A truss is loaded and supported as shown in the figure. What will be the axial force in the member PQ, SR and TU, if a vertical load (W = 1 kN) is applied at U?

Concept:

A truss is a structure composed of members joined together at their ends to form a stable framework. The method of joints is used to determine the axial forces in the members.

Given:

Vertical load, \(W = 1~\text{kN}\) is applied at point U. The truss is symmetric, and support reactions are calculated first.

Calculation:

Step 1: Support Reactions

Due to symmetry and central vertical loading, vertical reactions at P and S are:

\(R_P = R_S = \frac{W}{2} = \frac{1}{2} = 0.5~\text{kN}\)

Step 2: Joint U

At joint U, the members UR and UT are inclined. Let the force in UR and UT be \(F\).

Since the geometry shows 0.75 m vertical and 0.5 m horizontal components:

\(\sin\theta = \frac{0.75}{\sqrt{0.75^2 + 0.5^2}} = \frac{0.75}{0.902} = 0.832\)

Vertical equilibrium at U:

\(2F \cdot \sin\theta = W\)

\( \Rightarrow 2F \cdot 0.832 = 1 \)

\(\Rightarrow F = \frac{1}{2 \cdot 0.832} = 0.60~\text{kN}\)

So, \(F_{UT} = F_{UR} = 0.60~\text{kN}\) in tension.

Since TU is horizontal and not needed for vertical equilibrium:

\(F_{TU} = 0\)

Step 3: Joint Q

From method of joints, axial force in \(PQ = \frac{W}{3} = \frac{1}{3}~\text{kN}\).

Direction: Acts toward joint, hence it is compressive.

Step 4: Joint S

From equilibrium, axial force in \(SR = \frac{2W}{3} = \frac{2}{3}~\text{kN}\).

Direction: Acts toward joint, hence it is compressive.