-
8/3/2019 P3.1 Report
1/30
Page | 1
ABSTRACT
The engineers bending formula is only applicable to sections
bent about its principal axes.
For other cases, we have to use the generalized bending
equation. This generalized bendingequation makes use of the
assumption that plane section before bending will still be a
plane
even after bending. The other assumption is that the material
obeys the idealized HookesLaw. This equation is applicable to all
sections regardless their geometry and their load
direction.
-
8/3/2019 P3.1 Report
2/30
Page | 2
1. OBJECTVESThe objectives of this project are to verify the
generalized bending theory by applying
it in an experiment and to increase students knowledge on some
features ofunsymmetrical bending.
2. THEORYThe EulerBernoulli beam equation or better known as the
Generalized BendingEquation is the means of calculating the
load-carrying and deflection characteristics
of beams. It is largely used when the section bending is not
about its principle axes,
such as unsymmetrical bending. The load-stress formula can be
derive by assuming
that the equilibrium equations, compatibility conditions and the
stress-strain relations
(Hooks law) be satisfied. And that the plane section before
bending remains plane
after bending.
2.1Equilibrium EquationThe application of the equation of
equilibrium to the free body in Figure 7.9b yields:
0 = ZZ A
Mx= yZZ A
My= - xZZ AEq 1.1
where
ZZ is the normal stress in the beam due to bending A denotes an
element of area in the cross section x = the perpendicular distance
to the centroidaly-axis y = the perpendicular distance to the
centroidalx-axis My = the bending moment about they-axis Mx = the
bending moment about thex-axis
-
8/3/2019 P3.1 Report
3/30
Page | 3
2.2Geometry Deformation(Strain)By considering two plane
cross-section perpendicular to the bending axis before bending,
with distance Lzz between their centroids. And after bending the
extension LZZ betweenthe two planes can be represented as a linear
function of x & y:
LZZ = a' + b'x + c'y
since ZZ = LZZ/LZZ , therefore:
= + +
2.3Stress-Strain RelationsAccording to Hooks law:
ZZ = EZZ
where E is young's modulus.
Therefore
=
+ +
ZZ = a + bx + cyEq 1.2
where
a = Ea'/LZZ b = Eb'/LZZ c = Ec'/LZZ
-
8/3/2019 P3.1 Report
4/30
Page | 4
2.4Generalized Bending EquationBy substituting 1.2 into 1.1
0 = (a + bx + cy)A = aA + bxA+ cyA
Mx= y(a + bx + cy)A = ayA + bxyA+ cy2A
My= - x(a + bx + cy)A = -axA - bx2A - cyxA
since xA = 0&yA = 0
andy2A = Ix ,x2A = Iy & xyA = Ixy
0 = aA
Mx = bIxy + cIx
My = - bIy - cIxy
where Ix & Iy are the second moment of inertia with respect
to the x and y axis
respectively, and Ixy is the product moment of inertia.
Solving the equation to obtain the constants:
a = 0 (since A 0)
= +
= +
Eq 1.3
substituting Eq 1.3 into Eq 1.2
= +
+ ( + )
This is the generalized bending equation that will be used to
calculate the normal stress.
-
8/3/2019 P3.1 Report
5/30
Page | 5
2.5Principal Moment of Inertia
Figure 1 Principal Axes
2.5.1 Principal Axes
In order to find the principal axis of inertia, let Ix & Iy
are the second moment of inertia
with respect to the x and y axis respectively, and Ixy is the
product moment of inertia.
Let (u,v) be the principal axes with the same origin and in the
same plane as the (x,y)
axes, and is the angle (x,y) must rotate to coincide with (u,v),
is positive in thecounter clockwise direction.
u =xcos +ysin
v =ycos xsin
to find the second and product moment of inertia
Iu= v2A = (ycos xsin)2A
Iv= u2A = (xcos +ysin)2A
Iuv= uvA = (xcos +ysin)(ycos xsin)A
after integrating and knowing that: y2A = Ix ,x2A = Iy & xyA
= Ixy and using
trigonometric identities. The following equation cans be
obtain:
Iu= 0.5[Ix
+ Iy] + 0.5[I
xI
y]cos2 - I
xysin2
Iv= 0.5[Ix + Iy] - 0.5[IxIy]cos2 + Ixysin2
Iuv= 0.5[IxIy]cos2 + Ixysin2
From the above equations, it can be shown that
Iu +Iv= Ix+ Iy
=
+
-
8/3/2019 P3.1 Report
6/30
Page | 6
3. EQUIPMENTSThe equipments used for this experiment are listed
as follows.
3.1
Frame A
Figure 2 Frame A
3.2 Frame B
Figure 3 Frame B
-
8/3/2019 P3.1 Report
7/30
Page | 7
3.3 TDS 303 Data Logger
Figure 4 TDS 303 Data Logger
4. PROCEDUREThe procedures of the experiment are stated as
follows:
a)The power of the TDS 303 Data Logger and the frame readout is
turned on.b)Then the test rigs are unloaded and the strain gauge
circuits are balanced in order tozero the initial reading of each
channel.
c) It is followed by loading the beam according to case1 of
Table 1.d)When the load is stabilised the strain reading is
taken.e)The steps (c) and (d) are repeated for cases 2, 3 and
4.
Case Frame Px Py
1 A 300 N 0 N
2 A 300 N 200 N
3 B 20.5 kg 0 kg
4 B 20.5 kg 20.5 kg
Table1: Loading values for different cases
-
8/3/2019 P3.1 Report
8/30
Page | 8
5. EXPERIMENTAL RESULTS
Ix Ixy Iy E
OR
208.8 x 103
mm4
72.74 x 103
mm4
73.64 x 103
mm4
70 x 109
Pa
(Aluminium)208.8 x 10
-9m
472.74 x 10
-9m
473.64 x 10
-9m
4
Table 1 Properties of Material
*Note: Different load direction compared with Frame ATable 2
Loading of individual cases
Table 3 Moment Calculation
Table 4 Experimental Data for Strain
Case Frame Px Py
1 A 300N 0N
2 A 300N 200N
3 B 20.5kg * 0kg
4 B 20.5kg* 20.5kg
Case Formula Mx/Nm My/Nm
1 Mx = Py X
d
My= Px X
d
d=0.33m
0 -49.5
2 -33 -49.5
3 0 33.18
4 -33.18 33.18
Case 1 Case 2 Case 3 Case 4
Gauge
No
(horizontal
frame)
(vertical
frame)
(horizontal
frame)
(vertical
frame)
(horizontal
frame)
(vertical
frame)
(horizontal
frame)
(vertical
frame)
1 -190 -217 -281 -300 109 127 25 53
2 -120 -179 -195 -237 68 106 -1 55
3 -53 -141 -115 -177 25 82 -32 58
4 12 -104 -33 -115 -13 60 -61 59
5 79 -67 49 -53 -54 37 -90 60
6 149 -29 133 5 -95 14 -119 62
7 219 5 219 67 -135 -6 -147 64
8 281 43 296 130 -175 -28 -175 66
9 357 80 386 189 -217 -52 -205 67
-
8/3/2019 P3.1 Report
9/30
Page | 9
Table 5 Distances X and Y for calculation of theoretical
strain
Case 1 Case 2 Case 3 Case 4
Formula
= ( + ) ( + )( )
(horizontal
frame)
(vertical
frame)
(horizontal
frame)
(vertical
frame)
(horizontal
frame)
(vertical
frame)
(horizontal
frame)
(vertical
frame)
Gauge
No 9 411 94 454 218 -275 -63 -232 61
Corner
C -254 -254 -366 -366 171 171 58 58
Table 6 Theoretical Strain
Experimental () Theoretical () Difference ()Formula tan = [
Ixtan + Ixy ]/[ Iy + Ixytan ]Case 1 70 71 1
Case 2 63 64 1
Case 3 71 71 0
Case 4 97 91 6
Table 7 Experimental and Theoretical Neutral Angle,
Table 9 Theoretical Strain vs Experimental Strain Error
Percentage
Case 1 - 4
Horizontal
Frame
Vertical Frame
Gauge No
9
x1= 36.32 x 10-3
m
x2=-9.29 x 10-3
m
y1 = 23.23 x 10-3
my2= -45.17 x 10
-3
m
Corner C
x3=-9.29 x 10-3
m
y3= 23.23 x 10-3
m
Gauge
No 9 13.1% 14.9% 15.0% 13.3% 21.1% 17.5% 11.6% -9.8%
Corner
C 25.2% 14.6% 23.2% 18.0% 36.3% 25.7% 56.9% 8.6%
x3 x1C
x2
y3
y2
y1CENTROID
-
8/3/2019 P3.1 Report
10/30
Page | 10
6. DISCUSSION AND CONCLUSION6.1 Bok Chian Check
a) Without using the generalized bending equation, describe
briefly an alternative method
of calculating the normal stresses at any point on an
unsymmetrical beam in bending,
From the principal moment of inertia, the angle of the neutral
axis can be found using
tan2 = -(2Ixy)/( Ix + Iy). Therefore the principal axis can be
determine with respect to thex & y axis. Hence the symmetric
loading equation can be used:
ZZ = Mv V+ Mv U = 0
Iv Iu
the equation can be simplified to:
V = Iutan = tan
U Iv
b) For the existing experimental setup for Frame A and B,
determine the expressions for
Mxand My at the loading point, in terms of the applied
loads.
Moment = Force x Perpendicular Distance
Mx = Px/2 x 0.33 (Nm)
My = Py/2 x 0.33 (Nm)
c) For Frame A only: When Px is loaded first, observe the load
reading of Py is applied.
Explain why?
When Pyis applied, the load reading of Px decreases. This is
because when Pyis applied,
due to the unsymmetrical section of the beam, it cause a
increase in moment about the x-
axis, in the positive directions, which is the direction Px is
also acting. Therefore it results
in the sensor detecting a lower net force acting in the Px
direction.
d) Briefly explain why the NA is not 90 degrees to the applied
load.
Neutral axis is an axis in the cross section of a beam or shaft
along that have zero
longitudinal stresses or strains. It is perpendicular only to
the applied load if the structure
is symmetrical. However in this situation, since the beam is
unsymmetrical, the neutral
axis will be at an angle to the load. The angle of inclination
will depend on the magnitude
and direction of the applied load.
e) For the case, determine the theoretical strain distribution
at the strain gauged section.
By means of a graph, compare the theoretical strain distribution
with the experimental
strain distribution.
All of the experimental strain for data differ from the
theoretical strain by approximatelythe same percentage amount,
except for a few. The difference can be attributed to the
-
8/3/2019 P3.1 Report
11/30
Page | 11
material and experimental error, the beam have been subjected to
repeat loading and
unloading for a extensive period of time, and hence suffer from
metal fatigue. The
experimental strain are all lower than the theoretical strain in
terms of magnitude, which is
expected as due to the imperfect microstructure of the material
and the conservation of
energy, the experimental strain must be lower or equal to
calculated strain. Therefore the
experimental strain data can be said to be precise but not very
accurate.
f) For all the case, plot the experimental strain distribution
across the beam section and
hence, estimate the position of the neutral axes. Compare the
theoretical angles between
the neutral axis and x-axis with those of the experiment.
The experimental angles are found to be very close to the
theoretical angles, with all
having a difference by 1 with the exception of one. In case 4,
the difference between the
experimental and theoretical angle is found to be 6. This can be
human error as the
neutral axis is almost vertical with the y-axis. Which cause the
graph to be out of bound of
the paper, and the angle had to be estimated. Therefore the
experimental angles can be said
to be accurate and precise with the exception of case 4 as an
anomaly.
g) Explain why unsymmetrical sections are preferred in certain
structural designs.
In symmetrical beams, the neutral axis is fixed and is
perpendicular to an applied load.
Therefore if there is an second load perpendicular to the first
load, then one of the load
will result in stress acting on the weakest part of the
cross-section. Therefore
unsymmetrical sections are preferred when the bending moments
are greater on one axis
than the other. Another unintended benefit of unsymmetrical
sections is the less material
used to make it, which will reduce the weight of the
structure.
Conclusion
The experimental data are close to the theoretical values
calculated using the Generalized
Bending Equation, with differences mainly due to the metal
fatigue of the beam and
human errors, and therefore in agreement with the generalized
bending theory. The
features of unsymmetrical structure were also explored and were
found to be useful in
situation where the bending moment are significantly different
on different axis.
-
8/3/2019 P3.1 Report
12/30
Page | 12
6.2 Chia Yiwen Yvonne
a) Without using the generalized bending equation, describe
briefly an alternativemethod of calculating the normal stress at
any point on an unsymmetrical beam in
bending.
Under the earlier section of the report, it is derived that tan2
= 2
. With the
value of angle , we can determine the principal axes of bending
with respect to thereference axes.
Figure A Principle axes determined from
When the principle axes are defined, we can apply the general
stress equation for
symmetrical loading for the case of an unsymmetrical
loading.
The formula used is = [
] + [
].
Hence therefore the alternative equation to the generalized
bending equation would be:
= [ ] + [ ] , where Mu= M cos and Mv= M sin
In the case when bending stress is zero at the neutral axis
(NA), the equation becomes:
= [
] + [
] = 0
The above equation can then be further simplified to () = -
(
) tan = tan .
Figure B Illustration of angle
The angle is the angle of position of the neutral axis with
reference to the u-axis, asillustrated in the figure above.
b) For the existing experimental setup for Frame A and B,
determine the expressions forMx and My at the loading point, in
terms of the applied loads.
-
8/3/2019 P3.1 Report
13/30
Page | 13
The expressions of Mx and My are:
a. Mx = Py X db. My= Px X d
Where d=0.33m
The results for the individual moments for each case are:
Table A Calculation for Mx and My
c) For Frame A only: When Px is loaded first, observe the load
readings of Px when Pyis applied. Explain why?It had been observed
that after Py is loaded, the reading for Px will be reduced. A
moment is formed about the beam with the load is applied only in
the positive x-
direction. When Py is applied, the resultant moment increases in
the x-direction and
the beams natural rotates towards the positive x-direction. This
results in a reductionof magnitude of load Px as the beam now bends
toward the same direction where Px
is acting. On the side note, the resultant bending of the beam
is minuscule and we are
not able to observe any significant movement of the beam through
naked eye.
Another possible reason is the strain gauges which are attached
to the beam measuresthe instantaneous strain that the beam
experienced. Hence with results to the effect
explain above, the strain gauges read the in strain experienced
by the beam.
d) Briefly explain why the NA is not at 90 degrees to the
applied load.A neutral axis (NA) is defined as the axial that no
tension or compression is
experienced. In symmetrical bending, the neutral axis will lie
on the axis of symmetry.
For our experiment, the beam is not a symmetrical shape and
hence the neutral axis
will shift depending on the load applied to obtain an imaginary
axis where the forces
are zero.
In our experiment, the applied load is tension forces in both X
and Y direction, which
are perpendicular. Hence the neutral axis cannot be
perpendicular to Px or Py or the X
or Y direction. Depending on the magnitude of load applied, the
NA will vary it
inclination. Therefore, the NA will not be perpendicular to the
load applied.
Case Formula Mx/Nm My/Nm
1 Mx = Py X d
My= Px X d
d=0.33m
0 -49.5
2 -33 -49.5
3 0 33.18
4 -33.18 33.18
-
8/3/2019 P3.1 Report
14/30
Page | 14
e) For the cases, determine the theoretical strain distribution
at the strain gauged section.By means of a graph, compare the
theoretical strain distribution with the experimental
strain distribution.
Table B Experimental Vs Theoretical Strain Data
Graph A Experimental Vs Theoretical Strain
By comparison through the graph plotted, there are slight
discrepancies between the
experimental and theoretical strain among the four cases, which
is probably the results
of experimental errors.
As observed there is a common trend in all cases. As strain
value increases the error
increases. This may be due to the experimental set-up where the
load is applied
indirectly onto the beam, through a load cell for Frame A and
dead weights for Frame
B. Hence, there might be a slight misalignment in load passing
exactly through the
shear centre, resulting in negligible torsion of the beam. The
torsion experience is
highest at Corner C and Gauge No. 9, however the formula used to
calculate
theoretical strain does not take torsional stress into
consideration and hence results inhigher error between theoretical
and experimental strain values.
Formula
=( + ) ( + )
( ) Experimental Theoretical
Gauge
No
(horizontal)
(vertical)
(horizontal)
(vertical)
Case 1 1/C -190 -217 -254 -254
9 357 80 411 94
Case 2 1/C -281 -300 -366 -366
9 386 189 454 218
Case 3 1/C 109 127 171 171
9 -217 -52 -275 -63
Case 4 1/C 25 53 58 589 -205 67 -232 61
-1000
-500
0
500
1000
1500
1 2 3 4
Cases
Thoeretical vs Experiemental
Theoretical Horizontal
Gauge 9
ExperimentalHortizontal Gauge 9
Theoretical Vertical
Gauge 9
Experimental Vertical
Gauge 9
-
8/3/2019 P3.1 Report
15/30
Page | 15
f) For all cases, plot the experimental strain distributions
across the beam section andhence, estimate the positions of the
neutral axes. Compare the theoretical angles
between the neutral axis and x-axis with those of the
experiment.
Kindly refer to the group experimental strain distributions
across the beam section
graphs as attached.
Experimental () Theoretical () Difference ()
Formula tan = * Ixtan Ixy +/* Iy Ixytan +
Case 1 70 71 1
Case 2 63 64 1
Case 3 71 71 0
Case 4 97 91 6
Table B Experimental and Theoretical Neutral Angle,
The theoretical angle were calculated through the formula,
=
=
Sample Calculation, for Case 1,
= =
= = =
+ +
= As seen in Table B, the experimental angles between the
neutral axis and x-axies with
those of experiments have minimum errors. Hence the results
prove that the
generalized bending equations are accurate enough for
engineering calculation andprediction of neutral axis in
unsymmetrical bending.
Case 4 difference is of a greater value maybe due to the
experimental strain plots is
too huge to fit into an A4 graph paper and the team estimated
the neutral axis through
projection.
g) Explain why unsymmetrical sections are preferred in certain
structural design.By using unsymmetrical sections, the structural
weight can be minimized, and in any
structural design, lighter weight is always preferred.
A symmetrical beam has a fixed neutral axis and is perpendicular
to the applied load.
This means if there is a second loading perpendicularly to the
first loading, will results
in stresses in two directions, with one of it is acting at the
cross-section of the weakest
tolerance. On the other hand the neutral axis of an
unsymmetrical beam is dependent
on the loads applied which results in no particular
cross-section that are significantly
vulnerable to applied stresses. Therefore, while using
unsymmetrical section the main
concern will be designing it such that applied stresses act
along the shear centre to
prevent torsion.
-
8/3/2019 P3.1 Report
16/30
Page | 16
CONCLUSION
The experiment had demonstrated the behaviours of loadings and
bending moments with
respect to an unsymmetrical beam section. Where the neutral axis
location is dependent
upon the loading the beam experiences which is the major
advantage of unsymmetrical
beams in structural construction.
6.3 Christophorus Felix Kusnadi
a. Without using the generalized bending equation, describe
briefly an alternativemethod of calculating the normal stresses at
any point on an unsymmetric
beam in bending.
First, we find the angle value of the principal axes by using
the formula:=
+
Principle axes determined from value
On the principal axes, we are allowed to apply the generalized
bending equation for
symmetrical bending on unsymmetrical bending case. The formula
is:
= +
Where Mu= M cos and Mv= M sin
We can simplify the above equation and come up with the final
equation:
=
=
Illustration of angle
-
8/3/2019 P3.1 Report
17/30
Page | 17
is the angle between neutral axis and u-axis.
b.For the existing experimental setup for frame A and B,
determine the expressionsfor Mx and My at the loading point, in
terms of applied loads.
Mx = 0.5 Px d
My = 0.5 Py d
Where Px and Py are forces in x and y direction while d value is
0.33 metres.
c. For Frame A only: When Px is loaded first, observe the load
reading of Py isapplied. Explain why?
When Pyis applied, the load reading of Px decreases. This is
caused by the unsymmetrical
section of the beam, which causes moment about the x-axis when
Py is applied. Thismoment is on the same direction as Px which
causes the gauge to measure less net force
acting in the x direction.
d.Briefly explain why the Neutral Axis is not at 90 degrees to
the applied load.Neutral axis is an axis on a beam which has no
longitudinal stresses or strains. On
symmetrical structures, neutral axis is perpendicular to the
applied load. However, our
case deals with unsymmetrical beam, hence the neutral axis will
not be perpendicular to
the applied load. It will form an certain angle depending on the
magnitude and direction of
loads applied.
e. For the case, determine the theoretical strain distribution
at the strain gaugedsection. By means of a graph, compare the
theoretical strain distribution with the
experimental strain distribution.
From the results obtained, we can observe that some differences
between the experimental
and theoretical results do exist. We can also see that the error
percentage of each and every
case is nearly the same. (Graph attached in Appendix). This
error may be caused by some
experimental errors such as machine errors, material errors
(since the beam had been used
for years of loading and unloading process), the strain gauges,
and also the weight. Theerrors for each case is mainly around
20%.
f. For all the case, plot the experimental strain distribution
across the beam sectionand hence, estimate the position of the
neutral axes. Compare the theoretical
angles between the neutral axis and x-axis with those of the
experiment.
We can refer to table 6 in the Results section of this report,
from the table we can see that
the theoretical and experimental values of angles between the
neutral axis and x-axis are
quite close. For case 1 and 2, there are 1 degree difference.
For case 3, the experimental
and theoretical results are exactly the same. Meanwhile for case
4, there is a 6 degreedifference between the theoretical and
experimental values. (Graph attached in Appendix).
-
8/3/2019 P3.1 Report
18/30
Page | 18
g. Explain why unsymmetrical sections are preferred in certain
structural designs.Unsymmetrical sections are preferred in certain
designs mainly because of its neutral axis
nature. In symmetrical beams, the neutral axis is perpendicular
to the applied load. Hence
if there is a second load which is perpendicular to the first
one, the stress caused by the
second load will act upon the weakest part of the beam. This is
dangerous since it cancause the beam to break. In unsymmetrical
beam, the neutral axis is not fixed, it depends
on the load combination, which makes it more preferred in
structures undergone multiple
load. Another advantage of using unsymmetrical beams is the
material usage. In some
cases, making unsymmetrical beam requires less materials than
the symmetrical one.
Conclusion
The experiment verified the generalized bending theory as the
theoretical results are quite
close to the experimental ones. From the experiment, we can also
observe the behaviour of
unsymmetrical beams undergone various loads. We also observe
that for anunsymmetrical beam, its neutral axis position depends on
the loading experienced. Using
unsymmetrical beams can be advantageous in building a structure
undergoing various
loads.
6.4 Goh Zhi Wei
a) Alternative method of calculating the normal stresses
Besides using the generalized bending equation, another method
of calculation for the
normal stresses at any point on an unsymmetric beam in bending
is by moving the cross-
section of the beam to the principle centroidal axis of the beam
when doing the calculation.
Using this method, the IXY will be equal to zero as the
principal centroidal axis of the
cross-section will become the X and Y axes. Hence, the neutral
axis of the cross-section
will be the same as the axis of the couple M representing the
forces acting on the section,
provided that the couple vector M is directed along one of the
principal centroidal axes of
the cross-section.
Therefore, the whole calculation for the normal stresses at any
point on an unsymmetric
beam in bending can be simplified to using the same method that
is used when calculating
for the normal stresses on a symmetric beam in bending. In this
simplified calculation, the
principal axes of the cross-section can be determined by using
Mohrs circle oranalytically.
b) For the existing experimental setup for Frame A and B,
determine the expressions for Mx
and My at the loading point, in terms of applied load.
For both Frame A and B, the expressions for Mx and My are:
Mx = Py/2 x 330mm
My = Px/2 x 330mm
-
8/3/2019 P3.1 Report
19/30
Page | 19
c) In Frame A, load readings of PX changes when PY is
applied.
When PX is loaded first, the load readings of PX will decrease
when PY is subsequently
applied to the frame. This is because when PY is applied, there
will be a bending moment
created by PY, this causes the frame to bend in a way that will
move it away from the
pressure point of load PX. Therefore, the pressure cause by load
PX on the measuringgauge is reduced, resulting in a lesser than
actual load reading of PX.
d) Neutral Axis (NA) is not at 90 degrees to the applied
load
A Neutral Axis (NA) is defined as the axial does not undergoes
any tension or
compression. The neutral axis will only be at 90 degrees to the
applied load only if the
load is being applied to a symmetrical frame, as the neutral
axis will lie on one of the axis
of symmetry of the frame.
However, in this experiment, the load is applied on an
unsymmetric beam. Hence, the
neutral axis will change depending on the load applied in order
to obtain the imaginary
axial axis where the forces are zero. When the applied load is
added onto the unsymmetricframe, the forces on the load will result
in tension forces in both the X and Y directions.
This will result in the neutral axial not being at 90 degrees to
either applied loads PX or PY,
but at an inclination to the loads.
e) For the cases, determine the theoretical strain distribution
at the strain gauged section. By
means of a graph, compare the theoretical strain distributions
with the experimental strain
distributions.
The theoretical strain distribution at the strain gauged section
is shown in the table below.
From the 4 graphs for the 4 cases (attached to appendix), it can
be seen that there are slight
differences between the theoretical and experimental strain
values showed in all the cases.
This might be due to the experimental errors that happen during
the experiment, like the
wear and tear of the equipments used and the initialisation of
the data logger is not done
properly.
From all the 4 cases, it can be noted that there are similar
pattern in that the differences are
larger when the strain values increase. This is because the load
is applied at the bracket
that is attached to the beam. This causes slight misalignment in
that the load does not passexactly through the shear centre,
resulting in, close to negligible, torsion of the beam.
Form
ula = ( + ) ( + )( )
Experimental TheoreticalGaug
e No
(horizont
al)
(vertica
l)
(horizon
tal)
(verti
cal)
Cas
e 1
1/C -190 -217 -254 -254
9 357 80 411 94
Cas
e 2
1/C -281 -300 -366 -366
9 386 189 454 218
Cas
e 3
1/C 109 127 171 171
9 -217 -52 -275 -63Cas
e 4
1/C 25 53 58 58
9 -205 67 -232 61
-
8/3/2019 P3.1 Report
20/30
Page | 20
Since the torsional stress is the greatest at the two ends of
the cross sectional area of the
beam and that the method to calculate the theoretical strain
does not take torsional stress
into account, the differences in the strain values will be the
largest at the two ends of the
frame.
f)
For all cases, plot the experimental strain distributions across
the beam section andhence, estimate the positions of the neutral
axes. Compare the theoretical angles
between the neutral axis and x-axis with those of the
experiment.
Determination of experimental neutral axes:
For all the 4 cases, the graphs of experimental strain
distributions across the beam section
are plotted. Then, the neutral axes are being drawn by drawing a
straight line through 2
points, 1st
point is where the horizontal experimental strain line cuts the
x-axis and the 2nd
point is where the vertical experimental strain line cuts the
y-axis. All the graphs are
attached in the appendix.
Comparison of the theoretical neutral angles with the
experimental neutral angles:
The experimental neutral angles are found by measuring the
angles that the experimental
neutral axes make with the x-axis.
The theoretical neutral angles are found by using the following
method shown below:
By using the Generalized Bending Equation:
= ( + )( + ) =
= = + +
Hence, we are able to find the value of, which is the
theoretical neutral angle, from the
equation above.
The drawing of the neutral angles and the calculations of the
neutral angles can be found
in the appendix.
The values of both the theoretical and experimental neutral
angles are shown in the table
below:
From the table above, it can be seen that there are only a
slight differences between all the
Case Theoretical () Experimental () Difference ()
1 71 70 1
2 65 63 2
3 71 71 0
4 91 97 6
-
8/3/2019 P3.1 Report
21/30
Page | 21
theoretical neutral angles and the experimental neutral angles.
Therefore, it can be
concluded that the generalized bending equations are accurate
enough to do the
engineering calculations and predictions of neutral axes in
unsymmetrical bending.
However, in case 4, there is a larger difference of 6 between
the angles. This may be due
to the vertical experimental strain in case 4 being too
vertical, close to 90and so its
intercept with the y-axis are too high up to be able to fit and
draw in a normal A4 size
graph paper. Hence, the experimental neutral axis for case 4 is
drawn by an estimated
projection of the line resulting in a not so accurate drawing of
the neutral axis.
g) Explain why unsymmetrical sections are preferred in certain
structural design. By using unsymmetrical sections, the structural
weight can be minimized and this is useful
for structural designs that require the structural weight to be
as light as possible.
Furthermore, a symmetrical beam is at 90o
to the applied load and has a fixed neutral axis.
This will result in a risk that if a second load is applied
perpendicularly to the first loading,
it will result in stresses occurring in two directions with one
of it acting on the cross-section that has the weakest tolerance.
However, an unsymmetrical beam will not have this
problem. As the neutral axis for unsymmetrical beam is dependent
on the loads that are
applied to it, there will not be a particular cross-section that
is significantly vulnerable to
the applied stresses.
Conclusion
In this experiment, the behaviours of bending moments and
loadings undergo by an
unsymmetrical beam section is being demonstrated. It can also be
concluded that as the
position of the neutral axis is dependent on the loadings that
are applied on the beam, the
use of unsymmetrical beams can be preferred in certain
structural designs.
6.5 Jayakrishnan Radhakrishnan
a) Without using the generalized bending equation, describe
briefly an alternative method
of calculating the normal stresses at any point on an
unsymmetric beam in bending.
Previously, it is derived that tan2 = 2
. By knowing the value of the angle , the
principal axis is determined with respect to the reference axis
as shown in the figure below.
-
8/3/2019 P3.1 Report
22/30
Page | 22
While considering an unsymmetric loading, when the principal
axis is defined, the general
stress equation for a symmetrical loading = [
] + [
] is applied.
Therefore, alternative equation to find the normal stress is
given by:
= [ ] + [ ], where Mu= M cos and Mv= M sin
At the neutral axis, the bending stress will be zero.
= [
] + [
] = 0
which gives, () = - (
) tan = tan .
is the angle of position of the neutral axis with reference from
the U-axis.
b) For the existing experimental setup for frame A and B,
determine the expressions forMx and My at the loading point, in
terms of applied loads.
Mx = (Py/2) x 330mm My = (Px/2) x 330mmc) For Frame A only: When
Px is loaded first, observe the load readings of Px when Py is
applied. Explain why?
When Py is applied, the reading of Px is reduced. It results
from the moment created when
Py is applied. Therefore the resultant moment is increased in
the direction where Px is
applied causing the beam to bend in that direction. This leads
to the decrease in the
reading of Px.
d) Briefly explain why the NA is not at 90 degrees to the
applied load.
A neutral axis is at 90 degrees to an applied load only when the
frame is symmetrical
where the neutral axis is located at the axis of symmetry. The
frame is unsymmetrical for
this experiment. As a result its location cannot be at the axis
of symmetry. When the loads
are applied on the frame, it causes tension in the direction
where Px and Py is applied. As
a result the neutral axis will not be at 90 degrees for both the
loads.
-
8/3/2019 P3.1 Report
23/30
Page | 23
e) For the cases, determine the theoretical strain distribution
at the strain gauged section.
By means of a graph, compare the theoretical strain distribution
with the experimental
strain distribution.
FRAME A B
LOADCASE 1 2 3 4
Px(N,kg) 300 300 -20.50
-
20.50
Py(N,Kg( 0 200 0 20.5
Formulae My= Px X d
d= 330 mm Mx = Py X d
My (N,m) -49.5 -49.5 33.18 33.18
Mx (N,m) 0 -33 0
-
33.18
Theoretical Strain ( )
Formula = * (MxIy + MyIxy)y - (MyIx + MxIxy)y]/( IxIy - Ixy
2)
I mm4
Ix = 208.8 x103
Iy = 73.64
x103
Ixy =
72.74x103
Location (horizontal frame) (vertical frame)
Case 1 2 3 4 1 2 3 4
Gauge No
9 x = 36.21mm y = 23.23mm
x = -
9.29mm
y = -
45.17mm
Strain 411 454 -275
-
232 94 218 -63 61
Corner C x = -9.29mm y = 23.23mm
Strain -254 -366 171 58
-
254
-
366 171 58
From the graph (attached to appendix), it is seen that the
theoretical and experimental
strain values showed a slight variation for all the cases. It is
probably due to the
experimental errors. While observing the data it is found that
as strain value increases, the
error also increases. It is due to the indirect application of
the loads to the beam causing
misalignment of the load passing through the shear centre. As a
result the torsion isnegligible at the beam. But the torsional
stress is greater at the two ends of the beam. As
-
8/3/2019 P3.1 Report
24/30
Page | 24
the torsional stress is not considered for the calculation of
theoretical strain, it leads to an
increase in the error between the theoretical strain and the
experimental strain values.
f) For all cases, plot the experimental strain distributions
across the beam section and
hence, estimate the positions of the neutral axes. Compare the
theoretical angles between
the neutral axis and x-axis with those of the experiment.
From the bending equation,
= ( + )( + ) =
Therefore, the theoretical angle,
= ( + )( + ) =tan
The theoretical and experimental angles between the neutral axis
and x-axis are stated
below:
Case
Theoretical
angle, Experimental
angle, 1 71 70
2 65 63
3 71 71
4 91 97
For Case 1 and 3, it is observed that there is not much
deflection in Theoretical and
Experimental angles. For Case 2 and 4, a variation of 2 and 6
are found respectively. But
for Case 4, the neutral axis and the vertical frame strain
values are projected in order to get
theoretical and experimental values.
g) Explain why unsymmetrical sections are preferred in certain
structural designs
Unsymmetrical sections can reduce the weight. It is preferred in
considering the design of
a structure.
Conclusion
From this experiment it is found out the influence of loadings
as well as the bending
moments for an unsymmetrical section of beam. It is also seen
that the neutral axis
depends on the loading of the unsymmetrical beam. It is an
advantage in structuralconstruction by using unsymmetrical
beams.
-
8/3/2019 P3.1 Report
25/30
Page | 25
REFERENCES
Boresi, P Arthur and Richard J. Schmidt. Advanced Mechanics of
Materials, Oxford
University Press, 2003.
Ambrose, J., Patrick T.. Simplified Design of Concrete
Structures, Wiley, 2007.
-
8/3/2019 P3.1 Report
26/30
Page | 26
Appendix A Sample Calculation
Diagram
Calculation of Centroid
= = = = = + =
= +
=
= + +
=
Calculation of Ix
=
+ +
+
=
=
+ +
+
=
= + = + =
A1
A
2
O O
O1 O1
-
8/3/2019 P3.1 Report
27/30
Page | 27
Calculation ofand
Case 1,
= =
= = =
+ +
=
Case 2,
= =
=
=
=
+
+
=
Case 3,
= =
= = =
+ +
=
Case 4,
= =
= = =
+ +
=
-
8/3/2019 P3.1 Report
28/30
Page | 28
Sample Calculation for Theoretical Strain
For example CASE 2
HORIZONTAL GAUGE NO 9
= ( + ) ( + )( )
Sub the following to the equation above:
= = = = = = = =
= =
=
=
= =
VERTICAL GAUGE NO 9
= ( + ) ( + )( )
Sub the following to the equation above:
= = = = = = = = = =
= =
= =
-
8/3/2019 P3.1 Report
29/30
-
8/3/2019 P3.1 Report
30/30
Appendix B Graphs
