Question

A solid beam with a rectangular cross-section of breadth 0.5 m and depth 0.12 m, experiences a vertical shear force of 50 kN at a section.
The maximum shear stress in that section is ______ MPa (answer in two decimal places).

Hint:

Find the average shear stress V/A first, then multiply by 1.5 since a rectangular section has a parabolic shear distribution.

Answer 1

Correct Answer: 1.25

Step 1: Find the average shear stress.
For any cross-section carrying a shear force \(V\), the average shear stress is \( \tau_{avg} = V / A \), where \(A\) is the cross-sectional area.
Here \(A = 0.5 \times 0.12 = 0.06 \ \text{m}^2\), and \(V = 50 \ \text{kN} = 50000 \ \text{N}\), so \( \tau_{avg} = 50000 / 0.06 = 833333.33 \ \text{Pa} \).

Step 2: Apply the rectangular section shear factor.
For a solid rectangular cross-section the shear stress is not uniform, it follows a parabolic distribution across the depth and peaks at the neutral axis.
The standard result is \( \tau_{max} = 1.5 \, \tau_{avg} \) for a rectangle, so \( \tau_{max} = 1.5 \times 833333.33 = 1250000 \ \text{Pa} \).

Final Answer:
Convert to MPa by dividing by \(10^6\).
\[ \tau_{max} = 1250000 \ \text{Pa} = 1.25 \ \text{MPa} \] \[ \boxed{\tau_{max} = 1.25 \ \text{MPa}} \]