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## Q. 2.3

Determine the forces produced by the applied loads in the members of the truss shown in Figure 2.13.

## Verified Solution

The support reactions are calculated as shown. By inspection, the force in the vertical web members is obtained directly. Thus, the force in member 23 is given as:

\begin{aligned}P_{23} &=W_{3} \\&=10 \text { kips … tension } \\P_{45} &=0\end{aligned}

The force in the diagonal web member 12 is given by the magnitude of the shear force in the first panel multiplied by the coefficient l/h. Thus, the force in member 12 is:

\begin{aligned}P_{12} &=V_{1} \times l / h \\&=15 \times 14.14 / 10 \\&=21.21 \text{kips… compression }\end{aligned}

The force in the diagonal web member 25 is given by the magnitude of the shear force in the second panel multiplied by the coefficient l/h. Thus, the force in member 25 is:

\begin{aligned}P_{25} &=\left(V_{1}-W_{3}\right) \times l / h \\&=5 \times 14.14 / 10 \\&=7.07 \text { kips … tension }\end{aligned}

The force in the bottom chord member 13 is given by the magnitude of the moment at node 2 multiplied by the coefficient 1/h. Thus, the force in member 13 is:

\begin{aligned}P_{13} &=M_{2} \times 1 / h \\&=V_{1} \times a \times 1 / h \\&=150 \times 1 / 10 \\&=15 \text { kips …tension }\end{aligned}

Similarly, the force in the bottom chord member 35 is given by the magnitude of the moment at node 2 multiplied by the coefficient 1/h. Thus, the force in member 35 is:

\begin{aligned}P_{35} &=M_{2} \times 1 / h \\&=V_{1} \times a \times 1 / h \\&=150 \times 1 / 10 \\&=15 \text { kips … tension }\end{aligned}

The force in the top chord member 24 is given as the magnitude of the moment at node 5 multiplied by the coefficient 1/h. Thus, the force in member 24 is:

\begin{aligned}P_{24} &=M_{5} \times 1 / h \\&=\left(V_{1} \times 2 a-W_{3} \times a\right) \times 1 / h \\&=(300-100) \times 1 / 10 \\&=20 \text { kips … compression }\end{aligned}