Structural Concrete - Hassoun

21.8 V-Shape Beams Subjected to a Concentrated Load at the Centerline of the Beam 883
Solution
Given θ = π/2 and the span L = 2a = the distance between the two supports. The bending moment at
the centerline is
M c = Pa ( 1
=
4 1)
Pa
4 = PL =+90 K ⋅ ft
8
M A = M c − Pa
2 = PL
8 − P ( ) L
=− PL =−90 K ⋅ ft
2 2 8
T A = T c = 0
These values are similar to those obtained from the structural analysis of the fixed-end beam subjected
to a concentrated load at midspan.
Example 21.8
The beam shown in Fig. 21.11 has a V-shape in plan and carries a uniform dead load of 3.5 k/ft and a
live load of 3 K/ft. The inclined length of half the beam is a = 10 ft and θ = 60 ∘ . Design the beam for
shear, bending, and torsional moments using f c ′ = 4ksiandf y = 60 ksi.
Solution
1. w u = 1.2 D + 1.6 L = 1.2 × 3.5 + 1.6 × 3 = 9.0 K/ft.
2. Assuming a rectangular section with a ratio of long to short side of y/x = 1.75, the value of λ is
2.77 (from Table 21.2). For θ = 60 ∘ = π/3,
M c =
w u a 2 sin 2 θ
6(sin 2 θ + λ cos 2 θ) = 9(100)(0.75)
=+78 K ⋅ ft
6(0.75 + 2.77 × 0.25)
a
M A = M c − w 2 100
u = 78 − 9 ( )=−372 K ⋅ ft
2 2
T A = M c cot θ = 78 × 0.577 = 45 K ⋅ ft = 540 K ⋅ in.
T c (at x = 0) =M c cot θ = 45 K ⋅ ft = 540 K ⋅ in.
V A = 9 × 10 = 90 K
The bending moment is zero at M N = 0 = M c −w u x 2 /2. Hence, 78 − 9 2 x2 = 0andx= 4.16 ft
measured from c. The bending moment diagram is shown in Fig. 21.11.
3. Design for a bending moment, M u , equal to –372 K⋅ft.
a. For f c ′ = 4ksi, f y = 60 ksi, ρ max = 0.0018, choose ρ = 0.015, R u = 702 psi and φ = 0.9
(Appendix A).
bd 2 = M u 372 × 12
= = 6332 in. 3
R u 0.705
For a ratio,
y
x = d + 3 = 1.75
b
as assumed, then d = 21.4 in. and b = 13.8 in. Use a section 14 × 24 in.
A s = ρ max bd = 0.015(14 × 21.4) =4.5in. 2