Question

Two simply supported beams \( B_1 \) and \( B_2 \) have spans \( l \) and \( 2l \) respectively. Beam \( B_1 \) has a cross-section of \( 1 \times 1 \) units and \( B_2 \) has a cross-section of \( 2 \times 2 \) units. These beams are subjected to concentrated loads \( W \) each at the centre of their spans. The ratio of the maximum flexural stress in these beams is ............

• A. 2
• B. 4
• C. 0.5
• D. 0.25

Hint:

When comparing maximum flexural stresses in beams with different dimensions and spans, use the formula for bending stress and ensure you account for differences in the moment of inertia and span length.

Answer 1

The Correct Option is B

The maximum flexural stress \( \sigma_{\text{max}} \) in a beam is given by the formula: \[ \sigma_{\text{max}} = \frac{M_{\text{max}} \cdot y}{I} \] where: 
- \( M_{\text{max}} \) is the maximum bending moment, 
- \( y \) is the distance from the neutral axis to the outermost fiber, 
- \( I \) is the moment of inertia of the beam's cross-section.
For a beam subjected to a central load \( W \), the maximum bending moment is: \[ M_{\text{max}} = \frac{W \cdot l}{4} \] where \( l \) is the span of the beam. 
Now, let's calculate the moment of inertia for the two beams: 
1. For beam \( B_1 \), with a cross-section of \( 1 \times 1 \) units, the moment of inertia is: \[ I_1 = \frac{1 \times 1^3}{12} = \frac{1}{12} \] 2. For beam \( B_2 \), with a cross-section of \( 2 \times 2 \) units, the moment of inertia is: \[ I_2 = \frac{2 \times 2^3}{12} = \frac{16}{12} = \frac{4}{3} \] Next, the maximum flexural stress for each beam is given by: \[ \sigma_{\text{max},1} = \frac{M_1 \cdot y_1}{I_1} = \frac{\frac{W \cdot l}{4} \cdot \frac{1}{2}}{\frac{1}{12}} = \frac{W \cdot l}{4} \cdot 6 = \frac{6 W l}{4} \] \[ \sigma_{\text{max},2} = \frac{M_2 \cdot y_2}{I_2} = \frac{\frac{W \cdot 2l}{4} \cdot \frac{2}{2}}{\frac{4}{3}} = \frac{W \cdot 2l}{4} \cdot \frac{3}{4} = \frac{3 W l}{8} \] The ratio of the maximum flexural stresses in the two beams is: \[ \frac{\sigma_{\text{max},1}}{\sigma_{\text{max},2}} = \frac{\frac{6 W l}{4}}{\frac{3 W l}{8}} = 4 \] Thus, the ratio of the maximum flexural stress in these beams is 4.