Shaft Torsion: Solid and Hollow Shafts

Choose a shaft, solid or hollow, and what it has to carry: a power at a speed, or a torque. The page works out the shear stress, the twist and the safety factor, or sizes the smallest shaft of the chosen section that will do. The table under the drawing does the arithmetic.

Worked example — Shaft Torsion: Solid and Hollow Shafts

Angular speed	314.2 rad/s
Torque	477.5 N·m
Inner diameter	72.39 mm
Polar moment of area	614,000 mm⁴
Maximum shear stress	29.63 MPa
Shear stress at the bore	28.15 MPa
Angle of twist	0.835°
Twist per metre	0.557°/m

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Shaft Torsion: Solid and Hollow Shafts

A shaft carrying a torque T is in torsion. The shear stress grows in proportion to the radius, τ = T r ÷ J, largest at the surface. J is the polar moment of area, J = π(D⁴ − dᵢ⁴) ÷ 32, which sets how much of the section sits far from the axis. The shaft twists through φ = T L ÷ (G J), where G is the shear modulus. A shaft turning at angular speed ω carries power P = T ω.

What each symbol means

T: the torque, in N·m. From a power and a speed, T = P ÷ ω
J: the polar moment of area, in m⁴. The fourth power of the diameter is why a little more diameter buys a lot
τ: the shear stress, zero on the axis and largest at the surface
φ: the angle of twist, in radians. Proportional to the length
G: the shear modulus, the material's stiffness in shear. About 80 GPa for steel and 26 GPa for aluminium

The middle of a shaft carries almost nothing 🖖

The shear stress in a twisted shaft grows in proportion to the distance from the centre: zero on the axis, largest at the surface. The inner half of the radius carries only a sixteenth of the torque while holding a quarter of the metal. So take the middle out. Press Size a hollow shaft: for the same 150 kW at 3000 rpm, a solid shaft needs 32.30 mm and a hollow one with its bore at 80% of the diameter needs 38.50 mm, and the hollow one weighs 51.2% as much. The plot shows the whole trade: at the same outer diameter a bore of 80% keeps 59.0% of the torque capacity for 36.0% of the mass. A car's propshaft is a wide thin tube for a related reason this page does not compute: it must also be stiff in bending, or at its length and speed it would whirl, and a wide tube gives the most stiffness for its weight.

Spin a shaft ten times faster and it can carry ten times the power 🖖

Power is torque times angular speed. For a shaft of a given size the torque is capped by the stress the steel can take, so the power it can carry rises in direct proportion to how fast it turns. Press Car propshaft, 3000 rpm, then Same power, 300 rpm: the same 150 kW at a tenth of the speed needs ten times the torque, the stress climbs to 296.29 MPa and the safety factor falls from 4.87 to 0.49. To bring the stress back to where it was, and the design factor with it, the shaft must be thicker by the cube root of ten: 2.15 times the diameter. The Ship's propeller shaft is the extreme: 20 MW at 90 rpm is carried on a solid bar 650 mm across, where strength alone would accept 531 mm; real ones are sized for bending, fatigue and vibration too. It is also why the output shaft of a reduction gearbox, turning slowly, is usually the thickest one in it.

Twisting is shear, and shear is tension at 45° 🖖

Every point on the surface of a twisted shaft is in pure shear, and the results table shows what that means at 45° to the axis: one principal stress of +τ and one of −τ, the same size. Follow the link under the table to see this state in Mohr's circle. The same fact decides where a shaft breaks. Mild steel, like any ductile metal, gives way in shear, across the section. A brittle material such as cast iron gives way in tension, along the 45° helix. Where stiffness matters more than strength, the lighter metal can win: press Aluminium, same stiffness and a bar 52.98 mm across twists as much as the 40 mm steel one for 60.3% of its mass.

Problem solved in full

Sizing a drive shaft, solid and hollow 5 steps
A steel shaft must carry 150 kW at 3000 rpm with a design factor of 2. The steel yields at 250 MPa in tension. Find the smallest solid diameter, the smallest hollow outer diameter with its bore at 80% of the diameter, and how much the hollow one weighs against the solid one. Press Size a hollow shaft to check.

Open this problem in the tool
1. The torque follows from the power and the angular speed:
  
  $T = \frac{P}{ω} = \frac{150 000}{2 π \cdot 3000/60} = 477.5 N m$
2. The allowable shear stress is the shear yield, tensile yield ÷ √3, divided by the design factor:
  
  $τ_{allow} = \frac{σ _{y}}{3 n _{d}} = \frac{250}{3 \cdot 2} = 72.17 MPa$
3. A solid shaft reaches that stress at its surface when:
  
  $d = 3 \frac{16 T}{π τ _{allow}} = 3 \frac{16 \cdot 477 . 5}{π \cdot 72 . 17 \times 1 0 ^{6}} = 32.30 mm$
4. A hollow shaft with a bore of 0.8 D has 1 − 0.8⁴ of the solid J at the same D, so its outer diameter is the solid diameter divided by the cube root of 1 − 0.8⁴:
  
  $D = \frac{d}{3 1 - 0 . 8 ^{4}} = \frac{32 . 30}{3 0 . 5904} = 38.50 mm, d_{i} = 0.8 D = 30.80 mm$
5. Mass goes with the area of the section, so the ratio of the two masses is:
  
  $\frac{m _{hollow}}{m _{solid}} = (1 - 0 . 8^{2}) (\frac{38 . 50}{32 . 30})^{2} = 0.36 \cdot 1.4208 = 51 . 2%$
Answer

The solid shaft needs 32.30 mm and the hollow one 38.50 mm outside and 30.80 mm inside, and the hollow one weighs 51.2% as much for exactly the same strength. It is 19% wider, which is the price, and the reason hollow shafts appear wherever weight matters more than space.

$d = 32.30 mm, D = 38.50 mm, 51 . 2%$

Learning path

Where the load goes

References (1)

Where the four materials' modulus and yield figures would be checked. THEY WERE NOT TAKEN FROM HERE — see the note: A materials handbook or the relevant specifications: EN 10025 / ASTM A36 for mild steel, AISI 4140, 6061-T6 aluminium and Ti-6Al-4V.

By the 42Math team · Reviewed against published sources. Corrections welcome.