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Department of Aeronautical Engineering

AEROSPACE STRUCTURES

Laboratory Manual (2015-16)

III YEAR B.TECH

(AERONAUTICAL ENGINEERING)


MARRI LAXMAN REDDY INSTITUTIONS | 2

AEROSPACE STRUCTURES

III YEAR B.TECH

(AERONAUTICAL ENGINEERING)

Document No:

MLRIT/AERO/LAB MANUAL/AS Lab

VERSION 1.0.0

Date of Issue

30 JUNE 2015

Date of Revision

JUNE 2015

Compiled by

K.Veeranjaneyulu

( Professor)

Verified by

Dr.M.Satyanarayana

Gupta

Authorized by

HOD(Aero)



PREFACE

Aerospace structures are an important subject of Aeronautical engineering students in

the III year B.Tech of JNTU, Hyderabad.

This manual is a collective effort of the faculty teaching Third year AS lab Through AVS

subject.

This manual will need constant upgradation based on the student feedback and change in the

syllabus.

HOD (AERO) PRINCIPAL



LAB CODE

1. Student should report to the concerned as per the time table.

2. Students who turn up late to the labs will in no case be permitted to the

program schedule for the day.

3. After completion of the program, certification of the concerned staff in-charge

in the observation book is necessary.

4. Student should bring a note book of 100 pages and should enter the

readings/observations into the note book while performing the experiment.

5. The record of observations along with the detailed experimental Algorithm of

the experiment in the immediate last session should be submitted and certified

staff member in-charge

6. Not more than 3 students in a group are permitted to perform the experiment on

the set.

7. The group-wise division made in the beginning should be adhered to and no

mix up of students among different groups will be permitted.

8. The components required pertaining to the experiment should be collected from

stores in-charge after duly filling in the requisition form.

9. When the experiment is completed, should disconnect the setup made by them,

and should return all the components / instruments taken for the purpose.

10. Any damage of the equipment or burn –out components will be viewed

seriously either by putting penalty or by dismissing the total group of students

from the lab for the semester / year.

11. Students should be present in the labs for total scheduled duration.

12. Students are required to prepare thoroughly to perform the experiment before

coming to laboratory.

13. Algorithm sheets/data sheets provided to student‟s groups should be

maintained neatly and to be returned after the experiment.



INDEX

SNO Experiment

No

Name of the Experiment Page

No

Objective , Outcomes, Hardware Software Requirements

1 1 Determination of young’s modulus of mild

steel by deflection method

2 2 Verification of Castiglione’s theorem

3 3 Verification of Maxwell’s reciprocal theorem

4 4 Verification of principle of superposition

5 5 Shear center of open channel section

6 6 Shear center of open angle section

7 7 Buckling of column when both ends are hinged

8 8 Buckling of column when one end hinged and

other end fixed

9 9 Buckling of column when both ends-fixed



10 10 Preparation of riveted joint

11 11 Failure strength of riveted joint

12 12 Free longitudinal vibrations

13 13 Study of forced vibrations

14 14 NDT using magnetic particle test equipment

15 15 NDT using ultrasonic test equipment

16 16 Find the surface cracks on the given specimen

using dry penetration technique

17 17 The longitudinal and circumferential stresses

of thin walled pressure vessel



Objective

To find the deflection of beam subjected to different loading and support conditions.

To determine the shear centre of open and closed section beams

To find the Amplitude and frequency of free and forced vibration of beams

To find the longitudinal strain and circumfential strain of thin cylindrical shell

To find the crippling load of columns

To determine the failure strength of riveted joint

Outcome

Ability to calculate the deflection of beams and calculate the young‟s modulus

Student is able to identify the load path of the thin walled section beams

Student is able to understand the deflection of beams under different loads and support

conditions

Able to calculate the buckling load of columns subjected to different end conditions

Equipment required

UTM

Beam set up

Column test rig

Shear centre of open and closed section beams

Vibration of beams

Thin cylinder apparatus



Experiment No: 1

Experiment as given in the JNTUH curriculum.

DETERMINATION OF YOUNG’S MODULUS OF MILD

STEEL BY DEFLECTION METHOD

AIM:

To determine the Young‟s modulus of a given mild steel Beam

APPARATUS:

Beam test set up

Weights 500 Gms – 2 No.„s

Loading hooks – 1 No.

Mild steel bar

Measuring tape

Dial gauge – 1 No.

FORMULA:

The formula for young‟s modulus from the deflection of a rectangular beam which is simply

supported is given by

𝑬 =𝒎 × 𝒈 × 𝒍𝟑

𝟒𝟖 × 𝑰 × 𝒚

Where,

I = moment of inertia of the beam

g = acceleration due to gravity = 9.81m/sec2

y = deflection of the beam

m = mass of the load applied

l = distance between the two supports



PROCEDURE:

The beam is placed on the frame where both ends are simply supported.

The longitudinal and the cross-sectional dimensions of the beam are noted.

The mid-point of the beam is marked and the loading hook should be placed there at

the centre.

Now the dial gauge is mounted exactly on the middle of the loading hook.

The load is applied on the loading hook and the corresponding deflection is noted

down.

Now these values are substituted in the theoretical formula given and the young‟s

modulus of the material is found.

Fig: simply supported beam with load at centre.

TABULAR COLUMN:

S. No Weight

(Kg)

Deflection

(div)

Deflection

(mm)

Young‟s Modulus

PRECAUTIONS:

o Take the readings without parallax error

o While doing the experiment, see that no external loads are acting on the table or

frame.

RESULT:



The Young‟s modulus of given material is found and is of value ___________

Experiment conducted at MLRIT

Case 1: Determine Young’s modulus of simply supported Beam when the load on the

beam is 1000gms

Input data:

Mild steel Beam

25mmX6mmX990mm

Weight: 1000 Gms

FORMULA:

𝑬 =𝒎×𝒈×𝒍𝟑

𝟒𝟖×𝑰×𝒚

Result: 2.67X105N/mm

2

Case2: Determine Young’s modulus of simply supported Beam when the load on the

beam is 800gms

Result: 2.6X105N/mm

2


beam is 700gms


2


beam is 500gms


2



VIVA QUESTIONS

1. Define Young‟s modulus?

Young‟s modulus is a measure of stiffness of an elastic material and is a quantity used to

characterize materials.

2. The expression for the young‟s modulus is E=stress/strain=ζ/ε

3. The young‟s modulus of the stainless steel is 200 Gpa

4. The young‟s modulus of the iron is 190–210 Gpa

5. The young‟s modulus of the aluminum is 69 Gpa

6. Define moment of inertia?

Moment of inertia is a property of rotating bodies that defines its resistance to a change in

angular velocity about an axis of rotation. It is a measure of the resistance of a body to angular

acceleration about a given axis that is equal to the sum of the products of each element of mass

in the body and the square of the element's distance from the axis.



Experiment No: 2


VERIFICATION OF CASTIGLIANO’S THEOREM

AIM:

To verify Castigliano‟s theorem for a given simply supported beam on loading.

APPARATUS REQUIRED:

Simply supported beam,

Dial gauge – 1 no.,

Supporting structure

Load 500 gms – 2 nos.

FORMULA USED:

Moment of inertia,

I =bd3

12 mm

4

b = breadth of beam in mm

d = width of beam in mm

Deflection of beam 𝛿

W = load applied in kg = 9.81*W in N

𝛿 = deflection in mm

′l‟ = length of the beam – 1040mm

THEORY:

Beam subjected to a load system, deflection point „P‟ is given by partial differential

co-efficient of the total strain energy with respect to pressure force acting at point and in the

direction in which the deflection is designed.

The figure shows a beam of span „l‟ applying load „W‟.

Reaction at A, RA = wb/l. Reaction at B = wa/l.



Fig: simply supported beam with eccentric load

PROCEDURE:

Fix the given beam on the frame so that it‟s simply supported.

Mark the point where the beam is loaded measure the distance AC & CB (AC=a;

BC=b).

Load the beam and note down the deflection in dial gauge which as placed on the load

point.

Note down the corresponding deflection in dial gauge.

Calculate moment of inertia and deflection.

Fig: Practical Setup

Tabular column:

Sl. No. Load Deflection(div) Deflection(mm)



RESULT:

The deflection under loading on a simply supported beam where theoretical deflection

calculated as follows:

𝛿𝑐 =𝑊𝑎2𝑏2

3𝐸𝐼𝐿=

E = Young‟s Modulus of Mild steel from tables = 2.1*105 N/mm

2

Compare the experiment value to theoretical value of deflection.

VIVA QUESTIONS

1. State Castigliano's first theorem?

If the strain energy of an elastic structure can be expressed as a function of generalised

displacement qi; then the partial derivative of the strain energy with respect to generalised

displacement gives the generalised force Qi.

2. State Castigliano's second theorem?

If the strain energy of a linearly elastic structure can be expressed as a function of

generalised force Qi; then the partial derivative of the strain energy with respect to

generalised force gives the generalised displacement qi in the direction of Qi.

3. Expression for the deflection of the beam?

𝛿𝑐 =𝑊𝑎2𝑏2

3𝐸𝐼𝐿

4. Classification of beams?

Simply supported, cantilever, hinged, fixed

5. Formula for deflection of beam given by castigliano‟s theorem?




Case1: verification of Castiglione’s theorem using simply supported beam when the

load of 1000grms applied.

Result: 1.43mm


load of 800grms applied.

Result: 1.01mm


load of 800grms applied

Result: 0.84mm



Experiment 3:


VERIFICATION OF MAXWELL’S RECIPROCAL THEOREM

AIM:

To verify Maxwell‟s reciprocal theorem for a given simply supported bean

EQUIPMENT:

Beam test set up

Simply supported beam

Dial gauge – 1No.

Measuring tape

Weights 200gm – 3 No. „s

One loading hook

THEORY:

The following are the three versions of Maxwell‟s reciprocal theorem

1. The deflection at point B due to load at point A is equal to the deflection at point A

due to load at point B.

2. The slope at point B due to unit moment at point A is equal to the slope at the point A

due to unit moment at point B.

3. The slope at point B due to unit load at point A is equal to the slope at point A due to

unit load at point B

PROCEDURE:

Place the beam on the simply supported edges.

Measure the length of the beam with the measuring tape

Mark two points A & B which are equidistant from the supports.

At first the loading hook is mounted at point A and the dial gauge is mounted at point

B, now the load is applied at A and corresponding deflections are noted down.



Repeat the same procedure by changing three positions of the dial gauge and the

loading hook.

TABULAR COLUMN:

1. When load is applied at point A

S. No Load applied (gms) Deflection (div) Deflection (mm)

2. When load is applied at point B

S. No Load applied (gms) Deflection (div) Deflection (mm)

PRECAUTIONS:

Make sure that dial gauge tip is in touch with the beam.

The dial gauge needle should be adjusted to zero before taking the readings.

Take the readings without parallax error.

RESULTS:

Hence Maxwell‟s reciprocal theorem is verified.




Case1:

Verification of Maxwell’s reciprocal theorem on a simply supported beam when the

applied loads are 200grams to 600grams

Result:

Case2:


applied loads are 500grms

1000grams

1500grams

Result:

Case3:


applied loads are 600grms

900grams

1200grams

Deflection at

point A

0.03mm

0.05mm

0.09mm

Deflection at

point B

0.03mm

0.05mm

0.09mm

Deflection at

point A

0.52

1.08

1.69

Deflection at

point B

0.52

1.08

1.69

Deflection at

point B



Result:

Deflection at

point A



Experiment 4:


VERIFICATION OF PRINCIPLE OF SUPERPOSITION

AIM:

To verify the principle of superposition using simply supported beam.

APPARATUS REQUIRED:

Simply supported beam,

Meter scale,

Dial gauge -1 No.,

2 hooks,

Slotted weight 200gm - 6 Nos.

THEORY:

Deflection by combined loading is equal to sum of the deflection by individual loading.

The total deformation is equal to the algebraic sum of the deformation is equal of the

individual section. This principle of finding out the resultant deformation is known as

principle of super position.

PROCEDURE:

Both side simply supported condition.

Place the beam on the frame.

The distance between supports in measured and it is taken as span length „l „.

Dial gauge is mounted middle of the beam „C‟ and two loading hooks are mounted

equidistant from „C‟.

Two points „D„& „E„ are selected nearby mid span to get accurate readings.

First, at points „D„& „E„ load is applied gradually in terms of 200gm and for every

load corresponding deflection reading δc to be noted.

The above procedure is followed at „D‟ position and readings should be taken at point

„C‟ and corresponding deflection δc1 is noted.



The above procedure is followed at „E‟ position and readings should be taken at point

„C‟ and corresponding deflection δc2 is noted.

Now all the individual loads to determine the combined load by loading in C and D

and note down the deflection.

Deflection by combined loading is equal to sum of the deflection by individual

loading.

𝛅𝐜 = 𝛅𝐜𝟏 + 𝛅𝐜𝟐

PRECAUTIONS:

o Take the readings without parallax error.

o While doing the experiment, see that no external force should act on the table or

frame

TABULAR COLUMN:

Case A

Load (w) 𝛿𝑐 in div 𝛿𝑐 in mm

Case 1



Load(w1) 𝛿𝑐1 𝛿𝑐1

in mm

Case 2

Load(w2) 𝛿𝑐2 in Div. 𝛿𝑐2

in mm

Load W = Load W1 + Load W2

Deflection

𝜹𝒄 = 𝜹𝒄𝟏 + 𝜹𝒄𝟐

Result:

The principle of superposition is verified.



Case1: Verification of principle of superposition when the applied loads are 500-1500

Gms esult:


Gms


Gms

𝛿𝑐 in

mm

𝛿𝑐1 in

mm

𝛿𝑐2 in

mm

0.47 0.26 0.23

1.03 0.45 0.46

1.58 0.75 0.76

𝛿𝑐 in

mm

𝛿𝑐1 in

mm

𝛿𝑐2 in

mm

0.47 0.26 0.23

1.03 0.45 0.46

1.58 0.75 0.76

𝛿𝑐 in mm 𝛿𝑐1

in

mm

𝛿𝑐2 in

mm

0.47 0.26 0.23

1.03 0.45 0.46

1.58 0.75 0.76



Experiment: 5


SHEAR CENTER OF OPEN SECTION

(CHANNEL)

AIM:

To determine shear centre of the given channel section and validate with theoretical

value.

THEORY:

For any unsymmetrical section there exists a point at which any vertical force does not

produce a twist of that section. This point is known as shear center.

The location of this shear center important in the design of beams of open sections

when they should bend without twisting, as they are weak in resting torsion. A thin walled

channel section with its web vertical has a horizontal axis of symmetry and the shear center

lies on it. The aim of the experiment is to determine its location on this axis if the applied

shear to the tip section is vertical (i.e. along the direction of one of the principal axes of the

section) and passes through the shear center tip, all other sections of the beam do not twist.

APPARATUS REQUIRED:

Channel section

Weight hanger - 2 No.‟s (each of 100 Gms.)

Dial gauge with stand - 2 No.‟s

Weights of 200gms -6Nos.

PROCEDURE:

1. Mount two dial gauges on the flange at a known distance apart at the free end of the

beam (see fig). Set the dial gauge readings to zero.

2. Place a total of say 1.3 kilograms load at A (loading hook and six load pieces will

make up this value). Note the dial gauge readings (Hooks also weigh a 100gm each).



A-side dial gauge rotate in anticlockwise direction. B-side dial gauge rotates in

clockwise direction. Note down the dial gauge readings.

3. Now remove one load piece from the hook at A and place in hook at B. this means

that the total vertical load on this section remains 1.4 kilogram. Record the dial gauge

readings.

4. Transfer carefully all the load pieces to B one by one. Noting each time the dial gauge

reading. This procedure ensures that while the magnitude of the resultant vertical

force remains the same, its line of action shifts by a known amount along AB every

time a load piece is shifted. Calculate the distance „e‟ (see fig) of the line of action

from the web thus:

e =AB Wa−Wb

2Wv

where,

Wv = total load applied = (Wa+Wb)

Wa = load applied at point A

Wb = load applied at point B

e = location of shear centre from the web

5. For every load case calculate the algebraic difference between the dial gauge is (u-v)

readings as the measure of the angle of twist 0 suffered by the section.

6. Plot 0 against e and obtain the meeting point of the curve (a straight line in this case)

with the e-axis (i.e., 0, the twist of the section is zero for this location of the resultant

vertical load). This determines the shear center.

7. Though a nominal value of 1.2 kilograms for the total load is suggested it can be less.

In that event the number of readings taken will reduce proportionately.

TABULAR COLUMN:

Dimensions of the beam and the section::

Length of the beam (L) : 500mm

Height of the web (h) : 100mm

Width of the flange (b) : 50mm

Thickness of the sheet (t) : 1.6mm

Distance between the two hook stations (AB) : 300mm



Load at A (Wa) :

Load at B (Wb) :

Theoretical location of the shear center (e) : ethe= 3b/ [6+(h/b)]

S.No Wa Wb d1 d2 d1 – d2 e = AB(Wa - Wb) / 2(Wa+Wb)

1.

2.

3.

4.

5.

6.

7.

Plot e versus (d1-d2) curve and determine where this meets the e axis and locate the

shear center.

PRECAUTIONS

1. For the section supplied there are limits on the maximum value of loads to obtain

acceptable experimental results. Beyond these the section could undergo excessive

permanent deformation and damage the beam forever. Do not therefore exceed the

suggested values for the load.

2. The dial gauges must be mounted firmly. Every time before taking the readings tap

the set up (not the gauges) gently several times until the reading pointers one the

gauges settle down and do not shift any further. This shift happens due to both back

lash and slippages at the points of contact between the dial gauges and the sheet

surfaces can induce errors if not taken care of. Repeat the experiments with identical

settings several times to ensure consistency in the readings.



RESULT:

Hence the shear centre of the given channel section is found and is validated with

theoretical value by plotting a graph.

a) Theoretical method =

b) Experimental method =

c) Error percentage =




Case 1: The location of the shear center when the total acting on the section is

1400gms

Theoretical method =

Experimental method =



Error percentage =

Case2: The location of the shear center when the total acting on the section is

1000gms



Error percentage =


800gms



Error percentage =

VIVA QUESTIONS

1. Classify different types of sections?

Types of sections are I-section, Z- section, C- section, D- section, L-section etc.,

2. Define shear center?

The point where the twisting moment is zero is called Shear Center.

3. Define Shear force?

A shear load is a force that tends to produce a sliding failure on a material along a plane that

is parallel to the direction of the force.

4. Define Shear stress?

The stress developed in the body when a tangential force is applied on it.



Experiment: 6


SHEAR CENTER OF OPEN SECTION

(ANGLE)

AIM:

To determine shear centre of the given channel section and validate with theoretical

value.

THEORY:

For any unsymmetrical section there exists a point at which any vertical force does not

produce a twist of that section. This point is known as shear center.

The location of this shear center important in the design of beams of open sections

when they should bend without twisting, as they are weak in resting torsion. A thin walled

angle section with its web vertical has a horizontal axis of symmetry and the shear center lies

on it. The aim of the experiment is to determine its location on this axis if the applied shear to

the tip section is vertical (i.e. along the direction of one of the principal axes of the section)

and passes through the shear center tip, all other sections of the beam do not twist.

APPARATUS REQUIRED:

Channel section

Weight hanger - 2 No.‟s (each of 100 Gms.)

Dial gauge with stand - 2 No.‟s

Weights of 100gms -6Nos.

PROCEDURE:

1. Mount two dial gauges on the flange at a known distance apart at the free end of

the beam (see fig). Set the dial gauge readings to zero.



2. Place a total of say 0.7 kilograms load at A (loading hook and six load pieces will

make up this value). Note the dial gauge readings (Hooks also weigh a 100gm

each). A-side dial gauge rotate in anticlockwise direction. B-side dial gauge

rotates in clockwise direction. Note down the dial gauge readings.

3. Now remove one load piece from the hook at A and place in hook at B. this means

that the total vertical load on this section remains 0.8 kilogram. Record the dial

gauge readings.

4. Transfer carefully all the load pieces to B one by one. Noting each time the dial

gauge reading. This procedure ensures that while the magnitude of the resultant

vertical force remains the same, its line of action shifts by a known amount along

AB every time a load piece is shifted. Calculate the distance „e‟ (see fig) of the

line of action from the web thus:

e =AB Wa−Wb

2Wv

where,

Wv = total load applied = (Wa+Wb)

Wa = load applied at point A

Wb = load applied at point B

e = location of shear centre from the web

5. For every load case calculate the algebraic difference between the dial gauge is (u-

v) readings as the measure of the angle of twist 0 suffered by the section.

6. Plot 0 against e and obtain the meeting point of the curve (a straight line in this

case) with the e-axis (i.e., 0, the twist of the section is zero for this location of the

resultant vertical load). This determines the shear center.

7. Though a nominal value of 0.6 kilograms for the total load is suggested it can be

less. In that event the number of readings taken will reduce proportionately.

TABULAR COLUMN:

Dimensions of the beam and the section::

Length of the beam (L) : 500mm

Height of the web (h) : 50mm

Width of the flange (b) : 50mm



Thickness of the sheet (t) : 1.6mm

Distance between the two hook stations (AB) : 300mm

Load at A (Wa) :

Load at B (Wb) :

Theoretical location of the shear center (e) : ethe= 3b/ [6+(h/b)]

S.No Wa Wb d1 d2 d1 – d2 e = AB(Wa - Wb) / 2(Wa+Wb)

8.

9.

10.

11.

12.

13.

14.

Plot e versus (d1-d2) curve and determine where this meets the e axis and locate the

shear center.

PRECAUTIONS

1. For the section supplied there are limits on the maximum value of loads to obtain

acceptable experimental results. Beyond these the section could undergo excessive

permanent deformation and damage the beam forever. Do not therefore exceed the

suggested values for the load.

2. The dial gauges must be mounted firmly. Every time before taking the readings tap

the set up (not the gauges) gently several times until the reading pointers one the

gauges settle down and do not shift any further. This shift happens due to both back

lash and slippages at the points of contact between the dial gauges and the sheet

surfaces can induce errors if not taken care of. Repeat the experiments with identical

settings several times to ensure consistency in the readings.



RESULT:

Hence the shear centre of the given channel section is found and is validated with

theoretical value by plotting a graph.

d) Theoretical method =

e) Experimental method =

f) Error percentage =



g)


Case 1: The location of the shear center when the total acting on the section is

800gms





Error percentage =


800gms



Error percentage =


800gms



Error percentage =

VIVA QUESTIONS

1. Define Shear strain?

2. The distance „e‟ of the line of action from the web is given as



e =AB Wa − Wb

2Wv

3. Define Shear strength?

Shear strength is the strength of a material or component against the type of yield or

structural failure where the material or component fails in shear.

4. Define Shear Modulus?

Shear modulus or modulus of rigidity, denoted by G, or sometimes S or μ, is defined as the

ratio of shear stress to the shear strain.



Experiment: 7

BUCKLING OF COLUMN WHEN BOTH ENDS ARE HINGED

AIM:

To find buckling load of column using column test setup arrangement under both ends are

hinged condition.

APPARATUS REQUIRED:

Column test,

Load indicator,

Specimen rod and

Two hinged supports.

FORMULA USED:

𝐏 =𝛑𝟐𝐄𝐈

𝒍𝐞𝟐

P = Crippling load

E = Young‟s Modulus of Specimen

I = Moment of Inertia

le = Effective length

EULER’S COLUMN THEORY:

As per Euler‟s equation for buckling load of long column based on bending stress the

effect of direct stress is neglected .This may be adjusted justified with the statement ,direct

stress included in a column is negligible as compared to the bending stress.

ASSUMPTION:

Initially the column is perfectly straight and the load is truly axial.

The cross section of column is uniform throughout its length.



The column material is perfectly elastic homogenous and isotropic and obey hooks

law.

The length of is very large as compared with cross sectional dimensions and the

failure occurs due to buckling load.

PROCEDURE:

1. Consider a column AB of length “l” hinged at both of its end A&B.

2. The column is rotated by hand in order to ensure the hinged support.

3. It is positioned to have cone hinged support.

4. The load is gradually applied by rotating the loading wheel connected to digital meter.

5. The load indicator is viewed simultaneously from the display of digital load indicator.

6. Now the column just starts buckling.

7. Till the column deflection touches the specified position of span the load is given to

column.

8. Now shape of deflection of column occurs as shown in fig meanwhile applied load

value approximately coincides with the theoretical value. Deflection will be noted

from digital meter.

TABULAR COLUMN

Sl. No Specimen Young’s

modulus

Length

(mm)

Diameter

(mm)

Crippling

load

1) Stainless

steel 2*10

5 N/mm

2 1000

With this crippling load of the column is spring shot. This load is known as buckling

load of column.

Sl. No Load(kg) Load(N) Deflection(mm)



Result:

Thus buckling load of column was found and compared its value with theoretical

values.


Case 1: The buckling load of the column when diameter of the column is 4mm

Theoretical

value

Experimental

value


Theoretical

value

Experimental

value



Experiment : 8

BUCKLING OF COLUMN WHEN ONE END HINGED AND

OTHER END FIXED

AIM:

To find buckling load of column using column test setup arrangement under one end hinged

and one end fixed condition.

APPARATUS REQUIRED:

Column test,

Load indicator,

Specimen rod and

One hinged support & one fixed support.

FORMULA USED:


𝒍𝐞𝟐

P = Crippling load




EULER’S COLUMN THEORY:

As per Euler‟s equation for buckling load of long column based on bending

stress the effect of direct stress is neglected .This may be adjusted justified with the statement

,direct stress included in a column is negligible as compared to the bending stress.

ASSUMPTION:


The cross section of column is uniformed through its length.




law.

The length of is very large as compared with cross sectional dimensions and the


PROCEDURE:

1. Consider a column AB of length “l” with one end fixed other end hinged.

2. The column cannot be rotated because it is one end fixed end other end hinged.

3. It is positioned to have a complete supports.

4. The load is gradually applied by rotating the loading wheel connected to load cell

intern to digital meter.



7. Till the deflection of column occurs as shown in figure mean while applied load value

approximately coincides with the theoretical value.

TABULAR COLUMN:

When one end is hinged and other end is fixed before loading.

Sl. No Specimen Young’s

modulus

Length

(mm)

Diameter

(mm)

Crippling

load

1) Stainless

steel 2*10

5 N/mm

2 1000

Loading column is stopped at crippling load. This load is known as buckling load of column.

Sl.no Load(kg) Load(N) Deflection(mm)

RESULT:

Thus buckling load of column was found and compared its value with theoretical values.





Theoretical

value

Experimental

value


Theoretical

value

Experimental

value



Experiment: 9

BUCKLING OF COLUMN WHEN BOTH ENDS-FIXED

Aim:

To find buckling load of column using column test setup arrangement under both ends are

fixed condition.

Apparatus required:

Column test,

Load indicator,

Specimen rod and

Two fixed supports.

Formula used:


𝒍𝐞𝟐

P = Crippling load




Euler’s column theory:

As per Euler‟s equation for buckling load of long column based on bending

stress the effect of direct stress is neglected .This may be adjusted justified with the statement

,direct stress included in a column is negligible as compared to the bending stress.

ASSUMPTION:


The cross section of column is uniformed through its length.




law.

The length of is very large as compared with its cross-sectional dimensions and the


PROCEDURE:

1. Consider a column AB of length “l” fixed at both of its end A&B.

2. The column is rotated by hand in order to ensure the fixed support.

3. It is positioned to have completed fixed support.

4. The load is gradually applied by rotating the loading wheel connected to digital meter.

5. The load indicator is connected with column test setup.



8. Till the column deflection of touches the speared position of span , the load is being

given to the column.

9. Now shape of deflection of columns occurred as shown in fig. meanwhile applied

load value approximately coincides with the theoretical value. Deflection will be

noted from the digital meter

TABULAR COLUMNS:

When both ends are fixed before loading

Sl. No specimen

Young‟s

modulus

N/mm2

Diameter

(mm)

Length

(mm)

Crippling

Load

Loading column is stopped at this crippling load. This load is known as bucking load of

column.

Sl.no Load(kg) Load(N) Deflection(mm)



RESULT:

Thus bucking load of column was found and compared its value with theoretical value.



Theoretical

value

Experimental

value


Theoretical

value

Experimental

value

Case3:The buckling load of the column when the diameter of the column is

5mm

Theoretical

value

Experimental

value



Experiment: 10

PREPARATION OF RIVETED JOINT

Case1:

AIM:

To make a riveted joint (double riveted zig-zag lap joint) with two given Aluminum metal

pieces.

WORK MATERIAL REQUIRED:

1. Aluminum plates (60 x 90 x 3) mm

2. Aluminum rivets.

TOOLS REQUIRED:

Sheet metal cutter

Steel rule

Scriber

Mallet

Files

Bench Vice

Centre punch

Dolly and Snap

Ball peen hammer

Drilling machine.

THEORY:

Riveted joints are permanent fastening and riveting is one of the commonly used

method manufacture of boilers, storage tank etc., involving joining, of steel sheets by means

of riveted joints. A rivet is a round rod of circular cross section. It consists of two parts. V it

head and shank. HS, WI and Al alloys are some of the metals commonly used for rivets. The

choice of particular metal will depend upon the place of application. Riveting is the process



of forming a riveted joints for thus a rivet is first placed in the hole drilled through the two

parts to be riveted. Then the shank end is made into a rivet head by applying pressure when

it is either cold or hot condition.

The pressure may be applied to form the second rivet head, either, by direct

hammering or through hydraulic or through hydraulic or pneumatic means the commonly

used riveted joints are of two types.

LAP JOINT:

In the lap joint the plates to be connected overlap each other and they are placed in a

two different parallel planes when the joint is made only with one row of rivets it is called

single.

Riveted lap joint is said to be double riveted, triple riveted etc., according to the

no of row of rivets in it built joint the edges of the plates are connected against each other

and the joint between them is covered by butt plates.

TYPES OF RIVET HEADS

The rivet heads classified as follows:

i) Indian Standers rivets for general and structural purposes below 12 mm diameter.

ii) Indian standard rivets for general and structural purposes 12 to 40 mm diameter.

iii) Indian standards boiler rivets 12 to 48 mm diameter. Below 12 mm diameter are

generally and structural purposes. Below 12 mm diameter are generally made of

mild steel, brass, copper or aluminum depending upon the purpose and place

where to be used.

Rivet Hole Diameter:

On Structural and pressure vessel rivet hole diameter is usually 1.5 mm larger than

nominal diameter of the rivet.



Case2:

AIM:

To make a riveted joint (double riveted zig-zag but joint) with two given Aluminum metal

pieces.

Case3:

AIM:

To make a riveted joint (double riveted zig-zag but joint) with two given Aluminum metal

pieces.

VIVA QUESTIONS

1. Define riveting?

Riveting is a process of forming a riveted joints for thus a rivet is first placed in a hole

drilled through the two parts to be rivets.

2. Types of rivets

3. Applications of riveted joints

Ship building

Construction of steel buildings

Bridges

Boilers and tanks

4. Types of rivet joints

5. Riveted joint is permanent type of joint.



Experiment: 11

FAILURE STRENGTH OF RIVETED JOINT

AIM:

To determine the strength of a riveted joint in the Universal Testing Machine.

EQUIPMENT:

Universal testing machine,

Riveted joint specimen.

DIMENSIONS:

Aluminium pieces of size. (60 x 90 x 3) mm

Aluminium rivets diameter 3mm

THEORY:

A riveted joint fail any of following manner.

1. By bearing of the plate b/w the rivet hole & the edge of the plate.

2. By tearing of the plates b/w rivets. The safe tensile loads that the plate can with stand

for one pitch length is called the tearing strength.

Tearing strength per pitch length = P x t

Pt = Ft x net are of flatted.

Pt = Ft (P – d) t

Failure due to shearing of rivet for a lap joint if load /pitch is large it is possible that the rivet

may shear off.

Ps = Fs x

In general in a lap joint if rivets are covered load per pitch length would be



Ps = N x F x x

Failure by tearing or crushing of rivet or plate the shape load on rivet.

Pb = Fb x dt where

Pb = allowance bearing stress

Fb = bearing value of rivet

EFFICIENCY OF A JOINT.

Let Pt, Ps, P b be the safe load per pitch from bearing sheering & bearing considerations.

PROCEDURE:

Take the given specimen & load it in the UTM.

Apply the load on the joint by switch on the UTM

Observe the loading dial reading and joint.

Observe the failure mode of the joint (Shearing, Tearing or Crushing).

Take the corresponding dial reading that is the load bearing capacity of the present

joint.

PRECAUTIONS:

Load the specimen exactly b/w the gauge points

Stop the UTM at the exact time of failure observed.

Take the reading from the dial without any parallel axis error.

RESULT: The failure mode of the riveted joint is observed and the load bearing capacity is

predicted.



VIVA QUESTIONS

1. Define Shearing of rivets

2. Define Tearing of rivets

3. Define crushing of rivets

4. What is pitch diameter

5. Define different strengths of riveted joint

6. Define efficiency of riveted joint?

It is the ratio of the strength of the joint to the bearing strength of the unpunched plate.



Experiment: 12

FREE LONGITUDINAL VIBRATIONS

AIM:

To study the Free Longitudinal vibrations of a spring mass system.

APPARATUS:

Universal vibration beam setup,

Open coil helical spring,

Mass hanger,

Weights and

Measuring tape.

THEORY:

When particles of the shaft (or) disc move parallel to the axis of the shaft then the vibrations

are known as longitudinal vibrations.

Consider a spring of negligible mass is fixed at one end and carries a mass on the other end, then

the theoretical time period.

𝑓𝑛 𝑡𝑕𝑒𝑜 =1

2𝜋 𝑔𝑘𝐿2

𝑊𝐿12 𝐻𝑧

Where

g =specific gravity= 9.81 m/sec2

k = Stiffness of the spring = F

δB

δB = Static deflection of spring half of the amplitude in m.

L = Distance between fixed end to stiffness of the spring= _________ m

L1 = Distance between fixed to weight pan = ________ m



W = Weight in the hook pan= ________ kg

PROCEDURE:

1. In the cantilever beam spring is fixed by means of hook so that the mass hanger of

known mass is placed as shown fig.

2. Place the sensor at the end of the beam.

3. Move the top of the spring rod up to some level. Leave it suddenly for the vibration in

the beam.

4. Repeat the experiment for different length (L) and for different weights with hook

pan.

5. Note down the maximum value of frequency and amplitude directly in the digital

indicator.

6. Using the initial and final length the deflection δ is calculated.

7. The procedure is repeated for different masses attached to the spring.



TABULAR COLUMN:

Sl no

mass attached

(w kg)

Deflection (δ)

of spring

Stiffness

(K) x 103

w/δ

fnexp

1/Texp

Fntheo

1/Ttheo

RESULT:

It is found that the experimental and theoretical values differ. This is due to the fact that we

assume the spring to be weightless which causes amount of error.

Free vibration

Sl.no

Length l1 in

mts

Length l in

mts

Amplitude

δ B to mm

Frequency (exp)

1 0.28 0.405 2.64 14

Data:

Weight in hook pan including hook pan W = 0.3 kg

L = Distance between fixed end to stiffness of the spring = 0.405m

L1 = Distance between fixed to weight pan = 0.28m



F =WL1

L=

0.3 × 0.280

0.405= 0.21 Kg

Deflection of the weight ‘W’ at C δC =L1

LδB where,

L1

δC=

L

δB

δC =0.280

0.405× 0.00135 = 9.33 × 10−4m

δB = half of the amplitude

But stiffness of the spring K =F

δB=

0.21

0.00135= 155.55 Kg/m

But ωn =K

m=

9

δC=

9.81

9.33 × 10−4= 102.54 1/sec


2𝜋 𝑔𝑘𝐿2

𝑊𝐿12 𝐻𝑧

Where

g = 9.81 m/sec2

k = stiffness of spring=[ F/δB ] = 155.55kg/m

L = 0.405m

L1 = 0.208m

W = Weight including weighing pan = 0.3 kg

fn theo =1

2π

9.81 × 155.55 × 0.4052

0.3 × 0.2082 Hz

fn theo = 22 Hz

fn exp = 14 Hz



Case1:




Case2:




Case3:




VIVA QUESTIONS

1. Define vibration?

Vibration is a mechanical phenomenon whereby oscillations occur about an equilibrium

point. The oscillations may be periodic such as the motion of a pendulum or random such as

the movement of a tire on a gravel road.

2. Classify different types of Vibrations?

3. Define frequency

4. Define amplitude

5. Define eccentric load

6. The theoretical time period is given as


2𝜋 𝑔𝑘𝐿2

𝑊𝐿12 𝐻𝑧

http://en.wikipedia.org/wiki/Oscillation

http://en.wikipedia.org/wiki/Equilibrium_point



http://en.wikipedia.org/wiki/Periodic_function

http://en.wikipedia.org/wiki/Random



Experiment: 13

STUDY OF FORCED VIBRATIONS

AIM:

To study the un-damped forced vibrations of equivalent spring mass system

DESCRIPTION ON SET UP:

Experiment 1:

The arrangement is to study forced, undamped vibrations. It consists of MS rectangular beam

supported at one end by trunion pivot to the side member of the frame. The other end of the

beam Is supported by the lower end of helical spring. Upper end of the spring is attached to

the screw rod.

Experiment 2:

DC variable speed motor connected through flexible shaft to the exciter unit can be mounted

at any position along a beam.

Additional known weights may be replaced in the disc.

PROCEDURE:

1. Support one end of the cantilever beam in the slot of trunion/fixed plate and clamped

by means of screw

2. Attach the other end of the beam to the lower end of spring

3. Adjust the screw to which the spring is attached such that beam is horizontal in the

above position

4. Weigh the excited assembly along with the discs bearing and flexible shaft

5. Clamp the assembly at any convenient position

6. Allow system to vibrate freely

7. Neglect the initial readings of the sensor (amplitude and frequency) because due to

some sensitive vibration

8. Place the sensor at the end of the beam



9. Note down the frequency and amplitude with varying speed in the digital indicator

10. Repeat the experiment by varying speed and by also fixing different efficient weights

on the disc

11. For any doubt refer to the figure as shown

Fig. Schematic view for forced vibrations

NOTE:

It is necessary to screw properly the weight on the disc.

TABULAR COLUMN:

Forced vibrations:

RPM Frequency in

Hz Amplitude in mm



Table of readings

Forced vibrations:

RPM Frequency in

Hz Amplitude in mm

600 8 1.03

720 11 2.75

843 14 1.22

960 16 1.15

1040 17 1.18

1300 22 1.56

1500 27 2.69

Graph of frequency versus amplitude

Disturbing force F0 = M0eω2

F0 = 0.024 × 0.04 × 75.392

F0 = 5.46 kg − m/s2

Where,ω =2πN

60=

2 × π × 720

60= 75.39/s

M0 = Mass of the eccentricity weight = 24 gms = 0.024 kg



e = Distance between middle of the shaft to middle of the eccentricity.

K = stiffness of the spring = M × g

δ

M = Mass of the disc exciter = 3.7 Kg

g = 9.81 m s2

δ = Deflection of the i. e. , amplitude in mm = 1.03 mm

𝐾 =3.7 × 9.81

0.00103= 35239.80 𝑘𝑔/𝑠2

Natural Frequency = ωn = K

M=

35239.81

3.7= 97.59/𝑠

𝑇𝑕𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 = 𝑓𝑡𝑕𝑒 =1

2𝜋 𝐾

𝑀=

1

2𝜋

35239.80

3.7= 15 𝐻𝑧

𝐸𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡𝑎𝑙𝑙𝑦 𝑡𝑕𝑒 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑝𝑒𝑎𝑘 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 = 11 𝐻𝑧

Case1:


F0 = 0.024 × 0.04 × 75.392

F0 = 5.46 kg − m/s2

Case2:


Mass of the disc =3.5kg

Case3:


Mass of the disc =3.8kg



NON DESTRUCTIVE TESTING (NDT)

INTRODUCTION:

Non destructive testing methods are a reliable means for detecting the presence of

defects at the manufacturing stage and also the different types of defects formed during the

service life of an assembly or a component. This type of testing methods are comfortable and

less expensive when compared to tensile tests compressive tests, shear and impact tests as

such the latter are of destructive nature.

NDTS‟s provide information on the quality of a material or component and do not

alter (or) damage the components or assemblies which are to be tested.

RELIABILITY OF DEFECT DETECTION:

The reliability of any NDT technique is a measure of the efficiency of the technique in

detecting flaws of specific type, shape & size defects of many types and sizes may be

introduced to a material or a component during manufacture and the exact nature and size

of any defects will influence the subsequent performance of the component other defects,

such as fatigue cracks, or corrosion cracks, may be generated with in a material during

service.

NDT systems co –exists and depending on the application may be used singly or in

conjuction, may be used singly or in conjunction with another. All the best methods are

complementary to each other. The purpose and usage of a technique depends upon the

type of flow present and the shape and size of the component to be examined different

NDT‟s are listed below.

1. Liquid penetrant

2. Magnetic Flaw detection.

3. Electrical methods.

4. Ultrasonic testing.

5. Radiography.

Similary Non – Destructive testing is applied it can lead to serious errors of judgment of

component quality.

BENEFIT OF NDT:



The benefit of NDT is identification of defects which if remained. Undetected could

result in a catastrophic failure which would be very costly in money and life.



Experiment: 14

NDT USING MAGNETIC PARTICLE TEST EQUIPMENT

AIM:

To analyze the intensity of cracks on surface and sub layer flaws in a given specimen.

APPARATUS:

E.M.Yoke

Steel specimen

Iron oxide power

Swab etc.

THEORY:

Magnetic particle inspection is used for Ferro magnetic components, when a Ferro

magnet is magnetized, magnetic discontinuities that lie in a direction approximately

perpendicular to the field will result info of a strong Leakage field. This leakage filed is

present at above the surface of magnetized component and its presence can be visibly

detected by utilization of finely divided magnetic particles. The application of dry or wet

particles in a liwuid carrier, over surface of component results in a collection of magnetic

particles at a discontinuity. The magnetic bridge, so formed indicates the location, size and

shape of discontinuity.

Magnetization may be introduced in a component by using permement magnets (EMS) or by

passing high currents through or around component.

PROCEDURE:

Clean the surface of a given specimen (Steel)

Place the EM Yoke on specimen vertically with its foldable legs on it.

Switch on the power supply.

Poor the iron oxide powder on the specimen.

After the observation demagnetize the specimen.



We observe the lines of accumulation of powder on the surface of crack position

depending upon its intensity of crack

PRECAUTIONS:

The power should be distributed uniformly on the specimen.

The iron oxide powder should not be distributed during the experiment.

RESULTS:

The surfaces flaws are detected where the accumulation of iron oxide are observed.

Case1:

Steel specimen

Case2:

Mild steel specimen

Case3:

Aluminum Alloy

VIVA QUESTIONS

1. Define Non destructive testing

2. What is reliability of defect detection?

The reliability of any NDT technique is a measure of efficiency of the technique in

detecting flaws of specific type, shape and size.

3. Types of NDT techniques

Liquid penetrant

Magnetic flaw detection

Electrical methods

Ultrasonic testing

Radiography

4. Benefits of NDT

5. What is catastrophic failure

6. What is magnetic particle

detection



Experiment: 15

NDT USING ULTRASONIC TEST EQUIPMENT

AIM:

To analyze the intensity of cracks on surface and sub layer flaws in a given specimen

APPARATUS:

Ultrasonic testing machine

Transducer

Specimen

THEORY:

Ultrasonic Testing (UT) uses a high frequency sound energy to conduct examinations

and make measurements. Ultrasonic inspection can be used for flaw detection evaluation,

dimensional measurements, material characterization, and more. A typical UT inspection

system consists of several functional units, such as the pulser/receiver, transducer, and

display devices. A pulser/receiver is an electronic device that can produce high voltage

electrical pulse. Driven by the pulser, the transducer of various types and shapes generates

high frequency ultrasonic energy operating based on the piezoelectricity technology with

using quartz, lithium sulfate, or various ceramics. Most inspections are carried out in the

frequency rang of 1 to 25MHz. Couplants are used to transmit the ultrasonic waves from the

transducer to the test piece; typical couplants are water, oil, glycerin and grease.

The sound energy is introduced and propagates through the materials in the form of waves

and reflected from the opposing surface. An internal defect such as crack or void interrupts

the waves' propagation and reflects back a portion of the ultrasonic wave. The amplitude of

the energy and the time required for return indicate the presence and location of any flaws in

the work-piece. The ultrasonic inspection method has high penetrating power and sensitivity.

It can be used from various directions to inspect flaws in large parts, such as rail road wheels

pressure vessels and die blocks. This method requires experienced personnel to properly

conduct the inspection and to correctly interpret the results. As a very useful and versatile

NDT method, ultrasonic inspection method has the following advantages; sensitivity to both

surface and subsurface discontinuities, superior depth of penetration for flaw detection or

measurement, ability to single sided access for pulse-echo technique, high accuracy in

determining reflector position and estimating size and shape, minimal part preparation,

instantaneous results with electronic equipment, detailed imaging with automated systems,

possibility for other uses such as thickness measurements.



Its limitations; necessity for an accessible surface to transmit ultrasound, extensive skill and

training, requirement for a coupling medium to promote transfer of sound energy into test

specimen, limits for roughness, shape irregularity, smallness, thickness or not homogeneity,

difficulty to inspect of coarse grained materials due to low sound transmission and high

signal noise, necessity for the linear defects to be oriented parallel to the sound beam,

necessity for reference standards for both equipment calibration, and characterization of

flaws.

PROCEDURE:

1. The couplant should be applied on the inspected area.

2. For the circular test specimen, the prop will be placed in the corresponding space in the

supporting fitting tool. Enough couplant should be used between the probe and tool.

3. For the flat specimen, no tool is needed, couplant only applied between the inspected

surface and the probe.

4. Special attention should be paid on the location where possible cracks exist.

5. A discontinuity like a crack produces a peak on the screen.

6. Attention should also be given to the movement of the possible peak caused by the cracks

on the specimen.

Case1: Stainless steel specimen

Case2: Aluminum specimen

Case3: Mild steel



Experiment: 15

Aim: To find the surface cracks on the given specimen using dry

penetration technique

APPARATUS:

Cleaner

Penetrant

developer







Experiment: 16

AIM: To determine the longitudinal and circumferential stress and of thin

walled pressure vessel

DESCRIPTION:

Test bench comprises of a brass pressure vessel of 0.5mm thick wall.

A, B strain gauges are mounted in 90 deg apart and are connected to the digital meter

in control panel.

The display of strain is seen in the digital indicator meter.

A hand pump is bused for input pressure.

The mode of fluid is air.

Discharge valve is located to discharge the pressure which is displayed in the dial

gauge.

PROCUDURE:

Keep the system on a table.

Discharge valve in closed position.

Pump the air to cylinder in steps of 0.2 Kg/cm2- upto 1 Kg/cm2.

In each step of A, B and note down the readings.

HOW TO SET THE METER?

Press A : Bring the display to zero(0). By rotating the pot on either direction.

Press the button and release.

Zero display should be displayed in the meter.

Start pumping the air and note down the reading in tabular column.

휀𝑥 = 𝐴𝑥𝑖𝑎𝑙 𝑆𝑡𝑟𝑎𝑖𝑛

휀𝑌 = 𝐶𝑖𝑟𝑐𝑢𝑚𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑆𝑡𝑟𝑎𝑖𝑛

Included angle between this to strain gauge is 90 deg.



FORMULA:

1. Strain acting on axial direction εX =1

𝐸 𝜎𝑋 − 𝜐𝜎𝑌

2. Strain acting on axial direction εY =1

𝐸 𝜎𝑌 − 𝜐𝜎𝑋

3. Force acting on axial direction Faxial = 𝜎𝑋 × 𝜋 × 𝑑 × 𝑡

4. Force acting on circumferential direction FCircumferential = 𝜎𝑌 × 2𝑡𝐿

5. Poisson′s ratio =εY

εX

ζx- Stress acting on axial direction.

ζy- Stress acting on circumferential direction.

E - Young‟s modulus.

From the above equations first you find the strains in two(2) directions to various

pressures.

Then you find the load acting on two three directions.

Sl.

No.

Pressure

ρ

kg/cm2

εx εy ζx N/mm2 ζy

N/mm2

Faxial N Fcir

N

1 0.2 15 33

2 0.4 29 80



3 0.6 45 118

4 0.8 62 158 6232.53*10^3 1.83*10^3 1467.76 0.37

5 1.0 78 189

Calculation done for SL. No. 4 can also be repeated to calculate other values.

Case1:

Pressure range: 0.2-0.6kg/cm2

Case2:


Case3:


VIVA QUESTIONS

1. Define Pressure

2. Define Poisson‟s ratio

3. Define axial strain

4. Define circumferential strain.

5. Force acting on axial direction Faxial = 𝜎𝑋 × 𝜋 × 𝑑 × 𝑡

6. Force acting on circumferential direction FCircumferential = 𝜎𝑌 × 2𝑡𝐿

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