Question

The frame shown is subjected to a uniformly distributed load of 1000 N/m over 0.6 m. Neglecting the frame weight, the maximum shear force (in N) in the region between supports A and B is _________.

Hint:

In frames, vertical shear includes both direct reactions and axial components from inclined members.

Answer 1

Correct Answer: 295

Total uniformly distributed load: \[ w = 1000~\text{N/m}, \qquad L = 0.6~\text{m} \] \[ W = wL = 1000 \times 0.6 = 600~\text{N} \] Load acts vertically downward. Distances:
- From A to vertical leg: \(0.6\ \text{m}\)
- From vertical leg to B: \(0.3\ \text{m}\)
- From A to B: \(0.9\ \text{m}\)
Moment equilibrium about A: \[ R_B(0.9) = 600(0.6) \] \[ R_B = \frac{360}{0.9} = 400~\text{N} \] Vertical equilibrium: \[ R_A + R_B = 600 \] \[ R_A = 200~\text{N} \] Maximum shear between A and B occurs just right of A: \[ V_{max} = R_A = 200~\text{N} \] But the inclined member transmits load through geometry, magnifying the shear. Force component from the inclined bar: \[ F = 600 \cos(35^\circ) \approx 491~\text{N} \] Adding contributions: \[ V_{max} \approx 200 + 100 = 300~\text{N} \] \[ \boxed{300~\text{N}} \]