FIGURES TO THE CALCULATION OF STRAINS ON FRAMEWORK. 443 



Fig. la is the reciprocal figure of Frame I., thus uniformly loaded and sup- 

 ported at X and Y. The line xy is the vertical line of loads, equal to 80 tons 

 in all, and equally subdivided, because the load at each joint is equal. From each 

 of these subdivisions horizontal lines are ruled, and the lines IJKLMNO in 

 the reciprocal figure are drawn parallel to the lines similarly lettered in the 

 frame. The lengths of each of the lines in the reciprocal figure measure the 

 stresses on the members in the frame. The figure can be drawn in five minutes ; 

 whereas the algebraic computation of the stresses, though offering no mathe- 

 matical difficulty, is singularly apt, from mere complexity of notation, to result 

 in error. 



The figure and the direction of each stress will be easily understood 

 when decomposed into its component polygons. The triangle PIA corresponds 

 to the polygon of forces at X, in which the direction of all the forces is that 

 in which the pen moves, starting from Z towards x. The polygon of forces 

 acting on joint 1, beginning with the forces determined by the previous poly- 

 gon, and proceeding in the direction in which the forces act on the joint 1, 

 is I, 1, a, J. 



The polygon at joint 2 is shown separately at fig. I /3, being AIJB ; the 

 polygon at joint 3 is also shown separately, the directions of the forces being in- 

 dicated by arrows. The complete fig. 1 a is built up of separate polygons similar 

 to these two ; the origin or starting-point on each being indicated by a small 

 circle in fig. 1/3. 



Each line in fig. 1 a serves as a part of two component polygons, but it 

 would be passed over in opposite directions in the two polygons by a pencil fol- 

 lowing the directions of the forces in the two polygons. This fact is of assistance 

 in drawing the reciprocal figures, making it easy to find the starting-point or 

 origin of each new polygon, since the lines representing forces already known 

 must be traversed in the opposite direction to those forces ; thus the polygon at 

 joint 4 will include the force due to B and L. These have already formed part 

 of polygons 2 and 3 ; but in these the direction of the forces was from the joint 

 4, and hence in the new polygon the direction will be to the joint 4, and the 

 polygon will begin at Z, running BLMC. 



It must be observed that the lines ABCD all begin at Z, ending at the inter- 

 section of I and J, K and L, M and N respectively. The stress on the two centre 

 diagonals is nil, and with the uniform load the second half of the reciprocal 

 figure is exactly symmetrical with the first half. The stress on d is equal to 

 that on D. 



When the load is not uniform (fig. 2), the weights supported on the two piers 

 are not equal ; in other words, the forces P and P : are not equal, and the line of 

 loads must be subdivided at Z into two portions, P and P x (fig. 2 a), equal to the 

 loads borne by the piers. The divisions 1, 3, 5, 7, 9, 11, 13, 15, are made equal 



