A beam has a triangular cross-section having base 40 mm and altitude 60 mm. If this section is subjected to a shear force of 36000 N, then the maximum shear stress in the cross-section would be
- A60 N/mm²
- B36 N/mm²
- C45 N/mm²Correct
- D30 N/mm²
Solution
Concept
Maximum shear stress in a triangular section
Formula
τmax = 1.5 × τavg (at mid-height)
Step-by-step solution
1. Area = ½ × 40 × 60 = 1200 mm². 2. τavg = V/A = 36000/1200 = 30 N/mm². 3. τmax = 1.5 × 30 = 45 N/mm² (at h/2 from the apex).
Explanation
For a triangular cross-section, the maximum shear stress occurs at mid-height and equals 1.5 times the average shear stress. (Rectangle 1.5; circle 4/3; triangle 1.5; diamond 9/8.)
Shortcut
Triangle: τmax = 1.5 × V/A = 3V/(bh).
Common mistake
Using the rectangular formula 1.5 V/(b h) with the full area b × h.
Correct answer: C) 45 N/mm²