A plane frame has 3 members, 4 joints and is supported by 6 external reaction components (e.g., fixed supports at both bases). It contains one internal shear release (which removes 1 constraint – it allows relative shear displacement but transfers axial force and moment). The degree of static indeterminacy of the frame is
- A0
- B1
- C2Correct
- D3
Solution
Concept
Degree of static indeterminacy with internal releases
Formula
Ds = 3·m + r - 3·j - Σ rr
Step-by-step solution
1. Total static indeterminacy formula for a 2D rigid frame is Ds = 3·m + r - 3·j - Σ rr. 2. Given number of members m = 3, reaction components r = 6, and joints j = 4. 3. The internal shear release removes exactly 1 constraint (rr = 1). 4. Ds = 3·(3) + 6 - 3·(4) - 1 = 9 + 6 - 12 - 1 = 15 - 13 = 2. Answer: C
Explanation
The static indeterminacy of a 2D plane frame is given by Ds = 3m + r - 3j - sum(rr). For a frame with 3 members, 6 reactions, 4 joints, and 1 shear release constraint removed, Ds = 3(3) + 6 - 3(4) - 1 = 15 - 13 = 2.
Why the other options are wrong
A) 0 indicates a determinate frame. B) 1 ignores reaction components or miscounts joints. D) 3 represents the unreleased rigid frame without the shear release.
Shortcut
Ds = 3m + r - 3j - releases = 9 + 6 - 12 - 1 = 2.
Common mistake
Omitting the reduction for the internal release (rr = 1), which gives an incorrect indeterminacy of 3.
Correct answer: C) 2