A uniform beam of length 'l' and flexural rigidity 2EI is fixed at both the ends. What is the moment required for unit rotation at the centre of span?
- A4EI / L
- B8EI / L
- C2EI / L
- D32EI / LCorrect
Solution
Concept
Rotational Stiffness of Fixed Beam
Formula
M = 32EI / L
Step-by-step solution
1. Split the beam of length L into two symmetric half-spans of length L/2 with rigidity 2EI. 2. Apply stiffness formula M = 4EI'/L' for one half-span: M1 = 4(2EI) / (L/2) = 16EI/L. 3. Sum the moments from both left and right halves: M = 16EI/L + 16EI/L = 32EI/L.
Explanation
When a fixed-fixed beam of length L is rotated at its center, it acts as two separate half-span beams, each of length L/2 and flexural rigidity 2EI fixed at one end. The moment required to produce a unit rotation at the free end of one half-span is 4(2EI) / (L/2) = 16EI / L. Because both the left and right halves must be rotated together at the center, the total moment required is the sum of both halves, which is 16EI / L + 16EI / L = 32EI / L.
Shortcut
Split the beam into two symmetric half-spans of length L/2 with fixed ends. The moment for unit rotation at the near end of one half is 4(EIgiven) / (L/2) = 4(2EI) / (L/2) = 16EI/L. Multiply by 2 for both sides to get 32EI/L.
Common mistake
Students often forget to account for the half-length (L/2) of each symmetric segment and fail to sum the moments from both the left and right halves, leading them to incorrectly select 4EI/L or 8EI/L instead of 32EI/L.
Correct answer: D) 32EI / L