14 X sin 60 = 12.12 feet. The tangent of the angleIV Analysis Of Trus5es Method Of Joints Part 2 0300119 equals

12.12/21 = 0.577, 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 24 1/4 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).

Examples For Practice

1. Compute the apex loads due to weight for the truss represented in Fig. 27 if the roofing weighs 12 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 pints (4) and (7), 1,200 pounds.

For joints (1) and (3), 3,600 pounds.

For 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 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, compression members, and members subjected to bending stress combined 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 exerts a push on the joint as indicated. The forces acting at a joint are therefore concurrent and their lines 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.

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.

IV Analysis Of Trus5es Method Of Joints Part 2 0300120

Fig. 17.

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.

* How to ascertain the values of the reactions is explained in Art. 37. For the present their values in any given case are merely stated.

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 membersIV Analysis Of Trus5es Method Of Joints Part 2 0300121 and

IV Analysis Of Trus5es Method Of Joints Part 2 0300122 For clearness, we represent these forces so far as known in Fig. 17 (a); we can determine the two unknown 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 EC ( inch long with arrowhead pointing 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 = 0,000 pounds) represents the force exerted by the memberon the joint and DA (1.3 inches = 7,800 pounds) represents the force exerted by the memberon 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 acts toward the joint and is a push, and that exerted byacts away from the joint and is a pull. It follows thatis in compression andin 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