Question 12.P.4: A singly reinforced rectangular concrete beam of effective s...
A singly reinforced rectangular concrete beam of effective span 4.5 m is required to carry a uniformly distributed load of 16.8 kN/m. The overall depth, D, is to be twice the breadth and the centre of the steel is to be at 0.1 D from the underside of the beam. Using elastic theory find the dimensions of the beam and the area of steel reinforcement required if the stresses are limited to 8 N/mm² in the concrete and 140 N/mm² in the steel. Take m = 15.
From Eq. (12.13) \sigma_{\mathrm{s}}=-\sigma_{\mathrm{c}} \frac{E_{\mathrm{s}}}{E_{\mathrm{c}}}\left(\frac{d_{1}-n}{n}\right)=-\sigma_{\mathrm{c}} m\left(\frac{d_{1}-n}{n}\right) for a critical section
140=8 \times 15\left[\frac{0.9 d_{1}}{n}-1\right]
which gives
n = 0.451 d_{1}
Taking moments about the centroid of the steel reinforcement (see Fig. 12.6)
M=\left(\frac{8}{2}\right) \times 0.5 d_{1} \times 0.415 d_{1}\left(0.9 d_{1}-\frac{0.415 d_{1}}{3}\right)
from which
M = 0.632 d_{1}^{3}
Then
0.632 d_{1}^{3}=\frac{16.8 \times 4.5^{2} \times 10^{6}}{8}
so that
d_{1} = 406.7 mm
Then, equating the tensile force in the steel to the compressive force in the concrete
140 A_{\mathrm{s}}=\left(\frac{8}{2}\right) \times(0.5 \times 406.7) \times 0.415 \times 406.7
which gives
A_{s} = 980.6 mm².
