A cantilever AB of length 5 m (fixed at A) carries a uniformly varying load whose intensity is 20 kN/m at the fixed end A and zero at the free end B. EI is constant. The slope at the free end is:
Try it first: pick an option, then read the explanation.
Asked in SSC JE 04/12/2023 (9 am–11 am)
Detailed Explanation
For a cantilever carrying a triangular load with its peak w at the fixed end and zero at the free end, the standard results are θ = wL³/24EI and δ = wL⁴/30EI at the free end.
\[ \theta_B = \frac{wL^3}{24EI} = \frac{20\times5^3}{24EI} = \frac{2500}{24EI} \approx \frac{104.17}{EI} \]
Why “104.16/EI” Is Correct
Substituting w = 20 kN/m and L = 5 m gives 104.16/EI (kN·m² units). The total load here is only wL/2 = 50 kN, and it is concentrated near the support, which is why the slope is smaller than for a UDL of the same peak (wL³/6EI = 416.7/EI).
Why the Other Options Are Wrong
- A. 129.27/EI: 129.27/EI does not follow from any standard formula with these values.
- B. 64.64/EI: 64.64/EI is too small.
- D. 88.85/EI: 88.85/EI does not match wL³/24EI.
Key Point
Triangular load (peak at fixed end): θ = wL³/24EI, δ = wL⁴/30EI.
Key Concepts for Students
- If the peak is at the free end instead, θ = wL³/8EI and δ = 11wL⁴/120EI.
- UDL: θ = wL³/6EI, δ = wL⁴/8EI.
- Superposition: a UDL equals the sum of two opposite triangles.
Practice more: this is question 7 in Slope and Deflection MCQs – Part 1. Try the whole set and check every answer instantly.