This section is from the book "The Building Trades Pocketbook", by International Correspondence Schools. Also available from Amazon: Building Trades Pocketbook: a Handy Manual of reference on Building Construction.

This truss, shown in Fig. 43, is much used for pin-connected and structural-steel trusses.

Obtain the forces acting at each panel point and draw the frame diagram for the vertical loads, as in Fig. 43. Since the loading is symmetrical, the reactions R1 and R2 will each be one-half the load, or 27,200 lb. Draw in the stress diagram, Fig. 44, the load line abcde, etc., locating z midway between e and f. The polygon of external forces will be from a to b,b to c, c to d, etc., until j is reached; retracing the load line from j to z gives R2, and from z to a determines R1.

Analyzing each joint as before shown, no difficulty will be met until the joints C D O N M L and M N Q Z are reached; it is found that three unknown forces exist at each of these joints, and, as it is impracticable to determine the stresses graphically when more than two forces are unknown, the value of one must be otherwise obtained. Upon inspecting the frame diagram, it will be observed that the joint at B C is similar to DE; and it is reasonable to suppose that L K will have an equal stress with P 0. From this it would appear that there must be an equal and like stress exerted by L M, to retain the foot of L K in position, as is exerted by 0 N to keep P 0 in place. The stresses in M L and L C being known, that in D 0 may be determined by drawing a line from d parallel to I) 0, to such a point o that - having drawn on equal in length to Im - the line nm, parallel to NM, will close on to. The polygon of forces will be from to to I, I to c, c to d. d to o, o to n, and n to m, the starting point.

The joint MNQZ now offers no difficulty to solution, as ran has been previously determined. The final joint OPQN may be solved by drawing from p a line parallel to P Q, which will, if the diagrams have been correctly drawn, pass through n, and from this point will coincide with n q.

One half the vertical-load diagram being completed, and the loading being symmetrical, it is unnecessary to draw the other half unless as a check. The equality of the triangles Ikm and pon, and their resemblance to nmq, greatly assist. in drawing diagrams for trusses of this type.

If one unknown stress prevents the solution of the diagram, it may sometimes be found by trial. Thus, assume the amount of unknown stress and proceed with diagram; if it fails to close, again assume amount of unknown stress and try. Proceed thus until diagram closes.

First find the external forces and reactions, as in the Howe truss, page 143, and redraw the frame diagram, as in Fig. 45. The solution of the wind-stress diagram, Fig. 46, offers the same difficulties as in drawing the vertical-load diagram, and may be similarly solved. The last joint FRQPE is solved by drawing from f a line parallel to FR, and from q, a line parallel to R Q; their intersection is at r, and the polygon of forces is from e to f, f to r, r to q, q to p, and p to e.

Fig. 46.

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