For Wrought-iron.

( c/f) = 12000 pounds, per square inch. (t/f ) = 12000 pounds, per square inch. ( g/f ) = 8000 pounds, per square inch. (k/f) = 15000 pounds, per square inch.

For Steel.

(c/f) = 5000 pounds, per square inch. ( t/f) = 15000 pounds, per square inch. (g/f)= 10000 pounds, per square inch. (k/f)= 18000 pounds, per square inch.

Example I.

Lap Joint.

A wrought-iron plate, which cannot be over 12" wide, is required to be so long that it has to be made up from two lengths; the joint is to be a lap joint. The plate is in tension and is strained 65000 pounds. Design the joint.

We will assume that we propose to design the joint as shown in Figure 165, with staggered rivets, in that case the plate will only be weakened by one rivetdiole. We can readily see that the plate will not need to be very thick and decide to use 3/4" rivets, (that is 3/4" rivet-holes)1; we then shall have a net breadth of plate b = 12" - 3/4" = 11 1/4.

Of course s = 65000 in this case, and we know that (t/f) = 12000; inserting these values in Formula (114) we have:

Size of plate.

1 Before heating, in this case, the rivets would he 11/16".

h = 6500/11 1/4.12000 = 0.485"

Or we should use a plate one-half inch thick.

We next determine the number of rivets required.

In the first place there must be enough for bearing, that is not to crush the plate or get crushed by it. We use Formula (109) inserting the values, and have:

Required numberof Rivetsx = 65000/3/4.1/2.12000 = 14,44

Or we should need 15 rivets for bearing. Had we Figured without using the formula, we should have said, the bearing area of each rivet is 3/4" by 1/2" = 3/8 square inches, this at 12000 pounds, per square inch, would equal 4500 pounds safe-compression for each rivet, dividing this into 65000 pounds, the strain, would, of course, give the same result 14,44 or say 15 rivets.

We next see if there is any danger from shearing. The joint being a lap joint, the rivets will have, of course, only one sectional area to resist shearing, that is, will be in single shear, so that we use Formula (110) and inserting values have: x = 65000/0,7857. (3/4)2.(g/f) = 18,39

Or we must use 19 rivets to prevent the shearing.

Had we Figured without the use of formula, we should have said, area of a 3/4" rivet is = 0,4417 square inches. This multiplied by 8000 pounds (the safe shearing stress per square inch) = 3533,6 pounds, or each rivet could safely assume this amount of the strain without shearing. This amount being less than the safe compression on each rivet, would, of course, require a larger number of rivets, and should therefore be used, rather than the latter. We have, in effect 65000/3533,6 = 18,39 or say 19 rivets, being four more than required for bearing.

We next take up the question of bending; the joint being lapped the rivets will practically become short levers. We use Formula and from Table I, section No 7 the moment of resistance, r = 11/14. (3/8)3 = 0,04143 From the formula on page 49, Volume I, m/r = s we have the total extreme fibre strains on all the rivets, s= 16250/0,04143 =392228 pounds.

(112); inserting values, we have: x = 65000.1/2/0,1964.(3/4)3. = 26,15

Or we should have to use at least 26 rivets to prevent bending; which readily illustrates the great disadvantage of not transferring the strain in a direct plane, by using two cover-plates.

Had we not used the formula, we should have said, we have here a 3/4" circular lever, the free end projecting 1/2" and loaded uniformly with a load of 65000 pounds.

From Formula (25) or Table VII we have the bending-moment m =65000.1/2/2 = 16250 pounds-inch.

This divided by the safe strain, or safe modulus of rupture (k/f) = 15000 pounds, will give the number of rivets required, viz:

392228/15000 = 26,15 or 26 rivets as before.

We still have to decide the distance y (or A B, Figure 171). We use Formula (115). As we have more than one rivet we use in place of s the strain on each rivet or -s/26, which was the largest number required as above, therefore : s/26 = 65000/26 = 2500

Inserting this in Formula (115) we have: y = 2500/2. 1/2. 8000 = 0.3125"

This, however, being less than our rule which requires 11 /2 diameters from centre of hole to edge, we will stick to the rule. We now design the joint.

We have a plate 12" wide, 1/2" thick, lapping, and require 26 rivets.

They must be arranged not to weaken the plate by more than one rivet-hole.

If we arrange the rivet-holes as shown in Figure 172, we will find it the most economical arrangement. To be sure it allows for only

Desigmng the ioint

25 rivets, but that will probably be near enough, otherwise we should have to insert at least five more to keep them symmetrical. It will be readily seen that the weakest point is at section A where one rivet-hole is lost.

Section B is of same strength, two rivet-holes being lost, but the strain has been reduced by an amount equal to the value of one rivet-hole.

At section C we lose three rivet-holes, but the strain has been re-duced by the value of three rivet-holes, so that the plate practically has its full value here.

At sections D and E the plate is stronger to resist the remaining tension than required.

By Figuring out E (12" wide) it will be seen that the pitch on this line is more than required by the rule, Formula (108).

The pitch F G between two adjacent lines of rivets, measured on the slant from centre to centre of rivets, should be at least 2 1/2 diameters, 2 1/2. 3/4=1 7/8" or say 2".

It will be good practice for the student to carefully lay this joint out to scale.

Fig. 172.

Example II.

A steel plate 10" wide has to be pieced, and for local reasons this can only be done by a cover-plate on one side. The plate is subjected to a tensional strain of 135000 pounds. Design the joint.