Aim: The purpose of the five experiments is to measure the bending momentum of a beam

Aim:
The purpose
of the five experiments is to measure the bending momentum of a beam under several
loads that were applied at different locations plus measuring the deflection in
the beam. The aim of this experiment is to let ourselves awareof the effects on
internal forces of a bar through bending under two point loads by calculating
the deflection at middle of the bar.

Theory:
·
Strain gaugeis a device used to measure the.wikipedia.org/wiki/Deformation_(mechanics)#Strain” title=”Deformation (mechanics)”>strain of an object.
·
Normal
Force,
N: this force acts
perpendicular to the area and is developed whenever the external load tends to
push or pull the two segments of the body.
·
Shear
Force, V: this force is developed when
the external loads tend to cause the two segments of the body to slide over one
another.
·
Torque,
T: this effect is developed the external loads tend to twist one segment of the
body with respect to the other.
·
Bending
Moment,

M: this moment is caused by
the external loads that tend to bend the body about an axis within the plane of
the area.
·
A beam
is a horizontal.wikipedia.org/wiki/List_of_structural_elements” title=”List of structural elements”>structural
element
that is capable of withstanding.wikipedia.org/wiki/Structural_load” title=”Structural load”>load primarily by resisting.wikipedia.org/wiki/Bending” title=”Bending”>bending. The bending force induced into the
material of the beam as a result of the external loads, own weight,.wikipedia.org/wiki/Span_(architecture)” title=”Span (architecture)”>span and external reactions to these loads
is called a.wikipedia.org/wiki/Bending_moment” title=”Bending moment”>bending moment. Beams generally carry.wikipedia.org/wiki/Vertical_direction” title=”Vertical direction”>vertical.wikipedia.org/wiki/Gravitational” title=”Gravitational”>gravitational.wikipedia.org/wiki/Force” title=”Force”>forces but can also be used to carry.wikipedia.org/wiki/Horizontal_plane” title=”Horizontal plane”>horizontal loads (i.e., loads due to an.wikipedia.org/wiki/Earthquake” title=”Earthquake”>earthquake or wind). The loads carried by a beam
are transferred to.wikipedia.org/wiki/Column” title=”Column”>columns,.wikipedia.org/wiki/Wall” title=”Wall”>walls, or.wikipedia.org/wiki/Girder” title=”Girder”>girders, which then transfer the force to adjacent structural.wikipedia.org/wiki/Compression_member” title=”Compression member”>compression
members.
·
The bending moment
is the amount of bending that occurs in
a beam and the bending moment at any point along the beam is equal to the area
under the shear force diagram up to that point.
·
The moment of inertia
of
an object about a given axis describes how difficult it is to change its
angular motion about that axis. Therefore, it encompasses not just how much
mass the object has overall, but how far each bit of mass is from the axis. The
farther out the object’s mass is, the more rotational inertia the object has,
and the more force is required to change its rotation rate.
·
Stress in a beamsexperience.wikipedia.org/wiki/Compression_(physical)” title=”Compression (physical)”>compressive,.wikipedia.org/wiki/Tensile” title=”Tensile”>tensile and.wikipedia.org/wiki/Shear_stress” title=”Shear stress”>shear stresses as a result of the loads
applied to them. Typically, under gravity loads, the original length of the
beam is slightly reduced to enclose a smaller radius arc at the top of the
beam, resulting in compression, while the same original beam length at the
bottom of the beam is slightly stretched to enclose a larger radius arc, and so
is under tension. The same original length of the middle of the beam, generally
halfway between the top and bottom, is the same as the radial arc of bending,
and so it is under neither compression nor tension, and defines the neutral
axis.
·
The Forceis the component, perpendicular
to the surface of contact, of the.wikipedia.org/wiki/Contact_force” title=”Contact force”>contact force
exerted on an object by. The normal force is one of the components of the.wikipedia.org/wiki/Ground_reaction_force” title=”Ground reaction force”>ground reaction force and may coincide with it, for example considering
a person standing still on the ground, in which case the ground reaction force
reduces to the normal force. In another common situation, if an object hits a
surface with some speed, and the surface can withstand it, the normal force
provides for a rapid deceleration, which will depend on the flexibility of the
surface.
·
Shear force
is the force in the beam acting
perpendicular to its longitudinal (x) axis. For design purposes, the beam’s
ability to resist shear force is more important than its ability to resist an
axial force. Axial force is the force in the beam acting parallel to the
longitudinal axis.
Method:
vExperiment 1: Bending under a point
load:
First
of all, the steel bar has been selected that is equipped with strain sensors.
Then, the dimensions of the bar were measured and recorded. After that the
elastic modulus of the bar has been found from relevant references and
recorded. The bending moment inertia of the bar was calculate and recorded.
Next, the bar has insert on the supports of the experiment apparatus such that
the sensors are located in the middle of the bar and the strain gauges was set
to zero. After that, a point load of 500 g at 10 cm from one end of the bar was
applied as shown. The net load is the total load of the point load + the arm.
Then, the deflection at the middle of the bar has measured by using strain
gauge once and by using manual gauge. Lastly, the values were compared to find
the scale factor of the strain gauge as indicated.
.gif”>

vExperiment 2: Bending under two
point loads:
Firstly, the steel bar equipped with strain gauges was
selected. Then, the bar was put on the two supports of the experiment apparatus
such that the sensors are located in the middle of the bar. Then, the strain
gauge was set to zero. Next, the two point loads of 500 g at 10 cm were applied
from each end of the bar, as shown. After that, the deflection of the bar was
measured at its middle, by using manual gauge and recorded. Then, the scale
factor of the strain gauge and the actual deflection of the bar were obtained.
.gif”>
vExperiment 3: Bending under a
distributed load:
First,
the steel bar equipped with strain gauges was selected and the specifications
of bar applied were recorded. Then, the bar on the two supports of the
experiment apparatus was inserting such that the sensors are located in the
middle of the bar. Then, the strain gauge was set to zero. Next, a small block
on the bar was inserting and then applies a 500 g load on the block to mimic a
distributed load, as shown. After that, the deflection of the bar at its middle
was measured by using strain gauge and recorded. Then, the scale factor applied
which calculated in experiment 1 to determine the actual deflection. Then, the length
of the block was measured and recorded. Furthermore, the distributed load was
increased by 500 g intervals up to 2000 g and the steps were repeated and
recorded.
.gif”>

vExperiment 4: Bending under two
distributed loads:
Firstly,
the steel bar was selected equipped with strain gauges. Then, the bar on the
two supports was putting of the experiment apparatus such that the sensors are
located in the middle of the bar. Next, the straingauge was set to zero and the
two small blocks was putting on the bar as shown. After that, a 500 g load were
put ting on each block to mimic two distributed loads. Then the deflection of
the bar at its middle by using strain gauge were measured and recorded. After
that, the scale factor calculated in experiment 1 was applied to determine the
actual deflection.
.gif”>

vExperiment 5: Bending of a T-beam:
Firstly, the beam and load cell are
properly aligned. Turn the thumbwheel on the load cell counter clockwise to
apply a positive (down-ward). Set the digital force display to zero. Preload
the beam if about 100N and again set the digital display to zero. Next, take
the nine zero strain readings by choosing the number with the selector switch.
Fill the table with the zero force values. Then, increase the load to 100N and
note all nine of the strain readings. Report the procedure in 100N increments
to 500N. Finally gradually release the load and preload.

.jpg” alt=”Macintosh hd:users:abdullah:desktop:screen shot 2012-08-12 at 8.04.47 pm.png”>

.gif” alt=”Macintosh hd:users:abdullah:desktop:screen shot 2012-08-12 at 8.05.36 pm.png”>
Results:
vExperiment 1: Bending under a point
load:
Free body diagram:

.gif”>

1) The dimensions of the bar:Cross
section=594 mm.

2) The elastic modulus for reference =Elastic
modulus=200Gpa

3)
The deflection at the middle of the bar, using the gauge=55.3
mm/100.

4)
The deflection at the middle of the bar, using the strain
gauge system.=24mm

5)
The scale factor is

Scale factor strain gauge =.gif”> = 55.3/31 = 1.78

6)
The actual deflection is

Deflection actual= Deflection strain gauge×
Scale factor strain gauge = 31 x 1.78 = 55.2mm

7)
The bending moment of inertia is

I= (Bh3)/12
I= (24×43)/12
I=128 mm4

8)
The shear forces of the supports

740g
.gif”> 10cm

.gif”>

.gif”>

FyA 600mm FyB

∑Fy=0 i.e. F1
+ F2 = (740/1000) 9.81
F1=7.259-
F2……..(i)

∑Ma=
0 i.e.
0.74 x 9.81 x 0.1 – F2 x 0.594 = 0
F2=1.22N
Putting
FyB in (i)

F1=6.05N

9)
The shear diagram of the bar:

.gif”>

10)
The moment diagram of the bar:
.gif”>

11)
The deflection:

Deflection for
a three point bending is =[w L1x(L2-L12-x2]/6EIL
=7.259×0.1×0.296(0.5942-0.12-0.2962)/6x200x109x128x10-12×0.594

= 0.00060104m = 0.601mm

12)
Compare the measured and
calculated deflections and discuss about the error sources

Error= (55.3-55)/55 =5.45%
The deflection values of the calculated and measured
appear to be different than the errors we neglected while measuring the
deflection which include:
·
Correct positioning
of the gauge.
·
Correct point of
application of the load.
·
Exact point of
supports.
Thus there may be always a difference in the values
of measured and calculated values of deflection.
13)
Calculate the maximum
bending stress

σ= Mc/I
= (0.605 x0.002)/128×10-12
=9.45Mpa

vExperiment 2: Bending under two
point loads:
Free body diagram:

.gif” alt=”Text box: w1″>

.gif” alt=”Text box: w2″>

.gif”>

.gif” alt=”Text box: f1″>

.gif” alt=”Text box: f2″>

1)
The steel bar equipped with strain sensors has selected.

2)
The bar on the two supports of the experiment apparatus such
that the sensors are located in the middle of the bar was putting and the
strain gauge were set to zero.

3)
Two point loads of 500 g at 10 cm from each end of the
bar as shown below were applied.

·
NOW;

4) The deflection of
the bar at its middle using manual gauges =93mm/100

5)
The deflectionof the bar using strain gauge =60 mm/100

6)
The deflection at the middle of the bar, using the strain
gauge system.

Scale factor strain gauge =.gif”> = 93/ 60 = 1.55

7)
The found scale factor used to calculate the actual
deflection :

Deflectionactual= Deflectionstrain gauge×
Scale factorstrain gauge

=
60 x 1.55 = 92.8 mm/100

8)
Calculate the shear forces of the supports.

.gif”>

.gif”>

.gif”>.gif”> 10cm 740gm 740gm 10cm

.gif”>

.gif”>

FyA FyB
∑Fy=0 i.e.,
F1 + F2 = (0.74+0.74) 9.81
F1=14.52- F2…
(i)

∑Ma= 0
i.e. 0.74 x 9.81 x 0.1 + 0.74 x 9.81×0.5
– F2 x 0.592 = 0

F2 = 7.35 N

Putting
the value ofF2
in Eq (i)
F1=7.16N

9) The shear diagram of the bar:

.gif”>

10)
The
moment diagram of the bar:
.gif”>

11)
The
deflection at the middle of the bar:
Deflection for a three point bending
is =-[pbx(L2-b2-x2)]/6EIL
=26.5mm
Total deflection =26.5+44.3
=117.54
Deflection error =
[(117.54-92.8)/117.54] x100
=21%

12)
The
maximum bending stress:
σ= Mc/I

= (0.726×0.002)/ 128×10-12
=8.18 Mpa

vExperiment 3: Bending under a
distributed load:
Free body diagram:

.gif”>

1) The
deflection at the middle of the bar, using the strain gauge system= 27mm/100

2) The
scale factor calculated in experiment 1 to determine the actual deflection
Deflection actual = Deflection
strain gauge × Scale factor strain gauge
=27
x1.78=48.06mm/100
3)
The length of the
block= 52mm

Distributed
load(g)

Load Density(N/m)
=load/block
length

Measured
deflection
(mm/100)

Actual
deflection=measured deflection x scale factor

Estimated
deflection

620

11.92

26

48.88

-48.58

1120

21.53

46

86.48

-87.75

1620

31.15

68

127.84

-126.93

2120

40.76

87

163.56

-166.11

4)
Calculate the
shear forces of the supports at a load of 1000g.

.gif”>.gif”>10cm 11.04N

.gif”>.gif”>.gif”>.gif”>.gif”>.gif”>.gif”>7.35cm

.gif”>

.gif”>

.gif”>.gif”> 60cm
FyA FyB
∑Fy=0 i.e.,
F1 + F2 =(1.120×9.81)
F1=10.98- F2…………(i)
∑Ma= 0 i.e. (10.98×0.1)- FyB
x0.594=0
F2=1.86N
Putting
the value of FyB in (i)
F1=9.12N
5)
The shear diagram
of the bar at load of 1000g:
.gif”>
6)
The moment diagram of the bar at 1000gm:

.gif”>

.jpg” alt=”Macintosh hd:users:abdullah:desktop:screen shot 2012-08-12 at 7.59.26 pm.png”>
vExperiment 4: Bending under two
distributed loads:
Free body diagram:

.gif”>

F1

F2

1)
Two small blocks on the bar put.

2)
The deflection of the bar at the middle using strain gauge=50mm/100

3)
The scale factor calculated in experiment 1 to determine
the actual deflection

Deflection actual = Deflection
strain gauge × Scale factor strain gauge
= 50 x 1.76 = 88.2

7)
The shear forces of
the supports.

10cm .gif”>.gif”>.gif”>.gif”>.gif”>.gif”>.gif”>.gif”>.gif”>10cm
.gif”>.gif”>.gif”>.gif”>7.6cm

.gif”>.gif”>.gif”>.gif”>.gif”>.gif”>
.gif”>.gif”>
60cm
.gif”>

FyAFyB

∑Fy=0
i.e., F1 + F2
= (0.620+0.620) 9.81
F1=12.164-
F2……………… (i)

Now
as the system is symmetric so without further calculations
F1=F2=6.082N
8)
The shear diagram of
the bar:

.gif”>

9)
The moment diagram of the bar:

.gif”>

10)
The deflection at
the middle of the bar =72mm

11)
Compare your actual and estimated deflection:

% error =[(72.58-110.92)/110.92]x100=13%

11- The maximum bending stress of the bar:
σ= Mc/I
= (0.608×0.002)/ 177.5×10-12
=15.077Mpa

.gif” alt=”Macintosh hd:users:abdullah:desktop:screen shot 2012-08-12 at 7.58.08 pm.png”>

v
Experiment
5: Bending of a T-beam:
Moment:
= W/2 * 0.35= 100/2*0.35=17.5 N.m
= W/2 * 0.35= 200/2*0.35= 35 N.m
=
W/2 * 0.35= 300/2*0.35=52.5 N.m
=
W/2 * 0.35= 400/2*0.35=70 N.m
=
W/2 * 0.35= 500/2*0.35=87.5 N.m

L
O A D (N)

Gauge No

0

100

200

300

400

500

1

-077

-192

-303

-410

-520

-623

2

-081

-158

-236

-310

-385

-456

3

-060

-142

-218

-295

-370

-443

4

-003

-015

-029

-043

-056

-069

5

+003

-011

-025

-040

-053

-067

6

+005

+031

+050

+071

+091

+113

7

+029

+053

+076

+097

+121

+141

8

+042

+096

+142

+189

+237

+284

9

+051

+105

+154

+202

+253

+299

BENDING MOMENT (N.m)

Gauge No

0

17.5

35

52.5

70

87.5

1

0

-115

-226

-333

-443

-546

2

0

-077

-155

-229

-304

-375

3

0

-082

-158

-235

-310

-383

4

0

-012

-026

-040

-053

-066

5

0

-014

-028

-043

-056

-070

6

0

+026

+045

+061

+086

+108

7

0

+024

+047

+068

+092

+112

8

0

+054

+100

+147

+195

+242

9

0

+054

+100

+151

+202

+248

.gif” alt=”Macintosh hd:users:abdullah:desktop:screen shot 2012-08-12 at 8.34.46 pm.png”>

FORCE
(N)

Gauge no

0

17.5

35

52.5

70

87.5

1

-115

-226

-333

-443

-546

2,3

-079.5

-156.5

-232

-307

-397

4,5

-013

-027

-041.5

-054.5

-068

6,7

+025

+046

+064.5

+089

+110

8,9

+054

+100

+149

+198.5

+245

.jpg” alt=”Macintosh hd:users:abdullah:desktop:screen shot 2012-08-12 at 8.35.02 pm.png”>

Based on the
gradients from each of the lines, we make the x values = 0, the y intercepts are:

26.316+
26.313+ 26.390+ 26.317+ 25.963 / 5 = 26.2598

Therefore the
position of the neutral axis via the graph is 26.2598.

Shape

Area

y

Y(y*A)

1

2.4384*10-4

34.9

8.510016*10-6

2

20288*10-4

15.85

3.215648*10-6

total

4.4672*10-4

1.1725664*10-5

.jpg” alt=”Macintosh hd:users:abdullah:desktop:screen shot 2012-08-12 at 8.49.14 pm.png”>

Shape

D(mm)

b(mm)

h(mm)

I

D2

A(mm2)

1

8.652

38.1

6.4

832.3072

74.85710

243.84

2

10.398

6.4

31.7

16989.340

108.1184

202.88

I=I1+A1*d12+I2+A2*d22
= 58009.865mm4 = 0.058m4

Gauge

Neutral Axis

Theoretical (Mpa)

Experimental (Mpa)

Error (%)

1

26.598

40.126

38.22

4.75

2,3

18.598

28.057

26.53

5.44

4,5

3.598

5.428

4.76

12.31

6,7

5.102

7.697

7.7

0.04

8,9

11.502

17.352

17.15

1.16

Theoretical
stress: .jpg” alt=”Macintosh hd:users:abdullah:desktop:screen shot 2012-08-12 at 9.07.14 pm.png”>
1 = 87.5
*0.026598 / 0.058 = 40.1 Mpa
2,3
= 87.5 * 0.018598 / 0.058
= 28.057 Mpa
4,5
= 87.5 * 0.003598 / 0.058
= 5.428 Mpa
6,7
= 87.5 * 0.005102 / 0.058
= 7.697 Mpa
8,9
= 87.5 * 0.011502 / 0.058
= 17.352 Mpa
Experimental
Stress.jpg” alt=”Macintosh hd:users:abdullah:desktop:screen shot 2012-08-12 at 9.07.25 pm.png”>
1 = 70 *109 * 546 * 10-6
= 38.22MPa
2,3 = 70 *109 * 379 * 10-6
= 26.53MPa
4,5 = 70 *109 * 68 *
10-6 = 4.76MPa
6,7 = 70 *109 * 110 * 10-6
= 7.7MPa
8,9 = 70 *109 * 245 * 10-6
= 17.15MPa

Discussion
Experiment
1: Bending under a point load:
The deflection
of a beam depends in the length, cross-sectional shape and the material, where
the deflecting force is applied, and how the beam is supported. In addition,
the shear force is the force act on the beam and the bending moment is the
moment where the bending effect of any load is. While the strain gages
measuring the strain require, bonding adhesive, protective coating or coatings
and lead-wires to construct a sensor system. Based on the experiment result, it
shown that shear forces of the supports is F1=6.14N and F2=1.12N. Although, it
can be seen from figure1.1 that the shear force start with a constant line and
suddenly it droop and stay constants. However, figure1.2 illustrates the
bending moment and the graph is not linear. Furthermore, the deflection has
calculated as shown in the result. However, the relative error of the
deflection as indicated in table 1 and the error sources are human error while
taking the measurement or calculating the result.
v
Experiment
2: Bending under two point loads:
The
effects of two point loads have considered on the bar and from the experiment
found out the deflection of strain gauge are 58. The calculated deflection is
0.8062 mm. From the shear force graph, it can be seen that it is symmetric
about x-axis and at x = 0. It is constant till x = 100 mm and after that it
suddenly drops to 0 till x = 550 mm. after that it becomes negative. However,
from the moment graph it can seen that bending moment increases constantly from
x = 0 to the first point load. Then it remains constant until the next point
load. After that it decreases uniformly till zero and the maximum stress came
out to be 0.0103 GPa.

v
Experiment
3: Bending under a distributed load:
Here it can see that the
deflection by assuming a point load instead of the distributed load is coming
out to be a big difference. The reason could be general human error of noting
the correct values in decimals. Moreover, it can’t really mimic a distributed
load as we generally do mistakes in correct application of point load. So, from
the shear graph it can see that shear force remains constant from the starting
point till x = 75 mm, the value being equal to F1. After that it drops
uniformly to a negative value till the last point of distributed load. Then it
becomes constant, though the value remains negative and the moment graph
indicates that the bending moment increases uniformly till x = 75 mm. the
bending moment is highest at this point. After that it decreases irregularly.

v
Experiment
4: Bending under two distributed loads:
Using the strain gauge
system, to measure the deflection at the middle due to strain gauge is to be
44. Both the internal forces F1 and F2 came out to be 6.076 N. this shows the
symmetricity of the system and this is evident from the shear graph. The total
area under this graph cancels out to zero. It was expecting the same with the
moment graph but it came out totally different. This could be because of the
human errors which were made during the experiment. One possibility is while
noting the deflections, the table in which the apparatus was placed might have
got a little nudge. When calculated the deflection by using hypothetical point
loads, the values turned out to be different than the measured ones. As a
result the relative error came around 16 %.
The maximum stress value was 8.094 MPa.

v
Experiment
5: Bending of a T-beam:
When we look at
the relationship between bending moment and strain we can see that as weget
further and further from the neutral axis the values are getting larger and
larger. So ourneutral axis was at 26.248mm from the top of the bar and
corresponds with the larger valuesfor strain appropriately as the sensor
measuring this point is the furthest of all the sensorspresent on the bar. And
the sensors at positions 4 and 5 had exactly the opposite of theseresults with
the strain values being the smallest as they were the closest to the neutral
axis ofany of the other sensors.

We also notice
that the strain gauge readings at opposite sides are very close to each other
orexactly the same readings. When we were reading these values we would be
expecting that these values would be identical values but they aren’t in some
cases. This could be due tovarious reasons such as deformation or imperfections
within the bar due to the manufacturingprocess or through damage to the T-bar
from use or transportation. The attachment of thesensors is in slightly the
wrong spot or not on the mirror image side of the T-bar. The barwas made out of
aluminium which is quite soft and easier to change shape when hit withsomething
that is harder. There could also be slight temperature differences from side to
side of the bar and this could also result in non-identical readings on either
side of the bar as the
aluminium is
not a good insulator of heat transfer.

When we take a
good look at the values in table 10 we can see that the theoretical values and
experimental values aren’t identical. This could be possible due to the
attachment of the sensors to the T-bar. It could also be the T- bar dimensions
aren’t exactly the same all along the total length of the T-bar and the sensors
are good enough to measure this difference in shape as we are assuming a
perfect T-bar shape according to the dimensions on the device when working out
the moment of inertia for the theoretical stress. We are also assuming the
Young’s modulus of the aluminium to 70 Å~109GPa which it might not be as
it’s not a perfect piece of aluminium bar. But from the values in our
experiment we were quite within a safe working tolerance if we were to apply
this to a working environmental case study.

Conclusion
In conclusion, the moment of inertia of an object about a
given axis describes how difficult it is to change its angular motion about
that axis and the bending moment is the amount of bending that occurs in a
beam. While the force is the component, perpendicular to the surface of
contact, of the.wikipedia.org/wiki/Contact_force” title=”Contact force”>contact force exerted on an object and stress in a
beams experience.wikipedia.org/wiki/Compression_(physical)” title=”Compression (physical)”>compressive,.wikipedia.org/wiki/Tensile” title=”Tensile”>tensile and.wikipedia.org/wiki/Shear_stress” title=”Shear stress”>shear stresses as a result of the loads applied to
them. The Strain gauge is a device used to measure the.wikipedia.org/wiki/Deformation_(mechanics)#Strain” title=”Deformation (mechanics)”>strain of an object. However, from the four
experiments it can be seen that almost all the experiments are related to each
other. Although in this experiment, successfully used equations of equilibrium
to find the internal forces acting on the beam and used the strain gauge system
to measure the deflections. Also, from the experiments that a point load can
act as a distributed load only in theoretical values not in practical use as we
are bound to make small errors during experiment whatever it may be. Lastly,
experiment concludes the stress analysis practical it understood the effects of
two distributed loads on a bar.

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