Thus, if we take the pin and forces shown in Figure 189, we should change the notation to that adopted for the graphical method, which would be as shown in Figure 193. That is force E (22000 pounds) of Figure 189 would be known as A B in Figure 193 ; again force C1 (32000 pounds) of Figure 189 would be called force E F (not F E) in Figure 193.

Proceeding now to the calculation, we lay off along the vertical load line a e, the following forces :

Craphical method more simple.

Example solved graphically.

 a b = A B = 22000 pounds. bc = B C = 22000 pounds. c d = CD = 22000 pounds. d e = DE = 22000 pounds. and in the opposite direction, we lay off: ef = E F = 32000 pounds. fg = FG = 24000 pounds. ga = G A = 32000 pounds.

which will, of course, bring us back to the starting point a, as the opposing forces must aggregate the same sum.

We now select our pole o. This we remember can be arbitrarily located, or else at a distance o c = (k/f); in our case we will make the distance, say, 12000 pounds.

The distance (of pole from load line) being arbitrary we shall multiply the verticals v (inch-scale) in Figure 193, by this pole distance (pounds-scale) to obtain the bending-moments at the points of pin immediately below verticals.

If the pole distance from load line had been made = (k/f) then the length of verticals v measured at inch-scale would have been the required moments of resistance of the corresponding points of pin below verticals ; and each respective v multiplied by (k/f) would be the bending-moment at each point. We will, however, make in our case the pole distance, arbitrary, viz: c o = 12000 pounds. We now begin at any point of line A B and draw A I parallel b o till it intersects B C at I ; next draw I L parallel c o to intersection with C D at L ; and similarly draw L E parallel do; EK parallel oe; K J parallel of; J H parallel o g and H A parallel o a ; the last line must intersect the first at point of starting A or some error has been made.

Pole distance.

It will be noticed that the individual outlines of A I L E KJHA cover the capital letters in Figure 193, corresponding to small letters from which their respective parallel lines started in strain diagram. Thus, for instance A I covers letter B and is parallel to bo; similarly I L covers C and is parallel to c o ; L E covers D and is parallel do; K J covers F and is parallel of; JH covers G and is parallel o g ; similarly we can consider E K as covering E and it is parallel o e ; and HA as covering A and it is parallel o a.

We now measure the verticals through the Figure A IL E KJH A, the longest, of course, will give the greatest bending-moment. This happens to be the central one J J, it measures 1 9/16", therefore mJ = 1 9/16.12000 = 18750 which corresponds to mB of Figure 189. Similarly we should have :

Strain diagram.

Curve of bend-ing-moments.

 ml (formerly mEl) = 127/32.12000 = 10125 pounds-inch. m11 (formerly mc) = 1 17/32.12000 = 18563 pounds-inch.

Had we analyzed the strains on pin as shown in Figure 192 graphically, our verticals would have measured, at E:

 VE = 3/16" at C: Vc = 1 13/16" at E : VE1 = 31/32" and at B : VB = 1 3/8"

The corresponding bending-nioments would have been : at E:

 mE = 3/16. 12000 = 2250 pounds-mch. at C: mc = 1 13/16.12000 = 21750 pounds-inch. at E1 : mE1: = 31/32.12000 = 11625 pounds-inch. atB: mB = 1 3/8.12000 = 16500 pounds-inch.

which are very close to the correct moments, which we found arithmetically to be :

 mE = 2250 pounds-inch ; mc = 21637 pounds-inch; mE = 11617 pounds-inch; and mE = 16648 pounds-inch.

The simplest method of calculating pins, as a rule, will be - after calculating (or ascertaining from Tables) the safe bearing and shearing stresses of the pin, - to calculate the actual moment of resistance of the pin, see Table I, Section No. 7, fourth column. Now proceed graphically, being sure to make the pole distance in every case equal to the safe modulus of rupture (k/f) of the material of pin.

After this it will only be necessary to see that none of the verticals through the different polygonal Figures (corresponding to A IL E K JH A of Figure 193) - that none of these verticals measure at inch-scale more then the actual moment of resistance of the pin. If this is done the calculation and selection of the best arrangement becomes very simple and easy. After the final and best arrangement has been determined on, it would be well to calculate arithmetically the moments as per this final arrangement, thus checking the graphical solution.

The writer has frequently been told by contractors that owing to the friction due to the pressure of the nut and head, that it was impossible for any bending-moment to take place on a pin. As well it might be claimed that owing to the pressure of the walls, there is no bending-moment on a built-in beam. The safe modulus of rupture of a built-in beam can be assumed higher same as we do for pins, but the beam will break across if too heavily strained and so will the pin. Besides it must be remembered that the heads of eye-bars necessarily cannot fit the pins perfectly; and even if the argument were correct, (which it is not,) that friction offsets the bending-momont, the least rusting of the joints would diminish the pressure between nut and head, thus destroying its value.

Simplest method use curve of moments of resistance.

Contractors claim of no bending-moment ridiculous.

Again, contractors will admit that there exist bending-moments on pins, but will deny it in the case of rivets, though the cases are precisely analogous; this latter argument though, if sifted, will generally lead the contractor to admit that its real basis is the large number of rivets it frequently requires; and the less rivets he can get along with, the happier will your contractor be.