Consider a truss PQR loaded at P with a force F as shown in the figure. The tension in the member QR is

Concept:

To find the tension in a member of a truss, we can use either Lami’s Theorem or the Method of Joints. At joint P, three forces act in equilibrium: vertical load F, force in member PQ (45°), and force in member PR (30°). Since PR and QR are collinear, the force in PR is the same as the tension in QR.

Calculation:

Using Lami’s Theorem at joint P:

The angles between the forces are:

• Between PQ and PR: 45° + 30° = 75°
• Between F and PR: 60°
• Between F and PQ: 45°

Apply Lami's Theorem:

\( \frac{T_{PQ}}{\sin(60^\circ)} = \frac{T_{PR}}{\sin(45^\circ)} = \frac{F}{\sin(75^\circ)} \)

We are interested in:

\( T_{QR} = T_{PR} = \frac{F \cdot \sin(45^\circ)}{\sin(75^\circ)} \)

Now, plug in the values:

• \( \sin(45^\circ) = 0.7071 \)
• \( \sin(75^\circ) = 0.9659 \)

\( T_{QR} = \frac{F \cdot 0.7071}{0.9659} \approx 0.732 F \)

Alternate Method (Method of Joints):

Let T₁ be force in PQ (at 45°), and T₂ be force in PR (at 30°):

Vertical equilibrium:

\( T_1 \sin(45^\circ) + T_2 \sin(30^\circ) = F \)

Horizontal equilibrium:

\( T_2 \cos(30^\circ) = T_1 \cos(45^\circ) \)

\( T_1 = \frac{0.866}{0.7071} T_2 \approx 1.2247 T_2 \)

Substitute into vertical equation:

\( 1.2247 T_2 \cdot 0.7071 + T_2 \cdot 0.5 = F \)

\( T_2 (0.866 + 0.5) = F \Rightarrow T_2 = \frac{F}{1.366} \approx 0.732 F \)