The apex loads for this case are computed in Example 3, Page 26, to be as represented in Fig. 28. Supposing both ends of the truss to be fastened to the supports, then the reactions (due to the wind alone) are parallel to the wind pressure and the right and left reactions equal 3,600 and 7,200 pounds as explained in Example 2, Page 57.
To draw a clockwise polygon for the loads and reactions, we lay off BC, CF, and FF' to represent the loads at joints (1), (2), and (4) respectively; then since there are no loads at joints (5) and (7) we mark the point F' by C' and B' also; then lay off B'A to represent the reaction at the right end. If the lengths are laid off carefully, AB will represent the reaction at the left end and the polygon is BCFF'C'B'A B.
At joint (1) there are four forces, the reaction, the load, and the two stresses. AB and BC represent the first two forces, hence from C draw a line parallel to cd and from A a line parallel to ad and mark their intersection D. Then ABCDA is the polygon for the joint and CD and DA represent the two stresses. The former is 7,750 pounds compression and the latter 9,000 pounds tension.
At joint (2) there are four forces, the stress in cd (7,750 pounds compression), the load, and the stresses in fe and ed. As DC and CF represent the stress 7,750 and the load, from F draw a line parallel to fe and from D a line parallel to de, and mark their intersection E. Then DCFED is the polygon for the joint and FE and ED represent the stresses in fe and ed respectively. The former is 7,750 pounds and the latter 5,400, both compressive.
At joint (3) there are four forces, the stresses in ad (9,000 pounds), de (5,400 pounds), eg and ga. AD and DE represent the first two stresses; hence from E draw a line parallel to eg and from A a line parallel to ag and mark their intersection G. Then ADEGA is the polygon for the joint and EG and GA represent the stresses in eg and ga respectively The former is 5,400 and the latter 3,600 pounds, both tensile, At joint (4) there are five forces, the stresses in eg (5,400 pounds) and of (7,750 pounds), the load, and the stresses hire' and e'g. GE, EF and FF' represent the first three forces; hence draw from F' a line parallel to f'e' and from G a line parallel to e'g and mark their intersection E'. (The first line passes through G, hence
E' falls at G). Then the polygon for the joint is GEFF'E'G, and Fig. 29.
F'E' (6,250 pounds compression) represents the stress in fe'. Since E'G = 0, the wind produces no stress in member ge'.
At joint (5) three members are connected together and there is no load. The sides of the polygon for the joint must be parallel to the members joined there. Since two of those members are in the same straight line, two sides of the polygon will be parallel and it follows as a consequence that the third side must be zero. Hence the stress in the member e'd' equals zero and the stresses infe' and d' e' are equal. This result may be explained slightly differently: Of the stresses in e'd', and d'e' we know the first (6,250) and it is represented by E'F'. Hence we draw from F' a line parallel to e'd' and one from E' parallel to d'e and mark their intersection D'. Then the polygon for the joint is E'F'C'D'E; C'D' (6,250 pounds compression) representing the stress in e'd'. Since E' and D' refer to the same point, E'D' scales zero and there is no stress in e'd'.
The stress in ad' can be determined in various ways. Since at joint (6) there are but two forces (the stresses in ge' and eV' being zero), the two forces must be equal and opposite to balance. Hence the stress in d'a is a tension and its value is 3,600 pounds.
2. It is required to analyze the truss represented in Fig. 24 for wind pressure, the distance between trusses being 15 feet. The length 13 equals or Hence the area sustaining the wind pressure to be borne by one truss equals 24.4 X 15 = 366 square feet.
The tangent of the angle which the roof makes with the horizontal equals 14 20 = 0.7; hence the angle is practically 35 degrees. According to Art. 19, the wind pressures for slopes of 30 and .40 degrees are 32 and 36 pounds per square foot; hence for 35 degrees it is 34 pounds per square foot. The total wind pressure equals, therefore, 366 x 31 = 12,444, or practically 12,400 pounds.