Calculate shear stress and angle of twist for solid or hollow circular shafts under torque, or derive torque from transmitted power and speed.
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Worked examples
These are the solver's own reference problems — the answers come from published sources or independent hand computation, never from the solver itself. The solution below is the live solver output for each.
Example 1
SI solid shaft: d=50 mm, L=1.5 m, G=80 GPa, T=1 kN·m at the free end
Circular sections remain plane (Saint-Venant torsion of circular shafts); linear elastic material. Positive twist and torque are counterclockwise viewed from the free end toward the support.
Solution steps
Internal torque in each segment (method of sections)
Cut inside a segment and sum the applied torques between the cut and the free end.
T(segment) = Σ torques applied beyond the cut, ccw positive
segment 1: T = 1000 N·m
Maximum shear stress in each segment
Torsional shear stress peaks at the outer surface: τ = Tc/J with c the outer radius and J the polar moment of the (possibly hollow) circular section.
τmax = T·c/J, J = π(do⁴ − di⁴)/32
segment 1: J = 613600 mm^4, τ = 40.7 MPa
= 40.7 MPa
Angle of twist at the free end
Each segment twists by TL/JG; the free-end rotation is the sum over the segments. Positive means counterclockwise viewed from the free end.
φ = Σ(T·L/(J·G))
φ = 0.03056 rad = 0.03056 rad = 1.751 deg
= 1.751 deg
Results
Internal torque in segment 1 (ccw +)
1000N·m
Maximum shear stress in segment 1
40.7MPa
Governing (largest) shear stress
40.7MPa
Total angle of twist at the free end (radians; ccw +)
0.03056rad
Total angle of twist at the free end (degrees; ccw +)
1.751deg
Where this answer was checked
source
τ = Tc/J and φ = TL/JG for a solid circular shaft (NCEES FE Reference Handbook) — single solid segment, tip torque
Circular sections remain plane (Saint-Venant torsion of circular shafts); linear elastic material. Positive twist and torque are counterclockwise viewed from the free end toward the support.
Solution steps
Internal torque in each segment (method of sections)
Cut inside a segment and sum the applied torques between the cut and the free end.
T(segment) = Σ torques applied beyond the cut, ccw positive
segment 1: T = 3000 N·m; segment 2: T = 1000 N·m
Maximum shear stress in each segment
Torsional shear stress peaks at the outer surface: τ = Tc/J with c the outer radius and J the polar moment of the (possibly hollow) circular section.
Circular sections remain plane (Saint-Venant torsion of circular shafts); linear elastic material. Positive twist and torque are counterclockwise viewed from the free end toward the support.
Solution steps
Torque from transmitted power
Steady power equals torque times angular speed, so the shaft torque is the power divided by the rotational speed in radians per second.
T = P/ω with ω = 2πN/60
ω = 31.42 rad/s at 300 rpm; T = 40 hp / 31.42 rad/s = 0.7003 kip·ft
= 0.7003 kip·ft
Internal torque in each segment (method of sections)
Cut inside a segment and sum the applied torques between the cut and the free end.
T(segment) = Σ torques applied beyond the cut, ccw positive
segment 1: T = 0.7003 kip·ft
Maximum shear stress in each segment
Torsional shear stress peaks at the outer surface: τ = Tc/J with c the outer radius and J the polar moment of the (possibly hollow) circular section.
τmax = T·c/J, J = π(do⁴ − di⁴)/32
segment 1: J = 1.571 in^4, τ = 5.35 ksi
= 5.35 ksi
Angle of twist at the free end
Each segment twists by TL/JG; the free-end rotation is the sum over the segments. Positive means counterclockwise viewed from the free end.
φ = Σ(T·L/(J·G))
φ = 0.02233 rad = 0.02233 rad = 1.279 deg
= 1.279 deg
Results
Torque from transmitted power
0.7003kip·ft
Internal torque in segment 1 (ccw +)
0.7003kip·ft
Maximum shear stress in segment 1
5.35ksi
Governing (largest) shear stress
5.35ksi
Total angle of twist at the free end (radians; ccw +)
0.02233rad
Total angle of twist at the free end (degrees; ccw +)
1.279deg
Where this answer was checked
source
T = P/ω, τ = Tc/J, φ = TL/JG (NCEES FE Reference Handbook) — power-transmission torque then stress and twist
verified by
hand-recomputed
derivation
US route: P = 40·550 = 22000 ft·lbf/s; ω = 2π·300/60 = 31.4159 rad/s; T = 700.28 ft·lbf = 8403.4 lbf·in. J = π·2⁴/32 = 1.5708 in⁴. τ = 8403.4·1/1.5708 = 5350 psi = 5.35 ksi. φ = 8403.4·48/(1.5708·11.5e6) = 0.022330 rad = 1.2794°. SI cross-check: P = 29828 W, T = 949.45 N·m, J = 6.53807e-7 m⁴, τ = 949.45·0.0254/6.53807e-7 = 36.886e6 Pa = 5350 psi ✓, φ = 949.45·1.2192/(6.53807e-7·7.92897e10) = 0.022330 rad ✓.