A horizontal bar CBD having length of 2.4 m is supported and loaded as shown in Figure 2.17. The vertical member AB has cross sectional area of 550 mm². Determine the magnitude of load P so that it produces a normal stress equal to 40 MPa in member AB.

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The system shown in Figure 2.17 is in equilibrium. The nature of internal force developed in AB is essentially tensile. Taking moment of forces about the hinge C and equating to zero, we get

\begin{gathered} \sum M_c=0 \\ P \times 2.4=F_{ AB } \times 1.5 \\ P=\frac{F_{ AB } \times 1.5}{2.4}=\frac{40 \times 550 \times 1.5}{2.4}=13750 N =13.75 kN \end{gathered}

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