CHAP. XVI.] THE ETVEKTEP ATRCH. 337 



divided by the depth of truss, give at once the strains ir; the 

 flanges. Since the moment scale is 750 tons per inch and the 

 depth of truss 5 ft., the moment ordinates scaled off at 150 tons 

 per inch will give at once the strains in the flanges, without 

 division. 



For the shearing forces, we know from the preceding that 

 the maximum reaction at right end is for loads 6-11. This 

 reaction we have already found in the corresponding force line 

 by means of the closing line already drawn. We lay it off then 

 rigtot and left, half-way between the ends and first apex, that 

 being the effective length of load, the two half-end panels rest- 

 ing directly upon the abutments. 



The maximum shear at any point is evidently when the load 

 reaches from right support to that point, and is equal to the 

 sum of the chain reactions at the unloaded apices. Thus, max- 

 imum shear at 3 is equal to the interval III for the line 

 3-11; at 4, I III for line 4-11; at 5, I IV for line 5-11; 

 and at 6, I Y for line 6-11. Laying off the shear at 6, we 

 can draw the line 6-11, as indicated in -the diagram, and thus 

 determine the shear at 2. This we cannot find, as above, for 3, 

 4, etc., as for the load 2-11 ; owing to the shape of the chain 

 as represented, there is no upward reaction at 1, as there is no 

 angle of the chain at 6. 



The shear diagram is, of course, symmetrical on each side of 

 the centre. We can therefore construct it as represented, and 

 then the determination of the strain in the diagonals is easy. 

 We have only to multiply the shear at any apex by the secant 

 of the angle which the diagonals make with the vertical. This 

 we may do by properly changing the scale at once, and thus 

 scale off the strains directly. 



18. Analytical Determination of the Forces acting 

 upon the stiffening Truss. Assuming that the truss distrib- 

 utes the partial loading uniformly over the whole arc, we may 

 deduce very simple formulae for the forces acting upon the 

 truss. As we have already seen, for a maximum moment at 

 any point, the load must always extend out from one end. 



Let us represent, then, the ratio of the loaded part from left 

 to the whole span by k. 



Let the entire span be 2 1, then the loaded portion is 2 Tf I. 

 Let m be the load per unit of length ; then the whole load 

 [Fig. 108]. 

 22 



