A rolled steel joist, simply supported at its ends, carries a UDL that causes a maximum deflection of 10 mm and an end slope of 0.002 radian. The length of the joist is:
Try it first: pick an option, then read the explanation.
Correct answer: D. 16 m
Asked in SSC JE 23/01/2018 (Morning)
Detailed Explanation
Divide the deflection formula by the slope formula: w, E and I cancel, leaving a relation between δ, θ and L only.
\[ \frac{\delta}{\theta} = \frac{5wL^4/384EI}{wL^3/24EI} = \frac{5L}{16} \Rightarrow L = \frac{16\,\delta}{5\,\theta} = \frac{16\times10}{5\times0.002} = 16\,000\ \text{mm} \]
Why “16 m” Is Correct
The joist is 16 m long.
Why the Other Options Are Wrong
- A. 10 m: 10 m would give δ = 6.25 mm.
- B. 12 m: 12 m would give δ = 7.5 mm.
- C. 14 m: 14 m would give δ = 8.75 mm.
Key Point
Simply supported UDL: δmax = (5L/16) × θend.
Key Concepts for Students
- Central point load: δ/θ = (WL³/48)/(WL²/16) = L/3.
- Keep units consistent (mm with radians).
- Dividing two formulas removes unknown load and stiffness.
Practice more: this is question 29 in Slope and Deflection MCQs – Part 2. Try the whole set and check every answer instantly.