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Equilibrium of Non-Concurrent Forces 35

reactions, zero, equals, algebraic, moments and loads

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EQUILIBRIUM OF NON-CONCURRENT FORCES.

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.

Ģ 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 sys tem 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 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 erigins equal zero. Ģ 37- Determination of Reactions. The weight of a truss, its loads and the supporting forces or reactions are balanced and con stitute a system in equilibrium. After the loads and weight are ascertained, the reactions can be determined by means of condi tions 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 result ant 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 mo ments 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 R, and denote the left. and right-reactions respectively, then taking moments about the right end, 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 resented 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 and denote the left and right reactions respectively.

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