Example

First find component stiffnesses to relate the bolt load to the external load via ( 3a).

Bolt

The threaded length is not given, but from the sketch the runout coincides with the joint faces - though this exposed thread length is really excessive. Take half head and nut lengths to be 0.5 x 0.8 x 16 = 7 mm.
Shank length   = 20+7 = 27 mm         area = π/4 x 16² = 201 mm²
Thread length = 24+7 = 31 mm         area = A_s = 157 mm²   ( Table 1 )
1/k_b = Σ L/AE = ( 27/201 + 31/157 )/207         k_b = 624 kN/mm . . . watch units !

Flange

Assume lengths of effective conical frusta are each (20+24)/2 = 22 mm to equalise reactive areas as shown, and that ( 4) is applicable to cast iron with little error. So, for one frustum
k_f = 100 x 16 ( 0.702 + 0.654 x 16/22) / ( 1 - 0.12 x 16/22) = 2065 kN/mm
So, for the two flange frusta in series
1/k_j = 2 ( 1/2065 )         k_f = 1030 kN/mm

The proof load of the bolts is   F_p = A_s S_p = 157 x 590 = 93 kN
Assuming that the bolts are tightened initially to three-quarters of their proof load,   F_i = 0.75 x 93 = 70 kN
So applying ( 3a) with a design factor of 2   on the external load, to 'z' bolts presumably equally loaded :
        F_b = 70 + 2 x 200 x 389 / 1030 z     and   ≤ F_p = 93 kN     from which   z ≥ 6.3

Either 8 bolts would be used (7 are difficult to make equidistant) or 6 bolts might be employed with preload reduced to 73% rather than 75% of proof. But we have seen that it is impossible to expect 2% accuracy - so select 6 bolts. This decision is bolstered by the knowledge that since the joint is self-energising, a high preload is unnecessary from a joint operational point of view.

The example has demonstrated how to apply the theory, however the problem refers to the 'cylinder' and 'head' of a pump so presumably the pump is reciprocating and the loading on this joint will be alternating, therefore a fatigue rather than a static analysis would have been much more appropriate. We consider bolt fatigue in the next section.