APPENDIX.] PIVOT SPAN. 377 



The resultant shear thus obtained then, multiplied by the 

 secant of the angle with vertical, gives the strain in diagonal. 



If the sign of the result is negative ( ), it shows that the 

 strain on the diagonal is contrary to that indicated by its slope. 



To illustrate, let us again take the weight P 2 and consider 

 diagonal 3 4. 



The simple shear at apex 4 is 4.964 10 5.036. The 

 strain in C for P 2 we have found to be compression, and equal 

 to + 12.47. It is sloped as a strut, and its vertical component is 

 therefore to be subtracted from the shear above. Since its 

 angle is nearly 5 43' with the horizontal, this vertical compo- 

 nent is 12.47 x sin 5 43' = 1.24 



Since H is in this case horizontal, it has no vertical compo- 

 nent. The resultant shear is then 5.036 1.24 = 6.276. 



As the secant of the angle of 3 4 with the vertical is 1.49, 

 we have for the strain in 34, -- 6.276 x 1.49 = 9.35. 



This result being minus, and 3 4 being sloped as a strut, 

 the strain is 9.35 tons tension, agreeing closely with the value 

 found above by moments. 



The above method is preferable to the method by moments 

 for the diagonals, as we have only to determine the secants for 

 the verticals and the sines for the flanges, which is in most 

 cases easier than to find the lever arms for the diagonals and 

 the points of intersection of the upper and lower flanges in each 

 panel. It is, like the method of moments, of general applica- 

 tion to any framed structure whose outer forces are known. 



The method of diagram in Art. 124 will be found preferable 

 to both. 



It is unnecessary to pursue our example further. With the 

 mutual checks of the two methods of calculation explained 

 above, as well as the diagrams, correct results cannot fail to be 

 obtained. The diagrams should always be made first, as they 

 settle by mere inspection many points which may at first cause 

 trouble such as whether the shear in a piece is subtractive or 

 not according to our rule, the character of the strains in dif- 

 ferent pieces, etc. It is well to indicate on the diagrams com- 

 pressive strains by double or heavy lines. 



from right to left, thus iQ ; we shall have for any case vertical component = 

 strain x cos. 6. The cosine will change its sign according to the quadrant it 

 is in, according to the above rule, and the sign of the vertical component will 

 take care of itself. 



