Mechanics of materials / Sheet 02
Torsion
Formula reference for mechanical engineering interviews. The Tr/J form assumes a circular, prismatic, linear elastic shaft. Non-circular sections warp and need the separate relations in section 03.
Scope: Shafts, tubes, springsSections: 11Units: SI, N·m and mmRevision: A
42 formulas
The torsion equation
T / J = τ / r = G θ / L
One identity. Any two terms solve for the third.
Shear stress at radius r
τ = T r / J
Linear in r, so maximum at the outer surface
Max shear stress
τmax = T c / J = T / Zp
c is the outer radius
Polar section modulus
Zp = J / c = π d³ / 16
Solid round; the torsion analogue of S in bending
Angle of twist
θ = T L / G J
Result is in radians; multiply by 180/π for degrees
Torsional stiffness
kt = G J / L
GJ on its own is the torsional rigidity
Shear strain
γ = r θ / L
The compatibility relation behind the whole derivation
Strain energy
U = T² L / (2 G J) = ½ T θ
Used in Castigliano problems
Polar second moment of area
Solid round
J = π d⁴ / 32
Always exactly 2I for a circular section
Hollow round
J = π (do⁴ - di⁴) / 32
τmax still uses c = do/2
Thin-walled closed tube
J ≈ 2 π R³ t
R is the mean radius, t the wall thickness
Thin open section
J ≈ ⅓ Σ b t³
Channels, angles, slit tubes; b is the developed width
Perpendicular axis theorem
J = Ix + Iy
Valid for plane sections only
Non-circular sections
Rectangle, a > b
τmax = T / (α a b²)
α ≈ 0.208 for a square; from tables otherwise
Rectangle, twist
θ = T L / (β a b³ G)
β ≈ 0.141 for a square
Shear flow, Bredt-Batho
q = T / (2 Am)
Am is the area enclosed by the mean wall line
Closed thin wall stress
τ = T / (2 Am t)
Shear flow is constant, so τ is highest at the thinnest wall
Maximum shear in a rectangular shaft occurs at the midpoint of the longest side, and is zero at the corners. Corners are free surfaces on two faces, so shear must vanish there.
Power transmission
Power
P = T ω
ω in rad/s, T in N·m, P in watts
Speed conversion
ω = 2 π N / 60
N in rpm
Practical form
T = 9550 P / N
T in N·m, P in kW, N in rpm
Shaft sizing
d³ = 16 T / (π τallow)
Then check twist separately, it often governs
Combined bending and torsion
Equivalent torque
Te = √(M² + T²)
Max shear stress theory
Equivalent moment
Me = ½ [M + √(M² + T²)]
Max normal stress theory
von Mises
σv = √(σ² + 3 τ²)
Preferred for ductile shafts
Solid shaft shortcuts
σ = 32M / πd³, τ = 16T / πd³
Since J = 2I, bending stress is always double
Critical element is on the surface, at the fibre farthest from the neutral axis in bending. Torsional shear is uniform around the circumference, so the governing point is where bending tension peaks.
Hollow vs solid
Diameter ratio
k = di / do
Everything below is a function of k alone
Torque capacity retained
1 - k⁴
Also the fraction of stiffness retained
Mass retained
1 - k²
Compare the two before choosing a bore
Bore half the diameter
k = 0.5 → 93.75% capacity, 75% mass
Costs 6.25% of strength to save 25% of weight
Bore 70% of diameter
k = 0.7 → 76% capacity, 51% mass
Roughly the practical limit before walls get thin
The inner material sits at small r, where τ = Tr/J is small, so it earns almost nothing while carrying full mass. This is the standard answer to "how would you lighten this drive shaft."
Pure shear and failure planes
Stress state at the surface
σx = σy = 0, τxy = τ
Pure shear; Mohr's circle is centred on the origin
Principal stresses
σ1 = +τ, σ2 = -τ
At 45° to the shaft axis
Ductile failure
flat transverse plane
Fails in shear, and max shear is on the cross section
Brittle failure
45° helical fracture
Fails on the plane of maximum tensile principal stress
Shafts in series and parallel
Series
T is common, θ = Σ T Li / Gi Ji
Stepped shafts, shafts joined end to end
Parallel
θ is common, Ti ∝ GiJi / Li
Fixed at both ends, or concentric shafts
Indeterminate case
ΣT = Tapplied, θ1 = θ2
Equal twist is the compatibility equation that closes it
Helical spring, torsion of a wire
Spring index
C = D / d
D is mean coil diameter, d is wire diameter; keep C between 6 and 12
Shear stress
τ = Kw · 8 F D / (π d³)
Same 16T/πd³ with T = FD/2
Wahl factor
Kw = (4C - 1)/(4C - 4) + 0.615/C
Corrects for curvature and direct shear
Deflection
δ = 8 F D³ n / (G d⁴)
n is the number of active coils
Spring rate
k = G d⁴ / (8 D³ n)
Wire diameter dominates, to the fourth power
Numbers worth memorizing
| Material | G (GPa) | E (GPa) | ν |
|---|---|---|---|
| Steel | 79 | 200 | 0.29 |
| Aluminium | 26 | 69 | 0.33 |
| Titanium Ti-6Al-4V | 44 | 114 | 0.34 |
| Grey cast iron | 41 | 100 | 0.26 |
You never need to memorize G. Rebuild it from G = E / [2(1 + ν)] and it falls out to about 0.385E for steel. Common twist limit: 1 degree per 20 diameters of length, or per metre.
Concepts that decide the interview
- Torque is carried almost entirely by the outer material, because τ scales with r and J scales with r⁴. The inner half of the diameter gives 6.25% of J for 25% of the mass.
- Torsion produces pure shear, so principal stresses are ±τ at 45 degrees. That single fact explains both failure modes: flat transverse fracture in ductile shafts, 45 degree helix in brittle ones.
- Design is often governed by twist, not stress. Check both, and say which one sizes the shaft. Strength scales with d³, stiffness with d⁴.
- The Tr/J form applies only to circular prismatic sections. Non-circular sections warp out of plane and the derivation collapses.
- Closed sections beat open ones by roughly 3(R/t)², which is a factor near 1200 at R/t = 20. Slitting a tube lengthwise destroys torsional rigidity, which is why frames and robot arm links use closed tubes and why a single access cutout matters.
- Shear flow is constant around a closed thin wall, so the thinnest wall carries the highest stress.
- Real shafts fail at keyways, splines, shoulders and cross holes, with Kt around 2 to 3. Under reversed bending plus steady torque it is fatigue, not static yield, that sets the diameter.
Sheet 02 of 07 / Rev ACircular prismatic linear elastic unless noted