Torsion of shafts: practice problems with step-by-step solutions
Free Torsion of shafts practice problems with diagrams, hints and full worked solutions. New values every time, at three levels: shear stress, angle of twist, power transmission.
Free practice problemEasy
A solid circular shaft carries a torque T. Find the maximum shear stress τmax in the shaft.
A shaft is a rotating rod that passes on a turning effort: the drive shaft of a car, the spindle of a drill, the axle of a pump.
The turning effort is the torque T. It is a force times the arm it acts on, in N·m. Pushing 200 N on the end of a 0.5 m wrench gives T = 200 · 0.5 = 100 N·m.
A shaft loaded by a torque is in torsion. It twists a little, and its material carries shear stress τ (tau): neighbouring slices of the shaft try to slide past each other.
Two questions matter in design: is the shear stress low enough (strength), and is the twist small enough (stiffness)?
Self-check You push 200 N at the end of a 0.5 m wrench, at right angles. What torque do you apply?
100 N·m
200 N·m
400 N·m
50 N·m
Answer: 100 N·m. T = F·r = 200 · 0.5 = 100 N·m. Dividing by 0.5 gives 400; forgetting the arm gives 200.
2Shear stress and the polar moment
The centre of the shaft hardly deforms. The further a point is from the centre, the more it is sheared. So the shear stress grows linearly with the radius, from zero at the centre to its maximum at the surface.
τ = T · ρ / J τmax = T · r / J
Here ρ (rho) is the distance from the centre, r = d/2 is the outer radius, and J is the polar second moment of area. J measures how well the cross-section resists twisting:
For a solid shaft of diameter d, and for a hollow shaft with outer diameter do and inner diameter di:
J = π · d⁴ / 32 J = π · (do⁴ − di⁴) / 32
For a solid shaft the two formulas combine into one short one:
τmax = 16 · T / (π · d³)
Work in N·mm and mm, so the stress comes out in N/mm² = MPa. 1 N·m = 1000 N·mm.
Self-check In a solid shaft τmax = 60 MPa. What is the shear stress halfway between the centre and the surface?
0 MPa
15 MPa
30 MPa
60 MPa
Answer: 30 MPa. The stress grows linearly with ρ, so at ρ = r/2 it is half of τmax: 30 MPa. It is not the same everywhere, and it is zero only at the centre.
3Angle of twist and power
The free end of the shaft rotates by the angle of twist φ (phi). A longer or thinner shaft twists more; a stiffer material twists less:
φ = T · L / (G · J)
G is the shear modulus of the material: about 80 GPa for steel and 26 GPa for aluminium. The formula gives φ in radians. Multiply by 180/π for degrees.
Shafts often carry power from a motor. Power is torque times angular speed, where the speed in rpm must first be turned into rad/s:
P = T · ω ω = 2π · N / 60
So a question that gives kW and rpm starts with T = P/ω, with P in watts.
Self-check A shaft turns at ω = 100 rad/s and transmits 5 kW. What torque does it carry?
0.05 N·m
20 N·m
50 N·m
500 000 N·m
Answer: 50 N·m. T = P/ω = 5000 W / 100 rad/s = 50 N·m. 0.05 comes from using 5 instead of 5000 W; 20 divides ω by P.
4The method step by step
If power and speed are given, find the torque first: ω = 2πN/60, then T = P/ω.
Convert to N and mm: T in N·mm (N·m × 1000), L in mm, G in MPa (GPa × 1000).
Find J: π·d⁴/32 for a solid shaft, π(do⁴ − di⁴)/32 for a hollow one.
Shear stress: τmax = T·r/J with r the outer radius.
Twist: φ = TL/(GJ) in radians, then × 180/π if degrees are asked.
For a design question (find d), write τmax = 16T/(πd³) equal to the allowed stress and solve for d.
5A worked example
Follow every step. Try to predict the next one before you open it.
Worked example
A solid steel shaft (G = 80 GPa) of diameter 40 mm and length 1 m is fixed at one end. A torque of 500 N·m acts at the free end. Find τmax and the angle of twist φ in degrees.
1
Look at the shaft and convert units
T = 500 N·m = 500 000 N·mm, L = 1000 mm, G = 80 000 MPa, r = 20 mm.
2
Polar second moment of area
J = π · 40⁴ / 32 = 251 327 mm⁴
3
Maximum shear stress
τmax = T · r / J = 500 000 · 20 / 251 327 = 39.8 MPa
The short formula gives the same: 16 · 500 000 / (π · 40³) = 39.8 MPa.
A little under one and a half degrees over a metre of steel shaft is a sensible size.
Answer
τmax = 39.8 MPa and φ = 1.42°.
6Common mistakes
Using the diameter instead of the radius in τ = T·r/J. That doubles the stress.
Forgetting to convert N·m to N·mm, which makes the stress 1000 times too small.
Using rpm directly in P = T·ω. Convert to rad/s with ω = 2πN/60 first.
Giving φ in radians when degrees are asked, or the other way round.
For a hollow shaft: subtracting the diameters before raising to the fourth power. Use do⁴ − di⁴.
Self-check A hollow shaft with outer diameter 40 mm has J = 200 000 mm⁴. It carries T = 1000 N·m. What is τmax?
0.1 MPa
100 MPa
200 MPa
5 MPa
Answer: 100 MPa. τ = T·r/J = 1 000 000 · 20 / 200 000 = 100 MPa. Using r = 40 mm (the diameter) gives 200 MPa; leaving T in N·m gives 0.1 MPa; forgetting r gives 5.
7How exam questions look
Easy: shear stress. A solid shaft of diameter d carries T; find τmax. Variants ask for the stress at a given radius or the allowable torque.
Medium: angle of twist. You also get L and G; find φ in degrees. Variants use a hollow or stepped shaft.
Hard: power transmission. Power in kW and speed in rpm are given; find the torque, then τmax in a hollow shaft or the smallest diameter for an allowed stress.
Write the torque in N·mm next to the problem before anything else. It is the number every later step uses.
A solid shaft of diameter 40 mm carries a torque of 750 N·m. Find the shear stress τρ at a radius of 15 mm from the centre.
T = 750 N·md = 40 mmρ = 15 mm
Find τρ(MPa)
Show hints
Shear stress in a circular shaft is proportional to the distance from the centre: τ = T·ρ/J.
Use N and mm throughout: T in N·mm, L in mm, G in MPa (GPa × 1000), J in mm⁴.
J = 251327 mm⁴.
Show the answer and the step-by-step solution
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.
Polar second moment of area
J = π·40⁴/32 = 251327 mm⁴
Shear stress at ρ
For comparison, at the surface τmax = 59.68 MPa.
T = 750 N·m = 750000 N·mm
τρ = T·ρ/J = 750000·15/251327 = 44.76 MPa
Answer
τρ = 44.76 MPa
Example 2Medium
A solid brass shaft is fixed at the wall and twisted by a torque T at its free end. Find the angle of twist φ of the free end, in degrees.
T = 750 N·md = 40 mmL = 1.5 mG = 39 GPa
Find φ(°)
Show hints
Use φ = TL/(GJ), which gives radians.
Use N and mm throughout: T in N·mm, L in mm, G in MPa (GPa × 1000), J in mm⁴.
J = 251327 mm⁴.
Show the answer and the step-by-step solution
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.
Polar second moment of area
J = π·40⁴/32 = 251327 mm⁴
Angle of twist
φ = TL/(GJ) = 750·1000·1500/(39000·251327) = 0.1148 rad
φ = 0.1148·180/π = 6.576°
Answer
φ = 6.576°
Example 3Hard
A solid shaft must transmit 15 kW at 1000 rpm without the shear stress exceeding 75 MPa. Find the smallest diameter d that works.
P = 15 kWN = 1000 rpmτallow = 75 MPa
Find d(mm)
Show hints
Torque from power: T = P/ω with ω = 2πN/60 in rad/s.
Set τmax = 16T/(πd³) equal to the allowable stress and solve for d.
T = 143.2 N·m.
Show the answer and the step-by-step solution
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.
Torque
ω = 2πN/60 = 2π·1000/60 = 104.7 rad/s
T = P/ω = 15000/104.7 = 143.2 N·m
Diameter from the allowable stress
d = ∛(16T/(π·τallow)) = ∛(16·143239/(π·75)) = 21.35 mm
Answer
d = 21.35 mm
In practice you would round up to the next standard size.
Questions and answers
What is Torsion of shafts?
A shaft is a rotating rod that passes on a turning effort: the drive shaft of a car, the spindle of a drill, the axle of a pump.
The turning effort is the torque T. It is a force times the arm it acts on, in N·m. Pushing 200 N on the end of a 0.5 m wrench gives T = 200 · 0.5 = 100 N·m.
How do I solve Torsion of shafts problems step by step?
Find the polar moment of area: J = πd⁴/32 for a solid shaft.
Shear stress at the surface: τ = T·r/J, with r the outer radius.
Angle of twist: φ = TL/(GJ) in radians.
If power is given, find the torque first with P = T·ω, ω = 2πn/60.
What are common mistakes in Torsion of shafts?
Using the diameter instead of the radius in τ = T·r/J.
Forgetting to convert rpm to rad/s.
Giving the angle of twist in radians when degrees are asked.
What do the Torsion of shafts exercises cover?
Easy: shear stress
Medium: angle of twist
Hard: power transmission
Every problem has new values, so you learn the method instead of an answer. Stuck? Open a small hint first, and the full explanation only when you need it.
How practising works
Every problem has new values, so you learn the method instead of an answer.
Stuck? Open a small hint first, and the full explanation only when you need it.
Your progress, mistakes and saved problems are kept, so you can review before the exam.