lb/ft

Total: 1737510 in·lb/ft

F.S. RM

OTM

1737510 in·lb ft / 59261 in·lb ft /

29.320 > 2.00 OK = = =

Sliding Check

Sliding Force(s)Backfill pressure -7.8 lb/inWater pressure 127.6 lb/inManual lateral pressure 12 lb/inTotal: 131.8 lb/in

Resisting Force(s)Passive pressure @ toe 217.4 lb/inFriction 501.8 lb/inTotal: 719.3 lb/in

F.S. RFSF

719.3 lb in / 131.8 lb in /

5.457 > 2.00 OK = = =

Bearing Capacity Check

Bearing pressure < allowable (1221 psf < 6000 psf) - OKBearing resultant eccentricity < allowable (7.01 in < 51 in) - OK

Wall Top Displacement

(based on unfactored service loads)

Deflection due to stem flexural displacement 0.019 inDeflection due to rotation from settlement 0.022 inTotal deflection at top of wall (positive towards toe) 0.041 in

Stability Checks [1.0D + 1.0F + 1.0H + 0.6W]

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Page 223: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

Overturning Check

Overturning MomentsForce Distance Moment

Backfill pressure (horz) -4.68 lb/in 2.33 ft -1573.16 in·lb/ftWater pressure 127.6 lb/in 2.33 ft 42875 in·lb/ftManual lateral pressure 12 lb/in 11 ft 19008 in·lb/ft

Total: 60310 in·lb/ft

Resisting MomentsForce Distance Moment

Passive pressure @ toe 130.5 lb/in 1 ft 18788 in·lb/ftFooting Weight -425 lb/in 8.5 ft 520200 in·lb/ftStem Weight -203.13 lb/in 4.38 ft 127969 in·lb/ftKey Weight -31.25 lb/in 16.38 ft 73688 in·lb/ftBackfill Weight -354 lb/in 11 ft 560736 in·lb/ftSoil over toe Weight -110.63 lb/in 1.88 ft 29869 in·lb/ft

Total: 1331249 in·lb/ft

F.S. RM

OTM

1331249 in·lb ft / 60310 in·lb ft /

22.073 > 2.00 OK = = =

Sliding Check

Sliding Force(s)Backfill pressure -4.68 lb/inWater pressure 127.6 lb/inManual lateral pressure 12 lb/inTotal: 134.9 lb/in

Resisting Force(s)Passive pressure @ toe 130.5 lb/inFriction 393.4 lb/inTotal: 523.9 lb/in

F.S. RFSF

523.9 lb in / 134.9 lb in /

3.883 > 2.00 OK = = =

Bearing Capacity Check

Bearing pressure < allowable (957 psf < 6000 psf) - OKBearing resultant eccentricity < allowable (7.01 in < 51 in) - OK

Wall Top Displacement

(based on unfactored service loads)

Deflection due to stem flexural displacement 0.019 inDeflection due to rotation from settlement 0.022 inTotal deflection at top of wall (positive towards toe) 0.041 in

Stability Checks [1.0D + 1.0F + 0.6H + 0.6W]

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Page 224: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

13

11.7

10.4

9.1

7.8

6.5

5.2

3.9

2.6

1.3

0-50 -38.33 -26.67 -15 -3.33 8.33 20

Moment (ft·k/ft)

Offset (ft)

Moment

aAs fy

0.85 F'c

0.07 in² in / 60000 psi 0.85 4500 psi

1.15 in = = =

φMn φ As fy d a 2 / - 0.90 0.07 in² in / 60000 psi 11.63 in 1.15 in 2 / - 43.76 ft·k ft / = = =

Capacity (ACI 318-05 10.2) @ 0 ft from base [Negative bending]

aAs fy

0.85 F'c

0.03 in² in / 60000 psi 0.85 4500 psi

0.41 in = = =

φMn φ As fy d a 2 / - 0.90 0.03 in² in / 60000 psi 11.69 in 0.41 in 2 / - 16.02 ft·k ft / = = =

Capacity (ACI 318-05 10.2) @ 0 ft from base [Positive bending]

aAs fy

0.85 F'c

0.07 in² in / 60000 psi 0.85 4500 psi

1.15 in = = =

φMn φ As fy d a 2 / - 0.90 0.07 in² in / 60000 psi 11.63 in 1.15 in 2 / - 43.76 ft·k ft / = = =

Capacity (ACI 318-05 10.2) @ 11.66 ft from base [Negative bending]

aAs fy

0.85 F'c

0.03 in² in / 60000 psi 0.85 4500 psi

0.41 in = = =

φMn φ As fy d a 2 / - 0.90 0.03 in² in / 60000 psi 11.69 in 0.41 in 2 / - 16.02 ft·k ft / = = =

Capacity (ACI 318-05 10.2) @ 11.88 ft from base [Positive bending]

aAs fy

0.85 F'c

0 in² in / 60000 psi 0.85 4500 psi

0 in = = =

φMn φ As fy d a 2 / - 0.90 0 in² in / 60000 psi 11.63 in 0 in 2 / - 0 ft·k ft / = = =

Capacity (ACI 318-05 10.2) @ 13 ft from base [Negative bending]

aAs fy

0.85 F'c

0 in² in / 60000 psi 0.85 4500 psi

0 in = = =

φMn φ As fy d a 2 / - 0.90 0 in² in / 60000 psi 11.69 in 0 in 2 / - 0 ft·k ft / = = =

Capacity (ACI 318-05 10.2) @ 13 ft from base [Positive bending]

Stem Flexural Capacity

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Page 225: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

13

11.7

10.4

9.1

7.8

6.5

5.2

3.9

2.6

1.3

0-15 -10 -5 0 5 10 15

Shear (k/ft)

Offset (ft)

Shear

Vc 2 F'c d 2 4500 psi 11.63 in 18.72 k ft / = = =

φVn φ Vc 0.750 18.72 k ft / 14.04 k ft / = = =

Shear Capacity (ACI 318-05 11.1.1, 11.3.1) @ 0 ft from base [Positive shear]

Vc 2 F'c d 2 4500 psi 11.63 in 18.72 k ft / = = =

φVn φ Vc 0.750 18.72 k ft / 14.04 k ft / = = =

Shear Capacity (ACI 318-05 11.1.1, 11.3.1) @ 0 ft from base [Negative shear]

Vc 2 F'c d 2 4500 psi 11.63 in 18.72 k ft / = = =

φVn φ Vc 0.750 18.72 k ft / 14.04 k ft / = = =

Shear Capacity (ACI 318-05 11.1.1, 11.3.1) @ 13 ft from base [Positive shear]

Vc 2 F'c d 2 4500 psi 11.63 in 18.72 k ft / = = =

φVn φ Vc 0.750 18.72 k ft / 14.04 k ft / = = =

Shear Capacity (ACI 318-05 11.1.1, 11.3.1) @ 13 ft from base [Negative shear]

Stem Shear Capacity

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Page 226: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

ψe 1.0 uncoated hooked bars =

λ 1.0 normal weight concrete =

ldh 0.02 ψe λ fy

F'c

db 0.02 1.0 1.060000 psi

4500 psi0.75 in 13.42 in = = =

Factoring ldh by the 0.7 multiplier of 12.5.3 a : ldh 9.39 in =

8 db 8 0.75 in 6.0 minimum limit, does not control = =

Main vertical stem bars (bottom end) - Development Length Calculation (ACI 318-05 12.2.3, 12.5)

ψt 1.0 bars are not horizontal =

ψe 1.0 bar not epoxy coated =

ψs 0.80 bars are #6 or smaller =

λ 1.0 normal weight concrete =

s 2 / 6 in 2 / 3 in = =

cover db 2 / + 3 in 0.75 in 2 / + 3.38 in = =

cb 3 in lesser of half spacing, ctr to surface =

Ktr 0.0 no transverse reinforcement =

cb Ktr + db

3 in 0.0 + 0.75 in

4.0 = =

ld3.40

fy

F'c

ψt ψe ψs λ 2.5

db 3.40

60000 psi

4500 psi

1.0 1.0 0.80 1.0 2.5

0.75 in 16.1 in = = =

Main vertical stem bars (top end) - Development Length Calculation (ACI 318-05 12.2.3, 12.5)

ψt 1.0 bars are not horizontal =

ψe 1.0 bar not epoxy coated =

ψs 0.80 bars are #6 or smaller =

λ 1.0 normal weight concrete =

s 2 / 6 in 2 / 3 in = =

cover db 2 / + 3 in 0.75 in 2 / + 3.38 in = =

cb 3 in lesser of half spacing, ctr to surface =

Ktr 0.0 no transverse reinforcement =

cb Ktr + db

3 in 0.0 + 0.75 in

4.0 = =

ld3.40

fy

F'c

ψt ψe ψs λ 2.5

db 3.40

60000 psi

4500 psi

1.0 1.0 0.80 1.0 2.5

0.75 in 16.1 in = = =

Dowels for vertical stem bars (top end) - Development Length Calculation (ACI 318-05 12.2.3, 12.5)

ψe 1.0 uncoated hooked bars =

λ 1.0 normal weight concrete =

ldh 0.02 ψe λ fy

F'c

db 0.02 1.0 1.060000 psi

4500 psi0.63 in 11.18 in = = =

Factoring ldh by the 0.7 multiplier of 12.5.3 a : ldh 7.83 in =

8 db 8 0.63 in 5.0 minimum limit, does not control = =

2nd curtain vertical bars (bottom end) - Development Length Calculation (ACI 318-05 12.2.3, 12.5)

Stem Development/Lap Length Calculations

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Page 227: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

ψt 1.0 bars are not horizontal =

ψe 1.0 bar not epoxy coated =

ψs 0.80 bars are #6 or smaller =

λ 1.0 normal weight concrete =

s 2 / 12 in 2 / 6 in = =

cover db 2 / + 3 in 0.63 in 2 / + 3.31 in = =

cb 3.31 in lesser of half spacing, ctr to surface =

Ktr 0.0 no transverse reinforcement =

cb Ktr + db

3.31 in 0.0 + 0.63 in

5.30 = =

ld3.40

fy

F'c

ψt ψe ψs λ 2.5

db 3.40

60000 psi

4500 psi

1.0 1.0 0.80 1.0 2.5

0.63 in 13.42 in = = =

2nd curtain vertical bars (top end) - Development Length Calculation (ACI 318-05 12.2.3, 12.5)

Stem Development/Lap Length Calculations (continued)

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Page 228: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

Design moment Mu for toe need not exceed moment at stem base:

Mtoe 4.65 ft·k ft < Mstem / 10.62 ft·k ft / = =

Mu 4.65 ft·k ft stem moment does not control / =

Controlling Moment

aAs fy

0.85 F'c

0.05 in² in / 60000 psi 0.85 4500 psi

0.81 in = = =

φMn φ As fy d a 2 / - 0.90 0.05 in² in / 60000 psi 19.69 in 0.81 in 2 / - 53.8 ft·k ft / = = =

φMn 53.8 ft·k ft ≥ Mu / 4.65 ft·k ft / = = �

Flexure Check (ACI 318-05 10.2)

Vc 2 F'c d 2 4500 psi 19.69 in 31.7 k ft / = = =

φVn φ Vc 0.750 31.7 k ft / 23.77 k ft / = = =

φVn 23.77 k ft ≥ Vu / 1.42 k ft / = = �

Shear Check (ACI 318-05 11.1.1, 11.3.1)

β1 0.85 0.05F'c 4000 -

1000 - 0.85 0.05

4500 psi 4000 - 1000

- 0.8250 = = =

aAs fy

0.85 F'c

0.05 in² in / 60000 psi 0.85 4500 psi

0.81 in = = =

εt 0.003d

a β1 / 1 - 0.003

19.69 in0.81 in 14.0 /

1 - 1.0173 = = =

εt 1.0173 ≥ 0.004 = �

Minimum Strain Check (ACI 318-05 10.3.5)

φMn 53.8 ft·k ft ≥ 4 3 / Mu / 4 3 / 4.65 ft·k ft / 6.19 ft·k ft / = = =

Check is waived per ACI 10.5.3�

Minimum Steel Check (ACI 318-05 10.5.1)

ρST_provAST

t sST

1.58 in² in / 24 in 12 in

0.0055 = = =

ρST_min0.0018 60000

fy

0.0018 60000 60000 psi

0.0018 = = =

ρST_min 0.0018 =

ρST_prov 0.0055 ≥ ρST_min 0.0018 = = �18 inch limit governs

sST_max 18 in =

sST 12 in ≤ sST_max 18 in = = �

Shrinkage Temperature Steel (ACI 318-05 7.12.2)

Mu

φMn

4.65 ft·k ft / 53.8 ft·k ft /

0.0864 ratio to represent excess reinforcement = =

ψt 1.0 12 inches or less cast below 4.00 inches - =

ψe 1.0 bar not epoxy coated =

ψs 0.80 bars are #6 or smaller =

λ 1.0 normal weight concrete =

s 2 / 6 in 2 / 3 in = =

cover db 2 / + 4 in 0.63 in 2 / + 4.31 in = =

cb 3 in lesser of half spacing, ctr to surface =

Ktr 0.0 no transverse reinforcement =

cb Ktr + db

3 in 0.0 + 0.63 in

4.80 = =

ld3.40

fy

F'c

ψt ψe ψs λ 2.5

db 3.40

60000 psi

4500 psi

1.0 1.0 0.80 1.0 2.5

0.63 in 13.42 in = = =

Factoring ld by the excess reinforcement ratio 0.0864 per 12.2.5: ld 1.16 in =

12 inch minimum controls

ld_prov 155 in ≥ ld 12 in = = �

Development Check (ACI 318-05 12.12, 12.2.3)

Toe Unfactored Loads

24

in

#5 @ 6 in

Unfactored Loads

300 psf (Self-wt)

590 psf (Soil)

1221 psf 1129 psf

Toe Factored Loads

24

in

#5 @ 6 in

2.2D + 2.21F + 2.21H + 2.21W

660 psf (Self-wt)

1304 psf (Soil)

2692 psf 2489 psf2692 psf 2489 psf

2.35 k/ft

Toe Checks [2.2D + 2.21F + 2.21H + 2.21W]

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Page 229: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

Mhgov INF =

♦♦♦ Heel is bent 'upwards' QuickRWall does not handle this configuration. Check heel manually.�

Controlling Moment

aAs fy

0.85 F'c

0.07 in² in / 60000 psi 0.85 4500 psi

1.15 in = = =

φMn φ As fy d a 2 / - 0.90 0.07 in² in / 60000 psi 19.63 in 1.15 in 2 / - 75.44 ft·k ft / = = =

φMn 75.44 ft·k ft < Mu / INF = = �

Flexure Check (ACI 318-05 10.2)

Vc 2 F'c d 2 4500 psi 19.63 in 31.6 k ft / = = =

φVn φ Vc 0.750 31.6 k ft / 23.7 k ft / = = =

φVn 23.7 k ft ≥ Vu / 1.59 k ft / = = �

Shear Check (ACI 318-05 11.1.1, 11.3.1)

β1 0.85 0.05F'c 4000 -

1000 - 0.85 0.05

4500 psi 4000 - 1000

- 0.8250 = = =

aAs fy

0.85 F'c

0.07 in² in / 60000 psi 0.85 4500 psi

1.15 in = = =

εt 0.003d

a β1 / 1 - 0.003

19.63 in1.15 in 14.0 /

1 - 0.7135 = = =

εt 0.7135 ≥ 0.004 = �

Minimum Strain Check (ACI 318-05 10.3.5)

φMn 75.44 ft·k ft < 4 3 / Mu / 4 3 / INF INF = = =

As_min3 F'c

fyd

3 4500 psi 60000 psi

19.63 in 0.07 in² in / = = =

200 d fy / 200 19.63 in 60000 psi / 0.07 in² in / = =

As 0.07 in² in ≥ As_min / 0.07 in² in / = = �

Minimum Steel Check (ACI 318-05 10.5.1)

ρST_provAST

t sST

1.58 in² in / 24 in 12 in

0.0055 = = =

ρST_min0.0018 60000

fy

0.0018 60000 60000 psi

0.0018 = = =

ρST_min 0.0018 =

ρST_prov 0.0055 ≥ ρST_min 0.0018 = = �18 inch limit governs

sST_max 18 in =

sST 12 in ≤ sST_max 18 in = = �

Shrinkage Temperature Steel (ACI 318-05 7.12.2)

Mu

φMn

INF75.44 ft·k ft /

INF ratio to represent excess reinforcement = =

ψt 1.30 more than 12 inches cast below 19.25 inches - =

ψe 1.0 bar not epoxy coated =

ψs 0.80 bars are #6 or smaller =

λ 1.0 normal weight concrete =

s 2 / 6 in 2 / 3 in = =

cover db 2 / + 4 in 0.75 in 2 / + 4.38 in = =

cb 3 in lesser of half spacing, ctr to surface =

Ktr 0.0 no transverse reinforcement =

cb Ktr + db

3 in 0.0 + 0.75 in

4.0 = =

ld3.40

fy

F'c

ψt ψe ψs λ 2.5

db 3.40

60000 psi

4500 psi

1.30 1.0 0.80 1.0 2.5

0.75 in 20.93 in = = =

ld_prov 56 in ≥ ld 20.93 in = = �

Development Check (ACI 318-05 12.12, 12.2.3)

Heel Unfactored Loads

24

in#6 @ 6 in

Unfactored Loads

300 psf (Concrete self-wt)

590 psf (Soil weight)

1098 psf 803.3 psf

Heel Factored Loads

24

in#6 @ 6 in

2.2D + 2.21F + 2.21H + 2.21W

660 psf (Concrete self-wt)

1304 psf (Soil weight)

2422 psf 1772 psf

-1.76 ft·k/ft

1.59 k/ft

Heel Checks [2.2D + 2.21F + 2.21H + 2.21W]

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Page 230: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

Stem Internal Forces

-711.94 psf-690.63 psf

-66.3 psf

4.04 k/ft

-10.62 ft·k/ft

Stem Internal Forces

13

11.38

9.75

8.13

6.5

4.88

3.25

1.63

0-12 -9 -6 -3 0

Moment (ft·k/ft)

Moment

Stem Internal Forces

13

11.38

9.75

8.13

6.5

4.88

3.25

1.63

00 1.25 2.5 3.75 5

Shear (k/ft)

Shear

Stem Joint Force Transfer

Location Force@ stem base 4.04 k/ft

Stem Internal Forces

-711.94 psf -690.63 psf

-66.3 psf

Stem Forces [2.2D + 2.21F + 2.21H + 2.21W]

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Page 231: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

13

11.7

10.4

9.1

7.8

6.5

5.2

3.9

2.6

1.3

0-50 -38.33 -26.67 -15 -3.33 8.33 20

Moment (ft·k/ft)

Offset (ft)

Moment

φMn 43.76 ft·k ft ≥ Mu / 10.62 ft·k ft / = = �

Check (ACI 318-05 Ch 10) @ 0 ft from base

φMn 43.76 ft·k ft ≥ Mu / 0.06 ft·k ft / = = �

Check (ACI 318-05 Ch 10) @ 11.66 ft from base

φMn 42.83 ft·k ft ≥ Mu / 0.06 ft·k ft / = = �

Check (ACI 318-05 Ch 10) @ 11.69 ft from base

Stem Moment Checks [2.2D + 2.21F + 2.21H + 2.21W]

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Page 232: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

13

11.7

10.4

9.1

7.8

6.5

5.2

3.9

2.6

1.3

0-15 -10 -5 0 5 10 15

Shear (k/ft)

Offset (ft)

Shear

φVn 14.04 k ft ≥ Vu / 4.04 k ft / = = �

Shear Check (ACI 318-05 Ch 11.1.1) @ 0 ft from base

Stem Shear Checks [2.2D + 2.21F + 2.21H + 2.21W]

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Page 233: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

φMn 43.76 ft·k ft ≥ 4 3 / Mu / 4 3 / 10.62 ft·k ft / 14.16 ft·k ft / = = =

Check is waived per ACI 10.5.3�

Minimum Steel Check (ACI 318-05 10.5.1) @ 0 ft from base [Stem in negative flexure]

φMn 43.76 ft·k ft ≥ 4 3 / Mu / 4 3 / 3.27 ft·k ft / 4.36 ft·k ft / = = =

Check is waived per ACI 10.5.3�

Minimum Steel Check (ACI 318-05 10.5.1) @ 3.33 ft from base [Stem in negative flexure]

φMn 0 ft·k ft ≥ 4 3 / Mu / 4 3 / 0 ft·k ft / 0 ft·k ft / = = =

Check is waived per ACI 10.5.3�

Minimum Steel Check (ACI 318-05 10.5.1) @ 13 ft from base [Stem in negative flexure]

β1 0.85 0.05F'c 4000 -

1000 - 0.85 0.05

4500 psi 4000 - 1000

- 0.8250 = = =

aAs fy

0.85 F'c

0.07 in² in / 60000 psi 0.85 4500 psi

1.15 in = = =

εt 0.003d

a β1 / 1 - 0.003

11.63 in1.15 in 14.0 /

1 - 0.4214 = = =

εt 0.4214 ≥ 0.004 = �

Maximum Steel Check (ACI 318-05 10.3.5) @ 0 ft from base [Stem in negative flexure]

β1 0.85 0.05F'c 4000 -

1000 - 0.85 0.05

4500 psi 4000 - 1000

- 0.8250 = = =

aAs fy

0.85 F'c

0.07 in² in / 60000 psi 0.85 4500 psi

1.15 in = = =

εt 0.003d

a β1 / 1 - 0.003

11.63 in1.15 in 14.0 /

1 - 0.4214 = = =

εt 0.4214 ≥ 0.004 = �

Maximum Steel Check (ACI 318-05 10.3.5) @ 3.33 ft from base [Stem in negative flexure]

β1 0.85 0.05F'c 4000 -

1000 - 0.85 0.05

4500 psi 4000 - 1000

- 0.8250 = = =

aAs fy

0.85 F'c

0.07 in² in / 60000 psi 0.85 4500 psi

1.15 in = = =

εt 0.003d

a β1 / 1 - 0.003

11.63 in1.15 in 14.0 /

1 - 0.4214 = = =

εt 0.4214 ≥ 0.004 = �

Maximum Steel Check (ACI 318-05 10.3.5) @ 13 ft from base [Stem in negative flexure]

ρhAs_horz shorz /

t

0.62 in² 12 in / 15 in

0.0034 = = =

ρh_min 0.0020 bars No. 5 or less, not less than 60 ksi =

ρh 0.0034 ≥ ρh_min 0.0020 = = �3 twall 3 15 in 45 in = =

18 inch limit governs

smax 18 in =

shorz 12 in ≤ shorz_max 18 in = = �

Wall Horizontal Steel (ACI 318-05 14.3.3, 14.3.5)

Stem Miscellaneous Checks [2.2D + 2.21F + 2.21H + 2.21W]

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Page 234: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Luke HOUSTON ENGINEERING, INC.

Job # 7438-009Fargo Floodwalls

Mu

φMn

10.62 ft·k ft / 43.76 ft·k ft /

0.2426 ratio to represent excess reinforcement = =

ψe 1.0 uncoated hooked bars =

λ 1.0 normal weight concrete =

ldh 0.02 ψe λ fy

F'c

db 0.02 1.0 1.060000 psi

4500 psi0.75 in 13.42 in = = =

Factoring ldh by the 0.7 multiplier of 12.5.3 a : ldh 9.39 in =

Factoring ldh by the excess reinforcement ratio 0.2426 per 12.5.3 d : ldh 2.28 in =

8 db 8 0.75 in 6.0 minimum limit, does not control = =

6 inch minimum controls

ldh_prov 20 in ≥ ldh 6 in = = �

Development Check (ACI 318-05 12.12, 12.2.3)

ψt 1.0 bars are not horizontal =

ψe 1.0 bar not epoxy coated =

ψs 0.80 bars are #6 or smaller =

λ 1.0 normal weight concrete =

s 2 / 6 in 2 / 3 in = =

cover db 2 / + 3 in 0.75 in 2 / + 3.38 in = =

cb 3 in lesser of half spacing, ctr to surface =

Ktr 0.0 no transverse reinforcement =

cb Ktr + db

3 in 0.0 + 0.75 in

4.0 = =

ld3.40

fy

F'c

ψt ψe ψs λ 2.5

db 3.40

60000 psi

4500 psi

1.0 1.0 0.80 1.0 2.5

0.75 in 16.1 in = = =

llap 1.3 ld 1.3 16.1 in 20.93 in = = =

llap_prov 40 in ≥ llap 20.93 in = = �1 5 / llap 1 5 / 20.93 in 4.1859 ≤ 6.0 = =

strans 0 in ≤ 1 5 / llap 1 5 / 20.93 in 4.1859 = = = �

Lap Splice Checks (ACI 318-05 12.14.2.3, 12.15.1, 12.15.2) - #6 lap with #6, from 0 ft to 3.33 ft (from stem base)

Stem Miscellaneous Checks [2.2D + 2.21F + 2.21H + 2.21W] (continued)

QuickRWall 4.0 (iesweb.com) H:\Fargo\JBN\7400\7438\12_7438_009\006-I...\13' Usual.rwd Page 19 of 19 Tuesday 11/17/15 9:14 AM

Page 235: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Site: 7438-009 Owner:

Case:

User Data:A 15 Inches

B 17 Feet

C 12 FeetD 24 InchesD1 24 Inches

Hwall 13 Feet

Hwater 13 Feet

H2 5 FeetH3 5 Feet

Unit Weight of Soil (ϒsat):

Below Footing 0.118 k/cu. ft.Wet Side 0.118 k/cu. ft.Dry Side 0.118 k/cu. ft.

Ø= Below Footing 27 Degrees0.471 radians

Wet Side 27 Degrees0.471 radians

Dry Side 27 Degrees0.471 radians

SMF= 0.66667Unit Weight of Conc. (ρc) 0.15 k/cu. ft.

Unit Weight of Water (ρw) 0.0625 k/cu. ft.

Soil Capacity 6.2 k/sq. ft.2 Feet Ko Below Footing 0.55

15 Inches Wet Side 0.55Dry Side 0.55

Kp for sliding = 2.663fysteel 60 ksi Cohesion Factor (CF) 1 ksf

f'c(concrete) 4.5 ksi Frost Depth (Fr): 6 ft

Vertical Surcharge (Dry) (P3) 0 psf

Vertical Surcharge (Wet) (P4) 0 psf

Ice/Debris Load NoType of Loading Unusual

Toe Vertical Fill Width (F)= 3.75 ft

*All Dimensions in Calculations are converted to feet

City of Fargo

13' Floodwall Unusual to top of wall

Keyway Depth (G)Keyway Width (L)

Project Number:Fargo Floodwalls

1 − sin∅

1 − sin∅1 − sin∅

Page 236: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Force CalculationsFactor Unit Equation Description

EHW= 1.063 kips/ft Driving Wet Side Water Pressure

EHS= 0.500 kips/ft Driving Wet Side Soil Pressure

EV2= 13.080 kips Wet Side Vertical Component Weights

EV2d 11.000 ft from toe Moment arm for EV2

EV1= 2.213 kips Dry Side Vertical Component Weights

EV1d 1.875 ft from toe Moment arm for EV1

PehW= 9.031 kips Driving Wet Side Water Force

Pehwd 3.667 ft. above footing Moment arm for Pehw

PehS= 2.248 kips Driving Wet Side Soil Force

Pehsd 1.000 ft. above footing Moment arm for Pehs

PS1= 2.813 kips Vertical Weight of Stem and Key

PS1d 5.975 ft from toe Moment arm for PS1

PS2= 5.100 kips Vertical Weight of Footing

PS2d 8.500 ft from toe Moment arm for PS2

EH2= 1.062 kips Resisting Dry Side Soil Pressure

Peh2= 4.779 kips Resisting Dry Side Forces

Peh2d 1.000 ft from toe Moment arm for Peh2

Peh2s 2.248 kips Resisting dry side soil only

Peh2sd 1.000 ft from toe Moment arm for Peh2s

Pice/debris 0.000 kips .5 k/ft at Q100 Driving ice/debris Force

Pice/debrisd 15.000 ft. above footing Moment arm for Pice/debris

Pwind 0 kips Driving wind Force

Pwindd 15.000 ft. above footing Moment arm for Pwind

Applied Forces

(��� +�� + �)(ρ)

(�3 + �� + �)(ɤ�� − ρ)

��� � ρ +H3(C)(ɤ�� − ρ)

�2(�)(ɤ��)

��

2(��� + �� + �)

���

2(�3 + �� + �)

��� + �� + �

3− �

�3 + �� + �

3− �

��� � ρ� + �( )(ρ�)

! − � �� ρ� + �(�)(ρ�)

(�2 + �� + �)(ɤ��)

��"

2(�2 + �� + �)

! −�

2

2

��� � ρ� ! − � −�

2+ � ρ� ! −

2/$%�

! − � �� ρ�! − �

2+ � � ρ� ! −

2/$%"

�2 + �� + �

3− �

��� + ��

��� − ���

2+ ��� + ��

ρ� − ρ ∗ �2 + �� + � +�2 + �� + �

2

�2 + �� + �

3− �

.03(��� −��� )

Page 237: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Water Uplift

Force Calculation Factor Unit Equation DescriptionLZ-Z1= 17.117 ft Length of Seepage path from heel to toe

Δh= 8.000 ft Head differential between wet and dry side

Ldry side 7.000 ft Seepage path on Dry Side

LS 24.117 ft Total Seepage Path

uZ= 1.063 ksf Water Pressure at bottom of key (Wet side)

uZ1= 0.583 ksf Water pressure at bottom of footing (Dry Side)

HLZ-Z1= 5.678 ft Head Loss along Z-Z1

LSc 19.000 ft Length of concrete surface in sliding surface

HLLK = 0.374 ft Head Loss along key

ubottom key = 1.039 ksf Water pressure at bottom of key (Dry side)

HLUK = 0.971 ft Head Loss up key

utop key = 0.877 ksf Water pressure at top of key (Dry side)

P5 = 12.806 kips Water Uplift for Overturning

P5d 9.222 ft

P5sa 13.984 kips Water uplift for sliding along angle

P5sad -9.326 ft Moment Arm for P5sa

Lsss = 26.000 ft

Uf = 0.736 ksf

P5ss = 15.284 Water uplift of sliding along bottom of key

P5ssd -9.015 Moment Arm for P5ss

(!"+ �")

��� − �2

�2 + ��

)*)� + + ,�.+�

(��� +�� + �)(ρ)

��� + �� − ∆0 )*)� 1

(ρ)

∆0 )*)� �

! + �

%�(� )*)�)

(��� + �� + � − � 23)ρ

+ �

%�(� )*)�)

(��� +�� −� 43)ρ

5) + 567��789�,

2 +

5�7:9�, + 5)�2

(! − )

5) + 5)�2

(!)

5)�! −

2! − +

5�7:9�, − 5)�2

! − 2

3! − + 567��789�, ! −

2+5) − 567��789�,

2 ! −

3/$5

5)� !!

2+

5) − 5)�2

!2!

3/$5�

! + � + �� +�2

��� +�� + � − ��� −�2!

<��ρ

5= + 5>2

(!)

5> !!

2+5) − 5>

2!

2!

3/$5��

Moment Arm

for P5

Water pressure at key depth below

footing on dry side of footing

Page 238: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Distance Moment

to Toe(ft) About Toe (k-ft)

PS1 2.81 5.98 16.8 MPS1

PS2 5.10 8.50 43.4 MPS2

P3 0.00 1.88 0.0 MP3

P4 0.00 11.00 0.0 MP4

P5 -12.81 9.22 -118.1 MP5

-4.89 11.8 -57.9

EV2 - Backfill on Heel EV2 13.08 11.00 143.9 MEV2

EV1 - Fill on Toe EV1 2.21 1.88 4.1 MEV1

15.29 9.7 148.0

LL Live Load 0.0 0.00 0.0 MLL

0.0 0.0

EH Horiz. Earth Load PehS 2.25 -1.00 -1.2 MEH

WS Hydrostatic Pressure PehW 9.03 -3.67 -33.1 MPHW

ID Ice/Debris Force Pice/debris 0.00 -15.00 0.0 MID

W Wind Load Pwind 0.00 -15.00 0.0 MW

11.3 -34.3

Fargo Floodwalls

Ve

rtica

l Lo

ad

s

Applied Force Calculations

Total

P (kips)DescriptorLoad

Vertical Loads

DC

Horizontal Loads

EV

Total

Total

Total

Page 239: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Factor Equation Description

EM 1110-2-2502

Reference

PehW2 = 3.955 kips Water on Dry side for Overturning Calcs Figure 4-5

Pehw2d 0.705 ft Moment Arm for Pehw2 N/A

Peh2sb 1.360 ft Dry Side Soil to bottom of Footing Para. 4-8

Peh2sbd 2.333 kips Moment Arm for Peh2sb N/A

RS = 4.581 kips Resisting soil on Dry side to create equilibrium Para. 4-8

RSd -1 ft Moment Arm for RS N/A

Including Uplift

ΣVo 10.399 kips Total Vertical Load Figure 4-5

ΣMo 56.330 kip-ft 4-1

XR 5.417 ft Resultant Location 4-1

bl 16.25121 ft Length of Base in Compression 4-2

b% 95.595354 Percent of Base in Compression 4-2

Required b% 75 Required Base in Compression Appendix F

OK

Neglecting Uplift

ΣVon 23.205 kips Total Vertical Load Figure 4-5

ΣMon 174.430 kip-ft 4-1

XRn 7.517 ft Resultant Location 4-1

bln 22.550724 ft Length of Base in Compression 4-2

b%n 132.65132 Percent of Base in Compression 4-2

Required b%n 75 Required Base in Compression Appendix 4

OK

Global Stability

Overturning:

Fargo Floodwalls

$%� + $%" + $? + $@ + $A + �B1 + �B2

5)�2

�2 + �� +5�7:9�, + 567��789�,

2(�)

�2 + ��3

5)�2

�2 + ��3

�2 + �� + 5�7:9�,−�2

� +567��789�, − 5�7:9�,

2�

−2�3

/$CD"

−�2

EF1� +EF1" +EF? +EF@ +EFA +EGH� +EGH" +EFIJ +EKL +EJ + M% M%N + $CD"�6 $CD"�6N + $CD"($CD"N)(O7)(%E�)

ΣE7

ΣB7

3QR

S�!(100)

$%� + $%" + $? + $@ + �B1 + �B2

EF1� +EF1" +EF? +EF@ +EGH� +EGH" +EFIJ +EKL +EJ + M% M%N + $CD"�6 $CD"�6N + $CD"($CD"N)(O7)(%E�)

ΣE7T

ΣB7T

3QRT

S�T!

(100)

Total Overturning

Moment about

point o

(�2 + ��)(ɤ�� − ρ)(�2 + ��)/2

$CD − $CD" − $CD"�6(O7)(%E�)

Total Overturning

Moment about

point o

2/26/2016

Page 240: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Factor Equation Description

EM 1110-2-2502

Reference

Sliding Along Angle Between Bottom of Key and Bottom of Footing on Dry Side Figure 4-11

SBFa 1.722 kips Soil Below Footing on Sliding Surface Figure 4-11

Pwda 3.684 kips Water on Dry Side Figure 4-11

ΣVa 10.943 kips Sum of Vertical Forces on Sliding Surface Figure 4-11

αa 0.117 radians Angle of slip plane to horizontal plane Figure 4-11

6.710 degrees

Drained Condition

RFswad 5.142 kips Resisting Force from passive soil and water on dry side Figure 4-11

ΣHad 3.89 kips Sum of Horizontal Forces Figure 4-11

N'ad 11.323 kips Normal Force to Sliding Surface Figure 4-11

Tad 2.584 kips Tangential Force to Sliding Surface Figure 4-11

SSad 5.769 kips Drained Shear Strength 4-12

FSad 2.233 Factor of Safety 4-12

OK

Undrained Condition

RFswau 4.232 kips Resisting Force from passive soil and water on dry side Figure 4-11

ΣHau 4.800 kips Sum of Horizontal Forces Figure 4-11

N'au 11.429 kips Normal Force to Sliding Surface Figure 4-11

Tau 3.488 kips Tangential Force to Sliding Surface Figure 4-11

SSau 11.412 kips Undrained Shear Strength 4-12

FSau 3.272 Factor of Safety 4-12

OK

Sliding:

Fargo Floodwalls

Global Stability

� −�!

2(! − )(ɤ��)

5)2

�2 + �� +5) + 5)�

2(�)

%UV + �B1 + �B2 + $%� + $%" + $? + $@ − $A�

atan�!

$+ + $CD" − $+ Y: (.5)

$CD + $.��/+�6 .� + $.T+ − M��+

ΣB cos \ + Σ�+sin(\)

Σ�+ cos \ − ΣBsin(\)

]′+tan(∅)

%%+_+

$+ + $CD" − $+ (.5)

$CD + $.��/+�6 .� + $.T+ − M��`

ΣB cos \ + Σ�`sin(\)

Σ�` cos \ − ΣBsin(\)

��(%E�)√(!" + �")

%%`_`

2/26/2016

Page 241: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Factor Equation Description

EM 1110-2-2502

Reference

Sliding Along Horizontal Surface along Bottom of Key Figure 4-8

SBFs 3.717 kips Soil Below Footing on Sliding Surface

Pwds 3.310 kips Water on Dry Side Figure 4-11

ΣVs 11.638 kips Sum of Vertical Forces on Sliding Surface Figure 4-11

αs = 0 radians Angle of slip plane to horizontal plane Figure 4-11

0 degrees

Drained Condition

RFswsd 5.266 kips Resisting Force from passive soil and water on dry side Figure 4-11

ΣHsd 3.77 kips Sum of Horizontal Forces Figure 4-11

N'sd 11.638 kips Normal Force to Sliding Surface Figure 4-11

Tsd 3.765 kips Tangential Force to Sliding Surface Figure 4-11

SSsd 5.930 kips Drained Shear Strength 4-12

FSsd 1.575 Factor of Safety 4-12

OK

Undrained Condition

RFswsu 4.045 kips Resisting Force from passive soil and water on dry side Figure 4-11

ΣHsu 4.987 kips Sum of Horizontal Forces Figure 4-11

N'su 11.638 kips Normal Force to Sliding Surface Figure 4-11

Tsu 4.987 kips Tangential Force to Sliding Surface Figure 4-11

SSsu 11.333 kips Undrained Shear Strength 4-12

FSsu 2.273 Factor of Safety 4-12

OK

Global Stabilty

Fargo Floodwalls

�(! − )(ɤ��)

5>2(�2 + �� + �)

%UV� + �B1 + �B2 + $%� + $%" + $? + $@ − $A��

$+� + $CD" − $+� Y: (.5)

$CD + $.��/+�6 .� + $.T+ − M���+

ΣB� cos \� + Σ��+sin(\�)

Σ��+ cos \� − ΣB�sin(\�)

]′�+tan(∅)

%%�+_�+

.

$+� + $CD" − $+� (.5)

$CD + $.��/+�6 .� + $.T+ − M���`

ΣB� cos \� + Σ��`sin(\�)

Σ��` cos \� − ΣB�sin(\�)

�� %E� !

%%�`_�`

2/26/2016

Page 242: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Factor Equation Description

EM 1110-2-2502

Reference

Bearing Along Angle Between Bottom of Key and Bottom of Footing on Dry Side Eqns. From 1110-2-2502

ΣHadb 5.35 kips Sum of Horizontal Forces Figure 4-11

N'adb 11.493 kips Normal Force to Bearing Surface Figure 4-11

Tadb 4.032 kips Tangential Force to Bearing Surface Figure 4-11

SBFad 10.500005 ft Moment Arm for SBFa Figure 4-11

ΣMoaB 65.327363 kip-ft

XraB 5.970 ft Resultant Location Figure 4-11

eaB 2.589 ft Eccentricity of resultant Figure 4-11

B'aB 11.939256 ft Effective Base for Bearing Sec 5-2

Drained Condition

q0 = 0.333 k/ft2 Effective Overburden Pressure

y = 1.6007185

Nq = 13.175528 Bearing Capacity Factor 5-3a

Nc = 23.895844 Bearing Capacity Factor 5-3b

Nϒ = 9.4442971 Bearing Capacity Factor 5-3d

εcd = 1.16 Embedment Factor 5-4a

εqd = 1.08 Embedment Factor 5-4c

δd = 0.34 5-5

19.330524

εqi = 0.62 Inclination Factor 5-5a

εϒi = 0.08 Inclination Factor 5-5b

Q1 = 0.00 5-2

Q2 = 2.9267926 5-2

Q3 = 0.2731585 5-2

Qd = 38.205036 5-2

FS = 5-1

3.3241783 Eq. 5-1 Bearing Criteria Satisfied

Global Stabilty

Fargo Floodwalls

Bearing:

ΣE7U

ΣB )*)�2

− QbU

)*)� − 2(CU)

(C,) tan 45 +∅2

de∅ > 0, (]h−1)ijklm(∅)

]h − 1 tan(1.4∅)

1 + 0.2� !n tan 45 +

∅2

de∅ > 0, 1 + 0.1� !n tan 45 +

∅2

atanΣ�ΣB

1 −o+90

"

1 −o+∅

"

q�+(qh.)(rs)(]h)

qh+(qh.)(rs)(]h)

qh+ qh. ɤ�� − ɤ�� ]t /2

!′(u1 + u2 + u3)

(ɤ�� − ɤ�� )(� )

v(tan ∅)

Normal Component to the base of the structure of the

ultimate bearing capacity

EF1� +EF1" +EF? +EF@ + $A� $A�N + EGH� +EGH" +EGI +EFIJ +EKL +EJ + $CD"�($CD"�N)(O7)(%E�)+$+ $CD"N + %UV %UVN

$CD + $.��/+�6 .� + $.T+ + $CD�(O7)(%E�) − $+ − $CD"�(O7)(%E�)

ΣB cos \ + Σ�+6sin(\)

Σ�+6 cos \ − ΣBsin(\)

2(! − )3

2/26/2016

Page 243: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Factor Equation Description

EM 1110-2-2502

Reference

q0 = 0.333 k/ft2 Effective Overburden Pressure

y = 0

Nq = 1 Bearing Capacity Factor 5-3a

Nc = 5.14 Bearing Capacity Factor 5-3c

Nϒ = 0 Bearing Capacity Factor 5-3d

εcd = 1.10 Embedment Factor 5-4a

εqd = 1.00 Embedment Factor 5-4b

δd = 0.34 5-5

19.330524

εqi = 0.62 Inclination Factor 5-5a

εϒi = 0.08 Inclination Factor 5-5b

Q1 = 3.49 5-2

Q2 = 0.2053161 5-2

Q3 = 0 5-2

Qd = 44.088477 5-2

FS = 5-1

3.8360901

Drained q'max 1.30 Undrained q'max 1.30

Bearing Criteria Satisfied

Undrained Condition

Global Stabilty

Fargo Floodwalls

(ɤ�� − ɤ�� )(� )

v(tan ∅)

de∅ = 0, 5.14

de∅ = 0, 1

(C,)tan(45 +∅2)

]h − 1 tan(1.4∅)

1 + 0.2� !n tan 45 +

∅2

atanΣ�ΣB

1 −o+90

"

1 −o+∅

"

q�+(qh.)(rs)(]h)

qh+(qh.)(rs)(]h)

qh+ qh. ɤ�� − ɤ�� ]t /2

!′(u1 + u2 + u3)Normal Component to the base of the structure of the

ultimate bearing capacity

2/26/2016

Page 244: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Factor Equation Description

Load Factor 1.7

b 16.25121 Effective base in compression

qmin 0

Uws 0.676013 Uplift Pressure Wet side of Stem

Utws 9.659888 Total water uplift on wet side

Utwsd 6.391352 Moment arm for U tws

VLw 7.395112 Vertical load on wet side

Bws 0.897423 Bearing at wet side of stem

Btws 5.048546 Total Bearing uplift on wet side of footing

Btwsd 3.7504 Moment arm for B tws

Total Shear on Heel 3.989161

Total Moment on Heel 36.81552

Total Shear on Stem 10.27 Total Shear on Stem

Total Moment on Stem 38.90482

Uds 0.652666 Uplift Pressure Dry side of Stem

Utds 2.316169 Total water uplift on dry side

Utdsd 1.839562 Moment arm for U tds

VLd 1.021331 Vertical load on dry side

Bds 0.997126 Bearing at dry side of stem

Btds 4.30005 Total Bearing uplift on dry side of footing

Btdsd 1.956515 Moment arm for B tds

Total Shear on Toe -5.57382 Total Shear on Toe

Total Moment on Toe -10.9073 Total Moment on Toe

Utkw 0.9375 Water Pressure at top of key (wet side)

Total Shear on Key -7.65 Total Shear on Key

Total Moment on Key -7.66624 Total Moment on Key

Forces for Reinforcement/Shear

Heel Forces

Stem Forces

Toe Forces

Key Forces

$KL +$J + �

2��� +�3 ɤ�� − ρ

�32

Y7 −�2 ɤ�� − ρ�22

(Y7) jlN�likjb

$KL ��� + $J $.T+N −�12

+ �

2���

���

3+�3 ɤ�� − ρ

�32

�33

(Y7) − �2 ɤ�� − ρ�22

�23

(Y7) ( jlN�likjb)

x�7:9�, −x�7:9�, −x)�

! − � −

x�7:9�, −x�

2C − L +

x) +x67��789�,

2( )

x�� − 2

� − +x�7:9�, − x�

22 � −

3� − + x67��789�, � −

2

+x) −x67��789�,

2 � −

3

/x��

�B2+ � � ρ� +� ρ� +$@ − x��

r8z − r8z − r8.T

S(� + �)

(B − !��)( jlN�likjb)

�B2+ � � ρ� +� ρ� +$@�2

− x�� x��N − !�� !��N ( jlN�likjb)

x�7:9�, −x�7:9�, −x)�

! − � + � −

x+� +x)�

2(�)

[x)� ��2

+x+� −x)�

2(�3)(�)]/x�+�

�B1 + $? +� �� ρ� −x�+�

r8z − r8z − r8.T

S(�)

r8z + !+�2

(�)

!+� ��2

+r8z −!+�

22�3

� /!�+�

(B + −!�+�)( jlN�likjb)

�B1+ $? +� �� ρ��2

−x�+� x�+�N − !�+� !�+�N ( jlN�likjb)

}eS% ≥ 100,!, S�

}eS% ≥ 100,r8z

!1−

6C!

, 0

}er8.T = 0,!�

2S − � − � , (

!�+ r8.T

2)(�)

}er8.T = 0,S − � − �

3, r8.T �

�2

+!�− r8.T

2�

�3

/!��

x�9 +x=

2� −

x67��789�, +x�7:9�,

2� − M% jlN�likjb

x�9 ��2

+x= − x�9

2�

2�3

−x�7:9�, ��2

−x67��789�, −x�7:9�,

2�

2�3

− M%�2

( jlN�likjb)

(��� +�)(�)

Total

Moment

on Stem

Page 245: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Input: Values in red need to be entered by user Designer: LJB Date: 2/26/2016

Check: Date:Project Number:

Bridge No./Structure ID:

Owner:

Mu = 36.81552 k-ft 21.656 k-ft concrete wall thickness (t) 24 in.

Vu = 3.989161 k 2.347 k Clear Cover 4 in.

Nu = 0.0 k 0.0 k d dimension 19.625 in.

(factored) (service)150 psi

Bar Size = 6

Enter Reinforcement Spacing = 6 in. Reinforcement yield strength = 60 ksiAs = 0.88 in2

Concrete yield strength = 4.5 ksi

Enter design width = 12 in.Concrete yield strength = 4500 psi

Bar Size = None Ec = 4066.84 ksi

Enter Reinforcement Spacing = 6 in. Es = 29000 ksi

Av = 0.00 in2Lightweight Concrete? NO

Mod Factor ƛ = 1Does ACI 350 Apply? NO Flexure φ = 0.9

Shear φ = 0.75

Flexure Analysis: Mr = Phi(Mn)

For members without axial loads.

a = 1.150

Mn = 1,005.8 k-in

Mn= 83.8 k-ft

φMn = 75.44 k-ft

Minimum Reinforcement: (ACI 318 10.5)

As, min = 0.79 in2 not less than 0.785 in2

ACI 318 - 10.5 Requirement = Pass If Fail ------------>

1.33*Mu = 48.96 k-ft

As,req = 0.51 in2

As, min = 0.79 in2

She

ar

7438-009 Section Being Designed:Heel of Footing

Fargo Floodwalls Layer of Steel:

City of Fargo

Modulus of elasticity based on concrete unit weight of

Fle

xure

OK - Section and reinforcement sufficient for Flexural Resistance

*If the requirements of 10.5 are not satisfied then As needs to be at least 1/3 greater than that required by analysis. (10.5.3)

OK

��,8.T =? >��

>�SNS5kmjk�C<<kDlm200S

+

>�

ET = �� ∗ e, ∗ N −l2

l =�� ∗ e,

0.85 ∗ e′� ∗ S

��, �h =.85en�S

e,N − N" −

2E`

.85e′�S

Page 246: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Designer: LJB Date: 2/26/2016

Check: Date:

ACI 318

Governing Specification: USACE

Min. Steel Requirement:

As/Ag (Includes both faces) ≥ 0.0050 EM 1110-2-2104 2.8

Enter Reinforcing configuration of opposite mat:Bar Size = 5

Enter Reinforcement Spacing = 6 in.As = 0.62 in2

Required Min. Steel (both faces): 1.44 sq. in/ft

Max. Spacing Requirements:

s≤5h or s≤18.0 in. 120 in18 in Structural walls/footings:

or for componenets over 36.0 in Spacing can be ≤ 12 in. on faces > 18" thick

s≤12.0 12 in

Maximum Spacing Allowed: 18 in

ACI 350

Min. Steel Requirement: ACI 350-06 7.12.2.1 Minimum reinforcement ratio for Grade 60 steel

Length Between JointsRatio As/Ag 0.0030 ≤ 20 ft

20-30 ft

Steel in opposite face: 0.62 in230-40 ft≥ 40ft

Required Min. Steel: 0.86 sq. in/ft

Max. Spacing Requirements: ACI 350-065 7.12.2.2

Shrinkage and Temperature Reinforcement shall not be spaced greater than12in. apart

Maximum Spacing Allowed: 12 in

Layer of Steel:

Section Being Designed:

Heel of Footing

0

T&S Steel:

T&S Steel Requirement Met

T&S Steel:

Ratio (As/Ag)0.0030.003

0.0040.005

T&S Steel Requirement Met

Max Spacing Requirement Met

Max Spacing Requirement Met

Page 247: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Designer: LJB Date: 2/26/2016Check: Date:

Shear Analysis:

φVn ≥ Vu (ACI 318 Eq. 11-1)

Vn = Vc + Vs (ACI 318 Eq. 11-2)

Calculate d:

d = 19.625 (Loads placed at 'top' of member.) (ACI 318 11.1.3.1)

dv = 19.625 in

Calc. Vc

For members subject to shear & flexure only.Vc = 2* ƛ * √fc * bv * dv ƛ = 1.0 (ACI 318 Eq. 11-3)

bv = 12 in

Vc = 31.6 kips dv = 19.63 in

For members subject to axial compression.Vc = 2*(1 + Nu/(2000*Ag)) * ƛ * √fc * bv * dv Nu = 0 lb (ACI 318 Eq. 11-4)

Ag = 288 in2

Vc = 758.3 kips

Vc = 31.6 kips

Calc. Vs

Shear Reinforcement perpendicular to axis.

Vs = Av * fyt * d / s Av = 0.00 in2

s = 6 in

Vs = 0.0 kips

Vn = 31.6 kips

θVn = 23.7 kips Vu= 3.989161

θVn = 23.7 kips

Section Being Designed:Heel of Footing

Layer of Steel:0

OK - Section Adequate for Shear Resistance

Page 248: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Designer: LJB Date: 2/26/2016Check: Date:

Crack Control: ACI 318

Distribution of flexural reinforcement in beams and one-way slabs.

The spacing of reinforcement closest to the tension face, s, shall not exceed:

cc = 4 in (Clear cover)

fs = Calculated stress from unfactored Moment.

But not greater than Mserv = 259.9 kip-in

n = 7.13

x = 4.038 in NA to extreme comp fiber

y = 15.587 in NA to tension centroid

I = 1787.9 in4

fs = 16155.67 psi

s ≤ 27.1 in

Crack Control: ACI 350

Actual Stress:fs = 16155.67 psi

ACI 350-06 10.6.4.1, for normal environmental exposures as defined in 10.6.4.5

β= 1.28

s= 6 in.db= 0.75 in.

fs,max= 32.65 ksi26781.3 psi > 16155.67 psi

Section Being Designed:Heel of Footing

Layer of Steel:0

OK - Spacing Adequate

fs,max =320

β*SQRT(s2+4(2+

db∕2)

2)

OK - Spacing Adequate

< = 15@s,sss

>�− 2.5 i�

12(40,000

e�)

e� = m ∗E�� � ∗ �

}

� =�� ∗ �� ∗ m + 2 ∗ S ∗ N ∗ m

S−�� ∗ mS

� = N − �

} =13− S ∗ �? + m ∗ �� ∗ �"

Page 249: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Input: Values in red need to be entered by user Designer: LJB Date: 2/26/2016

Check: Date:Project Number:

Bridge No./Structure ID:

Owner:

Mu = 38.90482 k-ft 22.885 k-ft concrete wall thickness (t) 15 in.

Vu = 10.26603 k 6.039 k Clear Cover 3 in.

Nu = 0.0 k 0.0 k d dimension 11.625 in.

(factored) (service)150 psi

Bar Size = 6

Enter Reinforcement Spacing = 6 in. Reinforcement yield strength = 60 ksiAs = 0.88 in2

Concrete yield strength = 4.5 ksi

Enter design width = 12 in.Concrete yield strength = 4500 psi

Bar Size = None Ec = 4066.84 ksi

Enter Reinforcement Spacing = 6 in. Es = 29000 ksi

Av = 0.00 in2Lightweight Concrete? NO

Mod Factor ƛ = 1Does ACI 350 Apply? NO Flexure φ = 0.9

Shear φ = 0.75

Flexure Analysis: Mr = Phi(Mn)

For members without axial loads.

a = 1.150

Mn = 583.4 k-in

Mn= 48.6 k-ft

φMn = 43.76 k-ft

Minimum Reinforcement: (ACI 318 10.5)

As, min = 0.47 in2 not less than 0.465 in2

ACI 318 - 10.5 Requirement = Pass If Fail ------------>

1.33*Mu = 51.74 k-ft

As,req = 0.94 in2

As, min = 0.47 in2

She

ar

7438-009 Section Being Designed:Stem

Fargo Floodwalls Layer of Steel:

City of Fargo

Modulus of elasticity based on concrete unit weight of

Fle

xure

OK - Section and reinforcement sufficient for Flexural Resistance

*If the requirements of 10.5 are not satisfied then As needs to be at least 1/3 greater than that required by analysis. (10.5.3)

OK

��,8.T =? >��

>�SNS5kmjk�C<<kDlm200S

+

>�

ET = �� ∗ e, ∗ N −l2

l =�� ∗ e,

0.85 ∗ e′� ∗ S

��, �h =.85en�S

e,N − N" −

2E`

.85e′�S

2/26/201611:52 AM

Page 250: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Designer: LJB Date: 2/26/2016

Check: Date:

ACI 318

Governing Specification: USACE

Min. Steel Requirement:

As/Ag (Includes both faces) ≥ 0.0028 EM 1110-2-2104 2.8

Enter Reinforcing configuration of opposite mat:Bar Size = 5

Enter Reinforcement Spacing = 12 in.As = 0.31 in2

Required Min. Steel (both faces): 0.50 sq. in/ft

Max. Spacing Requirements:

s≤5h or s≤18.0 in. 75 in18 in Structural walls/footings:

or for componenets over 36.0 in Spacing can be ≤ 12 in. on faces > 18" thick

s≤12.0 12 in

Maximum Spacing Allowed: 18 in

ACI 350

Min. Steel Requirement: ACI 350-06 7.12.2.1 Minimum reinforcement ratio for Grade 60 steel

Length Between JointsRatio As/Ag 0.0030 ≤ 20 ft

20-30 ft

Steel in opposite face: 0.31 in230-40 ft≥ 40ft

Required Min. Steel: 0.54 sq. in/ft

Max. Spacing Requirements: ACI 350-065 7.12.2.2

Shrinkage and Temperature Reinforcement shall not be spaced greater than12in. apart

Maximum Spacing Allowed: 12 in

Layer of Steel:

Section Being Designed:

Stem

0

T&S Steel:

T&S Steel Requirement Met

T&S Steel:

Ratio (As/Ag)0.0030.003

0.0040.005

T&S Steel Requirement Met

Max Spacing Requirement Met

Max Spacing Requirement Met

2/26/201611:52 AM

Page 251: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Designer: LJB Date: 2/26/2016Check: Date:

Shear Analysis:

φVn ≥ Vu (ACI 318 Eq. 11-1)

Vn = Vc + Vs (ACI 318 Eq. 11-2)

Calculate d:

d = 11.625 (Loads placed at 'top' of member.) (ACI 318 11.1.3.1)

dv = 11.625 in

Calc. Vc

For members subject to shear & flexure only.Vc = 2* ƛ * √fc * bv * dv ƛ = 1.0 (ACI 318 Eq. 11-3)

bv = 12 in

Vc = 18.7 kips dv = 11.63 in

For members subject to axial compression.Vc = 2*(1 + Nu/(2000*Ag)) * ƛ * √fc * bv * dv Nu = 0 lb (ACI 318 Eq. 11-4)

Ag = 180 in2

Vc = 280.7 kips

Vc = 18.7 kips

Calc. Vs

Shear Reinforcement perpendicular to axis.

Vs = Av * fyt * d / s Av = 0.00 in2

s = 6 in

Vs = 0.0 kips

Vn = 18.7 kips

θVn = 14.0 kips Vu= 10.26603

θVn = 14.0 kips

Section Being Designed:Stem

Layer of Steel:0

OK - Section Adequate for Shear Resistance

2/26/201611:52 AM

Page 252: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Designer: LJB Date: 2/26/2016Check: Date:

Crack Control: ACI 318

Distribution of flexural reinforcement in beams and one-way slabs.

The spacing of reinforcement closest to the tension face, s, shall not exceed:

cc = 3 in (Clear cover)

fs = Calculated stress from unfactored Moment.

But not greater than Mserv = 274.6 kip-in

n = 7.13

x = 3.003 in NA to extreme comp fiber

y = 8.622 in NA to tension centroid

I = 574.8 in4

fs = 29374.05 psi

s ≤ 12.9 in

Crack Control: ACI 350

Actual Stress:fs = 29374.05 psi

ACI 350-06 10.6.4.1, for normal environmental exposures as defined in 10.6.4.5

β= 1.39

s= 6 in.db= 0.75 in.

fs,max= 30.05 ksi26781.3 psi < 29374.05 psi

Section Being Designed:Stem

Layer of Steel:0

OK - Spacing Adequate

fs,max =320

β*SQRT(s2+4(2+

db∕2)

2)

ACI 350 Requirements Not Met

< = 15@s,sss

>�− 2.5 i�

12(40,000

e�)

e� = m ∗E�� � ∗ �

}

� =�� ∗ �� ∗ m + 2 ∗ S ∗ N ∗ m

S−�� ∗ mS

� = N − �

} =13− S ∗ �? + m ∗ �� ∗ �"

2/26/201611:52 AM

Page 253: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Input: Values in red need to be entered by user Designer: LJB Date: 2/26/2016

Check: Date:Project Number:

Bridge No./Structure ID:

Owner:

Mu = 10.90726 k-ft 6.416 k-ft concrete wall thickness (t) 24 in.

Vu = 5.573823 k 3.279 k Clear Cover 4 in.

Nu = 0.0 k 0.0 k d dimension 19.6875 in.

(factored) (service)150 psi

Bar Size = 5

Enter Reinforcement Spacing = 6 in. Reinforcement yield strength = 60 ksiAs = 0.62 in2

Concrete yield strength = 4.5 ksi

Enter design width = 12 in.Concrete yield strength = 4500 psi

Bar Size = None Ec = 4066.84 ksi

Enter Reinforcement Spacing = 6 in. Es = 29000 ksi

Av = 0.00 in2Lightweight Concrete? NO

Mod Factor ƛ = 1Does ACI 350 Apply? NO Flexure φ = 0.9

Shear φ = 0.75

Flexure Analysis: Mr = Phi(Mn)

For members without axial loads.

a = 0.810

Mn = 717.3 k-in

Mn= 59.8 k-ft

φMn = 53.80 k-ft

Minimum Reinforcement: (ACI 318 10.5)

As, min = 0.79 in2 not less than 0.7875 in2

ACI 318 - 10.5 Requirement = Fail If Fail ------------>

1.33*Mu = 14.51 k-ft

As,req = 0.15 in2

As, min = 0.15 in2

She

ar

7438-009 Section Being Designed:Toe of Footing

Fargo Floodwalls Layer of Steel:

City of Fargo

Modulus of elasticity based on concrete unit weight of

Fle

xure

OK - Section and reinforcement sufficient for Flexural Resistance

*If the requirements of 10.5 are not satisfied then As needs to be at least 1/3 greater than that required by analysis. (10.5.3)

OK

��,8.T =? >��

>�SNS5kmjk�C<<kDlm200S

+

>�

ET = �� ∗ e, ∗ N −l2

l =�� ∗ e,

0.85 ∗ e′� ∗ S

��, �h =.85en�S

e,N − N" −

2E`

.85e′�S

2/26/201611:52 AM

Page 254: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Designer: LJB Date: 2/26/2016

Check: Date:

ACI 318

Governing Specification: USACE

Min. Steel Requirement:

As/Ag (Includes both faces) ≥ 0.0050 EM 1110-2-2104 2.8

Enter Reinforcing configuration of opposite mat:Bar Size = 6

Enter Reinforcement Spacing = 6 in.As = 0.88 in2

Required Min. Steel (both faces): 1.44 sq. in/ft

Max. Spacing Requirements:

s≤5h or s≤18.0 in. 120 in18 in Structural walls/footings:

or for componenets over 36.0 in Spacing can be ≤ 12 in. on faces > 18" thick

s≤12.0 12 in

Maximum Spacing Allowed: 18 in

ACI 350

Min. Steel Requirement: ACI 350-06 7.12.2.1 Minimum reinforcement ratio for Grade 60 steel

Length Between JointsRatio As/Ag 0.0030 ≤ 20 ft

20-30 ft

Steel in opposite face: 0.88 in230-40 ft≥ 40ft

Required Min. Steel: 0.86 sq. in/ft

Max. Spacing Requirements: ACI 350-065 7.12.2.2

Shrinkage and Temperature Reinforcement shall not be spaced greater than12in. apart

Maximum Spacing Allowed: 12 in

Layer of Steel:

Section Being Designed:

Toe of Footing

0

T&S Steel:

T&S Steel Requirement Met

T&S Steel:

Ratio (As/Ag)0.0030.003

0.0040.005

T&S Steel Requirement Met

Max Spacing Requirement Met

Max Spacing Requirement Met

2/26/201611:52 AM

Page 255: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Designer: LJB Date: 2/26/2016Check: Date:

Shear Analysis:

φVn ≥ Vu (ACI 318 Eq. 11-1)

Vn = Vc + Vs (ACI 318 Eq. 11-2)

Calculate d:

d = 19.688 (Loads placed at 'top' of member.) (ACI 318 11.1.3.1)

dv = 19.688 in

Calc. Vc

For members subject to shear & flexure only.Vc = 2* ƛ * √fc * bv * dv ƛ = 1.0 (ACI 318 Eq. 11-3)

bv = 12 in

Vc = 31.7 kips dv = 19.69 in

For members subject to axial compression.Vc = 2*(1 + Nu/(2000*Ag)) * ƛ * √fc * bv * dv Nu = 0 lb (ACI 318 Eq. 11-4)

Ag = 288 in2

Vc = 760.7 kips

Vc = 31.7 kips

Calc. Vs

Shear Reinforcement perpendicular to axis.

Vs = Av * fyt * d / s Av = 0.00 in2

s = 6 in

Vs = 0.0 kips

Vn = 31.7 kips

θVn = 23.8 kips Vu= 5.573823

θVn = 23.8 kips

Section Being Designed:Toe of Footing

Layer of Steel:0

OK - Section Adequate for Shear Resistance

2/26/201611:52 AM

Page 256: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Designer: LJB Date: 2/26/2016Check: Date:

Crack Control: ACI 318

Distribution of flexural reinforcement in beams and one-way slabs.

The spacing of reinforcement closest to the tension face, s, shall not exceed:

cc = 4 in (Clear cover)

fs = Calculated stress from unfactored Moment.

But not greater than Mserv = 77.0 kip-in

n = 7.13

x = 3.458 in NA to extreme comp fiber

y = 16.229 in NA to tension centroid

I = 1329.9 in4

fs = 6699.907 psi

s ≤ 79.6 in

Crack Control: ACI 350

Actual Stress:fs = 6699.907 psi

ACI 350-06 10.6.4.1, for normal environmental exposures as defined in 10.6.4.5

β= 1.27

s= 6 in.db= 0.625 in.

fs,max= 33.37 ksi26781.3 psi > 6699.907 psi

Section Being Designed:Toe of Footing

Layer of Steel:0

OK - Spacing Adequate

fs,max =320

β*SQRT(s2+4(2+

db∕2)

2)

OK - Spacing Adequate

< = 15@s,sss

>�− 2.5 i�

12(40,000

e�)

e� = m ∗E�� � ∗ �

}

� =�� ∗ �� ∗ m + 2 ∗ S ∗ N ∗ m

S−�� ∗ mS

� = N − �

} =13− S ∗ �? + m ∗ �� ∗ �"

2/26/201611:52 AM

Page 257: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Input: Values in red need to be entered by user Designer: LJB Date: 2/26/2016

Check: Date:Project Number:

Bridge No./Structure ID:

Owner:

Mu = 7.666244 k-ft 4.510 k-ft concrete wall thickness (t) 15 in.

Vu = 7.645077 k 4.497 k Clear Cover 3 in.

Nu = 0.0 k 0.0 k d dimension 11.5625 in.

(factored) (service)150 psi

Bar Size = 7

Enter Reinforcement Spacing = 12 in. Reinforcement yield strength = 60 ksiAs = 0.60 in2

Concrete yield strength = 4.5 ksi

Enter design width = 12 in.Concrete yield strength = 4500 psi

Bar Size = None Ec = 4066.84 ksi

Enter Reinforcement Spacing = 6 in. Es = 29000 ksi

Av = 0.00 in2Lightweight Concrete? NO

Mod Factor ƛ = 1Does ACI 350 Apply? NO Flexure φ = 0.9

Shear φ = 0.75

Flexure Analysis: Mr = Phi(Mn)

For members without axial loads.

a = 0.784

Mn = 402.1 k-in

Mn= 33.5 k-ft

φMn = 30.16 k-ft

Minimum Reinforcement: (ACI 318 10.5)

As, min = 0.47 in2 not less than 0.4625 in2

ACI 318 - 10.5 Requirement = Pass If Fail ------------>

1.33*Mu = 10.20 k-ft

As,req = 0.18 in2

As, min = 0.47 in2

She

ar

7438-009 Section Being Designed:Key

Fargo Floodwalls Layer of Steel:

City of Fargo

Modulus of elasticity based on concrete unit weight of

Fle

xure

OK - Section and reinforcement sufficient for Flexural Resistance

*If the requirements of 10.5 are not satisfied then As needs to be at least 1/3 greater than that required by analysis. (10.5.3)

OK

��,8.T =? >��

>�SNS5kmjk�C<<kDlm200S

+

>�

ET = �� ∗ e, ∗ N −l2

l =�� ∗ e,

0.85 ∗ e′� ∗ S

��, �h =.85en�S

e,N − N" −

2E`

.85e′�S

2/26/201611:52 AM

Page 258: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Designer: LJB Date: 2/26/2016

Check: Date:

ACI 318

Governing Specification: USACE

Min. Steel Requirement:

As/Ag (Includes both faces) ≥ 0.0050 EM 1110-2-2104 2.8

Enter Reinforcing configuration of opposite mat:Bar Size = 7

Enter Reinforcement Spacing = 12 in.As = 0.60 in2

Required Min. Steel (both faces): 0.90 sq. in/ft

Max. Spacing Requirements:

s≤5h or s≤18.0 in. 75 in18 in Structural walls/footings:

or for componenets over 36.0 in Spacing can be ≤ 12 in. on faces > 18" thick

s≤12.0 12 in

Maximum Spacing Allowed: 18 in

ACI 350

Min. Steel Requirement: ACI 350-06 7.12.2.1 Minimum reinforcement ratio for Grade 60 steel

Length Between JointsRatio As/Ag 0.0030 ≤ 20 ft

20-30 ft

Steel in opposite face: 0.60 in230-40 ft≥ 40ft

Required Min. Steel: 0.54 sq. in/ft

Max. Spacing Requirements: ACI 350-065 7.12.2.2

Shrinkage and Temperature Reinforcement shall not be spaced greater than12in. apart

Maximum Spacing Allowed: 12 in

Layer of Steel:

Section Being Designed:

Key

0

T&S Steel:

T&S Steel Requirement Met

T&S Steel:

Ratio (As/Ag)0.0030.003

0.0040.005

T&S Steel Requirement Met

Max Spacing Requirement Met

Max Spacing Requirement Met

2/26/201611:52 AM

Page 259: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Designer: LJB Date: 2/26/2016Check: Date:

Shear Analysis:

φVn ≥ Vu (ACI 318 Eq. 11-1)

Vn = Vc + Vs (ACI 318 Eq. 11-2)

Calculate d:

d = 11.563 (Loads placed at 'top' of member.) (ACI 318 11.1.3.1)

dv = 11.563 in

Calc. Vc

For members subject to shear & flexure only.Vc = 2* ƛ * √fc * bv * dv ƛ = 1.0 (ACI 318 Eq. 11-3)

bv = 12 in

Vc = 18.6 kips dv = 11.56 in

For members subject to axial compression.Vc = 2*(1 + Nu/(2000*Ag)) * ƛ * √fc * bv * dv Nu = 0 lb (ACI 318 Eq. 11-4)

Ag = 180 in2

Vc = 279.2 kips

Vc = 18.6 kips

Calc. Vs

Shear Reinforcement perpendicular to axis.

Vs = Av * fyt * d / s Av = 0.00 in2

s = 6 in

Vs = 0.0 kips

Vn = 18.6 kips

θVn = 14.0 kips Vu= 7.645077

θVn = 14.0 kips

Section Being Designed:Key

Layer of Steel:0

OK - Section Adequate for Shear Resistance

2/26/201611:52 AM

Page 260: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Designer: LJB Date: 2/26/2016Check: Date:

Crack Control: ACI 318

Distribution of flexural reinforcement in beams and one-way slabs.

The spacing of reinforcement closest to the tension face, s, shall not exceed:

cc = 3 in (Clear cover)

fs = Calculated stress from unfactored Moment.

But not greater than Mserv = 54.1 kip-in

n = 7.13

x = 2.537 in NA to extreme comp fiber

y = 9.026 in NA to tension centroid

I = 413.8 in4

fs = 8415.816 psi

s ≤ 63.8 in

Crack Control: ACI 350

Actual Stress:fs = 8415.816 psi

ACI 350-06 10.6.4.1, for normal environmental exposures as defined in 10.6.4.5

β= 1.38

s= 12 in.db= 0.875 in.

fs,max= 17.89 ksi26781.3 psi > 8415.816 psi

Section Being Designed:Key

Layer of Steel:0

OK - Spacing Adequate

fs,max =320

β*SQRT(s2+4(2+

db∕2)

2)

OK - Spacing Adequate

< = 15@s,sss

>�− 2.5 i�

12(40,000

e�)

e� = m ∗E�� � ∗ �

}

� =�� ∗ �� ∗ m + 2 ∗ S ∗ N ∗ m

S−�� ∗ mS

� = N − �

} =13− S ∗ �? + m ∗ �� ∗ �"

2/26/201611:52 AM

Page 261: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Input: Values in red need to be entered by user Designer: LJB Date: 2/26/2016

Check: Date:Project Number:

Bridge No./Structure ID:

Owner:

Mu = 0 k-ft 0.000 k-ft concrete wall thickness (t) 24 in.

Vu = 0 k 0.000 k Clear Cover 5 in.

Nu = 0.0 k 0.0 k d dimension 18.5 in.

(factored) (service)150 psi

Bar Size = 8

Enter Reinforcement Spacing = 12 in. Reinforcement yield strength = 60 ksiAs = 0.79 in2

Concrete yield strength = 4.5 ksi

Enter design width = 12 in.Concrete yield strength = 4500 psi

Bar Size = None Ec = 4066.84 ksi

Enter Reinforcement Spacing = 6 in. Es = 29000 ksi

Av = 0.00 in2Lightweight Concrete? NO

Mod Factor ƛ = 1Does ACI 350 Apply? NO Flexure φ = 0.9

Shear φ = 0.75

ACI 318

Governing Specification: USACE

Min. Steel Requirement:

As/Ag (Includes both faces) ≥ 0.0050 EM 1110-2-2104 2.8

Enter Reinforcing configuration of opposite mat:Bar Size = 8

Enter Reinforcement Spacing = 12 in.As = 0.79 in2

Required Min. Steel (both faces): 1.44 sq. in/ft

Max. Spacing Requirements:

s≤5h or s≤18.0 in. 120 in18 in Structural walls/footings:

or for componenets over 36.0 in Spacing can be ≤ 12 in. on faces > 18" thick

s≤12.0 12 in

Maximum Spacing Allowed: 18 in Max Spacing Requirement Met

T&S Steel:

T&S Steel Requirement Met

She

ar

7438-009 Section Being Designed:Longitudinal Footing Reinforcement

Fargo Floodwalls Layer of Steel:

City of Fargo

Modulus of elasticity based on concrete unit weight of

Fle

xure

Page 262: DESIGN DOCUMENTATION REPORT – APPENDIX F ......geotechnical report predicted settlements for shallow footings of no more than 1 to 1 ½ inches. 3.1.2 Unit Weights For the design,

Input: Values in red need to be entered by user Designer: LJB Date: 2/26/2016

Check: Date:Project Number:

Bridge No./Structure ID:

Owner:

Mu = 0 k-ft 0.000 k-ft concrete wall thickness (t) 15 in.

Vu = 0 k 0.000 k Clear Cover 3 in.

Nu = 0.0 k 0.0 k d dimension 11.6875 in.

(factored) (service)150 psi

Bar Size = 5

Enter Reinforcement Spacing = 12 in. Reinforcement yield strength = 60 ksiAs = 0.31 in2

Concrete yield strength = 4.5 ksi

Enter design width = 12 in.Concrete yield strength = 4500 psi

Bar Size = None Ec = 4066.84 ksi

Enter Reinforcement Spacing = 6 in. Es = 29000 ksi

Av = 0.00 in2Lightweight Concrete? NO

Mod Factor ƛ = 1Does ACI 350 Apply? NO Flexure φ = 0.9

Shear φ = 0.75

ACI 318

Governing Specification: USACE

Min. Steel Requirement:

As/Ag (Includes both faces) ≥ 0.0028 EM 1110-2-2104 2.8

Enter Reinforcing configuration of opposite mat:Bar Size = 5

Enter Reinforcement Spacing = 12 in.As = 0.31 in2

Required Min. Steel (both faces): 0.50 sq. in/ft

Max. Spacing Requirements:

s≤5h or s≤18.0 in. 75 in18 in Structural walls/footings:

or for componenets over 36.0 in Spacing can be ≤ 12 in. on faces > 18" thick

s≤12.0 12 in

Maximum Spacing Allowed: 18 in Max Spacing Requirement Met

T&S Steel:

T&S Steel Requirement Met

She

ar

7438-009 Section Being Designed:Longitudinal Stem Reinforcement

Fargo Floodwalls Layer of Steel:

City of Fargo

Modulus of elasticity based on concrete unit weight of

Fle

xure

2/26/201611:52 AM