Rcc 10

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Fifth EditionReinforced Concrete Design

• A. J. Clark School of Engineering •Department of Civil and Environmental Engineering

CHAPTER

6b

REINFORCED CONCRETEA Fundamental Approach - Fifth Edition

SHEAR AND DIAGONAL TENSION IN BEAMS

ENCE 454 – Design of Concrete StructuresDepartment of Civil and Environmental Engineering

University of Maryland, College Park

SPRING 2004By

Dr . Ibrahim. Assakkaf

CHAPTER 6b. SHEAR AND DIAGONAL TENSION IN BEAMS Slide No. 1ENCE 454 ©Assakkaf

Web Reinforcement Design Procedure for shear

The design of stirrups for shear reinforcement involves the determination of stirrup size and spacing pattern.A general procedure is as follows:

1. Determine the critical section, shear values based on clear span, and draw a shear diagram for the factored shear force Vu.

2. Check whether

( )dbfVV wccu ′+≤ 8φ

2


Where bw is the web width or diameter of the circular section. If this condition is not met, the cross section has to be enlarged.

3. Use minimum shear reinforcement Av if Vularger than one-half φVc with the following exceptions:(a) Concrete joist construction(b) Slabs and Footings(c) Small shallow beams of depth not

exceeding 10 in. or 2 ½ times flange thickness



Good construction practice dictates that some stirrups always be used to facilitate proper handling of the reinforcement cage.

4. If Vu > φVc, shear reinforcement must be provided such as that Vu ≤ φ(Vc + Vs), where


y

wv

y

wcv f

sbAf

sbfA 50or 75.0min =′= whichever is larger

( )

+=

stirrups inclinedfor cossin

stirrups alfor vertic

ααs

dfAs

dfA

Vyv

yv

s

3


If α = 450, then

5. Maximum spacing s must be

except that if , then the spacing should be


sdfA

V yvs

414.1=

in. 242≤=

ds

dbfV wcs ′> 4

in. 124≤≤

ds


The flowchart of Figure 14 contains the sequence of calculations needed for the design of vertical stirrups.Simple corresponding modifications of this chart can be made so that the chart can be used in the design of inclined web reinforcement steel.


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Figure 14. Flowchart for webreinforcement design procedure



Example 3A continuous reinforced concrete is 15 in. wide and 33 in. deep, and has an effective depth of 31 in. The beam is subjected to two concentrated live loads of 62.5 kips each as shown, and a distributed live load of 0.32 k/ft. The distributed load does not include the weight of the beam. Design the web reinforcement if = 4000 psi and fy = 60,000 psi. Assume normal weight concrete.

cf ′

5



Example 3 (cont’d)

spanclear 051 ′′−′

PP

0-5 ′′′0-5 ′′′0-5 ′′′wuA

A

Section A-A

51 ′′

13 ′′

As

33 ′′


Example 3 (cont’d)– Calculate the factored loads:

• Estimate self-weight of the beam

• Factored distributed load:

• Factored concentrated load


( )ft

kips 52.0ftlb 625.515150

1443315

==×

=sww

( )ft

Kips 0.132.052.02.1 =+=uw

( ) kips 1005.626.1 ==uP

6




Section A-A

51 ′′

13 ′′

As

33 ′′

spanclear 051 ′′−′

100 k100 k

0-5 ′′′0-5 ′′′0-5 ′′′wu = 1.0 k/ftA

A



Example 3 (cont’d)– Establish the shear force diagram for Vu:

100 k100 k

0-5 ′′′0-5 ′′′0-5 ′′′wu = 1.0 k/ft

107.5 k 107.5 k

( ) ( ) k 5.1072

151100221 =

+== RR

7


100 k100 k

0-5 ′′′0-5 ′′′0-5 ′′′wu = 1.0 k/ft

107.5 k 107.5 k

107.5 102.5

Vu (kips)2.5

107.5

2.5

+

-


See Figure 15 for enlargement


100 k100 k

0-5 ′′′0-5 ′′′0-5 ′′′wu = 1.0 k/ft

107.5 k 107.5 k


M

V

107.5 k

( ) 50for 15.107 ≤≤−= xxVu

M

V

100 k

( ) 105for 11005.107 ≤≤−−= xxVu107.5 k

( ) ( ) k 9.104158.25.10758.285.213

* =−==

′=′′=

uu VVd

x

x

8


Example 3 (cont’d)– Because of the symmetry, we will focus on

the left half of the shear diagram as shown in Fig. 2.

– Shear capacity:( ) ( )( )

( )( )

( ) ( ) kips 57.22027.23582.5875.08

kips 27.2351000

3115400088

kips 82.581000

3115000,4122

=+=′+

==′

==′=

dbfV

dbf

dbfV

wcc

wc

wcc

φ

λ




Example 3 (cont’d)– Shear Capacity (cont’d)


( )( )

( ) kips 12.4482.5875.0

required are stirrups 06.22219.104

OKsection -cross 57.2209.104

*

*

==

=>=

<=

c

cu

u

V

VV

V

φ

φ


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9.1045.107

*uV

cVφ

5.102

12.44

06.22cVφ21

*

*85.213 ′=′′=d

0.5 ′ 5.2 ′

Sym.CL

sVφ required

Vu(kips) 0

Figure 15


Web Reinforcement Design Procedure for shearExample 3 (cont’d)– Stirrups are required to the point where

– From Figure 15, this point is located at the first concentrated load and it is at distance 5 ft from the face of the support.

– Determine the “required φVs” on the Vu diagram:

kips 06.2221

== cu VV φ

( )52.58for 38.63 required

112.445.107 max required

≤≤−=−−=−−=

xxVx

wxVVV

s

cus

φ

φφ

10


Example 3 (cont’d)– Try No. 3 vertical stirrups (Av = 0.22 in2):

– Establish ACI Code maximum spacing requirements:

( )( )( )

in. 5 use

in. 05.512.449.104

316022.075.0 required

required ** =

−==

s

yy

VdfA

sφ

φ

( )( )

kips 04.8175.0

12.449.104

kips 6.1171000

3115400044

** =

−==

==′

φφ s

s

wc

VV

dbf


φφ cu

sVVV −

= :Eq.17


#3 #4 $5 #6 #7 #8 #9 #10 #111 0.11 0.20 0.31 0.44 0.60 0.79 1.00 1.27 1.562 0.22 0.40 0.62 0.88 1.20 1.58 2.00 2.54 3.123 0.33 0.60 0.93 1.32 1.80 2.37 3.00 3.81 4.684 0.44 0.80 1.24 1.76 2.40 3.16 4.00 5.08 6.245 0.55 1.00 1.55 2.20 3.00 3.95 5.00 6.35 7.806 0.66 1.20 1.86 2.64 3.60 4.74 6.00 7.62 9.367 0.77 1.40 2.17 3.08 4.20 5.53 7.00 8.89 10.928 0.88 1.60 2.48 3.52 4.80 6.32 8.00 10.16 12.489 0.99 1.80 2.79 3.96 5.40 7.11 9.00 11.43 14.0410 1.10 2.00 3.10 4.40 6.00 7.90 10.00 12.70 15.60

Number of bars

Bar numberTable 1. Areas of Multiple of Reinforcing Bars (in2)



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Example 3 (cont’d)– Since 81.04 kips < 117.6 kips, the

maximum spacing shall be the smallest of the following values (see Eq. 18):

( )( )

( )( )

in. 24

in. 5.15231

2

in. 6.181550

000,6022.0,15400075.0

000,6022.0max50

,75.0

max

max

max

max

=

===

=

=

′=

s

ds

bfA

bffA

sw

yv

wc

yv

Therefore, use a maximum spacing of 15 in.

controls


Example 3 (cont’d)– Determine the spacing requirements

between the critical section and the first concentrated load:

– The results of applying above equation for values of x ranges from 3 to 5 are tabulated as shown

( )( )( )xV

dfAs

s

yy

−==

38.63316022.075.0

required required

φφ

5.2655.1745.083

Required s (in)x (ft)


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Example 3 (cont’d)– Since no stirrups are required in the

distance between the concentrated loads, it is clear that the maximum spacing of 15 in. need not be used in that distance.

– A spacing of 5 in. will be used between the face of the support and the concentrated load.

– The center part of the beam will be reinforced with stirrups at a spacing slightly less than the maximum spacing of 15 in.



Example 3 (cont’d)Final Sketch for Shear Reinforcement:

67 ′′−′

56 @ spaces 10 ′=′′

3 ′′ ″−′

2111

″−′

2111

LCSym.


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Example 3 (cont’d)Final Sketch for Shear Reinforcement:


Section A-A

51 ′′

33 ′′clear

211″

stirrup 3#


Deep Beams

Deep beams are structural elements loaded as beams but having a large depth/thickness ratio.For deep beams, the shear span/depth ratio is not to exceed 2 for concentrated load and 4 for distributed load, where the shear span is the clear span of the beam for distributed load.

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Deep Beams

Because of the geometry of deep beams, they behave in a non-linear analysis as two-dimensional rather one-dimensional members and are subjected to a two-dimensional state of stress.As a result, plane sections before bending do not necessarily remain plane after bending.Strain distribution no longer considered linear.


Deep Beams

ln

h

Heavy loading2span/depth ≤

4 :load ddistribute

2 :load edconcentrat

<

<

dl

da

n

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Deep Beams

Design Criteria– Deep beam have a higher nominal shear

resistance Vc than do normal beams.– While the critical section for calculating the

factored shear force Vu is taken at a distance d from the face of the support in normal beams, the shear plane in the deep beam is considerably steeper in inclination and closer to the support.


Deep Beams

Design Criteria (cont’d)– If x is the distance of the failure plane from the

face of the support, the expression for distance become

– In either case, the distance x should not exceed the effective depth d of the beam.

axlx n

50.0 :loadedconcentrat15.0 :load uniform

==

16


Deep Beams

Design Criteria (cont’d)– The factored shear force Vu has to satisfy the

condition

– If not, the section has to be enlarged.– The strength reduction factor φ = 0.75 for

deep beams (the same as for normal beams).

( )

dbfV

dbfV

wcn

wcu

′≤

′≤

10

or 10φ


Deep Beams

Design Criteria (cont’d)– The nominal shear resistance force Vc of plain

concrete can be taken as

– Where

dbfdbM

dVfdV

MV wcwu

uc

u

uc ′≤

+′

−= 625009.15.25.3 ρ

5.25.25.30.1 ≤

−<

dVM

u

u

17


Deep Beams

Design Criteria (cont’d)– If Vu > φVc, shear reinforcement has to be

provided such that Vu ≤ φ(Vc + Vs), where Vs is the force resisted by the shear reinforcement:

dfdlsAdl

sAV y

n

h

vhn

v

vs

−

+

+

=12

/1112

/1

Av = total area of vertical stirrups spaced at sv in the horizontal directionat both faces of the beam

Avh = total area of horizontal stirrups spaced at sh in the vertical directionat both faces of the beam


Deep Beams

Design Criteria (cont’d)– Maximum and Minimum Requirements

• The shear reinforcement required at the critical section must be provided throughout the entire length of the beam.

vv

hvh

h

v

bsAbsA

ds

ds

0025.0min0015.0min

in. 12or 5

max

in. 12or 5

max

==

≤

≤

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