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Experiment no 1:
Apparatus:
Tinius Olsen Universal Testing Machine
Theory:
This universal testing machine is intended only for static type of
mechanical tests. The loading is done by hydraulic mechanism. This machine
can be adopted for many special applications and a large variety of tests
can be performed on this machine using proper attachments, for example
tensile test, bending test, shearing test, torsion test.
Components:
The loading application is accomplished by a hydraulic piston & cylinder located at the base. The hydraulic pressure is developed by a gear pump. The pump in combination with automatic valve provided uniform rates of loading. Safety valves protect the gauges from overload. The speed of loading is controlled by pilot hand wheel and due to this valve arrangement (load/unload), load may be easily applied and removed or held constant. Construction of entire loading unit is very rugged and compact. Hydraulic oil of Grade SAE 40 is used. The capacity of oil reservoir is 13 gallons.
Indicating Mechanism:
Loads on Universal testing machine are indicated on precision type hydraulic gauges, which are mounted on the instrument panel. These gauges are provided with maximum pointers.
Operation:
Loads on this machine are indicated on two hydraulic gauges. Application of all test loads (compression/tension) is accompanied by the upward movement of piston. The lower cross heads may be adjusted by using cross head adjusting crank to any desired position. For tensile loading the test specimen is mounted on the upper side on the lower head. For compressive loading the test specimen is mounted on the lower side of lower head.
All zero setting must be made with specimen, testing.
Shut Down:
Close the load completely. Open the unload valve slowly and gradually. After piston completely sets to its seat (the load pointer suddenly drops below the zero load mark). Close the unload valve. Stop the pump by pressing the stop button. Close the gauge valve.
EXPERIMENT NO 02
Apparatus:
Universal Testing Machine, given specimen
Theory:
In the course of operation all articles are subject to the action of
external forces, which creates stresses that inevitably cause deformation.
To keep these stresses within permissible limits, it is necessary to select
suitable material. A comprehensive knowledge of mechanical characteristics
of metal products (strength, ductility, malleability, creep etc.) is essential
for this purpose. The mechanical properties quoted in different books of
materials are based on different mechanical tests.
Mechanical tests are those in which specially prepared specimens of standard size are tested on special machine to obtain different mechanical properties of metals. Mechanical tests are conducted under various loading conditions. They may be:
After loading is applied, it will be noted that up to a certain stress, the strain is directly proportional to stress (obeys Hooke’s law).
After plotting strain on horizontal axis and stress on vertical axis, the prominent features of a stress-strain curve are:
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E =
Stress
Strain
Graphically
E = 4000/0.03
= 133.333 kips
For Mild Steel
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E = 39783/0.08
= 497.287 kips
EXPERIMENT NO. 3
OBJECT:
To determine the shear stress of given wooden specimen.
APPARATUS:
Universal testing machine, wooden shearing attachment, wooden specimen.
THEORY:
Shear stress: The stress that is caused by forces acting along or parallel to the area.
Or
Internal reacting shear force is expressed as force per unit area.
It is also termed as tangential stress.
A shearing stress is produced whenever applied loads cause one section of body to tend to slide past its adjacent section.
Properties of timber:
Timber has been used as a structural material since pre historic times. We use more wood than we do any other engineering material. Wood has three noticeable characteristics
Several factors influence the behavior of wood. The moisture content in timber, age of timber & type of timber. During drying, the wood shrinks & cracking may occur.
Wood is highly anisotropic. Because of the orientation of fibers, the strength in longitudinal (lengthwise) direction is may be about 25-50 times greater than the strength in the radial or tangential (sidewise) direction.
However clear wood composite of fiber & lignin & free from imperfection such as knots has longitudinal tensile strength of 10,000 psi to 20,000 psi. strength in tension is superior to either compression or shear. In compression fibers buckle, in shear the fibers sliding past one another.
SPECIMEN & ATTACHMENT:
A parallel to grain specimen is prepared to test as prescribed in ASTM methods. Attachment used in this test is very specialized & also used to determine shear strength of adhesives of bonding wood. The specimen is placed in the tool with lower under the flat blade. Minor misalignment is of wooden block is automatically compensated by semicircular blade which assures uniforms lateral distribution of load. In use tool is placed on the testing machine table & load is applied in compression against the blade. Provision is made to mount the blade on the lower crosshead.
Observation/calculation:
Shear load: 5480 lbs.
Shear area: 4 in².
Shear stress for given wooden specimen = shear load =1620 psi
shear area
EXPERIMENT NO 4
OBJECT:
To determine the single shear strength of given 1/2" dia aluminum, mild steel and brass bar.
APPARATUS:
Universal testing machine, Bar shearing attachment, Allen key, Specimen.
WORKING FORMULA:
Single share stress
t =F/A (lb/in2)
Where
t = Share stress
F = Applied Force
A = Cross sectional area
BAR SHEARING PROCESS:
The shearing process starts when blade (upper cutting edge) descends against bar, the metal first deformed plastically over the die (lower cutting edge). Due to small lateral clearance between the two cutting edges, the deformation is highly localized. The blade (upper cutting edge) penetrates into the bar and opposite surface bulges slightly. When penetration reaches 15% to 60% of diameter of bar (depending upon properties of material), the applied force exceeds the shear resistive force of the metal and metal suddenly shears or ruptures through remainder of its diameter. Due to non-homogeneity in the metal, the final phase of shearing doesn’t occur uniformly.
BAR SHEARING TOOL:
Bar shearing tool comprises of shearing blade (upper cutting edge) & bar holding base (contains lower cutting edge i.e. die also serve to hold bar firmly during course of practical). The shearing blade have precision notches in all four sides, are used for ¼", ½", 3/4" & 1" diameter bars for testing in single or double shear. Shearing blade &die are made of tempered tool steel. They are ground to true edge.
SPECIMEN & ATTACHMENT:
A parallel to grain specimen is prepared for test as prescribed in ASTM
Methods. Attachment used in this test is very specialized and also used to determine shear strength of adhesives of bonding wood. The specimen is placed in the tool with lower portion under the flat blade. Minor misalignment is of wooden block is automatically compensated by semicircular blade which assure uniform lateral distribution of load. In use tool is placed on the testing machine table and load is applied compression against the blade. Provision is made to mount the blade on the lower crosshead.
OBSERVATION:
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RESULT:
EXPERIMENT NO: 5a
OBJECT:
To determine hardness of different materials.
APPARATUS:
Hardness testing machine, Specimen, Indenter with different diameter balls.
THEORY:
Hardness is the resistance to penetration. Hardness tests give measure of the resistance of a metal to the penetration & resistance to scratching or abrasion.
The Brinell hardness test is of static indentation type. It is one of the most widely used hardness tests in engineering practice.
The Brinell hardness test is basically simple and consists of applying constant load, usually 500 to 3000 kg on a hardened steel ball type indenter, 10 mm in diameter, to the flat surface of a work piece. The 500-kg load is used for testing non-ferrous metals such as copper, aluminum alloys, whereas 3000-kg load is for testing harder metals such as steel, cast iron etc.. The load is held for a specified period of time (10 to 15 sec. for iron or steel, about 30 sec. for softer metals) after which the diameter of recovered indentation is measured in mm. This time period is required to ensure that plastic flow of the work metal has stopped.
Hardness is evaluated by taking the mean diameter of the indentation (two readings at right angles to each other) & calculating Brinell hardness number (HB) by dividing applied load by the surface area of indentation, as per formula:
HB = load .
( p D / 2 ) [ D - ( D2 – d2 ) ]
Where D = Diameter of the ball in mm.
d = Diameter of the indentation in mm.
For the selection for Brinell hardness test, following chart is very
helpful:
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| Steel & cast iron |
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| Copper, copper & Aluminum alloy |
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| Aluminum |
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Lead, tin & their alloy |
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PROCEDURE:
HB = load .
( p D / 2 ) [ D - ( D2 – d2 )½]
Where D = Diameter of the ball in mm.
d = Diameter of the indentation in mm.
OBSERVATION CHART:
| Material |
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| Diameter of
Ball (D)
mm. |
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| Diameter of Indentation (d) mm. |
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| Brinell Hardness No. |
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RESULT:
The Brinell hardness number of different materials is
found to be
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EXPERIMENT NO 5(b)
OBJECT:
To determine Hardness of different materials by " Rockwell Method."
APPARATUS:
Hardness tester, Specimen, Different indenter.
THEORY:
Hardness is the resistance to penetration. Hardness tests give measures of the resistance of a metal to the penetration & resistance to scratching or abrasion.
The Rockwell hardness test is of static indentation type. It is one of the most widely used hardness tests in engineering practice.
Rockwell hardness differs from Brinell hardness testing in that hardness is determine by the
depth of indentation made by a constant load impressed upon indenter rather than surface area of indentation. Rockwell test consists of measuring the additional depth to which an indenter is forced by heavy (major) load beyond the depth of previously light (minor) load. Application of minor load eliminates backlash in the load train and causes indenter to breakthrough slight surface roughness and to crush particles of foreign matter. Thus contributing greater accuracy in the test.
In regular Rockwell hardness test the minor load is always 10 kg. The major load however can be 60,000 or 150 kg. Depending upon the type of scale used (selection of scale is done according to type of material).
SELECTION OF ROCKWELL SCALE:
It is necessary to select the Rockwell scale to suit a given set of circumstances. Therefore knowledge of the factors that govern the proper choice of scale is mandatory. There are 15 different scale of regular Rockwell hardness is available. The influencing factors which governs the selection of Rockwell scale is:
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For ball indenter red dial graduation. For diamond cone indenter black dial graduations.
PROCEDURE:
The Rockwell hardness number can be represented as:
OBSERVATION:
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| High Carbon Steel |
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| Spring steel |
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RESULT:
The Rockwell hardness number of different material are found to be:
For High Carbon Steel = 43
For Spring Steel = 26
EXPERIMENT NO: 6
Object:
To determine modulus of rigidity of given material by method torsion.
Apparatus:
Torsion apparatus, load.
Theory:
Torsion is the shear produced by rotation or by a torque. In this type of shear,one layer of a material is made to rotate on an adjection layer. This type of shear is present in rotating shafts when transmitting power.
In torsion test, equal and opposing moments are applied at opposing ends of a suitable specimen in planes perpendicular to its longitudinal axis. Observation of applied torque and corresponding twist or rotation is used to determine modulus of rupture, modulus of rigidity and angle of twist of given material.
Torsion formula:
Torsion formula is used for determining Modulus of Rigidity of given material.
T/J = Gq /L = t /r
Where:
T = Torsion lb.-in
J = Polar moment of inertia in4
t = Shear stress . Psi.
G = Modulus of rigidity. Psi.
q = Angle of twist. Rad.
L = Length of shaft in..
r = Radius of shaft in..
The assumptions that are made for the formulation of torsion formula are as under :
Modulus of Rigidity is define as:
The relationship between shearing stress & shearing strain, assuming
HOOKE’S LAW to apply to shear is :
t = G g
where:
G represents modulus of elasticity in shear OR Modulus of rigidity.
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Calculations:
G 1 = 874.33 ksi
G 2 =691 ksi
Mean G = ( G 1+ G 2) /2 = 782.66
Result:
Modulus of elasticity in shear OR Modulus of rigidity of given material
Is found to be = 782.66 ksi