Shear Stress Calculator: Bolts, Pins, Punching and Shaft Torsion

Shear stress, shear strain and torsion calculator

Work out the shear stress in a bolt or pin in single or double shear, a glued or welded lap joint, or the force to punch a hole; the shear strain and shear modulus; and, on the Torsion tab, the shear stress and angle of twist of a solid or hollow shaft from a torque or from the power and speed of a motor. Each tab gives a safety factor against the material's shear strength and finds the pin or shaft size you need for the safety factor you want.

Direct shear

• Shear stress is the force across a plane divided by its area: \(\tau = V / A\). For a round pin \(A = \pi d^2 / 4\).
• Single shear: two plates overlap and the bolt could be cut at one plane. Double shear: a clevis holds the middle plate between two outer ones, the pin is cut at two planes and the stress is halved, \(\tau = V / (2A)\).
• Bearing stress is the pin pressing on the side of its hole, \(\sigma_b = V / (d\,t)\). Thin plates often fail this way first.
• Shear strain is the change of angle of a small square, \(\gamma = \tau / G\), with the shear modulus \(G = E / (2(1+\nu))\) for a material that is the same in every direction.
• Shear strength: a ductile metal yields in shear at about \(0.577\,\sigma_y\) (the von Mises rule, 1/√3); the ultimate shear strength used for punching is about 0.75 times the tensile strength for steel (you can change the factor).

Torsion of shafts

A torque T twists a shaft; the shear stress grows straight out from the middle and is highest at the surface: \(\tau_{max} = T c / J\) with \(c = D/2\) and the polar moment of area \(J = \pi D^4 / 32\) (solid) or \(J = \pi (D^4 - d^4) / 32\) (hollow). The angle of twist is \(\theta = T L / (G J)\) in radians. A motor's torque comes from its power and speed: \(T = P / \omega\) with \(\omega = 2\pi n / 60\) for n in rpm.

Worked examples

M10 bolt in single shear, 10 kN. A = π × 10² ÷ 4 = 78.54 mm², so τ = 10,000 ÷ 78.54 = 127.3 MPa. (Use the bolt's shank diameter only if the threads are not in the shear plane.)

16 mm clevis pin in double shear, 30 kN. Two planes of 201.1 mm² each: τ = 30,000 ÷ (2 × 201.1) = 74.6 MPa. In A36 steel (shear yield ≈ 0.577 × 250 = 144 MPa) the safety factor is about 1.9.

Punching a 20 mm hole in 3 mm A36 plate. The sheared area is the wall of the hole, π × 20 × 3 = 188.5 mm². With a shear strength of 0.75 × 400 = 300 MPa the punch needs 56.5 kN, about 5.8 tonnes-force.

50 mm solid steel shaft, 1 kN·m. J = π × 50⁴ ÷ 32 = 613,592 mm⁴, so τmax = 1,000,000 N·mm × 25 mm ÷ 613,592 mm⁴ = 40.7 MPa. Over 1 m with G = 79.3 GPa it twists 0.0206 rad, 1.18°.

A 7.5 kW motor at 1450 rpm turns at ω = 151.8 rad/s, so it gives T = 7,500 ÷ 151.8 = 49.4 N·m.

Questions

Single or double shear: which is stronger?

Double shear. The same pin carries twice the force at the same stress, because the load is shared by two cross-sections. It also keeps the pin from bending, which is why clevis joints are used for heavy loads.

Is the shear stress the same across the whole pin?

No: V/A is the average. In a beam the shear stress peaks in the middle of the section (4/3 of the average for a round bar). Pin and bolt design uses the average with a safety factor, as here.

Why is a hollow shaft lighter for the same strength?

The middle of a solid shaft carries almost no stress. Taking it away removes a lot of weight but little of J. A tube with an inside diameter of 80% of the outside is 64% lighter but has only 41% less J (so about 69% more stress at the same torque and outside diameter); make it a little bigger and it is both lighter and stronger than the solid shaft.

What is the shear modulus of steel?

About 77 to 80 GPa (11,200 to 11,600 ksi). From E = 200 GPa and ν = 0.3 the formula gives 76.9 GPa.