Torsion
Method of Sections
Torsion of Circular Elastic Bars
Basic Assumptions for Circular Members
• A plane section of material perpendicular to the axis of a circular member remains
after the torque is applied (no warping or distortion of parallel planes normal to the
axis of member takes place).
• In a circular member subject to a torque, shear strains 𝛾 vary linearly from the
central axis reaching 𝛾𝑚𝑎𝑥 at the periphery.
- This assumptions hold only for circular and tubular members. For this class of
members, these assumptions work so well that they apply beyond the limit of the
elastic behaviour.
Torsion formula
• Internal resisting torque caused by shear stresses must be in equilibrium with
externally applied torque 𝑇.
• In linear elastic range, shear stresses vary linearly from the central axis of the
member, since stress is proportional to strain, and the latter varies linearly from
the center.
Moment of Inertia of cross-sectional area:
A more general relation for shear stress at any point on cross-section at distance 𝜌 from
center can be written as:
• Application to Circular tubes:
Circular bar made from two different materials
• Two materials are bonded together at the interface; therefore, the same strain assumption
applies as for a solid member.
• The ratio of the shear stresses at the interface is equal to the ratio of the Shear Modulus for
inner and outer cores.
-No interacting shear force/stress is developed between the cores at the interface (illustrated
schematically in the next slide).
Shear stress orientation in pure torsion:
Torsion failure surfaces
Example:
The maximum internal torque is 30 Nm
Example (4-3):
Design of circular members in torsion
For a shaft rotating with a frequency of 𝑓 𝐻𝑧 and transmitting a constant torque 𝑇,
the work dene per second is: (1 ℎ𝑝 = 745.7 𝑊)
Example:
Stress Concentration
Any abrupt change in geometry results in stress concentration (large perturbation of
shear stresses).
- The stress is determined for the smaller shaft
- Shear stress at the step above the corner is
entirely zero.
Angle-of-Twist of Circular Members
Substituting into Eq. 4-13
Example:
• Great similarity is observed between the above torsion equation and the one we had
for uniaxial load-displacement.
• Torsion spring
Example:
Example:
- On every imaginary cylindrical surface with radius r
the applied torque is resisted by a uniformly distributed
shear stress.
- Shear strain is not linearly
varying with radius, unlike the torsion problem
- Geometric relation:
- What happens if the left surface of the rubber
bushing is fixed?!
Statically Indeterminate Problems
Global Equilibrium:
Geometric Compatibility
• The problem can be temporarily reduced to statically determinate problem by removing
one of the redundant supports. Then the required Boundary Condition is restored by
twisting the released end based on the superposition concept.
Strain Energy due to torsion
Elastic Strain Energy:
• Strain Energy Density (per volume)
• By using Hook’s law:
• Elastic Strain energy of a bar under torsion:
- If the external twisting torque is gradually applied to the bar, the external work is:
- From equality of the internal stored elastic strain energy and the externally applied
work:
Torsion of Inelastic Circular Bars
• As in elastic case, the twisting moment equilibrium at a cross section and the
deformation assumption of linear strain variation from the axis remains applicable.
• Only the difference in material properties affect the solution.
• After the stress variation is known, the resisting torque can be calculated as:
• The geometric strain-rotation relation is also applicable:
Example:
• The applied torque can be calculated using the moment equilibrium equation:
• Shear stress distribution due to elastic unloading:
• The initial rotation might be calculated from the twist of the elastic core.
at 𝜌 = 4𝑚𝑚: 𝛾 = 2 × 10−3
- For loading:
- For unloading:
- Residual twist:
Torsion of Solid Noncircular Members
• The two assumptions considered for circular members do not apply.
• Analytical solutions have been obtained for rectangular elastic members.
• Maximum shear stress and angle-of-twist relations:
- 𝑏 is the length of the long side and 𝑡 is the thickness of the cross section.
Torsion of Thin-Walled Tubular Members
• The product of wall thickness and shear stress is constant which is called
Shear Flow (𝑞).
• Equilibrium of the twisting moment:
- 𝐴 is the area enclosed by the center line of the wall’s contour.
- Eq. (4-34) is not applicable if the tube is slit. Eq. (4-30) might be used in that case.
• The angle of twist can be obtained as below for elastic material:
• Elastic shear strain energy:
• External work per unit length:
• Equilibrating the strain energy and the external work:
Example: