How to calculate bolt shear strength

A bolt in shear fails when the load across its cross-section exceeds the metal's shear strength. The calculation is one line, but three details decide the answer: which area is in the shear plane, how many planes there are, and that shear strength is only about 60% of tensile strength.

The formula

V = n · A · τ · τ ≈ 0.6 · Su

V is the ultimate shear capacity, n the number of shear planes, A the bolt area in the plane, and τ the shear strength — taken as 0.6·Su from the ultimate tensile strength (the von Mises value is 0.577). The bolt shear strength calculator runs this for any grade, size and plane.

Which area: threads or shank

If the shear plane cuts through the threads, use the tensile stress area A = (π/4)(d − 0.9382·P)²; if it crosses the unthreaded shank, use the full A = (π/4)·d². The shank is larger, so a joint detailed to keep the shank in the faying surface is stronger. When in doubt, assume threads in the plane — it is the conservative case.

Worked example — M12 class 8.8

M12 coarse (pitch 1.75 mm), class 8.8 so Su = 830 MPa, single shear, threads in the plane:

A = (π/4)(12 − 0.9382×1.75)² = 84.3 mm² · τ = 0.6×830 = 498 MPa · V = 84.3 × 498 ≈ 42.0 kN

Put a second shear plane through it (double shear — bolt cut in two places) and the capacity doubles to about 84 kN. Remember this is the ultimate load; divide by a safety factor of 2–5 for the allowable working shear.

Bolt shear strength table (single shear, threads in plane)

Ultimate shear capacity V = A·0.6·Su, computed with the formula above:

Metric (coarse)	8.8	10.9	12.9
M8	18.2 kN	22.8 kN	26.8 kN
M10	28.9 kN	36.2 kN	42.4 kN
M12	42.0 kN	52.6 kN	61.7 kN
M16	78.0 kN	97.8 kN	114.7 kN
M20	121.9 kN	152.8 kN	179.2 kN

Inch (UNC)	SAE Grade 5	SAE Grade 8
1/4-20	2.33 kip	2.91 kip
3/8-16	5.66 kip	7.07 kip
1/2-13	10.35 kip	12.94 kip
5/8-11	16.46 kip	20.59 kip
3/4-10	24.34 kip	30.43 kip

These are ultimate single-shear values for threads in the plane; shank in the plane runs higher, double shear doubles them, and the allowable working load is after a safety factor. Generate any size, grade or plane on the calculator.

Where it connects

Shear is one failure mode; a bolted joint also has to survive tension and the preload that clamps it — see the bolt preload & torque calculator — and the threads themselves must not strip, which is the thread engagement guide. For the bearing and tear-out at the hole, check the plate, not just the bolt.

Common mistakes

• Using tensile strength as shear strength. Shear is ~0.6 of tensile — using Su directly overstates capacity by about 65%.
• Forgetting double shear. A bolt through three plates (two faying surfaces) has two shear planes and twice the capacity — but only if both planes are loaded.
• Quoting ultimate as allowable. The formula is failure load; apply the code's safety factor (often 2–5) for the working load.
• Assuming shank when threads are in the plane. Short grips and thin plates usually put threads across the shear plane — use the smaller tensile stress area unless the joint is detailed otherwise.

Frequently asked questions

How do you calculate the shear strength of a bolt?

Shear capacity = shear area × shear strength × number of shear planes: V = n·A·(0.6·Su). The shear strength is taken as about 0.6 of the ultimate tensile strength Su, and the area is the tensile stress area when threads cross the shear plane, or the full shank area when they do not. An M12 class 8.8 bolt (At = 84.3 mm², Su = 830 MPa) carries 84.3 × 0.6 × 830 ≈ 42 kN in single shear.

What is the shear strength of a bolt as a fraction of tensile strength?

About 60%. The simple engineering rule is shear strength ≈ 0.6 × ultimate tensile (Su); the von Mises distortion-energy value is 0.577. So a 830 MPa class-8.8 bolt has roughly 498 MPa of shear strength. Design codes use their own calibrated factors, but 0.6·Su is the standard textbook figure.

Should the threads be in the shear plane?

It depends on how the joint is built, and it matters: threads-in-plane uses the smaller tensile stress area, shank-in-plane uses the larger full-diameter area, so a shanked joint is stronger. The conservative assumption — threads in the shear plane — is the usual default unless the grip is detailed to keep the shank across the faying surface.

How much does a safety factor reduce the allowable shear load?

The formula gives the ULTIMATE (failure) capacity. The allowable working load is that divided by a safety factor — commonly 2 to 5 depending on the code, load type and consequences. The 42 kN ultimate on an M12 8.8 becomes about 14–21 kN allowable at a factor of 2–3.