OpenSpecEngineering

Finds the peak shear stress and the angle of twist of a round shaft under torque, solid or hollow. For a solid shaft it can also work the other way: from the torque and an allowable shear stress, it gives the smallest diameter. The allowable can be your own number or one from the ASME transmission-shafting code [2]. Machine designers use it to size drive shafts and axles.

Equations

τ = T·r/J     θ = T·L/(G·J)

J = π·d_o⁴/32 (solid)     J = π·(d_o⁴ − d_i⁴)/32 (hollow)

T is the torque, r = d_o/2 the outer radius, J the polar moment of inertia, L the length between the points where the torque is applied, and G the shear modulus [1]. Pick a material to fill in G, or type your own. τ is the shear stress at the outer surface, where it peaks. θ is the twist of one end relative to the other. For the required diameter, the solid-shaft form is solved for d_o:

d_o = (16·T/(π·τ_allow))^1/3

The ASME code allows 8000 psi in commercial steel shafting, or 6000 psi with a keyway. For a listed steel [3] it allows the smaller of 0.30 S_y and 0.18 S_ut, less 25 % with a keyway [2].

Limits and common mistakes

• The equations hold for round shafts in the elastic range only. A square or rectangular bar warps and needs other formulas.
• The result is for torque alone. A shaft that also carries belt pull or gear loads is in bending too and needs a combined check.
• The ASME allowables are static. A shaft under fluctuating or reversing torque needs a fatigue check.
• Shoulders, grooves and keyways raise the local stress. The keyway option applies the code’s 25 % reduction; it is not a stress-concentration factor.

Worked example

A commercial steel line shaft with a keyway must carry 500 N·m. The pulleys are 800 mm apart. Find the smallest diameter the ASME code allows, and the twist between the pulleys.

1. Choose Shaft Type Solid and Solve for Required Diameter (given target Stress).
2. Enter Applied Torque (T) 500 N·m and Shaft Length (L) 800 mm.
3. Choose Material (G) Steel. The shear modulus fills in as 79 GPa.
4. Choose Allowable Shear Stress Basis Commercial steel shafting (ASME code) and Keyway in the Shaft With keyway. Press Solve. The Solution shows: Required Outer Diameter (d_o): 39.4842 mm
   Angle of Twist (θ) at this diameter: 1.2158 deg
   Polar Moment of Inertia (J): 238,612 mm⁴
   Allowable Shear Stress (τ_allow): 41.3685 MPa

Check by hand. 6000 psi is 41.37 MPa, and (16 × 500,000 N·mm / (π × 41.37))^1/3 = 39.5 mm. A 40 mm shaft works at 16 × 500,000 / (π × 40³) = 39.8 MPa, just under the allowable. The twist is 500,000 × 800 / (79,000 × 238,612) = 0.0212 rad, or 1.22°.