126. - Application Of Graphic Statics To Simple Triangular Frames With But One External Load

To show the application" of graphic statics in its simplest form, we will take a simple triangular frame, composed of a horizontal beam and two pairs of eye bars, with pin connections, sustaining a single load at the bottom, as in Fig. 27.1. For convenience of illustration we will assume a load of 100 lbs., the application being precisely the same for any load. Now, at the lower joint d, we have three forces applied, viz., the load, the stress in A and the stress in B. The amount of the load we know, and we also know that its direction is vertical and downward. To find the amount of the stress in A and B, draw a vertical line, 1 - 2, equal to one hundred lbs. at a scale of, say one hundred lbs. to the inch, with the arrow head at the bottom. Then by proposition 3, § 124, for the three forces at d to be in equilibrium they must be represented by the sides of a triangle, taken in order, and the line 1 - 2 must be one side of the triangle. To find the other sides from one end of the line, representing the load, draw a line parallel to A and from the other end a line parallel to B and continue the lines until they intersect. It will make no difference in the result whether we draw the line parallel to A from the top or bottom of the load line, as is illustrated in Fig. 271, the lines a. a. and b. b. being of the same length in both diagrams. For trusses, however, it is usually more convenient to draw the stress lines so that they will be to the left of the load line, as in the left half of the stress diagram, Fig. 271, and in future examples the stress diagram will be drawn in that way. Now, if we letter the lines in the stress diagram to correspond with the lettering of the frame, using small letters for the stress diagram and capital letters for the frame, then the length of the line a, measured by the scale of 100 lbs. to the inch, will give the stress in A, and the length of the line b, gives the stress in B.

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Fig. 271.

Moreover, the arrow heads must follow each other in rotation, and, as the arrow on the load line points down, those on a and b must point as shown in the stress diagram. Now, if we place arrow heads on A and B, pointing in the same way as those on a and b, we see that the heads point from the joint d, and hence, as explained in §125, these pieces must be in tension, as is also obvious from the manner of loading.

Stress in C. - If the beam C had a slot through its center, and the pins at P1 and P2 were free to move, it is obvious that the load would cause the pieces A and B to come together until P1 and P2 touch. Therefore C must exert a compressive resistance, to hold P1 and P2 in place. The amount of this compression is found by drawing a horizontal line from 3 to the load line, and measuring its length by the scale of 100 lbs. to the inch. The line c also divides the load line in the proportion of the vertical reactions at P1 and P2. To show that this is true we will consider the forces acting at P1. At this joint we have three forces, viz., the stress in A, compression in C, and the vertical reaction represented by the arrow, and these three forces must form a triangle in the stress diagram. From the triangle of forces acting at d, we obtained the stress a, so that we have one side of the triangle of forces for P1 one of the other sides of the triangle must be parallel to C and the third must be parallel to the reaction. Consequently the triangle of forces at P1 must be 3 - 2 - 4 and the distance from 2 to 4 must be the reaction. The arrow heads for this triangle must point in the opposite direction from those in the triangle 1 - 2 - 3, or from 3 to 2, 2 to 4 and 4 to 3, so that P1 and c are both pointing towards the joint and therefore indicate compression. Scaling the lines in the stress diagrams we obtain the following values for the different stresses, viz., a = 52, b = 73, c = 36 1/2, P1 = 36 1/2,P2=63 1/2 lbs.

If the beam C is inclined, as in Fig. 272, the stresses will be found in precisely the same way, the line c in the stress diagram being drawn parallel to the beam. (Note - The lines in the stress diagrams must always be drawn parallel to the corresponding lines of the frame.) If the beam is notched so as to have horizontal seats, and anchored at the top, the reactions at P1 and P2 will be vertical.

If the beam is supported at the bottom only and abuts against a corresponding beam at the top, as in the case of the rafters at the top, then the entire reaction at the top will be horizontal, and at the bottom there must be both a horizontal and vertical reaction, the vertical reaction being the full amount of the load, and the horizontal reaction being equal to the product of the load by one-half of the span, divided by the rise, both measurements being in inches.

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Fig. 272.

If, instead of a suspended load, the load is supported by two struts, notched into a beam at the bottom, as in Fig. 273, the stresses in the struts and beam are found in precisely the same way as in Fig. 271, but in this case it will be found that the arrow heads on a and b point towards the joint, and therefore indicate compression, while there will be a pull in C to hold the ends of the struts together.

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Fig. 273.