Interactive Lecture
Axial, Shear, Bending, and Torsional Deformations
by Dr Arnaldo Delli Carri
Structural members deform under load. A beam that sags too much or a shaft that twists too far fails in service even without fracture — so getting the deformations right in design is just as important as getting the stresses right.
Different internal loads cause different deformation modes. This lecture focuses on axial extension/compression, shear deformation, bending curvature, and torsional twist.
Normal Stress: \(\sigma = \frac{P}{A}\)
Normal Strain: \(\varepsilon = \frac{\sigma}{E} = \frac{P}{EA}\)
Axial Elongation: \(\delta = \frac{PL}{EA}\)
For non-uniform members: \(\delta = \int_0^L \frac{P(x)}{E(x)A(x)}dx\)
Shear Stress: \(\tau = \frac{V}{A}\) (average)
Shear Strain: \(\gamma = \frac{\tau}{G} = \frac{V}{GA}\)
Shear Angle: Cross-sections slip relative to each other by angle \(\gamma\)
For beams: \(\delta_{shear} = \int \frac{kV^2}{GA}dx\) where \(k\) is shape factor
For isotropic materials, shear modulus relates to Young's modulus through Poisson's ratio: \(G = \frac{E}{2(1+\nu)}\)
Curvature: \(\frac{1}{\rho} = \frac{M}{EI}\)
Bending Strain: \(\varepsilon = \frac{y}{\rho} = \frac{My}{EI}\)
Deflection: \(v = \int\int \frac{M}{EI}dx^2\)
Fibres on one side extend, fibres on the opposite side compress — the neutral axis keeps its original length but curves
Shear Stress: \(\tau = \frac{Tr}{J}\)
Angle of Twist: \(\phi = \frac{TL}{GJ}\)
For circular shaft: \(J = \frac{\pi d^4}{32}\)
Twist angle increases linearly from fixed end to free end
Explore how different loads deform a round cantilever bar shown in three orthogonal views. The grid mesh shows how the material deforms. Deformations scale with load magnitude, and use the magnification slider for dramatic visual effect.
| Load Type | Deformation | Key Parameters |
|---|---|---|
| Axial | \(\delta = \frac{PL}{EA}\) | E, A, L |
| Shear | \(\delta_{shear} = \int \frac{kV^2}{GA}dx\) | G, A, k |
| Bending | \(v = \int\int \frac{M}{EI}dx^2\) | E, I, M(x) |
| Torsion | \(\phi = \frac{TL}{GJ}\) | G, J, T |
A steel rod (E = 200 GPa, A = 500 mm²) of length 2 m carries a tensile load of 80 kN. Calculate the elongation.
Show Solution
\(\delta = \frac{PL}{EA} = \frac{80 \times 10^3 \times 2}{200 \times 10^9 \times 500 \times 10^{-6}} = 1.6 \text{ mm}\)
A solid circular shaft (d = 50mm, L = 1.5m, G = 80 GPa) transmits torque T = 2 kN·m. Find the angle of twist.
Show Solution
\(J = \frac{\pi d^4}{32} = \frac{\pi (0.05)^4}{32} = 6.136 \times 10^{-7} \text{ m}^4\)
\(\phi = \frac{TL}{GJ} = \frac{2000 \times 1.5}{80 \times 10^9 \times 6.136 \times 10^{-7}} = 0.0611 \text{ rad} = 3.5°\)
A cantilever beam (L = 3m, EI = 10,000 kN·m²) has a tip load W = 20 kN. Find tip deflection.
Show Solution
For cantilever with tip load:
\(\delta = \frac{WL^3}{3EI} = \frac{20 \times 3^3}{3 \times 10000} = 0.018 \text{ m} = 18 \text{ mm}\)