Question 4.SP.4: A concrete floor slab is reinforced by 5/8-in.-diameter stee...
A concrete floor slab is reinforced by \frac{5}{8}-in.-diameter steel rods placed 1.5 in. above the lower face of the slab and spaced 6 in. on centers. The modulus of elasticity is 3.6 × 10^{6} psi for the concrete used and 29 × 10^{6} psi for the steel. Knowing that a bending moment of 40 kip • in. is applied to each 1-ft width of the slab, determine (a) the maximum stress in the concrete, (b) the stress in the steel.
Transformed Section. We consider a portion of the slab 12 in. wide, in which there are two \frac{5}{8}-in.-diameter rods having a total cross-sectional area
A_{s}=2\left[\frac{\pi}{4}\left(\frac{5}{8} in. \right)^{2}\right]=0.614 in ^{2}
Since concrete acts only in compression, all the tensile forces are carried by the steel rods, and the transformed section consists of the two areas shown. One is the portion of concrete in compression (located above the neutral axis), and the other is the transformed steel area nA_{s}. We have
n =\frac{E_{s}}{E_{c}}=\frac{29 \times 10^{6} psi }{3.6 \times 10^{6} psi }=8.06
n A_{s} = 8.06\left(0.614 in ^{2}\right)=4.95 in ^{2}
Neutral Axis. The neutral axis of the slab passes through the centroid of the transformed section. Summing moments of the transformed area about the neutral axis, we write
12 x\left(\frac{x}{2}\right)-4.95(4-x)= 0 \quad x= 1.450 in.
Moment of Inertia. The centroidal moment of inertia of the transformed area is
I=\frac{1}{3}(12)(1.450)^{3}+4.95(4-1.450)^{2}=44.4 in ^{4}
a. Maximum Stress in Concrete. At the top of the slab, we have c_{1} = 1.450 in. and
\sigma_{c}=\frac{M c_{1}}{I}=\frac{(40 kip \cdot in. )(1.450 in. )}{44.4 in^{4}} \sigma_{c}= 1.306 ksi
b. Stress in Steel. For the steel, we have c_{2} = 2.55 in., n = 8.06 and
\sigma_{s}=n \frac{M c_{2}}{I}=8.06 \frac{(40 kip \cdot in. )(2.55 in. )}{44.4 in ^{4}} \sigma_{s}= 18.52 ksi
