The distance from centre to centre of the rivets can be calculated from the principle that the part of plate in between each pair of holes should be the same strength as one rivet. It may be put in the form of a rule as follows: "The area of plate between a pair of holes multiplied by the tensile strength of the plate material, is equal to the cross sectional area of rivet multiplied by the shearing strength of the rivet material," or d = diameter of hole.

(p - d) t X T = d2 X .7854 X S.

Where - p = pitch of rivets.

t = thickness of plate.

T = tensile strength of plate.

S = shearing strength of rivet.

 For iron T may be taken as 22 tons " " S " " " " 19 " " steel T " " " " 28 " " " S " " " " 23 "

It should be noticed that d represents the diameter of the hole in plate, and this will for punched holes be about 1/32 in. larger than the diameter of rivet for, say, § in. rivets, varying up to 1/16 in. for 1 in. rivets.

For the sake of clearness it will, perhaps, be as well to work out an example in the use of the above formula. Suppose we require to find the pitch of rivets for a single-riveted lap-joint, steel plates and rivets. If the plates are ½ in. thick the diameter of rivet should be 7/8 in. Adding 1/20 in. on to this to allow for clearance, it gives a finished rivet diameter of .9 in. So that we have -

(p - .9) x .5 x 28 = .92 X .7854 x 23, from which we obtain p = 2 in.

The above calculation is based upon the assumption that the holes have been drilled, and in cases where the plates are drilled in position, it will be an advantage to take the clearance as slightly less than that allowed.

For punched work it is important to remember that the operation of punching damages the plate for some small distance all around the walls of the holes. Investigation seems to show that the plate is fractured for a distance of about 1/10 in. from the edge of hole. So that, in using the above rules for punched plates, 1/5 in. must be deducted from the space between the holes before proceeding to use the equation to obtain the pitch. It thus becomes -

(p - d - 1/5)t x T = d2 X .7854 x S.

Suppose we want to find the pitch of rivets for a single-riveted lap-joint formed of iron plates and rivets. Plates | in. thick, punched holes, and rivets 1 in. diameter. Adding 1/16 in. on to rivet diameter for clearance, we have -

(P - 1 1/16 - 1/5 ) x 5/8 x 22 = (1 1/16)2 x .7854 x 19 or - (p - 1.06 - .2) x .625 x 22 = (1.06)2 x .7854 x 19 from which - p = 2.5in.

For boiler work little attention need be given to the construction of joints with punched holes, as all good work is now drilled in position, one or two small tacking holes only being first put in the plates in the flat, the remainder being drilled after the plates are rolled and bolted together. Indeed, with a spacing arrangement attached to a drilling machine, there is no need to mark off the holes with the exception of those needed for tacking. It might be here remarked that after drilling, the plates are separated, and the burr or aris cleaned off, so that the plate surfaces may come into dead contact in riveting.

There is no need to calculate the pitch for every thickness of plate, as the space in between a pair of holes is the same in each case. Thus, for a single-riveted lap-joint formed of iron plates and rivets, with punched holes, the pitch = l 1/2 in. + diameter of rivet, and a similar joint of steel will have a pitch = 1 1/8 in. + diameter of rivet.

The space between the holes in a double and treble-riveted lap-joint will work out to about twice and three times, respectively, that of a single joint as above.

In general work, which has not to be subjected to much pressure, the pitch of rivets is usually taken greater than that shown in the above calculations.