Arrangement of Heads.

Rivets Riveting And Pins 20026

Fig. 178.

Rivets Riveting And Pins 20027

Fig. 179.

-with a free end carrying load e; the beam being loaded with three loads, two each =c/2 and one = b.

To calculate the bending-moment at any point of this pin we have to consider the end of pin farthest from e as built in solidly (and after getting reactions) we should multiply all the forces to the right or left of the point into their distances from the point. If we select the forces on the right we deduct the sum of the moments of those (right hand forces) acting in one direction from the sum of the moments of those (right hand forces) acting in the opposite direction, the difference will be the bending-moment at the point. To check the calculation we take all the forces on the left side of the point. But the reactions will not be d/2 each as shown in the table unless e were divided and onehalf put at each end of pin.

In every case of pin calculation, excepting where the forces form simple beams, as in Figure 179, the forces are supposed to act through their central axes, that is through a line drawn at half the thickness of their heads and at right angles to the pin. The heads are frequently thickened up, that is, made thicker than the rod or flat bar, in order to get the necessary bearing surface on the pin, but it will be readily seen that this lengthens the pin greatly, thus largely increasing the bending-moments. It is better, as a rule, to get the extra bearing surface by dividing the rod or flat bar into two or more parts and then distributing them along the pin symmetrically and at the most favorable points to avoid bendingmoment.

Pin with Un-symmetrical Strains.

Forces act through Central Axes.

Rivets Riveting And Pins 20028

Fig 180.

If the pin is arranged as shown in Figure 180 and we call the different lengths along the pin, as shown, l, l, and l„ the reaction nearest e will be

Forces arranged Un-symmetrically.

= c+b/2+l/l1.e and the further reaction will be

= c+b/2 - 1l11/l1.e

In practice, however, e would probably be so small and the convenience so great in making the two bars - alike, that the unequal stresses caused by e in the two would probably be overlooked.

But in calculating the bending-moment on pin, they will have to be considered as unequal, other wise, the bending-moments calculated from the right hand or left hand of any point would not agree.

To put the above in formulae we should have the following :

If a pin p q is strained by a number of forces a;,, x11, etc., in one direction, the forces p and q in the opposite direction will equal the reactions of a beam, see FormulŠ (14, 15, 16 and 17). If a single force y is placed beyond one of these resisting forces (say q Figure 181) the additional reaction on the nearer force (7) will be

Nearer Reaction Force.

Nearer Reaction Force.

p = + l/l1.y

(116) and on the further force

Further Reaction Force.

P1=l11 /l1 .y

(117) or, we must add q, to q and substract p, from p to get the real strains or reactions in the tie-roils p and q.

Whereql= the strain, in pounds to be added to nearer strain q owing to force y being placed on the (q) nearer free end of pin.

Where p1 = the strain, in pounds, to be deducted from further strain p owing to force y being placed on the other further free end of pin.

Where q = the force (reaction) at q, in pounds, resisting the forces x„ x11, etc., (see Formulae 14, 15, 16 and 17) and to which q1 must be added.

Where p = the force (reaction) at p, in pounds, resisting the forces x1, x11, etc., (see Formulae 14, 15, 16 and 17) and from which p1 must be deducted.

Where x1 x11 = the forces, in pounds, acting in opposite direction top and q, and all projected to line p and q as shown in Figure 177.

Where l, l„ l1, = the respective lengths, or distances, in inches, measured along pin, from centre lines to centre lines of respective pins, as shown in Figure 181.

As the forces x1,x11, etc., should always, if possible, be located along pin to make the resisting forces p and q even, which is done by putting the smallest force (x111) in the middle and dividing the others, we should have, where this is the case:

Nearer Reaction Force.

q = ∑.x/2 + l/l1.y

(118)

Further Reaction Force.

p = ∑.x/2 - l11/l1.y

(119)

Where 2 x = the sum of all the opposing forces (x1, x11, x111 etc.,) to and between p and q, in pounds, provided that they are divided and the halves or fractions located symmetrically with respect to p and q.

Wherep = the total resisting force (reaction), in pounds, or strain on p, the reaction furthest from free end.

Where q = the total resisting force (reaction), in pounds, or strain on q, the reaction next to free end.

Where l,l1,l1, and y = same as in FormulŠ (116 and 117); of course, all the forces must be projected to one line as shown in Figure 177.

If one of the forces, x1, x11, etc., were = y and were placed to the left of p (Figure 181) the forces p and q would be equal and each = one - half of the sum of all the other opposing forces, provided always that the forces x1, x11, etc., are symmetrically located in respect to p and q. Where there are a number of forces on both sides of pin, the pin might be treated as a continuous girder (see Table XVII, Volume I), but the calculation would become very difficult and the parts of all rods (or compression piece) acting in the same lines and direction would become very irregular.