Shear

Shear Design Example CEE 3150 – Reinforced Concrete Design – Spring 2004 Design the shear reinforcement for the following beam: fc0 = 4 kip/in2 fy = 60 kip/in2 ` = 24 ft b = 14 in h = 27 in SDL = 2.5 kip/ft

d0 = 2.5 in d = 22.5 in dt = 24 in A0s = 2.40 in2 (4 – #7) As = 8 in2 (8 – #9) wL = 3.5 kip/ft

1. Find the shear envelope. SW = Self Weight = bhγc = 14 in(27 in)(150 lb/ft3 )(1 foot/12 in)2 (1 kip/1000 lb) = 0.39 kip/ft wD = SW + SDL = 0.39 kip/ft + 2.5 kip/ft = 2.89 kip/ft

(1) (2) (3)

By inspection, the dead load is less than eight times the live load, so w = 1.4DL need not be checked. wu = 1.2wD + 1.6wL = 1.2(2.89 kip/ft) + 1.6(3.5 kip/ft) = 9.07 kip/ft

(4) (5)

Now calculate the maximum factored shear at the support (Vu,sup ) and midspan (Vu,mid ). Vu,sup = wu `/2 = (9.07 kip/ft)(24 ft)/2 = 108.9 kip Vu,mid = 1.6(wL )`/8 = 1.6(3.5 kip/ft)(24 ft)/8 = 16.8 kip

(6) (7)

The formula for the shear as a function of the distance, x, from the support is given by: (Vu,sup − Vu,mid )x `/2 (108.9 kip − 16.8 kip)x = 108.9 kip − (24 ft)/2 = 108.9 kip − (7.68 kip/ ft)x

Vu (x) = Vu,sup −

(8) (9) (10)

At the critical section, distance d from the support: Vu (d) = Vu (22.5 in) = 108.9 kip − (7.68 kip/ ft)(22.5 in)(1 foot/12 in) = 94.5 kip = Vu,max

(11) (12)

The resulting shear envelope is shown in Figure 1. 2. Check maximum shear reinforcing steel requirement. p Vc = 2 fc0 bd p = 2 4, 000 lb/in2 (14 in)(22.5 in)(1 kip/1000 lb) = 39.8 kip p φ(Vc + 8 fc0 bd) p = 0.75(39.8 kip + 8 4, 000 lb/in2 (14 in)(22.5 in)(1 kip/1000 lb)) = 149 kip > Vu,max = 94.5 kip O.K. The section is of adequate size for the shear reinforcing. 1

(13) (14) (15) (16)

100 90 80

Vu (kip)

70 60 50 40 30 20 10 0 0

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6 x (ft)

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Figure 1: The shear envelope. 3. Pick stirrup size and set spacing. To begin, we determine the distance from the support to the end of Zone C. φVc = 0.75(39.8 kip) = 29.9 kip Vu (x) = 108.9 kip − (7.68 kip/ ft)x = φVc = 29.9 kip (108.9 kip − 29.9 kip) ⇒x = = 10.3 ft (7.68 kip/ft)

(17) (18) (19)

Zone C ends 10.3 feet from the support, which is shown graphically by the dotted line in Figure 1. Try a #3 stirrup (Ab = 0.11 in2 ⇒ Av = 0.22 in2 ). Use Vu,max to calculate the stirrup spacing, s. s =

Av fy d Vu,max − Vc φ 2

0.22 in (60 kip/in2 )(22.5 in) 94.5 kip − 39.8 kip 0.75 = 3.45 in =

(20) (21) (22)

Spacing is inadequate: it is best to use s ≥ 4 in. Try a #4 stirrup (Ab = 0.20 in2 ⇒ Av = 0.40 in2 ): 0.40 in2 (60 kip/in2 )(22.5 in) 94.5 kip − 39.8 kip 0.75 = 6.26 in

s =

(23) (24)

Use s = 6 in as the stirrup spacing. The shear capacity of the stirrups, Vs is then calculated as: Av fy d 0.40 in2 (60 kip/in2 )(22.5 in) Vs = = = 90 kip (25) s 6 in p This value must be compared to 4 fc0 bd to check maximum spacing: p p 4 fc0 bd = 4 4, 000 lb/in2 (14 in)(22.5 in)1 kip/1000 lb = 79.7 kip (26) 2

p Thus, because Vs > 4 fc0 bd, the spacing must satisfy s ≤ min(d/4, 12 in) = min((22.5 in)/4, 12 in) = (22.5 in)/4 = 5.63 in

(27)

Use an initial spacing of 5 inches to meet the maximum spacing of 5.6 inches. We should change spacing at most two or three times. If we increase spacing to s = 8 in: 0.40 in2 (60 kip/in2 )(22.5 in) = 67.5 kip (28) 8 in p Because the shear capacity, Vs , at s = 8 in is less than 4 fc0 bd = 79.7 kip, the maximum spacing is governed by: Vs =

s ≤ min(d/2, 24 in) = min((22.5 in)/2, 24 in) = (22.5 in)/2 = 11.25 in

(29)

Thus s = 8 in can be used as soon as Vu (x) drops low enough: φ(Vc + Vs ) = 0.75(39.8 kip + 67.5 kip) = 80.5 kip ≤ Vu (x)

(30)

Solving Vu (x) = 80.5 kip for x gives (see Equation 10): 108.9 kip − (7.68 kip/ ft)x = 80.5 kip 108.9 kip − 80.5 kip = 3.70 ft ⇒x = 7.68 kip/ ft

(31) (32)

We will also consider the maximum spacing, s = 11 in: 0.40 in2 (60 kip/in2 )(22.5 in) = 49.1 kip 11 in φ(Vc + Vs ) = 0.75(39.8 kip + 49.1 kip) = 66.7 kip Vs =

(33) (34)

Solving as before, this spacing may begin 5.5 feet from the support. Moving on to Zone B, the maximum spacing is s = d/2, or s = 11 in as shown above. We also need to check the minimum steel area from {ACI 11.5.5.3}: p √ 0.75 fc0 bw s 0.75 4, 000 lb/in2 (14 in)(11 in) Av ≥ = (35) fy 60, 000 lb/in2 ≥ 0.12 in2 (36) 50bw s 50 psi(14 in)(11 in) Av ≥ = (37) fy 60, 000 lb/in2 ≥ 0.13 in2 (38) The larger area (Av ≥ 0.13 in2 ) controls. A #3 would work (Av = 2Ab = 2(0.11 in2 ) = 0.22 in2 ), but we do not want to change bar size. Use a #4 at spacing s = 11 in. Lastly, determine where Zone A starts: φVc 0.75(39.8 kip) = = 14.9 kip < Vu,mid = 16.8 kip 2 2

(39)

Thus, stirrups are required everywhere since there is no point on the beam where the concrete alone is twice as strong as it needs to be. That is, there is no Zone A for this problem. 3

Rebar at Center 8 @ 5"

4 @ 8"

6 @ 11"

10"



 2"

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11

x (feet)

Figure 2: Stirrup design layout. 4. Sketch. The first stirrup will be placed a distance s/2 from the support. The initial spacing is 5 inches so that the first stirrup would be at 2.5 inches. But, we want to detail everything in whole inches so use an initial space of 2 inches. Nine additional stirrups at 5 inch spacing takes us to 47 inches past the support or 3.9 ft—a bit past the 3.70 feet calculated above in Equation 32. Four more stirrups at 8 inch spacing takes us to 71 inches (5.9 feet) and 6 more stirrups at 11 inches on center will leave a 10 inch distance to the reinforcing bar at the midspan.

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