Sm Lab Manual

TENSION TEST ON METALS

(IS 1608 – 1972 & IS 432 – 1966)

Expt. No:

Date:

AIM: - To determine the following elastic properties of the test piece and to study the type and character of fracture.

(i) Yield Stress(ii) Proof Stress(iii) Ultimate Tensile Strength(iv) Actual breaking Stress(v) Nominal breaking stress(vi) Ductility

(a) Percentage elongation(b) Percentage reduction in area

(vii) Modulus of elasticity.

TERMINOLOGY: -

Gauge length ( L o)

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 gauge length 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 ( F m)

It is the maximum load which the test piece with stands during the test.

Nominal Breaking Stress

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

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.

Proof Stress (Rp)

The stress at which a non-proportional elongation equal to a specified percentage of the original gauge length takes place. When a proof stress is specified the non-proportional elongation should be stated (say 0.3%) and the symbol used for the stress should be supplemented by an index giving this prescribed percentage of original length, for example Rp 0.2.

Permanent Set Stress

The stress at which after removal of a load, a prescribed permanent elongation, expressed as a % of the original gauge length results.

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

PRINCIPLE: -

A - Limit of proportionalityB - Limit of elasticityC - Upper yield pointC - Lower yield pointD - Point of ultimate stressE-Breakingpoint

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

Up to limit of proportionality A, the material obeys Hook’s law and so the curve will be a straight line. Point B is the limit of elasticity up to which bar can be loaded without any permanent set. ie. on removing the load, the whole deformation will vanish. Beyond point ‘B’ the rate of increase in strain 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 point ’D’ 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 as the curve reaches E, the bar breaks.

Modulus of elasticity E = __P l with usual notations. AoΔ

From the straight line graph between load and extension P/A can be determined. Measure l& A and calculate E

During loading at a particular point the load remains constant for few seconds and again goes on increasing. 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 100 Original length

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

PROCEDURE: - Clean the mild steel rod neatly with sand paper and measure the diameter of the rod at three places and find the mean of them (d). Calculate the original cross sectional area (So). Insert suitable jaws in the grips. Calculate the maximum load assuming the ultimate stress of mild steel as 500 N/mm2. Insert the test piece in the grips by adjusting the cross head of the machine after making zero correction. 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. 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).

Operate the lower grip operation handle and lift the lower cross head, up and grip fully lower part of the specimen. Lock the jaws in position by operating the jaw-locking handle. Turn the right control valve slowly in open position (anti clock wise) until a desired loading rate is reached. Start the electronic equipment along with hydraulic pump and note the extension corresponding to each load. Increase the load gradually until the test piece is broken. Close the right control valve; take out the broken piece of test pieces. Open the left control valve to

take the piston down. Place the broken pieces together so that the length between gauge lengths after elongation can be noted. Calculate the reduced cross sectional area by measuring the reduced diameter of the broken pieces.

Draw the Load Vs Extension graph and calculate the required quantities.

NOTE:1. If an extensometer is used for noting the extension, first of all calculate the gauge

length Lo = 5.65So. Mark the gauge length on the specimen such that central point is at middle of gauge length. Punch the marked points. Again mark the extensometer gauge length on either side of the center point. Place and tighten the specimen in the extensometer.

2. If the minimum elongation specified is not obtained, the result of the test should unless otherwise agreed be discarded if the distance between the fracture and the nearer gauge mark is less than one third of gauge length.

3. To avoid the possibility of rejecting test pieces due to fracture being outside the limits specified above, the following method might be employed. -

(a) Before testing subdivide the gauge length Lo into N equal parts.

(b) After testing, designate by ‘A’ the end mark on the shorter piece, on the longer piece designate by ‘B’. The graduation mark, the distance from which to the fracture is most nearly equal to the distance from the fracture to the end mark A.

(c) If ‘n’ is the number of intervals between A&B, the elongation after fracture is determined as follows.

If (N-n) is an even number (Fig.A) measure the distance between A&B and the distance from a graduation mark C, at (N-n)/2 intervals from B, then calculate the elongation after fracture from the formula:

A = AB + 2BC – Lo x 100 Lo

If (N-n) is an odd number (see Fig. B) measure the distance between A&B and distance from B to the graduation mark C ’ and C” at (N-n-1)/2 and (N-n+1)/2 intervals from B; then calculate the elongation after fracture from the formula:

A =AB+ BC ’ + BC ’’ – Lo x 100 Lo

A B C

n N-n 2

Fig(A)

A B C’ C’’

n N-n- 1 1 2

Fig(B)

Measurement of elongation:

Any statement of the result of percentage of elongation test should include the dimensions of the section of the test piece and its gauge length (where it is convenient, for economic or other reasons to use a fixed gauge length irrespective of the cross sectional area, the equivalent elongation on 5.65So, may if required be obtained by means of a formula or conversion chart given in IS 3803-1967. In case of dispute the elongation should be measured on a gauge length of 5.65So.

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:-

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

Calculations:

1. Yield Stress = Fy/So = N/mm2

2. Ultimate Stress = Fm/So = N/mm2

3. Nominal breaking stress = Fb/So = N/mm2

4. Actual breaking stress = Fb/Su = N/mm2

5. Percentage elongation on aGauge length of ________mm = A = (Lu-Lo)/Lo x 100 =--------%

6. Percentage reduction in area = (Su-So)/So x 100 = ------------%

7. Young’s Modulus = N/mm2.

Load in

Extension in

DOUBLE SHEAR TEST

(IS 5242 – 1969)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 following formula:

Fs = 1/2F = 2Fd2/4 d2

Where ‘d’ is the actual diameter of the specimen.

PROCEDURE :- Measure the diameter of the rod and calculate area of cross section ‘A’. Assuming ultimate shear stress as 350 N/mm2, determine the approximate load that the specimen will carry.

Place the specimen in the shear shackle firmly and apply the load on the specimen after adjusting the machine for zero error. Load the specimen to failure and note the failure load. The load at failure divided by area of cross section will give the ultimate shear strength.

RESULT:-

Shear Strength of given specimen = N/mm2

INFERENCE:-

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 = 2F/d2 =

SPRING TEST

ExptNo :Date:

AIM to determine stiffness of the given springs and 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.

D – Wire diameter

P – Pitch of coil

N – Number of coils.

W – Axial load on spring. P d

N – Modulus of rigidity for the spring material

Fs - Maximum shear stress induced in the spring wire.

F – Bending stress induced in the spring wire due to bending.

- Deflection of spring as a result of axial load.

- Angle of helix. R

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

i.e. T = WR cos , M = WR sin

θ = Angle of twist as a result of twisting moment WR cos

= Angle of bend, as a result of bending moment WR sin

We know that length of spring wire l = 2nR sec

Twisting moment T = /16 fs d3

WR cos = T/16fs d3

We know that M/I = f/y

f = My/I = WRsin d/2 = 32 WR sin /64 d4 d3

We know that T/J = Nθ/l

= Tl/JN = (WRcos*l )/JN

Angle of bend due to bending moment = Ml/El = (WR sin*l)/EI. Work done by the load in deflecting the spring is equal to strain energy of the spring.

½ W = 1/2Tθ+1/2 M

W = Tθ + M

= WRcos x(WR cosl)/JN + WR sin x (WRsinl)/EI

= WR* l(cos2/JN + sin2/EI)

Now substituting the values of l = 2nr sec, J = /32 (d4) and I = /64(d4) in the above equation.

= (64WR3n) sec{cos2)+2 sin2}d4 N E

In the case of closed coiled spring is very small so that cos = 1, sin = 0then = (64WR3n)/Nd4

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 of 1:5 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 be noted against the arrow on the lower compression plate.

PROCEDURE: - 1. Spring under tension :- suspend the spring between the tension hook and the hook provided by the side of the platform. Balance the machine and note initial readings. Place a weight of 5 Kg on the pan and note readings. Similar readings are taken by adding weights. Note the readings during unloading also. Calculate the average of two readings and plot a graph between load and deflection. From the graph determine the value of W/. Using vernier calipers determine the mean diameter of wire as well as external coil diameter. Calculate mean coil diameter by subtracting mean diameter of wire from external diameter of coil. Measure Distance between 10 turns of the spring from which determine pitch of the spring. Using the data collected, determine stiffness and modulus of rigidity of the material of the spring.

Spring under compression: Balance the machine carefully by means of sliding weight. Place the spring on the compression plate and lower the compression plate until the spring just touches the upper compression plate. Note the initial reading. Place 5 Kg on the pan and note reading. Repeat the process by adding extra weights. Readings are noted during unloading also.

RESULT :-

Spring UnderCompression Tension

Modulus of rigidity

Stiffness

INFERENCE :-

OBSERVATIONS AND TABULATIONS:-

Particulars Spring UnderCompression Tension

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)

=

Sl No Load Kg

Scale Reading Average reading

Loading Unloading

Deflection

Spring under TensionSpring under compre-ssion

Maximum load Wm = Maximum deflection m =

Calculations :- 1. Open coiled spring:-

Stiffness = W/

E= 2N (1+1/m) = 2N (1+0.3) = 2.6N

= 64WR3 n/d4 sec [cos2/N +2 sin2/E]

N = (64WR3n)/d4 sec [cos2 + 2sin 2/2.6]

= N/mm2

For closed coiled spring N = (64WR3n)/ d4

Torsional shear stress at maximum load Wm =qmax = 16Wm R/ d3)

Elastic strain energy stored = U= (Wm x m)/2

Volume of the spring V = [2R(d2)/4 x n]

Strain energy per unit volume = U/V =

Spring 2

Spring 3

Spring 4

BRINELL HARDNESS TEST

(IS – 1500-1968)

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 test piece 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 Kgf) by the curved surface area 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

h=12

(D−√D2−d2 )

The curved surface area of indention = Dh

=

πD2

(D−√D2−d2 )

Brinell Hardness HB = Applied loadArea of indentation

=

Fπ Dh

=

2FπD (D−√D2−d2)

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 in seconds. 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’. No deformation shall be visible at the back of the test piece after the test.

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

Ball diameterin mm

Load Kg

HB Values100 200 300 400 500

2.9 187.5 1.91 0.95 0.64 0.48 0.425.0 750 3.81 1.90 1.27 0.97 0.8410.0 3000 2.64 3.81 2.54 1.90 1.70

Load for testing: Ferrous Metals – Hardness between 140 – 450F/D2 = 30 (D in mm) time – 10 seconds

Non-ferrous metals: Brass, copper – HB between 35-140F/D2 = 10 (D in mm) time – 30 seconds

PROCEDURE:- Place the weights corresponding to the selected load on the suspender. Attach the ball of required diameter in the spindle sleeve and secure it in place with screw. Rotate the hand wheel so that the specimen located on the table of the machine is pressed against the ball until it bears against the limiter. The center of the ball when this is done should be at a distance of 2.5 times ball diameters from the edge of the specimen and at a distance of at least 4 times the diameter from the center of neighbouring impressions. By pressing down the starting button, the electric motor is switched on. Loading is over automatically on completion of the set duration.

Upon completion of the test, lower the table with the specimen by rotating the hand wheel. Then with the aid of microscope determine the dimensions of the impression in two directions, right angles to each other. The difference in the result of both measurements should not exceed 2%. Determine mean arithmetic value of ‘d’; Find

HB =

2FπD (D−√D2−d2)

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.

RESULT :-

Material Brinell Hardness Number

INFERENCE:-

BHN = P/Spherical area of indentation in mm²Where spherical area of indentation = area of projection on the ball circle =area abc =πDyTo find y, a b

Yy

oe = √( D2 )2−( d2 )

2

c

y =

D2

−√( D2 )2−( d2 )

2

d

BHN =

P

πD [D2 −√( D2 )2−( d2 )

2] =

PπD2

(D−√D2−d2)

=

2 PπD (D−√D2−d2)

OBSERVATIONS:-

Material of specimen

Load in Kg and duration

Diameter of indenter D mm

Diameter of indentation HB Value Mean d1 d2 d=(d1+d2)/2

ROCKWELL HARDNESS TEST

(IS- 1586-1968 & 3804 – 1966)

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 ball indenter.

GENERAL:- This is a direct reading hardness testing machine compared to Brinell hardness testing machine, 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 as Rockwell 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 used for 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 and 90 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 load of 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

PROCEDURE :- Adjust the pointer on the dial to the initial position, big needle on C0, B30. Plane and smoothen the face of the specimen using emery paper and, place it on the supporting platform. Fix the indenter in the fixing ring. Turn the hand wheel and make the specimen to press against the fixing ring, till the smaller pointer reaches the center of red dot (SET position) and the big needle to the B30, or C0 reading. Now apply the minor load fully. Now bring the lever arm slowly without jerk into the loading position. The penetration commences and is indicated by the dial indicator. The time of loading is approximately 6 seconds. When the pointer has come to stop, the hand lever is set back to the rest position and read the appropriate Rockwell value HRB or HRC on the dial. Repeat the experiment and take 3 sets of observations for each specimen.

Brinell hardness can also be found in this machine. The ball indenter is then 2.5mm and duration of loading is 15 sec. Loads for Brinell hardness are, 187.5 Kg for steel and cast iron and 62.5 Kg for non-ferrous alloys. After applying the final load, with the hand lever the specimen is taken out and the surface area of indentation produced is determined. Brinell hardness number is calculated as load divided by the surface area of indentation produced.

RESULT ;-

Material Rockwell Hardness Number

INFERENCE:-

OBSERVATIONS:-

Sl No. Material Test Loadin Kg

Penetrator used

Scale Used Rockwell Hardness Number

Mean

IMPACT TEST

(IS 1499 – 1977, 1598-1977 & 3766 – 1966)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 together with 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 is to 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 absorption of 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 or resilience.

PRINCIPLE :-

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

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 1410 47’ to the vertical and the initial energy 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+sin 1)

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 movethrough a high ‘h1’ making an angle θ2 with the rest position.

Residual energy E2 = Wl(1-cosθ2)

Energy absorbed is calculated for various values of θ2

and a graph is plotted between EL and θ2 which is thecalibration curve for the machine.

PROCEDURE:-

CHARPY TEST:-

Calculate the length of the pendulum by noting the period of oscillation using which the weight has to be determined. Mount the test piece in such a way that the edge of the hammer is in one line with the groove on the test piece. Then turn the groove away from the edge of the hammer. Adjust the support by proper means. Adjust the striker in such a way that the vertical edge hits the specimen. After placing the test piece on the support, lift the hammer by hand and place it by means of catch. Then release the hammer by drawing the safety device and raising the catch. The hammer hits the test piece. The pointer automatically moves with the hammer. The pointer reads the energy absorbed in breaking the specimen (The test set up is so arranged that the angle θ2 can be read from the dial from which the energy absorbed can be calculated using the relevant formulae).

For repeating the experiment, shift the pointer to the initial position.

IZOD TEST:- (Cantilever Test):-

The striker is fitted with the horizontal face in the striking position. Raise the pendulum to the izod position and lock it. The device for securing the specimen is screwed to the base plate of the pendulum stand. Then clamp the test piece into the vice in such a way that the middle of the groove (to be turned towards the hammer) is in one level with the upper face of the vice. Release the hammer. It hits the specimen. Note the readings as for the charpy 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

RESULT:-

Test No. Details of specimen

Energy Loss in Joules

Izod Charpy

From Graph

From Calculation

From Graph

From Calculation

Impact Strength

Impact modulus

INFERENCE:-

Description Izod CharpyWeight W

Length L

Initial energy E1

Initial Energy E1 = w1(1+sinθ1)

θ1 =

Energy loss EL = w1(sin θ1 + cos θ2)

θ2, Degrees Energy Loss (izod) Joules

Energy Loss (Charpy) Joules

0

10

20

30

40

50

60

70

80

90

100

110

120

140

14147’

Impact Strength

Impact modulus

Diagram showing specimen details

TESTS ON WOOD

(IS 1708 –1969 & IS 888 –1970)

Expt No:Date:

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

1. Static bending test2. Compression parallel to grain.3. Compression perpendicular to grain.

1. Static Bending Test.

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

EQUIPMENTS :- UTE – 40

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 and ‘Z’ is the modulus of section of the beam ie.bd2/6 for a rectangular cross-section. If we know the load at failure, (Wmax) and modulus of section, , from the above equation, f= w l/4Z. Assuming a maximum stress fmaxof about 600 N/mm2 then 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 equal 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 standard specimen cannot be obtained the dimensions of the test specimen should be such that the span is 14 times the depth).

PROCEDURE:- Measure the dimensions of the specimen and mark the span length the central point . Place the specimen on the supports such that the loading block is at the central point. Before applying the load, zero correction for the machine is to be done. Close the outlet valve and open the inlet valve slowly. Apply the load continuously such that the moving head of the machine moves at a rate of 0.00025 l2/d per minute where ‘l’; is the span and ‘d’ is the depth of the specimen. i.e. for a standard specimen the rate is 2.5mm/minute. The electronic equipment should be started along with hydraulic system. Measure the deflection at a load interval of 50 Kg. Load the specimen to failure and note the failure load. Draw a graph of load vs deflection.

2. Compression test parallel to grain:-

AIM :- To determine the compressive test of wood under compression parallel to grain using compressive 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:- Measure the dimensions of the specimen and calculate the cross sectional area. Assuming ultimate stress, determine the maximum load that can be applied on the specimen. Place the specimen on the compression plate such that the application of load will be parallel to grain. Select the dial for that particular range of load. Tightly close the outlet valve. Slowly open the inlet valve as well as valve corresponding to the selected dial. Apply the load continuously until the specimen fails. The pointer shows maximum reading and then turns back. Note the maximum load. Open the outlet vale to release the load on the specimen. Calculate the compressive strength of given specimen parallel to the grains.

3. Compression Perpendicular to Grains:-

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

PROCEDURE:- Measure the dimensions of the specimen. Place the specimen between compression plates such that axis of loading is perpendicular to the grain. Calculate the maximum load that can be applied on the specimen. Apply the load slowly by opening the inlet valve until specimen fails. Note down the maximum load open one outlet valve to release the load.

RESULT :-

1. Bending test

a) Fiber Stress at limit of proportionality =

b) Modulus of elasticity =

c) Modulus of rupture =

d) Elastic Resilience =

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

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

INFERENCE:-

Observations:-

1. Static bending test on timber

Load kNDeflectionmm

Figure showing specimen details

Span of the specimen, lmm =

Breadth of specimen, bmm =

Depth of specimen, dmm =

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

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

Maximum load, Wu =

Load at limit of proportionality =

(from graph)

Deflection at limit of proportionality =(from graph)

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

Equivalent fibre stress at maximum load = (W’l)/4Z =

(Modulus of rupture)

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

Elastic resilience = Work to limit of proportionality ---------------------------------------

Volume= Area under the curve upto limit of proportionality ------------------------------------------------------------

Volume

2.Compression test parallel to grain

Dimension of cross section =Crushing load =

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

C.S. area

1. Compression test perpendicular to grain :-

Dimension of cross section =Crushing load =Compressive strength perpendicular to grain = Crushing load

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

DEFLECTION TEST ON BEAMS

VERIFICATION OF MAXWELL’S RECIPROCAL THEOREM

Expt no;

Date:

AIM:- To verify Maxwell’s reciprocal theorem

EQUIPMENTS:- Magnetic Stand, Dial gauge etc.

PRINCIPLE:- Maxwell’s 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 the first 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:- Mark two points A & B on the fixed beam. Apply load at A and note the deflection at B. Increase the load and again note the deflection corresponding to each load. Interchange the position of load and deflection dial and note readings.

RESULT :- Maxwell’s reciprocal theorem is verified.

INFERENCE:-

Observations:-

Load at A Deflection at B

Load at B Deflection at A

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 by two 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 in the dial by drag pointer.

This strength-testing machine has three power measuring ranges. This permits finer graduations and hence 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 be tested 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 the center 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 lower grip goes back to its starting position for repeating the experiment. The difference between R2& R1 gives extension. The percentage elongation can be calculated using the formula (R2-R1)/R1 x 100

RESULT :-

Tensile strength of given wire =

Percentage elongation of given wire =

INFERENCE:-

OBSERVATION:-

Diameter of specimen =

Area of cross section of specimen A = πd2/4

Tensile load of specimen P =

Tensile strength of specimen = P/A

Initial extension scale reading R1 =

Final extension scale reading R2 =

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

FATIGUE TEST

Expt No,:

Date:

AIM:- To determine the number of cycles required for breaking the given specimen using Fatigue testing machine.

GENERAL:- While studying the mechanical properties of a material it is quite necessary to study the fatigue limit of the material also. In tension test, hardness test, torsion test etc. we have seen stress due to gradually applied load and without shock. In most cases, most of the members are subjected to efforts of loads that do not remain steady.

Some of the machine parts like axles, shafts, connect rods, pinion teeth are subjected to varying stress. Due to some loading in these parts, while they rotates, the stress in the member will change alternately (ie from tension to compression or viceversa). This alternating stresses are equal but opposite in sign and hence the mean stress becomes zero. This particular case of stress is known as reversed stresses or stress reversal.

Due to continuous effect of stress reversal, after sometime, the member will possibly get failed. For avoiding the failing due to stress reversal in the above said or similar machine parts, it is necessary to find the fatigue limit of the material.

Fatigue failure is a phenomenon in which a component fails due to repeated loading. Repeated loading condition in a component occurs when the stresses in it due to the load applied vary or fluctuate between maximum and minimum values. In the case of static loading conditions, the load is applied gradually, giving sufficient time for the strain to develop, whereas in the case of repeated loading this does not hold good. Hence, machine members subjected to repeated loading have been found to fail at stresses that are very much below the ultimate strength and very often below the yield strength.

Fatigue failure usually begins with a small surface crack undetectable with naked eyes, and grows rapidly deeper causing the component to fail. The stress concentration due to internal cracks, grooves, keyways etc. becomes more predominant after the surface crack develops.

Repeated loading can be applied in four fashions, namely, reversed axial loads, reversed bending loads, reversed torsional loads and combined loads. The subject machine applies load in reversed bending fashion so that the fibers of the test specimen are stressed once in tension and once in compression. The stresses vary in a sinusoidal form. The bending moment in the test cross-section of specimen is constant during the test. Counter records the number of revolutions at which the specimen fails.

PURPOSE:- The testing machine is mainly used for two purposes: 1. To study the behaviour of the material and to draw the S-N diagram and 2. To check the material for expected number of revolutions at specific stress.

In the first case, a number of identical test piece bars are made from the material to be tested. First fix one piece in the space provided in the machine. Then apply some load. Allow the electric motor to run so that it rotates the specimen also. The cycle counter will automatically record number of rotations made provided. After some time due to continuous stress reversal the specimen will fail. By knowing the diameter of the specimen, stress applied in the material can be calculated. Note the number of rotations. Then repeating the test for the remaining specimen by decreasing the load, the corresponding stresses as well as number of rotations are recorded. We can see as the load is decreasing, the number of rotations required for failing the specimen is increased. Then draw a graph with stress in the Y-axis and number of rotations in the X-axis. From the graph we can see after some tests, a limit is reached where the stress is not sufficient to break the material. This safe stress is known as Endurance limit.

Stress

Endurance Limit

No of cycles

In the second case, generally the bending stress to be applied is decided, depending upon the design requirements. Suppose the design requirement is such that it should withstand a bending stress of 400N/mm2, then the load to be applied is calculated as follows:

fb = (509.3 x P)/d³

P = (fb x d³)/509.3 = (400 x (8)³ )/509.3

P = 402 N

This load is applied and the number of revolutions at which the specimen fails are recorded and checked against the expected.

PRINCIPLE:-L = 10 cm L = 10 cm

SPECIMEN

P/2 P/2

P

M

The specimen loading arrangement (fig.) results in a constant bending moment PL/2 over the test length of specimen.

Where P = load applied over the specimen (N)L= 10cmNow Bending moment Mb = PL/2 = (Px100)/2 = 50P N-mmBending Stress fb = Mb/Z N/mm2

Where Z = Section modulus = πd³/32 for circular cross sectionsfb = (Mb x 32)/d³ = (50P x 32)/(π x d³)

= 509.3P/ d³ N/mm²

Where P is in N, fb in N/mm², Mb in N-mm, d in mm.

PROCEDURE:- Fix the specimen to the specimen pulling out stud in the tapping provided over face. Insert the specimen with stud into the bore of LH swiveling body and push it further till it gets inserted in the collet of RH swivelling body and rests against the specimen locator. In this position the specimen cannot be pushed further. By pressing down the locking rod such that is a\inserts into the slots of locking ring and prevents hollow shaft from rotating, tighten the specimen by rotating the clamping cum loosening ring with the help of special spanner. The locking rod is spring loaded and hence it will immediately come out of the slots, as soon as hand is released. In no case should be the locking enters the slots when machine is in running condition. Repeat the procedure for other side assembly. Take out the specimen pulling out stud by removing it from the tap in specimen. Select the load required, depending upon the bending moment to be imposed, by moving the loading weight and selecting proper set of additional weights. Lock the loading weight by locking screw. Use the pin and support while moving the loading weight so that the lever is not moved. Remove the pin from the support before starting the motor (otherwise the specimen will rotate without application of any bending moment). Check the direction of rotation. Reset the counter to show all zeroes before running the specimen. Start the motor, thus starting the test. The motor will stop after the specimen fails and the counter will have recorded the number of revolutions completed by the specimen.

RESULT ;-

Number of cycles required for breaking the specimen =

INFERENCE:-