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Solid shaft 300 kW; hollow shaft saving

Short answer

Torque T = 60×300000/(2π×120) = 23.87 kNm. Solid diameter = [16T/(π×100)]¹/³ = 106.7 mm. Hollow with d_i=0.6D_o: D_o = [16T/(π×100×(1-0.6⁴))]¹/³ = 111.8 mm. Weight saving = (8947-6281)/8947 = 29.8%.

A solid shaft is to transmit 300 kW at 120 rpm. If the shear stress is not to exceed 100 MPa, find the diameter of the shaft. What percent saving in weight would be obtained if this shaft were replaced by a hollow one whose internal diameter equals 0.6 of the external diameter, the length, material and maximum allowable shear stress being the same?

Data: P=300P = 300 kW, N=120N = 120 rpm, τmax=100 MPa\tau_{max} = 100\ \text{MPa}.

Torque

T=60P2πN=60(300×103)2π(120)=23873.2 N⋅m=2.3873×107 N⋅mmT = \frac{60P}{2\pi N} = \frac{60(300\times10^3)}{2\pi(120)} = 23873.2\ \text{N·m} = 2.3873\times10^{7}\ \text{N·mm}

Solid shaft

T=π16τd3⇒d3=16Tπτ=16(2.3873×107)π(100)=1.2159×106 mm3T = \frac{\pi}{16}\tau d^3 \Rightarrow d^3 = \frac{16T}{\pi\tau} = \frac{16(2.3873\times10^{7})}{\pi(100)} = 1.2159\times10^{6}\ \text{mm}^3 d=106.73 mmd = 106.73\ \text{mm}

Hollow shaft (di=0.6Dod_i = 0.6D_o, same TT and τ\tau)

T=π16τDo3(1−0.64)⇒Do3=16Tπτ(1−0.1296)=1.3969×106 mm3T = \frac{\pi}{16}\tau D_o^3(1-0.6^4) \Rightarrow D_o^3 = \frac{16T}{\pi\tau(1-0.1296)} = 1.3969\times10^{6}\ \text{mm}^3 Do=111.79 mm,di=0.6Do=67.07 mmD_o = 111.79\ \text{mm}, \qquad d_i = 0.6D_o = 67.07\ \text{mm}

Saving in weight

Weight ∝\propto area (same length and material).

As=π4(106.73)2=8947.0 mm2,Ah=π4(111.792−67.072)=6281.2 mm2A_s = \frac{\pi}{4}(106.73)^2 = 8947.0\ \text{mm}^2, \qquad A_h = \frac{\pi}{4}(111.79^2 - 67.07^2) = 6281.2\ \text{mm}^2 Saving=As−AhAs×100=29.8%\text{Saving} = \frac{A_s - A_h}{A_s}\times100 = 29.8\%

Answer: solid shaft d=106.7d = 106.7 mm; the hollow shaft (Do=111.8D_o = 111.8 mm, di=67.1d_i = 67.1 mm) saves about 29.8% of the weight.

✓ Answer checked by Claude Opus 5.5 (high effort)

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