264 THE BRACED AECH. [CHAP. XIY. 



desired.* A similar tangent to the hyperbola on the other side 

 determines the direction of the other reaction. We can then 

 resolve P in these two directions, and find at once V and H. 

 The problem, then, so far as a parabolic arc is concerned, is 

 sufficiently simple and easy of solution. We have only to draw 

 a straight line and two easily constructed curves. The formulae 

 for V and H and moment at crown M,, are for this case also 

 simple, and may be used for checking our results. They are : 



__ ! _. (a - x) 9 (3 a 9 - 10 a x - 5 a?) 



1Y1 = TS * - a z > 



H-iff p^ 2 -** 2 ) 2 y - i P ( a ~ X Y ( 2 a + g ) 



* ' 8 



where, as before, a is the half span, h the rise, and x the dis- 

 tance of weight P from crown. A negative moment always 

 indicates tension in lower or inner flange. 



2d. CIRCULAR ARC. 



In this case we have for the locus ILK [Fig. 92], for small 

 central angles a, the equation : 



7 P. (a 8 2 a; a?) a 8 ! 

 y = 4- h I 1 30 v - -, - ,, ,, } y I, 



(a + xf A 2 



a, h, and x being as above, and cp -% = the square of radius 



of gyration ; A being the area and I the moment of inertia of 

 cross- section. 



\Note. In att the cases hitherto considered, or which we 

 shall consider hereafter, the cross-section is assumed constant.] 



According to the exact formula, which is too complicated to 

 make it desirable to be given here, we have the following 



* For the construction of a tangent to a conic section, see Appendix, Fig. 4. 



