Mechanical Engineering Design Notes

Design Contents


In the diagram above assume that the one end of the bolt and part are welded to the support and the nut is just put on 'finger tight' to remove any play.

The deflection of a bar under axial load:

delta = P L / A E where P = load, L = length, A = cross section area and E = Young's modulus.
Stiffness = k = load/ delta, N/m, rearranging the above and with subscript p meaning part and b meaning bolt: kb = Ab Eb / Lb and kp = Ap Ep / Lp

Pb = part of load P carried in bolt and Pp = part of load carried in part.
P = Pb + Pp

The deflection of the bolt = Pb / kb and of the part = Pp / kp and these must be equal. Pb / kb = Pp / kp

Pb = kb(P - Pb) / kp

Pb kp = P kb - Pb kb

Pb (kb + kp) = P kb

Pb = P kb / (kb + kp) and similarly: Pp = P kp / (kb + kp)

Suppose that the nut is tightened so additional tensile load, Fo is placed in the bolt, then the force in the bolt: Fb = P kb / (kp + kb) + Fo This causes an equal and opposite compressive force in the part, so Fp = P kp / (kp + kb) - Fo
In practice Fo is applied on assembly, before the load P is applied. The initial force Fo must always be made large enough to keep the part in compression, ie Fp must be negative. This is needed to maintain a gasket seal or merely to reduce the effects of cyclic loading on the bolt.

Example calculation

David J Grieve, 10th February 2004.

Contact the Author:
Please contact me for comments and / or corrections or to purchase the book, at: