As Lab Manual New

AEROSPACE STRUCTURES – III – B.Tech I Sem

LIST OF EXPERIMENTS

1. Determination of Young’s Modulus of mild steel by deflection method

2. Verification of Castigliano’s theorem

3. Verifications of Maxwell’s reciprocal theorem

4. Verification of Principle of Superposition

5. Determination of Shear Center

a) Open section

i) Channel section

ii) Angle section

b) Closed Section

i) D-section

6. Buckling of column when both ends are hinged

7. Buckling of column when one end hinged and other end fixed

8. Buckling of column when both ends-fixed

9. Preparation of riveted joint

10. Failure strength of riveted joint

11. Vibrations

a) Free longitudinal vibration

b) Forced vibrations

c) Torsional vibration of single rotor shaft system

12. Non Destructive Testing (NDT)-magnetic particle detection

13. Thin walled pressure vessel test rig

MARRI LAXMAN REDDY INSTITUTONS, DUNDIGAL, HYD – 43 Page1


DETERMINATION OF YOUNG’S MODULUS OF MILD STEEL BY DEFLECTION METHOD

AIM:

To determine the Young’s modulus of a given steel specimen.

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. NoWeight

(Kg)

Deflectio

n

(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 ___________



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=σ/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.



VERIFICATION OF CASTIGLIANO’S THEOREM

AIM:

To verify Castingliano’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 mm4

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:

δ c=W a2 b2

3 EIL=¿

E = Young’s Modulus of Mild steel from tables = 2.1*105 N/mm2

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?

δ c=W a2 b2

3 EIL

4. Classification of beams?

Simply supported, cantilever, hinged, fixed

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



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.



VIVA QUESTIONS

1. State Maxwell’s Reciprocal Theorem?

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. What are the applications of Maxwell’s Reciprocal Theorem?

It has applications in structural engineering where it is used to define influence lines and derive

the boundary element method. It is used in the design of compliant mechanisms by topology

optimization approach.

3. Based on slope, define Maxwell’s reciprocal theorem?

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

Maxwell’s reciprocal theorem is also known as Betti's theorem.


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

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

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


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 δ c1 is noted.

The above procedure is followed at ‘E’ position and readings should be taken at point ‘C’

and corresponding deflection δ c2 is noted.

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

note down the deflection.


δ c=δ c1+δc 2

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) δ c in div δ cin mm



Case 1

Load(w1) δ c1δ c1

in mm

Case 2

Load(w2) δ c2 in Div. δ c2

in mm

Load W = Load W1 + Load W2

Deflection

δ c=δ c1+δc 2

Result:

The principle of superposition is verified.

VIVA QUESTIONS



1. State Principle of superposition?


δ c=δ c1+δc 2

2. Degrees of freedom for the hinged beam is

3. The principle of superposition is to find out resultant deformation.

4. For both ends hinged condition formula for deflection is

5. For principle of superposition Load W = Load W1 + Load W2



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 (W a−W b )

2W v

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 =





SHEAR CENTER OF OPEN SECTION

(ANGLE)

AIM:

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

value.

THEORY:



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





APPARATUS REQUIRED:

Channel section




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 (W a−W b )

2W v

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::



Width of the flange (b) : 50mm





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













RESULT:

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


d) Theoretical method =

e) Experimental method =

f) Error percentage =





VIVA QUESTIONS

1. Classify different types of sections?

2. Define shear center?

The point where a shear force can act without producing any twist in the section.

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?

5. Define Shear strain?

6. The distance ‘e’ of the line of action from the web is given as

e=AB (W a−W b )

2W v

7. 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.

8. 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.


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

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

http://en.wikipedia.org/wiki/Shearing_(physics)

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

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

http://en.wikipedia.org/wiki/Yield_(engineering)

http://en.wiktionary.org/wiki/twist

http://en.wiktionary.org/wiki/force

http://en.wiktionary.org/wiki/point


SHEAR CENTER OF CLOSED SECTION

AIM:

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

value.

THEORY:



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

when they should bend without twisting. The shear center is important in the case of closed

section like an aircraft wing, where the lift produces a torque about the shear center. Similarly

the wing strut of a semi cantilever wing is a closed tube of aerofoil section. A thin walled D-





APPARATUS REQUIRED:

Closed Section (D - Section)




PROCEDURE:

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

(see fig). Initial readings are note down consider that is zero when dial gauge tip touches

the flange.



2. Place a total of say 1.2 kilograms load at A (loading hook and five load pieces will make

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

down dial gauge reading.

3. Every time before taking readings tap the flange so as to clear the error in the dial gauge.

4. 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 1.2 kilogram. Record the dial gauge

readings.

5. Transfer carefully all the load pieces to B from A 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 (W a−W b )

2W v

6. For every load case calculate the algebraic difference between the dial gauge suffered by

the section.

Though a nominal value of two kilograms for the total load is suggested it can be less. In that

event the number of readings taken will reduce proportionately.

TABLE

Dimensions of the beam and the section :







Vertical load Wv = (Wa+Wb), 0.2 kgs – 5Nos.

S.No Wa Wb d1 d2 d1-d2 e=AB (W a−W b )

2W v

1.

2.

3.

4.

5.

6.

7.

Plot e versus (d 1-d 2) curve and determine where this meets the e axis and locate the

shear center.

PRECAUTIONS:













RESULT:

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


g) Theoretical method =

h) Experimental method =

i) Error percentage =





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= π2 EIle

2

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 SpecimenYoung’s

modulus

Length

(mm)

Diameter

(mm)

Crippling

load

1) Stainless steel

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

of column.

Result:

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



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:

P= π2 EIle

2

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.




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” 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 SpecimenYoung’s modulus

Length(mm)

Diameter(mm)

Crippling load

1) Stainless steel

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

RESULT:

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



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:

P= π2 EIle

2

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.




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

failure occurs due to buckling load.

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 specimenYoung’s modulusN/mm2

Diameter(mm)

Length(mm)

Crippling Load

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

RESULT:

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



For fixed-fixed condition

Load in Kg

Load in N

Deflection in mm δ

Deflection/loadδ/P

D(δ/P) d(δ) D(δ/P)/d(δ) Inverse

For hinged – fixed condition

Load in Kg

Load in N

Deflection in mm δ

Deflection/loadδ/P


FOR HINGED-HINGED CONDITION

Load in Kg

Load in N

Deflection in mm δ

Deflection/loadδ/P




VIVA QUESTIONS

1. Define Buckling?

2. Define buckling load?

3. Define Euler’s theory?

4. The expression for the crippling load P=

5. Buckling load is also known as

6. Effective length ‘Le’ for various types of end conditions is

7. Define Hooke’s law?

Hooke’s law states that the strain in a solid is proportional to the applied stress within the elastic

limit of that solid.



PREPARATION OF RIVETED JOINT

AIM:

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

pieces.

WORK MATERIAL REQUIRED:

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

2. Aluminium 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 aluminium 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.



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.



FAILURE OF RIVETED JOINTS

A riveted joint may fail in the following ways.

1. Shearing of rivets:

The rivets may fail in shear at the plane where two plates meet together. A rivet may fail

in single shear or double shear.

D = Rivet diameter.

Fg = Shear Stress of rivets.

A = Area shear

Where K = 1 for single shear and for double shear.

K = 2 (Theoretically)

= 1.75 to 1.875 (Practically)

Ps = Shear Strength = A x Fg

= n x k x d 2 x Fg.

N = Number of rivets in one pitch length

2. Tearing of rivets in one pitch length of a row of rivets.

The plate may bear along the line of min section that is along the line through the centre

of holes.

Ft = Tensile Stress of Plates.

A = Are of plate under bearing.

Pt = Tearing Strength = (p-d) t. Fx.

Where t = Thickness of plate.

P = Pitch.

3. Crushing of rivets & Plates:



The rivets may fail under crushing.

Fc = Crushing shares of rivets.

A = Area under crushing = n.dt.

Pc= Crushing strength.

A x Fc = n.d. Fc. T.

Strength of a riveted joint:

The least value out of shearing, bearing strength and crushing strength is called strength

of the riveted joint.

Rivets should rivets always, be placed at right angles to the acting forces and the

maximum stress induced in them should be either shear or crushing. For this reason, the length

of the rivet between the head should not exceed four or five 3 times its diameter.

Efficiency of a riveted joint:

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

P = Tearing strength of unpunched plate

= P x t x t

s= Shearing efficiency.

c = Crushing efficiency

t = Tearing efficiency

The lowest of the three efficiencies is called as the efficiency of the joint.

PROCEDURE:



Take the given Aluminium sheet and cut it to the required dimension. With the help of

the sheet metal cutter.

Place the cottoned aluminum sheet in the bench vice and files for the right angles.

Mark the dimensions of the rivet hole with the help of the scriber, dot punch & centre

punch.

Drill the holes on the two Aluminium plates with the help of the drilling machine.

Now take the rivets and punch them with the ball peen hammer in the required holes.

Before driving the rivets clamp them properly and make the riveting and be sure that the

plates are properly joined.

ANALYSIS:

The riveted joint are usually performed in the areas where other means of joints or

impossible it makes a strong joint with less material and less applied force.

APPLICATIONS:

The riveted joint usually finds the application in the

Slip building

Construction of steel buildings

Bridges

Boilers & tanks.

PRECAUTIONS:

Cut the Al sheet to the required size.

Clamp the plates properly.

Properly drill the holes in punched area.

RESULT: Thus the preparation of double riveted zig-zag lap joint is achieved.



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.



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.

f n (theo )=1

2π √ gk L2

W L12 Hz

Where

g =specific gravity= 9.81 m/sec2

k=Stiffness of the spring=[ FδB ]

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 nomass 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.noLength l1 in

mts

Length l in

mts

Amplitude

δ B to mmFrequency (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=W L1

L=0.3×0.280

0.405=0.21Kg

Deflection of the weight ‘W ’ at C (δ C )=L1

LδBwhere ,( L1

δC

= LδB

)



δC=0.2800.405

×0.00135=9.33×10−4 m

δB = half of the amplitude

But stiffness of the spring ( K )= FδB

= 0.210.00135

=155.55 Kg /m

But ωn=Km

=√ 9δC

=√ 9.819.33×10−4 =102.54 1/ sec

f n (theo )=1

2π √ gk L2

W L12 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

f n (theo )=1

2π √ 9.81×155.55×0.4052

0.3×0.2082Hz

f n (theo )=22 Hz

f n(exp)=14 Hz



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

f n (theo )=1

2π √ gk L2

W L12 Hz


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

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

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

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


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:

RPMFrequency in

HzAmplitude in mm



Table of readings

Forced vibrations:

RPMFrequency in

HzAmplitude 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=M 0 eω2

F0=0.024×0.04×75.392

F0=5.46 Kg−m/ sec2

Where ,ω=2πN60

=2×π×72060

=75.39/ sec

M 0=Mass of the eccentricityweight=24 gms=0.024 Kg

e=Distancebetweenmiddle of the shaft ¿middle of the eccentricity .

K=stiffness of the spring= M ×gδ

M=Mass of the disc (exciter )=3.7 Kg

g=9.81m/ sec2



δ=Deflection of thei . e . , amplitude∈mm=1.03mm

K=3.7×9.810.00103

=35239.80kg/ sec2

Natural Frequency=ωn=√ KM

=√ 35239.813.7

=97.59 /sec

Theoretical frequency=f the=1

2 π √ KM

= 12π √ 35239.80

3.7=15 Hz

Experimentally themaximum peak value of frequency=11Hz



VIVA QUESTIONS

1. What are forced vibrations

2. Define damping

3. What are free vibrations

4. Define longitudinal vibrations

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

vibrations are known as longitudinal vibrations.

5. Define stiffness of the spring



TORSIONAL VIBRATIONS OF SINGLE ROTOR SHAFT

SYSTEM

AIM:

To study the Torsional vibrations (un-damped) of single rotor shaft system

DESCRIPTION OF SETUP:

The general arrangement for carrying out the experiments is as follows:

One end of the shaft is gripped in the plate and heavy flywheel free to rotate in bush

bearing is fixed at the other end of the shaft.

The flywheel can be clamped at convenient position along the shaft. Thus, length of the

shaft can be varied during the experiments. Specially used plates are used for clamping top end

of the shaft. The bush bearing support to the flywheel shaft provides negligible lateral movement

during the experiments. The bush housing is fixed to side member of the mainframe.

Procedure:

Fix the flywheel at convenient position along the shaft

Grip one end of the shaft at the bracket by the flat plate

Place the frequency sensor on the side of the disc, so that the frequency and amplitude

can be seen in the digital indicator.

Twist the rotor-on the shaft up to 500 and allow oscillating.

Directly note down the frequency in digital meter

Make the following observations

a) Shaft diameter = 4 mm

b) Diameter of the disc = 200 mm

c) Weight of the disc (W) = 5.4 Kg

d) Modulus of rigidity of the shaft = 80 G Pa

Increase length, frequency will decrease and Decrease length, frequency will increase.



OBSERVATION TABLE

S. No ObservationsLength of the shaft

‘L’ cmFexp Fthe

SPECIMEN CALCULATIONS:

1. Determinetorsional stiffnessK t=GI p

L

L = Length of the shaft in meters

J=I p=Polarmoment of inertia of the shaft= π d4

32m4

d = Diameter of the shaft = 0.004 m

G = Modulus of rigidity of the shaft = 80 GPa

2. determine T-theoretical = The

¿2π √ IK t

where, I=moment of inertia of the disc=Wg

×D2

8

Where W = weight of the disc = 5.4 Kg

g = 9.81 m/sec2 = specific gravity

D = diameter of the disc = 0.2 meters

CALCULATIONS:

Natural frequency of the rotor = fn(the)

f n (the )=1

T the

= 12π √GJ

LI;T the=2π √ LI

GJ



Where,

D = diameter of the disc = 200mm

g = 9.81m/sec2

W = 5.4 Kg

I=5.4×0.22

9.81×8=2.75×10−3 Kg−m−sec2

T the=2π √ IK t

=2π √ 2.75×10−3 ×9.814.188

=0.50 sec

CONCLUSION:

Frequency (the )=1

T the

= 10.50

Frequency (the )=1.98/sec ¿Hz

Frequency (exp )=2

sec ¿HZ directly¿ observationtable .¿





VIVA QUESTIONS

1. Define torsion

2. Write the expression for the torsion equation

3. Define torsional vibrations

4. Write the expression for the torsional stiffness

5. Define polar moment of inertia



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.



MAGNETIC PARTICLE DETECTION

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.



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



THIN WALLED PRESSURE VESSEL TEST RIG

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.

ε x=Axial Strain

ε Y=Circumference Strain

Included angle between this to strain gauge is 90 deg.



FORMULA:

1.Strainacting onaxialdirection εX=1

E (σ X−υσY )

2.Strainacting onaxialdirection εY=1

E (σY−υ σ X )

3. Force actingonaxialdirection Faxial=σ X×π×d× t

4. Force actingoncircumferential direction FCircumferential=σY ×2 tL

5.Poisso n' s ratio=εY

εX

σx- Stress acting on axial direction.

σy- 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 σx N/mm2 σy

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.

VIVA QUESTIONS

1. Define Pressure

2. Define Poisson’s ratio

3. Define axial strain

4. Define circumferential strain.

5. Force actingonaxial directionFaxial=σ X ×π×d ×t

6. Force actingoncircumferential direction FCircumferential=σY ×2 tL


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