604 Dr. C. Chree: Applications of 



Between A and C, 



y = (W/E w * 2 )L(a 3 -3a 2 ^ + 3cx 2 -x*)A 

 %/^=(W/Ea)/c 2 )i(-a 2 + 2c^-^), I. . (87) 



d 2 y\dx 2 = ( W/Eo/e 2 ) (o- *). j 



Between C and B, 



y=(W/Ecotc 2 )±{a z -c d + Z{c 2 -a 2 )x}A 

 dylda±(WIE»i*Jl(<?-a*) 9 V. . (88) 



d 2 y/dx 2 = 0. ) 



To take account of the rod's own weight, we simply add 

 the values of y &c. given in (69) or (70) to the corresponding 

 values given above. 



If the weight W were between the centre and the support, 

 (86) would apply to the portion of the rod between the 

 centre and the weight, while (88) would apply to the portion 

 between the support and the end. For the portion between 

 the weight C and the support A we should have 



y=-(W/Ea>K 2 )^(3a^ 2 ~^ 3 -3A + c 3 ),*J 



dy/dx=-(W/Eo) K 2 )i(2ax-x 2 -c 2 ), \. . (89) 



d 2 y/dx 2 = - (W/Ea)* 2 ) {a-x) . ) 



In these last equations c is less than a. Equations (89) 

 are deducible from equations (87) by altering the sign of W 

 and interchanging a and c. We have spoken of W as a 

 iveight depending from the bar. By supposing W negative, 

 however, we have the effect of an upward pull, relieving part 

 of the weight of the bar. We may suppose this obtained 

 from a belt passing over a pulley fixed above the bar. 



§ 27. By the use of symmetrical auxiliary weights, we can 

 without altering the position of the supports vary the portion 

 of the bar whose length is unaffected by bending. Suppose, 

 for instance, the bar carried by rollers in the position 

 answering to a/Z = l/V3, for which the total length is un- 

 affected when the bar is unloaded. Then to secure that a 

 symmetrical central portion 2b shall have its length unaffected, 

 we select the auxiliary weights and their position so that one 

 of the following three equations holds : — 



W{c-a)+w{il 2 -al + ±b 2 \=Q, (90) 



W(-a 2 + 2cb-b 2 )+™.\l{l 2 -3a 2 )-{l-by\=0, (91) 



o 



W(c 2 -a 2 )+^Z(/ 2 -3a 2 )-(Z-6) 3 }=0. . . (92) 

 o 



