Question 12.P.7: Repeat P.12.4 using ultimate load theory assuming σcu = 24 N...

The maximum bending moment is

M_{\max }=\frac{16.8 \times 4.5^{2}}{8}=42.5 \mathrm{\ kN} \mathrm{\ m}

Assume that the neutral axis is 0.45d_{1} from the top of the beam, i.e. M = M_{u}. Then

42.5 \times 10^{6}=0.15 \times 24 \times 0.5 d_{1} \times\left(0.9 d_{1}\right)^{2}

which gives

d_{1} = 307.8 mm

From Eq. (12.24) M_{\mathrm{u}}=0.65 \sigma_{\mathrm{Y}} A_{\mathrm{s}} d_{1},

42.5 \times 10^{6}=0.65 \times 280 A_{\mathrm{s}} \times 0.9 \times 307.8

from which

A_{s} = 843.0 mm²