Torsion

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Torsion of Circular Cylinders[edit | edit source]

Torsion of a cylinder with a circular cross section

About the problem:[edit | edit source]

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

Assumptions:[edit | edit source]

  • 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 | edit source]

  • 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 | edit source]

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 | edit source]

Torsion of a noncircular cylinder

About the problem[edit | edit source]

  • 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 | edit source]

  • 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 | edit source]

  • Torsional rigidity ().
  • Maximum shear stress.

Solution:[edit | edit source]

Displacements[edit | edit source]

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

Strains[edit | edit source]

Therefore,

Stresses[edit | edit source]

Therefore,

Equilibrium[edit | edit source]

Therefore,

Internal Tractions[edit | edit source]

  • 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 | edit source]

  • 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 | edit source]

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

Therefore,

From equilibrium,

Hence,

The Green-Riemann Theorem[edit | edit source]

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 | edit source]

  • 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 | edit source]

Introduction to Elasticity