Elasticity

B. Sc. First Year Physics (Semester-III)

Units–I: Elasticity Syllabus: Twisting couple on a cylinder (wire), Torsional pendulum, Determination of coefficient of Rigidity (η) for a Wire by Horizontal torsion apparatus, Bending of Beams, Bending Moment, Cantilever (Weight of the beam is ineffective, Weight of the beam is effective),Depression of a Beam loaded at the centre, Determination of ‘Y’ by Bending of a Beam. ----------------------------------------------------------------------------------------------------------1. Elasticity: Elasticity is that property of matter by virtue of which a body regains its original shape or size on the removal of the deforming force. a) Stress: The restoring force per unit area, set up inside the body is called stress. It is measured by the magnitude of the deforming force acting on unit area of the body within the elastic limit. Stress =

F A

where, F = Restoring force & A = Area of the body.

S.I. unit = N/m2 C.G.S. unit = dyne/cm2. Dimensions = [M L-1 T-2] b) Strain: The change produced in the dimensions of a body by a system of forces or couples in equilibrium, is called strain. It is measured by the ratio of change in length, volume and shape to its original length, volume and shape of the body. i.e. (Longitudinal strain, Volumetric strain and Shearing strain). It is dimension less quantity and has no unit. c) Hook’s Law: Hook’s law states that within elastic limit, the stress is directly proportional to the corresponding strain of the body. i.e. stress ∝ strain

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Mr. A.V. Patil M.U. College, Udgir.


E=

Stress Strain

This constant of proportionality (E) is called modulus of elasticity. The modulus of elasticity divided into three types: 1) Young’s Modulus (Y): It is defined as the ratio of longitudinal stress to the longitudinal strain within the elastic limit is known as Young’s modulus. S.I. unit = N/m2 C.G.S. unit = dyne/cm2. Dimensions = [M L-1 T-2] 2) Modulus of Rigidity (η ): It is defined as the ratio of shearing stress to the shearing strain within the elastic limit is known as Modulus of rigidity. S.I. unit = N/m2 C.G.S. unit = dyne/cm2. Dimensions = [M L-1 T-2] 3) Bulk Modulus (Y): It is defined as the ratio of uniform and normal stress on the surface of the body to the volumetric strain within the elastic limit is known as Bulk modulus. S.I. unit = N/m2 C.G.S. unit = dyne/cm2. Dimensions = [M L-1 T-2] 2. Twisting couple on a cylinder (wire): Let us consider a cylindrical rod of length (l) and radius (r). The upper end of the rod is fixed and a twisting couple is applied at its lower end in a plane perpendicular to it’s as shown in Figure 2.1. Due to elasticity of the material of the rod, a restoring couple is set up in it to equal or oppose the twisting couple. Let the rod is twisted, a line AB on surface of the cylindrical shell takes up the position AB′ through an angle φ is the shear. To understand this, let the

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cylindrical shell where cut along AB. After twisting it acquires the shape of a parallelogram AB′ C′ D is as shown in Figure 2.3.

Figure: 2.1

Figure: 2.2

Figure: 2.3

The φ is the angle of shear and θ is the angle of twist at the lower end of the rod, then BB′ = AB φ lφ = xθ

φ=

xθ l

(1)

Let F be the tangential force acting on the base of this thin cylindrical base, producing a shell then, 3



F Tangential stress = Area of base of the shell But area of the base of the shell is equal to the circumference × thickness of the shell. Area of the shell = 2π x dx Therefore, Tangential stress =

F 2πxdx

If η is the modulus of rigidity of the material of rod then, Tangential (Shearing) stress η =

η=

Tangential (Shearing) strain

F 2πxdx φ

From equation (1) above equation becomes

η=

Fl 2πx 2θdx

Then the force is F =

2πηx 2θdx l

The momentum of this force about the axis OO′ of the rod is, = F.x =

2πηx 3θdx l

Integrating the above equation between the limit x = 0 to x = r, then we get, 2πηθ 3 ∫0 x dx l r

=

r

2π ηθ x 4  =   l  4 0

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= =

2πηr 4θ 4l

πηr 4θ 2l

This is the expression for total couple acting on the cylindrical rod. If θ = 1 then, the twisting couple per unit area is given by C=

π ηr 4

where, C is the tortional rigidity of the material of the rod.

2l

3. Tortional Pendulum:

Figure 3.1 Arrangement of Tortional Pendulum. If one end of fairly thin and long wire is clamped to a rigid support and the other end is attached to the center of a heavy body (i.e. disc or sphere), then this arrangement is called the Tortional Pendulum. If the disc to be turned in the horizontal plane to twist the wire and then released, it executes torstional vibrations of a definite period about the wire of the axis. Let θ be the angle of any instant through which the body be twisted. Then the restoring couple set up in the wire is given by, Cθ =

η πr 4 2l

θ where, C is the couple per unit twist of the wire;

The couple produces an angular acceleration

d 2θ

dt 2

in the disc. If I be

the moment of inertia of the disc about the wire of the axis.

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I

d 2θ + Cθ = 0 dt 2

d 2θ C + θ =0 dt 2 I This equation represents a simple harmonic motion of the time period. T = 2π

I C

This is the expression for the time period of a tortional pendulum. 3.1 Determination of modulus of rigidity: The apparatus consist of a long thin wire whose one end is attached to a rigid support. And other free end of the wire, a heavy disc is fixed in the center. The disc is turned in the horizontal plane and the released; it executes tortional vibrations, whose time period is given by T = 2π

I C

But, C =

ηπ r 4 2l

Put this value in above equation T = 2π

2lI where, η is the modulus of rigidity, I is moment of inertia, r is ηπ r 4

the radius and l is the length of material. If R is the radius of the disc and M is its mass, then the moment of inertia about the wire of the disc. MR 2 I= 2 Then the time period of the disc is, T = 2π

MR 2l ηπ r 4

Square on both sides of the above equation. T2 =

4π 2 MR 2l ηπ r 4 6



η=

4π MR 2l T 2r 4

This expression represents the modulus of rigidity of the given wire by using tortional pendulum.

4. Determination of coefficient of Rigidity (η) for a Wire by Horizontal torsion apparatus:

Figure 4.1: Horizontal tortional apparatus 4.1 Construction: The apparatus consist of the specimen rod about 50 cm. long and 0.25 cm. in radius. One end of the rod is fixed at point A and other end is attached to the steal axle which is fixed to the pulley B. A cord is wounded round the pulley and by the free end a load is suspended. This produces a couple, which tends to twist the rod about its own axis. To measure the twist produced at two points separated by a fixed distance it apart, pointers P1 and P2 are provided which move over the scale in the form of graduated circular areas S1 and S2 respectively. 4.2 Theory and Methods:

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Let R be the radius of the pulley and M is mass suspended to the free end of the cord. The moment of the couple acting on the rod due to load of the mass i.e. mg × R . If l is the length between two pointers P1 & P2 and θ1 & θ2 be the corresponding twist produced respectively, then,

ηπ r 4 ( θ 2 − θ1 ) where, r is the radius of the rod. mgR = 2l In practice we measure θ1 & θ2 in degrees, converting these into radians, we have, mgR =

η=

ηπ r 4 ( θ 2 − θ1 ) π × 2l 180

360mgRl π r (θ 2 − θ1 ) 2 4

Hence with the help of the formula, the value of modulus of rigidity () for the material can be determined. 5. Bending of Beams:

Figure 5.1 A rod of rectangular, whose length is very large in compression to its thickness is called as beam. A beam may be considered of number of thin plane layers, collected parallel to each other. Also each layer is regarded as to consist of number of parallel longitudinal fibers, which are called the longitudinal filaments i.e. Neutral surface.

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When a beam is fixed horizontal at one end and loaded at the other, it undergoes bending. The filament of outward side are length-ended and subjected to tension, while those of the inner side shortened and compressed. In between these portions, there is a layer of surface in which the filaments are neither lengthened nor shortened. Such surface is called neutral surface i.e. EFGH as shown in Figure 5.1. the line of intersection of the plane of bending with the neutral surface is called the neutral surface is called the neutral axis i.e. NN′ as shown in Figure 5.1. Plane of bending as shown in Figure 5.1, the beam has a plane of symmetry ABCD parallel to its length. If the bending is uniform, all longitudinal filaments are bend into circular arcs in planes parallel to the plane of symmetry ABCD. Then plane ABCD may be called the plane of the plane of bending. 6. Bending Moment:

Figure: 6.1 When a horizontal beam is fixed at one end and loaded at the other, a bending is produced due to moment of the load. Let ABCD represents as the section of the beam fixed rigidly in the wall at AD with the other end BC, loaded with a weight W, as shown in Figure 6.1, such beam is called cantilever. Let us consider the equilibrium of a portion BCEF of the beam cut by the transverse plane EF across it. A force W acts downwards at the end BC, hence an equal and opposite reaction force equal to its acting vertically upward along EF i.e. W′ . these two equal and opposite forces constitute a couple. These clockwise

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couple which bends the beam is called as bending couple and the moment of these couple is called as the bending moment. 6.1 Expression for Bending Beam: Let the small portion under consideration bend in the form of circular arc, subtending an angle θ at the center of curvature O. let R be the radius of curvature of the neutral surface EF′ . Consider a filament GH distant Z form EF in the unbounded or unbent position of the beam. Let us consider a small portion of the beam bounded by two transverse sections AB and CD close to each other as shown in Figure: 6.2. After bending as shown in Figure: 6.3 AC elongated to A′ C′ and BD to B′ D′ . Line EF represent the neutral surface, which is neither stretched nor shortened.

Figure: 6.2

Figure: 6.3

Consider a filament GH distant Z from EF in the unbend position of the beam. After bending its position as shown by G′ H′ in Figure: 6.3. Now, G′ H′ = (R+Z)θ Before bending GH = EF = E′ F′ But E′ F′ = Rθ ∴ GH = Rθ Change in the length of the filament = G′ H′ - GH = (R+Z)θ - Rθ 10



= Rθ + Zθ - Rθ = Zθ Change in length i.e. Strain = -------------------------Original length =

G ′H ′ − GH GH

=

Zθ Rθ

=

Z R

Now consider a small area ∂a of the section PQRS about a point A, distance Z from the neutral surface as shown in Figure: 6.4

The strain produced in a filament passing through this area will be Z Now Y =

Stress Strain

Stress = Y

× Strain

Hence, stress about the point A is Y × Z

R .

R , where Y is the Young’s

Modulus of the material. The force on the area ∂a is, = Y× Z

R × ∂a

The total moment of the forces acting on the filament in the section PQRS is given by, i.e.

Y∂aZ ∑ R

2

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=

Y ∑∂aZ R

2

Thus, the moment of the force about MN i.e. =

Y aK R

=

YI R

2

since I = ak2

This is restoring couple or the bending moment of the beam being equal and opposite to the moment of the bending couple due to the load in the position of equilibrium. Thus, Bending Moment =

YI . R

7. Cantilever: The horizontal beam fixed at the one end and other is loaded is called cantilever.

Figure: 7.1 A cantilever is a beam fixed horizontally at one end and loaded at the other end. There are two cases arises: a) When the weight of beam is itself produces no bending. b) When it does so i) When the weight of beam is ineffective. ii) When the weight of beam is effective. 7.1 When the weight of beam is ineffective:

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Consider a beam AB of length l whose one end is fixed at point A and other is loaded at point B. the end B is depressed to the position B′ . where the cut of the beam is not effective. ∂ is the depression produced in the beam. R is the radius of the curvature of bending of the beam. Consider a section P of the beam at a distance x from the fixed end A. the momentum of the external couple at this section due to the load W or the bending moment acting on it. i.e. = W = W

× PB′

× (l − x)

(1)

Since the beam is in equilibrium then, W

× (l − x) =

But,

R=

YI g

(2)

R

1 d2y dx 2

Put this value in above equation, we get W × (l − x ) = YI g

d2y dx 2

(3)

The moment of the load increases to move towards at the point A then the radius of curvature different at different points. Suppose another point Q at distance dx from point P. dθ is the angle at point O, then PQ is dx = Rθ R=

dx dθ

(4)

Then equation (3) becomes d 2 y W (l − x) = YI g dx 2

Integrating above equation, we get

∫

d2y W (l − x) =∫ 2 YI g dx

dy W = dx YI g

∫ (l − x)dx 13



dy W = dx YI g

 x2  lx −  + C 2 

(5)

where, C is constant of integration, it can be obtained from initial value i.e. x=0, dx=0 and dy=0, then equation (5) becomes, dy W  x2  = lx − dx YI g  2 

(6)

Integrating above equation, we get

∫

dy W = dx YI g

y=

 x2  lx − ∫  2

 dx 

W  lx 2 x 3  −  + C1  YI g  2 6

Since, C1 = 0, then y=

W YI g

 lx 2 x 3  −   6  2

(7)

When x = l and y = δ then above equation becomes, δ=

W l 3 l 3   −  YI g  2 6 

δ=

W l 3    YI g  3 

(8)

The equation (8) shows the depression of the beam when weight of the beam is ineffective. 7.2 When the weight of beam is effective: In this case the weight W at point B, the weight of the portion ( l − x ), the beam is also action at the mid-point or the center of gravity of the portion. So that, if W be the weight per unit length of the beam i.e. a × Wt × w ( l − x ) is acting at a distance

l−x from the section PQ. 2

= W ( l − x) + w( l − x) = W ( l − x) +

( l − x) 2

w 2 ( l − x) 2 14



Since, beam is in equilibrium W ( l − x) + dθ =

YI dθ w 2 ( l − x) = g 2 R dx

W ( l − x) + w

2( YI g

W ( l − x) + w 2

dy =

l − x ) Rdx 2

2(

But, dt = ( l − x ) dθ

l − x ) Rdx 2

YI g

Now taking the integration within the limit 0 to l . y=

W YI g

l

2

∫

w 2 ( l − x ) Rdx 2 YI g

W ( l − x) +

0

R is negligible. Wl 3 wl 4 y= + 3YI g 8YI g But wl = W1 y=

Wl 3 W1l 3 + 3YI g 8YI g

y=

l3 3YI g

3W1   W +  8  

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Elasticity

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