By J. S. C. Browne, N. Hiller and G. E. Walker (Auth.)

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Diagram is drawn under the load diagram in Fig. 11. f. diagram is then divided into convenient sections, and the area of each trapezium calculated. f. diagram. m. diagram, Fig. f. diagram using these areas. m. m. m. d. loading below the base line, as in Fig. 1 Id. m. at any section is then given by measuring the ordinate between the two diagrams. Position of Supports for Uniformly-Loaded Beam to give Maximum Load-carrying Capacity The maximum intensity of uniformly distributed loading >v that a given uniform beam of length / can carry is determined by its strength and the position of its supports, assuming that is rests on two simple supports.

Imagine now that the x, y and z co-ordinates of A, B, C and P have been measured in inches and that they are A (3, 5, 36), B (6, 36, 44), C (40, 19, 54) and P(20, 20, 18). The component of the force in a cord parallel to any axis is the tension in the cord multiplied by the cosine of the angle between the cord and any line parallel to the axis under consideration which intersects the cord. The cosine of this angle is the ratio of the projected length of the cord along the particular axis, to the length of the cord.

The forces must also be coplanar. This is easily shown to be so, since a plane can be constructed to contain the lines of 29 EQUILIBRIUM OF STRUCTURES action of any two of the forces, if the third force does not lie in this plane it will have a component perpendicular to the plane. But there is no force to counteract such a component and the particle could not be in equilibrium. Hence the third force cannot have a component perpendicular to the plane containing the other two forces. Thus all three forces must be coplanar.