A rectangular bar of cross-sectional area 20000 mm² is subjected to an axial load of 30 KN, section is inclined at an angle of 45° with normal cross-section of the bar, what will be the shear stress on a section? 

Concept:

The shear stress on an inclined section of a bar subjected to axial load can be calculated using the formula:

\(\tau = \frac{P \sin(2\theta)}{2A}\)

Calculation:

Given:

• Cross-sectional area, \(A = 20000 \, \text{mm}^2\)
• Axial load, \(P = 30 \, \text{kN} = 30000 \, \text{N}\)
• Angle of inclination, \(\theta = 45^\circ\)

The shear stress is given by:

\(\tau = \frac{P \sin(2\theta)}{2A}\)

Substitute the given values:

\(\tau = \frac{30000 \times \sin(90^\circ)}{2 \times 20000} = \frac{30000 \times 1}{40000} = 0.75 \, \text{N/mm}^2\)

The shear stress on the section is 0.75 N/mm²