Bending-moment Curve.

Bearing on pins.

The safe shearing value for a 2 inch wrought-iron pin, would be from Table XXXIX= 25000 pounds, and for steel 31670 pounds. By calculation we should have had for iron =25133 pounds and for steel = 31416 pounds.

If we had a 2 inch circular tie-rod its strength in tension from Table XXXIX would be, if of wrought-iron =37600 pounds and if of steel =47000 pounds. By calculation we should have had for iron = 37752 pounds and for steel =47190 pounds.

Pins are calculated similarly to rivets. The same formulae can be used for bearing area (109) h being the thickness of head of eye-bar, in inches, and for single shear (110) and double shear (111). For bending where there are but two eye-bars or rods, each pulling in opposite directions, Formula (112) could be used; using for h the thickness, in inches, of either rod ; where there are two eye-bars or rods pulling in one direction with only one between them pulling in the opposite direction Formula (113) might be used ; h in this case being the thickness, in inches, of the large (central) single rod, and twice the thickness of either of the smaller rods. In all of the formulas x should be= 1.

As a rule, however, there are several rods at each pin-connected joint, and the bending-monicnt has to be carefully calculated for each special case. In designing pin-joints this should be borne in mind, and the pieces with largest strains brought as closely together, as possible, to avoid excessive bending-moments, which will require very large pins, and necessitate large eyes and heads to the rods and bars, which means, of course, very great expense; pin-connections for small trusses are more expensive than riveted joints; for large trusses they are cheaper. They are very much better in both cases, more easy to transport and erect, and agree much nearer with the theoretical calculation, which assumes that all members around a joint are free to move, and not rigid, as is the case in riveted trusses. Then, too, in case of any movement in the truss, it can be readily adjusted by means of nuts, swivel-links, sleeve-nuts, etc. This cannot be done in a riveted truss, as the joints not being flexible, the tightening of any one part might throw strains on some joint not able to bear it.

Shearing of Pins- Tension on

Rods.

Calculation of Pins.

Designing Pin-joint.

Pin-joint best.

Fig. 177.

In calculating a pin-joint, the strength of pin, as a rule, should be calculated in the several lines or directions in which it is being strained, that is first along the line of one rod, then along the line of another rod (or compression piece) and so on. In doing this all the other strains must be projected (and reduced) to the same line. If this is correctly done, it will be found in every case, that the sum of all the strains acting in one direction along this line will be equal to the sum of all acting in the opposite direction along the line. Thus in Figure 177 we have the end elevation of a pin A strained by three forces in tension B, C and D and one force in compression E.

We will first examine the pin along the line A C. Make, at any convenient scale, the lines A C, A B, A D and A E, each in length just equal to the amount of strain they are subjected to. Project them all on to A C, then will A b represent the resultant or equivalent of strain A B along the line and in the direction A. C. Similarly A d will be the resultant of A D along the line of A C, but it will be in the opposite direction; e A will be the resultant of E A along the line A C, but it will be noticed that though on the opposite side of the pin it is in the same direction as A C.

Strains reduced to same line.

Were we to examine the pin along the line A D, we should have the strain of A D in one direction and in the opposite directions the resultant strains A c1 and .4 b1. It will he noticed that E A has no resultant along the line A D, being normal to it; and it need not be considered, when calculating the pin along the line A D. We might still calculate the pin along the line A E; we should then have the strain E A in one direction and in addition thereto the resultant of A C; in the opposite direction we should have the resultant of A B. The largest strain A D has no resultant alone; the line A E. It is evident, however, that all of these resultants will be very small as compared to those along the other lines, and therefore need not be calculated.

In Figure 178 we have a plan of the pin, showing in plan the strains and resultants acting along the line A C. As arranged in this Figure the pin evidently becomes a double-ended cantilever, supported at the fulcrum d, and loaded on one free end with the load c and on the other free end with two loads b and e.

If the strains were small, this would probably be the most economical arrangement, as A D could then be made in one piece. But if the strains were heavy it would require a tremendous pin.

In the latter case the arrangement shown in Figure 179 would be better. Here the pin becomes a beam, supported at both ends by d/2

(or one-half of the force A D) ; the beam carries three loads c, e and b. In this case A D would be made up of two rods, separated. The forces c, e and b must be so distributed as to make the least bending-momenl on the beam or pin, that is, the larger ones placed nearer the supports or outer edges, and the smaller ones in the centre. Frequently it would he more economical to divide c the larger force into two halves and place them immediately next to d/2. The small force e is frequently put on the outside, as it is not sufficient to affect the beam or pin seriously and only enlarges the span of beam, that is, length of pin or distance between supports d/2 thus greatly increasing the bending-moment. This arrangement is shown in Figure 180. Here we have a beam supported at two points