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Chapter-7 Torsion

Torque is a moment that tends to twist a member

about its longitudinal axis. Its effects is of primary

concern in the design of axles or drive shafts used

in vehicles and machinery.

We can illustrate physically what happens when a torque is applied to a

circular shaft by considering the shaft to be made of a highly

deformable material such as a rubber, Figure (a). When the torque is

applied, the circles and longitudinal grid lines originally marked on the

shaft tend to distort into the pattern shown in figure (b).

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Torsional deformation of a circular shaft

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By inspection, twisting causes the circles to remain circles, and each longitudinal

grid line deforms into a helix that intersects the circles at equal angles. Also, the

cross sections at the ends of the shaft remain flat and radial lines on these ends

remain straight during the deformation.

From these observations we can assume

that if the angle of rotation is small, the

length of the shaft and its radius will remain

unchanged.

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If the shaft is fixed at one end

and a torque is applied to its

other end, the shaded plane in

Fig c will distort into a skewed

form as shown. Here a radial line

located on the cross section at a

distance x from the fixed end of

the shaft will rotate through an

angle Φ(x). The angle Φ(x) is

called the angle of twist. It

depends on the position x and

will vary along the shaft as

shown.

Fig c

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In order to understand how this distortion strains the

material, we will now isolate a small element located

at a radial distance ρ from the axis of the shaft. Due

to the deformation, the front and rear faces of the

element will undergo a rotation. As a result, the

difference in these rotations, ΔΦ, causes the

element to be subjected to a shear strain.

To calculate this strain, note that before deformation

the angle between the edges AB and AC is 90Ο ;

after deformation, however, the edges of the element

are AD and AC and the angle between them is θ’.

'lim2

BAalongABCAalongAC

dxdBD

Therefore: dx

d

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Since dx and dΦ are the same for all elements located at points on the cross section

at x, then dΦ/dx is constant over the cross section, and this equation states the

magnitude of the shear strain for any of these elements varies only with its radial

distance ρ from the axis of the shaft. In other words, the shear strain within the shaft

varies linearly along any radial line, from zero at the axis of the shaft to a maximum

γmax at its outer boundary, Fig e. Since dΦ/dx = γ/ρ= γmax/c, then:

Fig e

max

c

The shear strain for the material

increases linearly with ρ.

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The torsion formula:

If the material is linear-elastic, then Hooke’s law applies, G and consequently

a linear variation in shear strain leads to a corresponding linear variation in shear

stress along any radial line on the cross section. We can write:

max

c

This equation expresses the shear-stress distribution as a function of the radial

position ρ of the element; in other words, it defines the stress distribution over the

cross section in terms of the geometry of the shaft. Using it, we will now apply the

condition that requires the torque produced by the stress distribution over the entire

cross section to be equivalent to the resultant internal torque T at the section,

which holds the shaft in equilibrium.

Each element of area dA, located at ρ, is subjected to a force of dAdF

The torque produced by this force is: dAdT

We therefore have for the entire cross section:

AA

dAc

dAT max

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Since: c

maxis constant,

A

dAc

T 2max

The integral to this equation depends only on the geometry of the shaft. It

represents the polar moment of inertia of the shaft’s cross sectional area computed

about the shaft’s longitudinal axis. We will symbolise its value as J, and therefore

the above equation can be written in a more compact form, namely,

J

cT max

Where: max = the maximum shear stress in the shaft, which occurs at the outer

surface.

T= the resultant internal torque acting at the cross section. Its value is

determined from the method of sections and the equation of

moment equilibrium applied about the shaft’s longitudinal axis.

J = the polar moment of inertia of the cross-sectional area.

c = the outer radius of the shaft.

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The shear stress at the intermediate distance ρ can be determined from a similar

equation:

J

T

“Torsion formula”

Solid shaft: 4

2cJ

Tabular shaft: If a shaft has a tabular cross section, with inner radius ci and

outer radius co, then the polar moment of inertia can be determined by

subtracting J for a shaft of radius ci from that determined for a shaft of radius

co. The result is:

44

02

iccJ

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Solution:

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Solution:

44

02

iccJ o = Outer

i = Inner

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Angle of twist:

Occasionally the design of a shaft depends on restricting the amount of rotation or

twist that may occur when the shaft is subjected to a torque.

T(x)

Angle of twist:

dx

GxJ

xTL

0

Here:

Φ = the angle of twist of one end of the

shaft with respect to the other end,

measured in radians.

T(x) = the internal torque at the arbitrary

position x, found from the method of

sections and the equation of moment

equilibrium applied about the shaft’s axis.

J(x) = the shaft’s polar moment of inertia

expressed as a function of position x.

G = the shear modulus of elasticity for the

material.

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Constant torque and cross-sectional area:

Usually in engineering practice the material is homogeneous so that G is constant.

Also, the shaft’s cross-sectional area and the applied torque are constant along the

length of the shaft.

In this case:

T(x) =T

J(x) = J

GJ

LT

.

.

If the shaft is subjected to several different torques, or the cross-sectional area or

shear modulus changes abruptly from one region of the shaft to the next then the

angle of twist from one end of the shaft with respect to the other is then found from

the vector addition of the angles of twist of each segment.

GJ

LT

.

.

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Sign convention:

In order to apply the above equation, we must develop a sign convention for the

internal torque and the angle of twist of one end of the shaft with respect to the other

end. To do this, we will use the right-hand rule, whereby both the torque and angle

will be positive, provided the thumb is directed outward from the shaft when the

fingers curl to give the tendency for rotation.

Using the method of sections, the internal torques are found for each segment.

GJ

LT

GJ

LT

GJ

LT CDBCABDA

.

.

.

.

.

. 321/

15

Table

ΦB/A =

Solution:

BC

CD DA

+X

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Stress concentration

J

cT maxThe torsion formula, can be applied to regions of a shaft having a

circular cross section that is constant or tapers slightly.

When sudden changes arise in the cross section, both the shear-stress and

shear-strain distributions in the shaft become complex and can be obtained

only by using experimental methods or possibly by a mathematical analysis

based on the theory of elasticity.

In order to eliminate the necessity for the engineer to perform a complex stress

analysis at a shaft discontinuity, the maximum shear stress can be determined

for a specified geometry using a torsional stress-concentration factor K.

23 J

cTK

..max