Bending Moment in Pins.

The only difficult part of the process of calculating the sizes of pins will generally be found in determining the bending moment. In cases where the strains all act in the same plane, the bending moment can generally be determined by multiplying the outside force by the distance from its centre to the centre of the next bar, as in the foregoing example. When, however, the forces act in several planes, as is generally the case, the process of determining the bending moment is more difficult, and can be best determined by a graphic process, first published by Prof. Chase Green, and included in his lectures to the students in engineering at the University of Michigan.

As the pieces acting on any well-designed joint are symmetrically arranged, it is unnecessary to consider more than one-half of their number. Fig. 4 shows a sketch of one-half the members of a joint in the lower chord of a Howe truss. The pieces are parallel to the plane of the paper, and the pin is perpendicular to the same, but drawn in cabinet perspective, at an angle of 45° with a horizontal.

The bars are assumed to be each one inch thick, and the channel to have one-half-inch web. The centre of the hanger is " from the centre of the channel.

The method of obtaining the bending moment is as follows:

Draw the line A B at an angle of 45° with a horizontal, and, commencing with e, lay off the distances between the centres of the bars to a scale (14" or 3" to the foot will be found most convenient); then draw the lines 1-2, 2-3, etc., parallel to the pieces which they represent in the truss, to a scale of pounds. Resolve the oblique forces into their horizontal and vertical components (in this example there is but one oblique force).

Next draw the stress diagram (Fig. 6) as follows: On a horizontal line lay off 1-2 equal to the first or outer force; 2-3, equal to the next, 3-4; and 4-1, being the horizontal component of the brace, closes the figure. In the same way, lay off the vertical forces 15, 56, 61. If the forces are correct, the sum of the forces acting in one direction will always equal those acting in the opposite direction. From 1 draw the line 10 at 45°, equal to the same scale of, say, 20,000 pounds, or any other convenient length. Draw 0 2, 03, 0 4, etc. Then, in Fig. 5, starting at the first horizontal force, draw c d parallel to 0 2, de parallel to 03, ef parallel to 0 4, and ƒ k parallel to 0 1.

In the same way, starting at the first vertical force, draw r s parallel to 0 5, s t parallel to 0 6, and t 2 parallel to 01. Then th

only necessary to draw the diagonal from the ends of the ordinates at that point. Thus, the resultant at X, Fig. 5, will be m-n, and it is evident that this is the longest hypothenuse which can be

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FIG. 5.

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line c d e fk will represent the boundary of the horizontal ordinates, and rs to the boundary of the vertical ordinates. And to find the resultant of these ordinates at any point on the pin, it is

drawn; and this hypothenuse, multiplied by 0-1 (20,000 pounds), gives 52,500 pounds as the maximum bending moment on the pin. To obtain the maximum bending moment, it is necessary to take the longest hypothenuse that can be drawn, no matter at what place it occurs.

If one desires to try the effect of changing the order of the bars on the pin, it can readily be done. Suppose the diagonal tie to change places with the next chord bar. The horizontal stress diagram then becomes 1-2, 2-3, 3-4, 4-1. The equilibrium polygons

will now be (Fig. 7) c d e f'k' and r' s't' w, and the longest hypothenuse, wx, or 33", which makes the bending moment 75,000 pounds, showing that the arrangement in Fig. 4 is the best.

As a rule, in arranging the bars on a pin, those forces which counteract each other should be close together.

To further illustrate this method of determining the bending moment on pins, we will determine the bending moment for the pin at the joint 4, Fig. 8. This is the same truss as worked out on page 535, the strains given in Fig. 8 being of the strains at the joint, as all the pieces are doubled. Fig. 9 shows the size and arrangement of the ties and strut. It is assumed that the web of

parallel to 0 2, b c In the same way,

the channel is reënforced to make it " thick. Drawing the line AB, Fig. 11, we lay off the outer force at a; then measuring off an inch the distance between centres of the two outer bars, we lay off the next force parallel to the direction in which it acts; and in the same way, the other two forces. The three inclined forces must be resolved into their horizontal and vertical components. We next draw the stress diagram (Fig. 10) to the same scale of pounds, making 10 equal 20,000 pounds. The lines 0 4 and 0 6 happen, in this case, to coincide. Then, in Fig. 11, we draw a parallel to 03, c d = 0 4, and d e parallel to 0 1. we obtain the line hjk B. In this case, it will be seen that the longest horizontal ordinate is hb, while at that point there is no vertical ordinate; also, that no hypothenuse can be drawn which will be as long as h b, so that we must take hb as the greatest resultant; and this, multiplied by 20,000 pounds, gives 31,800 pounds as the maximum bending moment on the pin. It will be seen that this is just the product of the outer force by its arm to the centre of the next bar, so that the greatest bending moment is at that point.

To determine the sizeof the pin, we find, from Table III., that for a steel pin to sustain this moment, allowing a fibre strain of 20,000 pounds, we shall need a 2" pin. This pin has a bearing value of 31,500 pounds for a bar an inch thick. The outer bar in this case is " thick, and has a strain of 31,800 pounds, equivalent to 42,400 pounds for a 1" bar. And we see, from Table II., that we shall need to use a 3" pin to meet this strain. The shearing strength of a 34" pin is 36 tons. or more than double the strain. Hence we must use a 34" pin, or, by increasing the thickness of the bars, we might reduce the pin to 3 inches.

PROPORTIONS OF CAST-IRON BEARING-PLATES FOR GIRDERS AND COLUMNS (1895).

If a heavily loaded column or girder should rest directly upon a wall or pier of masonry, the weight would be distributed over such a small area that in most cases there would be danger of crushing the masonry, particularly if it were of brick or rubble work. To

prevent this, it is customary to put a bearing-plate between the end of the beam or column and the masonry, the size of the plate being such that the load from the column or girder divided by the area of the plate shall not exceed the safe crushing-strength of the masonry per unit of measurement.

The load per square inch on different kinds of masonry should not exceed the following limits: