Exp.No: 01 DETERMINATION OF HARDNESS OF A GIVEN MATERIAL

S t r e n g t h o f M a t e r i a l s L a b o r a t o r y - P a g e | 1 |

AIM

To determine the Rockwell hardness number (RHN) of the given specimen.

APPARATUS REQUIRED

Rockwell testing machine

Specimens

PREREQUISITE QUESTIONS

Define hardness.

Differentiate hardness and toughness.

CONDUCT OF EXPERIMENT

Theory

The hardness of a material is

resistance to penetration under a localized

pressure or resistance to abrasion. Hardness

tests provide an accurate, rapid and

economical way of determining the

resistance of materials to deformation. There

are three general types of hardness

measurements depending upon the manner

in which the test is conducted:

a. Scratch hardness measurement,

b. Rebound hardness measurement

c. Indention hardness measurement.

In scratch hardness method the

material are rated on their ability to scratch

one another and it is usually used by mineralogists only.

In rebound hardness measurement, a standard body is usually dropped on to the

material surface and the hardness is measured in terms of the height of its rebound .The

general means of judging the hardness is measuring the resistance of a material to

indentation. The indenters usually a ball cone or pyramid of a material much harder than that

being used.

Hardened steel, sintered tungsten carbide or diamond indenters are generally used in

indentation tests; a load is applied by pressing the indenter at right angles to the surface being

tested. The hardness of the material depends on the resistance which it exerts during a small

Exp.No: 01 DETERMINATION OF HARDNESS OF A GIVEN MATERIAL USING ROCKWELL HARDNESS TESTING MACHINE

Date:


amount of yielding or plastic. The resistance depends on friction, elasticity, viscosity and the

intensity and distribution of plastic strain produced by a given tool during indentation.

Indenter Inner load (kgf) Major load (kgf) Total load (kgf) Load E

Steel ball 10 90 100 130

Steel cone 10 50 60 100

Diamond 10 140 150 100

Procedure

1. Place the specimen securely upon the anvil.

2. Elevate the specimen so that it come into contact with the penetrate and put the

specimen under a preliminary or minor load of 100+2N without shock

3. Apply the major load 900N by loading lever.

4. Watch the pointer until it comes to rest.

5. Remove the major load.

6. Read the Rockwell hardness number or hardness scale.

Precautions

1. Test should be performed on smooth, flat specimens from which dirt and scale have

been cleaned.

2. The test should not be made on specimens so thin that the impression shows

through the metal, nor should impression be made too close to the edge of a specimen.

Observation

S.No Material Scale Indenter

(mm)

Load

(kgf) Dia

RHN Mean

RHN Trial 1 Trial 2

1) Aluminum B 1.6 100 Red 106 106 106

2) Brass B 1.6 100 Red 107 111 109

3) Copper B 1.6 100 Red 109 112 110.5

VIVA QUESTIONS

How can you improve hardness of a material?

Which material has high hardness? Why?

Why is it necessary to check hardness?

STIMULATING QUESTIONS

1. Exemplify the applications which considers material’s hardness as an important

design consideration.

2. Interpret the importance of material’s hardness in gears.


RESULT

Thus the hardness of the material found to be

RHN of copper = 110.5

RHN of brass = 109

RHN of aluminum = 106


AIM

To determine the Brinell’s hardness number (BHN) of the given specimen.

APPARATUS REQUIRED

Brinell testing machine

Specimens

Ball indenter

Micrometer


List out the mechanical properties of a material

What is meant by red hardness or hot hardness?


Theory

The hardness of a material is resistance to penetration

under a localized pressure or resistance to abrasion. Hardness

tests provide an accurate, rapid and economical way of

determining the resistance of materials to deformation. There

are three general types of hardness measurements depending

upon the manner in which the test is conducted:

a. Scratch hardness measurement,

b. Rebound hardness measurement

c. Indention hardness measurement.

In scratch hardness method the material are rated on their ability to scratch one

another and it is usually used by mineralogists only.

In rebound hardness measurement, a standard body is usually dropped on to the

material surface and the hardness is measured in terms of the height of its rebound .The

general means of judging the hardness is measuring the resistance of a material to

indentation. The indenters usually a ball cone or pyramid of a material much harder than that

material being used.

Hardened steel, sintered tungsten carbide or diamond indenters are generally used in

indentation tests; a load is applied by pressing the indenter at right angles to the surface being

tested. The hardness of the material depends on the resistance which it exerts during a small

amount of yielding or plastic. The resistance depends on friction, elasticity, viscosity and the

intensity and distribution of plastic strain produced by a given tool during indentation.

Exp.No: 02 DETERMINATION OF HARDNESS OF A GIVEN MATERIAL USING BRINELL HARDNESS TESTING MACHINE

Date:


F= Applied load in kgf

D=Diameter of indentor

d=Diameter of the indentation

Where,

F= 30 D2 for mild steel

F= 10 D2 for brass and copper

F= 5 D2 for aluminum

Procedure

1. Place the specimen securely upon the anvil.

2. Elevate the specimen so that it come into contact with the penetrate and put the

specimen under a preliminary or minor load without shock

3. Apply the major load by loading lever.

4. Watch the pointer until it comes to rest.

5. Remove the major load.

6. Calculate the Brinell’s hardness number.

Precautions

1. Brinell test should be performed on smooth, flat specimens from which dirt and

scale have been cleaned.

2. The test should not be made on specimens so thin that the impression shows

through the metal, nor should impression be made too close to the edge of a specimen.

Observation

S.No Material

Dia of

indenter

(mm)

Load

(kg)

Dia of indentation (mm)

BHN Trial 1 Trial 2 Mean

1) Aluminum 2.5 187.5 1.0 0.9 0.95 28.43

2) Brass 2.5 187.5 1.0 0.9 0.95 56.58

3) Copper 2.5 187.5 0.95 0.85 0.9 91.70

VIVA QUESTIONS

1. Name any two material which has high hardness.

2. Name any two material which has low hardness.

3. Say something on Rigid Body


Enumerate the advantages of Rockwell Hardness test over Brinell hardness test.

Interpret the importance of material’s hardness in gears.


RESULT

Thus the hardness of the material found to be

BHN of aluminum = 28.43

BHN of brass = 56.58

BHN of copper = 91.70


AIM

To conduct a tensile test on a steel plate specimen and determine the following: (i)

Limit of proportionality (ii) Elastic limit (iii) Yield strength (iv) Ultimate strength (v)

Young’s modulus of elasticity (vi) Percentage elongation in length (vii) Percentage reduction

in cross sectional area. (viii) Stress-Strain curve.

APPARATUS REQUIRED

Universal Testing Machine (UTM)

Mild steel rod specimen

Graph paper

Measuring Scale

Vernier Caliper


Define Hooke’s Law.

Define yield point.

How can you measure tensile strength?


Theory

The tensile test is most applied one, of all

mechanical tests. In this test ends of test piece are fixed

into grips connected to a straining device and to a load

measuring device. If the applied load is small enough,

the deformation of any solid body is entirely elastic. An

elastically deformed solid will return to its original form

as soon as load is removed. However, if the load is too

large, the material can be deformed permanently. The

initial part of the tension curve which is recoverable

immediately after unloading is termed. As elastic and the

rest of the curve which represents the manner in which

solid undergoes plastic deformation is termed plastic.

The stress below which the deformations essentially

entirely elastic is known as the yield strength of material. In some material the onset of

plastic deformation is denoted by a sudden drop in load indicating both an upper and a lower

yield point. However, some materials do not exhibit a sharp yield point. During plastic

deformation, at larger extensions strain hardening cannot compensate for the decrease in

Exp.No: 03 DETERMINATION OF TENSILE STRENGTH OF A GIVEN MILD STEEL ROD USING UNIVERSAL TESTING MACHINE

Date:


section and thus the load passes through a maximum and then begins to decrease. This stage

the “ultimate strength”’ which is defined as the ratio of the load on the specimen to original

cross-sectional area, reaches a maximum value. Further loading will eventually cause ‘neck’

formation and rupture.

Procedure

1. Measure the original length and diameter of the specimen. The length may either be

length of gauge section which is marked on the specimen with a preset punch or the

total length of the specimen.

2. Insert the specimen into grips of the test

machine and attach strain-measuring

device to it.

3. Begin the load application and record

load versus elongation data.

4. Take readings more frequently as yield

point is approached.

5. Measure elongation values with the help

of dividers and a ruler.

6. Continue the test till Fracture occurs.

7. By joining the two broken halves of the

specimen together, measure the final

length and diameter of specimen.


Precautions

1. If the strain measuring device is an extensometer it should be removed before necking

begins.

2. Measure deflection on scale accurately & carefully

Observations

Original dimensions

Length = 400 mm

Diameter = 10 mm

Area = 78.5 mm2

Final dimensions

Length = 420 mm

Diameter = 6 mm

Area = 28.26mm2

S.No Load(N) Extension

(mm)

Stress

(N/mm2) Strain

Kgf N

1. 250 2452.5 1 31.24204 0.000156

2. 500 4905 1 62.48408 0.000312

3. 750 7357.5 2 93.72611 0.000469

4. 1000 9810 2 124.9682 0.000625

5. 1250 12262.5 3 156.2102 0.000781

6. 1500 14715 4 187.4522 0.000937

7. 1750 17167.5 4 218.6943 0.001093

8. 2000 19620 5 249.9363 0.00125

9. 2250 22072.5 6 281.1783 0.001406

10. 2500 24525 6 312.4204 0.001562

11. 2750 26977.5 7 343.6624 0.001823

12. 3000 29430 8 374.9045 0.002331

13. 3250 31882.5 9 406.1465 0.002487

14. 3500 34335 10 437.3885 0.002799

15. 3750 36787.5 11 468.6306 0.003112

16. 4000 39240 12 499.8726 0.003424

17. 4250 41692.5 13 531.1146 0.00358

18. 4500 44145 14 562.3567 0.003893

19. 4750 46597.5 15 593.5987 0.004349

20. 4250 41692.5 16 531.1146 0.004861

21. 4000 39240 18 499.8726 0.00503

22. 3750 36787.5 20 468.6306 0.005299


VIVA QUESTIONS

1. Why stress value is decreasing after ultimate point?

2. Why stress remains constant at yield point?

3. Is it a normal stress failure or shear stress failure?

4. How do you define strain energy?

5. Differentiate Tensile Strain and Tensile stress.

6. Purpose of UTM.

7. Define a Hydraulic jack.

8. Differentiate between pneumatic and hydraulic pumps.

9. What is Yield Strength?

10. What is working stress?

11. What is Ultimate strength?

12. What is factor of safety?

13. What is stress?

14. What is strain?

15. Tell something on elastic constants.


1. Tensile strength is always lower than compressive strength – tell your opinion

2. Mild steel fails due to shear – tell your opinion


RESULT

Elongation in length (%) : 5%

Reduction in Area (%) : 64%

Yield Strength : 343 MPa

Ultimate Tensile Strength : 593 MPa

Normal breaking stress : 468 MPa

Actual breaking stress : 1301 MPa

Young’s Modulus : 200 GPa


AIM

To determine the compression strength of a wooden specimen when load is applied

parallel and perpendicular to grains.

APPARATUS REQUIRED

Universal Testing Machine (UTM)

Wooden specimen

Measuring Scale


What do you mean by compressive strength of a material?

Define bulk modulus.

What is volumetric strain?


Theory

This is the test to know strength of a material under compression.

Generally compression test is carried out to know either simple compression characteristics o

f material or column action of structural members. It has been observed that for varying heigh

t of member, keeping crosssectional and the load applied constant, there is an increased tend

ency towards bending of a member. Member under compression usually bends along minor a

xis, i.e, along least lateral dimension. According to column theory slenderness ratio has

more functional value.

If this ratio goes on increasing, axial compressive stress goes on decreasing and member buck

les more and more. End conditions at the time of test have a pronounced effect on compressiv

e strength of materials. Effective length must be taken according to end conditions assumed, a

t the time of the test.

Procedure

1. Select some concrete block with uniform shape and size.

2. Measure its dimensions. (Length x breath x height)

3. Place the specimen on the lower platform of compression testing machine and

lower the spindle till the upper motion of ram is offered by a specimen the oil pressure

start increasing the pointer engineering start returning to zero leaving the drug pointer

that is maximum reading which can be noted down.

Exp.No: 04 DETERMINATION OF COMPRESSION STRENGTH OF A WOOD

Date:


Observations

Length of the specimen : 40 mm

Breath of the specimen : 40 mm

Height of the specimen : 40 mm

Specimen Nature of

applied load

Load

applied area

(mm2)

Breaking load Compressive

strength

(N/mm2) Kg N

Wood 1 Parallel to

grains 1600 5520 54151.2 33.84

Wood 2 Perpendicular

to grains 1600 4550 44635.5 27.89

VIVA QUESTIONS

1. Differentiate homogeneous, isotropic, orthotropic and anisotropic materials.

2. Differentiate Compressive Strain and Compressive stress.

3. What is Poisson’s ratio?

4. Differentiate Longitudinal and Lateral Strain.

5. Relation between Bulk Modulus and Young’s modulus.


1. Why compressive strength of a material is always greater than tensile strength?

2. How can you improve the compressive strength of a material?

RESULT

Compressive strength of wood when load is applied parallel to grains : 33.84

N/mm2

Compressive strength of wood when load is applied perpendicular to grains :

27.89 N/mm2


AIM

To determine the impact strength of a given materials by Izod test

APPARATUS REQUIRED

Impact testing machine

A steel specimen 75 mm X 10mm X 10mm


What is impact load?

What is Strain energy?


Theory

An impact test signifies toughness of material that is ability of material to absorb

energy during plastic deformation. Static tension tests of unnotched specimens do not always

reveal the susceptibility of a metal to brittle fracture. This important factor is determined by

impact test. Toughness takes into account both the strength and ductility of the material.

Several engineering materials have to withstand impact or suddenly applied loads while in

service. Impact strengths are generally lower as compared to strengths achieved under slowly

applied loads. Of all types of impact tests, the notch bar tests are most extensively used.

Therefore, the impact test measures the energy necessary to fracture a standard notch bar by

applying an impulse load. The test measures the notch toughness of material under shock

loading. Values obtained from these tests are not of much utility to design problems directly

and are highly arbitrary. Still it is important to note that it provides a good way of comparing

toughness of various materials or toughness of the same material under different condition.

This test can also be used to assess the ductile brittle transition temperature of the material

occurring due to lowering of temperature.

Exp.No: 05 DETERMINATION OF IMPACT STRENGTH OF A GIVEN MATERIAL BY IZOD IMPACT TEST

Date:


Procedure

1. With the striking hammer (pendulum) in safe test position, firmly hold the steel

specimen in impact testing machine’s vice in such a way that the notch face the

hammer and is half inside and half above the top surface of the vice.

2. Bring the striking hammer to its top most striking position unless it is already there,

and lock it at that position.

3. Bring indicator of the machine to zero, or follow the instructions of the operating

manual supplied with the machine.

4. Release the hammer. It will fall due to gravity and break the specimen through its

momentum, the total energy is not absorbed by the specimen. Then it continues to

swing. At its topmost height after breaking the specimen, the indicator stops moving,

while the pendulum falls back. Note the indicator at that topmost final position.

5. Again bring back the hammer to its idle position and back

Precaution

1. Measure the dimensions of the specimen carefully.

2. Locate the specimen in such a way that the hammer, strikes it at the middle.

3. Note down readings carefully.

Observations

S.No Specimen Initial Dial

Reading (J)

Final Dial

Reading (J)

Impact

value (J)

Izod Impact

strength

(J/mm2)

1 Mild Steel 168 146 22 0.275

VIVA QUESTIONS

1. Define modulus of resilience.

2. Explain Castigliano’s Theorem.

3. Explain sudden impact.


1. Ductility is the property of a material by virtue of which it can be drawn into wires

under the action of tensile force. Why?

2. Why impact strength measured in terms of joules?

RESULT

The energy absorbed for Mild Steel is found out to be 0.275 Joules/mm2.


AIM

To determine the impact strength of a given materials by Charpy test

APPARATUS REQUIRED

Impact testing machine

Specimen


What is impact load?

What is Strain energy?


Theory

An impact test signifies toughness of material

that is ability of material to absorb energy during

plastic deformation. Static tension tests of unnotched

specimens do not always reveal the susceptibility of

a metal to brittle fracture. This important factor is

determined by impact test. Toughness takes into

account both the strength and ductility of the

material. Several engineering materials have to

withstand impact or suddenly applied loads while in

service. Impact strengths are generally lower as

compared to strengths achieved under slowly applied

loads. Of all types of impact tests, the notch bar tests

are most extensively used. Therefore, the impact test

measures the energy necessary to fracture a standard

notch bar by applying an impulse load. The test measures the notch toughness of material

under shock loading. Values obtained from these tests are not of much utility to design

problems directly and are highly arbitrary. Still it is important to note that it provides a good

way of comparing toughness of various materials or toughness of the same material under

different condition. This test can also be used to assess the ductile brittle transition

temperature of the material occurring due to lowering of temperature.

Procedure

1. With the striking hammer (pendulum) in safe test position, firmly hold the steel

specimen in impact testing machines vice in such a way that the notch faces s the

hammer and is half inside and half above the top surface of the vice.

Exp.No: 06 DETERMINATION OF IMPACT STRENGTH OF A GIVEN MATERIAL BY CHARPY IMPACT TEST

Date:


2. Bring the striking hammer to its top most striking position unless it is already there,

and lock it at that position.

3. Bring indicator of the machine to zero, or follow the instructions of the operating

manual supplied with the machine. 4. Release the hammer. It will fall due to gravity

and break the specimen through its momentum, the total energy is not absorbed by the

specimen. Then it continues to swing. At its topmost height after breaking the

specimen, the indicator stops moving, while the pendulum falls back. Note the

indicator at that topmost final position.

5. The specimen is placed on supports or anvil so that the blow of hammer is opposite

to the notch.

Precaution

1. Measure the dimensions of the specimen carefully.

2 Locate the specimen in such a way that the hammer, strikes it at the middle.

3 Note down readings carefully.

Observations

S.No Specimen Initial Dial

Reading (J)

Final Dial

Reading (J)

Impact

value (J)

Charpy

Impact

strength

(J/mm2)

1 Mild Steel 300 144 156 1.95

VIVA QUESTIONS

1. Types of Loads.

2. What is Resilience?

3. Define proof of resilience.


1. A ductile material can be drawn into wires. Why?

2. Stress induced in suddenly applied load is twice the amount of stress induced in

gradually applied load. Why?

RESULT

The energy absorbed for Mild Steel is found out to be 1.95 Joules/mm2.


AIM

To determine the stiffness of the spring, maximum shear stress and modulus of

rigidity of the closed coil helical spring.

APPARATUS REQUIRED

Spring testing machine

Vernier caliper

Measuring scale

Closed coil helical spring


What is stiffness?

Differentiate between closed and open coil helical spring.


Theory

Springs are elastic member which distort under load and regain their original shape

when load is removed. They are used in railway carriages, motor cars, scooters, motorcycles,

rickshaws, governors etc. According to their uses the springs perform the following

Functions:

1) To absorb shock or impact loading as in carriage springs.

2) To store energy as in clock springs.

3) To apply forces to and to control motions as in brakes and clutches.

4) To measure forces as in spring balances.

5) To change the variations characteristic of a member as in flexible mounting of

motors.

The spring is usually made of either high carbon

steel (0.7 to 1.0%) or medium carbon alloy steels.

Phosphor bronze, brass, 18/8 stainless steel and metal

and other metal alloys are used for corrosion resistance

spring. Several types of spring are available for different

application. Springs may classify as helical springs, leaf

springs and flat spring depending upon their shape. They

are fabricated of high shear strength materials such as

high carbon alloy steels spring form elements of not only

mechanical system but also structural system. In several

Exp.No: 07 TEST ON CLOSED COIL HELICAL SPRING

Date:


cases it is essential to idealize complex structural systems by suitable spring.

Maximum shear stress = 8wD/ πd3

Stiffness of the spring = w/s

Modulus of rigidity = 8wnD3/δd4

Procedure

1) Measure the diameter of the wire of the spring by using the micrometer.

2) Measure the diameter of spring coils by using the Vernier caliper

3) Count the number of turns.

4) Insert the spring in the spring testing machine and load the spring by a suitable

weight and note the corresponding axial deflection in tension or compression.

5) Increase the load and take the corresponding axial deflection readings.

6) Plot a curve between load and deflection. The shape of the curve gives the stiffness

of the spring.

Observations

Diameter of the spring wire, d = 05 mm

Diameter of the spring coil, D1 = 38 mm

Mean coil diameter D = 38-5 =33 mm

Number of turns (n) = 18

Height of the spring (h) = 155 mm

Load

applied

w (kN)

Deflection s (mm) Modulus of

rigidity

(kN/mm2)

Maximum

shear stress

(kN/mm2)

Stiffness

(kN/mm) Loading Unloading Mean

0.2 16 13 14.5 114.21 0.1345 0.014

0.4 30 28 29 114.20 0.2689 0.0138

0.6 42 40 41 121.69 0.4034 0.0146

0.8 58 54 56 118.28 0.5378 0.0193

1.0 72 72 72 114.99 0.6723 0.0139

VIVA QUESTIONS

1. Define Stiffness of a helical spring.

2. Differentiate between closed and open coil helical spring.


1. Brittle materials generally fail by shearing along the planes inclined at 50 to 60

degree to the longitudinal axis. Why?

2. Enumerate the mechanical properties of materials. What is their significance?


RESULT

The value of spring constant k of closely coiled helical spring is found to be

0.014 kN /mm

Modulus of rigidity = 114.21 kN/mm2

Maximum shear stress = 0.1345 kN/mm2


AIM

To determine the stiffness of the spring, maximum shear stress and modulus of

rigidity of the open coil helical spring.

APPARATUS REQUIRED

Spring testing machine

Vernier caliper

Measuring scale

open coil helical spring


What is stiffness?

Differentiate between closed and open coil helical spring.


Theory

Springs are elastic member which distort under load and regain their original shape

when load is removed. They are used in railway carriages, motor cars, scooters, motorcycles,

rickshaws, governors etc. According to their uses the springs perform the following

Functions:

1) To absorb shock or impact loading as in carriage springs.

2) To store energy as in clock springs.

3) To apply forces to and to control motions as in brakes and clutches.

4) To measure forces as in spring balances.

5) To change the variations characteristic of a member as in flexible mounting of

motors.

The spring is usually made of either high carbon steel (0.7 to 1.0%) or medium carbon

alloy steels. Phosphor bronze, brass, 18/8 stainless steel and metal and other metal alloys are

used for corrosion resistance spring. Several types of spring are available for different

application. Springs may classify as helical springs, leaf springs and flat spring depending

upon their shape. They are fabricated of high shear strength materials such as high carbon

alloy steels spring form elements of not only mechanical system but also structural system. In

several cases it is essential to idealize complex structural systems by suitable spring.

Exp.No: 08 TEST ON OPEN COIL HELICAL SPRING

Date:


Maximum shear stress = 8wD/ πd3

Stiffness of the spring = w/s

Modulus of rigidity = 8wnD3/δd4

Procedure

1) Measure the diameter of the wire of the spring by using the micrometer.

2) Measure the diameter of spring coils by using the Vernier caliper

3) Count the number of turns.

4) Insert the spring in the spring testing machine and load the spring by a suitable

weight and note the corresponding axial deflection in tension or compression.

5) Increase the load and take the corresponding axial deflection readings.

6) Plot a curve between load and deflection. The shape of the curve gives the stiffness

of the spring.

Observations

Diameter of the spring wire, d = 08 mm

Diameter of the spring coil, D1 = 59.6 mm

Mean coil diameter D = 59.6-8 = 51.6 mm

Number of turns (n) = 10

Height of the spring (h) = 140 mm

Load

applied

w (kN)

Deflection s (mm) Modulus of

rigidity

(kN/mm2)

Maximum

shear stress

(kN/mm2)

Stiffness

(kN/mm) Loading Unloading Mean

0.2 7 8 7.5 71.56 0.0513 0.027

0.4 12 12 12 89.44 0.1026 0.0333

0.6 19 18 18.5 87.03 0.1539 0.0324

0.8 24 23 23.5 91.35 0.2053 0.6340

1.0 29 29 29 92.53 0.2566 0.0344

1.2 34 33 33.5 96.12 0.3079 0.6338

1.4 39 39 39 96.33 0.3593 0.0359

1.6 45 45 45 95.41 0.41062 0.0355

1.8 50 51 50.5 95.64 0.4619 0.0356

2.0 56 56 56 95.83 0.5133 0.0357

VIVA QUESTIONS

1. Define Stiffness of a helical spring.

2. Differentiate between closed and open coil helical spring.


1. Brittle materials generally fail by shearing along the planes inclined at 50 to 60

degree to the longitudinal axis. Why?

2. A brittle material can be drawn into


RESULT

The value of spring constant k of closely coiled helical spring is found to be 0.014 kN /mm

Modulus of rigidity = 114.21 kN/mm2

Maximum shear stress = 0.1345 kN/mm2


AIM

To conduct torsion test on mild steel specimen to find out modulus of rigidity.

APPARATUS REQUIRED

A torsion testing machine.

Specimen

Measuring scale

Vernier caliber


What is Torsional force?

What is torsional rigidity?


Theory

A torsion test is quite instrumental in determining the value of modulus of rigidity of

a metallic specimen. The value of modulus of rigidity can be found out thought observations

made during the experiment by using the torsion equation

Where,

MT = Applied torque

G = Shear Modulus or Modulus of rigidity

Ѳ = Angle of twist in radians

L = length of the cylindrical bar

J = polar moment of inertia

r = radius of the cylindrical bar

γ = shear strain

Exp.No: 09 DETERMINATION OF TORSIONAL STRENGTH OF A GIVEN MILD STEEL ROD

Date:


Procedure

Measure the length and diameter of the given specimen

Select an appropriate range in the torque scale for the rod specimen

Fix the specimen in the appropriate grips and measure the length of the specimen

Adjust the torque scale to zero position

Apply the torque to the specimen till it fails

Note down the maximum torque applied to the specimen

Calculate the maximum twist (Ѳmax), stress developed (τ) and modulus of rigidity (G).

Observations

Material of the specimen: Mild steel

S.No Load Angle of twist Shear stress (N/mm2) Modulus of rigidity

(N/mm2)

1. 134 180 309.88 1973.03

2. 157.5 860 364.38 1160.2

3. 185 720 427.72 908.02

4. 202 1080 467.44 744.3

VIVA QUESTIONS

1. Differentiate Shear Strain and Shear stress?

2. What is factor of safety?

3. What is Torsional force?

4. Polar moment of inertia.

5. What is torsional bending?

6. Explain about modulus of rigidity.


1. Enumerate the mechanical properties of materials. What is their significance?

2. Discuss the silent features of fatigue testing of a material.

RESULT

Maximum angle of twist = 25.12 radians

Maximum shear stress = 467.47 N/mm2

Modulus of rigidity = 744.3 N/mm2


AIM

To find the values of bending stresses and young’s modulus of elasticity of the

material of a beam simply supported at the ends and carrying a concentrated load at the

center.

APPARATUS REQUIRED

Deflection of beam apparatus

Pan

Weights

Beam


What is Inertia?

Types of beams.


Theory

If a beam is simply supported at the

ends and carries a concentrated load at its

center, the beam bends concave upwards. The

distance between the original position of the

beams and its position after bending at different

points along the length of the beam, being

maximum at the center in this case. This

difference is known as ‘deflection’ In this

particular type of loading the maximum amount

of deflection (δ) is given by the relation,

δ = W l3 /48 EI ………… (i)

E = W l3 /48 δI ------------- (ii)

Where,

W = Load acting at the center (N)

L = Length of the beam between the supports (mm)

E = Young’s modulus of material of the beam (N/mm2)

I = Second moment of area of the cross- section (i.e., moment of Inertia) of the beam,

about the neutral axis (mm4)

Exp.No: 10 DEFLECTION TEST ON METAL BEAM

Date:


Bending stress

As per bending equation,

𝜎 = 𝑀 × 𝑦

𝐼

Where,

M = Bending moment, N-mm

I = Moment of inertia, mm4

σ = Bending stress, N/mm2

y = Distance of the top fiber of the from the neutral axis

Procedure

1. Adjust cast- iron block along the bed so that they are symmetrical with respect to

the length of the bed.

2. Place the beam on the knife edges on the block so as to project equally beyond each

knife edge. See that the load is applied at the center of the beam

3. Note the initial reading of Vernier scale.

4. Add a weight of 20N (say) and again note the reading of the Vernier scale.

5. Go on taking readings adding 20N (say) each time till you have minimum six

readings.

6. Find the deflection (δ) in each case by subtracting the initial reading of Vernier

scale.

7. Draw a graph between load (W) and deflection (δ). On the graph choose any two

convenient points and between these points find the corresponding values of W and δ.

Putting these values in the relation

δ = WL3/48 EI

Calculate the value of E

8. Calculate the bending stresses for different loads.

OBESERVATION

Material of the specimen: Mild steel

Load Deflection (mm) Mean deflection

(mm)

Young’s Modulus

*103 (N/mm2) gms N Loading Unloading

500 4.905 0.66 0.68 0.67 237.58

1000 9.81 1.33 1.34 1.335 234.91

1500 14.715 2.09 2.10 2.095 224.54

2000 19.620 2.79 2.79 2.79 224.80


VIVA QUESTIONS

1. Say something on deformable solids?

2. Differentiate simple and compound stress

3. Tell About Moment of inertia.

4. Explain about Principal plane.

5. Explain about Principal axis.

6. Draw Shear force diagram for a cantilever beam with udl and point load.

7. Draw Shear force diagram for a SSB with udl and point load

8. Explain the equilibrium condition for a body.

9. Differentiate between Bar and column

10. Define Section modulus.

11. Unit of force, deflection, stress, strain, E, K, G.


1. A composite circular shaft is comprised of a steel core surrounded by an

aluminum annulus, perfectly bonded to each other as shown in the figure. If it

subjected to a pure torque, which one of the following statements is TRUE?

(A) Only shear stress is continuous across the steel–aluminum interface

(B) Only shear strain is continuous across the steel–aluminum interface

(C) Both shear stress and shear strain are continuous across the steel–

aluminum interface

(D) Both shear stress and shear strain are discontinuous across the steel–

aluminum interface

2. A horizontal rectangular plate ABCD is hinged at points A, B and C. AC and BD

are diagonals of the plate. Downward force P is applied at D. The upward

reactions RA, RB, and RC at points A, B and C, respectively, are

(A) indeterminate

(B) P, -P, P

(C) 0, P, 0

(D) P/3, P/3, P/3

RESULT

The Young’s modulus of the given beam is found to be 237.8 * 103 N/mm2.

Documents

Exp.No: 01 DETERMINATION OF HARDNESS OF A GIVEN MATERIAL