Prof Dr Bayan Salim Chapter 3: Shear & Diagonal Tension 1 3 Shear and Diagonal Tension in Beams Shear failure of RC beams, also called diagonal tension failure, if occurred, is sudden, with no advance warning of distress. RC beams are generally provided with special shear reinforcement to ensure that flexural failure would occur before shear failure if the member should be severely overloaded. Fig. below shows a shear-critical beam tested under third-point loading. With no shear reinforcement provided, the member failed immediately upon formation of the critical crack in the high-shear region near the right support. RC Beams without Shear Reinforcement a. Formation of Diagonal Cracks The average shear stress prior to crack formation is v = V / bd At locations of large shear force V and small bending moment M, diagonal cracks form mostly at or near the NA (why?) and propagate from that location, as shown in Fig. a below: These are called web-shear cracks. It was found that diagonal tension cracks form at an average shear stress vcr: vcr = Vcr / bd = 0.29 √ f’c
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Prof Dr Bayan Salim Chapter 3: Shear & Diagonal Tension 1
3 Shear and Diagonal Tension in Beams
Shear failure of RC beams, also called diagonal tension failure, if occurred, is sudden,
with no advance warning of distress.
RC beams are generally provided with special shear reinforcement to ensure that flexural
failure would occur before shear failure if the member should be severely overloaded.
Fig. below shows a shear-critical beam tested under third-point loading. With no shear
reinforcement provided, the member failed immediately upon formation of the critical
crack in the high-shear region near the right support.
RC Beams without Shear Reinforcement
a. Formation of Diagonal Cracks The average shear stress prior to crack formation is
v = V / bd
At locations of large shear force V and small bending moment M, diagonal cracks form
mostly at or near the NA (why?) and propagate from that location, as shown in Fig. a
below:
These are called web-shear cracks. It was found that diagonal tension cracks form at an
average shear stress vcr:
vcr = Vcr / bd = 0.29 √ f’c
Prof Dr Bayan Salim Chapter 3: Shear & Diagonal Tension 2
More commonly, when both the shear force V and bending moment M have large values,
flexure-shear cracks develop from top of flexural cracks and bend in a diagonal direction,
as shown in Fig. b below:
Tests show that, in the presence of large M, the nominal shear stress at which diagonal
tension cracks form and propagate is
vcr = Vcr / bd = 0.17 √ f’c
However, the shear at which diagonal cracks develop depends on the ratio of shear force
to bending moment (V / M), as shown in graph below from the evaluation of large
number of tests. The ACI 318-19 adopts that shear strength varies with of ρ1/3.
b. Behavior of Diagonally Cracked Beams Two types of behavior have been observed:
1. (in shallower beams) The diagonal crack, once formed, spreads immediately,
traversing the entire beam, splitting it in two and failing the beam. This process is
sudden.
Prof Dr Bayan Salim Chapter 3: Shear & Diagonal Tension 3
2. (in deeper beams) The diagonal crack, once formed, spreads toward into the
compression zone but stop short of penetrating to the compression face. The
failure load is higher and no sudden collapse occurs.
Figure below shows a portion of a beam in which a diagonal tension crack has formed:
Vext = Vint = Vuncracked conc. + Vsteel as a dowel + Vinterlocking
= Vcz + Vd + Viy
Vd and Viy are quite small and can be neglected with very little error. Upon the formation
of diagonal crack, stresses will increase in concrete (point a) and in steel (point b).
Equilibrium of the corresponding forces will establish itself and further load can be
applied before failure occurs.
In design procedures, RC beams are furnished with at least minimum web (shear)
reinforcement.
RC Beams with Web Reinforcement
a. Types of Web Reinforcement
Web reinforcement (or shear reinforcement) is provided in the form of vertical stirrups,
spaced at varying intervals along the axis of the beam depending on requirements, as
shown in (Fig a) below.
Small sized bars are used, generally Nos.10 to 16 (#3 to #5).
Simple U shaped bars similar to (Fig. b) are most common.
Multiple-leg stirrups (Fig. c) are sometimes necessary.
Prof Dr Bayan Salim Chapter 3: Shear & Diagonal Tension 4
Beam reinforcement cage at an edge column-beam joint, before lowering the cage into its
position down in the beam form. It shows the flexural and shear reinforcements. A
cantilever beam is shown also.
Prof Dr Bayan Salim Chapter 3: Shear & Diagonal Tension 5
b. Behavior of Web-Reinforced Concrete Beams
Web (shear) reinforcement is ineffective in the uncracked beam. After diagonal cracks
have developed, web reinforcement augments the shear resistance of a beam in four
separate ways:
1. Part of the shear force is resisted by the bars that traverse a particular crack.
2. The presence of these bars restricts the growth of diagonal cracks and reduces
their penetration into the compression zone.
3. The stirrups also counteract the widening of cracks.
4. The stirrups are arranged so that they tie the longitudinal reinforcement into the
main bulk of the concrete.
BEAMS WITH VERTICAL STIRRUPS In Fig. below, each vertical stirrup traversing the crack exerts a force Avfv.
Av is the cross sectional area of the stirrup (in case of the U-shaped stirrup it is twice the
area of one bar). fv is the tensile stress in the stirrup.
Vext = Vint = Vuncracked conc. + Vsteel as a dowel + Vinterlocking + Vstirrups
= Vcz + Vd + Viy + Vs
The sum of Vcz , Vd and Viy are considered as the contribution of concrete to the total shear
resistance, Vc.
Vext = Vc + Vs
Where Vs = n Avfv , n = no. of stirrups traversing the crack.
If s is the stirrup spacing and p the horizontal projection of the crack, then n = p / s. The
length p is assumed conservatively to be equal to d. Thus,
n = d / s.
Then, Vext = Vn = Vc + Vs
Vn = Vc + Avfy d / s In stress form (divide by bd): vn = vc + Avfy / bs
Prof Dr Bayan Salim Chapter 3: Shear & Diagonal Tension 6
ACI Code Provisions for Shear Design (ACI Code 318-’19)
According to ACI Code 318-’19:
Vu ≤ φ Vn φ = 0.75
Vu ≤ φ (Vc + Vs)
Vu ≤ φ Vc + φ Avfy d / s
(φ is conservative compared to 0.9 for bending. This reflects both the sudden nature of
shear failure and the large scatter of test results)
Shear Strength Provided by the Concrete For rectangular and T beams subject to shear and flexure, according to ACI Code 22.5,
Table 22.5.5.1 and 22.5.5.1.1 through 22.5.5.1.3. the concrete contribution to shear
strength Vc shall be calculated in accordance with
Vc (N); Shear strength. Should Vc ≤ 0.42 λ√ fc
' bwd : max allowed value of Vc (ACI
22.5.5.1.1)
Nu (N); Axial force. Should (Nu / 6Ag) ≤ 0.05 fc' (ACI 22.5.5.1.2)
Ag (mm2); fc' (MPa); bw & d (mm) fc
' < 70 MPa
ρw = As / bwd
λ = lightweight concrete factor
= 1.0 for normal weight concrete
= 0.75 for all-lightweight concrete
λs = size factor,
Eq. (a) is simple and conservative for Vc ; if Nu = 0 then: Vc = 0.17 λ√ fc
' bwd (N)
Prof Dr Bayan Salim Chapter 3: Shear & Diagonal Tension 7
Minimum Shear Reinforcement ACI Code 9.6 requires at least a minimum area of web (shear) reinforcement equal to
Av = 0.062 √f’c (bw s / fyt) ≥ 0.35 bw s / fyt
Where s = longitudinal spacing of web reinforcement, mm
fyt = yield strength of web steel, MPa,
Av = total cross-sectional area of web steel within distance s, mm2
This min Av shall be provided for all regions where Vu > ½[Vc] = ½[φ 0.17λ√fc′ bw d],
except for the cases shown below (Table 9.6.3.1). For these cases, at least min Av shall be
provided where Vu > φVc.
Critical section for shear design
Prof Dr Bayan Salim Chapter 3: Shear & Diagonal Tension 8
Example 1: A rectangular beam is to be designed to carry a shear force Vu of 100 kN. No web
reinforcement is to be used, and f’c = 28 MPa. What is the minimum cross section if