,Twist and torsion stiffness
- Compound shafts -
Fixed and simply
Shaft: The shafts are the machine elements which are
used to transmit power in machines.
Moment: The twisting moment for any
section along the bar / shaft is defined to be the algebraic sum of the
moments of the applied couples that lie to one side of the section under
consideration. The choice of the side in any case is of course arbitrary.
Strain: If a generator a ?? b is
marked on the surface of the unloaded bar, then after the twisting
moment 'T' has been applied this line moves to ab'. The angle ???' measured in
radians, between the final and original positions of the generators is defined
as the shearing strain at the surface of the bar or shaft. The same definition
will hold at any interior point of the bar.
of Elasticity in shear: The ratio of
the shear stress to the shear strain is called the modulus of elasticity
in shear OR Modulus of Rigidity and in represented by the symbol
of Twist: If a shaft of length L is
subjected to a constant twisting moment T along its length, than the
angle ? through which one end of the bar will twist relative to the other is
known is the angle of twist.
the differences in the forms of loading, we see that there are number of
similarities between bending and torsion, including for example, a linear
variation of stresses and strain with position.
In torsion the members are
subjected to moments (couples) in planes normal to their axes.
the purpose of desiging a circular shaft to withstand a given torque, we must
develop an equation giving the relation between twisting moment, maximum shear
stress produced, and a quantity representing the size and shape of the
cross-sectional area of the shaft.
all torsion problems, involve rotating machinery, however, for example some
types of vehicle suspension system employ torsional springs. Indeed, even coil
springs are really curved members in torsion as shown in figure.
torque carrying engineering members are cylindrical in shape. Examples are
drive shafts, bolts and screw drivers.
Torsion Theory or Development of Torsion Formula : Here we are basically interested to
derive an equation between the relevant parameters
st Term: It refers to applied loading
ad a property of section, which in the instance is the polar second
moment of area.
2 nd Term: This refers to stress, and the stress
increases as the distance from the axis increases.
rd Term: it refers to the deformation
and contains the terms modulus of rigidity & combined term ( ??? l)
which is equivalent to strain for the purpose of designing a circular shaft to
with stand a given torque we must develop an equation giving the relation
between Twisting moments max m shear stain produced and a quantity representing
the size and shape of the cross ??sectional area of the shaft.
to the figure shown above where a uniform circular shaft is subjected to a
torque it can be shown that every section of the shaft is subjected to a state
of pure shear, the moment of resistance developed by the shear stresses being
every where equal to the magnitude, and opposite in sense, to the applied
torque. For the purpose of deriving a simple theory to describe the behavior of
shafts subjected to torque it is necessary make the following base assumptions.
(i) The materiel is homogenous i.e of uniform
elastic properties exists throughout the material.
(ii) The material is elastic, follows Hook's law,
with shear stress proportional to shear strain.
(iii) The stress does not exceed the elastic limit.
(iv)The circular section remains circular
(v) Cross section remain plane.
(vi)Cross section rotate as if rigid i.e. every diameter rotates through the
now the solid circular shaft of radius R subjected to a torque T at one end,
the other end being fixed Under the action of this torque a radial line at the
free end of the shaft twists through an angle ? , point A moves to B, and AB
subtends an angle ??? ' at the fixed end. This is then the angle of distortion
of the shaft i.e the shear strain.
Since angle in radius = arc /
arc AB = R?
= L ? [since L and ? also
constitute the arc AB]
Thus, ? = R? / L (1)
From the definition of
Modulus of rigidity or Modulus of elasticity in shear
Let us consider a small strip of radius r and
thickness dr which is subjected to shear stress??'.
The force set up on each
= stress x area
= ?' x 2? r dr (approximately)
This force will produce a moment or torque
about the center axis of the shaft.
= ?' . 2 ? r dr . r
= 2 ???' . r2. dr
The total torque T on the
section, will be the sum of all the contributions.
?' is a function of r, because it varies with radius so writing down??' in
terms of r from the equation (1).
T = applied external Torque,
which is constant over Length L;
J = Polar moment of Inertia
[ D = Outside diameter ; d =
inside diameter ]
G = Modules of rigidity (or
Modulus of elasticity in shear)
? = It is the angle of twist
in radians on a length L.
Stiffness: The tensional
stiffness k is defined as the torque per radius twist
i.e, k = T /???= GJ / L
stepped solid circular shaft is built in at its ends and subjected to an
externally applied torque. T0 at the shoulder as shown in the figure. Determine
the angle of rotation ?0 of the shoulder section where T0 is applied ?
This is a statically indeterminate system
because the shaft is built in at both ends. All that we can find from
the statics is that the sum of two reactive torque TA and TB at the built ?? in
ends of the shafts must be equal to the applied torque T0.
TA+ TB = T0 ------ (1)
[from static principles]
TA ,TB are the reactive torque at the built in ends A and B. wheeras T0 is the
From consideration of consistent deformation,
we see that the angle of twist in each portion of the shaft must be same.
i.e ?a = ? b = ? 0
using the relation for angle of twist
modulus of rigidity G to be same for the two portions
So the defines the ratio of TA and TB
So by solving (1) & (2) we get
Torsion: The pure torsion refers to a
torsion of a prismatic bar subjected to torques acting only at the ends.
While the non uniform torsion differs from pure torsion in a sense that the bar
/ shaft need not to be prismatic and the applied torques may vary along the
Here the shaft is made up of
two different segments of different diameters and having torques applied at
several cross sections. Each region of the bar between the applied loads
between changes in cross section is in pure torsion, hence the formula's
derived earlier may be applied. Then form the internal torque, maximum shear
stress and angle of rotation for each region can be calculated from the
The total angle to twist of
one end of the bar with respect to the other is obtained by summation using the
If either the torque or the
cross section changes continuously along the axis of the bar, then the ?
(summation can be replaced by an integral sign ( ? ). i.e We will have to
consider a differential element.
After considering the differential element, we
Substituting the expressions
for Tx and Jx at a distance x from the end of the bar, and then integrating
between the limits 0 to L, find the value of angle of twist may be determined.
Application to close-coiled helical springs
Closed Coiled helical springs subjected to
Definition: A spring may be defined as an elastic member
whose primary function is to deflect or distort under the action of
applied load; it recovers its original shape when load is released.
Springs are energy absorbing units whose
function is to store energy and to restore it slowly or rapidly depending on
the particular application.
Important types of springs are:
There are various types of springs such as
spring: They are made of wire coiled
into a helical form, the load being applied along the axis of the helix.
In these type of springs the major stresses is torsional shear stress due to
twisting. They are both used in tension and compression.
springs: They are made of flat strip
of metal wound in the form of spiral and loaded in torsion. In this the major
stresses are tensile and compression due to bending.
In this the
major stresses are tensile and compression due to bending.
springs: They are composed of flat
bars of varying lengths clamped together so as to obtain greater
efficiency . Leaf springs may be full elliptic, semi elliptic or cantilever
types, In these type of springs the major stresses which come into picture are
tensile & compressive.
These type of springs are used in the
automobile suspension system.
Uses of springs :
apply forces and to control motions as in brakes and clutches.
(b) To measure forces as in spring balance.
store energy as in clock springs.
(d) To reduce the effect of shock or impact
loading as in carriage springs.
change the vibrating characteristics of a member as inflexible mounting of
Derivation of the Formula :
In order to derive a necessary formula which
governs the behaviour of springs, consider a closed coiled spring subjected to
an axial load W.
W = axial load
D = mean coil diameter
d = diameter of spring wire
n = number of active coils
C = spring index = D / d For circular wires
l = length of spring wire
G = modulus of rigidity
x = deflection of spring
q = Angle of twist
when the spring is being subjected to an
axial load to the wire of the spring gets be twisted like a shaft.
If q is the total angle of
twist along the wire and x is the deflection of spring under the action of load
W along the axis of the coil, so that
x = D / 2 .
again l =
D n [ consider ,one half turn of a close coiled helical spring ]
Maximum shear stress in spring section
including Wahl Factor
Assumptions: (1) The Bending & shear effects may be
(2) For the
purpose of derivation of formula, the helix angle is considered to be so small
that it may be neglected.
Any one coil of a such a spring will be
assumed to lie in a plane which is nearly r to the axis of the
spring. This requires that adjoining coils be close together. With this
limitation, a section taken perpendicular to the axis the spring rod becomes
nearly vertical. Hence to maintain equilibrium of a segment of the spring, only
a shearing force V = F and Torque T = F. r are required at any X -
section. In the analysis of springs it is customary to assume that the shearing
stresses caused by the direct
shear force is uniformly distributed and is negligible
so applying the torsion
Using the torsion formula i.e