Cornus Print

EXPERIMENT NO 01

Elastic constants of glass (Cornu’s method)

PHYS 43115 – ADVANCED PHYSICS LABORATARY ІІ

INDEX page

Introduction 03-06

Apparatus 07

Theory 08-12

Experimental procedure 13

Observations 14-17

Calculations and Graphs 18-39

Error calculations 40-41

Conclusions 42

Discussions 43

References 44

INTRODUCTIONElasticityElasticity is the physical property of a material that returns to its original shape after the stress (e.g. external forces) that made it deform is removed. The relative amount of deformation is called the strain. [1]

Young's modulus(E)In solid mechanics, Young's modulus (E) is a measure of the stiffness of an isotropic elastic material. It is also known as the Young modulus, modulus of elasticity, elastic modulus (though Young's modulus is actually one of several elastic moduli such as the bulk modulus and the shear modulus) or tensile modulus. It is defined as the ratio of the uniaxial stress over the uniaxial strain in the range of stress in which Hooke's Law holds.[1] This can be experimentally determined from the slope of a stress-strain curve created during tensile tests conducted on a sample of the material.Young's modulus is named after Thomas Young, the 19th century British scientist. However, the concept was developed in 1727 by Leonhard Euler, and the first experiments that used the concept of Young's modulus in its current form were performed by the Italian scientist Giordano Riccati in 1782 — predating Young's work by 25 years. [2} Young's modulus, E, can be calculated by dividing the tensile stress by the tensile strain:

Where,

E is the Young's modulus (modulus of elasticity) F is the force applied to the object; A0 is the original cross-sectional area through which the force is applied; ΔL is the amount by which the length of the object changes; L0 is the original length of the object.

figure (1)

Relationship between Stress and StrainRelationship between Stress and Strain are derived on the basis of the elastic behaviour of material bodies. [3]

A standard mild steel specimen is subjected to a gradually increasing pull by Universal Testing Machine. The stress-strain curve obtained is as shown below. [3]

Figure (2) [3]

A -Elastic LimitB - Upper Yield StressC - Lower Yield StressD -Ultimate StressE -Breaking Stress Elasticity and Elastic LimitElasticity of a body is the property of the body by virtue of which the body regains its original size and shape when the deformation force is removed. Most materials are elastic in nature to a lesser or greater extend, even though perfectly elastic materials are very rare.

The maximum stress upto which a material can exhibit the property of elasticity is called the elastic limit. If the deformation forces applied causes the stress in the material to exceed the elastic limit, there will be a permanent set in it. That is the body will not regain its original shape and size even after the removal of the deforming force completely. There will be some residual strain left in it. [3]

Modulus of Rigidity

Modulus of Rigidity (or Shear Modulus) is the coefficient of elasticity for a shearing force. It is defined as "the ratio of shear stress to the displacement per unit sample length (shear strain) . [9]

Modulus of Rigidity can be experimentally determined from the slope of a stress-strain curve created during tensile tests conducted on a sample of the material.

Definition of Modulus of Rigidity:

The ratio of shear stress to the displacement per unit sample length (shear strain)

G= shear stress shear strain

Figure (3) [11]

The bulk modulus The bulk modulus (K) of a substance measures the substance's resistance to uniform compression. It is defined as the pressure increase needed to cause a given relative decrease in volume. Its base unit is the pascal. [10]

The bulk modulus K can be formally defined by the equation:

Figure (4) [10]

Relationships between elastic moduli

shear modulus; sometimes G is used instead of

Poisson's ratio

Young's modulus

Bulk modulus

Where μ= G

APPARATUS

Glass beam

Na lamp

convex lens

flat glass plates

Traveling microscope. Two knife edges

A set of weights

String

Two lab-jacks.

Stands

Weight holders

Micrometer screw guage

Vernier caliper

Meter ruler

Mirror

Travelling microscope

Na lamp

Convex lens

Flat glass plate

Flat glass plate

Mass (M)Mass (M)

Mirror

Strip of glass plate

Knife edges

THEORY AND DIAGRAMS

A flat strip of glass plate is being applied weights at the both ends and bent to an arc as shown in the below figure.

Figure A

r1

r2

dp

dp+1

ap ap+1

The curvature obtained here is two perpendicular directions along its length and along its width. [8]

Where, r1 = radius of the curvature obtained due to its length r2 = radius of the curvature obtained across its width

figure B

Poisson’s ration is given by μ=Transversestrain

Longitudanal strain

It mentioned that the above strip of glass plate has a saddle shaped surface. The two radii of curvature produced are r1 & r2. Then Poisson’s ratio µ is given by [8]

μ=r1

r2 (1)

If a flat glass palate is rested on the centre of the curved glass beam and light form a sodium discharge lamp is directed at the air film between this plate and then beam by arrangement shown in figure A, interference fringes (which are hyperbolae) will appear. [8]

Figure C

r-e

e

r

rp

r2= (r−e )2+r p2

Figure D r2=r2−2ℜ+e2+r p

2

Since e<<2r, e2 is negligible, so 2ℜ=r p2 (2)

By the reflection the condition for the dark interference fringes (interspaced by bright fringes) to be formed is that

2 necos R=pλ (3)

Where n is the refractive index of the parallel-sided transparent plate of thickness e. R is the angle of refraction within the plate and p is an integer. [1]

Since incident and reflected rays are normal to the dark ring, R = 00

From eqn (3), cos o=1,

∴ne= pλ2 (4)

From eqn (2) & (4) r p2 =2ℜ=2 r ( pλ

2n )= rpλn

Therefore d p2= 4 rpλ

n (5)

Where λ is the wave length of the light and p is an integer and dp is the diameter of the pth dark fringe.If dp+q is the diameter of the (p+q)th dark fringe where q is an integer, then the equation (5) becomes

d p+q2 =

4 r ( p+q) λn

(6)

From the equations (5) and (6),

Where,

r – radius of the curvature

e –thickness of the film of refractive index n

between the flat glass and curve glass

rp – radius of the pth dark ring at plane surface

(6) – (5) d p+q2 −dp

2=4 rqλn

(7)

for air n = 1,

r1=d p+q

2 −dp2

4 λqWhere dp is the diameter perpendicular to the neutral axis of the pth dark fringe (figure C) and dp+q is the diameter of the (p+q) th dark fringe where p and q are integers. [8]

For the radius of the curvature r2,

r2=a p+q

2 −ap2

4 λq (8)

Where ap and ap+q now refer diameters along the neutral axis. [8]

a p+q2 −ap

2 =(4 λ r1 ) q (9)

Y = m1 X

From the gradient of the graph a p+q2 −ap

2 Vs. q, r1 can be calculated

Similarly to the perpendicular direction, d p+q

2 −dp2=(4 λ r2 ) q (10)

Y = m2 X

From the gradient of the graph (d2p+q-d2

p) Vs q, r2 can be determined.

A small initial curvature due to the weight of the glass strip will exit. To avoid error due to this, the mass M applied is plotted against 1/r1 and also 1/r2. Poissons ratio µ is then the ratio of the gradients of these graphs. [8]

The bending moment T for a plate is given by

T= EIr

(11)

Where E - Young modulus of elasticity I - The second moment of inertia of cross section of the plate about the line of neutral axis.

I=b t3

12 (12)

for the radius of curvature r1 appropriate to the longitudinal curvature and

I= l t 3

12 (13)

for r2 appropriate to lateral curvature, where l, b and t are the length, breadth and thickness respectively of the glass strip. As a bending moment T is easily calculated for a given mass. [8]

T= EIr

=Mgx

r=EIgx

1M

(14)

Y = m3 X

From the gradient of the graph r vs 1/M , the young modulus E for the glass can be calculated.

E=m3 gx

I (15)

From the relation between Poisson’s ratio(µ) & Young’s modulus(E) the Bulk modulus (K) and Shear modulus (G) can be evaluated.

The Bulk Modulus K is given by,

K= E3(1−2μ) (16)

The Shear Modulus G is given by,

G= E2(1+μ) (17)

EXPERIMENTAL PROCEDURE(1) The strip of glass and the other two small flat glass plates were cleaned by

using NaOH, HCl and finally with distilled water.

(2) The two knife edges were placed as shown in figure A in the theory part such that the distance between them is 51.0 × 10-2m and the strip of glass was placed on it symmetrically.

(3) A flat plate of glass was placed in the middle of the glass strip and masses of 0.060 kg were hanged at a distance of 31.0 × 10-2m from each knife edge.

31cm 51.0 cm 31cm

(4) The other flat glass plate was placed above this glass plate using a holder as shown in the figure A at an angle of 450 from the vertical axis and mirror was also placed above this second glass plate at an angle of 450 to its vertical axis as shown in the figure A in theory part.

(5) The convex lens was placed in front of the sodium lamp at a distance which is equal to its focal length and that the lamp, lens and the second glass plate all were on the same horizontal line and the travelling microscope was placed in front of the mirror as shown in figure A.

(6) Then the sodium lamp was switched on and the travelling microscope was adjusted and focused until a clear fringe pattern was observed from it as shown in the figure C.

(7) Then travelling microscope was focused to left and to 1st, 2nd , 3rd ……..etc up to 11th dark ring from the center and relevant microscope readings were recorded.

(8) Same procedure was repeated for right, up and down of the fringe pattern and relevant microscope readings were recorded.

(9) Above procedure (7) –(8) was repeated for different masses 0.100, 0.140, 0.180, 0.220, 0.260kg and relevant microscope readings were recorded.

(10) The thickness (t) and width (b) of the glass strip was recorded by using a vernier caliper and length (l) was measured by using the meter ruler.

(11) Least counts of the vernier caliper, micrometer screw gauge, travelling microscope and meter ruler were also recorded.

OBSERVATIONS for M = 60×10-3kg

for M = 100×10-3kg

for M = 140×10-3kg

d×10-2(m) a×10-2(m)

q dL dR aU aD

0 8.731 7.715 7.804 7.3981 9.085 7.400 8.000 7.2612 9.363 7.020 8.111 7.1003 9.637 6.700 8.201 6.9654 9.885 6.454 8.294 6.9005 10.250 6.242 8.370 6.8336 10.600 6.079 8.476 6.7377 8.731 7.715 8.540 6.6748 8.609 6.6149 8.671 6.56910 8.754 6.509

d×10-2(m) a×10-2(m)

q dL dR aU aD

0 6.136 5.473 7.609 7.0501 6.611 5.119 7.750 6.9132 6.965 4.820 7.856 6.7813 7.244 4.572 7.938 6.6814 7.512 4.357 8.025 6.6005 7.779 4.170 8.098 6.5086 8.044 4.014 8.161 6.4337 8.250 6.3758 8.291 6.3309 8.373 6.29310 8.448 6.234

for M = 180×10-3kg

for M = 220×10-3kg

d×10-2(m) a×10-2(m)

q dL dR aU aD

0 7.427 6.380 7.645 7.1781 7.758 6.125 7.721 7.0002 8.036 5.895 7.850 6.8743 8.270 5.718 7.920 6.7534 8.515 5.540 8.010 6.6745 8.730 5.385 8.104 6.6186 8.924 5.255 8.161 6.5137 8.220 6.4568 8.284 6.4159 8.350 6.34010 8.424 6.287

d×10-2(m) a×10-2(m)

q dL dR aU aD

0 7.350 6.372 7.632 7.3001 7.600 6.069 7.751 7.0112 7.816 5.793 7.853 6.9503 7.996 5.555 7.944 6.8504 8.140 5.384 8.019 6.7015 8.289 5.218 8.106 6.6546 8.416 5.096 8.187 6.5567 8.541 4.992 8.251 6.5008 8.663 4.890 8.301 6.422

for M = 260×10-3kg

The distance from the line of action of the weights to the knife edge (X) =25.5×10-2m The length of the glass strip (l) =121.8×10-2m The breadth of the glass strip (b) =5.78×10-2m

d×10-2(m) a×10-2(m)

q dL dR aU aD

0 7.000 6.450 7.539 7.3501 7.355 6.125 7.706 7.1002 7.579 5.824 7.824 6.9593 7.754 5.628 7.916 6.8674 7.891 5.450 8.004 6.7695 8.031 5.296 8.081 6.6706 8.132 5.181 8.152 6.6607 8.250 5.071 8.218 6.5288 8.352 4.977 8.300 6.4599 8.446 4.870 8.351 6.39710 8.540 4.795 8.406 6.358

d×10-2(m) a×10-2(m)

q dL dR aU aD

0 7.736 6.837 7.571 7.3501 7.903 6.613 7.703 7.1602 8.144 6.400 7.859 7.0053 8.321 6.247 7.934 6.9074 8.486 6.089 8.019 6.8055 8.616 5.962 8.130 6.7256 8.738 5.854 8.202 6.6447 8.837 5.741 8.249 6.6628 8.931 5.665 8.321 6.5089 9.016 5.585 8.374 6.45210 9.115 5.516 8.443 6.403

The thickness of the glass strip (t) =0.55×10-2 m

Least count of the traveling microscope = 1×10-5m Least count of the meter ruler = 0.5×10-3m Least count of the Vernier caliper =0.1×10-3m

CALCULATIONS

The wave length of Na light () ¿(5890+5896 ) ×10−10

2m

= 5893× 10 -10 m (A) For transverse fringe patteren

for M =60 ×10-3kg

q dL(×10-2m) dR(×10-2m) dp+q(×10-2m) d2p+q(×10-4m) d2

p+q -d2p

(x10-4m2)

0 8.731 7.715 1.016 1.0323 01 9.085 7.400 1.685 2.8392 1.80692 9.363 7.020 2.343 5.4896 4.45733 9.637 6.700 2.937 8.6260 7.59374 9.885 6.454 3.431 11.7718 10.73955 10.250 6.242 4.008 16.0641 15.03186 10.600 6.079 4.521 20.4394 19.4071

Linear Regression for DataY = A + B * X

Parameter Value Error

0 1 2 3 4 5 6

0

5

10

15

20

The graph of d 2 p+q -d 2 p vs. q (for 60 ×10 -3 Kg)

d p+q2 −dp

2=4 rT 1 λ

nq

d2p+q -d2

p

(×10-4 m2)

q

------------------------------------------------------------A -1.31124 0.70679B 3.24833 0.19603

The gradient of the graph (mT1) = 3.24833 × 10-4 m2

But according to the theory part mT1 = 4 r T 1 λ

n n = 1 = 5893× 10-10 m

rT 1=mT 1× n

4 λ

rT 1=3,24833 ×10−4 m2× 1

4 ×5893 ×10−10m

rT 1=137.81 m

for M =100 ×10-3kg


p+q -d2p

(x10-4m2)

0 6.136 5.473 0.663 0.4396 01 6.611 5.119 1.492 2.2261 1.78652 6.965 4.820 2.145 4.6010 4.16143 7.244 4.572 2.672 7.1396 6.70004 7.512 4.357 3.155 9.9540 9.51445 7.779 4.170 3.609 13.0249 12.58536 8.044 4.014 4.030 16.2409 15.8013

0 1 2 3 4 5 6-2

0

2

4

6

8

10

12

14

16

18


d p+q2 −dp

2=4 rT 2 λ

nq

d2p+q -d2

p

(×10-4 m2)


Parameter Value Error------------------------------------------------------------A -0.74528 0.37525B 2.65552 0.10408The gradient of the graph (mT2 ) = 2.65552 × 10-4 m2


n n = 1 = 5893× 10-10 m

rT 2=mT 2× n

4 λ

rT 2=2.65552 ×10−4 m2× 1

4 ×5893 ×10−10m rT 2=112.66m

for M =140×10-3kg

q


p+q -d2p

(x10-4m2)

0 7.427 6.380 1.047 1.0962 01 7.758 6.125 1.633 2.6667 1.57052 8.036 5.895 2.141 4.5839 3.48773 8.270 5.718 2.552 6.5127 5.41654 8.515 5.540 2.975 8.8506 7.75445 8.730 5.385 3.345 11.1890 10.09286 8.924 5.255 3.669 13.4616 12.3654


Parameter Value Error------------------------------------------------------------

0 1 2 3 4 5 6

0

2

4

6

8

10

12

14


d p+q2 −dp

2=4 rT 3 λ

nq

d2p+q -d2

p

(×10-4 m2)

A -0.44548 0.22323B 2.08598 0.06191 The gradient of the graph (mT3 = 2.08598 × 10-4 m2


n n = 1 = 5893× 10-10 m

rT 3=mT 3× n

4 λ

rT 3=2.08598× 10−4 m2 ×1

4 ×5893 ×10−10m rT 3=88.49 m

for M =180 ×10-3kg q dL(×10-2m) dR(×10-2m) dp+q(×10-2m) d2

p+q(×10-4m) d2p+q -d2

p

(x10-4m2)

0 7.350 6.372 0.978 0.9565 01 7.600 6.069 1.531 2.3440 1.38752 7.816 5.793 2.023 4.0925 3.13603 7.996 5.555 2.441 5.9585 5.00204 8.140 5.384 2.756 7.5955 6.63905 8.289 5.218 3.071 9.4310 8.47456 8.416 5.096 3.320 11.0224 10.06597 8.541 4.992 3.549 12.5954 11.63898 8.663 4.890 3.773 14.2355 13.2790


Parameter Value Error------------------------------------------------------------A -0.12208 0.07554B 1.68671 0.01587The gradient of the graph (mT4) = 1.68671 × 10-4 m2


n n = 1

0 2 4 6 8-2

0

2

4

6

8

10

12

14


d p+q2 −dp

2=4 rT 4 λ

nq

d2p+q -d2

p

(×10-4 m2)

q

= 5893× 10-10 m

rT 4=mT 4 × n

4 λ

rT 4=1.68671 ×10−4 m2 ×1

4× 5893× 10−10 m rT 4=71.56 m

for M =220×10-3kg q dL(×10-2m) dR(×10-2m) dp+q(×10-2m) d2

p+q(×10-4m) d2p+q -d2

p

(x10-4m2)

0 7.000 6.450 0.550 0.3025 01 7.355 6.125 1.230 1.5129 1.21042 7.579 5.824 1.755 3.0800 2.77753 7.754 5.628 2.126 4.5199 4.21744 7.891 5.450 2.441 5.9585 5.65605 8.031 5.296 2.735 7.4802 7.17776 8.132 5.181 2.951 8.7084 8.40597 8.250 5.071 3.179 10.1060 9.80358 8.352 4.977 3.375 11.3906 11.08819 8.446 4.870 3.576 12.7878 12.485310 8.540 4.795 3.745 14.0250 13.7225

0 2 4 6 8 10-2

0

2

4

6

8

10

12

14

d p+q2 −dp

2=4 rT 5 λ

nq

d2p+q -d2

p

(×10-4 m2)


Parameter Value Error------------------------------------------------------------A 0.02375 0.0695B 1.38696 0.01175

The gradient of the graph (mT5) = 1.38696 × 10-4 m2


n n = 1 = 5893× 10-10 m

rT 5=mT 5× n

4 λ

rT 5=1.38696 × 10−4 m2 ×1

4 ×5893 ×10−10m rT 5=58.84 m

for M =260 ×10-3kg q dL(×10-2m) dR(×10-2m) dp+q(×10-2m) d2

p+q(×10-4m) d2p+q -d2

p

(x10-4m2)

0 7.736 6.837 0.899 0.8082 01 7.903 6.613 1.290 1.6641 0.85592 8.144 6.400 1.744 3.0415 2.23333 8.321 6.247 2.074 4.3015 3.49334 8.486 6.089 2.397 5.7456 4.93745 8.616 5.962 2.654 7.0437 6.23556 8.738 5.854 2.884 8.3175 7.5093

7 8.837 5.741 3.096 9.5852 8.77708 8.931 5.665 3.266 10.6668 9.85869 9.016 5.585 3.431 11.7718 10.963610 9.115 5.516 3.599 12.9528 12.1446


Parameter Value Error------------------------------------------------------------

A -0.14327 0.09381B 1.24699 0.01586The gradient of the graph mT6 = 1.24699 × 10-4 m2


n n = 1 = 5893× 10-10 m

rT 6=mT 6× n

4 λ

0 2 4 6 8 10

0

2

4

6

8

10

12

14


d p+q2 −dp

2=4 rT 6 λ

nq

d2p+q -d2

p

(×10-4 m2)

q

rT 6=1.24699× 10−4 m2× 1

4 ×5893 ×10−10m rT 6=52.90 m

(B) For longitudinal fringe patterenFor 60 ×10-3kg

q aU(×10-2m) aD(×10-2m) ap+q(×10-2m) a2p+q(×10-4m) a2

p+q -a2p

(x10-4m2)

0 7.804 7.398 0.406 0.1648 0.00001 8.000 7.261 0.739 0.5461 0.38132 8.111 7.100 1.011 1.0221 0.85733 8.201 6.965 1.236 1.5277 1.36294 8.294 6.900 1.394 1.9432 1.77845 8.370 6.833 1.537 2.3623 2.19756 8.476 6.737 1.739 3.0241 2.85937 8.540 6.674 1.866 3.4820 3.31728 8.609 6.614 1.995 3.9800 3.81529 8.671 6.569 2.102 4.4184 4.253610 8.754 6.509 2.245 5.0400 4.8752

0 2 4 6 8 10

0

1

2

3

4

5

The graph of a 2 p+q -a 2 p vs. q (for 60 ×10 -3 Kg)

a p+q2 −ap

2 =4 r L1 λ

nq

a2p+q -a2

p

(×10-4 m2)

q

Linear Regression DataY = A + B * X

Parameter Value Error------------------------------------------------------------A -0.10603 0.04101B 0.48844 0.00693The gradient of the graph (mL1) = 0.48844 × 10-4 m2

But according to the theory part mL1 = 4 r L1 λ

n n = 1 = 5893× 10-10 m

r L1=mL 1× n

4 λ

r L1=0.48844 × 10−4 m2× 1

4 ×5893 ×10−10m r L1=20.72 m

For 100×10-3kgq aU(×10-2m) aD(×10-2m) ap+q(×10-2m) a2

p+q(×10-4m) a2p+q -a2

p

(x10-4m2)

0 7.609 7.050 0.559 0.3125 0.00001 7.750 6.913 0.837 0.7006 0.38812 7.856 6.781 1.075 1.1556 0.84313 7.938 6.681 1.257 1.5800 1.26754 8.025 6.600 1.425 2.0306 1.71815 8.098 6.508 1.590 2.5281 2.21566 8.161 6.433 1.728 2.9860 2.67357 8.250 6.375 1.875 3.5156 3.20318 8.291 6.330 1.961 3.8455 3.53309 8.373 6.293 2.080 4.3264 4.013910 8.448 6.234 2.214 4.9018 4.5893




n n = 1 = 5893× 10-10 m

0 2 4 6 8 10

0

1

2

3

4

5


a p+q2 −ap

2 =4 r L2 λ

nq

a2p+q -a2

p

(×10-4 m2)

q

r L2=mL 2× n

4 λ

r L2=0.45769 ×10−4 m2×1

4 × 5893 ×10−10 m r L2=19.42 m

For 140 ×10-3kgq aU(×10-2m) aD(×10-2m) ap+q(×10-2m) a2

p+q(×10-4m) a2p+q -a2

p

(x10-4m2)

0 7.645 7.178 0.467 0.2181 0.00001 7.721 7.000 0.721 0.5198 0.30172 7.850 6.874 0.976 0.9526 0.73453 7.920 6.753 1.167 1.3619 1.14384 8.010 6.674 1.336 1.7849 1.56685 8.104 6.618 1.486 2.2082 1.99016 8.161 6.513 1.648 2.7159 2.49787 8.220 6.456 1.764 3.1117 2.89368 8.284 6.415 1.869 3.4932 3.25719 8.350 6.340 2.010 4.0401 3.822010 8.424 6.287 2.137 4.5668 4.3487

0 2 4 6 8 10

0

1

2

3

4

5 a2p+q -a

(×10-4 m




n n = 1 = 5893× 10-10 m

r L3=mL 3× n

4 λ

r L3=0.43476 ×10−4 m2× 1

4 ×5893 ×10−10m r L3=18.44 m


p+q(×10-4m) a2p+q -a2

p

(x10-4m2)

0 7.632 7.300 0.332 0.1102 0.00001 7.751 7.011 0.740 0.5476 0.43742 7.853 6.950 0.903 0.8154 0.70523 7.944 6.850 1.094 1.1968 1.08664 8.019 6.701 1.318 1.7371 1.62695 8.106 6.654 1.452 2.1083 1.99816 8.187 6.556 1.631 2.6601 2.54997 8.251 6.500 1.751 3.0660 2.95588 8.301 6.422 1.879 3.5306 3.4204

q


Parameter Value Error------------------------------------------------------------A -0.08026 0.05033B 0.43063 0.01057

The gradient of the graph (mL4) = 0.43063 × 10-4 m2

But according to the theory part mL4 = 4 r L 4 λ

n n = 1 = 5893× 10-10 m

r L4=mL 4× n

4 λ

0 2 4 6 8-0.5

0.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

a2

(×10

r L4=0.43063× 10−4m2 ×1

4 ×5893 ×10−10m r L4=18.27 m


p+q(×10-4m) a2p+q -a2

p

(x10-4m2)

0 7.539 7.350 0.189 0.0357 0.00001 7.706 7.100 0.606 0.3672 0.33152 7.824 6.959 0.865 0.7482 0.71253 7.916 6.867 1.049 1.1004 1.06474 8.004 6.769 1.235 1.5252 1.48955 8.081 6.670 1.411 1.9909 1.95526 8.152 6.660 1.492 2.2261 2.19047 8.218 6.528 1.690 2.8561 2.82048 8.300 6.459 1.841 3.3893 3.35639 8.351 6.397 1.954 3.8181 3.782410 8.406 6.358 2.048 4.1943 4.1586

0 2 4 6 8 10

0

1

2

3

4


a p+q2 −ap

2 =4 r L5 λ

nq

a2p+q -a2

p

(×10-4 m2)




n n = 1 = 5893× 10-10 m

r L5=mL 5× n

4 λ

r L5=0.42491 ×10−4 m2×1

4 × 5893 ×10−10 m r L5=18.03 m


p+q(×10-4m) a2p+q -a2

p

(x10-4m2)

0 7.571 7.350 0.221 0.0488 0.00001 7.703 7.160 0.543 0.2948 0.2462 7.859 7.005 0.854 0.7293 0.68053 7.934 6.907 1.027 1.0547 1.00594 8.019 6.805 1.214 1.4738 1.42505 8.130 6.725 1.405 1.9740 1.92526 8.202 6.644 1.558 2.4273 2.37857 8.249 6.662 1.587 2.5186 2.46988 8.321 6.508 1.813 3.2870 3.23829 8.374 6.452 1.922 3.6941 3.645310 8.443 6.403 2.040 4.1616 4.1128

q




n n = 1 = 5893× 10-10 m

r L6=mL6 ×n

4 λ

r L6=0.4156 × 10−4 m2× 14×5893 ×10−10m

0 2 4 6 8 10

0

1

2

3

4


a p+q2 −ap

2 =4 r L6 λ

nq

a2p+q -a2

p

(×10-4 m2)

q

r L6=17.63 m

For transverse fringe patterenM (× 10-3 Kg) rT (m) 1/rT (× 10-3 m-1)

60 137.81 7.256100 112.66 8.876140 88.49 11.301180 71.56 13.974220 58.84 16.995260 52.90 18.904


Parameter Value Error------------------------------------------------------------A 3.13919 0.43559B 0.06091 0.0025

50 100 150 200 250 3006

8

10

12

14

16

18

20

The graph of 1/rT vs. M (for transverse fringe pattern)1/rT

(× 10-3 m-1)

M(× 10-3 Kg)

The gradient of the graph (mT) = 6.091 ×10-2 m-1.Kg-1

For longitudinal fringe patternM (× 10-3 Kg) rL (m) 1/rL (× 10-2 m-1)

60 20.72 4.826100 19.42 5.149140 18.44 5.423180 18.27 5.473220 18.03 5.546260 17.63 5.672

Linear Regression for Data1_B:Y = A + B * X

Parameter Value Error------------------------------------------------------------A 4.72291 0.11614B 0.00391 6.67599E-4 The gradient of the graph mL = 3.91 ×10-2 m-1.Kg-1

50 100 150 200 250 300

4.8

5.0

5.2

5.4

5.6

5.8

The graph of 1/rL vs. M (for longitudinal fringe pattern)1/rL

(× 10-2 m-1)

M(× 10-3 Kg)

The poison’s ratio (μ) = r1

r2

=mL

mT

= 3.916.091

=0.642

M (× 10-3 Kg) 1/M (Kg-1) rT (m)

60 16.67 137.81100 10.00 112.66140 7.14 88.49180 5.56 71.56220 4.55 58.84260 3.85 52.90


Parameter Value Error

The graph of rT vs.1/M rT (m)

rT=EIgx

1M

2 4 6 8 10 12 14 16 1840

60

80

100

120

140

------------------------------------------------------------A 33.93855 7.59229B 6.67006 0.83562

The gradient of the graph m3 = 6.67006 m.Kg

I=b t3

12

I=5.78× 10−2 × (0.55 × 10−2 )3

12

I=8.013 ×10−10m4

The Young modulus (E) is give by,

E=m3 gx

I

E=6.67006 m. Kg× 9.81 m. s−2 ×25.5 ×10−2m8.103 ×10−10m4

E=2.05917 ×1010 Kg. m−1 . s−2

E=2.059 ×1010 N .m−2

The Bulk modulus (K) is given by,

K= E3(1−2μ)

K=2.059 ×1010 N . m−2

3(1−2× 0.642)

K=2.4166 ×1010 N . m−2

The shear modulus (G) is given by,

G= E2(1+μ)

G=2.059 ×1010 N .m−2

2(1+0.642)

G=0.62698× 1010 N . m−2

ERROR CALCULATIONS

μ ¿mL

mT

lnμ=lnmL−ln mT

δμμ

=δ mL

mL

−δ mT

mT

For maximum error,δμμ

=δ mL

mL

+δ mT

mT

δμ=0.642×[ 66.7599× 10−4 m−1 . Kg−1

3.91× 10−2 m−1 . Kg−1 + 0.25 ×10−2 m−1 . Kg−1

6.091× 10−2 m−1 . Kg−1 ]δμ=0.642× [ 0.1707+0.0410 ]δμ=0.1359

E=m3 gx

Iln ( E )=ln (m3 )+ ln ( g )+ ln (x )−ln (I )δEE

=δ m3

m3

+ δxx

− δII

For maximum error,δEE

=δ m3

m3

+ δxx

+ δII

I=b t3

12lnI=ln (b t 3 )−ln12lnI=ln (b )+3 lnt−ln 12δII

= δbb

+3δtt

δEE

=δ m3

m3

+ δxx

+ δbb

+3δtt

δEE

=0.83562m .Kg6.67006 m . Kg

+ 1× 10−3 m255 ×10−3 m

+ 1×10−4 m578 ×10−4 m

+3×1×10−4 m

55 ×10−4 m

δE=2.059× 1010 N . m−2× [ 0.1491127 ]

δE=0.307023× 1010 N . m−2

K= E3(1−2μ)

ln K=lnE−ln 3(1−2μ)ln K=lnE−ln 3−ln(1−2 μ)

δKK

= δEE

−δ (1−2 μ)(1−2 μ)

δKK

= δEE

+2δμ

(1−2 μ)δKK

= δEE

+ 2 δμ(1−2 μ)

δKK

=0.307023 ×1010 N . m−2

2.059 ×1010 N .m−2 + 2× 0.1359(1−2× 0.642)

δ K=2.4166 ×1010 N . m−2× [ 0.807929 ]δK=1.95244 × 1010 N . m−2

G= E2(1+μ)

lnG=lnE−ln 2 (1+μ )lnG=lnE−ln 2−ln (1+μ )δGG

= δEE

−δ (1+μ )(1+μ )

δGG

= δEE

−δ ( μ )

(1+μ )For maximum error,δGG

= δEE

+δ ( μ )

(1+μ )

δG=G ×[ 0.307023 ×1010 N . m−2

2.059 ×1010 N . m−2 + 0.1359(1+0.642 ) ]

δG=0.62698 × 1010 N . m−2× [0.262576 ]

δG=0.164630 ×1010 N . m−2

CONCLUSION

The Experimental value of the Poisson’s ratio = (0.642 ± 0.136)

The Experimental value of the Young modulus = (2.059 ± 0.307) × 1010N.m-2

The Experimental value of the Bulk modulus = (2.416 ± 1.952) × 1010N.m-2

The Experimental value of the Shear modulus = (0.627 ± 0.165) × 1010N.m-2

Standard values for glass

Poisson’s ratio = 0.18-0.30

Young modulus = 5× 1010N.m-2 - 9× 1010N.m-2

Bulk modulus = 4× 1010N.m-2

Shear modulus = 2× 1010N.m-2 - 2× 1010N.m-2

DISCUSSION

REFERENCES

(1) http://en.wikipedia.org/wiki/Elasticity_(physics)(2) http://en.wikipedia.org/wiki/Young's_modulus(3) http://www.eduresourcecollection.com/civil_sm_Elasticity.php(4) http://chsfpc5.chem.ncsu.edu/~franzen/CH795N/dft_modules/

polymer_module/crystalline/elastic_constants.htm(5) http://jsomers.net/papers/feynman-lectures-on-physics/Vol%202%20Ch

%2038%20-%20Elasticity.pdf(6) http://www.springerlink.com/content/x230q281q3210725/fulltext.pdf

(7) http://www.maths.tcd.ie/~tharris/seniorfresh/CMReport.pdf

(8) Whittle, R. M. and Yarwood, J. Experimental Physics for Students

(9) http://www.engineeringtoolbox.com/modulus-rigidity-d_946.html

(10) http://en.wikipedia.org/wiki/Bulk_modulus

(11) http://en.wikipedia.org/wiki/Shear_modulus

(12) http://geophysics.ou.edu/geomechanics/notes/elast_const/elast_const.html

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