Strength of Materials and Structural Analysis
What the exam tests
This domain tests stress, strain, axial deformation, bending and shear, deflection, buckling, and the classification and approximate analysis of structures. You should be able to move between force, stress, deformation and capacity, and to recognise whether a structure is determinate, indeterminate or unstable.
Core rules
- Axial stress σ = N/A; strain ε = ΔL/L; within the elastic limit E = σ/ε, so ΔL = NL/(AE).
- Bending stress σ = My/I; section modulus S = I/y, so σ = M/S. Shear stress τ = VQ/(Ib).
- Deflection is a serviceability check; strength is an ultimate-limit-state check. Both must be satisfied.
- Euler buckling: Pcr = π²EI/(KL)², where KL is the effective length. K depends on end restraint, not on the actual length alone.
- For a planar truss, m + r = 2j is statically determinate if stable; for a rigid frame, 3m + r = 3j is determinate.
- Compatibility and equilibrium are both needed for indeterminate structures; simple equilibrium alone is insufficient.
Worked example
A steel tie of diameter 20 mm, length 2.0 m, carries a tensile load of 50 kN. E = 200 GPa. Find stress, strain and elongation. Radius = 10 mm = 0.010 m. Area A = π(0.010)² = 3.142 × 10⁻⁴ m². Stress σ = 50,000 / 3.142 × 10⁻⁴ = 159.2 × 10⁶ Pa = 159 MPa. Strain ε = σ/E = 159.2 × 10⁶ / 200 × 10⁹ = 7.96 × 10⁻⁴. Elongation ΔL = εL = 7.96 × 10⁻⁴ × 2.0 = 1.59 × 10⁻³ m = 1.59 mm. The tie is within elastic range only if the yield strength exceeds 159 MPa; the calculation does not prove that.
Traps
Using diameter as radius; converting mm² to m² incorrectly; entering E in GPa without converting to Pa; confusing stress with force; using the actual length instead of effective length in buckling; and assuming a structure is determinate because it looks simple. Another common error is treating a deflection limit as a strength limit.