Accordingly, after the loads and reactions on a truss, which is to be analyzed, have been ascertained*, we look for a joint at which only two members are connected (the end joints are usually such). Then we consider the forces at that joint and determine the two unknown forces which the two members exert upon it by methods explained in Arts. 16 or 17. The forces so ascertained are the direct stresses, or stresses, as we shall call them for short, and they are the values of the pushes or pulls which those same members exert upon the joints at their other ends.
Next we look for another joint at which but two unknown forces act, then determine these forces, and continue this process until the stress in each member has been ascertained. We explain further by means of Examples. 1. It is desired to determine the stresses in the members of the steel truss, represented in Fig. 16, due to its own weight and that of the roofing assumed to weigh 12 pounds per square foot. The distance between trusses is 14 feet.
The apex loads for this case were computed in Example 1, Art. 20, and are marked in Fig. 16. Without computation it is plain that each reaction equals one-half the total load, that is, of 12,000, or 6,000 pounds.
The forces at joint (1) are four in number, namely, the left reaction (6,000 pounds), the load applied there (1,500 pounds), and the forces exerted by members 12 and 13. For clearness, we represent these forces so far as known in Fig.
17 (a); we can deter mine the two un known forces by merely constructing a closed force polygon for all of them. To construct the polygon, we first represent the known forces; thus AB (1 inch long with arrowhead pointing up) represents the reaction and BC ( inch long with arrowhead point ing down) represents the load. Then from A and C we draw lines
parallel to the two unknown forces and mark their intersection D (or D'). Then the polygon is ABCDA, and CD (1.5 inches = 9,000 pounds) represents the force exerted by the member 12 on the joint and DA (1.3 inches = 7,800 pounds) represents the force exerted by the member 13 on the joint. The arrowheads on BC and CD must point as shown, in order that all may point the same way around, and hence the force exerted by member 12 acts toward the joint and is a push, and that exerted by 13 acts away from the joint and is a pull. It follows that 12 is in compression and 13 in tension.
If D' be used, the same results are reached, for the polygon is ABCD'A with arrowheads as shown, and it is plain that CD' and DA also D'A and CD are equal and have the same sense. But one of these force polygons is preferable for reasons explained later.
Since his in compression, it exerts a push (9,000 pounds) on joint (2) as represented in Fig. 18 (a), and since 13 is in tension it exerts a pull (7,800 pounds) on joint (3) as represented in Fig. 19 (a).
The forces at joint (2) are four in number, the load (3,000 pounds), the force 9,000 pounds, and the force exerted upon it by the members 24 and 23; they are represented as far as known in Fig. 18 (a). We determine the un known forces by constructing a closed polygon for all of them. Represent ing the known forces first, draw AB (1.5 inches long with arrowhead point ing up) to represent the 9,000 pound force and BC (i inch long with arrow head pointing down) to represent the load of 3,000 pounds. Next from A and C draw lines parallel to the two unknown forces and mark their inter section D; then the force polygon is