A couple.-A couple consists of two parallel forces equal in amount, but opposite in direction. The arm of the couple is the perpendicular distance between the forces. The moment of a couple is equal to one of the forces multiplied by the arm. The moment of a couple is constant about any point in the plane and may be represented graphically by twice the area of the triangle having one of the forces as a base and the arm of the couple as an altitude. The moment of a force about any point may be represented graphically by twice the area of a triangle, as shown in (c) Fig. 10.

It will be seen from the preceding discussion, that in order that a system of non-concurrent forces be in equilibrium it is necessary that the resultant of all forces save one shall coincide with the one and be opposite in direction. Three non-concurrent forces can not be in equilibrium unless they are parallel. The resultant of a system of non-concurrent forces may be a single force or a couple.

Equilibrium Polygon. First Method.-In Fig. 11 the resultant, a, of P1 and P2 acts through their intersection and is equal and parallel to a in the force polygon (a); the resultant, b, of a and P, acts through their intersection and is equal and parallel to b in the force polygon; the resultant, c, of b and P, acts through their intersection and is equal and parallel to c in the force polygon; and finally the resultant, R, of c and P, acts through their intersection and is equal and parallel to R in the force polygon. Ris therefore the resultant of the entire system of forces. If R is replaced by an equal and opposite force, E, the system of forces will be in equilibrium. Polygon (a) in Fig. 11 is called a force polygon and (b) is called a “funicular" or an "equilibrium" polygon. It will be seen that the magnitude and direction of the resultant of a system of forces is given by the closing line of the force polygon, and the line of action is given by the equilibrium polygon.

=

The force polygon in (a) Fig. 12 closes and the resultant, R, of the forces P1, P2, P3, P4, Pь is parallel and equal to Pé, and is opposite in direction. The system is in equilibrium for translation, but is not in equilibrium for rotation. The resultant is a couple with a moment = - Ph. The equilibrant of the system of forces will be a couple with a moment +Ph. From the preceding discussion it will be seen that if the force polygon for any system of non-concurrent forces closes the resultant will be a couple. If there is perfect equilibrium the arm of the couple will be zero.

Second Method. Where the forces do not intersect within the limits of the drawing board, or where the forces are parallel, it is not possible to draw the equilibrium polygon as shown in Fig. 11 and Fig. 12, and the following method is used:

The point o, (a) Fig. 13, which is called the pole of the force polygon, is selected so that the strings a-o, b-o, c-o, d-o and e-o in the equilibrium polygon (b), which are drawn parallel to the corresponding rays in the force polygon (a), will make good intersections with the forces which they replace or equilibrate.

In the force polygon (a), P, is equilibrated by the imaginary forces represented by the rays o-a and b-o, acting as indicated by the arrows within the triangle; P is equilibrated by the imaginary forces represented by the rays o-b and c-o, acting as indicated by the arrows within the triangle; P, is equilibrated by the imaginary forces represented by the rays o-c and d-o, acting as indicated by the arrows within the triangle; and P1 is equilibrated by the imaginary forces o-d and e-o, acting as indicated by the arrows within the triangle. The imaginary forces are all neutralized except a-o and o-e, which are seen to be components of the resultant, R.

To construct the equilibrium polygon, take any point on the line of action of P1 and draw strings o-a and o-b parallel to rays o-a and o-b, b-o is the equilibrant of o-a and P1; through the intersection of string o-b and P, draw string c-o parallel to ray c-o, c-o is the equilibrant of o-b and P1; through the intersection of string c-o and P, draw string d-o parallel to ray d-o, d-o is the equilibrant of c-o and P3; and through the intersection of string d-o and P, draw string e-o

parallel to ray e-o, e-o is the equilibrant of d-o and P4. Strings o-a and e-o acting as shown are components of the resultant, R, which will be parallel to R in the force polygon and acts through the intersections of strings o-a and e-o.

The imaginary forces represented by the rays in the force polygon may be considered as components of the forces and the analysis made on that assumption with equal ease.

It is immaterial in what order the forces are taken in drawing the force polygon, as long as the forces all act in the same direction around the force polygon, and the strings meeting on the lines of the forces in the equilibrium polygon are parallel to the rays drawn to the ends of the same forces in the force polygon.

The imaginary forces a-o, b-o, c-o, d-o, e-o are represented in magnitude and in direction by the rays of the force polygon to the same scale as the forces P1, P2, P3, P4. The strings of the equilibrium polygon represent the imaginary forces in line of action and direction, but not in magnitude.

Graphic Moments.—In Fig. 14 (b) is a force polygon and (a) is an equilibrium polygon for the system of forces P1, P2, P3, P4. Draw the line M-N = y, parallel to the resultant, R, and with ends on strings o-e and o-a produced. Let r equal the altitude of the triangle L-M-N, and H equal the altitude of the similar triangle o-e-a. H is the pole distance of the resultant, R.

Ry: Hr

= moment of resultant R about any point in the line M-N and therefore

The statement of the principle just demonstrated is as follows:

The moment of any system of coplanar forces about any point in the plane is equal to the intercept on a line drawn through the center of moments and parallel to the resultant of all the forces, cut off by the strings which meet on the resultant, multiplied by the pole distance of the resultant. It should be noted that in all cases the intercept is a distance and the pole distance is a force.

This property of the equilibrium polygon is frequently used in calculating the bending moments in beams and trusses which are loaded with vertical loads.

Bending Moments in a Beam.-It is required to find the moment at the point M in the simple beam loaded as in (b) Fig. 15. The moment at M will be the algebraic sum of the moments

of the forces to the left of M.

and the moment of R1

The moment of P1 = H X B-C, the moment of P2 = HX C-D - HX B-A. The moment at M will therefore be

=

M1 = HX B-C + H X C-D - H X B-A

=

- H X A-D

=

The moment of the forces to the right of M may in like manner be shown to be

H.y

In like manner the bending moment at any point in the beam may be shown to be the ordinate of the equilibrium polygon multiplied by the pole distance. The ordinate is a distance and is measured by the same scale as the beam, while the pole distance is a force and is measured by the same scale as the loads.

Equilibrium Polygon as a Framed Structure.—In (a) Fig. 16 the rigid triangle supports the load P1. Construct a force polygon by drawing rays a-1 and c-1 in (b) parallel to sides a-1 and

c−1, respectively, in (a), and through pole 1 draw 1-b parallel to side 1-b in (a). The reactions R1 and R2 will be given by the force polygon (b), and the rays 1-a, 1-c and 1-b represent the stresses in the members 1-a, 1-c and 1-b, respectively, in the triangular structure. The stresses in 1-a and 1-c are compression and the stress in 1-b is tension, forces acting toward the joint indicating compression and forces acting away from the joint indicating tension. Triangle (a) is therefore an equilibrium polygon, and polygon (b) is a force polygon for the force P1.

From the preceding discussion it will be seen that the internal stresses at any point or in any section hold in equilibrium the external forces meeting at the point, or on either side of the section.

Algebraic Moments. Stresses in a Roof Truss.-The reactions may be found by applying the fundamental equations of equilibrium to the structure as a whole. In the truss in (a) Fig. 17 by taking moments about the right reaction we have

To find the stresses in the members of the truss in (a) Fig. 17, proceed as follows: Cut the truss by means of plane A-A, as in (b), and replace the stresses in the members cut away with external forces. These forces are equal to the stresses in the members in amount, but opposite in direction, and produce equilibrium.

To obtain stress 4-x take center of moments at L,, and take moments of external forces

To obtain stress in 4-5 take center of moments at Lo, and take moments of external forces