EQQILIBRItTM OF AN ISOTROPIC ELASTIC ROD OF CIRCULAR SECTION. 927 



zx = fj(2a+ \)xy, 



zz= - 2E«?/. 

 Mp = < -^ + z{a-p^ - Ta^) > sin o), 

 Mo. = ] - o^^ + ^(o-p^ + Ta^) /> ( - cos oj), 



% = 



Z^p -I- pi ._ / (J- 4- 1 _ 7. j|^2p j, g|,i ^^ 



I Sin CO, 



pp = 0, pco = 0, pz = /xf o- + '- j(p2 _ a^) i 



The integral stress is a ^/-force through the origin of amount 



j jzydS = - K'Ea\ 

 S2s(this solution) ^ . 



III. 



% = - (TX, 



u, = - a-y, 

 u, = z. 



Up= - o-p, , 



The only stress is 22 = E. 



The integral stress is a z-force through the origin of amount 



SdS = TrEa^. 



S3^(thissolution)(-2--I-^) 



IV. 



Mj. = 2a-x|/, 



M, = 22 + o-(?/2-a;2), 



M, = - 2zy. 



The only stress is 22 = — 2E?/. 



The integral stress is a yz-GOU])le of amount 



Mp = (z^ + o-p^) sin (0, 



M„ = ( - 2^ + o-p2)( - cos Oj), 

 M^ = - 2zp sin w. 



{yzz - zzy)dS — — — TrEa*. 



S4s(this solution) 



rEa* 



V. 



M^ = 22 + o-(a;2-?/2), 



Uy^lvxy, 



u,= - 2zx. 



Up = (^2 + crp2) cos w, 

 Ww = ( - 2^ + crp2) sin 

 u^ = - 2r;p cos w. 



r 



The only stress is zz= - 2Ea;. 



TRANS. ROY. SCO. EDIN., VOL. XLIX. PART IV. (NO. 17). 



127 



