Of course, the rivets must be of steel too.

We will again assume that we can stagger the rivets, so that we shall lose only one rivet-hole. The plate will evidently have to be thick and we will decide to use 1" rivets ; this would leave us a net breadth of plate b = 10" -1" = 9" From Formula (114) we have : h = 135000/9. 15000 = 1" Or the plate will have to be just one inch thick. The cover-plate should be at least the same, if there were only one rivet, but as there will evidently be more than one and we propose staggering the rivets, the cover-plate will have to be considerably thicker ; we can therefore leave the cover-plate out of consideration for the present as, being thicker, or in case of one rivet equal to the plates to be joined, it will certainly be as strong, and not crush.

Butt joint single cover-plate.

Size of plate.

Now, as for the number of rivets, from Formula (109) we have for bearing : x = 135000/1.1.15000 = 900 Or nine rivets are required not to crush the plate or be crushed by it (each side of joint).

From Formula (110) we have for single shearing (as there is evidently only one area to each rivet to resist shearing) : x = 135000/0,7857.12.10000 = 17.2 Or seventeen rivets are required to resist the shearing (each side of joint). The rivets will evidently be levers in this case and we have from Formula (112) x = 135000/0,1964.13.18000 = 38,1

Or it will require thirty-eight rivets to resist the bending-moment (each side of joint). This again shows the advantage of using both a top and bottom cover-plate and so avoiding the great leverage on the rivets.

It must now be borne in mind that the joint is a butt joint, therefore, unlike the case of lap joint where the whole number of rivets bear on each plate, we must here use twice the number, as only one-half, or those each side of the joint bear on one plate, or we require in all 76 rivets.

The plate is so narrow that we cannot get more than three rivets on a line across the plate, without infringing on the rules for pitch. We can, therefore, stagger the end rivets and make the rest either chain riveted or zig-zag riveted. The chain riveting will require a little longer cover-plate, but in the case of a plate-girder flange would have the advantage of using the rivets that are needed for the angle-irons.

In that case we should not stagger the end rivets, for our plate would only be weakened by one additional hole, and, of course, the gross breadth of plate would have to be 12 inches instead of 10 inches to give same strength.

Required number of rivets.

Designing the joint.

Figure 173 shows this joint chain riveted, with end rivets, staggered, to correspond to our calculated example.

We see that it takes 39 rivets each side of joint for symmetry.

Figure 174 shows this joint zig-zag riveted. It has three advantages over the other, it is shorter, takes just the right number of only 7 inches and from Formula (114) would require a cover-plate of thickness h = 135000/7.15000 = 1,3 or say 1 5/16" thick.

Fig. 173.

rivets and requires a thinner cover-plate. For in Figure 173 at the first line of rivets next to the joint (C), the cover-plate loses three rivet-holes and bears the full strain, or its clear breadth would be

Whereas, in Figure 174, next to the line C we lose only two rivets and have consequently a char breadth of 8 inches and require from Formula (114) a cover-plate of thickness equal to h = 135000/8.15000 = 1,125 or only 1 1/8 inch cover-plate.

Had we not used the formula we should have Figured out our rivets, etc, as follows :

Required net or clear area of plate

135000/15000= 9 square inches. Net breadth being 9" gives, of course, 1" thickness.

Bearing area of each rivet = 1.1 = 1 square inch, which will safely bear 15000 pounds or we should need

135000/15000 = 9 rivers

Single shearing area of each rivet (or area of a circle 1" diameter)

= 0,7854 square inches, which at 10000 pounds per square inch, would give a resistance to shearing per rivet = 7854 pounds or we should need

135000/7854=17,2 rivets. For bending we should nave a one inch circular lever, projecting one inch and uniformly loaded with

135000 pounds.

The bending-moment would be Formula (25) m = 135000.1/2 = 67500 pounds.

The moment or resistance would be Table I, section No. 7 r = 11/14 (1/2)3 = 0,0982 The strain s, therefore, on all the rivets will be [page 49, Volume

I] s = 678500/0,0982 = 68737

This divided by the safe modulus of rupture for steel rivets ( k/f) = 18000 will give the required number, as before,

687373/18000 = 38,2

Example III.

Butt-joint Two Cover-plates.

Same problem as before, but two cover-plates to be used, one above and one below the joint.

We will again decide to stagger the rivets and losing only one rivet-hole will again require a 1" thick plate.

Now, for bearing we will have the same result as before, viz.: 9 rivets required, but in shearing it is evident that now each rivet has two resisting areas or is in double shear, so that we will need only one-half of the previous quantity or17.2/2 =8,6 or say, nine rivets. Had we used

Formula (111) we should have obtained this result, for : x = 135000/1,5714. 12. 10000 = 8,6

For the bending-moment we use Formula (113) and have x = 135000.1/0,7857. 13.18000 = 9,5 or say ten rivets required to resist cross-breaking. This shows how much better proportioned the different strains (shearing, bending and bearing) are to each other, where we use two cover-plates. Had we calculated directly without use of formula we should have obtained the same results. We have already calculated net area of plate and bearing value of rivets in Example II, also the single shearing value, which was 7854 pounds per rivet; as we now have two areas this would be doubled or the resistance of each rivet, to shearing would be 15708 pounds and the number required 135000/15708= 8,6 as before.