Examples. 1. It is required to compute the apex loads for the truss represented in Fig. 16, it being of steel, the roof such that it weighs 15 pounds per square foot, and the distance between adjacent trusses 14 feet.
The span being 42 feet, the formula for weight of truss (Art. 19) becomes 42 14 X 42 + 1) = 1,575.84 pounds.
The length 14 scales about 241 feet, hence the area of roofing sustained by one truss equals 48-i X 14 = 679 square feet, and the weight of the roofing equals 679 X 15 = 10,185 pounds.
The total load equals 1,575.84 + 10,185 = 11,760.84 pounds.
Now this load is to be proportioned among the five upper joints, but joints numbered (1) and (7) sustain only one-half as much load as the others. Hence for joints (1) and (7) the loads equal As the weight of the truss is only estimated, the apex loads would be taken as 1,500 and 3,000 pounds for convenience.
2. It is required to compute the apex loads due to a snow load on the roof represented in Fig. 16, the distance between trusses being 14 feet.
The horizontal area covered by the roof which is sustained by one truss equals 42 X 14 = 588 square feet.
If we assume the snow load equal to 10 pounds per horizontal square foot, than the total snow load borne by one truss equals at the joints (2), (4), and (5).
3. It is required to compute the apex loads due to wind pressure on the truss represented in Fig. 16, the distance between trusses being 14 ft.
The inclination of the roof to the horizontal can be found by measuring the angle from a scale drawing with a protractor or by computing as follows : The triangle 346 is equilateral, and hence its angles equal 60 degrees and the altitude of the triangle equals and hence the angle equals 30 degrees.
According to Art.19, 32 pounds per square foot is the proper value of the wind pressure. Since the wind blows only on one side of the roof at a given time, the pressure sustained by one truss is the wind pressure on one half of the area of the roof sustained by one truss, that is 14 X 24i X 32 = 10,864 pounds.
One half of this pressure comes upon the truss at joint (2) and one fourth at joints (1) and (4).
1. Compute the apex loads due to weight for the truss represented in Fig. 27 if the roofing weighs 10 pounds per square
foot and the trusses (steel) are 12 feet apart.
Ans. As shown in Fig. 27.
2. Compute the apex loads due to a snow load of 20 pounds per square foot on the truss of Fig. 25, the distance between trusses being 15 feet.
For joints (4) and (7), 1,200 pounds. Ans. For joints (1) and (3), 3,600 pounds.
joint (2) , 4,800 pounds.
3. Compute the apex loads due to wind for the truss of Fig. 26, the distance between trusses being 15 feet.
( Pressure equals practically 29 pounds per Ans. ) square foot. Load at joint (2) is 4,860 and at joints (1) and (3) 2,430 pounds.
21. Stress in a Member. If a truss is loaded only at its joints, its members are under either tension or compression, but the weight of a member tends to bend it also, unless it is vertical. If purlins rest upon members between the joints, then they also bend these members. We have therefore tension members, com pression members, and members subjected to bending stress com bined with tension or compression. Calling simple tension or compression direct stress as in " Strength of Materials," then the process of determining the direct stress in the members is called "analyzing the truss." 22. Forces at a Joint. By "forces at a joint" is meant all the loads, weights, and reactions which are applied there and the forces which the members exert upon it. These latter are pushes for compression members and pulls for tension members, in each case acting along the axis of the member. Thus, if the horizontal and inclined members in Fig. 15 are in tension, they exert pulls on the joint, and if the vertical is a compression member, it ex erts a push on the joint as indicated. The forces acting at a joint are therefore concurrent and their lifftes of action are always known.
23. General Method of Procedure. The forces acting at a joint constitute a system in equilibrium, and since the forces are concurrent and their lines of action are all known, we can determine the magnitude of two of the forces if the others are all known; for this is the important problem mentioned in Art. 16 which was illustrated there and in Art. 17.