130 STRESSES IN TWO-HINGED ARCH 



A* P 1 U L P 1 UL 



A = 2 - - =H3 (68) 



Now for equilibrium, the values of A as given in equations (67) 

 and (68) must be equal, and we have, after solving for H 



P U L 



2 r (69) 



which is an expression for computing the horizontal thrust in any two- 

 hinged arch due to external loads. This formula holds for any system 

 of loading as long as P is the unit stress due to that loading, U is the 

 stress in the member and P 1 is the unit stress in the member due to a 

 unit load acting at the point at which the deformation is desired, and 

 parallel to the direction in which the deformation is to be measured. 



The method of finding the correct value of the horizontal reaction, 

 H, is as follows: (i) calculate the stresses in the arch for the given 

 loading on the assumption that it is a simple truss with one end sup- 

 ported on frictionless rollers ; (2) calculate the stresses in the arch for 

 an assumed horizontal reaction, H 1 :=, say, 20000 Ibs. on the assumption 

 that it is a simple truss on frictionless rollers ; (3) calculate the defor- 

 mation, A. of the free end of the truss for the given loads by means 

 of formula (67) ; (4) calculate the deformation, A' of the free end of 

 the truss for the assumed horizontal reaction H 1 = 20000 Ibs. by means 

 of formula (68) . The true value of H is then by formula (69) given 

 by the proportion 



H\ H 1 :: A: A' 

 * H 1 A 20000 A (70) 



-[ ^S . ,. __ m . __ 



The calculation of the horizontal reaction, H, and the stresses can 

 be made by algebraic methods alone or by a combination of graphic 

 and algebraic methods. The first requires less work, while the second 



