35. Conditions of Equilibrium of Non=Concurrent Forces Not Parallel may be stated in various ways; let us consider four. First:

1. The algebraic, sums of the components of the forces along each of two lines at right angles to each other equal zero.

2. The algebraic sum of the moments of the forces about any origin equals zero.

Second:

1. The sum of the components of the forces along any line equals zero.

2. The sums of the moments of the forces with respect to each of two origins equal zero.

Third:

The sums of the moments of the forces with respect to each of three origins equals zero.

Fourth:

1. The algebraic sum of the moments of the forces with respect to some origin equals zero.

2. The force polygon for the forces closes.

It can be shown that if any one of the foregoing sets of conditions are fulfilled by a system, its resultant equals zero. Hence each is called a set of conditions of equilibrium for a non-concurrent system of forces which are not parallel.

The first three sets are "algebraic" and the last is "mixed," (1) of the fourth, being algebraic and (2) graphical. There is a set of graphical conditions also, but some one of those here given is usually preferable to a set of wholly graphical conditions.

Like the conditions of equilibrium for concurrent forces, they are used to answer questions arising in connection with concurrent systems known to be in equilibrium. Examples may be found in Art. 37.

36. Conditions of Equilibrium for Parallel Non-Concurrent Forces. Usually the most convenient set of conditions to use is one of the following:

First:

1. The algebraic sum of the forces equals zero, and

2. The algebraic sum of the moments of the forces about some origin equals zero.

Second:

The algebraic sums of the moments of the forces with respect to each of two origins equal zero.

37. Determination of Reactions. The weight of a truss, its loads and the supporting forces or reactions are balanced and constitute a system in equilibrium. After the loads and weight are ascertained, the reactions can be determined by means of conditions of equilibrium stated in Arts. 35 and 36.

The only cases which can be taken up here are the following common ones: (1) The truss is fastened to two supports and (2) The truss is fastened to one support and simply rests on rollers at the other.

Case (1) Truss Fastened to Both Its Supports. If the loads are all vertical, the reactions also are vertical. If the loads are not vertical, we assume that the reactions are parallel to the resultant of the loads.

The algebraic is usually the simplest method for determining the reactions in. this case, and two moment equations should be used. Just as when finding reactions on beams we first take moments about the right support to find the left reaction and then about the left support to find the right reaction. As a check we add the reactions to see if their sum equals the resultant load as it should.

Examples. 1. It is required to determine the reactions on the truss represented in Fig. 20, it being supported at its ends and sustaining two vertical loads of 1,800 and 600 pounds as shown.

The two reactions are vertical; hence the system in equilibrium consists of parallel forces. Since the algebraic sum of the moments of all the forces about any point equals zero, to find the left reaction we take moments about the right end, and to find the right reaction we take moments about the left end. Thus, if R1 and R2 denote the left- and right-reactions respectively, then taking moments about the right end,

(R1X 24) - (1800 X 15) - (600 X 15) = 0, or 24R1= 27,000 + 9,000 = 36,000;

„ hence R1 = 36,000/24 = 1,500 pounds.

Taking moment about the left end,

- R2X 24 + 1,800 X 9 + 6OO X 9 = 0, or 24R2 = 16,200 + 5,400 = 21,600; hence R2= 900 pounds.

As a check, add the reactions to see if the sum equals that of the loads as should be the case. (It will be noticed that reactions on trusses and beams under vertical loads are determined in the same manner.)

2. It is required to determine the reactions on the truss represented in Fig. 28 due to the wind pressures shown (2,700, 5,400 and 2,700 pounds), the truss being fastened to both its supports.

The resultant of the three loads is evidently a single force of 10,800 pounds, acting as shown in Fig. 34. The reactions are parallel to this resultant; let R1 and R2 denote the left and right reactions respectively. Fig. 34.

The line abc is drawn through the point 7 and perpendicular to the direction of the wind pressure; hence with respect to the right support the arms of R1 and resultant wind pressure are ac and bc, and with respect to the left support, the arms of R2 and the resultant wind pressure are ac and ab. These different arms can be measured from a scale drawing of the truss or be computed as follows: The angle equals the angle, and was shown to be 30 degrees in Example 3, Page 26. Hence ab = 14 cos 30°, bc = 28 cos 30°, ac = 42 cos 30°.

Since the algebraic sums of the moments of all the forces acting on the truss about the right and left supports equal zero,

R1 X 42 cos 30° = 10,800 X 28 cos 30°, and R2 X 42 cos 30° = 10,800 X 14 cos 30°.

From the first equation,

R1 10,800 X 28/42 = 7,200 pounds, and from the second,

R2=10,800 X 14/42 -------= 3,600 pounds.

Adding the two reactions we find that their sum equals the load as it should.

Case (2) One end of the truss rests on rollers and the other is fixed to its support. The reaction at the roller end is always vertical, but the direction of the other is not known at the outset unless the loads are all vertical, in which case both reactions are vertical.