smlab manual - mechanical

8/9/2019 Smlab Manual - Mechanical

1/46


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INDEX

Expt.

No.

Date of

Conducting

Title of Experiment Date of

submission

Mark Signature Page

No:

1

Tensile Test on Metals

2Spring Test1. Open

2. Closed

3Hardness Test1. Brinell

2. Rockwell

4Impact Test - 1. Charpy

2. Izod

5Static Bending Test on woods

6

Test On Wood1. Compression Test

Parallel To Grain2. Compression Test

Perpendicular To Grain

Double Shear test

7

1. Tensile Test on Thin

Wires

2. Deflection Test-

Verification Of MaxwellsReciprocal Theorem


3/46

TENSION TEST ON METALS

(IS 16081972 & IS 4321966)

Expt. No:

Date:

AIM: -To determine the following elastic properties of the material and to study the type and characterof fracture.

(i) Yield Stress

(ii) Ultimate Tensile Strength(iii) Actual breaking Stress

(iv) Nominal breaking stress

(v) Ductility(a) Percentage elongation

(b) Percentage reduction in area(vi) Modulus of elasticity.

EQUIPMENTS: Universal Testing Machine (UTE-40), extensometer, gauge marking tools, screw

gauge, meter scale etc.

GENERAL

Gauge length (Lo)

It is the prescribed part of the cylindrical or prismatic portion of the test piece on which elongation is

measured at any moment during the test.

Percentage elongation after fracture (A)

It is the variation of the gauge length of test piece subjected to fracture expressed as a percentage of the

original gauge length Lo

(If the gauge length is other than 5.65So, A should be supplemented by a suffix indicating the gaugelength used. For e.g. A100 means, percentage elongation after fracture measured on a gauge length of

100 mm).

So = The original cross sectional area of specimen.

Ultimate Load (Fm)

It is the maximum load which the test piece withstands during the test.

Nominal Breaking Stress


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It is the breaking load divided by the original area of the section

Actual Breaking Stress

It is the breaking load divided by the actual area of cross section

Tensile strength (Rm)

It is the ultimate load divided by the original cross sectional area of test piece.

Yield Stress (fy)

In steel, which exhibits a yield phenomenon a point is reached during the test at which a plastic

deformation continues to occur at nearly constant stress.

Permanent Set Stress

The stress at which after removal of a load, a prescribed permanent elongation, expressed as a

percentage of the original gauge length results.

PRINCIPLE: -Typical stress-strain curve for an M.S. bar of uniform cross section as shown in figure

.

A - Limit of proportionality

B - Limit of elasticityC - Upper yield point

C - Lower yield point

D - Point of ultimate stressE - Breakingpoint

Up to limit of proportionality A, the material obeys Hooks law and so the curve will be astraight line. PointB is the limit of elasticity up to which bar can be loaded without any permanent set.


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ie. on removing the load, the whole deformation will vanish. Beyond point B the rate of increase instrain will be more till the point C is reached, where the material undergoes additional strain without

increase in stress and undergoes plastic deformation. This is known as Yield point and the stress is

known as yield stress. Actually at this point there is a drop in stress and yielding commences.

After yielding any further increase in stress will cause considerable increase in strain and curve

raised till pointD is reached which is known as point of ultimate stress. The deformation in this range is

partly elastic and plastic. From this moment neck formation takes place. On continuing the loading asthe curve reachesE, the bar breaks.

Modulus of elasticity E = _Pl

Ao

where l is the extensometer gauge length and Ao represents the original area of cross section ofthe specimen. From the straight-line graph between load and extension P/ can be determined as the

slope.

During loading at a particular point the load remains constant for few seconds and again goes onincreasing. This point corresponds to yield point. Stress at that point gives yield stress. Tensile

strength can be calculated by dividing maximum load by original cross sectional area of the test piece.

Percentage elongation = Final length - original length x 100Original length

Percentage reduction of area = Original area - Final area x 100Original area

PROCEDURE: -


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7/46

OBSERVATIONS:-

1. Mean diameter (d) mm =

2. Original cross sectional area, So =

3. Approximate ultimate load = 500 So =

4. Original gauge length, Lo =

5. Extensometer gauge length, Le =

6. Reduced diameter, Du =

7. Reduced cross sectional area, Su =

8. Final gauge length, Lu =

9. Load at yield point, Fy =

10. Ultimate load, Fm =

11. Breaking load, Fb =

Least count of extensometer =

Load- Extension Table

Load in

kN

Extensionin mm

Load in

kN

Extensionin mm

Load inkN

Extensionin mm

Calculations:

1. Yield Stress fy =o

y

S

F= N/mm

2

2. Ultimate Stress fu =o

u

S

F= N/mm

2


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3. Nominal breaking stressfn =o

b

S

F= N/mm2

4. Actual breaking stressfact =u

b

S

F= N/mm2

5. Percentage elongation on a

Gauge length of ________mm =

100

o

ou

L

LL= = %

6. Percentage reduction in area =

100

o

ou

S

SS= = %

7. Slope of load vs extension curve, =P

= = N / mm

8. Youngs Modulus =A

Pl= = N /mm

2

RESULT: -

1. Yield Stress =

2. Proof Stress =

3. Ultimate tensile strength =

4. Actual breaking stress =

5. Nominal breaking stress =

6. Percentage elongation =

(Gauge length________)

7. Percentage reduction in area =

8. Modulus of elasticity =

INFERENCE:


9/46


10/46

BRINELL HARDNESS TEST

(IS1500-1968)

Expt No. :

Date :

AIM :- Top determine the Brinell Hardness Number of the material of the given specimen.

EQUIPMENTS : Brinell Hardness Testing Machine, Microscope etc.

PRINCIPLE :- The test consists in forcing a steel ball of diameter D under a load F into the testpiece and measuring the mean diameter d of the indentation left in the surface after removal of the

load. The Brinell Hardness HB is obtained by dividing the test load F (in kg(f)) by the curved surfacearea of the indentation (in square millimeters). The curved surface is assumed to be a portion of the

sphere of diameter D. The depth of indentation h is given by 22

2

1dDDh

The curved surface area of indention = Dh

= 222

dDDD

Brinell Hardness HB = Applied loadArea of indentation

=Dh

F

= 22

2

dDDD

F

Usually Brinell Hardness HB is supplemented by an index giving at the first place the diameter of the

ball in mm., at the second place the test load in Kg and at the third place the duration of the load inseconds. For example, the symbol: HB 5/750/20 indicates that the test was conducted using a steel ball

5mm diameter under a test load of 750 Kg, which was maintained for 20 seconds.

Normally a ball of 10mm nominal diameter shall be used. Balls of diameters 1, 2, 2.5 and 5mm are also

used but in no case the nominal diameter of the ball shall be less than one millimeter unless otherwise

specified.

The surface of the piece to be tested shall be sufficiently smooth and even to permit the accurate

determination of the diameter of the indentation. It shall be free from oxide scale and foreign matter.

The thickness of the test piece shall not be less than 8 times the depth of the indentation h. Nodeformation shall be visible at the back of the test piece after the test.


11/46

The following table shows the minimum thickness of various ball diameters, loads and hardness values:-

Balldiameter

in mm

LoadKg

HB Values

100 200 300 400 500

2.9 187.5 1.91 0.95 0.64 0.48 0.42

5.0 750 3.81 1.90 1.27 0.97 0.84

10.0 3000 2.64 3.81 2.54 1.90 1.70

Load for testing: Ferrous MetalsHardness between 140450F/D2 = 30 (D in mm) time10 seconds

Non-ferrous metals: Brass, copperHB between 35-140F/D2 = 10 (D in mm) time30 seconds

NOTE:- For most metals, Brinell hardness increases linearly with the tensile strength values of the

metal.

Tensile Strength = k x Brinell Number in tonnes/sq.inchFor mild steel, k = 0.23, for plain carbon steel, k = 0.22

For wrought light alloys, tensile strength = (BHN/4)-1

It should be noted that the same analysis of metals or alloy will give a variation in hardness

values in the forged, hot or cold rolled, extruded, cast or heat treated conditions.It is recommended that the Brinell Test as specified in IS 1500-1968 should not be used for steels with a

Brinell hardness exceeding 450. For harder steels, a test with harder indenter, for example, tungsten

carbide and diamond may be substituted. But the hardness number would then be on a different scale

In cases when a tungsten carbide ball is used, the test shall be termed as Modified Brinell Hardness

Test and the symbol HBW should be used.

PROCEDURE:-


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OBSERVATIONS:-

Material ofspecimen

Load in Kgand duration

Diameter ofindenter

D mm

Diameter of indentation HB Value Mean

d1 d2 d = (d1+d2)/2

CALCULATIONS:-

BHN = P/Spherical area of indentation in mm

Where spherical area of indentation = area of projection on the ball circle

= area abc= Dy

To find y,

oe =

22

22 dD

h =

22

222

dDD

BHN =

22

222

dDDD

P

=

222

dDDD

P

= 22

2

dDDD

P

=

O

e


13/46

RESULT :-

Material Brinell Hardness Number

INFERENCE:


14/46

ROCKWELL HARDNESS TEST

(IS- 1586-1968 & 38041966)

Expt No.:

Date:

AIM :- To determine the Rockwell hardness number of the material of the given specimen.

EQUIPMENTS :- Rockwell hardness testing machine, diamond cone penetrator, 1/16 steel ballindenter.

GENERAL:- This is a direct reading hardness testing machine compared to Brinell hardness testingmachine, testing is quicker with a much smaller permanent indentation. This method of test is well

suited to finished or machined parts of simple shape. Various models of Rockwell machines are

available for testing inside cylindrical surfaces, thin strip metal, wire, safety razor blades etc.

PRINCIPLE:- The hardness of a material can be defined as the resistance to penetration/indentation.The test consists in forcing an indenter of standard type (cone or ball) into the surface of the test piece in

two operations and measuring the permanent increase of depth of indentation e of the indenter under

specified conditions. The unit of measurement of e is 0.002mm from which a number known asRockwell hardness is derived.

The method is used for testing of hardness over a wide range of material hardness. The hardness of a

material is measured by the depth of penetration of the indenter in the material. The depth of

penetration is inversely proportional to hardness. Both ball and diamond type of indenters is used in this

test. This test gives direct hardness readings on a large dial provided with two scales. Scale B is usedfor tests on unhardened steel, phosper, bronze, aluminium and magnesium, light alloys etc. For readings

on this scale a 1/16 (1.5875mm) diameter steel ball is used for indentation with a 10 Kg minor load and90 Kg major loads. The minor load is applied to overcome the film thickness on the metal surface

which may have formed in due course of time. Minor load also eliminates error in the depth

measurements due to springing of the machine frame or setting down of specimen and table attachments.

Scale C is used with a 120 cone angle diamond indenter with a minor load of 10 Kg and a major loadof 140 Kg. This is applicable to test the harder metals such as hardened steels or hard alloys.

The Rockwell hardness with reference to these two scales is written as HRB, HRC followed by values of

the hardness. For example HRB45 means the Rockwell hardness corresponding to the scale B is 45

The Rockwell hardness is derived from the measurements of the depth of impression.

HRB = 130(depth of penetration (mm))

0.002

HRC = 100- (depth of penetration (mm))

0.002


15/46

PROCEDURE :- .

OBSERVATIONS:-

Sl No. Material Test Loadin Kg

Penetratorused

Scale Used RockwellHardness

Number

Mean


16/46

RESULT ;-

Material Rockwell Hardness Number

INFERENCE:


17/46

SPRING TEST

Expt No:

Date:

AIM:- To determine stiffness, the modulus of rigidity, of the material of the springs.

EQUIPMENTS: - Spring testing machine, Screw gauge, Vernier calipers.

PRINCIPLE:-

R - Mean radius of spring coil.

DWire diameter

PPitch of coil

NNumber of coils.

WAxial load on spring.

NModulus of rigidity for the spring material

Fs - Maximum shear stress induced in the spring wire.

FBending stress induced in the spring wire due to bending.

- Deflection of spring as a result of axial load.

- Angle of helix.

Moment M at any point on the spring due to axial W load W is WR. Component of M along theaxis of the wire will produce torsion and component perpendicular to the axis will produce bending.

i.e. T = WRcos , M = WRsin

= Angle of twist as a result of twisting moment WRcos

= Angle of bend, as a result of bending moment WRsin

We know that length of spring wirel = 2nRsec

Twisting moment T =3

16 dfs

WRcos = 316

dfs


18/46

We know thaty

f

I

M

f=4)

64(

)2

(sin

d

dWR

I

My

=

3

sin32

d

WR

We know thatl

N

J

T

JN

lWR

JN

Tl

cos

Angle of bend due to bending momentEI

lWR

EI

lM

sin I.

Work done by the load in deflecting the spring is equal to strain energy of the spring.

MTW2

12

1)(2

1

W= T+ M

]sincos

[

sinsin

coscos

222

EIJNlWR

EI

lWRWR

JN

lWRWR

Now substituting the values ofl =2nrsec, 44

6432dIanddJ

in the above equation.

ENd

nWR

22

4

3 sin2cossec

64

In the case of closed coiled spring is very small so that cos = 1, sin = 0, then

4

364

Nd

nWR

Stiffness =W/ where w is the load and is the deflection.

GENERAL: - In this machine the weighing mechanism is located in the upper housing and has a lever

ratio of1 : 5 Balancing weight is placed at one end and the loading pan on the other side of the lever

There is a vertical graduated scale fixed on the right stand from which the deflection of the spring can benoted against the arrow on the lower compression plate.

The Modulus of Rigidity of both springs are obtained from the relationship shown above as

ENd

nWR

22

4

3 sin2cossec

64

but E = 2N (1+1/m) = 2N (1+0.3) = 2.6N as 1/m = 0.3 for steel


19/46


20/46

OBSERVATIONS AND CALCULATIONS:-

Particulars Spring

Open coiled Closed coiled

Diameter of the wire d

Outer diameter of coil D

Effective radius of spring R

No of turns n

Pitch P = L/n

tan = P/(2R)

=

Maximum load Wm

Maximum deflectionm

Sl No Load kg Scale Reading Average

reading

Deflection

Loading Unloading

Spring under

Tension

Spring under

compression


21/46

CALCULATIONS :-

1. Open coiled spring:-

Stiffness = W/ = = N/mm

6.2

sin2cossec

64 224

3 d

nWRN =

= N/mm2

Torsional shear stress at maximum load Wm , qmax= 316

d

RWm

=

= N/mm2

Elastic strain energy storedU= Wmm)/2 = = N mm

Volume of the spring V = n

d

R 42

2

= = mm3

Strain energy per unit volume = U/V = = Nmm/mm3

2. Close coiled spring

Stiffness = W/ = = N/mm

4

364

d

nWRN

= = N/mm2

Torsional shear stress at maximum load Wm , qmax= 316

d

RWm

=

= N/mm2

Elastic strain energy storedU= Wmm)/2 = = N mm

Volume of the spring V = nd

R4

22

= = mm3

Strain energy per unit volume =U/V = = Nmm/mm3


22/46

RESULT :-

Spring under Compression

(open coiled spring)

Spring under Tension

(closed coiled spring)

Stiffnesss

Modulus of Rigidity

Torsional shear stress

Strain energy per unit volume

DISCUSSION :-


23/46


24/46


25/46

IMPACT TEST

(IS 14991977, 1598-1977 & 37661966)

Expt No. :

Date:

AIM :- To find the impact strength (energy required to rupture the specimen) in izod and charpy tests.

EQUIPMENTS:- Impact testing machine (Model IT-30)

The principal features of a single blow pendulum impact testing machine are

1. A moving mass whose kinetic energy is great enough to cause rupture of the test specimen

placed in its path.

2. An anvil and a support on which the specimen is placed to receive the blow and

3. A means of measuring the energy required to rupture the specimen and residual energy of the

moving mass after the specimen is broken.

GENERAL:-

The ordinary tensile and bending tests are no true criterion of the impact resisting qualities of a material.Satisfactory performance of certain machine parts such as parts of percussion drilling equipments, parts

of automotive engines, parts of rail road equipments - track and buffer devices; depends upon the

toughness of the parts under shock loading. Some materials will withstand great deformation togetherwith high stress without fracture. Such materials have great toughness. Some materials under tension

can be drawn out to a considerable elongation without fracture. Such materials are ductile. A ductile

material that can be stretched out only under high stress is tough. One way of determining toughness isto fracture the specimen by a single blow from a moving mass of metal and determining the energy

absorbed in fracturing the specimen. The impact test measures energy required for fracture not force.

In the design of many machine parts subject to impact loading the aim is to provide for the absorptionof as much energy as possible through elastic action and then dissipate that elastic energy by some

damping device. In such cases the elastic energy capacity derived from static loading may be adequate.

The impact test gives energy capacity at rupture. This is different from the elastic energy capacity orresilience.

PRINCIPLE :-

The charpy test consists of measuring the energy absorbed in breaking by one blow from a swinginghammer, under prescribed conditions, a test piece V notched in the middle and supported at each end.


26/46

The izod test consists of breaking by one blow from a swinging hammer under specified conditions, a

V notched test piece gripped vertically with the bottom of the notch in the same plane as the upper face

of the grips. The blow is struck at a fixed position on the face having the notch. The energy absorbed is

determined.

CALIBRATION OF THE MACHINE :-

The pendulum in its highest position is inclined at an angle of 141 0 47 to the vertical and the initialenergy in this position is 300J for conducting the charpy test. In the case of izod test, it is inclined at an

angle of 90o

and the initial energy is 168J

Initial Energy E1=wh= Wl (1+sin1)

Considering the pendulum as a simple pendulum, l

can be determined and from the above formula,weight of the pendulum can be determined.

After breaking the specimen, the pendulum will move

through a high h1 making an angle 2 with the rest position.

Residual energy E2 = Wl(1-cos2)

Energy absorbed is calculated for various values of2and a graph is plotted between EL and 2 which is the

calibration curve for the machine.

PROCEDURE:-

CHARPY TEST:-


27/46

IZOD TEST:- (Cantilever Test)

Calculate,

Impact Strength = Impact value

------------------------------------------Area of cross section of the specimen

Below notch in m2

Impact modulus = Impact value

---------------------------------------------

Volume of cross section of specimen

Below notch in m3


28/46

OBSERVATION:

Description Izod Charpy

Weight W

Length L

Initial energy E1

Initial Energy E1 = w1(1+sin1)

1 =

2 =Energy loss EL = w1(sin 1 + cos 2)

2, Degrees Energy Loss (izod)

Joules

Energy Loss (Charpy)

Joules0

10

20

30

40

50

60

70

80

90

100

110

120

140

14147

CALCULATIONS:-

Charpy TestEnergy loss = EL = w1(sin 1 + cos 2) =

= J


29/46


------------------------------------ = = J/ mm2

(Area of cross section of the

specimen below notch in m2)

Impact modulus = Impact value------------------------------------ = = J/mm3

Volume of cross section of

specimen below notch in m3

Izod Test

Energy loss = EL = w1(sin 1 + cos 2) == J


------------------------------------ = = J/ mm2

(Area of cross section of thespecimen below notch in m2)

Impact modulus = Impact value------------------------------------ = = J/mm3

Volume of cross section of

specimen below notch in m3


30/46

RESULT:-

INFERENCE:

Test No. Details of

specimen

Energy Loss in Joules

Izod Charpy

FromGraph

FromCalculation

FromGraph

FromCalculation

Impact Strength

Impact Modulus


31/46


32/46


33/46

STATIC BENDING TEST.

AIM:- To determine the modulus of elasticity and modulus of rupture of the given timber specimen.

EQUIPMENTS :- UTE40

Theory:-

For a beam, simply supported at the ends with a central concentrated load W, the bending moment is M

=Wl/4 = fz where l is the span of the beam, f is the extreme fibre stress andZ is the modulus ofsection of the beam ie.bd

2/6 for a rectangular cross-section. If we know the load at failure, (Wmax) and

modulus of section, , from the above equation, f= Wl/4Z. Assuming a maximum stress fmax of abou

600 N/mm2then we get Wmax= (4fmax Z)/1

For simply supported beam with central concentrated load, the deflection at center = W13/(48 EI). From

the equation we can find the value of modulus of elasticity E. I is the moment of inertia which is equa

to (bd3)/12 for a rectangular section. To find the modulus of rupture fu , load the specimen to failure and

note the load as Wu. Then from the above equation, modulus of rupture fu = (Wul)/4Z.

The test specimen should be of size 50 x 50 x 750 mm that should be absolutely free from the defect and

shall not have a slope of grain more than 1 in 20 parallel to its longitudinal edges. (Where a standardspecimen cannot be obtained the dimensions of the test specimen should be such that the span is 14

times the depth).

PROCEDURE:-


34/46

OBSERVATION AND CALCULATIONS:-

Span of the specimen,lmm =

Breadth of specimen, b mm =

Depth of specimen, d mm =

Modulus of section, Z = (bd2)/6 =

Moment of inertia, I = (bd3)/12 =

Maximum load, Wu =

Load at limit of proportionality =

(from graph)

LoadkN

Deflection

mm

Load

kN

Deflection

mm

LoadkN

Deflection

mmLoad

kN

Deflection

mm


35/46

Deflection at limit of proportionality =(from graph)

Fibre stress at limit of proportionality = M/Z = (Wl)/4Z =

=

Equivalent fibre stress at maximum load = (Wl)/4Z =(Modulus of rupture) =

Modulus of elasticity, E = (W13)/48 I =

=

Elastic resilience = Work upto limit of proportionality

---------------------------------------

Volume= Area under the curve up to limit of proportionality

------------------------------------------------------------

Volume

= = Nmm/mm3

RESULT :-

a) Fiber Stress at limit of proportionality =

b) Modulus of elasticity =

c) Modulus of rupture =

d) Elastic Resilience =

INFERENCE:


36/46


37/46

COMPRESSION TESTS ON WOOD

(IS 17081969 & IS 8881970)

Expt No:

Date:

AIM:-

To study the behaviour of wood and to determine the strength under following types of loading.

1. Compression parallel to grain.

2. Compression perpendicular to grain.

Compression test parallel to grain

AIM :- To determine the compressive test of wood under compression parallel to grain usingcompressive testing machine.

EQUIPMENTS:- Compression testing machine

PRINCIPLE:- The test consists of subjecting a wooden piece to compressive load and recording the

maximum load P at failure. Then the compressive strength shall be calculated using the formula P/A

where A is the cross-sectional area of the given specimen.

PROCEDURE:-


38/46

Compression Perpendicular to Grains:-

AIM:- To find out the compressive strength of specimen perpendicular to grain.

PROCEDURE:-

OBSERVATIONS:-

Compression test parallel to grain

Dimension of cross section =Crushing load =

Compressive strength parallel to grain = Crushing load

------------------- =C.S. area

Compression test perpendicular to grain: -

Dimension of cross section =

Crushing load =

Compressive strength perpendicular to grain = Crushing load------------------ =

C.S. area


39/46

RESULT :-

1.Compressive strength of given timber specimen parallel to grain =

2.Compressive strength of given timber specimen perpendicular to grain =

INFERENCE:


40/46

DOUBLE SHEAR TEST

(IS 52421969)

Expt No. :

Date:

AIM: - To determine the shear strength of the given material subjecting the specimen to fail under

double shear.

EQUIPMENTS: Universal Testing machine, Shear shackle, Screw gauge etc.

PRINCIPLE: The test consists of subjecting a suitable length of steel specimen in full cross section to

double shear, using a suitable test rig, in a testing machine under a compressive load or tensile pull and

recording the maximum loaf F to fracture. The shear strength Fs shall be calculated from the followingformula:

Fs = 222

4

2

d

F

d

F

where d is the actual diameter of the specimen.

PROCEDURE: -


41/46

OBSERVATIONS:-

Diameter of the specimen = mm

Approximate ultimate shear strength = N/mm2

Area of cross section in double shear =

Approximate load =

Failure load F =

Shear strength = F/A=2

2

d

F

= = N/mm2

RESULT :-

Shear Strength of given specimen = N/mm2

INFERENCE:


42/46

VERIFICATION OF MAXWELLS RECIPROCAL THEOREM

Expt no;

Date:

AIM:- To verify Maxwells reciprocal theorem

EQUIPMENTS:- Magnetic Stand, Dial gauge etc.

PRINCIPLE:- Maxwells reciprocal theorem states that for a linearly elastic body, the vertical

displacement of a point B of the beam due to force P at another point A is equal to the vertical

displacement of point A; due to the same force at point B. Or in other words, the work done by thefirst system of loads due to displacement caused by a second system of loads equals the work done by

the second system of loads due to displacement caused by the first system of loads.

PROCEDURE:-


43/46

OBSERVATIONS:-

Load at A Deflection at B

Load at B Deflection at A

RESULT :- Maxwells reciprocal theorem is verified.

INFERENCE:-


44/46

TENSILE TEST ON THIN WIRES

Expt No:-

Date:-

AIM:- To determine the tensile strength and elongation of the given wire using tensile tester.

PRINCIPLE :- In this test the strength is determined in such a manner that test specimen is gripped bytwo grips vertically arranged one below the other and continuously tensile stressed until it breaks. At

the same time elongation is also indicated on a scale.

GENERAL :- The machine is for determining the tensile strength and elongation of various fibrous and

generic materials, textile, rubber, plastic, leather, cardboard, plywood, paper, asbestos, cables and

conductors etc. The machine consists of a base and a vertical column, which supports the load-measuring unit. The base houses the drive unit. The drive is effected by electric motor whose stroke is

transmitted through the set of pulleys to the lead screw. When pull is applied to specimen, the pendulum

gets deflected from its vertical position in proportion to pull applied and the tensile force is indicated inthe dial by drag pointer.

This strength-testing machine has three power measuring ranges. This permits finer graduations andhence betters reading accuracy for the lower ranges. The measuring ranges are set by attaching weight

disks on the pendulum rod stud. For preventing sudden fall of pendulum rod and rupture of specimen, a

damping unit is provided which ensures that the pendulum rod slowly goes back to its vertical position.

PROCEDURE:- Depending on the materials to be tested, mount the appropriate grips. Select the load

range in accordance with the strength of wire. Mount the required weight disc on the stud for setting the

appropriate machine range. Set the machine for required gripping length and if all the specimens to betested are having constant gripping length, then set the position of the adjustable collar so that for the

subsequent tests the gripping length is not required to be adjusted again and again. To prevent the

sudden fall of the pendulum rod, adjust the setscrew of dashpot unit. Grip the test specimen in thecenter of the two vertical grips, which are arranged one below the other. While fixing the specimen

lock the load cell with the help of locking device.

Note the initial extension scale reading R1 with the help of pointer. Unlock the load cell, switch on

power supply and operate machine in forward direction till specimen breaks. Note the extension scale

reading R2 with the help of pointer when specimen just ruptures. Note the load from the dial, which

gives the tensile force of specimen. Press the stop button and the reverse direction button so that lowergrip goes back to its starting position for repeating the experiment. The difference between R2 & R1gives extension. The percentage elongation can be calculated using the formula (R2-R1)/R1 x 100

OBSERVATION:-

Diameter of specimen =

Area of cross section of specimenA = d2/4

Tensile load of specimen P =


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Tensile strength of specimen = P/A

Initial extension scale reading R1 =

Final extension scale reading R2 =

Percentage elongation = [(R2-R1)/R1] x 100

RESULT :-

Tensile strength of given wire =

Percentage elongation of given wire =

INFERENCE:-


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smlab manual - mechanical

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