CHAP. V.] SUPPLEMENT TO CHAP. XTV. 305 



J 



24. Arch with three Hinges. If there are three hinges, the mo- 

 ment Mo at the crown must be zero, and therefore M = Hr(l cos $). 

 But for = , M must also be zero, hence H r (1 cos a) = 0, and therefore 

 H is zero. Then for any point G = 0, and M = 0, and N = 0. That is, 

 for the arch with three hinges there are for a change of temperature no outer 

 forces, and hence no strains. 



25. Arch hinged at Ends. Here, since for $ = a, M = 0, we 

 have from (54) 



Mo = Hr (1 cos a) \ 



M = - H r (cos <j> - cos a) V ..... (58) 



From (56), since f or ^ = 0, A = 0, 



A9 = ~*|r / (cos^-coscO^ + ^LS !? /<^..(59) 

 Jo EA Jo 



From (57), since f or <j> = 0, A * = 0, 



r* c* i 



(1 cos0) I (co8<j>coaa)d<j> I (1 cos 0) (oos^ coaa)d<j> I 

 ft 



i / d*- 



Jo 



H r cos a cos 



For = a, this becomes zero, and we have for the horizontal thrust 



sin a 



r* / " r 2 cosa I*. cos* a / ' 



= I cos $ (cos <j> cos a) a $ = I (cos <j> cos a)a^H ^ I i 



* / I / A / 



c/0 /O t/0 



.... (60) 

 Performing the integrations indicated (A?t. 7), and putting, for brevity, 



ic = - t we have 

 Ar* 



H= _ 2EIf<sina _ .. (61) 

 r 2 (a 8 sin a COS a + 2 a cos* a) + 2 icr* a cos* a 



By series (Art. 20), we obtain the approximate formula 



. . 



s 151 



The above are the expressions given in Art. 165 without proof. The less 

 A, the greater for equal dimensions is H. For h = 0, we have 

 H = E A e , as we should have for a straight beam. 



26. Arch without Hinges. In this case we have the general equa- 

 tions (54) and (55), which apply directly without change. 



From (56), since for = 0, A $ = 0, we have 



M r A Hr* ** 



20 



