Question

A simply supported rectangular beam of span 6 m having width of 100 mm and depth of 200 mm is subjected to a uniformly distributed load of 20 kN/m over the entire span. The maximum shear stress induced in the beam is

• A. 3 N/mm$^2$
• B. 4.5 N/mm$^2$
• C. 6 N/mm$^2$
• D. 9 N/mm$^2$

Hint:

To find max shear stress, always follow the two steps:
1. Find $V_{max}$ from the beam's loading diagram (for SS beam with UDL, it's at the supports).
2. Apply the correct formula for $\tau_{max}$ based on the cross-section shape (for a rectangle, it's $1.5 \times V/A$).

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
Simply supported beam 6m, UDL 20kN/m, 100×200mm. Find τ_max.

Step 2: Key Formula (Alternate):
V_max = wL/2 (at supports). τ_max = 1.5×V_max/A.

Step 3: Detailed Explanation:
Total load=20×6=120kN. R_A=R_B=60kN. V_max=60kN=60000N. A=100×200=20000mm². τ_avg=60000/20000=3. τ_max=1.5×3=4.5 N/mm². Note: If answer key shows 9, it may have used total load instead of reaction.

Step 4: Final Answer:
Maximum shear stress is 4.5 N/mm².