This document discusses shear stresses and design of shear reinforcement in reinforced concrete beams. It covers key topics such as: 1) The components that resist shear in an RC beam including aggregate interlock, dowel action, and shear reinforcement when used. 2) Design procedures for shear as outlined in ACI 318, including calculating the nominal shear provided by concrete (Vc) and shear reinforcement (Vs), and ensuring the shear capacity exceeds the factored shear demand (Vu). 3) Types and placement of common shear reinforcement such as stirrups. Minimum requirements for stirrups are also covered. 4) An example problem is worked through demonstrating shear design of a simply supported beam section using stirrup

1. REINFORCED CONRETE (1)
(CE411)
Chapter 4
Shear and Diagonal Tension in
Beams
Instructor:
Eng. Abdallah Odeibat
Civil Engineer, Structures , M.Sc.
1

2. A
2

3. 3

4. SHEAR STRESSES IN CONCRETE BEAMS
4
Diagonal tension – Mohr’s circle
Principal stresses
2
2
2 2
p
f f
f v
 
= + +
 
 
Flexural stress
Shear stress
2
tan 2
v
f
 =
VQ
v
Ib
= Principal angle

5. SHEAR STRENGTH OF RC BEAMS WITHOUT
WEB REINFORCEMENT
 Total Resistance = vcz + vay + vd (when no
stirrups are used)
5
vcz - shear in
compression zone
va - Aggregate
Interlock forces
vd = Dowel action from
longitudinal bars
Note: vcz increases
from (V/bd) to (V/by) as
crack forms.

6. DESIGNING TO RESIST SHEAR
 Shear Strength (ACI 318 Sec 11.1)
6
demand
capacity 
 u
n V
V

( ) factor
reduction
strength
shear
0.75
Strength
Shear
Nominal
section
at
force
shear
factored
−
=
=
=

n
u
V
V

7. DESIGNING TO RESIST SHEAR
 Shear Strength (ACI 318 Sec 11.1)
7
s
c
n V
V
V +
=
=
=
s
c
V
V
Nominal shear provided by the shear reinforcement
Nominal shear resistance provided by concrete

8. TYPES OF STIRRUPS
8

9. TYPES OF STIRRUPS
9

10. TYPES OF STIRRUPS
10

11. TYPES OF STIRRUPS
11

12. BEHAVIOR OF BEAMS WITH WEB
REINFORCEMENT
 Truss analogy
 Concrete in compression is top chord
 Longitudinal tension steel is bottom chord
 Stirrups form truss verticals
 Concrete between diagonal cracks form the truss
diagonals
12

13. Truss analogy
13

14. DESIGNING TO RESIST SHEAR
 Shear Strength (ACI 318 Sec 11.1)
14
s
c
n V
V
V +
=

15. DESIGNING TO RESIST SHEAR
 Shear Strength (ACI 318 Sec 11.1)
15
s
c
n V
V
V +
=

16. 16
s
c
n V
V
V +
=

17. ACI CODE PROVISIONS FOR SHEAR
DESIGN
17

18. MINIMUM WEB REINFORCEMENT
18

19. DESIGN OF SHEAR REINFORCEMENT
a) Where the factored shear Vu is less than one-
half øVc, it shall be permitted to waive the use
of shear reinforcement.
b) Where the factored shear Vu exceeds one-half
øVc and is less than øVc, a minimum amount
of shear reinforcement shall be employed as
specified in previous slide, where maximum
spacing s shall not exceed d/2 and 600 mm.
19

20. DESIGN OF SHEAR REINFORCEMENT
c) Where the factored shear Vu exceeds øVc, the
difference (Vu- øVc) shall be provided by shear
reinforcement and the following limitations shall
be employed:
➢ If øVs, calculated in slide 15 is less than 2 øVc,
the spacing limits given (b) shall govern.
➢ If øVs, calculated in slide 15 is more than 2 øVc,
the spacing limits shall be half of that given (b)
shall govern.
➢ øVs, calculated in slide 15 shall not exceed 4 øVc. 20

21. 21
Note1 : If Vu< ø Vc/2 , no need for shear reinforcement
Note 2: if Vu > 5 ø Vc , alter the section
300 mm

22. CRITICAL SECTION
22

23. CRITICAL SECTION
23

24. CRITICAL SECTION
24

25. EXAMPLE
 A rectangular beam is to be designed for shear, with
effective depth (d) =350 mm, total depth (h)=400 mm, and
width (b) = 450 mm , to carry two factored point loads as
shown below . If fy =420 MPa, fyt =420 MPa, and fc’ =28
MPa. Use 2 legs ø 10 mm stirrups. (Neglect the own weight
of the beam).
25

26. 26
Location (m) 0<x<2 2<x<4.5 4.5<x<7 7<x<9
Vu (kN)
Ø Vc (kN)
Ø Vs (kN)
Av
s max
s req

27. EXAMPLE
 Design for shear reinforcement using stirrups for
simply supported rectangular beam given that :
27

28. SOLUTION
28
1
2

29. SOLUTION
29
SFD
3 Shear at distance d
348

30. SOLUTION
30
4 Calculate
Comparison
limits
3 Design of stirrups
use ø10 mm stirrups , (2 legs) , Av =157 mm2

31. 31
8 Calculate S for each interval
538 mm
s= 119 mm

32. 32
9 Final arrangement of stirrups
1523.5 mm 964 mm
17 ø 10 @ 100 mm 5 ø 10 @ 250 mm
HW: Redesign this
beam using ø8 stirrups