Thread Engagement Calculator: Minimum Engagement Length, Thread Shear Area and Stripping Strength (FED-STD-H28, ISO 965-1)

Method

Thread limits. The shear areas are taken at minimum material, the weakest combination the tolerances allow. For metric threads the 6H/6g limits come from the ISO 965-1 grade 6 tolerances and the g fundamental deviation (M12 × 1.75: d min 11.701, d2 min 10.679, D1 max 10.441, D2 max 11.063 mm). The "basic" option uses basic diameters with no tolerance.

Inch thread limits. For unified inch threads the 2A/2B limits come from the ASME B1.1 formulas, with D the basic major diameter and P the pitch. The pitch diameter tolerance is 0.0015 D^(1/3) + 0.0015 √D + 0.015 P^(2/3), and the allowance is 0.3 × that tolerance. The major diameter tolerance is 0.060 P^(2/3), and the 2B pitch tolerance is 1.3 × the 2A value. The 2B minor tolerance is 0.25 P − 0.4 P² (0.05 P^(2/3) + 0.03 P/D − 0.002 below 1/4 in).

Shear areas. The formulas come from FED-STD-H28/2B, Appendix B, Table 11.B.1, and are also printed in Machinery's Handbook. Here n = 1/P, L_E is the engagement length, E_n,max is the maximum pitch diameter of the internal thread and E_s,min the minimum pitch diameter of the external thread. The internal thread shears on a cylinder at the minimum major diameter of the bolt, D_s,min: A_Sn = π n L_E D_s,min [1/(2n) + 0.57735 (D_s,min − E_n,max)]. The external thread shears on a cylinder at the maximum minor diameter of the tapped hole, K_n,max: A_Ss = π n L_E K_n,max [1/(2n) + 0.57735 (E_s,min − K_n,max)].

Loads. The bolt breaks at F_b = Rm × As, with Rm the minimum tensile strength of the grade (ISO 898-1, SAE J429, ASTM) and As the tensile stress area (ISO 898-1, ASME B1.1). Each thread strips at its shear strength times its shear area. The shear strength is τ = k × Rm, where k is the shear to tensile strength ratio, so F_n = k Rm,nut A_Sn and F_s = k Rm,bolt A_Ss. Rm,nut and Rm,bolt are the tensile strengths of the tapped part and the bolt.

Minimum engagement. Both stripping loads grow in proportion to L_E, so the minimum length is L_min = margin × F_b / (k Rm A_S per unit length). It is worked out for each thread, and the larger governs. With a margin of 1 the bolt and the governing thread fail at the same load. That is the FED-STD-H28/2B design aim: the screw breaks rather than strips. The ratio J = (A_Ss τ_s) / (A_Sn τ_n) of Machinery's Handbook shows which thread is weaker. Above 1 the tapped part governs.

Helical coil insert. The tapped hole is an STI thread, the standard thread with every basic diameter larger by 2 × 0.649519 P (ASME B18.29.1 tapped-hole limits follow this). The parent material then shears at D_s = d + 1.299 P with E_n,max = d2 + 1.299 P + the class pitch tolerance, a larger area than the standard thread. The bolt thread still strips against the standard-class internal thread the insert forms.

Assumptions

• Full-form threads over the whole engagement length. Chamfers, countersinks and the incomplete threads at the end of a tapped hole do not count. FED-STD-H28/2B suggests counting about half the countersink depth.
• The formulas treat the load as shared evenly along the engagement. It is not: the first engaged threads carry most of it. The shear-area method is still the accepted design approach because it was calibrated against stripping tests.
• Shear strength τ = k × Rm, default k = 0.5. FED-STD-H28/2B sizes the shear area at twice the tensile stress area, which is the same as k = 0.5. The 0.6 preset, typical for ductile metals, is less conservative and gives a shorter engagement.
• Minimum-material limits for 6H/6g or 2A/2B. Inch limits come from the ASME B1.1 formulas and differ from the rounded tables by up to about 0.0002 in.
• Aluminium and brass strengths are typical values (ASM). Minimum values from a material specification are lower. Enter them as a custom material for design.
• Solid parts: no nut dilation, thin walls or hollow studs, which need a longer engagement (FED-STD-H28/2B 70.7.8). When both parts have similar strength they can strip together at a lower load; check critical joints by test.
• With a helical insert, only the parent material and the bolt thread are checked. The insert itself and its installation are not.

Standards and references

Worked example: M12 × 1.75 class 8.8 screw in 6061-T6, 18 mm engagement

An M12 × 1.75 class 8.8 screw (Rm 800 MPa) goes into a tapped hole in 6061-T6 aluminium (Rm 310 MPa typical), 6H/6g, with 18 mm (1.5 d) of full thread. Shear/tensile ratio 0.5 (FED-STD-H28/2B), margin 1.0, no insert. This is the default case loaded in the calculator.

Frequently asked questions

What is the minimum thread engagement for a steel bolt in aluminium?

For an M12 class 8.8 screw in 6061-T6 with typical strengths and the FED-STD-H28/2B shear ratio of 0.5, the tapped thread matches the bolt strength at 16.66 mm, about 1.39 d. The less conservative ratio of 0.6 gives 13.88 mm, 1.16 d. The common 2 × d rule for aluminium gives 24 mm, which covers lower minimum strengths, the uneven load sharing and repeated assembly.

Why not just use 1 × d in steel and 2 × d in aluminium?

Rules of thumb ignore the bolt grade, the pitch and the tolerance class. A class 12.9 screw in mild steel needs far more than 1 × d, and a fine pitch changes the shear areas. The calculator shows the rule-of-thumb length next to the calculated minimum so you can see how much margin the rule gives in your case.

Which diameters do the thread shear area formulas use?

Minimum material. The internal thread shears at the smallest major diameter the bolt may have, with the largest pitch diameter of the hole. The external thread shears at the largest minor diameter the hole may have, with the smallest pitch diameter of the bolt. That is FED-STD-H28/2B section 70.2, and it gives the weakest fit the tolerances allow.

How deep should a tapped hole be?

At least the required full-thread engagement plus the incomplete threads the tap leaves at the bottom, plus any countersink (FED-STD-H28/2B counts only about half the countersink depth). The drilled hole goes further still. Typical tap chamfers leave about 2 incomplete threads with a bottoming tap and 3 to 5 with a plug tap.

Does a helical coil insert make the thread stronger?

In the parent material, yes. The tapped STI thread is larger by 2 × 0.649 P on every diameter, so the parent shears on a bigger cylinder and the required length drops. Tick the insert option to see it. The bolt thread strength does not change, and the insert must be long enough on its own. Standard lengths are 1 to 3 × d.

What shear strength should I use for the threads?

The calculator uses τ = k × Rm with k = 0.5 by default: FED-STD-H28/2B sizes the shear area at twice the tensile stress area, which is the same as k = 0.5. The 0.6 preset, typical for ductile steels and aluminium, is less conservative and gives a shorter engagement. Grey cast iron is stronger in shear than in tension, so either preset is conservative for it.

Do all engaged threads carry the same load?

No. Elastic stretch of the bolt and compression of the tapped part put most of the load on the first few engaged threads. The shear-area method still assumes an even share, because its factors come from stripping tests to failure, where yielding spreads the load. For fatigue or critical joints, test.

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