Theory of Elasticity. 279 



pal sections. Under these circumstances we find by considera- 

 tion of the equations of equilibrium* that the radial stress will 

 fall off from the interior to the exterior according to a simple 

 hyperbolic law. The ordinary solution of this problem, based 

 on the supposition of no stress when there is no pressure 

 on the tube, results in a circular stress much greater at the 

 interior than at the exterior, a varying radial stress, and an 

 axial stress constant at all distances. 



If we examine the strains we find the actual strainsf to be a 

 circular extension decreasing from the inside out, a radial con- 

 traction decreasing from the inside out and an* axial strain, 

 changing from extension at the inside to contraction at the out- 

 side. The shearing strains are all zero. 



*Let p x be the pressure inside the cylinder and Q the circular or tangential 

 stress constant at all distances from the axis. Let R, the axial stress, be the 

 same as the axial stress on the supposition of no primary stress, or R = R and 

 let the shearing stresses S T U all be zero. Then the conditions of equilibrium 

 simply require that the radial stress P shall satisfy the differential equation (in 

 circular coordinates) 



dP P—Q , . „ .,. . Pi= — Pi (at inner surface) 



1 = and the surface conditions „ „ „ 



dr . r P = (at outer surface) 



These yield at once the relation P = Q l\ J (r being the outer radius) 



whence Pi = —p x =Q ( l — -- ) (r x being the inner radius) furnishes Q =p x — — 



and P =»i —, °—. The ordinary solution for no primary stress (see Love, 



r{r -ry) 



Theo. of Bias., vol. I, art. 130) 



Aft 7*i Tit / I V* n \ 



is e ^ r—= — ( — : — — ) (radial strain — actually contraction) 



or 7V— ?i \ 2(A + /u) 2/z r 2 / 



f = — = — r~(— 7^ 1- — ) (tangential strain — actually extension) 



r r„ 2 — ?v \2(A + ju) 2u r 2 / 



g = (axial strain) a = b = c = (shearing strains) 



Tl'Pi 1 



A =e+f+g = — —i- — (dilatation — positive and constant) 



?V— n 2 A + fi 



P= AA + 2fj,e= I (\ J (= — »j at r = r x of course) 



T 2 —r{- \ r 2 / 



r^—ri' \ r 2 / 



Ti^Vi A 9'i^Pl 



R — A& + 2aq = — — — = 2 a — (a constant tension) 



™ ro *-r } * % + n rf-rj 1 v 



f Our actual strains are — 



P 7? + R~ px r x f „ Tx r \ . ,. . , . . 



s= a — = — — - — ( I — a— 2a- ) (radial strain — actu- 



E E E r — ri\ ro + rx r / 



ally contraction) 



