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Solid shaft 120 Nm: diameter from limits

Short answer

Strength condition: d = (16T/(πτ))¹/³ = (16×1.2×10⁵/(π×100))¹/³ = 18.28 mm. Stiffness condition: using twist 0.5236 rad over 10d gives d = (320T/(πGθ))¹/³ = (320×1.2×10⁵/(π×10⁵×0.5236))¹/³ = 6.16 mm. Use larger diameter 18.28 mm, adopt 20 mm.

A solid circular shaft is subjected to a torque 120 Nm. Determine the diameter if the allowable shear stress is 100 N/mm2100\ \text{N/mm}^2 and the allowable angle of twist is 30∘30^\circ per 10 diameter length of the shaft. G=105 N/mm2G = 10^5\ \text{N/mm}^2.

Data: T=120 N⋅m=1.2×105 N⋅mmT = 120\ \text{N·m} = 1.2\times10^5\ \text{N·mm}, τallow=100 N/mm2\tau_{allow} = 100\ \text{N/mm}^2, G=105 N/mm2G = 10^5\ \text{N/mm}^2, allowable twist θ=30∘=0.5236\theta = 30^\circ = 0.5236 rad over a length L=10dL = 10d (taken as stated in the question). Find the diameter satisfying both strength and stiffness; the larger value governs.

(a) Strength condition

T=π16τd3⇒d3=16Tπτ=16(1.2×105)π(100)=6111.5 mm3T = \frac{\pi}{16}\tau d^3 \Rightarrow d^3 = \frac{16T}{\pi\tau} = \frac{16(1.2\times10^5)}{\pi(100)} = 6111.5\ \text{mm}^3 d1=18.28 mmd_1 = 18.28\ \text{mm}

(b) Stiffness condition

θ=TLGJ=T(10d)Gπ32d4=320 TπGd3⇒d3=320(1.2×105)π(105)(0.5236)=233.44 mm3\theta = \frac{TL}{GJ} = \frac{T(10d)}{G\frac{\pi}{32}d^4} = \frac{320\,T}{\pi G d^3} \Rightarrow d^3 = \frac{320(1.2\times10^5)}{\pi(10^5)(0.5236)} = 233.44\ \text{mm}^3 d2=6.16 mmd_2 = 6.16\ \text{mm}

Design diameter

The larger of the two governs, so the shaft is designed for strength: d=18.28d = 18.28 mm.

Answer: provide a solid shaft of diameter d=18.28 mmd = 18.28\ \text{mm}, adopt 20 mm.

✓ Answer checked by Claude Opus 5.5 (high effort)

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