Question

A beam of triangular section has base width 200 mm and height 100 mm. The maximum shear stress in the beam section due to a shear force of 20 kN is ............

• A. 1.5 MPa
• B. 1.33 MPa
• C. 2.66 MPa
• D. 3.0 MPa

Hint:

For a triangular beam section under shear, the maximum shear stress occurs at the base, and it can be calculated using the formula involving the shear force, area, and the distance from the neutral axis to the maximum shear stress location.

Answer 1

The Correct Option is C

The maximum shear stress \( \tau_{\text{max}} \) in a beam with a triangular cross-section is given by the formula: \[ \tau_{\text{max}} = \frac{3V}{2A} \cdot \frac{d}{y} \] Where:
- \( V \) is the shear force applied to the beam,
- \( A \) is the cross-sectional area of the beam,
- \( d \) is the height of the triangular section,
- \( y \) is the distance from the neutral axis to the point where the maximum shear stress occurs.
For a triangular section, the cross-sectional area \( A \) is: \[ A = \frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} \cdot 200 \cdot 100 = 10,000 \, \text{mm}^2 = 10 \, \text{cm}^2 \] The distance \( y \) from the neutral axis to the maximum shear stress location (at the base of the triangle) is: \[ y = \frac{d}{3} = \frac{100}{3} \approx 33.33 \, \text{mm} \] Now, using the formula for shear stress and substituting the given values:
- \( V = 20 \, \text{kN} = 20,000 \, \text{N} \),
- \( A = 10,000 \, \text{mm}^2 \),
- \( d = 100 \, \text{mm} \),
- \( y = 33.33 \, \text{mm} \),
We get: \[ \tau_{\text{max}} = \frac{3 \cdot 20,000}{2 \cdot 10,000} \cdot \frac{100}{33.33} = 2.66 \, \text{MPa} \] Thus, the maximum shear stress in the beam section is 2.66 MPa.