600 Dr. C. Chree : Applications of 



vanishes at the centre. It also vanishes at a point (real or 

 imaginary) between the centre and support whose abscissa is 

 given by 



A« s = Z{3(2a/Z-l)}* (74) 



To apply to our problem, x 2 must be real and less than a ; 

 this is true only when a/l lies between 0'5 and 3— n/6, or 

 •5505. Between the support and the end, dy/dx vanishes for 

 a point whose abscissa is given by 



a- 3 = 1-1(1 -3a 2 // 2 ) 1 " ( 75 ) 



This applies to our problem only when «//lies between 3—^6 

 (or -5505) and VT/3 (or 0'57735). 



In all cases d 2 y/dx 2 vanishes at the end of the bar. It 

 vanishes at the centre when a/l=l/2. Between the centre 

 and the support it vanishes at a point whose abscissa is 

 given by 



* x =l(2a/!-l)* ; (76) 



and at this point we find 



[dy/dx) x ^ = - (wF/EflMc*) I (*i/*) 8 E- ('^ 3 /Ea>K 2 )l (2a//- 1)*. (77) 



These values of x and dyjdx are real, and apply to our 

 problem whenever a/l exceeds 1/2. When ajl>}/2, we have 

 dyjdx negative and d^yjdx 2 positive at x=a? 1 ; and it follows 

 that dyjdx, numerically considered, is here a maximum . When 

 a/l is less than 1/2, dyjdx is positive between the centre and 

 the support, and its value at the support is greater than at 

 any point nearer the centre. At the support we have 



(dyldx) a = (ivl 3 IRcofc*yi(ajl) { (ajl) 2 - 6(ajl) + $}.-. (78) 



This is positive or negative according as a/l is less or greater 

 than 0'5505. Beyond the support d 2 y/dx 2 can vanish only at 

 the end, where we have 



(dyjdx) l ={tcPj'Eco f c 2 )^\l-3{all) 2 \. . . (79) 



This is positive or negative according as a/l is less or greater 

 than 1/^3. When dyjdx is positive for a certain value of x,. 

 the total length of a longitudinal fibre between the central 

 section a? = and the section considered is increased when 

 the fibre lies above the neutral plane ; and the element of the 

 fibre answering to this value of x has also its length increased 

 if d 2 yjdx 2 be positive. If, on the other hand, while dyjdx is 

 positive, d' 2 y/dx 2 is negative, the element is shortened though 

 the total length is still increased. Answering to d 2 y/dx2z=0, 

 we get, included between the corresponding value of oc and 



