184] TORSION OF PRISMS. 485 



V, F 15 F 2 are uniquely determined, each vanishing at the origin. The 

 origin is any point on the line of centers of mass of the cross- 

 sections. 



The most general solution of de Saint -Venant's problem now 

 contains six constants, a, a 1? a 2 , /3, & 1? & 2 . We may examine separately 

 the corresponding simple strains, by putting in each case all the 

 constants but one equal to zero. 



I. a 4= 0. 



u = 



Z 2 = Ea, 



This is a stretch -squeeze of ratio ?y, which has been already treated 

 in 176. 



II. /3+0. 



Ill) u = pyz, v = pxs, w = pV. 



These equations represent a torsion, whose rotation 



is proportional to the distance from the origin, or fixed section. 



We have u = v = w = for x = y = the line of centers. For 

 the stresses we find 



112) Z.-E%-0, Z x = 



so that the stress on any cross -section is completely tangential. 

 If the contour of the bar is circular, we have 



cos (nx) : cos (ny) = x : y, 



dV 

 so that ^ n = and V = tv = 0, and accordingly we see that the 



plane cross -sections remain planes. For any other form of cross- 

 section than a circle, V does not vanish, so that the cross -sections 

 buckle into non- plane surfaces. This buckling was neglected in the 

 old Bernoulli -Euler theory of beams, and constitutes, as was shown 

 by de Saint -Venant, a serious defect in that theory, which is still 

 largely used by engineers. 



In order to produce this torsion, we should apply at the end 

 cross -section the stresses 



113) Z x =X, = iipy, Z y = Y z = -iipx, 



that is at every point a tangential stress perpendicular to the radius 

 vector of the point in the plane and proportional to its distance 

 from the line of centers. As in practice it would be impossible to 



