CHAP. VI.] SUPPLEMENT TO CHAP. XIV. 309 



any desired portion of the span. As we shall in every case suppose the 

 load to extend from the left end up to any point, we shall take the limits 

 of z = a and 2, and therefore obtain 





For z = a, this is zero, as it should be, since then the load has not come 

 on. For = a, the load extends over the whole span, and V =pa, or 

 half the whole load, as it should. We might have obtained this result at 

 once by moments. Thus, 



_,. ' n ,, a z. Sa t 2az s t 



V x 2a=p (a e) (a + z-\ -- ) = - 



a 2 



but have preferred the above method as showing how uniform loading is 

 deduced directly from concentrated by inserting pdz for P and inte- 

 grating. 



29. Horizontal Thrust. In precisely similar manner we have 



from (21), Art. 4, for the thrust due to concentrated load P, H = P ^ s '. 



& fl 



Put P =p d z and integrate between the limits z = a and 2, and we have 

 m=p a '-**' + * ...... (72) 



For z = a, this is zero, as should be. For z = a, or for full load over 

 whole span, H = ^rj-. We may also deduce the above equation directly 



a III 



by moments. 



The above formulae (71) and (72) are all that we need either for calcula- 

 tion or diagram. They apply evidently equally well, whether the arch be 

 circular or parabolic, or, in general, whatever its shape may be. The form 

 has no influence upon either the thrust or the reaction. 



For the moment at any point whatever, whose distance horizontally 

 from crown is * and vertically below crown y, we have at once 



M = H (h - y) - V (a- as) + (a - x)*. 



2 



If this point is an apex, then the moment divided by depth of arch at 

 this point is the strain in flange opposite that apex. A positive moment 

 throughout this work always indicates compression in the inner or lower 

 flange. 



B. ARCH HINGED AT ENDS ONLY. 



30. Reaction. The vertical reaction at the end is precisely the same 

 as before for three hinges, and is given by equation (71). This reaction is 

 evidently independent of the shape of the arch, and the above formulae 

 holds good generally. 



31. Horizontal Thrut Parabolic Arch. We must here 

 distinguish the shape of the arch, and treat first the parabola. We have 

 already from eq. (27), Chapter HI., Art. 8, for a single load, 



H _ 5 _ 5 a 4 - 6 a* z 1 + * 

 64 " a 8 A ' 



