The maximum shear force in a rectangular beam section of depth 150 mm is 50 kN. If the permissible maximum shear stress is 5 MPa, the width of the section is:
Try it first: pick an option, then read the explanation.
Asked in SSC JE 14/11/2022
Detailed Explanation
In a rectangular section the maximum shear stress, at the neutral axis, is 1.5 times the average shear stress V/(BD). Set it equal to the permissible value and solve for B.
\[ \tau_{max} = \frac{3}{2}\,\frac{V}{BD} \Rightarrow 5 = \frac{3}{2}\times\frac{50\times 10^3}{B\times 150} \]
\[ B = \frac{3\times 50\times 10^3}{2\times 5\times 150} = 100\ \text{mm} \]
Why “100 mm” Is Correct
The required width is 100 mm. Using the average stress alone (V/BD = 5) would give 66.67 mm, which is unsafe because the peak stress at the neutral axis would then be 7.5 MPa.
Why the Other Options Are Wrong
- B. 66.67 mm: 66.67 mm uses τavg = V/(BD) and ignores the factor 1.5.
- C. 200 mm: 200 mm is double the required width.
- D. cannot be predicted using the given data: Depth, shear and permissible stress are all given, so B can be found.
Key Point
Rectangle: τmax = 1.5 V/(BD).
Key Concepts for Students
- 1 MPa = 1 N/mm²; 1 kN = 1000 N.
- The shear stress is zero at the top and bottom fibres.
- Shear rarely governs timber and steel rectangular beams, but it must be checked.
Practice more: this is question 7 in Bending and Shear Stress MCQs – Part 1. Try the whole set and check every answer instantly.