A uniform circular bar of diameter d and length l extends by an amount e under a tensile pull W. When it is simply supported and carries the same load W at mid-span, its deflection is:
Try it first: pick an option, then read the explanation.
Correct answer: C. el²/(3d²)
Asked in SSC JE 24/01/2018 (Morning)
Detailed Explanation
For a circle, I = πd⁴/64 and A = πd²/4, so I = A·d²/16. This lets the bending deflection be written using the axial extension.
\[ e = \frac{Wl}{AE},\qquad \delta = \frac{Wl^3}{48EI} = \frac{Wl^3}{48E\,(Ad^2/16)} = \frac{Wl^3}{3AEd^2} = \frac{Wl}{AE}\cdot\frac{l^2}{3d^2} = \frac{e\,l^2}{3d^2} \]
Why “el²/(3d²)” Is Correct
The central deflection is el²/(3d²).
Why the Other Options Are Wrong
- A. el/(2d): el/2d has the wrong power of l/d.
- B. e²l/(3d²): e² makes the expression dimensionally wrong (length² ÷ length...).
- D. √e/(3d²): √e/(3d²) is not dimensionally a length.
Key Point
Circular bar: δ(central load) = e·l²/(3d²); rectangular bar: δ/e = (l/2d)².
Key Concepts for Students
- I = Ad²/16 for a solid circle; I = Ad²/12 for a rectangle (about the depth).
- Check dimensions of every option — two can be rejected at once.
- Radius of gyration of a circle = d/4.
Practice more: this is question 27 in Slope and Deflection MCQs – Part 2. Try the whole set and check every answer instantly.