5.1 A rectangular beam secƟon is 10 in. wide and has a structural depth of 13.5 in. Given that the
concrete strength is 4000 psi and there is no shear reinforcement in the secƟon, determine the
nominal shear capacity allowed by the ACI Code.
5.2 A rectangular beam secƟon is 14 in. wide and has a structural depth of 20.5 in. Given that the
secƟon has No. 3 (No. 10) sƟrrups at 9 in., a specified concrete strength of 3000 psi, and fyt of 60,000
psi, calculate the factored shear capacity of the secƟon.
5.3 A rectangular beam secƟon is 16 in. wide and has a structural depth of 26 in. Given that the
secƟon has No. 3 (No. 10) sƟrrups at 12 in., a specified concrete strength of 4000 psi, and fyt of 60,000
psi, calculate the factored shear capacity of the secƟon.
≔Vu=⋅ϕV
n59.3 kip
5.4 A rectangular beam secƟon is 16 in. wide and has a structural depth of 26 in. Given that the secƟon
has No. 4 (No. 13) sƟrrups at 13 in., a specified concrete strength of 4000 psi, and fyt of 60,000 psi,
calculate the factored shear capacity of the secƟon.
Note: Both problem 5.3 and 5.4 have the maximum allowable spacing of the stirrups equal to
d
/2. Going from a No. 3 (No. 10) to a No. 4 (No. 13) stirrup increases the sectional capacity 27
5.5 A Tsecon shown in Figure P5.5 has a structural depth d= 22 in. a web width bw= 6 in. and a
flange width of bf= 36 in. If the secon has a specified concrete strength of 5000 psi and No. 3 (No.
10) srrups at 10 in, determine the factored shear capacity of the secon.
5.6 A rectangular beam secƟon is 16 in. wide and has a structural depth of 26.5 in. Given that the
factored shear on the secƟon is 90 kips, and a specified concrete strength of 4000 psi, determine t he
required spacing of No. 4 (No. 13) sƟrrups.
≔b16in ≔d 26.5 in ≔λ1.0 ≔f’c4000 psi ≔fyt 60000 psi
20.40 in2≔ϕ0.75
5.7 A Tsecon shown in Figure P5.5 has a structural depth d = 24 in. a web width bw= 8 in. and a
flange width of bf= 36 in. If the secon has a specified concrete strength of 5000 psi and a factored
shear of 50 kips, determine the spacing of No. 3 (No. 10) srrups.
5.8 A simple span rectangular beam has and effecve length of 18 , a width of 14 in. and a struc tural
depth of 24 in. It is reinforced with 3 No. 9 (No. 29) bars longitudinally and No. 3 (No. 10) srrups at
12 in. on center over the enre length. Determine the maximum factored load the beam can carry in
plf. The specified concrete strength is 5000 psi, fy= 60,000 psi, and and fyt = 40,000psi.
≔b14in ≔d24in ≔l18ft ≔f’c5000 psi ≔λ1.0
≔As3.0 in2≔Av⋅20.11in
2≔fy60000 psi ≔fyt 40000 psi
Flexural capacity ≔ϕ0.90
⋅Asfy
5.9. A beam is to be designed for loads causing a maximum factored shear of
60.0 kips, using concrete with f c = 5000 psi. Proceeding on the basis that the
concrete dimensions will be determined by diagonal tension, select the appropriate
width and effective depth ( a ) for a beam in which no web reinforcement
is to be used, ( b ) for a beam in which only the minimum web reinforcement is
provided, as given by Eq. (5.13), and ( c ) for a beam in which web reinforcement
provides shear strength V s = 2 V c . Follow the ACI Code requirements,
and let d = 2 b in each case. Calculations may be based on the more approximate
value of V c given by Eq. (5.12 d ).
Vu60kip
f’c5000psi fy60000psi fyt 60000psi ϕ0.75 λ1.0
vc2λf’cpsi141psi vuϕvc
106psi
Reinforcement details
SOLUTION
a) No web reinforcement
3vu
5.10. A rectangular beam having b = 10 in. and d = 17.5 in. spans 15 ft face to face of
simple supports. It is reinforced for flexure with three No. 9 (No. 29) bars that
continue uninterrupted to the ends of the span. It is to carry service dead load
D = 1.27 kips/ft (including self-weight) and service live load L = 3.70 kips/ft,
both uniformly distributed along the span. Design the shear reinforcement,
using No. 3 (No. 10) vertical U stirrups. The more approximate Eq. (5.12 d )
for V c may be used. Material strengths are f c = 4000 psi and f y = 60,000 psi.
l15ft d 17.5in b10in
L 3.70 kip
ft
 D 1.27 kip
ft

λ1.0 ϕ0.75
f’c4000psi fy60000psi fyt 60000psi
Reinforcement details
SOLUTION
wu1.2 D1.6 L7.44 kip
ft
 Av2As3 0.22 in2

Vu
wul
255.83 kip
7.5′