Structural Concrete - Hassoun

314 Chapter 8 Design of Deep Beams by the Strut-and-Tie Method
NO. 4 @ 12 in. c/c
NO. 4 @ 12 in. c/c
40" 40"
NO. 8 bars
6- NO. 8 bars
12"
16"
10' = 120" 16"
Figure 8.19
Reinforcement details.
Development length of No. 8 bars required = 47.5(1) = 47.5 inch (table 7), which is greater
than 19 in. Use a standard 90 ∘ hook enclosed with column reinforcement:
l dh
(0.02ψ e f y )d b
λ √ f ′ c
Reinforcement details are shown in Fig. 8.19.
= 0.02(1.0)(60000)(1.0)
(1.0) √ 4000
= 19 ≤ 19 in. OK
Example 8.3 AASHTO LRFD Method
A simply supported deep beam has a depth of 4 ft. The clear span is 4 ft, and the width is 1 ft (Fig. 8.20).
The beam is carrying a total factored load of 250 K at midspan. Design the beam using the strut-and-tie
model. Assume f c ′ = 4ksi,f y = 60 ksi, and bearing pads of 8 in. × 8in.
250 K
A
4"
4'
B
60°
C
4"
4'
R A
R A = R B = 125 KR B
Solution
Figure 8.20
Example 8.3: Idealized deep beam.
1. Calculate reactions. Total factored load given is 250 K. The reaction is
R A = R B = 125 K