CivilSolve

Statics & Mechanics of Materials

Torsion of Circular Shafts

Maximum shear stress and angle of twist for solid or hollow stepped circular shafts fixed at one end, under concentrated torques (τ = Tc/J, J = π(do⁴−di⁴)/32, φ = Σ TL/JG) or under transmitted power at a given rpm (T = P/ω). Use for: 'find the max shear stress in the shaft', 'angle of twist at the free end', 'shaft transmitting power at rpm'. Not for: non-circular sections or shafts fixed at both ends (statically indeterminate).

Direct solving is free and needs no account. Have a word problem instead? Submit it as text.

Inputs

Load a sample problem:
segmentsShaft segments in order starting AT the fixed support: segment 1 touches the wall, the last segment ends at the free end. Nodes are numbered 0 at the support, 1 at the end of segment 1, and so on to the free end.
segments 1
node torquesConcentrated torques at nodes. Give EITHER nodeTorques OR powerTransmission, not both.
node torques 1
power transmissionPower mode: steady power at the given speed, with the resulting torque T = P/ω applied at the free end. Give EITHER this OR nodeTorques.

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

Given

segments
  1. 1.length 1.5 mouter diameter 50 mmshear modulus 80 GPa
node torques
  1. 1.node 1torque 1 kN*msense ccw

Assumptions

  • 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

  1. 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

  2. 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

  3. 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
verified by
hand-recomputed
derivation
J = π·0.05⁴/32 = 6.13592e-7 m⁴. τ = 1000·0.025/6.13592e-7 = 40.744e6 Pa = 40.74 MPa. φ = 1000·1.5/(6.13592e-7·80e9) = 1500/49087.4 = 0.030558 rad = 1.7509°.

Example 2

SI stepped shaft: 60 mm then 40 mm, torques at both nodes

Given

segments
  1. 1.length 1 mouter diameter 60 mmshear modulus 80 GPa
  2. 2.length 0.8 mouter diameter 40 mmshear modulus 80 GPa
node torques
  1. 1.node 1torque 2 kN*msense ccw
  2. 2.node 2torque 1 kN*msense ccw

Assumptions

  • 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

  1. 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

  2. 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 = 1272000 mm^4, τ = 70.7 MPa; segment 2: J = 251300 mm^4, τ = 79.6 MPa

    = 79.6 MPa

  3. 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 = 0.02947 rad + 0.03979 rad = 0.06926 rad = 3.968 deg

    = 3.968 deg

Results

Internal torque in segment 1 (ccw +)

3000N·m

Maximum shear stress in segment 1

70.7MPa

Internal torque in segment 2 (ccw +)

1000N·m

Maximum shear stress in segment 2

79.6MPa

Governing (largest) shear stress

79.6MPa

Total angle of twist at the free end (radians; ccw +)

0.06926rad

Total angle of twist at the free end (degrees; ccw +)

3.968deg

Where this answer was checked
source
Method of sections + φ = Σ TL/JG (NCEES FE Reference Handbook) — two solid segments, internal torque changes at node 1
verified by
hand-recomputed
derivation
2 kN·m ccw at node 1, 1 kN·m ccw at node 2. Segment 2 carries 1 kN·m; segment 1 carries 3 kN·m. J1 = π·0.06⁴/32 = 1.272345e-6; τ1 = 3000·0.03/1.272345e-6 = 70.735e6 = 70.74 MPa. J2 = π·0.04⁴/32 = 2.513274e-7; τ2 = 1000·0.02/2.513274e-7 = 79.577e6 = 79.58 MPa. φ = 3000·1/(1.272345e-6·80e9) + 1000·0.8/(2.513274e-7·80e9) = 0.029473 + 0.039789 = 0.069262 rad = 3.9684°.

Example 3

US power mode: 40 hp at 300 rpm, solid 2 in shaft, L=4 ft, G=11.5e6 psi

Given

segments
  1. 1.length 4 ftouter diameter 2 inshear modulus 11500 ksi
power transmission
power 40 hprpm 300

Assumptions

  • 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

  1. 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

  2. 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

  3. 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

  4. 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 ✓.