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## Q. 2.4

Determine the forces produced by the applied loads in members 49, 39, and 89 of the complex truss shown in Figure 2.15.

## Verified Solution

The modified truss shown in Figure 2.16 is created by removing member 39 , adding the substitute member 38 , and applying the 20 kips load. The member forces in the modified truss may now be determined; the values obtained are:

$P_{49}^{\prime}$ = 7.51 kips … tension

$P_{89}^{\prime}$ = -13.98 kips … compression

$P_{38}^{\prime}$ = 21.54 kips … tension

The 20 kips load is removed from the modified truss, and unit virtual loads are applied at nodes 3 and 9 in the direction of the line of action of the force in member 39. The member forces for this loading condition may now be determined; the values obtained are:

$u_{49}$= -0.81 kips … compression

$u_{89}$= -0.50 kips … compression

$u_{38}$= -1.93 kips … compression

The multiplying ratio is given by:

\begin{aligned}P_{38}^{\prime} / u_{38} &=21.54 / 1.93 \\&=11.16\end{aligned}

The final member forces in the original truss are:

\begin{aligned}P_{49} &=7.51+11.16(-0.81) \\&=-1.53 \text { kips … compression }\\ P_{89} &=-13.98+11.16(-0.50) \\ &=-19.56 \text { kips … compression } \\ P_{39} &=P_{38}^{\prime} / u_{38} \\ &=11.16 \text { kips … tension } \end{aligned}