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.
The deflection of the bolt = Pb / kb and of the
part = Pp / kp and these must be equal.
Pb / kb = Pp / kp
P = Pb + Pp
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.
David J Grieve, 10th February 2004.