Determine the force in the member AC of the given pin jointed plane frame.

Concept:

In a pin-jointed plane frame in equilibrium, if three forces act at a joint and are in equilibrium, Lami’s Theorem is applicable:

\( \frac{F_1}{\sin\alpha} = \frac{F_2}{\sin\beta} = \frac{F_3}{\sin\gamma} \)

Given:

Joint C is under equilibrium with three forces:
• Force in member AC
• Force in member BC
• A vertical downward load of 6 kN

Coordinates for triangle:
• AB = 8 m
• AC = BC = $\sqrt{4^2 + 3^2} = 5$ m
So, angle θ = tan⁻¹(3/4)

At joint C, all three forces meet:

AC and BC are inclined at θ = tan⁻¹(3/4) = 36.87°

Angle between AC and BC = 180° – 2θ = 106.26°

Angle between AC and vertical load = 90° + θ = 126.87°

Angle between BC and vertical load = 90° + θ = 126.87°

Using Lami's Theorem at Joint C:

\( \frac{T_{AC}}{\sin(126.87°)} = \frac{T_{BC}}{\sin(126.87°)} = \frac{6}{\sin(106.26°)} \)

Using:

• sin(126.87°) = sin(53.13°) ≈ 0.799
• sin(106.26°) ≈ 0.961

Now, calculate the force in member AC:

\( T_{AC} = \frac{6 \times \sin(126.87°)}{\sin(106.26°)} = \frac{6 \times 0.799}{0.961} = 4.99 \approx 5 \, \text{kN} \)

Force in member AC = 5 kN (Tensile)