Transcript Shafts - Srm University

PH0101 UNIT 1 LECTURE 2
• Shafts
• Torsion Pendulum-Theory and Uses
• Worked Problem
PH0101 UNIT 1 LECTURE 2
1
Shafts
• Any rotating member which is transmitting
torque is called shaft.
• It is an arrangement for the transmission of
a couple applied at one end to appear at the
other end without any appreciable twist in it.
• For example, in automobile such as buses,
lorries and vans, the driving shaft (axle)
transmits the power torque form engine to
the wheel.
PH0101 UNIT 1 LECTURE 2
2
Requirements for a good shaft
• Shaft must transmit the couple without any appreciable twist
in it
• The twist in the shaft should be very small when the large
couple is applied to it.
• It is found that the efficiency of a shaft varies as
πNr4
2l
where N - rigidity modulus of the material
r - radius of shaft
l - length of shaft
PH0101 UNIT 1 LECTURE 2
3
• Hence, it is better to use shafts of large
radius and the material is of high rigidity
modulus.
• Hollow shaft is stronger and better than
the solid shaft of the same length, mass
and material.
PH0101 UNIT 1 LECTURE 2
4
Note:Hollow shaft is better than solid shaft
• If two cylinders of the same length and same
mass and of the same material are made such
that one cylinder is solid of radius r and the other
cylinder is hollow of inner radius r1 and outer
radius r2 then,
• Mass of solid cylinder = Mass of hollow cylinder
πr2  = π (r22 – r12) 
r 2 = r 22 – r 12
PH0101 UNIT 1 LECTURE 2
5
In the case of a solid cylinder, twisting couple per
unit twist,
C
 N r4

2l
 N (r2 2  r1 2 ) 2
2l
For a hollow cylinder, twisting couple per unit twist
C1 
 N (r2 4  r1 4 )
2l

N (r2 2  r 1 2 ) (r2 2 - r1 2 )
2l
Dividing C1 by C
C 1 r2  r1
 2
C r2  r1 2
2
2
PH0101 UNIT 1 LECTURE 2
6
i.e., C1 > C
• Thus twisting couple for a hollow cylinder is
greater than that for a solid cylinder of the
same mass, length and material.
• Due to this reason, hollow shaft is better than
solid shaft.
• Driving shafts (axles) in automobiles are of
hollow tubes and not of solid rods.
PH0101 UNIT 1 LECTURE 2
7
Torsion Pendulum
A torsion pendulum is an oscillator for which the restoring
force is torsion
•The device consists of a disc or
other body of large moment of
inertia mounted on one end of a
torsionally flexible elastic rod wire
whose other end is held fixed;
• If the disc is twisted and
released, it will undergo simple
harmonic motion, provided the
torque in the rod is proportional to
the angle of twist.
PH0101 UNIT 1 LECTURE 2
Fixed End
Torsionally flexible elastic wire
Disc
8
Theory
• When the disc is rotated in a horizontal plane so
as to twist the wire, the various elements of the
wire undergo shearing strains.
• Restoring couples, which tend to restore the
unstrained conditions, are called into action.
• Now when the disc is released, it starts
executing torsional vibrations.
• If the angle of twist at the lower end of the wire is
θ, then the restoring couple is C θ, where C is
the torsional rigidity of the wire,
PH0101 UNIT 1 LECTURE 2
9
This couple acting on the disc produces in it an
angular acceleration given by equation.
d 
Cθ= I
2
dt
2
where I is the moment of inertia of the disc about
the axis of the wire.
The minus sign indicates that the couple C θ
tends to decrease the twist.
PH0101 UNIT 1 LECTURE 2
10
d  C


2
I
dt
2
The above relation shows that the angular
acceleration is proportional to the angular
displacement θ and is always directed towards
the mean position.
Hence the motion of the disc is simple
harmonic motion
PH0101 UNIT 1 LECTURE 2
11
The time period of the vibration will be given by
T = 2π
 2
or T = 2 π
Displacement
Acceleration

C



 I

I/C
PH0101 UNIT 1 LECTURE 2
12
Uses of Torsion Pendulum
(1)For determining the moment of inertia of an
irregular body
•First, the time period of pendulum is determined when it is
empty and then the time period of the pendulum is
determined after placing a regular body on the disc and
after this the time period is determined by replacing the
regular body by the irregular body whose moment of inertia
is to be determined.
•It is ensured that the body is placed on the disc such that
the axes of the wire pass through the centre of gravity of
the body placed on the disc.
PH0101 UNIT 1 LECTURE 2
13
• If I, I1 and I2 are the moments of inertia of the
disc, regular body and irregular body and T, T1
and T2 are the time periods in the three cases
respectively, then
• T=2π
•
T1 = 2 π
• T2 = 2 π
I
C
I  I1
C
I  I2
C
we have
T12 – T2 =
4 2 I 1
C
and
2
4

I2
T22 – T2 =
C
PH0101 UNIT 1 LECTURE 2
14
T1  T
2
T2  T
2
2

2
4 I1 / C
I1


2
4 I 2 / C I 2
or
2
T2  T
2
T1  T
2
2
I 2  I1 
2
The moment of inertia of the regular body I1 is
determined with the help of the dimensions of the
body, thus the moment of inertia of the irregular
body is calculated.
PH0101 UNIT 1 LECTURE 2
15
(2) Determination of Torsional Rigidity
For determining the modulus of rigidity N, the
time period of the pendulum is found
(i) when the disc is empty, and
(ii) when a regular body is placed on the disc
with axis of wire passing through the centre
of gravity of the body.
PH0101 UNIT 1 LECTURE 2
16
If T is the time period of the pendulum in first case
and T1 in the second case, then we have
T=2π
I
C
and T1= 2 π
I  II
C
where I is the moment of inertia of the disc and I1
the moment of inertia of the regular body placed
on the disc.
PH0101 UNIT 1 LECTURE 2
17
4 I 1
C
2
T12 – T2 =
or
C
4 I1
2
T1  T
2
2
For a wire of modulus of rigidity N, length l and
radius r, we have
C
Nr
4
2l
PH0101 UNIT 1 LECTURE 2
18
4 I1
2
T1  T
2
N
2

Nr 4
2l
8lI1
(T1  T )r
2
2
4
Thus, the value of N can be determined.
PH0101 UNIT 1 LECTURE 2
19
Worked Problem
A torsion pendulum is made using a steel
wire of diameter 0.5mm and sphere of
diameter 3cm. The rigidity modulus of steel
is 80 GPa and density of the material of the
sphere is 11300 kg/m3. If the period of
oscillation is 2 second, find the length of the
wire.
8I
N 2 4
T r
PH0101 UNIT 1 LECTURE 2
20
For sphere, I = 2/5 MR2
M = volume × density
M = 4/3π (3/2 × 10-2)3 × 11300 = 0.1598 kg
I = 2/5 × 0.1598× (3/2 × 10-2 )2
= 0.14382 × 10-4 kgm2
PH0101 UNIT 1 LECTURE 2
21
9
2  0.5
3 
10 
2 4 80  10  2  
NT r
2


l

4
8I
8    0.14382  10
PH0101 UNIT 1 LECTURE 2
4
 5.531m
22
THANK YOU
THANK YOU
PH0101 UNIT 1 LECTURE 2
23