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)

Question

The primary purpose of mine ventilation is to:

Answer 1

Let's find the same force by starting the method of joints at P instead of R, which is an equally valid check on member PR.

The triangle's third angle, at Q, is $180-60-30=90^\circ$. With $PR=5\ \text{m}$, the sine rule gives $PQ=5\sin(30^\circ)/\sin(90^\circ)=2.5\ \text{m}$ and $QR=5\sin(60^\circ)/\sin(90^\circ)=4.33\ \text{m}$.

Taking moments about P to find the roller reaction at R: the horizontal distance from P to Q is $PQ\cos60^\circ=1.25\ \text{m}$, so $R_R \times 5 = 10 \times 1.25$, giving $R_R=2.5\ \text{kN}$ upward. By vertical equilibrium of the truss, $R_P=10-2.5=7.5\ \text{kN}$ upward, with no horizontal reaction needed at P since there is no horizontal load anywhere on the truss.

Now isolate joint P. Two members meet there: PQ, rising toward Q at $60^\circ$ above the horizontal, and PR, running horizontally toward R. The support reaction $R_P=7.5\ \text{kN}$ also acts upward at this joint.

Vertical equilibrium at P: $R_P + F_{PQ}\sin60^\circ = 0 \Rightarrow F_{PQ} = -7.5/0.8660 = -8.66\ \text{kN}$, so PQ is in compression.

Horizontal equilibrium at P: $F_{PQ}\cos60^\circ + F_{PR} = 0 \Rightarrow F_{PR} = -F_{PQ}\cos60^\circ = 8.66 \times 0.5 = 4.33\ \text{kN}$.

Let's summarize:

The truss's third angle works out to 90 degrees, fixing the geometry through the sine rule.
Moments about P give the roller reaction at R, and overall vertical balance gives the reaction at P.
Resolving forces at joint P instead of joint R gives the same tension of 4.33 kN in member PR.

So member PR carries a tensile force of about 4.33 kN.

Answer 2

Let's find the same force by starting the method of joints at P instead of R, which is an equally valid check on member PR.

The triangle's third angle, at Q, is $180-60-30=90^\circ$. With $PR=5\ \text{m}$, the sine rule gives $PQ=5\sin(30^\circ)/\sin(90^\circ)=2.5\ \text{m}$ and $QR=5\sin(60^\circ)/\sin(90^\circ)=4.33\ \text{m}$.

Taking moments about P to find the roller reaction at R: the horizontal distance from P to Q is $PQ\cos60^\circ=1.25\ \text{m}$, so $R_R \times 5 = 10 \times 1.25$, giving $R_R=2.5\ \text{kN}$ upward. By vertical equilibrium of the truss, $R_P=10-2.5=7.5\ \text{kN}$ upward, with no horizontal reaction needed at P since there is no horizontal load anywhere on the truss.

Now isolate joint P. Two members meet there: PQ, rising toward Q at $60^\circ$ above the horizontal, and PR, running horizontally toward R. The support reaction $R_P=7.5\ \text{kN}$ also acts upward at this joint.

Vertical equilibrium at P: $R_P + F_{PQ}\sin60^\circ = 0 \Rightarrow F_{PQ} = -7.5/0.8660 = -8.66\ \text{kN}$, so PQ is in compression.

Horizontal equilibrium at P: $F_{PQ}\cos60^\circ + F_{PR} = 0 \Rightarrow F_{PR} = -F_{PQ}\cos60^\circ = 8.66 \times 0.5 = 4.33\ \text{kN}$.

Let's summarize:

The truss's third angle works out to 90 degrees, fixing the geometry through the sine rule.
Moments about P give the roller reaction at R, and overall vertical balance gives the reaction at P.
Resolving forces at joint P instead of joint R gives the same tension of 4.33 kN in member PR.

So member PR carries a tensile force of about 4.33 kN.

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)

Show Hint

Correct Answer: 4.33

Solution and Explanation

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