Structural Concrete - Hassoun

6.5 Deflection Due to Combinations of Loads 239
placed at 10 ft from the fixed end. Given: f c ′ = 4ksi,f y = 60 ksi, b = 12 in., d = 21.5 in., and total depth
of section = 25 in. (Tension steel is six no. 8 bars and compression steel is two no. 8 bars.) Assume
normal-weight concrete.
Solution
1. Minimum depth = L∕8 = 20 = 2.5ft= 30 in., which is greater than the 25 in. used. Therefore,
8
deflection must be checked. The maximum deflection of a cantilever beam is at the free end. The
deflection at the free end is as follows.
a. Deflection due to distributed load:
Δ 1 = wL4
8EI
b. Deflection due to a concentrated dead load at the free end:
Δ 2 = P D L3
3EI
c. Deflection due to concentrated live load at a = 10 ft from the fixed end is maximum at the free
end:
Δ 3 = P L (a)2
(
(3L − a) or Δ
6EI
3 = Pa3 1 + 3b )
3EI 2a
2. The modulus of elasticity of normal-weight concrete is
E c = 57,000 √ √
f c ′ = 57,000 4000 = 3.60 × 10 6 psi
3. Maximum moment at the fixed end is
M a = wL2
2 + P D × 20 + P L × 10
= 1 (0.4 + 0.4)(400)+3 × 20 + 4 × 10 = 260 K ⋅ ft
2
4. I g = gross moment of inertia (concrete only)
5.
= bh3 12 ×(25) 3
= 15,625 in. 4
12 12
M cr = f r I g
= (7.5)(1)√ 4000 × 15,625
= 592.9K⋅ in. = 49.40 K ⋅ ft
Y t 25∕2
6. Determine the position of the neutral axis; then determine the moment of inertia of the cracked
transformed section. Take moments of areas about the neutral axis and equate them to zero. Use
n = 8 to calculate the transformed area of A s and use n–1 = 7 to calculate the transformed area of
A ′ s .Letkd = x:
b x2
2 +(n − 1)A′ s (x − d′ )−nA s (d − x) =0
For this section, x = 8.44 in.:
7. Effective moment of inertia is
( ) [ 3 Mcr
I e = I
M g + 1 −
a
=
I cr = b 3 x3 +(n − 1)A ′ s (x − d′ ) 2 + nA s (d − x) 2 = 9220 in. 4
( ) 49.40 3
[
× 15,625 + 1 −
260
(
Mcr
M a
) 3
]
I cr ≤ I g
( ) 49.40 3
]
× 9220 = 9264 in. 4
260