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Shaft diameter for power and bending

Short answer

Torque T =60P/(2πn) =663.1 N·m. Allowable shear: min(0.30×330,0.18×560)=99 MPa, reduced by25% for keyway →74.25 MPa. Equivalent twisting moment Te =√((1.5×450)²+(1.2×663.1)²) =1043.5 N·m. Then d³ = (16 Te)/(π τ_allow) gives d=41.5 mm, use42 mm standard.

A solid shaft transmits 25 kW at 360 rpm. The bending moment at the critical section is 450 N.m. The shaft material has ultimate tensile strength 560 MPa and yield strength 330 MPa, and the shaft has a keyway. The load is applied with minor shocks, so take Kb = 1.5 and Kt = 1.2. Using the ASME code, find the diameter of the shaft.

Use the ASME code equation d3=16πτallow(KbM)2+(KtT)2d^3 = \dfrac{16}{\pi \tau_{allow}}\sqrt{(K_b M)^2 + (K_t T)^2}.

Torque

T=60P2πn=60×25 0002π×360=663.1 N.mT = \frac{60P}{2\pi n} = \frac{60 \times 25\,000}{2\pi \times 360} = 663.1\ \text{N.m}

Allowable shear stress

0.30 Syt=0.30×330=99 MPa0.18 Sut=0.18×560=100.8 MPa\begin{aligned} 0.30\,S_{yt} &= 0.30 \times 330 = 99\ \text{MPa} \\ 0.18\,S_{ut} &= 0.18 \times 560 = 100.8\ \text{MPa} \end{aligned}

The smaller value is 99 MPa. With a keyway the stress is reduced by 25 percent:

τallow=0.75×99=74.25 MPa\tau_{allow} = 0.75 \times 99 = 74.25\ \text{MPa}

Equivalent moment

KbM=1.5×450=675 N.mKtT=1.2×663.1=795.8 N.mTe=6752+795.82=1043.5 N.m\begin{aligned} K_b M &= 1.5 \times 450 = 675\ \text{N.m} \\ K_t T &= 1.2 \times 663.1 = 795.8\ \text{N.m} \\ T_e &= \sqrt{675^2 + 795.8^2} = 1043.5\ \text{N.m} \end{aligned}

Diameter

d3=16×1043.5×103π×74.25=71 572 mm3d^3 = \frac{16 \times 1043.5 \times 10^3}{\pi \times 74.25} = 71\,572\ \text{mm}^3 d=41.5 mmd = 41.5\ \text{mm}

Adopt the next standard size, 42 mm.

Answer: d = 41.5 mm, so use a 42 mm shaft.

✓ Answer checked by Claude Opus 5.5 (high effort)

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