This section is from the book "The Building Trades Pocketbook", by International Correspondence Schools. Also available from Amazon: Building Trades Pocketbook: a Handy Manual of reference on Building Construction.
In Fig. 34, forces a b of 50 lb. and c b of 100 lb. act at b in the directions shown. To find their combined action, draw c 6, to any scale, equal to 100 lb., and a b equal to 50 lb. Thus, if the scale is Jin. to 101b., c 6 = 2 1/2in.. and a b = 1 1/4in. Draw a d parallel to c 6, and c d to a 6, intersecting at d; then d b, called the resultant, gives the direction, and, by scaling, the amount of their combined action, 145 lb. The figure abcd forms a parallelogram of forces.
Assume forces c a of 1,000 lb., and a b of 800 lb., acting at a, Fig. 35. Make distance c a, to any scale, equal to 1,000 lb., and a b equal to 800 lb.; draw resultant c 6, which, by scaling, is found to be 1,550 lb: it is opposed to the direction in which forces ca and ab act around the triangle. This figure cab forms a triangle of forces.
The preceding diagrams may be called polygons of forces, but the term is usually applied to diagrams mining the resultant of several forces. When a number of forces act as at d. Fig. 36, their resultant is obtained thus: Draw a line parallel to and having the same direction (as indicated by the arrow points) and magnitude as one of the forces. At the end of this line, draw one parallel to a second force, having the same direction and magnitude as this second force. Continue thus until all the forces have been plotted; a straight line ioining the free ends of the first and last lines will be the closing side of the polygon; mark it opposite in direction to the other forces, of which it will he the resultant. Thus, the resultant of the 4 forces acting at d is obtained by drawing 1-2, 2-3, 3-4,4-5, parallel and equal to forces a d, bd, cd, and e d, respectively. Connecting 5 and 1, the resultant is obtained.
Since the effect of several forces may be determined by a single resultant, so may one force be resolved into several. For example, the force a b, Fig. 37, may be resolved into any two directions by drawing components parallel to those directions. Thus, from a draw ac vertically, and from 6 draw cb horizontally, intersecting at c; then a c is the vertical component, and c b, the horizontal component of a b.
(a), Fig. 38, 1,000 lb. is supported at c by cords c a and c b, secured at a and 6. This figure, drawn
to scale, accurately represents the outline of the structure, and is called a frame diagram. To obtain the stresses in c a and c b, draw 1-2, in (b), making its length to any scale and direction represent the magnitude and action of W. Thus, if 1 in. = 400 lb., 1-2 = 2 1/2 in. long, and, its direction being vertical in (a), it is so drawn in (6). From 1 draw 1-3 parallel to a c, and from 2 draw 2-3 parallel to cb; they intersect at 5, forming with 1-2 a triangle. If 1-3 and 2-3 are measured, using same scale as for 1-2, the stresses ca and cb may be obtained. Diagram (6) is called a stress diagram.