Water Resources HubFE section 8 of 16 · free theory
Mechanics of Materials
How members stretch, twist, bend, and buckle: axial stress and strain, torsion of shafts, beam bending and shear, combined loading, stress transformation with Mohr's circle, standard deflection cases, and Euler buckling.
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Axial stress, strain, and Hooke's law
Pull (or push) on a prismatic bar and two things happen: an internal stress develops, and the bar gets longer (or shorter). Within the elastic range, stress and strain are proportional — that proportionality constant is Young's modulus, E.
σ = P / A
| σ | normal (axial) stress — positive in tension |
| P | axial force |
| A | cross-sectional area |
ε = δ / L
| ε | normal strain, dimensionless (mm/mm) |
| δ | change in length |
| L | original length |
σ = Eε δ = PL / (AE)
| E | modulus of elasticity (steel ≈ 200 GPa, aluminium ≈ 70 GPa) |
The exam loves the unit shortcut: with P in newtons and A in mm², P/A comes out directly in N/mm² = MPa. And δ = PL/(AE) only applies while the material stays elastic — past yielding, the linear relationship is gone.
Worked example Stretch of a steel tie rod
Given:
- Steel rod, diameter 20 mm, length 1.5 m.
- Axial tensile load P = 45 kN.
- E = 200 GPa.
Solution:
- Area: A = πd²/4 = π(20)²/4 = 314.2 mm².
- Stress: σ = P/A = 45,000 N / 314.2 mm² = 143.2 N/mm² = 143.2 MPa (tension).
- Strain: ε = σ/E = 143.2 MPa / 200,000 MPa = 7.16×10−4.
- Elongation: δ = εL = 7.16×10−4 × 1500 mm = 1.07 mm.
Answer: σ ≈ 143 MPa tension; the rod stretches about 1.07 mm.
Torsion of circular shafts
Twist a shaft and shear stress varies linearly from zero at the centre to a maximum at the outer surface. Two formulas do nearly all the work: one for the stress, one for the angle of twist. Both use the polar moment of inertia J — for a solid shaft, J = πd4/32.
τ = Tr / J
| τ | shear stress at radius r (maximum at r = c, the outer radius) |
| T | applied torque |
| J | polar moment of inertia about the shaft axis |
φ = TL / (GJ)
| φ | angle of twist, in radians |
| G | shear modulus (steel ≈ 75–79 GPa) |
| L | shaft length |
Jsolid = πd4/32 Jhollow = π(D4 − d4)/32
Power transmission ties in neatly: P = Tω, so a shaft's required diameter follows from the power and the rotational speed. When a problem gives both a stress limit and a twist limit, size the shaft for each and take the larger diameter.
Worked example Sizing a shaft for strength and stiffness
Given:
- Solid steel shaft carrying T = 1.8 kN·m.
- Allowable shear stress 40 MPa; allowable twist 1.5° per metre.
- G = 79 GPa.
Solution:
- Strength: τ = Tc/J = 16T/(πd³) ≤ 40 MPa gives d³ ≥ 16(1800)/(π × 40×106), so d ≥ 61.2 mm.
- Stiffness: φ/L = T/(GJ) ≤ 1.5°/m = 0.02618 rad/m gives J ≥ 1800/(79×109 × 0.02618) = 8.70×10−7 m4, so d ≥ 54.6 mm.
- Strength governs. Choose d = 65 mm (next practical size up).
- Check: J = π(0.065)4/32 = 1.753×10−6 m4; τ = 1800(0.0325)/J = 33.4 MPa ≤ 40 MPa ✓; φ/L = 1800/(GJ) = 0.0130 rad/m = 0.745°/m ≤ 1.5°/m ✓.
Answer: A 65 mm solid shaft satisfies both limits, with strength controlling the design.
Beam bending and shear stress
Bending puts the outer fibres in the most stress and the neutral axis in none — stress varies linearly through the depth. Shear stress does the opposite: it peaks at the neutral axis and vanishes at the top and bottom. The flexure formula and the shear formula are the pair to memorise.
σ = My / I σmax = Mc / I
| M | bending moment at the section |
| y | distance from the neutral axis |
| I | second moment of area about the neutral (bending) axis — rectangle: bh³/12 |
| c | distance from neutral axis to the outer fibre (h/2 for a symmetric section) |
τ = VQ / (It)
| V | shear force at the section |
| Q | first moment of the area above (or below) the point about the neutral axis |
| t | width of the section at the level where τ is evaluated |
For a rectangle, the shear formula collapses to τmax = 3V/(2A) at the neutral axis; for a solid circle, τmax = 4V/(3A). And note the symbols: bending uses I, torsion uses J — they are different quantities, and swapping them is a favourite exam trap.
Worked example Simply supported beam under uniform load
Given:
- Simply supported beam, span L = 5 m, uniform load w = 12 kN/m.
- Rectangular section 150 mm × 350 mm (depth vertical), E = 12 GPa.
Solution:
- Reactions: R = wL/2 = 12(5)/2 = 30 kN each end. Maximum shear Vmax = 30 kN at the supports.
- Maximum moment at midspan: Mmax = wL²/8 = 12(25)/8 = 37.5 kN·m.
- I = bh³/12 = 0.15(0.35)³/12 = 5.359×10−4 m4; c = 0.175 m. Bending stress: σmax = Mc/I = 37,500(0.175)/5.359×10−4 = 12.2 MPa.
- Shear stress: τmax = 3V/(2A) = 3(30,000)/(2 × 0.0525) = 0.857 MPa at the neutral axis.
- Deflection: δmax = 5wL4/(384EI) = 5(12,000)(625)/(384 × 12×109 × 5.359×10−4) = 0.0152 m = 15.2 mm downward at midspan.
Answer: σmax ≈ 12.2 MPa, τmax ≈ 0.857 MPa, midspan deflection ≈ 15.2 mm.
Combined axial load and bending
Real members rarely see one load at a time. When axial load and bending act together, the stresses simply add at each fibre — compression plus bending compression on one face, and possibly net tension on the other even when the axial load is compressive.
σ = ±P/A ± Mc/I
| σ | combined normal stress at the outer fibre — evaluate both faces and keep the worst |
Sign discipline matters: take tension as positive, compression as negative, and add. The exam likes columns where the bending stress slightly exceeds the uniform compression, leaving one face in (small) tension — always compute both faces.
Stress transformation and Mohr's circle
A stress element can be rotated, and the normal and shear stresses on the rotated faces change with the angle. The principal stresses are the maximum and minimum normal stresses (shear is zero on those planes); the maximum in-plane shear stress acts on planes at 45° to them.
σ1,2 = (σx + σy)/2 ± √[((σx − σy)/2)² + τxy²]
| σ1, σ2 | principal stresses (σ1 ≥ σ2) |
| τxy | shear stress on the x-face, with the standard sign convention |
τmax, in-plane = √[((σx − σy)/2)² + τxy²] tan 2θp = 2τxy / (σx − σy)
Quick check with σx = 80 MPa, σy = −40 MPa, τxy = 30 MPa: the centre is (80 − 40)/2 = 20 MPa, the radius is √(60² + 30²) = 67.1 MPa, so σ1 = 87.1 MPa, σ2 = −47.1 MPa, and τmax = 67.1 MPa at θp = 13.3°. A useful special case: a shaft surface in pure torsion (τ only) has principal stresses of ±τ at 45° — which is why torsion failures in brittle materials spiral at 45 degrees.
Beam deflection — the standard cases
The exam rarely asks you to integrate the moment-curvature relationship; it hands you a standard loading and expects the handbook formula. Learn these four cold, including where the maximum occurs:
Simply supported, centre point load P: δmax = PL³ / (48EI) (at midspan)
Simply supported, uniform load w: δmax = 5wL4 / (384EI) (at midspan)
Cantilever, end point load P: δmax = PL³ / (3EI) (at the free end)
Cantilever, uniform load w: δmax = wL4 / (8EI) (at the free end)
The matching maximum moments are PL/4, wL²/8, PL, and wL²/2. Notice the pattern: a cantilever deflects 16 times more than a simply supported beam under the same centre/end point load (1/3 vs 1/48) — end fixity matters enormously. Try the beam stress & deflection calculator to check your hand calculations.
Euler column buckling
A long, slender column can buckle sideways well below the material's crushing strength. The critical load depends on stiffness (EI), length, and — critically — the end conditions, folded into the effective length factor K.
Pcr = π²EI / (KL)²
| Pcr | critical (Euler) buckling load |
| K | effective length factor: 1.0 pinned–pinned, 0.5 fixed–fixed, 0.7 fixed–pinned, 2.0 fixed–free |
| I | second moment of area about the axis of buckling — use the smallest I (weak axis) |
Euler's formula assumes a long column (elastic buckling). Short, stocky columns crush at P = σyA instead — the exam sometimes asks which mode controls, so check both when the slenderness is in doubt. Also watch for K: a flagpole (fixed–free, K = 2) buckles at one-quarter the load of the same column pinned at both ends.