Bolt and Nut Tightening Loads

Bolt forces can be applied to bolt and nut geometries in SimSolid.

Bolts and Nuts

In SimSolid, bolts are automatically identified by their geometric attributes. Bolts are required to have cylindrical bodies and a head with a hexahedral based shape. The hex shape can be on an outer or inner diameter in the bolt head. Nuts are identified in a similar manner using this hex based geometric signature.

In SimSolid, tightening loads can be applied to a variety of geometries, including the following:

Blind bolts
Bolts with nuts
Nuts on threaded rods
Nuts on a generic post or handle

Relationship Between Torque M and Axial Force F

M is the maximum moment realized at the end of the tightening and it is equilibrated by moment from friction forces in contact between nut and the structure.

Assume for simplicity that normal forces in contact are distributed evenly, so the contact pressure is as follows:(1)

P = \frac{F}{C o n t a c t A r e a}

(2)

P = \frac{F}{π (R 1^{2} - R 0^{2})}

R0 and R1 are inner and outer radii of the contact spot. Friction distributed force will be $T = f * P$ where f is a friction coefficient.

In a polar coordinate system, the elementary moment of the friction force with respect to the bolt axis is:

d M = T * r^{2} * d R * d T e t

Where r is the distance to axis while dR and dTet are radius and angle differentials respectively.

Integrate the elementary moment over the contact area to obtain the following:

M = \frac{2 * F * f * (R 1^{3} - R 0^{3})}{3 * (R 1^{2} - R 0^{2})}

This equation relates applied torque, M, and axial force.

Axial Force

Axial force depends on the structure and bolt stiffness, and on nut placement relative to the bolt:(3)

F = K * D

K is structure stiffness factor, and D is relative displacement.

Relative displacement can be expressed by the following:

D = N * H

Here, N is number of nut turns and H is thread pitch. Therefore,

F = K * H * N

(equation A)

Assume that at first analysis pass one nut turn is described (N(1)=1), and corresponded axial force F(1) is found from the analysis. The structure stiffness factor in this case can be defined as the following:(4)

F (1) = K * H * 1

This implies: $F = F (1) * N$ .

Now you can relate torque to the number of turns:(5)

M = \frac{2 * N * F (1) * f * (R 1^{3} - R 0^{3})}{3 * (R 1^{2} - R 0^{2})}

Therefore, in order to realize prescribed torque M, after the first analysis is done with N=1, a second analysis (second convergence pass) must be performed using the following equation: (6)

N (2) = M / |\frac{2 * N * F (1) * f * (R 1^{3} - R 0^{3})}{3 * (R 1^{2} - R 0^{2})}|

In general, at pass (i+1) the number of turns applied is as follows: (7)

N (i + 1) = M / |\frac{2 * N (i) * F (i) * f * (R 1^{3} - R 0^{3})}{3 * (R 1^{2} - R 0^{2})}|

Here, N(i) is the number of turns applied at previous passes, and F (i) is result axial force evaluated at previous pass. These corrections for number of turns applied are important because in the course of passes solution is refined, which changes structure stiffness factor K in equation A above. So, K is not constant, but depends on pass K(i)