How to use this calculator
- Pick the bolt grade. Choose the grade (SAE 5/8 or ISO 8.8/10.9/12.9); this sets the ultimate tensile strength Su.
- Enter the diameter and pitch. Enter the nominal diameter d and thread pitch P (the pitch is only used when the shear plane is in the threads).
- Choose the shear plane. Select whether the shear plane passes through the threads (tensile stress area) or the shank.
- Set the shear planes. Enter 1 for single shear or 2 for double shear, then read the ultimate shear capacity.
How it works
A bolt loaded across its axis fails in shear over its cross-section. The ultimate shear capacity is V = n · A · τult, where n is the number of shear planes, A is the shear area, and the shear strength is taken as τult = 0.6 · Su — about 60% of the ultimate tensile strength (the von Mises value is 0.577).
The area depends on where the shear plane crosses the bolt. Through the threads, use the tensile stress area At = (π/4)·(d − 0.9382·P)²; through the unthreaded shank, use the full Ashank = (π/4)·d². Because MPa × mm² = N, the result comes straight out in newtons. This is the failure load — apply a safety factor to get an allowable load.
Worked example
Verified against the live calculator
An M10×1.5 class 8.8 bolt in single shear, with the shear plane in the threads:
the tensile stress area is
At = (π/4)·(10 − 0.9382 × 1.5)² = (π/4)·8.5927² ≈ 57.99 mm²
(the standard At for M10 is 58.0 mm²). With Su = 830 MPa the shear
strength is τ = 0.6 × 830 = 498 MPa, so
V = 1 × 57.99 × 498 ≈ 28,880 N — about
28.9 kN at failure. Put the shank in the shear plane instead and
A rises to 78.54 mm² (≈ 39.1 kN); make it double shear and the capacity doubles.
The calculator returns exactly these numbers.
Frequently asked questions
How do you calculate the shear strength of a bolt?
Take the shear area A at the failure plane, multiply by the shear strength of the material (about 0.6 × the ultimate tensile strength Su), and by the number of shear planes n: V = n · A · 0.6·Su. For an M10×1.5 class 8.8 bolt sheared through the threads, A = 58 mm², 0.6·830 ≈ 498 MPa, so V ≈ 28.9 kN in single shear.
Should I use the thread area or the shank area?
Use whichever the shear plane actually crosses. If the joint faces line up so the cut passes through the threaded part, use the tensile stress area At = (π/4)(d − 0.9382·P)² — the smaller, conservative area. If the unthreaded shank spans the joint, use the full shank area (π/4)·d², which is larger, so the bolt is stronger.
What is single shear vs double shear?
Single shear cuts the bolt on one plane (a simple lap joint) — one shear area carries the load. Double shear cuts it on two planes (a clevis or double-lap joint), so two shear areas share the load and the capacity doubles. Set the number of shear planes to 1 or 2 above.
Why is shear strength 0.6 times the tensile strength?
It is a common engineering approximation: a ductile metal shears at roughly 60% of its ultimate tensile strength. The theoretical von Mises (distortion-energy) ratio is 0.577, and design codes use values in that range. We use 0.6 as the everyday estimate.
Is this the allowable load I can apply?
No — this is the ULTIMATE (failure) shear capacity. Divide it by a safety factor to get an allowable working load. The factor depends on your code, loading and consequences of failure; bolted connections are often designed with a factor of 2 or more, and proof/preload also matter for slip-critical joints.
What ultimate strengths do the grades use?
SAE Grade 5 = 827 MPa, SAE Grade 8 = 1034 MPa, ISO class 8.8 = 830 MPa, 10.9 = 1040 MPa, 12.9 = 1220 MPa. The first ISO digit roughly gives the tensile strength in 100 MPa, the second the yield-to-tensile ratio.
Does this work in metric and imperial?
Yes — enter the diameter and pitch in mm or inches; the shear area shows in mm² or in², and the capacity in N, kN, tonne or lbf. Toggle SI/Imperial in the header.
Method & assumptions
- Reports the ultimate (failure) shear capacity — V = n · A · 0.6·Su. Divide by a safety factor for an allowable working load.
- Shear strength is the common 0.6 × ultimate-tensile approximation; the von Mises (distortion-energy) ratio is 0.577. Codes use values in this range.
- Threads-in-plane uses the tensile stress area At = (π/4)(d − 0.9382·P)²; shank-in-plane uses the nominal shank area (π/4)·d².
- Direct shear only — no bending, prying, bearing on the plate, thread tear-out or combined tension+shear interaction, and no fatigue.
- Friction-grip (slip-critical) joints are governed by preload and slip, not bolt shear; this sizes the bolt against gross shear failure.