Assume a shear stress t₁ at the corner of the thickness t₁ and a shear stress of t₂ at the corner of thickness t₂ .  Because shear stress always occurs in a set of four stresses, we know the stress in the wall along the length of our piece.

 

Where the thickness is t₁ , the shear stress is t₁, and where the thickness is t₂ , the shear stress is t₂ .  Writing  S F_x = 0,

 

             t₁ t₁ L - t₂ t₂ L = 0                     or                       t₁ t₁ = t₂ t₂

 

We made no assumptions about the size of the piece so L and s are arbitrary.  This means that

 

             t t = constant

 

We define this product as a new quantity called shear flow, q .  Note this is a force per unit length.

 

Looking back at the cross section, the shear flow around the circumference of the wall must be statically equivalent to the applied torque.

 

 

Over a small piece of the circumference, q creates a net force

 

             dF = q ds

 

which in turn creates a torque around the axis of the shaft

 

             dT = q ds h

 

 

The total torque is then found by integrating around the circumference.

 

             T =   ∫  dT =   ∫  q h ds

 

But q is constant, so

 

             T = q  ∫  h ds

 

 

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