Aci - Shear in Beam

7/26/2019 Aci - Shear in Beam

1/15

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

Fifth Edition

CHAPTER

4b

Reinforced Concrete Design

ENCE 355 - Introduction to Structural DesignDepartment of Civil and Environmental Engineering

University of Maryland, College Park

SHEAR IN BEAMS

Part I Concrete Design and Analysis

FALL 2002By

Dr Ibrahim Assakkaf

CHAPTER 4b. SHEAR IN BEAMS Slide No. 1ENCE 355 Assakkaf

Shear Analysis Procedure

The shear analysis procedure involves

the following:

Checking the shear strength in an existing

member

Verifying that the various ACI code

requirements have been satisfied and met.

Note that the member may reinforced or

plain.


2/15



Example 1

A reinforced concrete beam of rectangular

cross section shown is reinforced with

seven No. 6 bars in a single layer. Beam

width b = 18 in., d= 33 in., single-loop No.

3 stirrups are placed 12 in. on center, and

typical cover is 1 in. Find Vc, Vs, and the

maximum factored shear force permitted

on this member. Use = 4,000 psi andfy= 60,000 psi.



Example 1 (contd)

33

bars6#7

COV.2

11

stirrup21@3#

81


3/15



Example 1 (contd) The force that can be resisted by concrete

alone is

The nominal shear force provided by the

steel is

The maximum factored shear force is

( )( )kips1.75

1000

3318000,422 === dbfV wcc

( )( )( )kips3.36

12

336011.02=

==

s

dfAV

yv

s

( )

kips7.94

3.361.7585.0maximum

=

+=+= scu VVV


ACI Code Provisions for Shear Design

According to the ACI Code, the design of

beams for shear is based on the following

relation:

Shear Reinforcement DesignRequirements

un VV Where

= strength reduction factor (= 0.85 for shear)

Vn = Vc + VsVs = nominal shear strength provided by reinforcement

stirrupsinclinedfors

dfA yv=

(1)

(2)


4/15



Symbols

Av= total cross-sectional area of web reinforcement within

a distances, for single loop stirrups,Av = 2AsAs = cross-sectional area of the stirrup bar (in

2)

bw = web width = b for rectangular section (in.)

s = center-to-center spacing of shear reinforcement in a

direction parallel to the longitudinal reinforcement (in.)

fy = yield strength of web reinforcement steel (psi)

Shear Reinforcement Design

Requirements



For inclined stirrups, the expression for

nominal shear strength provided by

reinforcement is

For = 450, the expression takes the form

( )

s

dfAV

yv

s

cossin +=

s

dfAV

yv

s

414.1=

(3)

(4)



5/15



The design for stirrup spacing can be

determined from


Requirements

stirrups)45(for414.1

required

and

stirrups)cal(for vertirequired

0

s

yv

s

yv

V

dfAs

V

dfAs

=

=

where

cus

VVV

=

(5)

(6)

(7)



According to the ACI Code, the maximum

spacing of stirrups is the smallest value of


in.24

2

50

max

max

max

=

=

=

s

ds

b

fAs

w

yv

(8)

If Vs exceeds , the maximum spacing must

not exceed d/4 or 12 in.

dbf wc4


6/15



It is not usually good practice to space

vertical stirrups closer than 4 in.

It is generally economical and practical to

compute spacing required at several

sections and to place stirrups accordingly

in groups of varying spacing. Spacing

values should be made to not less than 1-in. increments.


Requirements



Critical Section

The maximum shear usually occurs in this

section near the support.

For stirrup design, the section located a

distance dfrom the face of the support is called

the critical section Sections located less than a distance dfrom

the face of the support may be designed for the

same Vu as that of the critical section.



7/15



Critical Section


Requirements

d

d

Critical Section

Figure 1

L



Critical Section (contd)

The stirrup spacing should be constant from the

critical section back to the face of the support

based on the spacing requirements at the

critical section.

The first stirrup should be placed at a maximumdistance ofs/2 from the face of the support,

wheres equals the immediately adjacent

required spacing (a distance of 2 in. is

commonly used.



8/15


The design of stirrups for shearreinforcement involves the

determination of stirrup size and

spacing pattern.

A general procedure is as follows:

1. Determine the shear values based on

clear span and draw a shear diagram for

Vu.

2. Determine if stirrups are required.

Stirrup Design Procedure


3. Determine the length of span over which

stirrups are required.

4. On the Vu diagram, determine the area

representing required Vs. This will

display the required strength of the

stirrups to be provided.

5. Select the size of the stirrups. Find thespacing required at the critical section ( a

distance dfrom the face of the support.



9/15


6. Establish the ACI Code maximum

spacing requirements.

7. Determine the spacing requirements

based on shear strength to be furnished

by web reinforcing.

8. Establish the spacing pattern and show

detailed sketches.



Example 2

A continuous reinforced concrete beam

shown in the figure is 15 in. wide and has

an effective depth of 31 in. The factored

loads are shown, and the factored uniform

load includes the weight of the beam.

Design the web reinforcement if = 4000psi andfy = 60,000 psi.


cf


10/15


Example 2 (contd)


51

13

As

spanclear051

100 k100 k

0-5 0-5 0-5

wu = 1.0 k/ftA

A

Section A-A


Example 2 (contd)

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

( ) ( )k5.107

2

151100221 =

+==RR


11/15


100 k100 k

0-5 0-5 0-5

wu = 1.0 k/ft

107.5 k 107.5 k

107.5102.5

Vu (kips)2.5

107.5

2.5

+

-

Example 2 (contd)

See Fig. 2 for enlargement


100 k100 k

0-5 0-5 0-5

wu = 1.0 k/ft

107.5 k 107.5 k

Example 2 (contd)

M

V

107.5 k

50for5.107 = xxVu

M

V

100 k

105for1005.107 = xxVu107.5 k

( ) k9.10458.25.10758.2

85.213

*

===

==

uu VV

d

x

x


12/15


Example 2 (contd)

Because of the symmetry, we will focus on

the left half of the shear diagram as shown

in Fig. 2.

Determine if stirrups required:


( ) ( )( )

( ) kips25502

1

2

1

kips50100

3115000,4285.02

==

===

c

wcc

V

dbfV

Since ( =104.9 k) > (1/2 Vc = 25 k), stirrups are required.*

uV


Example 2 (contd)


9.104

5.107*

uV

cV

5.102

50

25cV2

1

*

*

85.213 ==d

0.5 5.2

Sym.CL

sVrequired

Vu(kips) 0

Figure 2


13/15


Example 2 (contd) Stirrups are required to the point where

From Fig. 2, 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 Vs on the Vu diagram:


kips252

1== cu VV

52.58for5.57required

505.107maxrequired

=

==

xxV

xwxVVV

s

cus


Example 2 (contd)

Assume No. 3 vertical stirrups (Av = 0.22 in2):

Establish ACI Code maximum spacing

requirements:


( )( )( )

in.6use

in.3.6509.104

316022.085.0

requiredrequired

*

* =

==s

yy

V

dfAs

( )( )

kips6.6485.0

509.104

kips6.1171000

3115400044

** =

==

==

ss

wc

VV

dbf


14/15


Example 2 (contd) Since 64.6 kips < 117.6 kips, the maximum

spacing shall be the smallest of the

following values (see Eq. 8):


( )( )

in.24

in.5.152

31

2

in.6.171550

000,6022.0

50

max

max

max

=

===

===

s

ds

b

fAs

w

yv

Therefore, use a maximum spacing of 15 in.

controls


Example 2 (contd)

Determine the spacing requirementsbetween the critical section and the firstconcentrated load:

The results of applying above equation for

values ofx

range from 3 to 5 are tabulatedas shown


( )( )( )xV

dfAs

s

yy

==

5.57

316022.085.0

requiredrequired

6.65

6.54

6.43

Requireds (in)x (ft)


15/15


Example 2 (contd) Since no stirrups are required in the

distance between the first concentrated

load , it is clear that the maximum spacing

of 15 in. need not be used in that distance.

A spacing of 6 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 2 (contd)

Final Sketch for Shear Reinforcement:


67

56@spaces10 =

3

2

111

2

111

LCSym.

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