Structural Concrete - Hassoun

242 Chapter 6 Deflection and Control of Cracking
Solution
1. The beam AB is subjected to a positive moment that causes a deflection downward at midspan
and negative moments at the two ends, causing a deflection upward at midspan. As was explained
earlier, the deflection is a function of the effective moment of inertia, I e . In a continuous beam,
the value of I e to be used is the average value for the positive- and negative-moment regions.
Therefore, three sections will be considered: the section at midspan and the sections at the two
supports.
2. Calculate I e : For the gross area of all sections, kd = 13.5 in. and I g = 114,300 in. 4 Also,
f r = 7.5λ √ f c ′ = 410 psi and E e = 57,000 √ f c ′ = 3.12 × 10 6 for all sections. The values of kd, I cr ,
and M cr for each cracked section, I e for dead load only (using M a of dead load), and I e for dead
and live loads (using M a for dead and live loads) are calculated and tabulated as follows:
Section
kd
(in.)
I cr
M cr
(in. 4 ) y t
(K⋅ft)
I e
(in. 4 )
(Dead Load)
I e
(in. 4 )
(D + L)
Midspan 6.67 48,550 24.5 159.4 86,160 50,960
Support A 10.9 34,930 13.5 289.3 114,300 60,880
Support B 12.6 44,860 13.5 289.3 114,300 55,415
Note that when the beam is subjected to dead load only and the ratio M cr /M a is greater than
1.0, I e = I g .
3. Calculate average I e from Eq. 6.8:
I e1 (average) =0.7(50,960)+0.15(60,880 + 55,415)
For dead and live loads,
= 53,116 in. 4
Average I e for end sections = 1 (60,880 + 55,415)
2
= 58,150 in. 4
I e2
(average) = 1 (50,960 + 58,150) =54, 550 in.4
2
For dead loads only,
Average I e for end sections = 114,300 in. 4
I e3
(average) = 1 (86,160 + 114,300) =100,230 in.4
2
4. Calculate immediate deflection at midspan:
Δ 1 (due to uniform load) = 5wL4
384EI e
Δ 2 (due to a moment at A, M A )= M A L2
16EI e
Δ 3 (due to a moment at B, M B )=− M B L2
16EI e
(downward)
(upward)
(upward)
Total deflection Δ = Δ 1 − Δ 2 − Δ 3