
MECHANICS  THEORY

Torsion of a Circular Bar 

Torsion, like a linear force, will produce both stress and the strain. However,
unlike linear stress and strain, torsion causes a twisting stress, called shear
stress (τ), and a rotation, called shear strain (γ).
The animation at the left illustrates as the torsion moment increases, the shear
strain (angle γ) changes linearly. This concept
will help develop equations that can be used to solve for the shear stress, strain
and even the angle of twist. 



Torsional Stress and Strain

The Change in Angle, γ, and
the Angle of Rotation, θ 

If a torque (or moment) is applied to the end of a circular bar as shown, the
bar will twist an angle θ. This angle will be a
function of the bar length, L, and stiffness, G (shear modulus).
The twist angle, θ, starts at 0 and increases linearly
as a function of x. On the other hand, the change of angle, γ,
is constant along the length.
If a small differential element, dx, is sliced from the bar, the two angles
must be compatible at the outside edge (arc length AA in red). This gives the
relationship,
Arc Length AA = r dθ = γ dx
The change of angle, γ, is also the shear strain.
Using Hooke's Law, the this angle is
γ = τ/G
Combining these two equations gives,
τ = G r dθ/dx
Generally, the rate of twist, dθ/dx, is not used
but it is related to the torque, T. The following section will use this fact
to develop a more useful form of this equation. 





Angle of Twist

Shear Stress Distribution Due
to Torque on a Circular Bar 

The first step is finding a relationship between the rate of twist, dθ/dx
and the applied torque, T. From the previous equation, τ =
G r dθ/dx, the shear stress is a linear function
of the radius.
The shear stress on all the differential elements, dA, generate a moment around
the center,
dM = r (τ dA)
All these differential moments must total the applied moment or torque, T, giving
Substituting the shear stress, τ, equations gives,
Note that G and dθ/dx are constant, giving
The integral term is a special form that is commonly seen in mechanic topics
and has a special name, the polar moment of inertia, J. The equation for J depends
only on the geometry and is commonly listed in engineering handbooks.
Using the term J and rearranging, gives
This can now be integrated along its length to find the total twist angle.
If the torque, stiffness, and cross sectional area are constant, then this simplifies
to




Angle of Twist for Shaft Constructed
from Various
Segments 

In cases when the circular shaft has different crosssections along its length,
the angle of twist equation can be applied on each section, and then summed.
Each segment can have a different load, length, stiffness and crosssection
area. The total twist is simply the sum of all the individual segment twist angles. 




Torsional Shear Stress



The equation for the rate of twist, dθ/dx = T/(GJ),
can also be combined into the shear stress equation,
τ = G r dθ/dx,
to give the torsional shear stress as function of the radius. This simplifies to just,
This relationship assumes the G, J, and T are constant along the rod length. 





Polar Moment of Inertia for Circular Rods

Polar Moment of Inertia Calculation 

Recall, the polar moment of inertia is defined as
For a circular cross section, dA is the radius times the element thickness,
giving dA = 2πr dr. Substituting into the integral
gives,
Solving gives,




Polar Moment of Inertia for
a Hollow Shaft 

This relationship can also be used for a hollow shaft. Simply use the outside radius, r_{o}, to find the polar moment of inertia for a solid shaft, and then subtract the polar moment of inertia from the hollow section using the inside
radius, r_{i}. The relationship becomes,






If the difference between r_{o} and r_{i} is small, less than 5% of the r_{i}, then the tube can be considered a thinwalled tube. The polar moment for thin tubes can be approximated by
J = 2πr^{3}t (t << r)
where r is generally assumed to be the average radius of r_{o} and r_{i}. 



