8.2 Strength of Materials & Mechanics of Deformable Bodies

8.2 Strength of Materials & Mechanics of Deformable Bodies

Strength of Materials extends rigid-body mechanics by accounting for material deformation under applied loads. For electrical engineers, this discipline provides the essential framework for designing motor drive shafts, transmission tower structural members, conduit supports, transformer tanks, and high-voltage cable structures.

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1. Simple Stress, Strain, Hooke's Law & Thermal Stresses

Axial Stress and Strain

• Normal (Axial) Stress ($\sigma$): Internal force per unit cross-sectional area: $\sigma = \frac{P}{A}$
• Normal Strain ($\varepsilon$): Unit elongation or linear deformation: $\varepsilon = \frac{\delta}{L}$
• Hooke's Law: Within the elastic proportional limit, stress is directly proportional to strain: $\sigma = E \varepsilon \implies \delta = \frac{P L}{A E}$ where $E$ is the Modulus of Elasticity (Young's Modulus) (typically $200\ \text{GPa}$ for structural steel).

Poisson's Ratio & Elastic Constants

When a bar is subjected to axial tension, it elongates longitudinally and contracts laterally:

$\nu = -\frac{\varepsilon_{\text{lateral}}}{\varepsilon_{\text{axial}}}$

• Modulus of Rigidity (Shear Modulus $G$): Relates shear stress to shear strain ($\tau = G \gamma$): $G = \frac{E}{2(1 + \nu)}$
• Bulk Modulus ($K$): Relates volumetric stress to volumetric strain: $K = \frac{E}{3(1 - 2\nu)}$

Thermal Deformation & Thermal Stress

Temperature changes induce free thermal expansion or contraction:

$\delta_{\text{th}} = \alpha L \Delta T$

where $\alpha$ is the coefficient of thermal expansion ($/^\circ\text{C}$). If thermal expansion is fully constrained by unyielding supports, compressive thermal stress develops:

$\sigma_{\text{th}} = E \alpha \Delta T$

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2. Torsion of Circular Shafts & Power Transmission

When a circular shaft is subjected to an applied torque $T$, internal shear stresses are generated on cross-sectional planes.

Elastic Torsion Formula

$\tau = \frac{T r}{J}$

where $r$ is the radial distance from the central axis, and $J$ is the Polar Moment of Inertia of the cross-section:

• Solid Circular Shaft ($d$): $J = \frac{\pi d^4}{32}$ $\tau_{\text{max}} = \frac{16 T}{\pi d^3}$
• Hollow Circular Shaft (Outer $D$, Inner $d$): $J = \frac{\pi (D^4 - d^4)}{32}$ $\tau_{\text{max}} = \frac{16 T D}{\pi (D^4 - d^4)}$

Angle of Twist ($\phi$)

The angular rotation of one end of a circular shaft of length $L$ relative to the other:

$\phi = \frac{T L}{G J} \quad (\text{in radians})$

Power Transmission in Rotating Shafts

Electrical motor shafts transmit mechanical power $P$ related to torque $T$ and rotational speed $N$ (in rpm) or $\omega$ (in rad/s):

$P = T \omega = T \left(\frac{2\pi N}{60}\right) \implies T = \frac{60 P}{2\pi N}$

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3. Shear & Bending Moment Diagrams, Flexural & Transverse Shear Stresses

Beams are structural members designed to support lateral transverse loads.

Differential Relations for Beams

$\frac{dV}{dx} = -w(x) \quad (\text{slope of shear curve = negative distributed load})$

$\frac{dM}{dx} = V(x) \quad (\text{slope of moment curve = shear force})$

Maximum bending moments occur at cross-sections where the shear force is zero ($V(x) = 0$).

Flexure Formula (Bending Stress)

Under pure bending, normal flexural stress $\sigma_b$ varies linearly across the beam depth from the neutral axis:

$\sigma_b = -\frac{M y}{I}$

• Maximum flexural stress occurs at extreme fibers ($y = c$): $\sigma_{\text{max}} = \frac{M c}{I} = \frac{M}{S}$ where $S = \frac{I}{c}$ is the Section Modulus of the cross-section.

Transverse Shear Stress Formula

$\tau = \frac{V Q}{I b}$

where $V$ is internal shear force, $I$ is moment of inertia, $b$ is width at the evaluation section, and $Q = A' \bar{y}'$ is the first moment of area above or below the plane of interest.

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4. Beam Deflection & Thin-Walled Pressure Vessels

Elastic Curve Differential Equation for Beam Deflection

$E I \frac{d^2 y}{dx^2} = M(x)$

Integrating once yields beam slope $\theta(x) = \frac{dy}{dx}$, and integrating twice yields deflection $y(x)$.

Common Standard Beam Deflection Formulas

Thin-Walled Pressure Vessels ($t < D/20$)

Under internal gage pressure $P$:

• Cylindrical Vessel:
  • Circumferential (Hoop) Stress: $\sigma_1 = \frac{P D}{2 t}$
  • Longitudinal Stress: $\sigma_2 = \frac{P D}{4 t}$
• Spherical Vessel:
  • Uniform Membrane Stress: $\sigma = \frac{P D}{4 t}$

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5. Combined Stress Transformations & Mohr's Circle

For a two-dimensional state of plane stress $(\sigma_x, \sigma_y, \tau_{xy})$, stress components on an inclined plane rotated counterclockwise by angle $\theta$ are:

$\sigma_{x'} = \frac{\sigma_x + \sigma_y}{2} + \frac{\sigma_x - \sigma_y}{2} \cos(2\theta) + \tau_{xy} \sin(2\theta)$

$\tau_{x'y'} = -\frac{\sigma_x - \sigma_y}{2} \sin(2\theta) + \tau_{xy} \cos(2\theta)$

Principal Stresses and Maximum Shear Stress

$\sigma_{1,2} = \frac{\sigma_x + \sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2}$

$\tau_{\text{max, in-plane}} = R = \sqrt{\left(\frac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2}$

$\tan(2\theta_p) = \frac{2\tau_{xy}}{\sigma_x - \sigma_y}$

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Solved Board Exam Examples

Example 1: Motor Shaft Torsion & Power Sizing

Problem: A solid steel electric motor drive shaft rotates at $N = 1800\ \text{rpm}$ and transmits $P = 75\ \text{kW}$ of mechanical power. The allowable shear stress of the steel material is $\tau_{\text{allow}} = 50\ \text{MPa}$. Determine the minimum required shaft diameter $d$.

Solution:

1. Calculate transmitted torque $T$: $\omega = \frac{2 \pi N}{60} = \frac{2 \pi \times 1800}{60} = 60\pi \approx 188.496\ \text{rad/s}$ $T = \frac{P}{\omega} = \frac{75,000}{188.496} = 397.887\ \text{N}\cdot\text{m}$
2. Apply elastic solid shaft torsion formula: $\tau_{\text{max}} = \frac{16 T}{\pi d^3} \le \tau_{\text{allow}}$
3. Rearrange to solve for minimum diameter $d$: $d^3 \ge \frac{16 T}{\pi \tau_{\text{allow}}} = \frac{16 \times 397.887}{\pi \times 50 \times 10^6} = \frac{6366.192}{157,079,632} = 4.0528 \times 10^{-5}\ \text{m}^3$ $d \ge (4.0528 \times 10^{-5})^{1/3} = 0.03435\ \text{m} = 34.35\ \text{mm}$
4. Select standard commercial shaft diameter: 35 mm.

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Example 2: Flexural Stress & Midspan Deflection of a Timber Beam

Problem: A simply supported timber beam of length $L = 6\ \text{m}$ supports a uniform dead load $w = 12\ \text{kN/m}$. The beam cross-section is rectangular with base $b = 150\ \text{mm}$ and height $h = 300\ \text{mm}$. Compute the maximum bending stress $\sigma_{\text{max}}$ and maximum midspan deflection $\delta_{\text{max}}$. Take $E = 12\ \text{GPa}$.

Solution:

1. Calculate maximum bending moment for simply supported uniform loading: $M_{\text{max}} = \frac{w L^2}{8} = \frac{12,000 \times 6^2}{8} = 54,000\ \text{N}\cdot\text{m}$
2. Calculate section modulus $S$ and centroidal moment of inertia $I$: $I = \frac{b h^3}{12} = \frac{0.15 \times (0.30)^3}{12} = 3.375 \times 10^{-4}\ \text{m}^4$ $S = \frac{b h^2}{6} = \frac{0.15 \times (0.30)^2}{6} = 2.25 \times 10^{-3}\ \text{m}^3$
3. Calculate maximum flexural stress $\sigma_{\text{max}}$: $\sigma_{\text{max}} = \frac{M_{\text{max}}}{S} = \frac{54,000}{2.25 \times 10^{-3}} = 24.0 \times 10^6\ \text{Pa} = 24.0\ \text{MPa}$
4. Compute maximum midspan deflection $\delta_{\text{max}}$: $\delta_{\text{max}} = \frac{5 w L^4}{384 E I} = \frac{5 \times 12,000 \times 6^4}{384 \times (12 \times 10^9) \times (3.375 \times 10^{-4})} = \frac{77,760,000}{1,555,200} = 0.050\ \text{m} = 50.0\ \text{mm}$