The shear centre of a thin half-circular (semicircular) section of radius r and constant thickness lies at a distance x from the centre, where x is:
Try it first: pick an option, then read the explanation.
Asked in SSC JE 23/01/2018 (Morning)
Detailed Explanation
The shear centre is the point through which the load must pass so that the section bends without twisting. For an open thin arc, take moments of the shear flow about the centre of curvature.
Measure θ from one tip; a point on the arc is at \(y = r\cos\theta\) from the neutral axis, and \( I = \frac{\pi r^3 t}{2} \).
\[ A\bar y = \int_0^{\theta} (t\,r\,d\phi)\,r\cos\phi = t r^2\sin\theta,\qquad q = \frac{V\,t r^2 \sin\theta}{I} \]
\[ V\,e = \int_0^{\pi} q\,r\,(r\,d\theta) = \frac{V t r^4}{I}\int_0^{\pi}\sin\theta\,d\theta = \frac{2Vtr^4}{\pi r^3 t/2} \Rightarrow e = \frac{4r}{\pi} \]
Why “4r/π” Is Correct
The shear centre is 4r/π ≈ 1.27r from the centre — outside the section itself, beyond the crown of the arc.
| Thin open section | Position of shear centre |
|---|---|
| Semicircular arc, radius r | 4r/π from the centre of curvature |
| Channel | Outside the web, on the side opposite the flanges |
| Angle, T | At the intersection of the legs / of flange and web |
| I-section, rectangle, circle | At the centroid (double symmetry) |
Why the Other Options Are Wrong
- A. r/π: r/π is too small.
- B. 2r/π: 2r/π is the centroid distance of a thin semicircular arc, not the shear centre.
- C. 3r/π: 3r/π is not a standard result.
Key Point
Thin semicircular arc: shear centre at 4r/π from the centre (centroid at 2r/π).
Key Concepts for Students
- Shear centre and centroid coincide only for doubly symmetric sections.
- Loads not through the shear centre cause twisting.
- For sections made of thin intersecting legs, the shear centre is at the intersection.
Practice more: this is question 53 in Bending and Shear Stress MCQs – Part 4. Try the whole set and check every answer instantly.