The radial line CA will rotate to a new position CA’.  The angle df is the angle of twist of the cross section and the deformation angle of the line PA is the max shear strain, g_max .  The arc length AA’ can be written two ways.

 

             | AA’ | = dx  g_max  = R df

 

 

 

If we consider a point B a distance r from the center, it moves to position B’.  Because the angle df if the same, the shear strain of an element on this smaller cylinder will be smaller.  We will call this angle g .  The arc length BB’ can be written as

 

             | BB’ | = dx  g  = r df

 

From these equations, we see two expressions for  df/dx .

 

             df/dx =  g_max / R =  g / r

 

So,        g  =  g_max  ( r / R )

 

From this we see that shear strain varies linearly with the distance r measured from the center of the cylindrical member.  Now, assume that the material remains within the linear, elastic range of its behavior.  We can apply Hooke’s Law.

 

             t = G g = G g_max ( r / R ) = t_max ( r / R )

 

We see then that the shear stress varies linearly from the center of the cross section.  Now, lets remember that the stress distribution on the cross section must be statically equivalent to the net torque, T.

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