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.
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}