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

Free shear stress calculator: shear stress and strain in bolts and pins in single or double shear, glued joints and punching, shear modulus from E and Poisson's ratio, and torsion of solid and hollow shafts with the angle of twist and motor power. Safety factors and sizes.

Material

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True stress–strain Shade areas Plateau serrations

Speed

Yield point or smooth yielding?

Mild and structural steels (A36, 1020) have a yield point. Carbon and nitrogen atoms pin the dislocations, so the stress climbs a little past the yield strength until a band of metal breaks free: that is the upper yield point. The force then drops to the lower yield point and the bar stretches at nearly constant stress while Lüders bands spread along it (the Lüders strain, about 1 to 3% for A36), with small serrations as each band starts. Only then does strain hardening begin, up to the tensile strength (58 to 80 ksi, 400 to 550 MPa for A36, at about 15 to 20% strain) and necking. The upper yield is usually only a few % to about 10% above the lower one, and its height depends on the test: a stiff machine, a fast pull and a smooth, well-aligned specimen give a taller, sharper peak; a soft machine or a slightly bent bar can hide it. The line also bends a little just before the peak (some grains yield early), so the top is rounded rather than a sharp corner. Codes therefore use the lower yield (or 0.2% offset) as the yield strength: 250 MPa (36 ksi) for A36.

Aluminium alloys (2014-T6, 6061-T6), copper, stainless steel and most other metals yield continuously: no peak, no drop, no plateau. The curve just bends away from the straight line, so the yield strength is defined by the 0.2% offset rule: draw a line with slope E starting at 0.2% strain, and where it meets the curve is the yield strength (414 MPa, 60 ksi, for 2014-T6).

Engineering stress\(\displaystyle \sigma = \frac{F}{A_0},\)\(\displaystyle \varepsilon = \frac{L - L_0}{L_0}\)

True stress\(\displaystyle \sigma_t = \frac{F}{A} = \sigma\,(1+\varepsilon),\)\(\displaystyle \varepsilon_t = \ln(1+\varepsilon)\)

Hardening\(\displaystyle \sigma_t = K\,\varepsilon_p^{\,n}\)

Necking starts\(\displaystyle \frac{d\sigma_t}{d\varepsilon_t} = \sigma_t \;\Rightarrow\; \varepsilon_t \approx n\)

Resilience\(\displaystyle U_r = \frac{\sigma_y^2}{2E}\)

Toughness\(\displaystyle U_t = \int_0^{\varepsilon_f} \sigma\,d\varepsilon\)

The curves are a model fitted to each material's typical Young's modulus, yield strength, tensile strength and elongation (Hollomon / Ramberg–Osgood hardening, a Lüders plateau for mild steel). Specimen: a standard round bar, 12.5 mm across with a 50 mm gauge length (0.500 in and 2.000 in in US units). Real curves also depend on the test speed, temperature and the exact alloy.

Supports

Loads drag them along the beam; drag a handle to change the size

Cross-section

Deflection limit

Bending stress (flexure formula)\(\displaystyle \sigma = \frac{M\,c}{I} = \frac{M}{S},\)\(\displaystyle S = \frac{I}{c}\)

Shear stress\(\displaystyle \tau = \frac{V\,Q}{I\,t}\)

Deflection curve\(\displaystyle E I\,\frac{d^2 y}{dx^2} = M(x),\)\(\displaystyle \frac{dM}{dx} = V\)

Rectangle\(\displaystyle I = \frac{b\,h^3}{12},\)\(\displaystyle S = \frac{b\,h^2}{6}\)

Simply supported, load P at the middle\(\displaystyle M_{max} = \frac{P L}{4},\)\(\displaystyle \delta = \frac{P L^3}{48 E I}\)

Simply supported, spread load w\(\displaystyle M_{max} = \frac{w L^2}{8},\)\(\displaystyle \delta = \frac{5 w L^4}{384 E I}\)

Cantilever, load P at the end\(\displaystyle M_{max} = P L,\)\(\displaystyle \delta = \frac{P L^3}{3 E I}\)

Fixed both ends, spread load w\(\displaystyle M_{ends} = \frac{w L^2}{12},\)\(\displaystyle \delta = \frac{w L^4}{384 E I}\)

The beam is solved exactly for any mix of loads (singularity functions: equilibrium plus the deflection conditions at the supports), so the fixed and propped cases, which statics alone cannot solve, come out right too. It assumes a straight, elastic beam with small deflections (Euler–Bernoulli), bending about the strong axis, no buckling sideways (long, slender I-beams without bracing can twist out of the way: lateral-torsional buckling) and no self-weight unless you add it as a spread load.

Words used here

Bending moment M: How much the loads try to bend the beam at a point: force × distance, summed over everything on one side. Sagging (smiling, top squeezed) is drawn upward here; hogging (frowning, top stretched, over supports) downward.
Shear force V: The force trying to slide one part of the beam past the next. It jumps at every point load and support, and the slope of the moment diagram equals it, so M is largest where V crosses zero.
Neutral axis: The line through the section that is neither stretched nor squeezed. For these symmetric sections it sits at mid-depth; the stress grows in a straight line away from it.
Second moment of area I: How the area is spread away from the neutral axis (m⁴, in⁴). Depth counts cubed, which is why I-beams put their metal in the flanges and why a plank is far stiffer on edge.
c and section modulus S = I/c: c is the distance from the neutral axis to the outermost fibre, where the stress is largest. S rolls I and c into one number: σmax = M/S.
Flexure formula σ = Mc/I: The largest bending stress in the section, valid while the material stays elastic.
Q and τ = VQ/(It): Q is the first moment of the area above the neutral axis and t the width there. The shear stress is largest at the neutral axis; in an I-beam the thin web carries nearly all of it.
Deflection δ and L/360: How far the beam sags. Codes limit it as a fraction of the span (L/360 for floors under live load, L/240 or L/180 for roofs) so floors do not feel bouncy and plaster does not crack.
Simply supported, cantilever, fixed, propped, overhang: A pin and a roller at the ends; built into a wall at one end only; built in at both ends; built in at one end and resting on a support at the other; resting on two supports with part of the beam sticking out past one of them.
Point load, spread load (UDL), moment: A force at one spot (kN, kip); a force spread evenly along a length (kN/m, kip/ft), such as a floor's weight; a twisting couple applied at one spot (kN·m, kip·ft).
EI (flexural rigidity): Young's modulus times I: how hard the beam is to bend. Deflection is inversely proportional to it.

For learning and first estimates. Real parts need certified material values, the design code that applies (AISC, Eurocode, ASME…) and a check by a qualified engineer.

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.

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

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