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:
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 | 2 |
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.
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 | 3 |
RESULT
Thus the hardness of the material found to be
RHN of copper = 110.5
RHN of brass = 109
RHN of aluminum = 106
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 | 4 |
AIM
To determine the Brinell’s hardness number (BHN) of the given specimen.
APPARATUS REQUIRED
Brinell testing machine
Specimens
Ball indenter
Micrometer
PREREQUISITE QUESTIONS
List out the mechanical properties of a material
What is meant by red hardness or hot hardness?
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
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:
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 | 5 |
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
STIMULATING QUESTIONS
Enumerate the advantages of Rockwell Hardness test over Brinell hardness test.
Interpret the importance of material’s hardness in gears.
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 | 6 |
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
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 | 7 |
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
PREREQUISITE QUESTIONS
Define Hooke’s Law.
Define yield point.
How can you measure tensile strength?
CONDUCT OF EXPERIMENT
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:
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 | 8 |
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.
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 | 9 |
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
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 | 10 |
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.
STIMULATING QUESTIONS
1. Tensile strength is always lower than compressive strength – tell your opinion
2. Mild steel fails due to shear – tell your opinion
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 | 11 |
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
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 | 12 |
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
PREREQUISITE QUESTIONS
What do you mean by compressive strength of a material?
Define bulk modulus.
What is volumetric strain?
CONDUCT OF EXPERIMENT
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:
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 | 13 |
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.
STIMULATING QUESTIONS
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
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 | 14 |
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
PREREQUISITE QUESTIONS
What is impact load?
What is Strain energy?
CONDUCT OF EXPERIMENT
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:
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 | 15 |
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.
STIMULATING QUESTIONS
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.
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 | 16 |
AIM
To determine the impact strength of a given materials by Charpy test
APPARATUS REQUIRED
Impact testing machine
Specimen
PREREQUISITE QUESTIONS
What is impact load?
What is Strain energy?
CONDUCT OF EXPERIMENT
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:
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 | 17 |
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.
STIMULATING QUESTIONS
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.
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 | 18 |
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
PREREQUISITE QUESTIONS
What is stiffness?
Differentiate between closed and open coil helical spring.
CONDUCT OF EXPERIMENT
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:
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 | 19 |
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.
STIMULATING QUESTIONS
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?
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 | 20 |
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
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 | 21 |
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
PREREQUISITE QUESTIONS
What is stiffness?
Differentiate between closed and open coil helical spring.
CONDUCT OF EXPERIMENT
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:
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 | 22 |
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.
STIMULATING QUESTIONS
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
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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
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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
PREREQUISITE QUESTIONS
What is Torsional force?
What is torsional rigidity?
CONDUCT OF EXPERIMENT
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:
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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.
STIMULATING QUESTIONS
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
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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
PREREQUISITE QUESTIONS
What is Inertia?
Types of beams.
CONDUCT OF EXPERIMENT
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:
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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
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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.
STIMULATING QUESTIONS
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.