Structural Concrete - Hassoun

176 Chapter 4 Flexural Design of Reinforced Concrete Beams
b. Design of the section at D:Forf c ′ = 4ksi and f y = 60 ksi, ρ max = 0.01806 and ρ min = 0.00333, and
the design steel ratio of 1.5% is within the limits. For ρ = 0.015, R u = 700 psi (from Table A.2) or
from Eq. 3.22:
M u = R u bd 2 or 825.5 × 12 = 0.7(20)d 2
Solve to get d = 26.6 in.:
A s = 0.015 × 20 × 26.6 = 7.98in. 2
Choose eight no. 9 bars in two rows (area 8 in. 2 ), five in the lower row plus three in
the upper row. Minimum b for five no. 9 bars in one row is 14 in. (Table A.7). Total depth
h = 26.6 + 3.5 = 30.1 in. Use h = 30 in. Actual d = 30 −3.5 = 26.5 in. Check the moment capacity
of the section, a = 8 × 60/(0.85 × 4 × 20) = 7.06 in.:
26.5 − 7.06∕2
φM n = 0.9 × 8 × 60 = 826.9K⋅ ft
12
which is greater than 825.5 K ⋅ ft. Check that A s = 8in. 2 is less than A s,max :
A s,max = 0.01806 × 20 × 26.5 = 9.57 in. 2
which exceeds 8 in. 2 The final section is shown in Fig. 4.12.
Example 4.12
The two-hinged frame shown in Fig. 4.13 carries a uniform service dead load (including estimated
self-weight) of 2.33 K/ft and a uniform service live load of 1.5 K/ft on frame beam BC. The moment at
the corner B (or C) can be evaluated for this frame dimension, M b = M c =−wL 2 /18.4, and the reaction
at A or D equals wL/2. A typical section of beam BC is shown; the column section is 16 ×21 in. It is
required to:
a. Draw the bending moment and shear diagrams for the frame ABCD showing all critical values.
b. Design the beam BC for the factored moments, positive and negative, using f c ′ = 4ksi and
f y = 60 ksi. Show reinforcement details.
Solution
a. Calculate the forces acting on the frame using a computer program or the values mentioned previously.
Factored load (w u ) = 1.2(2.33) + 1.6(1.5) = 5.2 K/ft.
Because of symmetry M B = M C =−wL 2 /18.4 =−5.2(40) 2 /18.4 =−452.2 K⋅ft.
Positive moment at midspan (E) = w u L 2 /8 + M B = 5.2(40) 2 /8 −452.2 = 587.8 K⋅ft.
Vertical reaction at A = R A = R D = w u L/2 = 5.2(40)/2 = 104 K. Horizontal reaction at A = H A =
M B /h = 452.2/16 = 28.26 K.
The moment and shear diagrams are shown in Fig. 4.13.
Determine the location of zero moment at section F on beam BC by taking moments about F:
104(y)−28.26(16)−5.2(y) 2 ∕2 = 0 y = 4.963 ft say, 5 ft from joint B
b. Design of beam BC:
1. Design of section E at midspan: M u =+587.8 K⋅ft. Assuming two rows of bars, d = 21
−3.5 = 17.5 in. Calculate the moment capacity of the flange using a = 5.0 in.:
(
φM n (flange) =φ(0.85f c ′ )ab d − 5 )
2
17.5 − 2.5
= 0.9(0.85 × 4)×(5 × 60)× = 1147.5K⋅ ft
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which is greater than the applied moment; therefore, a < 5.0 in.