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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
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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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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:
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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.
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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
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RESULT :-
Material Brinell Hardness Number
INFERENCE:
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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
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PROCEDURE :- .
OBSERVATIONS:-
Sl No. Material Test Loadin Kg
Penetratorused
Scale Used RockwellHardness
Number
Mean
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RESULT ;-
Material Rockwell Hardness Number
INFERENCE:
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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
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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
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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
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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
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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 :-
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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.
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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:-
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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
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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
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Impact Strength = Impact value
------------------------------------ = = 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
Impact Strength = Impact value
------------------------------------ = = 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
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RESULT:-
INFERENCE:
Test No. Details of
specimen
Energy Loss in Joules
Izod Charpy
FromGraph
FromCalculation
FromGraph
FromCalculation
Impact Strength
Impact Modulus
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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:-
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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
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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:
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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:-
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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
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RESULT :-
1.Compressive strength of given timber specimen parallel to grain =
2.Compressive strength of given timber specimen perpendicular to grain =
INFERENCE:
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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: -
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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:
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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:-
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OBSERVATIONS:-
Load at A Deflection at B
Load at B Deflection at A
RESULT :- Maxwells reciprocal theorem is verified.
INFERENCE:-
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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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