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Chapter 2

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.

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

Step-by-Step

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