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3.3 Mechanics of Materials

Key Takeaways

  • Axial normal stress σ = P/A and elastic axial deformation δ = PL/AE; Hooke's law gives σ = Eε.
  • Flexure (bending) stress is σ = Mc/I, maximum at the extreme fiber; transverse shear stress is τ = VQ/(Ib).
  • Torsional shear stress in a circular shaft is τ = Tr/J with J = πr⁴/2 (πd⁴/32) for a solid shaft.
  • The slope of the shear diagram equals the load; the slope of the moment diagram equals the shear (dM/dx = V).
  • Euler critical buckling load is Pcr = π²EI/(KL)², with K = 1.0 pinned-pinned, 0.5 fixed-fixed, 2.0 fixed-free.
Last updated: May 2026

The Analytical Core of FE Civil

Mechanics of Materials carries 6–9 questions and underpins every Structural Engineering problem. The skill tested is matching a loading (axial, torsion, bending, shear, combined) to the correct stress equation and the right section property.

Stress, Strain, and Hooke's Law

Normal stress is force per unit area, σ = P/A. Normal strain is ε = δ/L. In the elastic range they follow Hooke's law:

σ=Eε\sigma = E\varepsilon

where E is the modulus of elasticity (≈ 29,000 ksi / 200 GPa for steel). Poisson's ratio ν relates lateral to axial strain (ν ≈ 0.3 for steel). The shear modulus is G = E / [2(1 + ν)].

Axial Loading

Elastic axial deformation of a prismatic bar:

δ=PLAE\delta = \frac{PL}{AE}

Thermal elongation is δ = αLΔT. If the bar is restrained, a thermal stress σ = EαΔT develops.

Torsion

For a circular shaft, the torsional shear stress is:

τ=TrJ,ϕ=TLJG\tau = \frac{Tr}{J}, \qquad \phi = \frac{TL}{JG}

The polar moment of inertia J = πr⁴/2 = πd⁴/32 (solid). Stress is maximum at the outer surface and zero at the center.

Shear and Moment Diagrams

The load, shear, and moment are linked by calculus:

RelationshipMeaning
dV/dx = −wslope of shear = −distributed load
dM/dx = Vslope of moment = shear
ΔV = −∫w dxchange in shear = area under load
ΔM = ∫V dxchange in moment = area under shear

Maximum bending moment occurs where shear crosses zero. For a simply supported beam of span L with a central point load P, M_max = PL/4; with a uniform load w, M_max = wL²/8.

Bending and Transverse Shear Stress

Flexure stress varies linearly through the depth:

σ=McI\sigma = \frac{Mc}{I}

maximum at the extreme fiber (distance c from the neutral axis). Transverse shear stress is τ = VQ/(Ib), maximum at the neutral axis where Q (first moment of area) is largest.

Beam Deflection

Standard cases are tabulated in the NCEES FE Reference Handbook — look them up, do not derive:

  • Simply supported, central point load P: δ_max = PL³/(48EI)
  • Simply supported, uniform load w: δ_max = 5wL⁴/(384EI)
  • Cantilever, end point load P: δ_max = PL³/(3EI)

Combined Stress and Mohr's Circle

When axial, bending, and torsion act together, superpose normal stresses and use Mohr's circle to find principal stresses and maximum shear:

σ1,2=σx+σy2±(σxσy2)2+τxy2\sigma_{1,2} = \frac{\sigma_x + \sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2}

The maximum in-plane shear stress equals the circle's radius, τ_max = √[((σx − σy)/2)² + τxy²].

Columns and Buckling

Long, slender columns fail by elastic (Euler) buckling, not crushing:

Pcr=π2EI(KL)2P_{cr} = \frac{\pi^2 EI}{(KL)^2}

The effective-length factor K depends on end conditions: K = 1.0 (pinned–pinned), 0.5 (fixed–fixed), 0.7 (fixed–pinned), 2.0 (fixed–free cantilever). Buckling controls when the slenderness ratio KL/r is large. These K values are tabulated in the handbook — confirm the support condition before selecting K.

Loading diagram...
Load–Shear–Moment Relationship
Test Your Knowledge

A simply supported beam spans 6 m and carries a uniformly distributed load of 10 kN/m over its entire length. What is the maximum bending moment?

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Test Your Knowledge

A pinned-pinned steel column is 4 m long with I = 8 × 10⁶ mm⁴ and E = 200 GPa. What is the Euler critical buckling load? (Use K = 1.0.)

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