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1.UNSYMMETRICAL BENDING OF BEAMS
AIM:To determine the principal axes of an unsymmetrical section.
THEORY:
The flexure formula = (My/I) based on the elementary theory of
bending of beams assumes that the load is always applied through one of the
principal axes of the section. Actually, even if the applied load passes through
the centroid and/or the shear center of the section the plane of bending and the
plane of loading need not necessarily be the same. Therefore, a knowledge of
the location of the principal axes is required for the determination of the stress
distribution in beams (of any arbitrary cross section) using flexure formula. The
determination of the principal axes experimentally is described here.
If Ix, Iy and Ixy are the moments and product of inertia of any section about
an arbitrary orthogonal centroidal axes OX and OY then the inclination of one
of the principal axes to OX is given by
2Ixy 1 tan-1 (2 Ixy) tan 2 = ------------ (1) = --- ----------- (Iy - Ix) 2 (Iy – Ix)
The experimental determination of the principal axes of a given section is
based on the fact that when the load passes through the shear center and is in the
direction of one of the principal axes of the section, the entire section under the
load deflects in the direction of the load only.
APPARATUS REQUIRED:
A thin uniform cantilever Z section as
At the free and extension pieces are attached on either side of the web to
facilitate vertical loading.
A steel support structure to mount the channel section as cantilever.
A string and pulley arrangement to apply the horizontal load WH.
1
Two hooks are attached to the extension pieces to apply the vertical load
Wv .
Two dial gauges.This enables the determination of displacements u and v.
PROCEDURE:
1. Mount two dial gauges on the section to measure the horizontal and
vertical deflections of a point (u/v).
2. Apply the vertical load Wv
3. Read u and v the horizontal and vertical deflections respectively, and note
down in the tabular column.
4. Increase the load WV in steps of about 200 noting down in each case the
values of u and v. Repeat the procedure and check for consistency in
measurements.
5. Calculate the inclination using Eqn (1).
GRAPH
1. Plot the graphs (u/v) vs. (WV/WH) and find the intersection of this curve
with a straight line through the origin at 45 (Note: The X and Y scales
must be chosen to be same for the graph).
FORMULAE
t x h13 b x t3
Ixx = + 2 + (t x b)(x)2 where, x=(h/2)-(t/2)12 12
h1 x t3 t x b3 Iyy = + 2 + (t x b) (y) 2
12 12 where, y=(b/2)-(t/2)
Ixy = 2 (bt) (+b/2) (-x11) +2 bt+ (-b/2) (x11)x11= x1 + x1
2
2 Ixy
tan 2 theor = Iyy - Ixx
theor =(1/2) tan-1 2 IXY
Iyy -Ixx
RESULT
Thus the principal axes of an unsymmetrical section has been determined.
OBSERVATION
b =50 mm,
h = 80 mm,
t = 2 mm,
h1 = 76 mm
(where, h1=(h-2t))
TABULATION
S. No.WV
(Kg)
Dial gauge readings WV/WH u/v WH
(Kg)u v123456
2.SHEAR CENTER OF CHANNEL SECTIONS AIM:
To determine the shear centre of an open channel section.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.
3
The location of this shear center is important in the design of beams of
open sections when they should bend without twisting, as they are weak in
resisting 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:
A thin uniform cantilever beam of channel section.At the free end
extension pieces are attached on either side of the web to facilitate
vertical loading.
Two dial gauges
A steel support structure to mount the channel section as cantilever.
Two loading hooks each weighing about 0.1Kg.
PROCEDURE:
1. Mount two dial gauges on the flange at a known distance apart at the free
end of the beam.Set the dial gauge readings to a fixed point.
2. Place a total of say 1.2 kilograms load at A (loading hook and six load
pieces will make up this value). Note down the dial gauge reading.
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 1.2
kilogram. Record the dial gauge readings.
4. Transfer carefully all the load pieces to B one by one. Noting each time
the dial gauge readings. 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’ of the line of action from the web thus:
FORMULA
4
1.Experimental location of the shear center (e):
AB (Wa-Wb)
e = 2Wv
2.Theoretical location of the shear center (e):
e= 3b / [6+ (h/b)]
GRAPH
Plot e versus (d1-d2) curve and determine where this meets the e axis and
locate the shear center.
RESULT Thus the shear centre of an open channel section has been determined.
OBSERVATIONS
Length of the beam (L) : 50cm
Height of the web (h) : 10cm
Width of the flange (b) : 5cm
Thickness of the sheet (t) : 1.6mm
Distance between the two hook stations (AB) : 30cm
Vertical load Wv = (Wa+Wb)
S. No.Wa
(Kg)Wb
(Kg)d1
(mm)d2
(mm)d1-d2
(mm)e
(mm)1234567
3.SHEAR CENTER OF I SECTIONS AIM
To determined the shear centre for I section
5
THEORY
The horizontal elements of the "I" are known as flanges, while the vertical element is termed the "web". The web resists shear forces, while the flanges resist most of the bending moment experienced by the beam. Beam theory shows that the I-shaped section is a very efficient form for carrying both bending and shear loads in the plane of the web. On the other hand, the cross-section has a reduced capacity in the transverse direction, and is also inefficient in carrying torsion, for which hollow structural sections are often preferred.
APPARATUS REQUIRED
A thin uniform cantilever beam of I section at the free end extension
pieces are attached on either side of the web to facilitate vertical loading.
Two dial gauges are mounted firmly on this section, a known distance
apart, over the top flange. This enables the determination of the twist, if
any, experienced by the section.
A steel support structure to mount the Angle section as cantilever.
Two loading hooks each weighing about 0.1Kg.
PROCEDURE:
5. Mount two dial gauges on the flange at a known distance apart at the free
end of the beam.Set the dial gauge readings to a fixed point.
6. Place a total of say 1.2 kilograms load at A (loading hook and six load
pieces will make up this value). Note down the dial gauge reading.
7. 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.
8. Transfer carefully all the load pieces to B one by one. Noting each time
the dial gauge readings. This procedure ensures that while the magnitude
of the resultant vertical force remains the same its line of action shifts by a
6
known amount along AB every time a load piece is shifted. Calculate the
distance ‘e’ of the line of action from the web.
FORMULA
1.Experimental location of the shear center (e):
AB (Wa-Wb)
e = 2Wv
GRAPH
Plot e versus (d1-d2) curve and determine where this meets the e axis and
locate the shear center.
RESULT Thus the shear centre of I section has been determined.
OBSERVATIONS
Length of the beam (L) : 50cm
Height of the web (h) : 10cm
Width of the flange (b) : 2cm
Thickness of the sheet (t) : 4mm
Distance between the two hook stations (AB) : 30cm
Vertical load Wv = (Wa+Wb)
S. No.Wa
(Kg)Wb
(Kg)Dial gauge readings
d1-d2e
(mm)d1 d2
1234567
4.SHEAR CENTER OF L SECTIONS
AIM
7
To determined the shear centre for L section.
APPARATUS REQUIRED
A thin uniform cantilever beam of L section at the free end extension
pieces are attached on either side of the web to facilitate vertical loading.
Two dial gauges are mounted firmly on this section, a known distance
apart, over the top flange. This enables the determination of the twist, if
any, experienced by the section.
A steel support structure to mount the Angle section as cantilever.
Two loading hooks each weighing about 0.1Kg.
PROCEDURE
9. Mount two dial gauges on the flange at a known distance apart at the free
end of the beam.Set the dial gauge readings to a fixed point.
10.Place a total of say 0.5 kilograms load at A. Note down the dial gauge
reading.
11.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 0.5
kilogram. Record the dial gauge readings.
12.Transfer carefully all the load pieces to B one by one. Noting each time
the dial gauge readings. 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’ of the line of action from the web.
FORMULA
1.Experimental location of the shear center (e):
AB (Wa-Wb)
e = 2Wv
GRAPH
8
Plot e versus (d1-d2) curve and determine where this meets the e axis and
locate the shear center.
RESULT Thus the shear centre of L section has been determined.
OBSERVATIONS
Length of the beam (L) : 50cm
Height of the web (h) : 10cm
Width of the flange (b) : 5cm
Thickness of the sheet (t) : 1.6mm
Distance between the two hook stations (AB) : 30cm
Vertical load Wv = (Wa+Wb)
S. No.Wa
(Kg)Wb
(Kg)Dial gauge readings
d1-d2e
(mm)d1 d2
12345
5.SHEAR CENTER OF CLOSED SECTION
AIMTo determine the shear center of a closed section .
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 a closed section like an aircraft wing, where the lift
produces a torque about the shear center. Similarly the wing strut of a semi
9
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
A thin uniform cantilever beam of ‘D’ section. At the free end extension
pieces are attached on either side of the web to facilitate vertical loading.
Two dial gauges are mounted firmly on this section, a known distance
apart, over the top flange. This enables the determination of the twist, if
any, experienced by the section.
A steel support structure to mount the ‘D’ section as cantilever.
Two loading hooks each weighing about 0.1 Kg.
PROCEDURE
13.Mount two dial gauges on the flange at a known distance apart at the free
end of the beam.
14.Place a total of say 1.2 kilograms load at A (loading hook and 6 load
pieces will make up this value). Note the dial gauge readings (nominally,
hooks also weigh 100 grams each). Note down dial gauge reading.
15.Now remove one load piece from the hook at A and place it at hook B.
The total vertical load on this section remains 1.2 kilogram. Record the
dial gauge readings.
16.Transfer carefully all the load pieces to B from A one by one. Note each
time the dial gauge readings. 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 when a load piece
10
is shifted. Calculate the distance ‘e’ (see fig) of the line of action from
the web thus:
(AB) x (Wa-Wb)
eexp --------------------- where WV = (Wa + Wb)
2Wv
For every load case calculate the algebraic difference between the dial
gauge readings suffered by the section.
GRAPH
Plot e Vs (d1-d2) curve and determine where this meets the axis and
locate the shear center.
RESULT
Thus the shear center of a closed section has been determined.
OBSERVATION
Length of the beam (L) : 500mm
Height of the web (h) : 100mm
Thickness of the sheet (t) : 1.6mm
Distance between the two hook stations (AB) : 300mm
Weights – 6 Nos. : 0.2kg
TABULATION
S. No.
Wa Wb
Dial gauge readings
(d1-d2)Algebraic difference
e d1 d2
6.CONSTANT STRENGTH BEAM
11
AIMTo determine the stress at various locations along the length of a
constant strength beam to show that they are equal and compare with theoretical values.
THEORYThe aerospace structures engineer is constantly searching for types
of structures which will save structural weight and still provide a structure which is satisfactory from a fabrication and economic standpoint. One such structure is constant strength beam.A beam in which section modulus varies along the length of the beam in the same proportion as the bending moment is known as constant strength beam. In this case the maximum stress remains constant along the length of the beam.
APPARATUS REQUIREDAluminium strength beam,Strain gauges, strain indicator and
weights with hook.
FORMULA USED1. Stress (experimental)
(N/m2)
Where σ = stressE=Young’s modulusε = strain(The system is in half bridge; therefore the stress value should be divided by 2).
2. Stress (theoretical)
Where M= moment for a load
(Nm){W= weight and L= length corresponding each load}. I=moment of inertia
12
(m4 )
PROCEDURE1. The constant strength beam is fixed as a cantilever. 2. Strain gauges are fixed.3. The strain gauges are fixed on the surfaces at each location to
increase the circuit sensitivity of the strain gauge circuit.4. The beam is loaded gradually by 200gm upto 1000gm placing the
weights slowly in the hook near the tip of the cantilever. 5. The strain gauge readings are noted for every load at locations at
A, B, C position and tabulated.
GRAPHGraphs are plotted for the following cases:
1. Stress V/S theoretical and actual strain for point A2. Stress V/S theoretical and actual strain for point B3. Stress V/S theoretical and actual strain for point C
(Stress on Y-axis and strain on X-axis).
RESULTThe experimental values of the stress are compared with the theoretical values and found to agree well with the experimental results.
OBSERVATIONSYoung’s modulus of aluminium,E = 70GPaDistance of point A from loading point, LA = 15cm = 0.150 mDepth of the beam, hA = d1 = 0.050 mWidth of the beam, t = 0.005mDistance of point A from loading point, LB = 23cm = 0.23 mDepth of the beam, hB = d2 = 0.060 mDistance of point A from loading point, LC = 39cm = 0.39 mDepth of the beam, hC = d3 = 0.080 m
13
DIAGRAM
TABULATION
S. No. Weight( Kg)A
(10-6)B
(10-6)C
(10-6)1 0.2
2 0.4
3 0.6
4 0.8
5 1
CALCULATION
1.Experimental stress2.Theoretical stress.
S. No. Weight( Kg) A(exp)
MPaB(exp) MPa
C(exp)
MPa A(the)
MPaB(the) MPa
C(the)
MPa
1 0.2
2 0.4
14
3 0.6
4 0.8
5 1
7.FLEXIBILITY MATRIXAIM:
To determine the flexibility matrix of a cantilever beam.
APPARATUS REQUIRED:1. Cantilever beam setup2. Weighing hook, weights.3. Dial gauge with stand (quantity 2).
THEORY:
Flexibility and its converse, known as stiffness matrix are important properties which characterize the response of structure by means of force displacement. The flexibility of a structure is defined as the displacement by a unit force and the stiffness is defined as the force required for unit displacement.
A flexibility matrix relates that force “p” at “K1” and “K2” of an element to the displacement at “K1” & “K2”.
The stiffness matrix is important factor of structure. It can be readily shown in the flexibility matrix.
FORMULA:
Theoertical method
1 2 free end11 12 1
Flexibility matrix = L
21 22 2
Flexibility coefficient
15
11 = Displacement along 1 when unit load is applied along 1.12 = Displacement at 1 when unit moment is applied at 2.21 = Displacement at 2 when unit load is applied at 1.22 = Displacement at 1 when unit moment is applied at 2.
L = Length
b bd3 11
d = I = ----- 12
21
11 12 1 unit load L3 L2
----- ----3EI 2EI
L2 L
---- ----2EI EI 12
21 22
22
1 unit
Experimental method
(a) Deflection matrix D
D11 D12
D21 D22 from dial gauge
(b) Co factor matrix of Deflection matrix is given by A
0.2 x 9.81 0.2 x 9.81 D11 D12
0.2 x 9.81 0.2 x 9.81 D21 D22
A11 A12
16
A =A21 A22
(c ) Adj A = A11 - A12
-A21 A22
(d ) Determinant matrix A = (A11x A22) - ( A12 xA21)(e) Flexibility Matrix A-1 = 1 x adj A
A
PROCEDURE:
1. Initially the total length of the given cantilever beam is divided into 2 equal parts K1, K2.
2. Now place the weight hook loading and dial gauge at K1.3. To find K12 keep the hook at 2 K2 & deflection at K1.4. To find K2, deflection at K2 unit load at K1. 5. To find K22 deflection K2 unit load at K2.6. Note down the reading of the dial gauge .
RESULTThus the flexibility matrix of a cantilever beam has been determined
OBSEVATIONS Material used =E=L=
TABULATIONS.No Load
in KgK11
in mmK12
in mmK21
in mmK22
in mm1
8.COMBINED BENDING AND TORSIONAIMTo determine the principal stress and principal planes of a hollow circular shaft due to combined loading.
THEORYThe most common combined load system encountered in structural design is probably due to bending and torsion. In an Aircraft wing the lift acting at the
17
center of pressure produces a torque about the elastic axis and varying bending moment along the wing span. To understand their combined effect a simpler specimen, namely a hollow cylinder is subjected to a bending and torsion.
APPARATUS REQUIREDHollow circular shaft fixed as a cantilever, weight hanger with slotted weights, strain gauges, connecting wires, strain indicator.
FORMULA USED1.Bending stress (experimental)
(N/m2)
Where σ = stressE=Young’s modulusε = strain(The system is in half bridge; therefore the stress value should be divided by 2).
2.Polar moment of inertia
(mm4)
3.Moment of Inertia
(mm4)
4.Bending Moment
(Nm)
5.Torque
(Nm)
6.Shear stressτmax (N/m2)
18
7.Principal Stress
σx = (N/m2)
8.Max. and Min. principal stress
σ1.2 =σx+ σy + (σx- σy )2 + (τxy)2
2 2 σy = 0
For root Aσ1.2 =σAX +0 + (σAX- 0 )2
+ (τxy)2 2 2
9.Principal angle
2xy 2max
Tan2 = ------------ = ----- x - y max
2max
Principal angle =2 = Tan-1 ------- = direction of angle max
PROCEDURE
1. At position A two strain gauges are fixed on the tube fixed as a cantilever, one on the top fiber and the other at the bottom to measure the bending stress, ().
2. Another strain gauge is fixed at the location C on the neutral axis at 45 to measure the shear stress ().
3. Similarly two more strain gauges are at B fixed (top & bottom) for verify the result at various locations of the tube.
4. The strain gauges on the tube are connected to half bridge circuit in the strain indicator to increase the circuit sensitivity, since the tension and compression get added up.
5. The strain gauge at 45 is connected to the half bridge of the strain indicator to measure the shear stress.
6. The outside diameter of the tube is measured.
19
7. Weights are added to the hook attached to the lever in steps and the strain gauge readings are noted from the strain indicator for each load.
8. From the strains the bending stress () shear stress () are calculated and hence principal stresses xy and principal angle () are calculated.
9. These values are compared with theoretical values.
GRAPHGraphs are plotted for the following cases:
4. Experimental Stress V/S εA
5. Experimental Stress V/S εB
6. Experimental Stress V/S εC
(Stress on Y-axis and strain on X-axis).
RESULTThus d the principal stress and principal planes of a hollow circular shaft due to combined loading has been determined.
OBSERVATIONSYoung’s Modulus of the tube (steel) = 200G PaThickness of the tube = 0.002mLength of the tube = L = 0.5mOutside diameter of the tube = do = 0.05mInside dia of tube = di = 0.046mDistance of the strain gauges near to tip = (LA) = 0.08mDistance of the strain gauges at the middle from tip = (LB) = 0.21mDistance of the Strain gauges far to tip = (LC) = 0.41mDistance from the center of the tube to the center of the hook (L0) = 0.15m
DIAGRAM
20
TABULATION
Sl No.
Weight ( kg)
ε
(10-6)ε
(10-6)εc
(10-6)
bending A
( N/m2)
bending
N/m2)
bending C
(N/m2)
MA
(Nm)MB
(Nm)MC
(Nm)T
(Nm)
max
N/m2)1 0.52 13 1.5
4 2
Sl No.x
( N/m2)x (N/m2)
Cx
(N/m2)
max
(N/m2)
min (N/m2)
max (N/m2)
min (N/m2)
Cmax (N/m2)
Cmin (N/m2)
C
1234
CALCULATIONS
9.FREE LONGITUDINAL VIBRATION
AIMTo study the Free Longitudinal Vibrations of a spring mass system.
THEORYWhen particles of the shaft (or) disc move parallel to the axis of the shaft then the vibrations are known as longitudinal vibration.
APPARATUSUniversal vibration beam setup, open coil helical spring, mass hanger, weights and tape.
PROCEDURE1) In the cantilever beam spring is fixed by means of hook so that the
mass hanger of known mass is placed.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.
21
4) Note down the maximum value of frequency and amplitude directly in the digital indicator.
RESULT:Thus the free longitudinal vibration of a spring mass system is studied and found to be Hz.
OBSERVATIONSDistance between fixed end to stiffness of the spring in m (L) =Distance between fixed to weight pan in m (L1) =
TABULATION
Sl. No. Mass attached (W) (Kg)
Deflection () of spring (mm)
Frequency(exp)
in Hz
1. 0.2DIAGRAM
10.FORCED VIBRATION
AIM: To study the un-damped forced vibration of equivalent spring mass system.
APPARATUSUniversal vibration beam setup, open coil helical spring, mass hanger, weights and tape.
FORMULA USED1.Angular speed () 2N = -------- (s-1)
60
22
2. Disturbing Force ( Fo ) Fo = mo e 2 (kgm/s2)
mo = mass of eccentricity weight = 24 gms = 0.024 Kge = Distance between middle of the shaft to middle of the eccentricity =0.035m.
3. Siffness of the spring (K)
M × g K= -------------
Where,
M = mass of the disc (exciter) = 3.7 Kgg = 9.81 m/sec2
= Deflection of beam i.e., amplitude in m
4. Theoretical Frequency
1 KTheoretical frequency = fthe = ------ -------- (Hz)
2 M
PROCEDURE
1. Support one end of the cantilever beam in the slot of trunion/fixed plate and clamp it by means of knob.
2. Attach the other end of 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 exciter assembly along with a discs bearing & flexible shaft.
(3.7kg)5. Clamp the assembly at any convenient position in the cantilever beam.6. Allow system to vibrate freely.7. Neglect the initial reading 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.
23
10. Repeat the experiment by varying speed and by also fixing different eccentric weight on the disc.
RESULTThe un-damped forced vibration of equivalent spring mass system is studied. TABULATION
Motor speed,N
Experimental Frequency, fnexp
(Hz)
Amplitude ( deflection,) Angular
speed,s-1)
Disturbng force ,Fo(kgm/s2)
Stiffness of spring,K(N/m)
Theoritical Frequency, fnthe
(Hz)mm m
11.TORSIONAL VIBRATION OF SINGLE ROTOR SHAFT SYSTEMSAIM:To determine the torsional vibration of single Rotor Shaft system.DESCRIPTION OF SETUP: The general arrangement for carrying out the experiments. 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 designed plates are used for clamping top end of the shaft. The bush bearing support to the flywheel shaft provides negligible lateral movement during experiments. The bush housing is fixed to side member of mainframe.
PROCEDURE:1) Fix the flywheel at convenient position along the shaft.2) Grip one end of the shaft at the bracket by the flat plate.3) Place the frequency sensor side the disc. So that the frequency and
amplitude can be seen in digital indicator.4) Twist the rotor-on the shaft up to 50 and allow oscillate.5) Directly note down the frequency in digital meter.
FORMULA: G.IP
1. Determine Torsional Stiffness kt= -----
24
L
L = length of shaft in m =0.003m
d4
J = IP = Polar M.I. of shaft = ------ m4
32
d = Shaft dia 3mm = 0.003m
G = Modulus of rigidity of shaft = 80 GPa = 80 × 109 N/m2.
2. Determine T theoretical = Tthe
I W r2
= 2 --- Where, I = M.I. of Disc = -- --- Kt g 4
Where, W = wt of the disc = 5.4 Kgs g = 9.81 m/sec2 = specific gravity D = Dia of disc = 0.200m. , r=d/2=0.1m
Natural frequency of rotar is = fn (the)
1fn (the) = ----------
Tthe
RESULT Thus the Torsional vibration of single Rotor Shaft system has been determined.
OBSERVATION Shaft Dia = 3mmDia of disc = 200mmWt. of the Disc,W = 3.7Kg
25
Modulus of rigidity for shaft = 80 Gpa = 80 × 109 N/m2.
TABLE:Sl. No.
Length of Shaft ‘L’ cm
Frequency(exp)
1
12.STRUCTURAL BEHAVIOUR OF A SEMI – TENSION FIELD BEAM(WAGNER BEAM)
AIM: To investigate and study the behaviors of a semi-tension field beam.
THEORY:
The development of a structure in which buckling of the web is permitted with the shear loads being carried by diagonal tension stresses in the web is a striking example of the departure of the design of Aerospace structures from the standard structural design methods in other fields of structures, such as beam design involving diagonal semi-tension field action in beam webs was done by Wagner and hence it is know as Wagner beam.
As thin sheets are weak in compression. The webs of the Wagner beam will buckle at a low value of the applied vertical load. The phenomena of buckling may be observed by noting the wrinkles that appear on the thin sheet. As the applied load further increased the stress in the compression direction. This method of carrying the shear load permits the design of relatively thin webs because of high allowable stresses in tension.
According to the theory developed by Wagner, the diagonal tensile stress +ve values or Compression stress –VE values.t in the thin web is given by the expression.
2Wt= -------------- in MPa (1) d x t sin 2
Where, W= shear load in Nd = Distance between the flanges in mmt = Thickness of the web in m = Angle at which wrinkling occurs
i.e.,
26
1 + td / 2AF
Tan4 = ------------------ (a) 1 + tb / AS
b = Distance between stiffeners in mmAF = Area of flange in m2
AS = Area of stiffener in m2
WStress in the Flange F = -------------- in MPa (2) 2AF tan
Wx b Stress in the Stiffener S = ---------- tan in MPa (3) AS x d
APPARATUS REQUIRED: A stiffened thin-webbed cantilever beam in a frame, Strain gauges, Strain indicator, Hydraulic jack, Load cell and Load indicator.
PROCEDURE:o The wrinkling angle is calculated using the equation (a) and a strain gauge is
fixed at this angle in the web.
o Strain gauges are also fixed on the flanges and a stiffener to measure their respective stresses (strain gauges are fixed).
o For individual strains load is applied gradually in steps using the hydraulic jack.
o For each load the load indicator reading, strain indicator reading corresponding to each strain gauge is noted.
RESULT:t, F and S values are calculated theoretically using equations (1), (2)
and (3) and compared with the experimental values given in the table.
SPECIFICATIONS:t = 0.8mm = 0.008mb = 225mm2 =0.225md = 270m =0.27mArea of flanges = AF = 390mm2 = 0.00039m2
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Area of stiffness = AS = 390mm2 = 0.00039m2
E = 70GPaTABULATION
Weight applied(W)
Strain(10-6)
Experimental stress(MPa)
Theoretical stress(MPa)
kg N εt σt σt
Weight applied(W)
Strain(10-6)
Experimental stress(MPa)
Theoretical stress(MPa)
kg N εf σf σf
Weight applied(W)
Strain(10-6)
Experimental stress(MPa)
Theoretical stress(MPa)
kg N εs σs σs
Schematic view of Wagner beam
SIDE VIEW: b
CF=F
CS=S
Ct=t
d
HYDRAULIC JACK
LLOAD CELL
13. STUDY ON PHOTOELASTICITY
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Introduction Photoelasticity is an experimental method to determine the stress distribution in a material. The method is mostly used in cases where mathematical methods become quite cumbersome. Unlike the analytical methods of stress determination, photoelasticity gives a fairly accurate picture of stress distribution, even around abrupt discontinuities in a material. The method is an important tool for determining critical stress points in a material, and is used for determining stress concentration in irregular geometries.Principle
The method is based on the property of birefringence exhibited by certain transparent materials. Birefringence is a property where a ray of light passing through a birefringent material experiences two refractive indices. The property of birefringence (or double refraction) is observed in many optical crystals. Upon the application of stresses, photoelastic materials exhibit the property of birefringence, and the magnitude of the refractive indices at each point in the material is directly related to the state of stresses at that point. Information such as maximum shear stress and its orientation are available by analyzing the birefringence with an instrument called polariscope.
When a ray of light passes through a photoelastic material, its electromagnetic wave components gets resolved along the two principal stress directions and each of these components experiences different refractive indices due to the birefringence. The difference in the refractive indices leads to a relative phase retardation between the two components. Assuming a thin specimen made of isotropic materials, where two-dimensional photoelasticity is applicable.,the magnitude of the relative retardation is given by the stress-optic law
where Δ is the induced retardation,
C is the stress-optic coefficient,
t is the specimen thickness,
σ1 and σ2 are the first and second principal stresses, respectively.
The retardation changes the polarization of transmitted light. The polariscope combines the different polarization states of light waves before and after passing the specimen. Due to optical interference of the two waves, a fringe pattern is revealed. The number of fringe order N is denoted as
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which depends on relative retardation. By studying the fringe pattern one can determine the state of stress at various points in the material.
For materials that do not show photoelastic behavior, it is still possible to study the stress distribution in such materials. The first step is to build a model using photoelastic materials, which has similar geometry as the real structure to be investigated. The loading is then applied in the same way to ensure that the stress distribution in the model is similar to the stress in the real structure.
Isoclinics and isochromatics
Isoclinics are the loci of the points in the specimen along which the principal stresses are in the same direction.
Isochromatics are the loci of the points along which the difference in the first and second principal stress remains the same. Thus they are the lines which join the points with equal maximum shear stress magnitude
Two-dimensional photoelasticity
Photoelasticity can be applied both to three-dimensional and two-dimensional state of stress. But the application of photoelasticty to the three-dimensional state of stress is more involved as compared to the state of two-dimensional or plane-stress systemThe experimental setup varies from experiment to experiment. The two basic kinds of setup used are plane polariscope and circular polariscope.
1. Plane polariscope
The setup consists of two linear polarizers and a light source. The light source can either emit monochromatic light or white light depending upon the experiment. First the light is passed through the first polarizer which converts the light into plane polarized light. The apparatus is set up in such a way that this plane polarized light then passes through the stressed specimen. This light then follows, at each point of the specimen, the direction of principal stress at that point. The light is then made to pass through the analyzer and we finally get the fringe pattern
2. Circular polariscope
In a circular polariscope setup two quarter-wave plates are added to the experimental setup of the plane polariscope. The first quarter-wave plate is placed in between the polarizer and the specimen and the second quarter-wave plate is placed between the specimen and the analyzer. The effect of adding the quarter-wave plate after the source-side polarizer is that we get circularly polarized light passing through the sample. The analyzer-side quarter-wave plate converts the circular polarization state back to linear before the light passes through the analyzer.
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The basic advantage of a circular polariscope over a plane polariscope is that in a circular polariscope setup we only get the isochromatics and not the isoclinics. This eliminates the problem of differentiating between the isoclinics and the isochromatics.
Applications Photoelasticity has been used for a variety of stress analyses and even for
routine use in design, particularly before the advent of numerical methods, such as for instance finite elements or boundary elements
Digitization of polariscopy enables fast image acquisition and data processing, which allows its industrial applications to control quality of manufacturing process for materials such as glass and polymer
Dentistry utilizes photoelasticity to analyze strain in denture materials. Dynamic photoelasticity integrated with high-speed photography is
utilized to investigate fracture behavior in materials. Another important application of the photoelasticity experiments is to
study the stress field around bi-material notches. Bi-material notches exist in many engineering application like welded or adhesively bonded structures
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