EXAMPLE 13 - VEHICLE COLLISION ON A PIER 1 · CT1 F X =579.56 kip F Z = 155.29 kip Figure 2 - CT Load Application. EXAMPLE 13 - VEHICLE COLLISION ON A PIER 4 2018

EXAMPLE 13 - VEHICLE COLLISION ON A PIER 1

CDOT Bridge Design Manual January 2019

Design Example 13

GENERAL INFORMATION

References and Software Used:

AASHTO LRFD 8th edition CDOT Bridge Structural WorksheetsLEAP Bridge Concrete CONNECT Edition, Version 16.02.00.01, Substructure Module

LEAP Bridge Concrete Model Description:

The following were also assumed in modeling the pier in the LEAP Bridge Substructure program: → The end of the column is fixed at the top of the drilled shaft→ The drilled shaft point of fixity is located at 3x drilled shaft diameter = 13.5 ft.→ Total drilled shaft length is 5x drilled shaft diameter = 23 ft.

Material properties used (refer to BDM Section 5.3):Pier cap concrete strength f'c = psiColumn concrete strength f'c = psiDrilled shaft concrete strength f'c = psiConcrete density γc = pcfSteel yield strength fy = ksi

4000.00150.0060.00

EXAMPLE 13 - VEHICLE COLLISION ON A PIER

The pier under design is a middle support of a two-span, 60-ft.-wide (out to out) bridge. Pier cap is 60 ft.long, 4.5 ft. wide, and 4 ft. deep, supported on three 20 ft. tall columns, spaced at 22 ft. Thesuperstructure consists of six BT54 girders spaced at 10.5 ft. with an 8 in. deck. Columns are round, 4 ft.in diameter, supported on drilled shafts, 4.5 ft. in diameter. Refer to Figure 1 for details.

The Designer should use the project geotechnical information and a suitable design tool to determine thedrilled shaft point of fixity and required drilled shaft total length and enter it in the LEAP BridgeSubstructure program.

4500.004500.00

Example 13 illustrates pier design to provide structural resistance to withstand the vehicular collisionusing LEAP Bridge Substructure software. The only load case investigated in this example is ExtremeEvent II. Other load cases are not discussed for this example but should be investigated in the completepier design.

APPENDIX A



Applied Loads:

DC - dead load of structural components and nonstructural attachmentsDW - dead load of wearing surfaces and utilitiesCT - vehicular collision force

DC: 1. Slab and girder dead loads - autogenerated from superstructure input, γ = 150 pcf2. Barrier dead loads - total load per foot = 486 plf (see Structural Worksheets B-606-7B)

DW: 1. Wearing surface total load per foot = 36.67 psf *(57 ft clear roadway width) = 2090.19 plf

In this example, the only loads included in the analysis are collision loads and dead loads, due to theimprobable coincidence of other loads (BDM Section 3.5.2). Designer may choose to include live loads,but in most cases shear from vehicle impact will control the design.

If the LEAP Bridge Superstructure model is available, DC and DW loads may be imported to pier model.Otherwise, they can be autogenerated as shown below.

(refer to BDM Section 3.4.2)

Figure 1 - Design Section

xy



CT:

Column length in LEAP Bridge Substructure model (includes drilled shaft length to fixity)L = 1/2 x 4 ft. cap depth + 20 ft. + 13.5 ft. = ft.

Point of application of CT from drilled shaft point of fixityy1 = 5 ft. + 2 ft. drilled shaft cover + 13.5 ft. = ft.

Case No. 1Case No. 2Case No. 3Case No. 4Summary of CT Load Cases

35.50

20.50

Z0.00

155.29155.29155.29

Location y1 / LCase Number ϕ

0151515

CT2CT3

Col. No.

1

MagnitudeX

600.00

CT4

Equivalent static load = 600 kip, applied in a direction of 0 to 15 degrees, 5 ft. above theground (AASHTO 3.6.5.1)

In the case of a multi-column pier, the Designer must investigate the collision force CT actingon each column separately and select the one with the maximum shear force. The Designershould then check the shear capacity of the column. The Designer is responsible fordetermining the most conservative load cases taking into account both directions of travelunder the bridge and the geometry of the bridge.

Note: The critical design section for a column is at the point of impact. The Designer shouldadd additional check points near the impact from the 'Structure Model' menu to get information needed for design.

0.58

0.58

0.580.58

123

579.56579.56579.56

CaseNameCT1

FX =579.56 kip

FZ = 155.29 kip

Figure 2 - CT Load Application



→ Summary of load combinations used in the design→ Controlling column design results→ Detailed shear design calculation for controlling column

Summary of Design:Column reinforcement - main rebar 22 #10 bars, equally spaced

shear reinforcement #4 ties @ 6"

Analysis of the columns is performed using the P-delta method. See below for the following outputs fromLEAP Bridge Substructure:

Design of a drilled shaft in the Extreme Event collision case is similar to the design of a drilled shaft inthe Strength cases and will not be shown for this example. The Designer must account for the collisionload in the drilled shaft design by applying Extreme Event loads from the bottom of the column to the topof the caisson. It is recommended that the Designer use a suitable design tool to analyze shaft-soilinteraction to determine stability and strength requirements.



Total Pool - USA | SHEET 1 OF 1PROGRAM: LEAP® Bridge Concrete CONNECT Edition-v16.2.0.1 | JOB NO.Bentley Systems, Inc. - www.bentley.com | BY DATE MAY 2017PHONE : TOLL-FREE 1-800-778-4277 | CKD. DATE------------------------------------------------------------------------------------------PROJECT: EXAMPLE 13 - VEHICLE COLLISION ON A PIER

LOAD COMBINATIONS - AASHTO LRFD 7 2016 Interims

Comb # 1 (EXT GP 2 ) = 1.00 ( 1.25 DC1 + 1.50 DW1 + 1.00 CT1 )Comb # 2 (EXT GP 2 ) = 1.00 ( 1.25 DC1 + 1.50 DW1 + 1.00 CT2 )Comb # 3 (EXT GP 2 ) = 1.00 ( 1.25 DC1 + 1.50 DW1 + 1.00 CT3 )Comb # 4 (EXT GP 2 ) = 1.00 ( 1.25 DC1 + 1.50 DW1 + 1.00 CT4 )Comb # 5 (EXT GP 2 ) = 1.00 ( 1.25 DC1 + 0.65 DW1 + 1.00 CT1 )Comb # 6 (EXT GP 2 ) = 1.00 ( 1.25 DC1 + 0.65 DW1 + 1.00 CT2 )Comb # 7 (EXT GP 2 ) = 1.00 ( 1.25 DC1 + 0.65 DW1 + 1.00 CT3 )Comb # 8 (EXT GP 2 ) = 1.00 ( 1.25 DC1 + 0.65 DW1 + 1.00 CT4 )Comb # 9 (EXT GP 2 ) = 1.00 ( 0.90 DC1 + 1.50 DW1 + 1.00 CT1 )Comb # 10 (EXT GP 2 ) = 1.00 ( 0.90 DC1 + 1.50 DW1 + 1.00 CT2 )Comb # 11 (EXT GP 2 ) = 1.00 ( 0.90 DC1 + 1.50 DW1 + 1.00 CT3 )Comb # 12 (EXT GP 2 ) = 1.00 ( 0.90 DC1 + 1.50 DW1 + 1.00 CT4 )Comb # 13 (EXT GP 2 ) = 1.00 ( 0.90 DC1 + 0.65 DW1 + 1.00 CT1 )Comb # 14 (EXT GP 2 ) = 1.00 ( 0.90 DC1 + 0.65 DW1 + 1.00 CT2 )Comb # 15 (EXT GP 2 ) = 1.00 ( 0.90 DC1 + 0.65 DW1 + 1.00 CT3 )Comb # 16 (EXT GP 2 ) = 1.00 ( 0.90 DC1 + 0.65 DW1 + 1.00 CT4 )

Load Combinations for Columns only:

Comb # 1C (EXT GP 2 ) = 1.00 ( 1.25 DC1 + 1.50 DW1 + 1.00 CT1 )Comb # 2C (EXT GP 2 ) = 1.00 ( 1.25 DC1 + 1.50 DW1 + 1.00 CT2 )Comb # 3C (EXT GP 2 ) = 1.00 ( 1.25 DC1 + 1.50 DW1 + 1.00 CT3 )Comb # 4C (EXT GP 2 ) = 1.00 ( 1.25 DC1 + 1.50 DW1 + 1.00 CT4 )Comb # 5C (EXT GP 2 ) = 1.00 ( 1.25 DC1 + 0.65 DW1 + 1.00 CT1 )Comb # 6C (EXT GP 2 ) = 1.00 ( 1.25 DC1 + 0.65 DW1 + 1.00 CT2 )Comb # 7C (EXT GP 2 ) = 1.00 ( 1.25 DC1 + 0.65 DW1 + 1.00 CT3 )Comb # 8C (EXT GP 2 ) = 1.00 ( 1.25 DC1 + 0.65 DW1 + 1.00 CT4 )Comb # 9C (EXT GP 2 ) = 1.00 ( 0.90 DC1 + 1.50 DW1 + 1.00 CT1 )Comb # 10C (EXT GP 2 ) = 1.00 ( 0.90 DC1 + 1.50 DW1 + 1.00 CT2 )Comb # 11C (EXT GP 2 ) = 1.00 ( 0.90 DC1 + 1.50 DW1 + 1.00 CT3 )Comb # 12C (EXT GP 2 ) = 1.00 ( 0.90 DC1 + 1.50 DW1 + 1.00 CT4 )Comb # 13C (EXT GP 2 ) = 1.00 ( 0.90 DC1 + 0.65 DW1 + 1.00 CT1 )Comb # 14C (EXT GP 2 ) = 1.00 ( 0.90 DC1 + 0.65 DW1 + 1.00 CT2 )Comb # 15C (EXT GP 2 ) = 1.00 ( 0.90 DC1 + 0.65 DW1 + 1.00 CT3 )Comb # 16C (EXT GP 2 ) = 1.00 ( 0.90 DC1 + 0.65 DW1 + 1.00 CT4 )

C:\Users\9365\AppData\Local\Temp\RCPD3A2.tmp\BR01_PR01.rcp Last modified: May 31, 2017



COLUMN DESIGN

COLUMN DESIGN - Column: 1 Code: AASHTO LRFD 7 2016 InterimsUnits: US Pier View: Upstation. Design/Analysis Method: P - Delta.

Column Type: Round D = 48.00 in

Column Section Properties

1 12.57 260576.25 260576.25Sec. Area

ft^2Ix

in^4Iz

in^4

DESIGN PARAMETERSf'c = 4500.0 psi fy = 60000.0 psiphi tens = 0.90 phi comp = 0.75Tens above = 0.005 Comp below = 0.002phi shear = 0.90 fy shear = 60000.00 psiEc = 4435.3 ksi Es = 29000 ksiConcrete Type : Normal Weight.

ReinforcementRebar Pattern: Circular Rebar Orientation: Face Parallel

Sheet # 1

Job #

Program: LEAP® Bridge Concrete CONNECT Edition Total Pool - USA Designed

Module: Substructure Copyright © Bentley Systems, Inc. 2016 Date MAY 2017Version: 16.02.00.01 www.bentley.com Phone: 1-800-778-4277 CheckedFile Name: BDM Example 13 - Vehicle collision on a pier... .lbc Date

Units: US (English) Design Code: AASHTO LRFD 7 2016 Interims



Main Reinforcement Schedule

1 US#10[M32] 22 3.00 0.00 22.00 NoneLayer Size No. bars Bar Dist.

inFrom

ftToft Hook

Note:Bar Dist refers to the distance between center of reinforcement and concrete surface.

Lateral Reinforcement Schedule

1 US#4[M13] Ties 6.00 0.00 22.00Layer Size Type Pitch

inFrom

ftToft

Design values used after P-Delta Analysis ( e-min effect included ).

0.00 13 -386.5 442.7 0.0 -75.3 0.0 78.76.00 13 386.5 432.5 0.0 -73.5 -0.0 2247.37.00 13 386.5 430.8 0.0 -73.2 -0.0 2634.68.00 13 -213.5 429.1 0.0 -72.9 -0.0 2475.8

22.00 13 -213.5 408.8 0.0 -69.5 -0.0 -512.00.00 15 -95.8 384.1 -24.2 -413.4 53.1 781.16.00 15 95.8 373.9 24.2 266.7 53.1 -202.27.00 15 95.8 372.2 24.2 242.2 53.1 -105.78.00 15 95.8 370.5 24.2 217.7 53.1 -63.0

22.00 15 95.8 350.2 24.2 -125.8 53.1 1340.20.00 16 -97.2 318.2 1.3 -176.2 126.0 722.46.00 16 97.2 308.0 -1.3 183.6 126.0 -136.17.00 16 97.2 306.3 -1.3 184.9 126.0 -52.18.00 16 97.2 304.6 -1.3 186.1 126.0 59.4

22.00 16 97.2 284.2 -1.3 202.4 126.0 1425.7

Locft Comb Fx

kipsFy

kipsFz

kipsMx

kips-ftMy

kips-ftMz

kips-ft

COLUMN DESIGN

0.00 15 384.1 413.4 781.1 2737.7 62.11 1.00 3.09774 6.00 13 432.5 73.5 2247.3 2787.1 88.13 1.00 1.23953 7.00 13 430.8 73.2 2634.6 2786.8 88.41 1.00 1.05734 8.00 13 429.1 72.9 2475.8 2784.7 88.31 1.00 1.12426

22.00 16 284.2 202.4 1425.7 2622.3 81.92 1.00 1.82108

Bot/Top Elev.ft Comb Pu

kipsMux

kips-ftMuz

kips-ftpMn

kips-ftIncldeg pPn/Pu pMn/Mu

Sheet # 2

Job #




Controlling moment case



COLUMN DESIGN

0.00 18.32 144.76 27.94 6.00 18.32 144.76 27.94 7.00 18.32 144.76 27.94 8.00 18.32 144.76 27.94

22.00 18.32 144.76 27.94

Bot/Top Elev.ft

As_minin^2

As_maxin^2

As_provin^2

SHEAR DESIGN:

0.00 L 396.2 14 503.9 0.64 0.80 340.05 219.82 3.14 31.47 48.00 33.63 0.0007 24.00 6.006.00 R 396.2 14 433.1 0.64 0.80 275.23 206.00 2.54 33.15 48.00 33.63 0.0012 24.00 6.00

L 396.2 14 433.1 0.64 0.80 275.23 206.00 2.54 33.15 48.00 33.63 0.0012 24.00 6.007.00 R 396.2 14 413.1 0.73 0.80 257.61 201.34 2.38 33.75 48.00 33.63 0.0014 24.00 6.00

L 396.2 14 413.1 0.73 0.80 257.61 201.34 2.38 33.75 48.00 33.63 0.0014 24.00 6.008.00 R 214.2 1 462.7 0.64 0.80 301.85 212.23 2.79 32.37 48.00 33.63 0.0010 24.00 6.00

L 214.2 1 462.7 0.64 0.80 301.85 212.23 2.79 32.37 48.00 33.63 0.0010 24.00 6.0022.00 R 214.2 1 632.7 0.64 0.80 465.64 237.36 4.30 29.54 48.00 33.63 0.0002 24.00 6.00

Locft Pos Vu

kips Comb phi*Vnkips

Av/sin^2/ft

Aprv/sin^2/ft

Vckips

Vskips Beta Theta

deg bin

dvin Eps_s Smax

inSprv

in

Column Design : NotesMin reinforcement = 1.0125 % of Ag.Shear Design : Notes- Pos is the design position. L suggests the calculation is done at immediate left of "Loc" and R suggests at immediate right of it.- Av/s is the total required area of steel per unit length due to shear plus torsion.- Aprvs/s is the total provided area of transverse steel reinforcement.- Vc is the nominal shear resistence of concrete.- Vs is the nominal shear resistence of transverse reinforcement.- Beta is the factor indicating ability of diagonally cracked concrete to transmit tension and shear.- Theta is the angle of inclination of diagonal compressive stress.- # Vu is greater tha phi*Vn.

Sheet # 3

Job #




Controlling shear case



5.8.3.3Kips201.34Nominal shear resistance from reinforcement, V

cots

v y vs

A f dV

s

θ⋅ ⋅ ⋅=

5.8.3.3Kips257.61'

Nominal shear resistance from concrete, V

0.0316

c

c c v vV f b dβ= ⋅ ⋅ ⋅ ⋅

Shear capacity calculations

5.8.3.4.22.38( ) ( ) ( )

4.8 4.8 51= or

1 750 1 750 39

Factor of longitudinal strain on shear,

s s esβ β

ε ε

β

= ⋅+ ⋅ + ⋅ +

5.8.3.4.233.75Angle of inclination of diagonal compression stress,

= 29+3500 s

θθ ε⋅

5.8.3.4.20.001357

Net tensile strain at the centroid of tension reinforcement

0.5

,

uu u p ps po

v

s s p ps

MN V V A f

ds E A E A

s

ε

ε

+ ⋅ + − − ⋅=

⋅ + ⋅

5.8.3.4.2Kip-ft2574.73Moment for the net tensile strain , us MεCalculations of and θ β

5.5.4.20.75For axial load,φ5.5.4.20.9For shear,φ5.5.4.20.9For bending, φ

Resistance factorsKips-430.81Factored axial force, Nu

Kips396.2Factored shear, Vu

Kip-ft2574.73Factored moment, Mu

Loads5.8.2.9in33.63Effective shear depth, dv

5.8.2.9in48Effective width of the web, bv

Section geometryin^20.8Area of transverse reinforcement within distance s, Av

in6Spacing of transverse reinforcement, sin^227.94Area of nonprestressed tension reinforcement, sAksi29000Modulus of elasticity of reinforcement, sEksi60Specified yield strength of reinforcement, yfpsi4500

'Specified compressive strength of concrete, cf

MaterialsReferencesUnitsValuesDescription

Controlling load combination 14Shear/torsion design details for Left

Column Design Details

Column 1Shear/TorsionPOI at location 7ft

16.2.0.1LEAP Bridge Concrete(AASHTO LRFD Specifications)5:07 PMTime:BDM Example 13 - Vehicle collision on a pier... .lbc5/31/2017Date:



OKCheck if the provided spacing of transverse steel is no greater than the max spacing allowed

5.8.2.7in24'

max'

max

Max spacing of transverse reinforcement, sIf 0.125 then 0.8 24.0 in

If 0.125 then 0.4 12.0 inu c v

u c v

v f s d

v f s d

< ⋅ = ⋅ ≤

> ⋅ = ⋅ ≤

5.8.2.9ksi0.27

Shear stress, vu

u pu

v v

V Vv

b d

φφ

− ⋅=

⋅ ⋅

Torsion reinforcement calculations

OKCheck if the provided spacing of transverse steel is no greater than the maximum allowed spacing

5.8.3.3in24Max allowed spacing of shear reinforcement, s

5.8.3.3in^2/ft0.06( )cot cot sin

Required shear reinforcement, A /

v y vs

vA f d

Vs

sθ α α⋅ ⋅ ⋅ + ⋅

=

5.8.3.3Kips182.61Required shear resistance from shear reinforcement, Vs

5.8.2.5in^2/ft0.06'0.0316

Min required shear reinforcement, A /

c

vv

y

vb s

Af

s

f ⋅≥ ⋅ ⋅

Shear reinforcement calculations

OKCheck if factored shear capacity is no less than

factored shear force ( V )

(V )n

u

φ5.8.3.3Kips413.055Factored shear resistance, Vnφ

5.8.3.3Kips1816.134'

Max allowed nominal shear resistance,

0.25n

n c v v p

V

V f b d V= ⋅ ⋅ ⋅ +

5.8.3.3Kips458.95Nominal shear resistance of the section, Vn

n c s pV V V V= + +

16.2.0.1LEAP Bridge Concrete(AASHTO LRFD Specifications)5:07 PMTime:BDM Example 13 - Vehicle collision on a pier... .lbc5/31/2017Date:

Documents

EXAMPLE 13 - VEHICLE COLLISION ON A PIER 1 · CT1 F X =579.56 kip F Z = 155.29 kip Figure 2 - CT Load Application. EXAMPLE 13 - VEHICLE COLLISION ON A PIER 4 2018