190 GRAPHIC AND ANALYTIC [CHAP. XII 



pression, and E 10 compression; while F9 would be tension 

 From 9 10 to the right the diagonals would alternate in strain, 

 the compressed upper flanges, as also the tensile lower flanges, 

 would diminish towards D, and the last diagonal should pass 

 t-victly through new position of F 2 , thus .closing the strain 

 diagram and checking the work. The reader will do well to 

 construct the diagram. 



The strains should be found for both cases, and the maximum 

 strains taken from each, which, compared with the permanent 

 strains due to the dead load, will give the total maximum 

 strains. 



We have taken for convenience of size too small a scale for 

 the frame to ensure good results. With a large and accurately 

 constructed frame diagram, dealing as we do with only single 

 weights, and consequently small strains, the above force scale 

 of 4 tons per inch would give very accurate results. 



If the strains due to uniform load (no end reactions) are 

 found by addition of the strains for each apex load diagramed 

 separately, the same scale may be employed ; but if all the 

 loads are taken as acting together (Fig. 5, b\ a smaller scale 

 for strains will have to be adopted, as the force line will other- 

 wise be too long. [See Art. 16 of Appendix for the method of 

 calculation.] 



125. Method of paing direct from one Span to next. 



By inspection of Fig. 86 we see that we might find the strains 

 in the intermediate span B C without first going through the 

 whole of A B or CD. Thus, if we knew the moment at B, 

 this moment, divided by depth of truss at B, would give the 

 strain in flange F 7 for the system of triangulation indicated 

 by the full lines. .If then we knew also the shear at B = 

 P A B = Ej F! (Fig. 86, a\ we could at once lay off F l 7 

 and E! F t (Fig. 86, a), and then proceed to find E 8 and 7 8, 

 just as before. In the same way the moment at C, divided by 

 height of truss at C, would give us strain in F 9, and with 

 shear at C = P A B C = E 2 F t = D, we could find E 10 

 and 9 10, as before. As we know already, a load anywhere 

 upon a beam causes positive moments at a fixed end i.e., 

 makes upper flange over support tension and lower flange com- 

 pression. But as we see from the last case, owing to the tri- 

 angulation, the last upper flange may also be compression (see 

 E 6 in -Fig. 86) if the inflection point lies between diagonal 5 6 



