torsion (moment along the longitudinal...

Torsion (Moment along the longitudinal axis)

� In this section we will be studying

what happens to shafts under

torsional effects. We will limit

ourselves with elements having

circular cross sections.

� How do we calculate stresses at

the section under torsion, what is its

distribution, and what is angle of

twist?

� How do we analyze statically

indeterminate shafts?

TorsionTorsional deformation of circular shafts

� Torque or torsion is the moment which acts along the longitudinal axis of

the shaft. Torsion is the governing force in the design of vehicle shafts.

� We can investigate what happens to a shaft under torsional effects by

studying the shaft made of highly deformable material.


Before Deformation

After Deformation

Assumption:

�No local deformation occurs where the torsion acts on the

shaft (Saint Venant’s principle holds)

� If the angle of twist is small, then the length and the

diameter of the shaft remain the same (unchanged).


� Consider the shaft which is fixed at one end, and the torque of T is applied to the

other end (free end). Notice that the dark-green longitudinal plane is deformed as

follows:

�The radial line at a distance

x from the fixed end rotates

about an angle ϕ(x).

� The angle ϕ(x) is called the

angle of twist, and it is a

function of the distance x.

� ϕ(x) increases with

distance x.


� In order to see what T does to the material, let’s look at a small element at a radial

distance ρ (rho) from its center :

� If the back face rotates ϕ(x), then the

front face rotates about ϕ(x) + Δϕ,

� The difference Δϕ between the faces

exerts on the element shear deformations,

� In order to calculate this deformation,

one needs to calculate the angle change

between the sides AB and AC.

� Before deformation, the angle between

them was 90o then the angle becomes θ’:

θ’

lim2 C A

B A

πγ θ

→→

′= −

γ

Torsion generated shear

deformations.


� The angle γ (gama) can be related to Δx (the segment’s thickness) and the angle Δϕ,

� At the limit Δx -> dx and Δϕ -> dϕ, then we can write the following:

BD d dxρ φ γ= =

Therefore

d

dxγ ρ

φ=

Notice that the term in red fonts is the same for every point at the section; it can be

said that it is constant. Therefore, the shear deformation is proportional only with the

radial distance ρ. In other words, shear deformations linearly change with the radial

distance.


� The following expressions can be written:

max

max

d

dx c

c

γφ γρ

ργ γ

= =

=

The shear strains on a cross

section

increase linearly with ρ.

d

dx

φ: constant


TorsionTorsion Formula

� In this section, we will develop a relationship between internal torsional moment

and shear stresses.

� Shear strains are caused by the shear stresses. Notice that since the change in

shear strains is linear therefore the shear stress change will be linear as well.

If we assume that the material remains

linear elastic under torsional effects, we

can write the following based on the

Hooke’s law:

Gτ γ=

By referring to the figure on the left in a

full circular cross-section, the shear

stresses would start from zero and

linearly increases with the radial

distance, and becomes the largest at

the periphery .

� From the Hooke’s law, also the following expression can be written:

max

c

ρτ τ =

This expression emphasizes that also the

shear stress is a function of ρ.

� Due to the equilibrium, internal torque must be equal to the external torque T.

We can write the force on a infinitesimally small area dA as dF = (τ)dA. The

torque due to this force is dT = ρ(τdA). If we integrate that

( ) 2max

maxA A A

T dA dA dAc c

τρρ τ ρ τ ρ = = =

∫ ∫ ∫


2max

A

T dAc

τρ= ∫

� The integral given above is related to the geometry of the cross-section, and

it is known as the polar moment of inertia of the cross-section. This value is

designated as J. The formula can be rewritten as follows:

max

TcJ

τ =

Maximum shear

stress at the

cross-section

Internal torque at

the cross-section

Radius of the

cross-section.


� By using the formulas given below, we can derive a general expression to

find shear stress at any point on the cross-section.

max

c

ρτ τ =

max

TcJ

τ =

T

Jτ ρ=

This formula is known to be the torsion formula.

It can be used for circular shafts and

if the material is homogenous and linear elastic.


TorsionPolar moment of inertia (for full circular sections)

� Polar moment of inertia of a full circular cross-section can be found as follows:

( )2 2 3

0

4

02

2 2

c c

A

d cJ A d dρ ρπ

πρ ρ π ρ ρ= = = =∫ ∫ ∫

TorsionPolar Moment of Inertia (Circular tube type cross-

sections)

� For circular cross-section with an hollow part (tube type) with inner and outer

radius ci and co, respectively, the polar moment of inertia can be found as follows:

( )4 4

2o i

J c cπ

= −

Shear stresses increases

linearly along the radial

path.

Failure due to Torsion

EXAMPLE 13.1

Example 5.3

� Free Body Diagram

� Polar moment of inertia of the hollow tube:

Example 5.3 (larger figures etc.)

� Shear Stresses:

At the outer most radial distance (@co):

At the inner radial distance (@ci):

EXAMPLE 13.1 Hint:

P = T × ω

where

P: power in Watt (N.m/s)

T: Torque in N.m

ω: frequency in rad/sec

EXAMPLE 13.1

torsion (moment along the longitudinal...

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