Slices and Shear

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Lecture 21Lecture 21 Splices and Splices and Shear Shear

February 5, 2003

CVEN 444

Lecture Goals Lecture Goals

Shear Design

B ar Splices B ar Splices

Why do we need bar splices? -- for long spans

Types of Splices

1. Butted &Welded

2. Mechanical Connectors

3. Lay Splices

Must develop 125%

of yield strength

T ension Lap Splices T ension Lap Splices

Why do we need bar splices? -- for long spans

Types of Splices

1. Contact Splice

2. Non Contact Spice (distance 6´ and 1/5

splice length)

Splice length is the distance the two bars areoverlapped.

Ty pes of Splices Ty pes of Splices Class A Spice (ACI 12.15.2)

When over entire splicelength.

and 1/2 or less of total reinforcement is

spliced win the req¶d lay length.

2dreq's

provideds

Ty pes of Splices Ty pes of Splices

Class B Spice (ACI 12.15.2)

All tension lay splices not meetingrequirements of Class A Splices

T ension Lap SpliceT ension Lap Splice(ACI 12.15)(ACI 12.15)

whereAs (req¶d) = determined for bending

ld = development length for bars (not

allowed to use excess reinforcementmodification factor)

ld must be greater than or equal to 12 in.

T ension Lap Splice (ACI 12.15)T ension Lap Splice (ACI 12.15)

Lap Spices shall not be used for bars larger than No. 11.

(ACI 12.14.2)

Lap Spices should be placed in away from regions of

high tensile stresses -locate near points of inflection(ACI 12.15.1)

Compression Lap SpliceCompression Lap Splice

(ACI 12.16.1)(ACI 12.16.1)

Lap, req¶d = 0.0005f y d b for f y < 60000 psi

Lap, req¶d = (0.0009f y -24) d b for f y > 60000 psiLap, req¶d 12 in

For f c 3000 psi, required lap splice shall be multiply

by (4/3) (ACI 12.16.1)

Compression Lap SpliceCompression Lap Splice(ACI 12.17)(ACI 12.17)

In tied column splices with effective tie area throughout

splice length 0.0015 hs factor = 0.83

In spiral column splices, factor = 0.75 But final splice

length 12 in.

Ex ampleEx ample Splice T ension Splice T ension

Calculate the lap-splice length for 6 #8 tension

bottom bars in two rows with clear spacing 2.5 in.

and a clear cover, 1.5 in., for the following cases

When 3 bars are spliced and As(provided) /As(required) >2

When 4 bars are spliced and As(provided) /As(required) < 2

When all bars are spliced at the same location.

f c= 5 ksi and f y = 60 ksi

For #8 bars, db =1.0 in. and E = ? F =? K = ?P =?

tr b c

c K d f

EFKP¨ ¸© ¹© ¹!

The As(provided) /As(required) > 2, class ? splice applies;

The As(provided) /As(required) < 2, class ? splice applies;

Ex ampleEx ample SpliceSpliceCompression Compression

Calculate the lap splice length for a # 10

compression bar in tied column when f c= 5 ksi and

when a) f y = 60 ksi and b) f y = 80 ksi

For #10 bars, d b =? in.

0.020.003

d f ! u

Check ls > 0.005 d b f y

For #10 bars, d b =? in. The ld = 2? in.

Check ls > (0.0009 f y ±24) d b

So use ls= ? in.

Shear Design Shear Design

U ncracked E lastic B eamU ncracked E lastic B eamB ehavior B ehavior

Look at the shear and

bending moment

diagrams. The actingshear stress distribution

on the beam.

The acting stresses distributed across the

cross-section.

The shear stressacting on the

rectangular beam.

IbVQ!X

The equation of the shear stress for a rectangular beam

is given as:

Note: The maximum1st moment occurs at

the neutral axis (NA).

IbVQ!X

avemax

Inertiaooment12

X X !¹ º ¸©

!¹ º

¸©ª

¨¹ º

¸©ª

The ideal shear stress distribution can be described as:

A realistic description of the shear distribution is shown

The shear stress acting along the beam can be described

with a stress block:

Using Mohr¶s circle, the stress block can be

manipulated to find the maximum shear and the crack

formation.

Inclined Cracking in Inclined Cracking in Reinforced Concrete B eams Reinforced Concrete B eams

Typical Crack Patterns for a deep beam

Flexural-shear crack -

Starts out as a flexural

crack and propagates due

to shear stress.

Flexural cracks in beams

are vertical

(perpendicular to thetension face).

For deep beam the cracks are given as:

The shear cracks Inclined (diagonal) intercept crack

with longitudinal bars plus vertical or inclined

reinforcement.

For deep beam the cracks

are given as:

The shear cracks fail due

two modes:

- shear-tension failure- shear-compression

failure

Shear Strength of RC B eams Shear Strength of RC B eams without Web Reinforcement without Web Reinforcement

vcz - shear in compression

va - Aggregate Interlock

forces

vd= Dowel action from

longitudinal bars

Note: vcz increases from

(V/bd) to (V/by) as crack

forms.

Total Resistance = vcz + vay + vd (when no stirrups are used)

Strength of Concrete in Shear Strength of Concrete in Shear (No Shear Reinforcement)(No Shear Reinforcement)

(1) Tensile Strength of concrete affect inclined

cracking load

(2) Longitudinal Reinforcement Ratio, Vw

2:0025.00075.0or

cracksrestrains

Strength of Concrete in Shear Strength of Concrete in Shear

(No Shear Reinforcement)(No Shear Reinforcement)(3) Shear span to depth ratio, a/d (M/(Vd))

a Deep shear spansmore detail design

required

Ratio has littleeffect

(4) Size of Beam

Increase Depth Reduced shear stress at

inclined cracking

(5) Axial Forces

- Axial tension Decreases inclined cracking load

- Axial Compression Increases inclined crackingload (Delays flexural

cracking)

F unction and Strength of F unction and Strength of Web Reinforcement Web Reinforcement

Web Reinforcement is provided to ensure that

the full flexural capacity can be developed.

(desired a flexural failure mode - shear failure

is brittle)

- Acts as ³clamps´ to keep shear cracks from

widening

Function:

F unction and Strength of F unction and Strength of Web Reinforcement Web Reinforcement

Uncracked Beam Shear is resisted uncracked

concrete.

Flexural Cracking Shear is resisted by vcz, vay, vd

bars.allongitudinromctiono l orcenterlock ggregateocom onentertical

zonencom ressioinShear

Designing to Resist Shear Designing to Resist Shear

Shear Strength (ACI 318 Sec 11.1)

capacity demand

V V J u

factored shear force at section

Nominal Shear Strength

0.75 shear strength reduction factor

Designing to Resist Shear Designing to Resist Shear

Shear Strength (ACI 318 Sec 11.1)

n c sV V V !

! Nominal shear provided by the shear reinforcement

Nominal shear resistance provided by concrete

Shear Strength Provided by Shear Strength Provided by ConcreteConcrete

Bending only Bending only

Simple formula

More detailed

Eqn [11.5]

Eqn [11.3]

d b f V

25001.9

u e¹¹

Shear Strength Provided by Shear Strength Provided by ConcreteConcrete

Bending and Axial Compression Bending and Axial Compression

Nu is positive for

compression and

are in psi.

Simple formula

Eqn [11.4]

Eqn [11.7]

50013.5

200012

A N d b f

d b f A

¹¹¹ º

©©©ª

Ty pical Shear Reinforcement Ty pical Shear Reinforcement

Stirrup - perpendicular to axis of members

(minimum labor - more material)

ACI Eqn 11-15

EE cossinyv

o90 !!E

Ty pical Shear Reinforcement Ty pical Shear Reinforcement

Bent Bars (more labor - minimum material) see req¶d

in 11.5.6

ACI 11-5.6

EE cossinyv

d f V yv

o 41.145 !!E

Stirrup Anchorage Requirements Stirrup Anchorage Requirements

Vs based on assumption stirrups yield

Stirrups must be well anchored.@

each bend must enclose a long bar

# 5 and smaller can use standard hooks 90o,135o, 180o

#6, #7,#8( f y = 40 ksi )

#6, #7,#8 ( f y > 40 ksi ) standard hook plus aminimum embedment

Refer to Sec. 12.13 of ACI 318 for development of web

reinforcement. Requirements:

Also sec. 7.10 requirement for minimum stirrups

in beams with compression reinforcement, beams subject to stress reversals, or beams

subject to torsion

Design Procedure for Shear Design Procedure for Shear

(1) Calculate Vu

(2) Calculate JVc Eqn 11-3 or 11-5 (no axial force)

(3) Check

°¯®

pu cu V V J

If yes, add web reinforcement (go to 4)

If no, done.

(4) pee cuc V V V J J

¹¹ º

¸©©ª

¨!! v

sb A minfor

50or 50 maxmin

(Done) 11.5.4 "24

Provide minimum

shear reinforcement

V V V V

J d)(req'calulate, I

Check:

11.5.4 illegalother wise, 8 d b f V wc sde

Solve for required stirrup spacing(strength)

Assume # 3, #4, or #5 stirrups

d f A s e

from 11-15

(7) Check minimum steel requirement (eqn 11-13)

b f A s !

(8) Check maximum spacing requirement (ACI 11.5.4)

illegal 8 I:ote

"242 4 I

d b f V

d sd b f V

d b f V

(9) Use smallest spacing from steps 6,7,8

Note: A practical limit to minimum stirrup

spacing is 4 inches.

L ti f M i Sh fL ti f M i Sh f

Location of Max imum Shear for Location of Max imum Shear for B eam Design B eam Design

Non-pre-stressed members:

Sections located less than a distance d from face of

support may be designed for same shear, Vu, as the

computed at a distance d.

Compression fan

carries load directly

into support.

Location of Max imum Shear for B eam Design

The support reaction introduces compression

into the end regions of the member.

No concentrated load occurs with in d from

face of support .

L ti f M i Sh f

Location of Max imum Shear for B eam Design

Compression from support at bottom of

beam tends to close crack at support

H omework H omework Determine the development length required for the bars

shown . f c =4-ksi and f y = 60-ksi. Check the anchorage

in the column. If it is not satisfactory, design an

anchorage using a 180o

hook and check adequacy.

H omework H omework Considering the anchorage of the beam bars into a

column, determine the largest bar that can be used with

out a hook. f c = 3-ksi and f y= 40ksi

H omework H omework A simple supported uniformly loaded beam carries a

total factored design load of 4.8 k/ft (including self-

weight) on a clear span of 34 ft. f c =3 ksi and f y=40 ksi.

Assume that the supports are 12 in wide and assumethat the bars are available in 30 ft lengths.

Design a rectangular beam

Determine bar cutoffs.Locate splices and determine the lap length.

Slices and Shear

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