Pin & Bolt Joint Calculator
Verify a transversely loaded pin or bolt joint: from the transverse force, pin diameter, arrangement and plate thicknesses the tool derives shear, bearing pressure at fork and rod, and pin bending, each with a safety factor against the yield strength, live with every input.
Pin/Bolt calculator (shear, bearing, bending)
Model: static check of a transversely loaded pin/bolt joint against the yield strength (shear τ = F/(n·A), bearing p = F/(d·t), bending σ_b = M_b/W). Clearance-free, uniform support assumed. No fatigue and no friction-grip check. Sizing tool for mechanical engineering.
Results
Key values
- Shear area A = π·d²/4
- 314.16 mm²
- Number of shear planes n
- 2
- Shear stress τ
- 15.92 N/mm²
- Fork bearing p_G
- 50 N/mm² (governs)
- Rod bearing p_S
- 25 N/mm²
- Section modulus W = π·d³/32
- 785.4 mm³
- Bending moment M_b
- 37.5 Nm
- Bending stress σ_b
- 47.75 N/mm²
Joint sketch
Formulas and fundamentals
The joint transfers a transverse force F perpendicular to the pin axis. The shear check compares the mean shear stress τ = F/(n·A) with the shear yield limit. Here A = π·d²/4 is the pin cross-section and n is the number of shear planes: single shear n = 1, double shear (fork and rod) n = 2. Double shear halves the shear stress and is therefore the preferred arrangement. The limit is the shear yield stress from the distortion energy hypothesis τ_lim = R_e/√3 ≈ 0.577·R_e; the available safety factor is S = τ_lim/τ.
The bearing pressure check (contact / hole bearing) verifies that the pin does not press unacceptably into the hole wall. Related to the projected area d·t, p = F/(d·t) per plate; the pressure is computed separately at the fork (effective thickness t_G) and the rod (thickness t_S). The governing value is the higher pressure, i.e. the smaller effective thickness. The limit is the onset of yielding at R_e, so S = R_e/p. For the double-shear fork, t_G is the sum of both cheeks so that p = F/(d·t_G) carries the full force.
The pin is additionally loaded in bending because fork and rod introduce the force at offset positions. Using the beam model (fork as support, centrally loading rod) the maximum bending moment is M_b = F/8·(t_S + t_G). With the section modulus of the circular section W = π·d³/32 the bending stress is σ_b = M_b/W, verified against the yield strength (S = R_e/σ_b). For thick plates and a short pin the bearing pressure usually governs, for a slender pin the bending does.
Worked example
Given: A double-shear fork-and-rod joint transfers a transverse force F = 10,000 N. The pin has d = 20 mm (material S355, R_e = 355 N/mm²), the effective fork thickness is t_G = 10 mm, the rod t_S = 20 mm. Required safety factor S_erf = 1.5.
Shear: A = π·20²/4 = 314.2 mm², with n = 2 gives τ = 10,000/(2·314.2) = 15.9 N/mm². Against τ_lim = 355/√3 = 205 N/mm² the safety factor is very high. Bearing pressure: at the fork p = 10,000/(20·10) = 50 N/mm², at the rod p = 10,000/(20·20) = 25 N/mm² – the fork governs, S = 355/50 = 7.1.
Bending: M_b = 10,000/8·(20 + 10) = 37,500 Nmm, W = π·20³/32 = 785.4 mm³, so σ_b = 37,500/785.4 = 47.7 N/mm² and S = 355/47.7 = 7.4. All three checks are well above S_erf = 1.5, the joint is adequately sized; the fork bearing pressure governs.
Frequently asked questions
What do single shear and double shear mean?
The number refers to the shear planes: in single shear two plates with one interface lie against each other and the pin is sheared at one location (n = 1). In double shear a fork straddles the central rod, giving two shear planes (n = 2). The double-shear arrangement halves the shear stress and avoids the eccentric bending of the single-shear interface.
Which check usually governs?
For typical dimensions shear is rarely critical; instead the bearing pressure at the thinnest plate or the bending of a slender pin governs. The calculator reports all three separately and marks the governing value, so the smallest safety factor is immediately visible.
Why is the bearing pressure computed separately for fork and rod?
Because the effective thicknesses differ. Related to the full force, the smaller thickness gives the higher pressure. The thinnest effective plate is therefore the weakest link of the bearing contact and determines the safety factor.
How is the shear limit derived?
As the shear yield stress from the distortion energy hypothesis (von Mises): τ_lim = R_e/√3 ≈ 0.577·R_e. This is the shear stress at which a ductile material starts to yield. The available safety factor is the ratio of this limit to the acting shear stress.
Does the calculator cover fatigue and fit?
No. The check is static against the yield strength and assumes a clearance-free, uniform support. Under cyclic load a fatigue check is required; with clearance or eccentric assembly bending and local pressure increase. Friction grip from preload is not considered.