NOTE TO ART. 124. [APPENDIX. 



\\V .an ulso, if we wish, apply the method of moments. 

 Thus, if \vo determine the point of intersection in the present 

 case of the inclined upper flange with the horizontal lower 

 flange, this point will be a common centre of moments for the 

 diagonals. The lever arms of the diagonals with reference to 

 this ]M)int must next be determined, and then we are ready. 



This point above for centre of moments is easily found ; thus 

 4 : 40 : : 10 : 100. 



It is therefore 60 ft. to the left of A, or 100 ft. left of B. 

 Take now any diagonal, as 3 4. Its angle with the horizontal 

 is very nearly 42, and with the vertical 48. Its lever arm is 

 then 80 sin 42 = 53.5, and sec. of angle with vertical is 1.49. 



Now take the weight P 2 . Its upward reaction at A is 4.964, 

 P 2 being 10. We have then 



[str. in 3 4] x 53.5 = 10 x 80 4.964 x 60 = + 502.16. 



The resultant rotation is then positive, or from left to right. 

 The point P 2 then sinks and 4 rises, and 34 is in tension and 



502.16 



= -^~ = 9.38 tons. 

 5o.o 



This is sufficient to illustrate the method. 



For the first method referred to above, viz., that by resultant 

 shear, the following points are to be observed : 



When a piece slopes towards the nearest support, we say it ia 

 sloped as a strut, whatever the real strain in it may be. When 

 it slopes away from the nearest support, it is sloped as a tie. 



The simple shear is the reaction at the support minus the 

 weights between any point and that support. 



If any three strained pieces are cut by a section through the 

 structure, the strains in these pieces are in equilibrium with the 

 simple shear at this section. Hence the algebraic sum of the 

 vertical components of these pieces must be equal and opposite 

 to the shear itself. 



In order to add these vertical components with proper signs, 

 we must remember that if a flange is in tension and sloped as a 

 strut, or in compression and sloped as a tie, we add the vertical 

 component of the strain in it to the simple shear already 

 obtained. If in compression and sloped as a strut, or tension 

 and sloped as a tie, we subtract. 



* In general, if we take compression as plus and tension as minus, and then 

 measure the angle made by the piece with the vertical always from that vertical 



