Question 18.7: The propped cantilever of Fig. 18.11(a) is 10 m long and is ...
The propped cantilever of Fig. 18.11(a) is 10 \mathrm{~m} long and is required to carry a load of 100 \mathrm{kN} at mid-span. If the yield stress of mild steel is 300 \mathrm{~N} / \mathrm{mm}^{2}, suggest a suitable section using a load factor against failure of 1.5.
The required ultimate load of the beam is 1.5 \times 100=150 \mathrm{kN}. Then from Eq. (18.19)
W_{\mathrm{U}}={\frac{6M_{\mathrm{P}}}{L}} (18.19)
the required plastic moment M_{\mathrm{P}} is given by
M_{\mathrm{P}}=\frac{150 \times 10}{6}=250 \mathrm{kN} \mathrm{m}
From Eq. (18.5)
M_{\mathrm{P}}=\sigma_{\mathrm{Y}}Z_{\mathrm{P}} (18.5)
the minimum plastic modulus of the beam section is
Z_{\mathrm{P}}=\frac{250 \times 10^{6}}{300}=833 333 \mathrm{~mm}^{3}
Referring to an appropriate handbook we see that a Universal Beam, 406 \mathrm{~mm} \times 140 \mathrm{~mm} \times 46 \mathrm{~kg} / \mathrm{m}, has a plastic modulus of 886.3 \mathrm{~cm}^{3}. This section therefore possesses the required ultimate strength and includes a margin to allow for its self-weight. Note that unless some allowance has been made for self-weight in the estimate of the working loads the design should be rechecked to include this effect.