EQUILIBRIUM OF AN ISOTROPIC ELASTIC ROD OF CIRCULAR SECTION. 923 



19. Solutions for a unit force at any internal point (/, w' , z'). 

 V force. Sp = S(p^) + SJ>. 

 S^p^^ consists of, for z>z', 



/■-, _ 1^-4 j/ \ 



^^' '''"4^/x{8-.)^^' 



(ii) U, = { y (^ - ^'f + ~{^ - ^')P' } COS (co - <.'), 

 Wa> = I - y (2 - zy + -^{^ - z')p'^ I sin (w - w'), 



1 



ir/x(8-v) a* 



(iii) 2/.p = (2 - 2') cos (o) - o)'), 

 M„ = - (2 - 2') sin (o) - oj'), 



Mj= — pCOs(a) — <!)'), 



I 



4-v p'2 v2-20v + 92 1 



27r/,t(8 - v) a* 37r/^(8 - vf (i^ 



SCO = J_ I ls<') + r.A - ^P ^V^') + f- - A)-,S(f) i . 



Itt/xv \ dp "^ \ dp dp dp, 



Q force. s„ = s;^'>+s;;^ 



S^^ consists of (for z';>z') 



(i) 21^ = -p'(2-2>, 



TTfx a* 



(ii) Up = I _(z - z'f + -^{z - 2>2 I sin (oj - o)'), 



M„ = I - 1(2 - 2')3 + 1^(2 - 2'y I ( - ) COS (a. - 0,'), 



M, = i - (2 - z'y-p + — p3 + M^p i sin ((o - co'), 



(2-2')sin(o)-o;),^ ^_ 



(2 -2') cos (a, -.'),[ {2;^^^^ + 



7r|ot(8 - v) a* ' 



2 a2 J 



(iii) Ma = - (2 - 2') sin (to - o/), "1 



' ' ' ( 4-v p'2 V2-20.. + 92 1 



Mm = - (Z - 2 ) cos (O) - O) ), W i—r -I- i-r r 



: , , ^ ' I) 27r/x(8 - v) a-4 37r/x(8 - v) 



u, = p sin (o) — (o ), I 



^(0 1/13 „(«) / V 3 \ 3 c;(0 / 9 1 9^ \q(«> I 



Z/orce?. Sz = Sf + 8^^'^ 

 S^^ consists of (for z'>z'). 



... 4-v 

 (1) "P= 4 P> 



•( ) ^ ^ 



^^. = 0, 



^ V(8-v)a2 



M, = 2 - 2', J 





in) lu = < (z - 2")" 



4-v , 1 , 



A 0'' > COSlcu 



, = I - (2 - 2')2 + ^p2 j sin ((o - w') . 



It^ = - 2(2 - 2')p cos (to - O)'), 



(iii) Up = cos (w - w'), 



M„ = - sin (co — o)'), 

 u, =0, 



,r/x(8 - v) a* 



/ 2 1/1 ,3 v-10 o A v^-20v + 92 p' ( 



i ~7rM8-v)^VT^ "^ 4 "^^y 3,rM« - v)2 a^i 



q(0 1 / 3q(0 ,3 d^(t) 3 3^(f)) 



^2=4^^1 37' * ~^V3^ ' ~9i;^'9^^^ ^ ■ ■ ■ 



(0 



(2) 



(3) 



(4) 



(5) 



(6) 



