92 Mr. C. Chree on Bars and 



Combining these two we should get, to determine A or B, a 

 differential equation of the second order. To solve this in all 

 its generality might be a very difficult process ; but when the 

 variation in the bar is small, an approximate solution is easily 

 obtained. We should only have to regard A and B as con- 

 stants when multiplied by -j- and — ~ -, i.e. on the right- 

 hand side of the above equations, and (m + n)a as constant 

 on the left-hand side. The variables m, n, &c. being ex- 

 panded in Fourier's series, there would be only simple 

 integrations to carry out. 



As a special and Very interesting case, we may suppose the 

 constitution of the bar to vary regularly from one end to the 

 other. Thus, let us assume 



■P =>o(l + «**), " 



m = m n (l + vz)i ^ (51) 



,=?n (l+pz), V 

 = n (l + qz); J 



where p , m 0) n are the values at the end = 0, while a f , p, q 



are constants so small that terms in (a f l) 2 &c. mav be neglected. 



Thus, 



M: 



m ( dm — n ) 



and 



{m + n)ot,= (m +n )a \ 1 + ^ j o* 



p(6m 2 — 3m n —n 2 )—qm (3 ni — 5n ) ) 1 

 (?n + n ){3m — n ) ) J _ 



For shortness write these in the form 



« = a (l+Kz), | ^ ^ 



{m + n)a=(ni + n )a (l + Hz), J 



where the values of K and H are those of the last expressions. 

 Employing these values in (49) and (50), with the simplifica- 

 tions obtained by writing A for A &c. in the coefficients of 

 small terms, we get 



= — \ — A (l — cos 2a z) f B sin (2a z) \ — KB o « o 0. 



dz 



dl 



dz 



d - = i? { A sin 2a z + B (1 - cos 2«o*) \ + K A a z. 



