Axial stress & strain
- Normal stress
- σ = F/A
- Strain
- ε = ΔL/L
- Hooke’s law
- σ = E·ε
- Elongation
- ΔL = FL/(EA)
- Thermal strain
- εT = α·ΔT
Formula sheets, the key method steps for every topic and 14 practice problems, with full solutions at the back.
Find the axial strain ε in the aluminium bar, in microstrain (µε = 10⁻⁶).
The stepped bar is fixed at the wall and loaded at its free end. Find the total elongation δ.
A solid aluminium shaft 1.5 m long carries a torque of 200 N·m. The angle of twist between its ends must not exceed 1.5°. Find the smallest diameter d that meets this limit.
A hollow shaft must transmit 75 kW at 3000 rpm. Its inner diameter is to be 0.6 times its outer diameter, and the shear stress must not exceed 60 MPa. Find the smallest outer diameter do.
Find the bending moment Mx at the section x = 6 m from support A. Use sagging moments positive and upward shear on the left face positive.
The marks A to G split the beam into six equal parts. At which mark is the bending moment largest in size, sagging or hogging? Click it in the diagram.
A symmetric I-section has flanges 130 mm wide and 18 mm thick, an overall depth of 220 mm and a web 14 mm thick. Find its elastic section modulus Sx for bending about the horizontal centroidal axis x.
A cast-iron beam has the T-section shown, with the flange on top. The allowable stresses are 35 MPa in tension and 150 MPa in compression. Find the largest sagging bending moment Mallow the beam can carry.
A point has σₓ = −30 MPa and σᵧ = 0 MPa. How large can the shear stress τxy (magnitude) become before the largest principal stress reaches 20 MPa?
A point in a loaded part has σₓ = 10 MPa, σᵧ = −60 MPa and τₓᵧ = 30 MPa (tension positive). Find the normal stress σθ on a plane whose normal is rotated 30° counter-clockwise from the x-axis.
A cantilever carries a uniformly distributed load w over its whole length. Find the deflection δ at the free end.
A simply supported timber beam of rectangular section, 150 mm wide, carries a uniformly distributed load w over its whole span. Its midspan deflection may not exceed span/360. Find the smallest depth hreq of the section.
A column of length L made of steel is fixed at both ends. Using Euler’s formula, find its critical buckling load Pcr.
A solid round column of length L made of aluminium is fixed at the base and free at the top. It must carry 65 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.
Answer: ε = 10913 µε
Hooke’s law: within the elastic range σ = E·ε. Combining with σ = P/A and ε = δ/L gives δ = PL/(AE).
E for aluminium = 70 GPa = 70000 MPa.
Answer: δ = 8.484 mm
Axially loaded members: each prismatic segment obeys δ = PL/(AE). For members in series the elongations add.
The free end carries P, so every section of the bar carries the same axial force P. Only the areas and materials change.
Answer: d = 46.03 mm
Torsion of circular shafts: shear stress grows linearly from zero at the centre to τmax = T·r/J at the surface, with J = πd⁴/32 for a solid shaft and J = π(do⁴ − di⁴)/32 for a hollow one. The angle of twist is φ = TL/(GJ) in radians. Power and torque are linked by P = T·ω.
In practice you would round up to the next standard size.
Answer: do = 28.55 mm
Torsion of circular shafts: shear stress grows linearly from zero at the centre to τmax = T·r/J at the surface, with J = πd⁴/32 for a solid shaft and J = π(do⁴ − di⁴)/32 for a hollow one. The angle of twist is φ = TL/(GJ) in radians. Power and torque are linked by P = T·ω.
A solid shaft would need d = 27.26 mm; the hollow one is a little larger but much lighter.
Answer: Mx = 120 kN·m
Method of sections: cut the beam where you want the internal forces and keep one side. The shear force V balances the vertical forces on that side, and the bending moment M balances their moments about the cut.
Equilibrium of the whole beam.
Only forces left of the cut appear. A UDL portion acts as one force at the middle of that portion.
Answer: F (x = 10 m)
Largest bending moment: M changes slope wherever the shear force changes, so its peaks sit where the shear force crosses zero: under a point load, inside a distributed load, or over a support of an overhang. At a fixed support of a cantilever the moment is largest at the wall. Compare the size of M at those places, sagging or hogging.
Take moments about one support, then use the vertical force balance.
Sum the moments of all forces on one side of the cut. Sagging is positive, hogging negative.
Answer: Sx = 501.2 × 10³ mm³
Second moment of area: for a rectangle about its own centroidal axis I = bh³/12, and about its base bh³/3 (parallel-axis theorem I = Ic + A·d²). A hole is handled by subtracting its I, as long as both parts share the same axis. The elastic section modulus is S = I/c, where c is the distance to the extreme fibre.
Both cut-outs together are (B − tw) wide and hw = H − 2tf high, and share the axis x.
Answer: Mallow = 11.48 kN·m
Composite areas: split the section into rectangles. The centroid is the area-weighted average ȳ = ΣAᵢyᵢ/ΣAᵢ. The second moment of area about the centroidal axis adds each part’s own bh³/12 and its parallel-axis term A·d², where d is the distance from the part’s centroid to the overall centroid. Bending stress is σ = M·y/I.
Sagging: the top is in compression, the bottom in tension.
Answer: τxy = 31.62 MPa
Plane stress transformation: on a plane rotated by θ, σ = (σₓ + σᵧ)/2 + (σₓ − σᵧ)/2·cos 2θ + τₓᵧ·sin 2θ. Mohr’s circle shows all planes at once: its centre is at C = (σₓ + σᵧ)/2 and its radius is R = √(((σₓ − σᵧ)/2)² + τₓᵧ²). The principal stresses are C ± R, and the largest in-plane shear stress is R.
Answer: σθ = 18.48 MPa
Plane stress transformation: on a plane rotated by θ, σ = (σₓ + σᵧ)/2 + (σₓ − σᵧ)/2·cos 2θ + τₓᵧ·sin 2θ. Mohr’s circle shows all planes at once: its centre is at C = (σₓ + σᵧ)/2 and its radius is R = √(((σₓ − σᵧ)/2)² + τₓᵧ²). The principal stresses are C ± R, and the largest in-plane shear stress is R.
Answer: δ = 54.79 mm
Beam deflection: deflections come from integrating the curvature M/(EI) twice. Standard cases are tabulated, for example PL³/(3EI) for a cantilever with an end load and 5wL⁴/(384EI) for a simply supported beam under a full uniform load. Deflections from separate loads add (superposition).
Work in N and mm: P in N, w in N/mm (kN/m = N/mm), L in mm, E in N/mm² (GPa × 1000), I in mm⁴.
Answer: hreq = 204.3 mm
Beam deflection: deflections come from integrating the curvature M/(EI) twice. Standard cases are tabulated, for example PL³/(3EI) for a cantilever with an end load and 5wL⁴/(384EI) for a simply supported beam under a full uniform load. Deflections from separate loads add (superposition).
Work in N and mm: P in N, w in N/mm (kN/m = N/mm), L in mm, E in N/mm² (GPa × 1000), I in mm⁴.
In practice, choose the next standard depth above this value.
Answer: Pcr = 67.38 kN
Euler buckling: a slender column becomes unstable at Pcr = π²EI/(KL)², where KL is the effective length: K = 1 for pinned–pinned, 2 for fixed–free, 0.7 for fixed–pinned and 0.5 for fixed–fixed. It buckles about the axis with the smallest I. The formula holds for slender columns, whose critical stress stays below the yield stress.
A rectangle buckles about its weak axis, using the smaller dimension cubed.
Answer: d = 81.04 mm
Euler buckling: a slender column becomes unstable at Pcr = π²EI/(KL)², where KL is the effective length: K = 1 for pinned–pinned, 2 for fixed–free, 0.7 for fixed–pinned and 0.5 for fixed–fixed. It buckles about the axis with the smallest I. The formula holds for slender columns, whose critical stress stays below the yield stress.
Check: the slenderness KL/r is 148.1, so the column is slender and Euler applies.
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