Bending Stress and Torsion

Bending (Flexure) Stress

For a beam in pure bending, the normal stress varies linearly from the neutral axis:

$\sigma = -\frac{My}{I}$

where:

• M = bending moment at the section
• y = distance from the neutral axis (positive upward)
• I = moment of inertia about the neutral axis

Maximum bending stress occurs at the extreme fiber (y = c): $\sigma_{max} = \frac{Mc}{I} = \frac{M}{S}$

where S = I/c is the section modulus.

Section Modulus (S) for Common Shapes

Key Points:

• Stress is zero at the neutral axis (y = 0)
• Stress is maximum at the top and bottom fibers
• Top fiber: compression when M > 0 (sagging)
• Bottom fiber: tension when M > 0 (sagging)

Transverse Shear Stress

For beams subjected to transverse loads:

$\tau = \frac{VQ}{Ib}$

where:

• V = shear force at the section
• Q = first moment of area above (or below) the point of interest
• I = moment of inertia of the entire cross-section
• b = width of the cross-section at the point of interest

For a rectangular cross-section: $\tau_{max} = \frac{3V}{2A} = 1.5 \times \tau_{avg}$

Maximum shear stress occurs at the neutral axis (not at the extreme fiber!).

For a circular cross-section: $\tau_{max} = \frac{4V}{3A} = 1.33 \times \tau_{avg}$

Torsion of Circular Shafts

Shear Stress

$\tau = \frac{Tr}{J}$

where:

• T = torque (torsional moment)
• r = radial distance from the center
• J = polar moment of inertia

Polar Moment of Inertia

Maximum shear stress occurs at the outer surface (r = c = d/2): $\tau_{max} = \frac{Tc}{J}$

Angle of Twist

$\phi = \frac{TL}{GJ}$

where:

• L = length of the shaft
• G = shear modulus
• φ is in radians

Power Transmission

$P = T\omega = T \cdot \frac{2\pi n}{60}$

where:

• P = power (watts)
• T = torque (N·m)
• ω = angular velocity (rad/s)
• n = rotational speed (rpm)

Example: A motor delivers 50 kW at 1,500 rpm. What torque does the shaft transmit?

T = P/ω = 50,000 / (2π × 1,500/60) = 50,000 / 157.08 = 318.3 N·m