Closed and Open Tubes. 533 



since the return lines are merely the outward Hues continued. 

 The boundary itself is one shear line, and the stresses at 

 P ami P' act in opposite directions. From the preceding 

 argument it is easy to convince oneself that a piece such 

 as PP'Q'Q is stressed in much the same way as a piece ofc' a 

 long rectangle. Let us then rind the stresses in an infinitely 

 long rectangle. Let the //-axis be taken parallel to the 

 dimension which has an infinite length and along the middle 

 of the strip. Then the shear lines are, except near the ends 

 of the rectangle, parallel to the //-axis. Therefore ' 



Integrating this and dividing by n } we get 



/(.i') being any function of x. In consequence of the 

 equation 



Ox OjJ 



f(x) must satisfy the equation 



whence we get 



Therefore 



^=i T (// 2 - l u 2 ) + M l r-! r N. 



If t is the width of the rectangle the boundary conditions 

 along the two sides are 



*-W+(i0 2 }=c, 



which makes yjr the same along the two boundaries. 

 This shows that the constant M is zero. Then 



f = K/-^ 2 )+^ (48) 



Thus the resultant shear stress is 



= 2nrx (49) 



This gives the shear stress at any point of a long thin 

 rectangle not too near the end.-. 



