Structural Concrete - Hassoun

16.13 Summary of Limit Design Procedure 595
Solution
1. Factored uniform load w u = 5.5 K/ft. Factored concentrated load P u = 48 K.
2. The plastic hinges will develop at A, B,andC, causing the mechanism shown in Fig. 16.30. Using
the virtual work method of analysis and assuming a unit deflection at C, then the external work is
equal to
(
W e = 48 × 1 + 5.5 24 × 1 )
= 114 K ⋅ ft
2
The internal work absorbed by the plastic hinges is
W i = M u θ(at A)+M u θ(at B)+M u (2θ) at C
( ) 1
= 4M u θ = 4M u = M u
12 3
Equating W e and W i gives M u = 342 K ⋅ ft. The general analysis gives directly
M u = w u L2
16 + P L
u
8 = 5.5
16 (24)2 + 48 24 = 342 K ⋅ ft
8
3. Design the critical sections at A, B,andC for M u = 342 K ⋅ ft.FromtablesinAppendixAandfor
f c ′ = 3ksi, f y = 40 ksi, and a steel percentage ρ = 0.013, R u = 420 psi (ρ max = 0.0203).
Where
M u = R u bd 2
342 × 12 = 0.42 × 14(d) 2
d is 26.4in. and the total depth h = 26.4 + 2.5 = 28.9in., say, 29 in.,
A s = ρbd = 0.013 × 14 × 26.4 = 4.8in. 2
Use five no. 9 bars in one row; A s provided = 5.0 in. 2 , b min = 13.875 in. < 14 in.
a =
A s f y
0.85 f ′ c b = 5.0 × 40
0.85 × 3 × 14 = 5.6in.
c =
a
0.85 = 6.6in. λ = c d = 6.6
26.4 = 0.25
4. The required rotation of plastic hinges is as follows:
a.
θ A =
L
6E c I [2(M A − M FZ )+(M B − M FB )]
E c = 57,400 √ f ′ c = 3.144 × 10 6 psi
E s = 29 × 10 6 psi and n = E s
E c
= 9.2
b. Determine the fixed end moments at A and B using factored loads:
M FA = M FB = w u L2
12 (uniform load)+P u L (concentrated load)
8
= 5.5 (24)2
12
+ 48 ×
24
8
Plastic M A = plastic M B = 342 K ⋅ ft
= 408 K ⋅ ft