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## Q. 1.2

Determine the angle of inclination and magnitude of the support reaction at end 1 of the pin-jointed truss shown in Figure 1.11 . End 1 of the truss has a hinged support, and end 2 has a roller support.

## Verified Solution

Taking moments about support 1 gives:

$V_{2}$ = 8 × 20/16

= 10 kips … acting vertically upward

The triangle of forces is shown at (ii), and the magnitude of the reaction at support 1 is given by:

\begin{aligned}R^{2} &=\left(V_{2}\right)^{2}+\left(F_{3}\right)^{2}-2 V_{2} F_{3} \cos f_{r} \\&=10^{2}+20^{2}-2 \times 10 \times 20 \cos 90^{\circ} \\R &=(100+400)^{0.5} \\&=22.36 \text{kips}\end{aligned}

The angle of inclination of R is:

\begin{aligned}\theta &=\operatorname{atan}(10 / 20) \\&=26.57^{\circ}\end{aligned}

Alternatively, since the three forces are concurrent, their point of concurrency is at point 6 in Figure 1.11 (i), and:

\begin{aligned}\theta &=\operatorname{atan}(8 / 16) \\&=26.57^{\circ}\end{aligned}

and

\begin{aligned}R &=20 \sin 90^{\circ} / \sin 63.43^{\circ} \\&=22.36 \text {kips}\end{aligned}

The reaction R may also be resolved into its horizontal and vertical components:

\begin{aligned} H_{1} &=R \cos \theta \\ &=20 \text { kips } \\ V_{1} &=R \sin \theta \\ &=10 \text { kips } \end{aligned}