3 shear and diagonal tension in beams

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


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.


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.


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.


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


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)


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


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

controlled by shear? Take ρ = 0.0080

Solution: Av < Av,min , then

Vc =

Vu = (φVc)

100,000 = φ[0.66(√(2 / 1 + 0.004d))(1)(0.0080)1/3 √28 bw d ]

One solution is bw = 350 mm and d = 800 mm. Check λs:

= √ 0.48 = 0.69 < 1.0 Ok

If min Av is used, then

concrete shear strength = φVc = φ 0.17 λ√ fc' bwd, then bw = 300 mm and d = 500 mm

will be sufficient (check this).

Maximum spacing of shear reinforcement

Max spacing of legs of shear reinforcement along the length of the member and across

the width of the member shall be in accordance with Table 9.7.6.2.2:

In no case, Vs > ⅔ √ f’c bw d, regardless the amount of Av used.


Max s item ensures prevention of potential cracks between legs across length of member;

and provides a more uniform transfer of diagonal compression across the beam width,

enhancing shear capacity. To satisfy this item for wide beams, provide multiple stirrup

legs (see Fig. c, p.4) across these wide beams.

Design of Web Reinforcement In design, it is usually convenient to select a trial web-steel area Av based on standard

stirrup sizes (usually from Nos. 10 to 16), for which the required spacing s can be found.

For vertical stirrups:

s =φ Avfyt d/(Vu – φ Vc)

It is undesirable to space stirrups closer than 100 mm; the size of stirrups should be

chosen to avoid closer spacing.

It is good practice to space stirrups uniformly over the entire distance. If shear varies

throughout this distance, it is more economical to compute spacing required at several

sections and to place stirrups accordingly, in groups of varying spacing.

The portion of any span through which web reinforcement is necessary can be found

from the shear diagram for the span, superimposing a plot of the shear strength of the

concrete. Where Vu > φVc , shear reinforcement must provide for the excess. The

additional length through which at least minimum web steel is needed can be found by

superimposing a plot of (φVc / 2). See Example 2.

Example 2: (Regions of shear reinforcement) A simply supported beam 400 mm wide having an effective depth of 550 mm carries a

total factored load of 137 kN/m on a 6 m clear span. It is reinforced with 6 No. 29 bars

(3870 mm2) of tensile steel throughout. If f’c = 28 MPa, what part of the beam is web

reinforcement required?

Solution:

Vu = 137×6/2 = 411 kN

Vu, d = 411 - 137×0.55 = 336 kN (critical section at distance d from the support)

See (Fig. a) for the linear variation of Vu to zero at midspan.

1. Using simple equation (a) Vc = 0.17 √ f’c bw d

Vc = 0.17 √ f’c bw d = 0.17√28 × 400 × 550 = 197,900 N

φVc = 0.75× 197.9 = 148.4 kN. Superimpose this on shear diagram, and from geometry,

the point at which web reinforcement is no longer required is

3 [(411 – 148.4)/411] = 1.92 m from the support face.

According to ACI Code, at least min Av is required wherever Vu > (φVc / 2)

In this case, φVc / 2 = 74.2 kN.

From geometry, this applies to a distance 3 [(411 – 74.2)/411] = 2.46 m from the support

face.


2. Using more accurate equation (b) Vc =

ρw = 3870/(400×550) = 0.0176 (< ρmax = ?)

Vc = [0.66 (0.0176)1/3 √28 + 0] bw d ≤ 0.42√28 bw d

= 0.9097 bwd ≤ 2.2224 bwd OK

Vc = 200,125 N; φVc = 150,094 N

See the plot in (Fig. b).

From graph, the point at which web reinforcement is no longer required is 1.90 m from

the support face.

From plot of φVc / 2, at least min Av is to be provided within a distance of 2.45 m from

the support face.


Example 3: (Design of shear reinforcement) Using U stirrups with fyt ≤ 420 MPa, design the web reinforcement for the beam in

Example 2. Try No.10 U stirrups. [HW: Try No.13 U stirrups]

Solution:

See shear diagram in Ex.2, Fig. a. The stirrups must be designed to resist that part of the

shear shown shaded.

Apply Table 9.7.6.2.2 for max spacing criteria:

Across length of beam:

Check φ Vs ≤ ⅓ φ √ f’c bw d,

φ Vs = (Vu – φ Vc) = 336 – 148.4 = 187.6 kN

187,600 N < ⅓×0.75×√28×400×550 = 291,000 N OK

smax = Avfyt / [0.062√ f’c bw] ≤ Avfyt / (0.35bw)

= 142×420/[0.062×√28×400 ≤ 142×420 /(0.35×400)

= 454 mm > 426 mm

smax = d/2 = 550 / 2 = 275 mm

smax = 600 mm

Smallest smax = 275 mm is imposed.

Across width of beam:

It should s ≤ d = 550 mm OK (already b = 400 mm < d)

However, available s (across b) = 400 – 2(40) – 10 = 310 mm < d = 550 mm.

At distance d from support face:

s =φ Avfyt d/(Vu – φ Vc)

= 0.75×142×420×550 /187,600 = 131 mm

USE No.10 U stirrups @125 mm

For smax = 275 mm,

(Vu – φ Vc) = φ Avfyt d / s

= 0.75×142×420×550 / 275 = 89,460 N

From Fig. a, Ex.2: A distance x1 from point of zero excess shear,

x1 = 1.92 × 89,460 / 262,600 = 0.65 m. This is (1.92 – 0.65) = 1.27 m from support face.

The 1st stirrup is usually placed at a max distance s/2 from the support.

The following spacing pattern is satisfactory:

1 space @ 50 mm = 50 mm

10 spaces @ 125 mm = 1250 mm → (50 + 1250 = 1300 mm = 1.3 m > 1.27 m OK)

4 spaces @ 275 mm = 1100 mm

Total = 2375 mm = 2.375 m from support face.

See the resulting stirrup pattern in Fig. below:


Effect of Axial Forces RC beams may also be subjected to axial forces, acting simultaneously with shear and

flexure. The main effect of axial load is to modify the diagonal cracking load of the

member. A term (Nu/6Ag ≤ 0.05f’c) is introduced in shear strength equations. See Table

22.5.5.1. (page 6) where:

Nu = axial thrust taken positive for compression, and negative for tension (N)

Ag = gross area of the concrete (mm2)

Example 4: A beam with dimensions: b = 300 mm, d = 600 mm, and h = 700 mm, with f’c = 28

MPa. Find the shear strength of the concrete Vc (a) if no axial forces are present, (b) if

axial compression of 270 kN acts, (c) if axial tension of 270 kN acts.

Solution:

a) Nu = 0;

Vc = 0.17 √ f’c bw d = 0.17√28 × 300 × 600 (10– 3) = 161.9 kN

b) Nu = 270 kN compression;

Vc = [0.17 √ f’c + Nu / 6Ag)] bw d

= [0.17√28 + 270,000/(6×300×700)] 300 × 600 (10– 3)

= [0.900 + 0.214] 300 × 600 (10– 3) = 200.5 kN

Check Nu/6Ag = 0.214 ≤ 0.05f’c = 1.4 MPa OK

c) Nu = 270 kN tension;

Vc = [0.17 √ f’c + Nu / 6Ag)] bw d (Nu is negative)

= [0.900 – 0.214] 300 × 600 (10– 3) = 123.5 kN

3 shear and diagonal tension in beams

Documents