Mechanics of Materials: Stress, Strain, and Torsion
Key Takeaways
- Mechanics of materials begins with internal load resultants: axial force, shear, moment, and torque.
- Stress and strain formulas require consistent area, length, modulus, and unit systems.
- Axial deformation is a compatibility problem when members are connected or constrained.
- Torsion formulas apply directly to circular shafts; noncircular sections need the handbook's appropriate relationships.
- Stress-strain curve questions test yield, ultimate strength, elastic modulus, ductility, and factor of safety.
Start with the internal load
Before using any stress formula, cut the member and identify the internal load resultant. Axial force produces normal stress P/A. Transverse shear produces shear stress models. Bending moment produces normal stress varying over the section. Torque produces torsional shear stress. The FE Mechanical exam often rewards candidates who identify the load type faster than candidates who memorize many formulas.
Use tension as positive unless you state otherwise. Compression stresses are usually negative by convention, but design checks often use magnitudes. Keep the sign convention separate from the failure comparison: a compressive stress of -12 ksi has a magnitude of 12 ksi.
Axial stress, strain, and deformation
Normal stress is sigma = P/A. Normal strain is epsilon = delta/L. In the linear elastic range, Hooke's law is sigma = E epsilon. For a prismatic member under axial load, deformation is delta = PL/(AE). If area, load, or modulus changes by segment, sum each segment's PL/(AE) with sign.
| Quantity | Common unit check |
|---|---|
| Stress | N/m^2, MPa, psi, ksi |
| Strain | Dimensionless, in/in, mm/mm |
| Modulus E | Same units as stress |
| Axial deformation | Length units |
| Thermal strain | alpha Delta T |
Thermal strain is alpha Delta T if expansion is free. If expansion is restrained, thermal strain creates stress. For a fully restrained uniform bar, the thermal stress magnitude is E alpha Delta T, with compression for heating when the bar wants to expand but cannot.
Stress-strain curve interpretation
A stress-strain curve gives more than one number. The initial slope is elastic modulus. The yield strength marks the start of significant plastic deformation. Ultimate tensile strength is the maximum engineering stress. Fracture strength occurs at rupture. Ductility is tied to elongation or reduction in area; toughness is related to area under the curve.
For FE-style factor of safety checks, identify the failure mode. A ductile member under static normal stress is often checked against yield strength. A brittle member may be checked against ultimate strength. Fatigue, buckling, and stress concentration require different models, so do not force every problem into sigma = P/A.
Torsion and circular shafts
For a circular shaft in elastic torsion, maximum shear stress is tau_max = Tc/J and angle of twist is phi = TL/(JG). Here c is outer radius, J is polar moment of inertia, and G is shear modulus. Hollow shafts use J based on outer and inner diameters. The handbook provides polar moment formulas; choose radius or diameter form carefully.
Power transmission connects torque and angular speed: P = T omega. Convert rpm to rad/s before calculating torque in SI units. In USCS, be consistent with hp, ft-lbf/s, and rpm. A common sequence is find torque from power, compute J from geometry, then check tau_max or angle of twist.
Handbook lookup choices
Use the FE Reference Handbook for section properties, material property tables when provided, and torsion formulas. Still write the model first: axial, thermal, torsional, or combined. If the shaft is not circular, the simple tau = Tc/J formula may not be valid, and the handbook's noncircular torsion relationships or problem-provided factors must control.
A 20 mm diameter steel rod carries 12 kN of axial tension. Which setup correctly finds the average normal stress?
A bar is fully restrained and heated. What mechanical effect does the restraint create?
A solid circular shaft transmits torque T. If the diameter is doubled while T stays the same, how does maximum elastic torsional shear stress change?