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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
AEROSPACE STRUCTURES – III – B.Tech I Sem
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
MARRI LAXMAN REDDY INSTITUTONS, DUNDIGAL, HYD – 43 Page2
AEROSPACE STRUCTURES – III – B.Tech I Sem
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 ___________
MARRI LAXMAN REDDY INSTITUTONS, DUNDIGAL, HYD – 43 Page3
AEROSPACE STRUCTURES – III – B.Tech I Sem
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.
MARRI LAXMAN REDDY INSTITUTONS, DUNDIGAL, HYD – 43 Page4
AEROSPACE STRUCTURES – III – B.Tech I Sem
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.
MARRI LAXMAN REDDY INSTITUTONS, DUNDIGAL, HYD – 43 Page5
AEROSPACE STRUCTURES – III – B.Tech I Sem
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:
MARRI LAXMAN REDDY INSTITUTONS, DUNDIGAL, HYD – 43 Page6
AEROSPACE STRUCTURES – III – B.Tech I Sem
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.
MARRI LAXMAN REDDY INSTITUTONS, DUNDIGAL, HYD – 43 Page7
AEROSPACE STRUCTURES – III – B.Tech I Sem
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?
MARRI LAXMAN REDDY INSTITUTONS, DUNDIGAL, HYD – 43 Page8
AEROSPACE STRUCTURES – III – B.Tech I Sem
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.
MARRI LAXMAN REDDY INSTITUTONS, DUNDIGAL, HYD – 43 Page9
AEROSPACE STRUCTURES – III – B.Tech I Sem
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.
MARRI LAXMAN REDDY INSTITUTONS, DUNDIGAL, HYD – 43 Page10
AEROSPACE STRUCTURES – III – B.Tech I Sem
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.
MARRI LAXMAN REDDY INSTITUTONS, DUNDIGAL, HYD – 43 Page11
AEROSPACE STRUCTURES – III – B.Tech I Sem
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.
MARRI LAXMAN REDDY INSTITUTONS, DUNDIGAL, HYD – 43 Page12
AEROSPACE STRUCTURES – III – B.Tech I Sem
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.
Deflection by combined loading is equal to sum of the deflection by individual loading.
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
MARRI LAXMAN REDDY INSTITUTONS, DUNDIGAL, HYD – 43 Page13
AEROSPACE STRUCTURES – III – B.Tech I Sem
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
MARRI LAXMAN REDDY INSTITUTONS, DUNDIGAL, HYD – 43 Page14
AEROSPACE STRUCTURES – III – B.Tech I Sem
1. State Principle of superposition?
Deflection by combined loading is equal to sum of the deflection by individual loading.
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
MARRI LAXMAN REDDY INSTITUTONS, DUNDIGAL, HYD – 43 Page15
AEROSPACE STRUCTURES – III – B.Tech I Sem
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.
MARRI LAXMAN REDDY INSTITUTONS, DUNDIGAL, HYD – 43 Page16
AEROSPACE STRUCTURES – III – B.Tech I Sem
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
MARRI LAXMAN REDDY INSTITUTONS, DUNDIGAL, HYD – 43 Page17
AEROSPACE STRUCTURES – III – B.Tech I Sem
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.
MARRI LAXMAN REDDY INSTITUTONS, DUNDIGAL, HYD – 43 Page18
AEROSPACE STRUCTURES – III – B.Tech I Sem
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 =
MARRI LAXMAN REDDY INSTITUTONS, DUNDIGAL, HYD – 43 Page19
AEROSPACE STRUCTURES – III – B.Tech I Sem
MARRI LAXMAN REDDY INSTITUTONS, DUNDIGAL, HYD – 43 Page20
AEROSPACE STRUCTURES – III – B.Tech I Sem
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).
MARRI LAXMAN REDDY INSTITUTONS, DUNDIGAL, HYD – 43 Page21
AEROSPACE STRUCTURES – III – B.Tech I Sem
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::
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
MARRI LAXMAN REDDY INSTITUTONS, DUNDIGAL, HYD – 43 Page22
AEROSPACE STRUCTURES – III – B.Tech I Sem
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.
MARRI LAXMAN REDDY INSTITUTONS, DUNDIGAL, HYD – 43 Page23
AEROSPACE STRUCTURES – III – B.Tech I Sem
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 =
MARRI LAXMAN REDDY INSTITUTONS, DUNDIGAL, HYD – 43 Page24
AEROSPACE STRUCTURES – III – B.Tech I Sem
MARRI LAXMAN REDDY INSTITUTONS, DUNDIGAL, HYD – 43 Page25
AEROSPACE STRUCTURES – III – B.Tech I Sem
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.
MARRI LAXMAN REDDY INSTITUTONS, DUNDIGAL, HYD – 43 Page26
AEROSPACE STRUCTURES – III – B.Tech I Sem
SHEAR CENTER OF CLOSED SECTION
AIM:
To determine the shear centre of the given closed 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 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-
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:
Closed Section (D - 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). Initial readings are note down consider that is zero when dial gauge tip touches
the flange.
MARRI LAXMAN REDDY INSTITUTONS, DUNDIGAL, HYD – 43 Page27
AEROSPACE STRUCTURES – III – B.Tech I Sem
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 :
Length of the beam (L) : 405mm
Height of the web (h) : 80mm
Thickness of the sheet (t) : 0.8mm
Distance between the two hook stations (AB) : 300mm
MARRI LAXMAN REDDY INSTITUTONS, DUNDIGAL, HYD – 43 Page28
AEROSPACE STRUCTURES – III – B.Tech I Sem
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:
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.
MARRI LAXMAN REDDY INSTITUTONS, DUNDIGAL, HYD – 43 Page29
AEROSPACE STRUCTURES – III – B.Tech I Sem
RESULT:
Hence the shear centre of the given closed section is found and is validated with
theoretical value by plotting a graph.
g) Theoretical method =
h) Experimental method =
i) Error percentage =
MARRI LAXMAN REDDY INSTITUTONS, DUNDIGAL, HYD – 43 Page30
AEROSPACE STRUCTURES – III – B.Tech I Sem
MARRI LAXMAN REDDY INSTITUTONS, DUNDIGAL, HYD – 43 Page31
AEROSPACE STRUCTURES – III – B.Tech I Sem
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.
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AEROSPACE STRUCTURES – III – B.Tech I Sem
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.
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AEROSPACE STRUCTURES – III – B.Tech I Sem
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
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 uniformed through its length.
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AEROSPACE STRUCTURES – III – B.Tech I Sem
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” 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.
5. The load indicator is viewed simultaneously from the display of digital load indicator.
6. Now the column just starts buckling.
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.
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AEROSPACE STRUCTURES – III – B.Tech I Sem
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
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 uniformed through its length.
The column material is perfectly elastic homogenous and isotropic and obey hooks law.
MARRI LAXMAN REDDY INSTITUTONS, DUNDIGAL, HYD – 43 Page36
AEROSPACE STRUCTURES – III – B.Tech I Sem
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.
6. The load indicator is viewed simultaneously from the display of digital load indicator.
7. Now the column just starts buckling.
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.
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AEROSPACE STRUCTURES – III – B.Tech I Sem
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
D(δ/P) d(δ) D(δ/P)/d(δ) Inverse
FOR HINGED-HINGED 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
MARRI LAXMAN REDDY INSTITUTONS, DUNDIGAL, HYD – 43 Page38
AEROSPACE STRUCTURES – III – B.Tech I Sem
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.
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AEROSPACE STRUCTURES – III – B.Tech I Sem
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
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AEROSPACE STRUCTURES – III – B.Tech I Sem
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.
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AEROSPACE STRUCTURES – III – B.Tech I Sem
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.
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AEROSPACE STRUCTURES – III – B.Tech I Sem
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:
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AEROSPACE STRUCTURES – III – B.Tech I Sem
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:
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AEROSPACE STRUCTURES – III – B.Tech I Sem
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.
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AEROSPACE STRUCTURES – III – B.Tech I Sem
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
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AEROSPACE STRUCTURES – III – B.Tech I Sem
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.
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AEROSPACE STRUCTURES – III – B.Tech I Sem
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.
MARRI LAXMAN REDDY INSTITUTONS, DUNDIGAL, HYD – 43 Page48
AEROSPACE STRUCTURES – III – B.Tech I Sem
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
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AEROSPACE STRUCTURES – III – B.Tech I Sem
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.
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AEROSPACE STRUCTURES – III – B.Tech I Sem
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
)
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AEROSPACE STRUCTURES – III – B.Tech I Sem
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
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AEROSPACE STRUCTURES – III – B.Tech I Sem
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
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AEROSPACE STRUCTURES – III – B.Tech I Sem
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
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AEROSPACE STRUCTURES – III – B.Tech I Sem
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
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AEROSPACE STRUCTURES – III – B.Tech I Sem
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
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AEROSPACE STRUCTURES – III – B.Tech I Sem
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
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AEROSPACE STRUCTURES – III – B.Tech I Sem
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
MARRI LAXMAN REDDY INSTITUTONS, DUNDIGAL, HYD – 43 Page58
AEROSPACE STRUCTURES – III – B.Tech I Sem
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.
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AEROSPACE STRUCTURES – III – B.Tech I Sem
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
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AEROSPACE STRUCTURES – III – B.Tech I Sem
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 .¿
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AEROSPACE STRUCTURES – III – B.Tech I Sem
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AEROSPACE STRUCTURES – III – B.Tech I Sem
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
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AEROSPACE STRUCTURES – III – B.Tech I Sem
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.
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AEROSPACE STRUCTURES – III – B.Tech I Sem
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
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AEROSPACE STRUCTURES – III – B.Tech I Sem
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.
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AEROSPACE STRUCTURES – III – B.Tech I Sem
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
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AEROSPACE STRUCTURES – III – B.Tech I Sem
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.
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AEROSPACE STRUCTURES – III – B.Tech I Sem
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
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AEROSPACE STRUCTURES – III – B.Tech I Sem
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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