Thread Engagement Length & Stripping Strength Calculator
Minimum thread engagement so the bolt breaks before the threads strip, using the FED-STD-H28 / Machinery’s Handbook shear areas of the internal and external threads — metric and UNC/UNF sizes, 6g/6H or 2A/2B limits, ISO and SAE bolt grades, 13 tapped materials or your own, the strength ratio J, stripping loads by depth and a check of the depth you have.
How it works
Bolt load = At × Sut (metric At = π/4 (d − 0.938194 P)², inch At = π/4 (D − 0.974279 P)²) · bolt-thread shear area As = π Le Kn max [½ + 0.57735 (Es min − Kn max) ÷ P] · hole-thread shear area An = π Le Ds min [½ + 0.57735 (Ds min − En max) ÷ P] · FED-STD-H28: Le = 2 At ÷ (As per unit length), J = As × Sut,bolt ÷ (An × Sut,material), hole length = J × Le when J > 1 · minimum engagement = max of the two × safety factor
A bolted joint should fail in the bolt — a visible, predictable event — never by stripping the threads in the part, which is hidden and hard to repair. Each thread strips by shearing along a cylinder. The bolt’s thread shears at the minor diameter of the hole (Kn), the hole’s thread at the bolt’s major diameter (Ds), and the width of thread being sheared at each is set by the pitch-diameter limits (Es for the bolt, En for the hole). FED-STD-H28/2B and Machinery’s Handbook turn this into the shear areas used here, taken at the weakest legal thread sizes: the smallest bolt (6g or 2A) in the largest hole (6H or 2B). The calculator compares the bolt’s tensile breaking load with those areas times each material’s shear strength. Under the H28 rule that is half the tensile strength, and you can switch to typical shear strengths instead. It reports the length each thread needs and which one governs, plus the strength ratio J: when it is above 1 the tapped material is the weak side and needs J times the bolt-thread length. Pick a standard metric or UNC/UNF size, any ISO or SAE bolt grade and 13 tapped materials, or enter your own figures and limit diameters; under More options you can also check the depth you actually have. Because the first few threads carry a disproportionate share of the load, engagement beyond about 2.5 diameters buys little; thin parts get inserts or a weaker bolt class instead.
Worked example
A class 8.8 M10 × 1.5 (stress area 58.0 mm², 46.4 kN) into 6061-T6 aluminium. At 6g/6H minimum material the hole’s thread has 21.6 mm² of shear area per mm of engagement and the bolt’s 15.5 mm². Under the H28 rule (aluminium 155 MPa, bolt 400 MPa) the hole thread needs 13.9 mm to match the bolt and the bolt thread 7.5 mm, so J = 1.86 and the hole governs. With a 1.25 factor, tap at least 17.3 mm, about 1.7D. Into mild steel the same bolt needs 10.8 mm (13.4 mm with the factor).
Sources: FED-STD-H28/2B, Screw-Thread Standards for Federal Services (Unified inch threads), Appendix A2 — length of engagement · Machinery’s Handbook — “Length of Engagement” and thread shear areas · Fastenal Engineering — Screw Thread Design (internal thread shear area) · ISO 965-1 tolerance formulas — ISO general-purpose metric screw threads
Frequently asked questions
Should engagement be measured from the top of the chamfer?
Count only full threads: subtract the chamfer/countersink and, for a blind hole, the incomplete threads at the bottom (2–3 pitches for a plug tap).
Why does a stronger bolt need deeper threads?
The hole has to hold the bolt’s full break load; a 12.9 bolt in aluminium needs about 50% more engagement than an 8.8 — or you accept that the bolt is the stronger part and the thread strips.
What are Es, En, Ds and Kn?
They are the limit diameters FED-STD-H28 uses: Es min is the smallest bolt pitch diameter, Ds min the smallest bolt major diameter, En max the largest hole pitch diameter and Kn max the largest hole minor diameter. Using the smallest bolt in the largest hole gives the least thread to shear, so the answer is safe for any parts within tolerance.
What is the J factor?
J = (As × bolt tensile strength) ÷ (An × tapped-material tensile strength). If J is above 1 the internal thread is the weaker side, and Machinery’s Handbook multiplies the bolt-thread engagement length by J to get the length the hole needs.
Why is the result different from the old “0.5 × circumference” rule?
That shortcut treats half of the cylinder at the mating diameter as sheared. The H28 areas follow the real thread geometry and tolerances, which gives roughly 0.55–0.85 of the cylinder depending on the size and the fit, so most sizes need a little less length than the shortcut suggested.
When should I use a threaded insert?
When the part is too thin for the calculated length, in soft materials such as aluminium, magnesium or plastic, or when the bolt will be removed often. A steel insert moves the shear to its larger outside diameter and wears far better than the parent material.
Related calculators
Shop reference values; adjust for tool maker recommendations, machine rigidity, coolant and the actual material condition, and verify fits and tolerances against the applicable standard (ISO 286 / ASME B4.1, ISO 965 / ASME B1.1).