material testing lab manual - ammini.edu.in 407(P)_ Material Testing...ME09 407(P) Material testing Lab Dept. of Mechanical Engineering 2 CONTENT Sl No. Name of Experiment Page No

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ME09 407(P) Material testing Lab Dept. of Mechanical Engineering

2

CONTENT Sl No. Name of Experiment Page No.

1 STUDY OF STRESS STRAIN CHARACTERISTICS 3 - 4

2 TENSION TEST 5 - 12

3 DOUBLE SHEAR TEST 13 - 15

4 TORSION TEST 16 - 20

5 TORSION TEST ON WIRES 20 - 25

6

SPRING TEST

(a) Open coiled spring 26 - 29

(b) Close coiled spring

30 - 32

7 IMPACT TEST

(a) Izod test 33 - 36

(b) Charpy test 37 - 40

8 HARDNESS TEST (a) Brinell Hardness test 41 - 43

(b) Rockwell Hardness test 44 – 46

(c) Viker’s Hardness test 47 – 49

9 BENDING TEST ON WOODEN BEAMS 50 - 53

10 FATIGUE TEST 54 - 57



3

STRESS STRAIN CHARACTERISTICS

Study: 1

Stress Strain characteristics of different materials determine their behaviour under different

loading conditions. The behaviour of materials depends upon the rate of loading and the type

of testing used to find the stress strain character. Stress-strain curves are an extremely

important graphical measure of a material’s mechanical properties. However, they are not

without some subtlety, especially in the case of ductile materials that can undergo substantial

geometrical change during testing. This will provide an introductory discussion of several

points needed to interpret on these curves, and in doing so will also provide a preliminary

overview of several aspects of a material’s mechanical properties.

One of the simplest tests for determining mechanical properties of a material is the tensile

test. In this test, a load is applied along the longitudinal axis of a circular test specimen. The

applied load and the resulting elongation of the member are measured. In many cases, the

process is repeated with increased load until the desired load levels are reached or the

specimen breaks. Load-deformation data obtained from tensile and/or compressive tests do

not give a direct indication of the material behavior, because they depend on the specimen

geometry.

However, using the relationships we previously discussed, loads and deformations may be

converted to stresses and strains.

σ PA

, ε δL

σ = normal stress on a plane perpendicular to the longitudinal axis of the specimen

P = applied load ,A = original cross sectional area

ε = normal strain in the longitudinal direction

δ = change in the specimen’s gage length ,L = original gage length

The resulting stress-strain curve or diagram gives a direct indication of the material

properties.


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5

TENSION TEST

Expt. No.: 2

Date :

AIM:

To conduct the tensile test on a mild steel specimen and determine the following 1. Yield strength 2. Ultimate strength 3. Young’s modulus of elasticity 4. Percentage

elongation and 5.Percentage reduction in Area

APPARATUS AND ACCESSORIES:

400KN Universal Testing Machine, Extensometer, mild steel specimen, Punching tool, hammer, Verniercaliper, Meter Scale.

400KN Capacity U.T.M:

The instrument can be used for various tests such as tension, compression, shear and bending. It can be operated under four different ranges: 40KN, 100KN, 200KN and 400KN, depending on the type of test, strength of the material and the accuracy required. The machine consists of 2 main units, the loading unit and the control panel.

Loading Unit

This consists of a robust base. The main hydraulic cylinder is fitted in the centre of the base. A geared motor is fitted to the base and the chain and sprocket driven by the motor rotate the two screwed columns mounted in the base. The lower table is connected to the main piston through a ball and seat joint. This joint ensures axial loading.

Control Panel

This consists of a hydraulic power pack which contains the hydraulic oil. Oil level sight glass is fitted to the oil tank to check the oil tank. A positive displacement type piston pump is provided to assure a continuous high pressure non- pulsating oil flow for the smooth application of the load on the specimen. A pendulum dynamometer is fitted to measure and indicate the force coming on the specimen. A big size load indicating dial fitted with a glass cover is mounted at the side of the control panel. The range indicating dial (located at the back side of the load indicating dial) is to be adjusted for the particular range selected.

Extensometer:

This instrument is used to measure the extension of the bar when it is loaded. It mainly consists of two vertical rods with dial gauges. The extension of the specimen is noted from the two dial gauges.


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7

MACHINE DETAILS:

Machine : Universal Testing Machine(UTM)

Capacity : 400 KN

Gauge Length adjustable from : 30 - 120 mm ( 30 - 70 mm can also be given )

Extensometer : Least Count = ............... mm.

THEORY:

Various machine and structure components are subjected to tensile loading in numerous applications. For safe design of these components, their ultimate tensile strength and ductility has to be determined before actual use. Tensile test can be conducted on UTM. A material when subjected to a tensile load resists the applied load by developing internal resisting force. This resisting force per unit cross-sectional area is known as stress. The value of stress in material goes on increasing with an increase in applied tensile load, but it has a certain maximum (finite) limit too. The minimum stress, at which a material fails, is called ultimate tensile strength.

The end of elastic limit is indicated by the yield point (load). This can be seen during experiment with increase in loading beyond elastic limit, original cross-section area goes on decreasing and finally reduces to its minimum value when the specimen breaks.

PROCEDURE:

1) Clean the given specimen with sand paper and measures its diameter at three different points and obtain the mean diameter.

2) Calculate the gauge length of the specimen using the empirical formula 5.65√A, where A is the mean area of cross section of the specimen. Mark the gauge length on the specimen by punch marks leaving equal distance at both ends for gripping.

3) The permissible tensile stress for mild steel bars may be taken as 140N/mm2. Assuming a factor of safety of 3 to 4, the ultimate tensile stress can be calculated. The ultimate load that can be taken by the specimen equals the product of area of cross section and the ultimate stress.

4) Insert the test piece in the grips by adjusting the cross head of the machine after making zero correction.

5) Keep the left valve (outlet valve) in fully closed position and the right valve (inlet) in normal open position. Open the right valve and close it after the lower table is slightly lifted.



8

OBSERVATIONS:

1) Diameter of the specimen

Least count of the Vernier = …………………....

SL NO. Main Scale

Reading

( MSR)

Vernier Scale

Reading

(VSR)

Diameter of the bar (mm)

MSR+(VSR x LEAST COUNT)

1

2

3

2)Load and extensometer dial gauge readings

Least count of the extensometer dial gauge = …………………...

Average diameter of the specimen = ……………mm

Load P (KN) Extensometer Reading

Average extensometer Reading

Re (divisions) Left dial gauge

Right dial gauge

0

5

10

15

20

25

30

35

40

45

50

55

60



9

6) Now adjust the load to zero by Tare push button. (This is necessary to remove the dead weight of lower table, upper cross head and other connecting parts from the load).

7) The extensometer is properly fixed to the middle of the specimen and its dial gauges are set to zero.

8) Start the machine and pump some oil into the cylinder by opening the inlet valve.

9) The outlet valve is closed and inlet is opened slightly and adjusted so as to ensure a uniform rate of loading. As the load increase, the extension also increases. The readings of both the dial gauges of the extensometer corresponding the load are noted and entered. Just at the yield point, extensometer must be removed from the specimen. At the yield point, the pointer will remain stationary on the dial for some time or it may kick back slightly. It is important that the inlet valve should be closed while removing the extensometer from the specimen. The load at the yield point when the pointer remain stationary or when it kicks back may also be noted.

10) The loading is then continued by opening the inlet valve. The pointer moves forward. On reaching the maximum load, it will be found that a neck is formed on the specimen. The pointer now moves backwards. Carefully watch the pointer and note the load at the time of breaking the specimen. The maximum load is indicated by the dummy pointer, which is also noted. After the specimen is broken, the loading or inlet valve is closed and the outlet is opened. The two pieces of the broken specimen is removed and the machine is switched off.

11) The two broken piece are placed together on a horizontal surface and the distance between the punch marks is measured and noted.

12) The reduced diameter of the specimen at the point of breaking is measured and reduced cross sectional area is calculated.

13) Finally a graph is plotted with stress on Y-axis and strain on X-axis in arithmetic

scale. The slope of the stress strain graph within elastic limit gives the modulus of

elasticity of the material.

RESULT:

1) Yield point Stress = …………………..N/mm2

2) Ultimate Stress = …………………..N/mm2

3) Nominal breaking stress = …………………..N/mm2

4) Actual breaking stress = …………………..N/mm2

5) Percentage elongation = ………………….. (%)



10

Length between punch marks after breaking = ………………mm

Diameter of the neck after breaking = …………….mm

CALCULATION:

Average diameter of the specimen = …………………. mm

Original cross sectional area (A) = ………………… mm2

Gauge length ( ) = 5.65√A = ………………… mm

Permissible tensile stress = 140 N/mm2

Factor of safety = 4

Ultimate stress = 140 x 4 = 560N/mm2

Ultimate load = 560 x A = …………….N

Range Selected = ……………..N

Load P (KN) Average

extensometer

Reading Re

(divisions)

Extension

δ =

(mm)

Stress = (P/A)

(N/mm2)

Strain = /



11

6) Percentage reduction area = ………………….. (%)

7) Young’s modulus = ……………………N/mm2

INFERENCE:

VIVA QUESTIONS:

1. Which steel have you tested?

ANS:

2. In what region of a stress vs. strain graph do you find Young’s Modulus?

ANS:

3. Why do we remove the extensometer after yielding occurs?

ANS:

4. What general information is obtained from the tensile test regarding the properties

of the material?

ANS:

5. Which stress have you calculated: Nominal stress or true stress?

ANS:

6. What kind of fracture has occurred and why?

ANS: Metals with good ductility normally exhibit a cup and cone fracture.

7. Which is the most ductile material? What is its elongation?

ANS:



12

GRAPH:

STRESS (N/mm2)

STRAIN

Yield point stress = L O C S

N/mm2 =

=

Ultimate Stress =

N/mm2 =

=

Nominal breaking stress =L BO

N/mm2 =

=

Actual breaking stress = L BF

N/mm2 =

=

Percentage elongation = O F O

100 (%) =

=

Percentage reduction area = O

x 100 (%) =

=

Young’s modulus (from graph) = N/mm2 =

=



13

DOUBLE SHEAR TEST

Expt. No.: 3

Date :

AIM:

To determine the shear strength of given material subjecting to fail under double shear.


Universal Testing machine of 40T,Shear shackle,mild steel specimen,Vernier calliper.

Double Shear tool:

In Double Shear test, the specimen is allowed to shear off at two cross sections under a compressive load. The schematic diagram for shear testing equipment under compressive load is shown in figure.

The shear tool consists of a suitable fabricated shackles with the specimen inserted as a connecting pin between the shackles. The fork plates of the shackles are firmly held together to eliminate any bending in the specimen under heavy loads. Separate dies are used to hold of different diameters.

THEORY:

When the cross section of a structural member is subjected to two equal and opposite forces acting tangentially across the section, shear stress is developed. The equal and opposite tangential forces acting across the section is called the shear force. When there exists only uniform shear stress on the plane it is called pure shear stress. The maximum resistance offered by the material under the action of shear force is called ultimate shear stress.

The shear stress developed across the section, τ =F/A

Where F = Shear force acting over the area

A= Area of cross section

If the shear force acts over two area of cross sections of the member, then the shear Stress =F/2A



14

OBSERVATIONS AND CALCULATIONS:

1) Diameter of the specimen Least count of the Vernier = …………………....

SL

NO.

Main Scale

Reading

( MSR)

Vernier Scale

Reading

(VSR)

Diameter of the bar (mm)

MSR+(VSR x LEAST COUNT)

Average

Diameter

(mm)

1

2

3

Area of cross section (A) = …………….. mm2

Assumed Ultimate Shear Strength = 500 kg/cm2 = 490.5 N/mm2

Maximum expected load = permissible shear stress X factor of safety X 2 X A

=

Range selected = …………………… KN

Breaking load, F = …………………... KN

Ultimate shear stress,τ = (F/ 2A) x1000 = …………………. N/mm2

Average diameter of the specimen = ……………mm



15

PROCEDURE:

1) Clean the given specimen with sand paper and measures its diameter at three different

points and obtain the mean diameter.

2) For mild steel, Maximum shear stress may be taken as 5000kg/cm2.Double shear

strength of the specimen can be calculated as F = 5000 x 2 x Area of cross section.

The range of the machine may be selected suitably according to the double shear

strength calculated.

3) Insert the specimen into the shear tool fitted with the appropriate dies. The specimen

should project equally on either side of the tool.

4) Place the tool with the specimen centrally on the lower cross head and the middle

cross head is adjusted just to touch with the top surface with the shear tool, using the

quick setting arrangement.

5) Close the outlet valve and open the inlet valve slowly and the load is applied to the

specimen. When the specimen breaks, note the breaking load.

RESULT:

Ultimate shear stress = ...................N/mm2

INFERENCE:

VIVA QUESTIONS:

1. What is single & double shear?

ANS:

2. What is find in shear test?

ANS:

3. What is the unit of shear strength?

ANS:



16

TORSION TEST

Expt. No.: 4

Date :

AIM:

To conduct torsion test on mild steel specimen for determining the modulus of rigidity.


Torsion testing machine, Standard specimen, Vernier calliper, Meter Scale.

Torsion testing machine:

It consists of two units namely the load applying unit and the load-indicating unit. The load applying unit is a horizontal platform consisting of two circular cross heads provided with rectangular shaped grips to hold the ends of the specimen which is specially prepared for the test. One of the cross head is fixed directly to the load indicating unit which is connected to a pendulum dynamometer. A pointer on a dial, which gives the torque in kg/cm, indicates the torque supplied on the specimen.

The other cross head can be moved in or out to fix the specimen. It can be rotated manually to provide twisting moment to the specimen. The angle of twist can be noted on a circular drum, which is graduated in degrees. Both the straining unit and the indicating unit are the attached at right angles to each other.

Test Specimen:

The cross section of the specimen is circular. The ends of the specimen are flattened to avoid slipping during twisting. The diameter near the ends is made larger than that at the centre. Gauge length of the specimen is the distance of the cylindrical portion. The dimensions of the gripped ends may be made according to the end grips of machine of used.

THEORY:



17

DIAGRAM:

GRAPH:



18

Let a cylindrical shaft or length L and radius R is subjected to a torque T. The torsion formula

is = = =

Where T = Torque in Nm

J = Polar moment of inertia =

D = Diameter of the shaft

F = Shear stress developed at the outer radius R

q = Shear stress at a radius r

N = Modulus of Rigidity

= Angle of twist in radians

L = length of the shaft

The relation between Modulus of Elasticity E, Modulus of Rigidity N and Poisson’s ratio 1/m

is given as ⎟⎠⎞

⎜⎝⎛ +=

mNE 112

Poisson’s ratio = 1

PROCEDURE:

1) Measure the diameter and the gauge length of the given specimen, which is specially

made to suit for the end grips.

2) Pull out the outer cross head and tightly fix one end of the specimen between the grips

of the fixed cross head attached to the load indicating unit.

3) The outer cross head is pushed in and the other end of the specimen is also fixed

tightly between the grips.

4) The reading to measure the angle of twist may be set to zero. The pointer on the dial

of the indicating unit is also set to zero.

5) Apply torsion to the specimen by rotating the hand wheel slowly in the anticlockwise

direction till the angular deformation is 1o. The corresponding twisting moment is

directly read from the position of the pointer on the dial in Kg-cm.

6) Repeat the procedure from 2 to 5 and the respective twisting moments are noted.



19

OBSERVATIONS:

Diameter of the specimen = .............................mm

Gauge length of the specimen = ..............................mm

Angle of twist in

(Degrees)

Angle of twist in (Radians)

Torque applied (Kg-cm) Average Torque

(Kg-cm)

Loading Unloading

0

1

2

3

4

5

6

CALCULATION:

JT =

Rf =

LNθ =

rq

==∴θJ

TLN

J (mm) = =

T from graph =...................... Nmm/rad

Modulus of Rigidity, N =......................... N/mm2

E for steel = 2.1 x 105 N/mm2 (Assumed)

Poisson’s ratio = ⎥⎦⎤

⎢⎣⎡ −= 12

1NE

m = ..........................



20

7) By rotating the hand wheel in the clockwise direction the torque applied on the

specimen can be released. The twisting moment for every degree of rotation should

also be noted while releasing the torque.

8) Finally when the angle of twist reaches zero the torque applied to the specimen will

also be zero. Now the specimen can be removed from the machine.

9) Finally draw a graph showing angle of twist along X-axis and average torque along

Y-axis.

RESULT:

Rigidity modulus = ...................N/mm2

Poisson’s ratio = .........................

INFERENCE:

VIVA QUESTIONS:

1. What are the assumptions made in this experiment?

ANS:

2. Define Poisson’s ratio? ANS:



21

TORSION TEST ON WIRES

Expt. No.: 5

Date :

AIM:

To determine the Modulus of rigidity of given wire.


Torsion Pendulum, cylindrical wire, vernier callipers, screw gauge, stop watch, metre scale, etc.

Torsion Pendulum Apparatus:

The torsion pendulum apparatus consists of

(1) Torsion pendulum in the form of disc, ring and a thin wire

(2) A bracket for suspending for suspending the Pendulum.

THEORY:

A disc suspended by elastic wire in equilibrium position, when twisted slightly in one direction, begins to oscillate about its equilibrium position i.e., it becomes a torsion pendulum.

To find modulus of Rigidity of wire:

The time of oscillation of a torsion pendulum with no weight is given by,

t1 = = 2

Where, I1 = mass moment of inertia of the suspended disc about the vertical axis.

L = Effective length of the wire

N = Modulus of rigidity of the material of the wire

J = Polar moment of inertia of the wire =

d = diameter of the wire N= Rigidity Modulus



22

DIAGRAM:

OBSERVATIONS:

Diameter of given wire (D) = .............................mm

Length of the wire (L) = ..............................mm

weight of central disc or ring (W) = ..............................gm

Outer radius of ring (R0) = .............................. cm

Inner radius of ring (Ri) = .............................. cm

CALCULATION:

Polar moment of inertia (J) = = ......................... mm3

Mass moment of inertia of ring, I2= WR2/2g =

= ......................... gcm2

Modulus of rigidity (Shear modulus)

= ......................... dyne/cm2

M.I of ring = W/2g [R02 + Ri

2] =



23

Now place the central weight on the disc and note the period of oscillation. Let it be t2.

t2 = 2 1 = 2 2 1

Where I2 = MR2/2 =Mass M.I of weight about axis of rotation.

Or I2= WR2/2g (W = weight of central disc or ring)

Hence, =

Or Modulus of rigidity (Shear modulus)

To find moment of inertia of ring:

Let W0 = Weight of larger disc

Wi = Weight of smaller disc

R0 = Radius of larger disc = Outer radius of ring

Ri = Radius of smaller disc = Inner radius of ring

ρ = Density of material of ring

t = Thickness of ring

M.I of ring = M.I of larger disc – M.I of smaller disc

= 1/2g [W0 R02 - Wi Ri

2] = 1/2g [ ρtπ R02R0

2 - ρtπ Ri2Ri

2]

= ρtπ /2g [ R04 - Ri

4] = ρtπ /2g [(R02 + Ri

2) (R02 - Ri

2)]

= (R02 + Ri

2) [ρtπ /2g (R0 + Ri) (R0 - Ri)

M.I of ring = W/2g [R02 + Ri

2] Where W= Weight of ring



24

Mod

ulus

of

rigid

ity, N

(N/m

m2 ) M

ean

10

(sec

)

10

(sec

)

Tim

e fo

r 10

osci

llatio

ns w

ith

wei

ght (

T 2) s

ec

mea

n

3 2

1

Tim

e fo

r 10

osci

llatio

ns w

ithou

t

wei

ght (

T 1) s

ec

mea

n

3 2

1

Leng

th

L

(mm

)

Spec

imen

Stee

l



25

PROCEDURE:

1) Measure the diameter of the wire using screw gauge.

2) Suspend the disc by the wire and clamp it on the stand.

3) Measure the length of the wire from the point of oscillation.

4) Apply a twist to the disc and allow it to oscillate. When the oscillation becomes

steady, start the stopwatch and note the time interval for 10 oscillations (T1).

5) Place the weight on the disc and find the time interval (T2) for 10 oscillations.

6) Repeat the experiment with three different lengths of the pendulum.

RESULT:

Rigidity modulus of the wire (Steel) = ...................N/mm2

INFERENCE:

VIVA QUESTIONS:

1. What are the tests that can be used for finding modulus rigidity of materials?

ANS:

2. The period of oscillation increases as the length of pendulum. Is this statement

correct or not?

ANS:

3. What is the different between plane moment of Inertia and polar moment of

inertia?

ANS:

4. What is the use of cylindrical weights in Torsion Pendulum?

ANS:



26

SPRING TESTS

Expt. No.: 6

Date :

AIM:

To determine the stiffness of the given spring and the Modulus of Rigidity of the Spring

material.


Spring testing machine, vernier calipers, Meter scale, open coiled and close coiled spring etc.

THEORY:

A helical spring is a piece of wire coiled in the form of helix. The helix of the coil is

so small that the bending effects can be neglected, then the spring is called a Close coiled

spring. In such a spring only torsional shear stress are introduced. On the other hand, if the

slope of the helix of the coil is quite appreciable, then both bending as well as torsional

stresses are introduced in the spring and a spring of this type is called an Open coiled spring.

Close Coiled Springs:

Close Coiled helical spring is a torsion spring and the length of the spring may be

taken equal to 2Rn, where ‘n’ is the number of turns and ‘R’ is the mean radius of the coil. If

W is the axial load applied to the spring, then the corresponding deflection,

δ =

Where, N is the Modulus of rigidity of the material of the spring.

d is the diameter of the wire forming the helix.

Open Coiled Springs:

In the case of open coiled helical spring, the coils are not close together and the helix

angle ‘α‘cannot be treated as small. The bending couple cannot be considered as negligible in

comparison with torsion couple. Thus, an open coiled helical spring falls under categories of

springs that is the torsion spring and the bending spring.



27


Tension Test On Close Coiled Helical spring

Main scale

Reading

MSR(mm)

Vernier scale

Reading

VSR(mm)

VSR x LC

(mm)

MSR+(VSR x LC)

(mm)

Diameter of

the wire

External

diameter of

the coil

Average external diameter of the coil, D1 = mm

Average diameter of wire, d = mm

Internal diameter of the coil, D2 = D1-2d

= --------------mm

Average diameter of the coil =

= ---------------mm

Average radius of the coil, R =

= -------------mm

Load in ‘N’

Deflection in‘mm’ Average deflection in

mm Loading Unloading

2

4

6

8

10



28

Let, W = Load applied axially to the spring

R = Mean radius of the coil

d = diameter of the coil wire.

n = number of turns

α = helix angle

N = modulus of rigidity of the material of the spring

E = modulus of elasticity of the of the material of the spring

δ = deflection of the spring.

1/m = Poisson’s ratio

Then, δ =

E = 2 1

Stiffness of the spring:

Stiffness of a spring is defined as the load required producing unit deflection for it.

Let, W = load acting on the spring in ‘N’

δ = deflection corresponding to the load in ‘mm’.

then,

Stiffness of the spring, k = N/mm

Tension Test On Close Coiled Helical spring

Procedure:

The procedure for tension test on the close coiled spring is given as follows:

1) Using the vernier calipers measure the external diameter of the coil and the coil wire

of the spring. Number of full turns on the spring is also counted.

2) The holder pins with hooks are attached on the middle movable head and the load

measuring attachment provided on the base box and the spring is suspended between

the hooks.



29

CALCULATION:

Modulus of Rigidity of close coiled spring, N = =

= ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ N/mm2

Stiffness of the close coiled spring, k = = = ‐‐‐‐‐‐‐‐‐‐‐‐ N/mm

Compression Test On Open Coil Spring

Main scale

reading(mm)

MSR

Vernier scale

reading(mm)

VSR

VSRxLC

MSR+(VSRxLC)

(mm)

Diameter of

the wire (d)

External

diameter of

the coil D1

Average external diameter of the coil,D1 = mm

Average diameter of wire,d = mm

Internal diameter of the coil, D2 = D1-2d

= --------------mm

Average diameter of the coil = D D

= ---------------mm

Average radius of the coil, R = D D

=-------------mm



30

3) By rotating the three legged loading head a small tensile load is applied on the

specimen. Adjust the tare weight so that the display for the load and deflection are

zero.

4) Rotate the loading head slowly so that the defection on the display scale is 2N. Note

the corresponding load from the dial gauge.

5) The loading may be continued up to a maximum deflection of 10N, noting down the

deflection and the load at every 2N intervals.

6) Repeat the experiment by unloading the specimen, noting down the deflection and

load at every 2N interval.

RESULT:

Rigidity modulus of close coiled spring = ...................N/mm2

Stiffness of the spring = ...................N/mm

Compression Test On Open Coil Spring

Procedure:

The procedure for conducting the compression test on the open coil spring is given below.

1) Measure the external diameter of the coil of the spring and also the diameter of the

wire at least three points. Count the number of full turns on the spring. Measure also

the length of the spring using the scale.

2) Fix the circular plates in their respective positions and the upper plate is raised by

rotating the handle. The spring is placed on the lower plate and the upper plate is

lowered to touch the specimen by rotating the handle.

3) Apply a small load on the specimen and adjust the tare button so that the display for

both load and deflection are set to zero.

4) Operate the straining handle and load the specimen. Note the load corresponding to

2N increment of deflection up to a maximum deflection of 10N.

5) Release the load by rotating the handle in the reverse direction. Note the deflection

corresponding to every 2N deflection.

6) Draw a graph with load (‘w’ N)on the Y-axis and deflection (‘δ’mm) on the X-axis.



31

Length of the spring, L = mm

Number of full turns, n =

Pitch of the coil, p = L/n =

Helix angle = tan α =

α in degrees =

Load in ‘N’

Deflection in‘mm’ Average deflection in

mm Loading Unloading

2

4

6

8

10

CALCULATION:

Modulus of rigidity of the material of the spring,

N =

Stiffness of the spring, k =



32

RESULT:

Rigidity modulus of close coiled spring = ...................N/mm2

Stiffness of the spring = ...................N/mm

INFERENCE:



33

IMPACT TESTS Expt. No.: 7

Date :

AIM:

To find the impact strength (energy required to rupture the specimen) in Izod and Charpy tests. Also to draw calibration curve of the machine used.


Impact testing machine, Standard gauges, Standard specimen

A. Izod Impact Test THEORY:

The test consists of breaking by one blow of a swinging hammer under prescribed conditions,

a standard test piece with V- notch in the specimen. The difference in height of free fall

before impact and height of rebound after the impact of the hammer gives the energy

absorbed by the specimen.

Let W be the weight of the hammer

L be the length of the pendulum

θ1 be the angle measured from the vertical up to the initial position of the pendulum

θ2be the Angle measured from the vertical up to the position of the pendulum after breaking the specimen

θ1

θ2



34


Weight of hammer =

Length of the pendulum =

Initial energy of pendulum =

Final energy of pendulum =

Energy lost due to impact = Initial energy – Final energy

Energy lost due to impact = WL(cosθ2 – cosθ1)

= WL(cosθ2 – cos90 )

Final angle after impact (θ2) in degrees = .................



35

Potential Energy possessed by the weight W before impact = Wh1

= W( 1- Lcosθ1)

Potential Energy possessed by the weight W after impact = Wh2

= W(l- Lcosθ2)

Loss in Potential Energy during impact = Initial P.E. – final P.E

=W(h1-h2)

=WL(cosθ2 – cosθ1 )

For Izod Impact test the initial energy during impact should be 168J. For a hammer

weighing 21.2 Kg and length 0.81m the initial angle of pendulum should be 900. Assuming

different values for θ2 a calibration graph can be plotted with Energy Lost along the Y-axis

and Final Angle θ2 along the X- axis.

Test Specimen:

The test specimen is generally square or circular in section. It consists of a V notch on

the surface. The presence of the notch weakens the section so that under impact loading,

breaking occurs at this section only. The plane of symmetry of the notch is kept normal to the

longitudinal axis of the specimen.

The length of the specimen shall be 75 mm of asqure or circular section of 10 mm size

or 11.4 mm diameter. At 28 mm from the one end , The V notch of 2 mm is cut for square

section and 3.5mm depth for circular section. The included angle of the notch is 45o.

Description of the machine:

The Izod Impact testing machine consists of a vertical post supported on a platform.

The vertical post carries a pendulum of specified length and weight. It can be raised and

locked at a definite height up. A lever is provided for releasing the pendulum.

The specimen can be fixed between wedge grips provided on the platform. Locking

arrangements are also there to keep the specimen in the correct position. Square and circular

shaped dies can be used for holding the respective specimens.



36

A pointer pivoted at the centre moves along with the pendulum. The final energy of the

pendulum after impact can be read from the semi-circular graduated scale provided on the

machine.

PROCEDURE:

The procedure for conducting the Izod impact test is given below:

1) The pendulum hammer is raised to its initial position and locked there. The pointer is

set to read zero on the graduated semicircular scale.

2) The specimen is fixed and locked between the grips to the correct height using the

gauge provided. The V notch should face the striking side

3) After releasing the lock, the hammer is allowed to fall down by pulling the lever. It

will strike on the projecting part of the specimen and break the specimen. Then the

hammer will shoot up to the other side of the machine due to the residual energy

possessed it.

4) Using the hand operated clutch, the swinging of the hammer can be stopped. It is

raised up and kept at the initial position and locked there.

5) The reading shown by the pointer on the graduated disc is noted as the final energy of

the pendulum.

6) The broken test piece is removed from the support by releasing the locking

arrangement.

7) Repeat the test with three other specimens and takes its average to get the result.

The result can also be verified from the calibration curve.

RESULT:

Izod impact value =.................

Final angle after impact (θ2) in degrees

(a) From calculation = .................

(b) From graph = .................



37

B. Charpy Impact Test

THEORY

A pendulum type single blow impact test, in which the specimen is supported at both

ends, as a simple beam, and broken by a falling pendulum on the face opposite to and

immediately behind the notch. The energy absorbed as determined by the subsequent of rise

of pendulum is a measure of impact strength or notch toughness and is expressed as N-m.

Toughness of a material is its ability to absorb energy and deform plastically before fracture.

It is usually measured by the energy absorbed in a notched beam impact test. The energy lost

due to impact can be calculated as explained below.

Potential Energy possessed by the weight W before impact = Wh1

= W( 1- Lcosθ1)

Potential Energy possessed by the weight W after impact = Wh2

= W( l- Lcosθ2)

Loss in Potential Energy during impact = Initial P.E. – final P.E

=W(h1-h2)

= WL(cosθ2 – cosθ1 )

For Charpy impact test the initial energy during impact should be 300J. for a hammer

weighing 21.2kg and length of 0.81 m the initial angle of the pendulum should be 141.8o.

Assuming different values for θ2 a calibration graph can be plotted with Energy Lost along

the Y-axis and Final Angle θ2 along the X- axis.

Test Specimen:

For the Charpy impact test, specimen may be prepared with three types of notches:

1. V notch

2. U notch

3. Key-hole notch.



38


Weight of hammer =

Length of the pendulum =

Initial energy of pendulum =

Final energy of pendulum =

Energy lost due to impact = Initial energy – Final energy = ................

Energy lost due to impact = WL(cosθ2 – cosθ1)

= WL(cosθ2 – cos141.8 )

Final angle after impact (θ2) in degrees = .................



39

The test piece shall have a length of 55 mm and of square section of size 10 mm. The notch is

made at the centre of the test specimen. The plane of symmetry of the notch should be

perpendicular the longitudinal axis of the test piece. The distance between the plane of

symmetry of the notch from the ends of the test piece shall be 27.5mm.

Description of the Machine:

The Charpy Impact testing machine consists of a vertical post supported on a plat

form. The Vertical post carries a pendulum of specified length and weight. It can be raised

and locked at a definite height up. A lever is provided for releasing the pendulum. The

specimen can be placed between the two supports provided on the platform. Centring gauges

are also there to keep the specimen in the correct position. Square dies can be used for

holding square and circular shaped specimen. A pointer pivoted at the centre moves along

with the pendulum. The final angle of the pendulum after impact can be read from the

semicircular graduated scale provided on the machine.

PROCEDURE:

The step-by-step procedure for conducting the Charpy test is given below:

1) The pendulum hammer is lifted up with its tool on it and kept in its initial position

and locked.

2) Adjust the pointer to coincide with the zero reading on the graduated Charpy scale.

3) Place the specimen horizontally on the Charpy support in such a way that the notch is

on the opposite side of the striking face and centered using the centring gauge.

4) After releasing the lock, the lever is pulled and allows the pendulum hammer to fall

and strike with the specimen. With the residual energy, it will swing to the other side.

The pointer also swings back. Note the reading on the Charpy scale indicated by the

pointer. The swinging of the pendulum can be stopped by using handle of the clutch.

5) The pendulum is raised up and kept at its initial position and locked. The broken

specimen is removed.

6) The experiment is repeated with two more specimen and note the reading for each

test.



40

RESULT:

Charpy impact value =.................

Final angle after impact (θ2) in degrees

(a) From calculation = .................

(b) From graph = .................

INFERENCE:



41

HARDNESS TEST Expt. No.: 8

Date :

AIM:

To determine the hardness number using Vickers, Brinell and Rockwell Hardness test.


Vickers Hardness testing machine, Rockwell Hardness testing machine, Brinell Hardness testing machine and Standard specimen.

A. Brinell- Hardness Test

THEORY:

The test consists in forcing a steel ball of diameter ‘D’ under the load ‘F’ into the test piece

and measuring the diameter of indentation left in surface (d). The Brinell hardness is obtained

by dividing the test load F in kgf by the curved surface area of indentation in square mm.

HBS or HBW = T LS

HBS or HBW = FπD D √D

HBS Brinell hardness in case where Steel ball is used for materials whose HB is

not exceeding 450.

HBW Brinell hardness in case where hard metal is used for materials whose HB is

not exceeding 650.

Normally, a ball indenter of 10mm nominal diameter shall be used. Balls of diameter 1, 2, 2.5

and 5mm are also used. It shall be of hardened and tempered steel with a hardness of atleast

850 hardness values.

The thickness of the test piece shall not be less than 8 times the indentation, the distance of

indentation from the edge of the test piece shall be atleast 2.5 times the diameter of

indentation and the distance between the centres of two adjacent indentations shall be atleast

4 times the diameter of indentation. The test load shall be maintained for 10 t0 15 seconds.



42


Material Load

(Kg)

Diameter Of indenter D (mm)

Diameter of

Indentation (mm)

d1(mm) d2 (mm)

d =

HBS

Mean

HBS

Mild Steel

Cast Iron

Aluminium

Diameter of the ball indenter (D) =...................................... mm

Diameter of the indentation (d) = ......................................mm

Load, F =....................................... Kg

HBS = FD D √D

=............................



43

PROCEDURE:

The procedure for conducting the test is given below

1) Depending upon the size of the sample and ball indenter select the suitable load

2) Insert the steel ball indenter and place the specimen on the work table. Raise the

specimen by twinning the hand wheel until the contact with the steel ball is obtained.

3) Close the valve and smoothly pump oil without causing any shock using hand lever

until desired load is obtained.

4) Maintain the load for 10 to 15 seconds. Then slowly open the valve thereby releasing

the oil pressure and the load.

5) Lower the specimen by twining the hand wheel. Remove the specimen and measure

the diameter of the indentation in two perpendicular directions (d1 and d2)

6) The average of d1 and d2 gives the diameter of indentation ‘d’. Calculate the HBS

using the formula.

RESULT:

Brinell Hardness Number of

1) Mild Steel =...................................

2) Cast Iron =..................................

3) Aluminum =..................................

INFERENCE:



44

B. Rockwell hardness Test

THEORY:

Hardness of a material may be defined as its resistance against plastic deformation,

usually by indentation. However, the term may also be defined as the resistance of a material

against abrasion, scratching or cutting. Among the various methods to determine the hardness

of metals, one important one is the Rockwell hardness test.

Rockwell hardness test is an indentation test using calibrated machines to force a

diamond cone or a hard steel ball under specified conditions into the surface of the material

under test in two operations (Preliminary and total loads ) and to measure the depth of the

impression under the specified load conditions.

Rockwell hardness test is a direct reading test. The indentation and the load used in this

are smaller than that is used in Brinell hardness test. The indenter is allowed to penetrate

through the metal specimen. The hardness number is read directly from the dial gauge. The

indenter used may be hardened steel ball of 1/16” (1.5875mm) diameter or may be 1200

diamond cone.

First of all, the indenter is allowed to touch the metal surface and preliminary or minor

load (usually 10 Kgf) is placed on it to seat the specimen. The dial is set to zero. Then the

major load is placed over the indenter. After the equilibrium is attained, the major load is

removed. The hardness number is read directly from the dial.

PROCEDURE:

The procedure for conducting the test is as given below:

1) Fix the indenter on the machine.

2) Rotate the load scale to set major load on it.

3) Place the specimen on the anvil of the machine. Raise anvil by rotating the wheel

until the specimen touches the indenter. The wheel is turned slowly till the small

pointer on the dial reaches the centre of red dot and the big needle to the B 30 or C

0 reading. Now the specimen is subjected to a minor load of 10 Kgf.



45


MATERIAL INDENTOR LOAD HARDNESS

SCALE ‘B’ OR

‘C’

ROCKWELL

HARDNESS NUMBER

1 2 MEAN



46

4) Push the loading handle forward to apply the major load to the specimen and allow

the total load to act on the specimen for 6 - 8 seconds.

5) When the pointer has come to a stop, the hand lever is set back to the test position

and the appropriate Rockwell value HRC can be read on the dial.

RESULT:

Rockwell hardness number of Tool steel = .................HRC

INFERENCE:



47

C. Vickers hardness Test

THEORY:

Hardness of a material may be defined as its resistance against plastic deformation, usually

by indentation. However, the term may also be defined as the resistance of a material against

abrasion, scratching or cutting.

The test consist in forcing a diamond indenter in the form of right pyramid with a square base

and specified angle with opposite faces at the vertex(136o) into the metal under a load P and

measuring the diagonal d of the indentation left in the surface of the test piece after removal

of the load.

The Vickers hardness number, HV, is given by

HV = Applied Load/ area of indentation = 2PSin 136o /2 = 1.854 P/ d2

d2

where P being the applied load (measured in kilograms-force) and d2 the area of the indentation (measured in square millimeters)

In case of Vickers hardness test conducted while using applied load of 30kg for a period of 15 seconds ,the following symbol shall be used, 250HV30/15.

PROCEDURE:

To perform the Vickers test, the specimen is placed on an anvil that has a screw threaded

base.

1) The indenter is pressed into the sample by an accurately controlled test force. 2) The force is maintained for a specific dwell time, normally 10 to15 seconds. 3) After the dwell time is complete, the indenter is removed leaving an indent in the

sample that appears square shaped on the surface.



48


SPECIMEN Load ‘P’ (Kg)

Dia of indentation (d) (mm) HV = 2PSin 136o /2

d2 d1 d2 d= d1+ d2

2 Steel

Aluminium

Copper



49

4) The size of the indent is determined optically by measuring the two diagonals of the square indent.

5) The Vickers hardness number is a function of the test force divided by the surface area of the indent. The average of the two diagonals is used in the following formula to calculate the Vickers hardness. The operation of applying and removing the load is controlled automatically.

Measurement of d The measuring of the impression is done by means of two dashes which are displaced by rotating screws located on the microscope head. Readings are taken by means of micrometrical screws, one scale division reading one micron. Illumination is provided by a 12 volt lamp fitted on the side of the top housing. The magnified image of the impression is got on the screen of the microscope by turning the anvil, without changing the position of the specimen, provided that the penetration is in the optical centre line. By turning the screw of the microscope , anyone of the large graduation mark is made to coincide, with the left edge of the ball impression or the left corner of the diamond image. The right hand edge or corner of the image can then be made to coincide with another of the large graduations by turning the micrometer screw. Having measured the image in this way the glass screen is rotated through 90 o and the measurement is repeated. RESULT:

Vickers Hardness Number for a) Steel = .................

b) Aluminium = ................. c) Copper = .................

INFERENCE:



50

BENDING TEST

Expt. No.: 9

Date :

AIM:

To find values of bending stress and young’s modulus of the material of a beam, which is simply supported at ends and carrying a concentrated load at centre.


UTM machine, Steel rule, Beam material

THEORY:

In a bending test, the test specimen is subjected to transverse loading, so as to produce pure bending of the beam. According to the theory of simple bending the normal stress and strain across any cross section of beam are linear. According to the theory

The strain is related to radius of curvature of the deflected beam and indirectly to

deflection. Hence for a given beam, the deflection can be expressed as a function of

loading ‘w’, material property and geometric property.

The test is done in UTM, the timber beam is supported at a convenient space and loading

is applied centrally. Standard size of specimen is 50*50*75mm, where a standard

specimen cannot be obtained dimensions of specimen shall be

Span(L) = 24*d, where d= depth

PROCEDURE:

1) Measure the size of the specimen and fix the span.

2) Select a suitable loading range and mount the beam supports over the cross heads

at correct span and place the specimen.

3) Fix the central loading device, start the motor and make the ram in the floating

stage.

4) Adjust the pointer to zero, rise the crosshead, so that the central loading device

just touches the top of beam specimen.


ME09 40

OBSER

Width o

Depth o

Length

Momen

Bendin

Bendin

07(P) Materi

RVATION

of beam,

of beam,

h of beam b

nt of inertia

ng Moment

ng Stress

ial testing La

NS AND C

etween sup

a

b

UTM

CALCULA

pports

for Bendin

ATIONS:

b = .......

d = ........

l = ........

I =

=

M =

=

=

ng Test

.......mm

...... mm

....... mm

Dept. of Meechanical Eng

gineering

51



52

5) Adjust the deflection dial to zero loading and open the inlet valve and load the

specimen.

6) Note down the deflection corresponding to load increments until specimen fails.

Note the maximum load.

RESULT:

Bending Stress =......................................

Young’s Modulus = ................................

INFERENCE:



53

Sl. No Load (W) (KN)

δ (mm) M =

W (N-mm)

(KN/mm2)

1

2

3

4

5

Young’s Modulus E = W

I

=...................... N/mm2



54

FATIGUE TEST

Expt. No.: 10

Date :

AIM:

To find the fatigue life of the given sample specimen and plot the S-N curve.


Fatigue testing Machine, Steel rule, Sample specimen, Stop watch, etc

THEORY:

Fatigue is the condition whereby a material cracks or fails as a result of repeated (cyclic)

stresses applied below the ultimate strength of the material. Fatigue failures generally involve

three stages: 1.) Crack Initiation, 2.) Crack Propagation and 3.) Fast Fracture

Fatigue failures often occur quite suddenly with catastrophic results and although most

insidious for metals, polymers and ceramics are also susceptible to sudden fatigue failures.

Fatigue causes brittle like failures even in normally ductile materials with little gross plastic

deformation occurring prior to fracture. The process occurs by the initiation and propagation

of cracks and, ordinarily, the fracture surface is close to perpendicular to the direction of

maximum tensile stress. Applied stresses may be axial (tension-compression), flexural

(bending) or torsional (twisting) in nature. In general there are three possible fluctuating

stress-time modes possible.

The simplest is completely reversed constant amplitude where the alternating stress varies

from a maximum tensile stress to a minimum compressive stress of equal magnitude. The

second type, termed repeated constant amplitude, occurs when the maxima and minima are

asymmetrical relative to the zero stress level. Lastly, the stress level may vary randomly in

amplitude and frequency which is merely termed random cycling.

PROCEDURE:

Fatigue testing can be done in different ways out of this the standard method is what we are

going to use here.

1) Prepare the sample in to the standard size of a fatigue test

2) Mount the sample on the machine and start the machine.

3) Run the sample at a particular stress level and note the point of breaking.



55

The Cyclic loading condition of the Material

Sample specimen for Fatigue testing

S-N curve



56

Fatigue testing Machine.

4) Repeat the test with a different load and note down the failure stress and fatigue life.

5) Plot the S-N curve using the Stress value and the No. of. Cycles required for failure.

RESULT:

1) The S-N Curve was plotted.

2) The Fatigue life of the Sample = ……………….

INFERENCE:



57


Sl.No

Load

Speed(rpm)

No. Of cycles for crack initiation (Ni)

No. Of cycles for crack Propagation (Np)

Fatigue Life (Nf = Ni+Np)

1 2 3 4 5

Mean Stress σ S S

2

=………………..

Stress Range σ σ σ

=……………….

Stress amplitude σ S S

2

=………………...

Stress Ratio R SS

=………………….

Fatigue Life (Nf) = No. Of cycles for crack initiation (Ni) + No. Of cycles for crack

Propagation (Np)

=…………………….


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material testing lab manual - ammini.edu.in 407(P)_ Material Testing...ME09 407(P) Material testing Lab Dept. of Mechanical Engineering 2 CONTENT Sl No. Name of Experiment Page No