Question

A truss PQR carries a vertical load of \(10\ \text{kN}\) at Q as shown. The force in member PR, in \(kN\), is . (rounded off to two decimal places)

Hint:

Work out the missing angle of the triangle to place joint Q, find the support reactions first, then apply the method of joints at one of the supports.

Answer 1

Correct Answer: 4.33

Step 1: Work out the geometry of the truss.
The figure shows the base PR is \(5\ \text{m}\), with the member at P making \(60^\circ\) with PR and the member at R making \(30^\circ\) with PR. Since the three angles of triangle PQR add to \(180^\circ\), the angle at Q is \(180-60-30=90^\circ\).
By the sine rule, \(PQ=PR\dfrac{\sin(30^\circ)}{\sin(90^\circ)}=5 \times 0.5=2.5\ \text{m}\) and \(QR=PR\dfrac{\sin(60^\circ)}{\sin(90^\circ)}=5 \times 0.8660=4.33\ \text{m}\).

Step 2: Find the support reactions.
P is a pin support (can push in any direction) and R is a roller (vertical reaction only). The horizontal distance of Q from P is \(PQ\cos60^\circ=2.5 \times 0.5=1.25\ \text{m}\).
Taking moments about P: \(R_R \times 5=10 \times 1.25\), so \(R_R=2.5\ \text{kN}\) (upward).
Vertical equilibrium of the whole truss gives \(R_P=10-2.5=7.5\ \text{kN}\) (upward); there is no applied horizontal load, so the horizontal reaction at P is zero.

Step 3: Apply the method of joints at R.
At joint R, member QR pulls up and to the left at \(30^\circ\) above the horizontal, and member PR runs horizontally toward P. Taking tension as positive:
Vertical: \(F_{QR}\sin30^\circ + R_R=0 \Rightarrow F_{QR}=-\dfrac{2.5}{0.5}=-5\ \text{kN}\) (compression).
Horizontal: \(-F_{QR}\cos30^\circ - F_{PR}=0 \Rightarrow F_{PR}=-F_{QR}\cos30^\circ=5 \times 0.8660=4.33\ \text{kN}\).

Step 4: Final Answer.
The positive sign shows member PR is in tension, carrying \(4.33\ \text{kN}\). \[ \boxed{4.33} \]