It is customary, therefore, to locate the forces along the pin symmetrically, regardless of their true resistances as they would be if treated as continuous girders, and to consider, that each takes its proportionate load (according to the thickness of its head) of the whole load along its respective line and direction. In each case it will require special study to obtain the most economical arrangement.

An example will more fully illustrate the method of calculating pins.

Locate Forces Symmetrically.

Example IV.

The joint A of a pin-connected wrought-iron truss is strained by the following members: The tie-rod B = - 28000 pounds: the tie-rod C = - 70000 pounds; the strut D = +20000 pounds; and the tie-rod E = - 88000 pounds. All as shown in Figure 182; design the joint.

We will assume that for certain reasons we wish to use a 2 3/4" diameter pin. Now the first thing to do is to settle the thickness of the (eye) heads. These, of course, must have sufficient bearing against pins not to crush the pin or be crushed by it. We use therefore the following formula :

Calculation of Pin-joint.

t = s/d.(c/f)

(120)

 Where t = the necessary thickness of eye-bar head, in inches. Where s = the strain on each eye-bar, in pounds. Where d = the diameter of pin, in inches.

Where (c/f)= the safe-compression stress, per square inch, of the material of eye-bar or pin (whichever is the weaker in resisting crushing should used.)

Accordingly we should have for thickness of the different eye-bar heads in our example the following :

AB = 28000/2 3/4.12000 = 0.85 or say 3/4"

AC = 7000/2 3/4.12000= 2.12 or say 2"

DA = 20000/2 3/4.12000 = 0,6 or say 5/8"

AE=88000/2 3/4.12000 = 2,67 or say 2 3/4"

The values for above thicknesses have been rather too broadly rounded off, but this is done to simplify the subsequent calculations.

Fig. 182.

Had we used Table XXXVI we should have had the same results. For A B (28000 pounds) the horizontal line 2 3/4" and vertical line 28000 (from above for iron) meet between the heavy bearing lines 13/16" and 7/8", for convenience, however, we will call it 3/4" though this should not, of course, be done in a real calculation.

For A C (70000 pounds) horizontal line 2 3/4" and vertical line 70000 from above (the second to the right of 68000) meet just beyond the 2" heavy bearing line. It should be, therefore, a little over 2" thick, but we will call it even 2".

Thickness by Table XXXVI.

For D A (20000 pounds) the horizontal line 2 3/4" and vertical line 20000 from above, meet between heavy bearing lines 9/16" and 5/8", we will call it 5/8".

To find A E (88000 pounds) which is larger than A C, we shall evidently have to divide it in halves, and, of course, double the result. We find that horizontal line 2 3/4" and vertical line 44000 from above, meet between the heavy bearing lines 1 5/16" and 1 3/8", or we will say 1 3/8"; this doubled or 2 3/4" is the required thickness therefore, for head A E.

Of course, if we use more than one bar for either of the strains we will divide the required thickness of head accordingly. Thus, if we decide to use two bars along the line A C, each would be strained also pulling to the right; also the projection of D A = 11000 pounds pushing to the right ; the projection of A E= 73500 pounds pulling to the left, thus opposing the other three.1

=70000/2 = 35000 pounds, and the thickness of head required for each would be only = 2.12/2 = 1,06 or say 1".

Now to arrange the different heads along the pin, we first lay off along each line (Figure 182) the amount of strain (measuring all at the same scale) and project these strains on to the different lines. We measure these projections and have along the line A B the strain A B= 28000 pounds pulling to the right; the projection of A C= 34500 pounds are shown in Figure 185; it will be noticed that in the latter case

Fig 183.

If our measurements are right the sum of the forces acting in one direction along line A B, must equal the sum of their opponents, or, A B + A C + D A = A E and we have in effect,

1These Figures have been rounded off to simplify the calculations.

28000 + 34500 + 11000 = 73500

The strains on the pin along line A B are shown in Figure 183. Those along line A C are shown in Figure 184 and those along A E

D A becomes = O. In Figure 186 are shown the strains along D A.

in this case A E becomes = 0.

As the largest strain in one direction (88000 pounds) is along line A E, we will select Figure 185 to design the joint and when we have settled the arrangement of heads along the pin to suit these strains, we will see how it affects the pin according to the strains along the other lines.

The simplest arrangement of the parts would be evidently that shown in Figure 185. We will first consider the shearing. The largest shearing strain will be between C and E and will be = 64000 pounds. From Table XL we find for wrought-iron, single shear (at 8000 pounds per square inch) that we should need a 3 3/16" diameter pin to resist 64000 pounds single shear (as each of the vertical spaces at the top evidently represent 10000 pounds), we must pass down vertically not quite half-way between the second and third fines to the right of 50000, till we strike the single shear iron curve and then pass along the horizontal line to the left to find diameter of pin, which is between 3 1/8" and 3 1/4" or say 3 3/16". Had we calculated arithmetically we should have had, area of cross-section required from Formula (7) transposed a =640000/8000= 8 square inches.