Examples. 1. It is required to construct a stress diagram for the truss represented in Fig. 24 supported at its ends and sustaining three loads of 2,000 pounds as shown. Evidently the reactions equal 3,000 pounds.
Following the directions of Art. 26, we letter the truss diagram, then draw the polygon for the loads and reactions. Thus, to the scale indicated in Fig. 24 (b), AB, BC, and CD repre sent the loads at joints (2), (3) and (5) respectively and DE and EA represent the right and the left reactions respectively. Notice that the polygon (ABCDEA) is a clockwise one.
At joint (1)there are three forces, the left reaction and the forces exerted by the members of and fe. Since the forces exerted by these two members must be marked AF and EF we draw from A a line parallel to of and from E one parallel to of and mark their intersection F. Then EAFE is the polygon for joint (1), and since EA acts up (see the polygon), AF acts down and FE to the right. We, therefore, place the proper arrowheads on of and fe near (1), and record (see adjoining table) that the stresses in those members are compressive and tensile respectively. Measuring, we find that AF and FE equal 6,150 and 5,100 pounds respectively.
We may next draw the polygon for joint (6) or (2) since there are but two unknown forces at each. At joint (2) for instance, the unknown forces are those exerted by fy and bg, and the known are the load ab (2,000 pounds) and the force exerted by af. Since the unknown forces must be marked FG and GB, draw from F a line parallel to fy, from B a line parallel to bg, and mark their intersection G. Then the poly gon for the joint is FABGF, and since AB acts down (see the polygon) BG and GF act down and up respec tively. Therefore, place the proper arrowheads on by and gf near (2), and record that the stresses in those mem bers are both compressive. Measur ing, we find that BG and GF scale 4,100 and 1,875 pounds respectively.
Now draw a polygon for joint (3) or (6) since there are but two un known forces at each joint. At (3) for instance, the unknown forces are those exerted by ch, and gh, the known forces being the load (2,000 pounds) and the force 4,100 pounds, exerted by by. Since the unknown forces must be marked CH and GH, draw from C a line parallel to ch, from G one parallel to gh, and mark their intersection H. Then