Torsion of Circular Cylinders edit

 
Torsion of a cylinder with a circular cross section

About the problem: edit

  • Circular Cylinder.
  • Centroidal axis thru the center of each c.s.
  • Length  , Outer radius  .
  • Applied torque  .
  • Angle of twist  .

Assumptions: edit

  • Each c.s. remains plane and undistorted.
  • Each c.s. rotates through the same angle.
  • No warping or change in shape.
  • Amount of displacement of each c.s. is proportional to distance from end.

Find: edit

  • Shear strains in the cylinder ( ).
  • Shear stress in the cylinder ( ).
  • Relation between torque ( ) and angle of twist ( ).
  • Relation between torque ( ) and shear stress ( ).

Solution: edit

If   is small, then

 

Therefore,

 

If the deformation is elastic,

 

Therefore,

 

The torque on each c.s. is given by

 

where   is the polar moment of inertia of the c.s.

 

Therefore,

 

and

 

Torsion of Non-Circular Cylinders edit

 
Torsion of a noncircular cylinder

About the problem edit

  • Solution first found by St. Venant.
  • Tractions at the ends are statically equivalent to equal and opposite torques  .
  • Lateral surfaces are traction-free.

Assumptions: edit

  • An axis passes through the center of twist (  axis).
  • Each c.s. projection on to the   plane rotates,but remains undistorted.
  • The rotation of each c.s. ( ) is proportional to  .  :  where   is the twist per unit length.
  • The out-of-plane distortion (warping) is the same for each c.s. and is proportional to  .

Find: edit

  • Torsional rigidity ( ).
  • Maximum shear stress.

Solution: edit

Displacements edit

 

where   is the { warping function}.\\ If   (small strain),

 

Strains edit

 

Therefore,

 

Stresses edit

 

Therefore,

 

Equilibrium edit

 

Therefore,

 

Internal Tractions edit

  • Normal to cross sections is  .
  • Normal traction  .
  • Projected shear traction is  .
  • Traction vector at a point in the cross section is { tangent} to the cross section.

Boundary Conditions on Lateral Surfaces edit

  • Lateral surface traction-free.
  • Unit normal to lateral surface appears as an in-plane unit normal to the boundary  .

We parameterize the boundary curve   using

 

The tangent vector to   is

 

The tractions   and   on the lateral surface are identically zero. However, to satisfy the BC  , we need

 

or,

 

Boundary Conditions on End Surfaces edit

The traction distribution is statically equivalent to the torque  . At  ,

 

Therefore,

 

From equilibrium,

 

Hence,

 

The Green-Riemann Theorem edit

If   and   then

 

with the integration direction such that   is to the left.

Applying the Green-Riemann theorem to equation (17), and using equation (16)

 

Similarly, we can show that  .   since  .

The moments about the   and   axes are also zero.

The moment about the   axis is

 

where   is the torsion constant. Since  , we have

 

If  , then  , the polar moment of inertia.

The Torsion Problem Summarized edit

  • Find a warping function   that is harmonic. and satisfies the traction BCs.
  • Compatibility is not an issue since we start with displacements.
  • The problem is independent of applied torque and the material properties of the cylinder.
  • So it is just a geometrical problem. Once   is known, we can calculate
    • The displacement field.
    • The stress field.
    • The twist per unit length.

Related Content edit

Introduction to Elasticity