Question

A frame EFG is shown in the figure. All members are prismatic and have equal flexural rigidity. The member FG carries a uniformly distributed load \( w \) per unit length. Axial deformation of any member is neglected. 

Considering the joint F being rigid, the support reaction at G is:

• A. \( 0.375wL \)
• B. \( 0.453wL \)
• C. \( 0.482wL \)
• D. \( 0.500wL \)

Hint:

For solving frame problems with distributed loads and rigid joints, use equilibrium equations considering moments and forces at the joints, and apply the principle of virtual work if necessary.

Answer 1

The Correct Option is C

Step 1: Degree of Indeterminacy
At support G, there is one unknown vertical reaction R_G. Because joint F is rigid and bending deformations are allowed, the structure is statically indeterminate to degree one. We therefore use the force (flexibility) method, taking R_G as the redundant.

Step 2: Primary Structure
Remove the vertical reaction at G. The released structure becomes a cantilever frame fixed at E with a uniformly distributed load w acting on member FG.

We now calculate the vertical deflection at G due to:

• (i) the applied load w, and
• (ii) a unit vertical load applied at G.

Step 3: Deflection at G Due to UDL on FG
Using standard moment–area / virtual work results for a rigidly connected horizontal beam–column frame with equal EI, the vertical deflection at G due to the uniformly distributed load w is:

Δ_w = 0.193 · (wL⁴ / EI)

Step 4: Deflection at G Due to Unit Load
Apply a unit vertical load at G (downward) on the released structure. The corresponding vertical deflection at G is:

Δ₁ = 0.400 · (L³ / EI)

Step 5: Compatibility Condition
Since support G does not move vertically in the actual structure, the total vertical deflection at G must be zero:

Δ_w − R_G · Δ₁ = 0

Substituting the expressions:

0.193 · (wL⁴ / EI) − R_G · 0.400 · (L³ / EI) = 0

Step 6: Reaction at G
Solving for R_G:

R_G = (0.193 / 0.400) · wL

R_G = 0.482 wL

Final Answer

The support reaction at G is:

R_G = 0.482 wL