Example.—Let the plate thickness be as in the last example 9⁄16, decimal equivalent = .5625, and the rivet diameter be 7⁄8 inch = decimal equivalent .875, the area of one rivet being .6013 square inch, and the pitch is calculated as follows:
| .6013 | = area of one rivet. | |||
| 2 | = the rivets in one pitch. | |||
| Plate thickness = .5625 ) | 1.2026 | ( 2.1377 | ||
| 1.1250 | ||||
| 776 | 0 | 2.137 | ||
| 562 | 5 | .875 | = rivet diameter. | |
| 213 | 50 | 3.012 | = pitch. | |
| 168 | 75 | |||
| 43 | 750 | |||
| 39 | 375 | |||
| 4 | 3750 | |||
| 3 | 9375 | |||
| 4375 | ||||
We find, therefore, that the pitch is 3.012, or 3 inches (which is near enough for practical purposes), and we may now make it clear that this is correct.
Fig. 3257.
In [Fig. 3257] the joint is shown drawn one-half full size, and the length a of plate left between the rivet holes measures (as nearly as it is necessary to measure it) 25⁄32 inches, or 2.156, and if we multiply this by the thickness of the plate = .5625 inch, we get 1.2 square inches as the area of the plate left between the rivet holes.
Now there are two rivets in a pitch (as one-half of b, one-half of c, and the whole of f), and as the area of each rivet is .6, therefore the area of the two will be 1.2, and the plate section and rivet section are shown to be equal.
The area at a is obviously the same as that at a, because the pitches of both rows of rivets are equal, this being an ordinary zigzag riveted joint.
We may now consider the diagonal pitch of the rivets, using the rule below.