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EGME 306A The Beam
Page 1 of 18
Group 2
EXPERIMENT 3:The Beam
Group 2 Members:
Ahmed Shehab
Marvin Penaranda
Edwin Estrada
Chris May
Bader Alrwili
Paola Barcenas
Deadline Date: 10/23/2015
Submission Date: 10/23/2015
EGME 306A – UNIFIED LABORATORY
EGME 306A The Beam
Page 2 of 18
Group 2
Abstract (Bader):
The main objective for this experiment was to determine the
stress, deflection, and strain of a supported beam
under loading, and to experimentally verify the beam stress and
flexure formulas. Additionally, maximum
bending stress and maximum deflection were determined. To
accomplish this, a 1018 steel I-beam with a strain
gage bonded to the underside was utilized in conjunction with a
dial indicator to monitor beam deflection. In
order to determine the values for strain and deflection, the beam
underwent testing utilizing the MTS Tensile
Testing machine, which applied a controlled, incrementally
increasing load to the beam. This data was then
utilized along with calculations for the beams neutral axis,
moment of inertia, and section modulus to determine
the required objective values. Final values of 12,150 psi for the
maximum actual stress (vs. 12,784.8 psi for
theoretical stress), and 0.0138 in for the maximum actual
deflection (vs. .0130 in for theoretical deflection)
correlated closely with each other, and successfully verify
established beam stress and flexure formulas.
EGME 306A The Beam
Page 3 of 18
Group 2
Table of Contents:
List of Symbols and Units 4
Theory 5
Procedure and Experimental Set-up 8
Results 9
Sample Calculations and Error Analysis 12
Discussion and Conclusion 15
Bibliography 16
Appendix 17
EGME 306A The Beam
Page 4 of 18
Group 2
List of Symbols and Units (Chris):
List of Symbols and Units Name of variables (units) Units
� Stress psi
� Applied load lbf
� Moment of Inertia in.4
� Strain in/in
� Length of the bar in
Z Section Modulus of Beam in3
� Distance to Beam Neutral Axis in
� Modulus of Elasticity psi
EGME 306A The Beam
Page 5 of 18
Group 2
Theory (Edwin):
There are two main objectives for this experiment: to determine
maximum bending stress values in
the beam and to determine the deflection in the beam. To help
visualize this phenomena, imagine
cutting a section of a symmetrically loaded beam:
Now, examine diagrams of this section before (Fig. A) and after
bending (Fig. B):
(Fig. A)
(Fig. B)
The main points to take away from the above diagrams are as
follows: When the moment, M is applied
as shown in Fig. A, forces will be in compression near the top
(positive moment) and in tension near
the bottom (negative moment). The effects from this moment
are seen in Fig. B.
For determining max stress values, one concept to note is that
our bending moment M can help
calculate bending stress. First, we recall our basic definition of
normal strain:
� =
�′ − �
�
EGME 306A The Beam
Page 6 of 18
Group 2
Since the beam is bending, however, we will need to alter the
above formula by taking into account
the radius of curvature, r, and the differential angle, �. Since
there are locations where forces are
compressive and locations where forces are tensile, there must
be a location where forces are neither
compressive nor tensile. This axis is called the Neutral Axis and
is labeled in Fig. B as N-N. We can
now alter the above formula by measuring the strain at distance
+� from the neutral axis:
� =
�′ − �
�
=
(� + �)∆� − �∆�
�∆�
=
�
�
The strain equation from above can be converted to stress by
using Hooke’s Law:
� = �
�
�
Now, if we allow c to be the max distance from the neutral axis,
we can build upon the above
expression as follows:
� = ��
�
�
To obtain the beam stress formula, we still need to define where
the neutral axis is located. We can do
this by relating the radius of curvature and the bending moment,
which will be determined by applying
a moment equilibrium equation about the neutral axis:
�� = ∫ �
�
�� =
��
�
∫ �
�
�� = 0
Since moment due to all forces is summed up by the product of
their forces and the moment arm about
the neutral axis, we deduce:
� = ∫ ��� = ∫ ���� =
��
�
∫ �2��
We will then let I be defined as the second moment of area
about the neutral axis, commonly called
the moment of inertia:
� = ∫ �2��
Finally, we deduce the moment equation as:
�� =
��
�
=
�
�
Our final objective involved determining beam deflection.
Using similar methods from bending stress,
we will examine the relationship between the radius of
curvature, r and the moment, M, at any given
point on a beam. We begin with the radius of curvature for any
point of a function from calculus:
EGME 306A The Beam
Page 7 of 18
Group 2
�
�
=
�2�
��2
(1 + �
2�
��2
)
3
2
We will note that for small deflections, the first derivative or
slope is small and becomes even smaller
when it is squared and can therefore be neglected. This will
greatly simplify the radius of curvature
to:
1
�
=
�2�
��2
By combing the above equation with the bending stress
equation, we develop the standard moment
curvature equation:
��
�2�
��2
= �(�)
The differential equation will only be useful if we apply it to
the beam with specific boundaries, or
boundaries of integration. Integrating with M(x)=Px/2 for 0≤x≤a
andM(x)=Pa/2 for a≤x≤a+b, we get:
�(�) =
�
2��
(
�3
6
−
��(� + �)
2
) , ��� 0 ≤ � ≤ �
�(�) =
�
2��
(
�3
6
−
��(2� + �)
2
+
��2
2
) , ��� � ≤ � ≤ � + �
The max deflection at x = a + b/2 is:
−�� =
��
2��
(
�2
3
+
��
2
+
�2
8
)
Lastly, by allowing a=b=L/3:
−�� =
23��3
48��
=
23��3
1296��
EGME 306A The Beam
Page 8 of 18
Group 2
Procedure and Experimental Set-Up (Marvin):
This experiment requires the use of MTS Insight Tensile
Testing Machine, Beam Deflection Dial indicator,
digital caliper, a 1018 steel beam and two separate data
acquisition software. The experiment begins with the
measurement of the length and the cross-sectional dimensions of
the 1018 steel beam with the use of the digital
caliper. The beam is then carefully placed into the MTS
machine while making sure that the beam's black marks
are aligned with the roller supports of the bending fixture. Once
the beam is squarely nested on the roller
supports, TestWorks 4 software is accessed in order to zero the
load reading. Now that the parameters have been
set up, the MTS' handset is used to position the upper bending
fixture over the beam. While the crosshead
lowers, the digital load readout from the software is observed.
When the load increases as the upper loading
component of the fixture makes contact with the beam, the
fixture is raised until a preload of 0.2 lb is applied.
The Beam Deflection Dial indicator is positioned in the center
of the beam on the bottom side. Do note, the dial
indicator needs to be set to zero. Now, the LabVIEW software is
accessed. On the software, the strain indicator
is zeroed by pressing “zero strain”. The software also requires
the inputs dialog for the thickness and width, a
value of 0.125 is used for thickness and a value of 0.500 is used
for width. The MTS machine loads the beam
up to 1000 lb in increments of 100 lb. When the digital load
readout reaches approximately 100 lb, the “Pause”
button is pressed since the data readings need to be taken for
every 100 lbs. Pausing for every 100 lb increments
are continued up until 1000 lbs. The procedure is repeated again
since the average of the 2 sets of readings will
be used for experimental calculations. Keep in mind, the dial
indicator and the strain indicator needs to be re-
zeroed again for the next set of experiment.
Note: for a detailed procedure, see Appendix A.1
EGME 306A The Beam
Page 9 of 18
Group 2
Results (Chris):
-TEST 1 / TEST 2 - - AVERAGE OF TEST 1 & 2 -
LOAD
( lbf )
STRAIN
GAGE
ELONGATIO
N ( ε )
INDICATOR
DEFLECTIO
N (in.)
LOA
D
( lbf )
STRAIN
GAGE
ELONGATIO
N ( ε )
INDICATOR
DEFLECTIO
N (in.)
100/99 47/46 .0011/.0016 99.5 46.5 .00135
202/19
8
91/88 .0023/.0029 200 89.5 .0026
316/29
5
144/131 .004/.0041 305.5 137.5 .00405
396/39
5
179/174 .005/.0059 395.5 176.5 .00545
497/50
0
222/224 .0064/.0071 498.5 223 .00675
596/59
8
268/268 .008/.0087 597 268 .00835
695/69
6
312/310 .0093/.010 695.5 311 .00965
797/80
0
354/355 .0109/.0114 798.5 354.5 .01115
897/89
8
318/400 .0122/.0129 897.5 359 .01255
996/99
6
365/445 .0134/.0142 996 405 .0138
TABLE 1. Strain gauge, indicator, and load data
(reference Appendix for calculations)
EGME 306A The Beam
Page 10 of 18
Group 2
LOAD
( lbf )
ACTUAL VALUES / DEFLECTION
( in. )
THEORETICAL VALUES / DEFLECTION
( in. )
99.5 0.00135 0.0013
200 0.0026 0.0026
305.5 0.0041 0.0039
395.5 0.0055 0.0051
498.5 0.00675 0.0065
597 0.00835 0.0077
695.5 0.00965 0.0091
798.5 0.01115 0.0104
897.5 0.01255 0.0117
996 0.0138 0.013
TABLE 2. Actual and Theoretical Values for Beam Deflection
(reference Appendix for calculations)
Figure 2-1. Actual vs. Theoretical Deflection using dial
indicator and calculated values.
The Load vs. Deflection curve consists of the actual
experimental values obtained during the
experiment compared with theoretical calculated values to
measure deflection in relation to
applied load.
0
200
400
600
800
1000
1200
0 0.005 0.01 0.015
Lo
ad
(
lb
f )
Deflection ( in. )
Actual vs. Theoretical Deflection
Actual Deflection
Theoretical Deflection
Linear (Actual
Deflection)
Linear (Theoretical
Deflection)
EGME 306A The Beam
Page 11 of 18
Group 2
LOAD
( lbf )
ACTUAL STRESS
( psi )
THEORETICAL STRESS
( psi )
99.5 1,410 1,277.20
200 2,685 2,567.23
305.5 4,125 3,921.44
395.5 5,295 5,076.70
498.5 6,690 6,398.82
597 8,040 7,663.18
695.5 9,330 8,927.54
798.5 10,635 10,249.66
897.5 10,770 11,520.44
996 12,150 12,784.80
TABLE 3. Actual and Theoretical Values for Beam Stress
(reference Appendix for calculations)
Figure 2-1. Actual vs. Theoretical Stress curve using
experimental and calculated values.
The Load vs. Stress curve consists of the actual experimental
values obtained during the
experiment compared with theoretical calculated values to
measure stress in relation to applied
load.
0
200
400
600
800
1000
1200
0 5,000 10,000 15,000
Lo
ad
(
lb
f )
Stress ( psi )
Actual vs. Theoretical Stress
Actual Stress
Theoretical Stress
Linear (Actual Stress)
Linear (Theoretical
Stress)
EGME 306A The Beam
Page 12 of 18
Group 2
Sample Calculations and Analysis (Paola):
E=30x106 psi
Load =P= 996 lb
Dimensions: a = b= 4”
Bending Moment:
� = � × �
� = 996 �� × 4"
� = 3984 lb-in
Shear force:
� =
�
2
� =
996 ��
2
� = 498 ��
Height of the beam: 0.278+0.438+0.290= 1.006”
Neutral axis (c):
� =
ℎ
2
0.278”
0.438”
0.290”
0.992”
0.191”
12”
4” 4”
4”
ym
V
M
X
X
+
-
498 lb
3984 lb-in
1
2
3
C
996 ��
2
498 ��
996 ��
2
498 ��
EGME 306A The Beam
Page 13 of 18
Group 2
� =
1.006
2
= 0.503"
Moment of inertia:
Ai xi Ai xi Ici di Ix
1 0.275776 0.867 0.2390977 0.00177608 0.367397
0.0389556913
2 0.083658 0.509 0.04258192 0.001337 0.009694 0.001345
3 0.28768 0.145 0.0417136 0.002016 0.354603 0.038189
∑ 0.647114 0.32339322 0.0784896913
� =
∑ �� ��
∑ ��
=
0.32339322
0.647114
= .499746 ��.
Moment of inertia:
�� = ∑(�� + �� ��
2) = ∑ ��
Ic= 0.07837 in4
Section modulus of the beam cross-section:
� =
�
�
=
0.07837 ��4
0.503"
= 0.15581 ��4
Maximum Bending Stress:
���� =
��
�
=
(3984 �� − ��)(0.503 ��)
0.07837 ��4
= 25570.4 ���
Maximum Deflection:
−�� =
23 ��3
1296 ��
=
23(996��)(123)
1296(30 × 106���)(0.07837��4)
= 0.01299 ��
Error Analysis:
A) Determining the maximum bending stress values in the beam
experimentally and theoretically
B) Determining the deflection in the beam by experimental and
theoretical methods.
A-1) Experimental Method:
���� =
��
�
=
(3984 �� − ��)(0.503 ��)
0.07837 ��4
= 25570.4 ���
���� = ����� = (30 × 106���)���� = 25570.4 ���
���� = 8.523 × 10−4
��
��
Bending stress uncertainty:
Δ�
�
=
Δ�
�
=
10−6
8.523 × 10−4
��
��
= 0.0011732
EGME 306A The Beam
Page 14 of 18
Group 2
Bending stress uncertainty: 0.0011732
A-2) Theoretical Method:
Bending stress uncertainty:
Δ�
�
=
Δ�
�
+
Δℎ
ℎ
+
ℎ3∆� + 3�ℎ2∆ℎ + 2�3∆� + 6��2∆�
�ℎ3 − 6��3
Δ�
�
=
0.0001
996
+
0.001 in
0.438
+
0.4383 (0.001) + 3(0.191)(0.4382 )0.001 + 2(0.2783 )0.001 +
6(.992)(0.2782 )0.001
(0.191)(0.4383) − 6(0.992)(0.278)3
Δ�
�
= -0.023679
B-1) Experimental Method for finding the deflection:
The deflection value, which is directly the uncertainty of the
dial indicator, is 10-4in.
B-2) Theoretical Method:
Δ�
�
=
Δ�
�
+
3Δ�
�
+
ℎ3∆� + 3�ℎ2∆ℎ + 2�3∆� + 6��2∆�
�ℎ3 − 6��3
Δ�
�
=
0.0001
996
+
3(0.001)
12
+
0.4383 (0.001) + 3(0.191)(0.4382 )0.001 + 2(0.2783 )0.001 +
6(.992)(0.2782)0.001
(0.191)(0.4383) − 6(0.992)(0.278)3
Δ�
�
= −0.0054985
EGME 306A The Beam
Page 15 of 18
Group 2
Discussion and Conclusion (Ahmed):
In this experiment we examined stress, deflection and strain of
the load applied to a 1018 steel I-beam. The
beam was placed horizontally and a force P was applied to it by
the MTS Insight Tensile Testing Machine. A
load was applied and readings were taken incrementally every
100 lbf until the test was concluded at 1000 lbf.
This process was repeated twice and the average of the two tests
was taken. This data was then utilized along
with calculations for the beams neutral axis, moment of inertia,
and section modulus to determine the required
objective values. Final values of 12,150 psi for the maximum
actual stress (vs. 12,784.8 psi for theoretical
stress), and 0.0138 in for the maximum actual deflection (vs.
.0130 in for theoretical deflection) correlated
closely with each other, and successfully verified established
beam stress and flexure formulas.
EGME 306A The Beam
Page 16 of 18
Group 2
BIBLIOGRAPHY (Edwin):
1) Beckwith, T. G., Buck, N. L. and Marangoni, R. D.,
Mechanical Measurements, Addison-Wesley.
2) Popov, E.P. Mechanics of Materials, Prentice-Hall Inc.
3) Sharma, P. The Beam (EGME 306A lab manual). CSU-
Fullerton.
4) Thomas, G.B., and Finney R.L., Calculus and Analytical
Geometry, Addison-Wesley,
EGME 306A The Beam
Page 17 of 18
Group 2
Appendix (All):
A.1 Procedures
1. Measure the cross-sectional dimensions of the beam by using
the caliper and measure the location of
the loading and support pins.
2. Place the beam into the MTS machine while making sure that
beam's black marks aligns with the
roller supports of the lower bending fixture.
3. Access the TestWorks 4 software by double clicking its icon
on the desktop
4. If prompted, make sure that the name field under the User
Login says "306A_lab" then click OK to
login
5. Under the Open Method dialog, select "exp-3 4 Point Flex
Mod X'
6. Select the Motor Reset button in the bottom right corner by
clicking on it
7. Zero the "load" readout by right clicking on the "Load cell"
icon and selecting "zero channel"
8. Do not close the TestWorks 4 software as it will be used
again in the later parts of the experiment.
Leave the software running.
9. Next, use the MTS' handset to position the upper bending
fixture over the beam
10. Enable the handset by pressing "unlock" button at the top
right of the handset
11. Slowly lower the crosshead using the down arrow until
fixture is nearly touching the beam. Make
sure not to pinch the strain gauge lead wires between the fixture
and the beam while lowering.
12. Once the crosshead is near the beam, use the thumb wheel of
the handset to lower/raise the fixture
onto the beam.
13. Observe the digital load readout from the screen while
lowering/raising the fixture and aim for a pre-
load value of 0.2 lb
14. Once finished, return control to the computer software by
locking the handset by pressing the
"unlock" button.
15. Take the magnetic base holding the dial indicator and
position it with the dial indicator in the center
of the beam on the bottom side. Make sure the dial indicator is
not touching the strain gauge.
16. Once the dial indicator is in position, lock it to the MTS
frame by activating the magnetic base.
17. Zero the dial indicator by turning the dial of the indicator.
18. Start the LabVIEW software by double clicking its icon on
the desktop. NOTE: LabVIEW and
TestWorks 4 should be open side-by-side.
19. In LabVIEW, select "Open" and double click "exp2&3-
Strain Mod 9-15 LV7.1"
20. Press the white arrow near the upper left corner of the
screen to start the strain gauge acquisition
EGME 306A The Beam
Page 18 of 18
Group 2
21. Press the "Zero Strain" button to zero-out the strain
indicator
22. Press the Green Arrow on the TestWorks 4 window
23. Name a sample ID for the test
24. Under the required inputs dialog, input 0.125 for thickness
and 0.500 for width.
25. Press OK when ready to start
26. When the digital load readout reaches approximately 100 lb,
press the "Pause" button to pause the
test.
27. Record the actual load from the digital readout, also record
the strain reading from the LabVIEW
window, and record the deflection of the beam from the dial
indicator.
28. Continue taking readings in increments of 100 lb until
reaching approximately 1000 lb (steps 26-27
will be repeated 10 times)
29. Steps 18-29 will be repeated again for the next set of
experiment.
III-1
Experiment III
The Beam
OBJECTIVES
The objectives of this experiment are (a) to determine the stress,
deflection and strain of a simply
supported beam under load, and (b) to experimentally verify the
beam stress and flexure
formulas.
THEORY
Structural members are usually designed to carry tensile,
compressive, or transverse loads. A
member which carries load transversely to its length is called a
beam. In this experiment, a beam
will be symmetrically loaded as shown in Fig. III-1(a), where P
is the applied load. Note that at
any cross section of the beam there will be a shear force V (Fig.
III-1(b)) and moment M (Fig.
III-1c). Also, in the central part of the beam (between the loads
P/2) V is zero and M has its
maximum constant value. Notice the sign convention of a
positive moment, M, causing a
negative (downward) deflection, y.
If in this part a small slice EFGH of the beam is imagined to be
cut out, as shown, then it is clear
that the external applied moment, M, must be balanced by
internal forces (stresses) at the
sections (faces) EF and GH. For M applied as shown in Fig. III-
2(a), these forces would be
compressive near the top, EG, and tensile near the bottom, FH.
Since the beam material is
considered elastic, these forces would deform the beam such
that the length EG would tend to
become shorter, and FH would tend to become longer. The first
fundamental assumption of the
beam theory can be stated as follows:
“Sections, or cuts, which are plane (flat) before deformation
remain plane after
deformation.”
Thus, under this assumption, the parallel and plane sections EF
and GH will deform into plane
sections E F′ ′ and G H′ ′ which will intersect at point O, as
shown in Fig. III-2(b). Since E F′ ′
and G H′ ′ are no longer parallel, they can be thought of as
being sections of a circle at some
radial distance from O. Convince yourself of this by drawing a
square on an eraser and observe
its shape when you bend the eraser. Since the forces near E G′ ′
are compressive, and those near
F H′ ′ are tensile, there must be some radial distance r where the
forces are neither compressive
nor tensile, but zero. This axis, N-N, is called the neutral axis.
Notice that N-N is not assumed to
lie in the center of the beam.
Consider an arc of distance +η, from the neutral axis, or
distance r + η from O (Fig. III-2(b)). At
this radius, the length of arc is l′ =(r + η) Δθ. As shown in Fig.
III-2(a), the length of the arc was
l before the deformation. This length is also equal to rΔθ
(because at N-N there are no forces to
change the length). Thus, the strain at distance +η from the
neutral axis can be found by:
ε
η θ θ
θ
η
=
l - l
l
=
(r + ) - r
r
=
r
′ Δ Δ
Δ
(III-1)
III-2
(a)
(b)
(c)
ym
b
-
Figure III-1. Symmetrically Loaded Beam (a), with Shear Force
Diagram (b)
and Bending Moment Diagram (c)
L
b c a
2
P
2
P
V
x
M
x
2
Pa
2
P
+
III-3
Figure III-2. Stresses and Strains of a Beam
III-4
In other words, the axial strain is proportional to the distance
from the neutral axis. It is remarked
that this strain is positive, because positive η was taken on the
tensile side of N-N in Fig. III-
2(b). Had η been taken in the opposite direction, then the strain
would have been negative, as
appropriate for the compressive side.
The second fundamental assumption is that Hooke’s Law
applies both in tension and
compression with the same modulus of Elasticity. Thus, from
Eqs. (I-3) and (III-1),
σ
η
= E
r
(III-2)
If c is the maximum distance from the neutral axis (largest
positive or negative value of η), then
the maximum stress (compressive or tensile) is given by σm =
Ec/r, and Eq. (III-2) can also be
written as
σ σ
η
=
cm
(III-3)
That is, the stress at a section EF or GH, due to applied moment
M, varies linearly from zero at
the neutral axis to some maximum value σm (positive or
negative) when η = c. To obtain the
beam stress formula, it remains to define where the neutral axis
is located, and to relate σm to M.
To locate the neutral axis, it is observed that the tensile and
compressive forces on a section are
equal to the stress times a differential element of area, as shown
in Fig. III-2(c). For static
equilibrium, the sum (or integral) of all these internal forces
must be zero. That is,
m
A A
dF = dA = dA = 0
c
σσ η∫ ∫
where, the integrals are over the whole cross-sectional area.
Thus, it is seen that the neutral axis
is located such that the first moment of area about it is zero;
that is, the neutral axis passes
through the centroid of the cross-sectional area. In Fig. III-2(c),
a rectangular area was used for
illustration; however, any shape of vertically symmetric cross-
sectional area is valid for the area
integral.
In a similar fashion, the moment due to all the forces is the sum
(or integral) of the forces times
their moment arms about the neutral axis, and this must be
equal to the external applied moment.
Thus,
M = dF = dA =
c
dAm 2∫ ∫ ∫η ησ
σ
η (III-4)
If I is defined as the second moment of area about the neutral
axis, commonly called the moment
of inertia,
I = dA2∫η (III-5)
then Eq. (III-4) can be written as:
m =
Mc
I
=
M
Z
σ (III-6)
III-5
where Z = I/c is the section modulus, which depends only on the
cross-sectional geometry of the
beam. Equation (III-6) is the beam stress equation which relates
the maximum (compressive or
tensile) stress to the applied moment. Notice its similarity to
Equation (I-1), the stress equation
for uniaxial tension. It is understood, of course, that σm is the
maximum bending stress at a
particular location, x, along the beam. In general, both σm and
M are functions of x, and are
related by Eq. (III-6).
The remaining question about the beam concerns its degree of
deformation, or flexure. That is,
how is the radius of curvature, r, related to the moment M (or
load P)? From calculus, it can be
shown that the curvature of a function y(x) is given by
1
r
=
d y
dx
(1 +
d y
dx
)
2
2
2
2
3
2
Thus, if x is the distance along the beam, y will be the
deflection as indicated in Fig. III-1(a). For
most beams of practical interest, this deflection will be small,
so that the slope dy/dx will be very
small compared to 1. Hence, a very good approximation is
1
r
=
d y
dx
2
2
But, since σm = Ec/r = Mc/I, there results the differential
equation of the elastic curve:
EI
d y
dx
= M(x)
2
2 (III-7)
To obtain the elastic curve of the beam, y(x), and the maximum
deflection, ym, it is necessary to
integrate Eq. (III-7) using the moment function M(x) in Fig. III-
1(c). Thus, using M(x) = Px/2 for
0 ≤ x ≤ a and M(x) = Pa/2 for a ≤ x ≤ a + b, it is found that
for
3P ax(a + b)xy(x) = - , 0 x a
2EI 6 2
⎛ ⎞
≤ ≤⎜ ⎟
⎝ ⎠
( ) for
3 2P ax(2a + b)a ax y x = - + , a x a + b
2EI 6 2 2
⎛ ⎞
≤ ≤⎜ ⎟
⎝ ⎠
and that the maximum deflection at x = a + b/2 is
)(
8
b +
2
ab
+
3
a
2EI
Pa
= y-
22
m (III-8)
In particular, for a = b = L/3,
- y =
23 Pa
48EI
=
23 PL
1296EIm
3 3
(III-9)
III-6
Although the above stress and flexure formulas are quite
simple, it took some of the best minds
of the 17th and 18th centuries to derive them correctly. Part of
the difficulty in obtaining the
correct results at that time was that there were no methods of
verifying the results
experimentally. Today, with the advent of sensitive
displacement dial gauges, the verification is
more convenient.
Figure III-3. MTS Insight Tensile Testing Machine
PROCEDURE
1. The test will be conducted on a 1018 steel beam (E = 30x106
psi) using the MTS testing
machine. The position of the beam in the testing machine is
shown in Fig. III-3.
2. Observe that the cross-section of the I-beam is not a
symmetric; one flange of the beam is
thicker than the other. Measure carefully the cross-sectional
dimensions and the location of
the loading and support points.
3. Carefully place the beam into the MTS machine if it is not
already installed. Align the 12-
inch black marks on the beam with the roller supports of the
lower bending fixture. Make
sure the beam is squarely resting and centered on the lower
support with the strain gauge
facing down.
Beam
Deflection
Dial indicator
Beam to
Be Tested
III-7
4. Enter the TestWorks 4 software by double clicking on the
icon on the desktop.
a. When prompted, make sure the Name field under the User
Login says “306A_lab”.
b. Click OK to login.
c. Under the Open Method dialog, select “exp-3 4 Point Flex
Mod X”.
d. Now, select the Motor Reset button in the bottom right corner
by clicking on it.
5. Zero the “load” readout by right clicking on the “Load cell”
icon and selecting “zero
channel”.
6. Next we will use the handset to position the upper bending
fixture over the beam.
a. Enable the handset by pressing the “unlock” button at the top
right of the handset
b. Slowly lower the crosshead using the down arrow until the
fixture is NEARLY touching
the beam.
c. Make sure not to pinch the strain gauge lead wires between
the fixture and the beam.
d. While observing the digital load readout on the screen, use
the thumb wheel of the
handset to lower the fixture onto the beam. Watch for the load
reading to increase when
the upper loading component of the fixture makes contact with
the beam.
e. Now, slowly raise the fixture with the thumb wheel until only
a very slight pre-load of
approximately 0.2 lb is applied.
f. When finished, return control to the computer software by
locking the handset using the
same button as before.
Note that the beam has a strain gauge bonded to its surface. See
Figure III-4. This strain
gauge has a strain gauge factor of 2.14 ± 0.5% and 350 ohm
resistance. Connect the strain
gauge lead wires to the “#1 Strain” channel of the grey DAQ
box. Data acquisition is
conducted by the LabVIEW software.
7. Take the magnetic base holding the dial indicator, and
position it with the dial indicator in
the center of the beam on the bottom side. Make sure the dial
indicator is not touching the
strain gauge. Also, be VERY CAREFUL not to damage the dial
indicator while doing this!
When the dial indicator is in position, lock it to the MTS frame
by activating the magnetic
base. Finally, zero the dial indicator by turning the dial of the
indicator. Note the divisions
on the dial indicator and the accuracy with which it can be read.
8. Start the LabVIEW software.
a. Double click the LabVIEW icon on the desktop, or follow the
procedure suggested for
Experiment II.
b. Select “Open” and double click “exp2&3-Strain Mod 9-15
LV7.1”.
c. Press the white arrow near the upper left corner of the screen
to start the strain gauge
acquisition.
d. Press the “Zero Strain” button to zero-out the strain
indicator.
9. You are now ready to run the experiment.
a. Press the Green Arrow on the TestWorks 4 GUI.
b. Give the sample ID for your group.
III-8
c. Under the Required Inputs dialog, enter 0.125 for the
thickness and 0.500 for the width.
DO NOT press OK yet!
10. You will load the beam up to 1000 lb in increments of 100
lb. To do this:
a. “OK” the Required Inputs dialog from above and the test will
begin.
b. When the digital load readout reaches approximately 100 lb,
press the “Pause” button to
temporarily stop the test.
c. Record the actual load on your data sheet from the digital
readout.
d. Toggle over to the LabVIEW screen and record the strain
reading from the software.
e. Record the deflection of the beam from the dial indicator.
f. Return to the TestWorks 4 screen and press the “Pause”
button again to resume the test.
g. Continue pausing and taking readings approximately every
100 lb from 0 to 1000 lb.
11. At 1000 lb the test will stop. Collect your last data point
before pressing “OK” at the last
dialog.
12. Before repeating the experiment, make sure your load
readout, strain indicator, and dial
indicator are all re-zeroed.
13. Repeat steps 7 through 9.
Figure III-4. Beam with Bonded Strain Gauge
REPORT REQUIREMENTS
1. Determine the neutral axis, moment of inertia, and section
modulus of the beam cross-
section.
2. Draw shear force and bending moment diagrams for the
beam for the maximum load.
3. Compute the maximum bending stress and deflection of the
beam for the maximum load.
4. Plot deflection versus load, and also stress versus load. Each
graph should contain a curve
based completely on theoretical calculations, and another using
the experimental data points.
III-9
5. By error calculation, determine if the beam theory is
vindicated within the precision of the
instruments.
REFERENCES
[1] Beckwith, T. G., Buck, N. L. and Marangoni, R. D.,
Mechanical Measurements, Addison-
Wesley.
[2] Popov, E.P. Mechanics of Materials, Prentice-Hall Inc.
[3] Thomas, G.B., and Finney R.L., Calculus and Analytical
Geometry, Addison-Wesley,
19
Experiment 3
The Beam
Ali Almoslim
EGME-306A
Mo-8:30a.m
ABSTRACT
The main objectives of this experiment are to determine
the stress, deflection, and the strain of a supported beam under
load. Also to experimentally verify the beam stress and flexure
formulas. The experiment was accomplished by using the
machine which applies a load to a supported beam and measure
the deflection and the strain of it. The moment of inertia was
0.0845 and deflection of 0.0083%.
TABLE OF CONTENTS
Abstract
Table of Contents
Introduction and Theory
Procedures
Summary of Important Results
Sample Calculations and Error Analysis
Discussion and Conclusion
References
Appendix
INTRODUCTION AND THEORY
Structural members are usually designed to carry tensile,
compressive, or transverse loads. A member which carries load
transversely to its length is called a beam. In this experiment, a
beam will be symmetrically loaded as shown in Fig. III-1(a),
where P is the applied load. Note that at any cross section of the
beam there will be a shear force V (Fig. III-1(b)) and moment M
(Fig. III-1c). Also, in the central part of the beam (between the
loads P/2) V is zero and M has its maximum constant value.
Notice the sign convention of a positive moment, M, causing a
negative (downward) deflection, y.
If in this part a small slice EFGH of the beam is imagined
to be cut out, as shown, then it is clear that the external applied
moment, M, must be balanced by internal forces (stresses) at the
sections (faces) EF and GH. For M applied as shown in Fig. III-
2(a), these forces would be compressive near the top, EG, and
tensile near the bottom, FH. Since the beam material is
considered elastic, these forces would deform the beam such
that the length EG would tend to become shorter, and FH would
tend to become longer. The first fundamental assumption of the
beam theory can be stated as follows:
“Sections, or cuts, which are plane (flat) before deformation,
remain plane after deformation.”
Thus, under this assumption, the parallel and plane section EF
and GH will deform into plane sections E’F’ and G’H’ which
will intersect at point O, as shown in Fig. III-2(b). Since E’F’
and G’H’ are no longer parallel, they can be thought of as being
sections of a circle at some radial distance from O. Convince
yourself of this by drawing a square on an eraser and observe its
shape when you bend the eraser. Since the forces near E’G’ are
compresiive, and those near F’H’ are tensile, there must be
some radial distance r where the forces are neither compressive
nor tensile, but zero. This axis, N-N, is called the neutral axis.
Notice that N-N is not assumed to lie in the center of the beam.
Consider an arc of distance +η, from the neutral axis, or
distance r + η from O (Fig. III-2(b)). At this radius, the length
of arc is l’=(r + η) Δθ. As shown in Fig. III-2(a), the length of
the arc was l before the deformation. This length is also equal to
rΔθ (because at N-N there are no forces to change the length).
Thus, the strain at distance +η from the neutral axis can be
found by:
(III-1)
In other words, the axial strain is proportional to the distance
from the neutral axis. It is remarked that this strain is positive,
because positive η was taken on the tensile side of N-N in Fig.
III- 2(b). Had η been taken in the opposite direction, then the
strain would have been negative, as appropriate for the
compressive side.
The second fundamental assumption is that Hooke’s Law
applies both in tension and compression with the same modulus
of Elasticity. Thus, from Eqs. (I-3) and (III-1),
(III-2)
If c is the maximum distance from the neutral axis (largest
positive or negative value of η), then
the maximum stress (compressive or tensile) is given by σm =
Ec/r, and Eq. (III-2) can also be written as
(III-3)
That is, the stress at a section EF or GH, due to applied moment
M, varies linearly from zero at
the neutral axis to some maximum value σm (positive or
negative) when η = c. To obtain the
beam stress formula, it remains to define where the neutral axis
is located, and to relate σm to M.
To locate the neutral axis, it is observed that the tensile and
compressive forces on a section are equal to the stress times a
differential element of area, as shown in Fig. III-2(c). For static
equilibrium, the sum (or integral) of all these internal forces
must be zero. That is,
where, the integrals are over the whole cross-sectional area.
Thus, it is seen that the neutral axis is located such that the first
moment of area about it is zero; that is, the neutral axis passes
through the centroid of the cross-sectional area. In Fig. III-2(c),
a rectangular area was used for illustration; however, any shape
of vertically symmetric cross-sectional area is valid for the area
integral.
In a similar fashion, the moment due to all the forces is the sum
(or integral) of the forces times their moment arms about the
neutral axis, and this must be equal to the external applied
moment.
Thus,
(III-4)
If I is defined as the second moment of area about the neutral
axis, commonly called the moment
of inertia,
(III-5)
then Eq. (III-4) can be written as:
(III-6)
where Z = I/c is the section modulus, which depends only on the
cross-sectional geometry of the beam. Equation (III-6) is the
beam stress equation which relates the maximum (compressive
or tensile) stress to the applied moment. Notice its similarity to
Equation (I-1), the stress equation for uniaxial tension. It is
understood, of course, that σm is the maximum bending stress at
a particular location, x, along the beam. In general, both σm and
M are functions of x, and are related by Eq. (III-6).
The remaining question about the beam concerns its degree of
deformation, or flexure. That is, how is the radius of curvature,
r, related to the moment M (or load P)? From calculus, it can be
shown that the curvature of a function y(x) is given by
Thus, if x is the distance along the beam, y will be the
deflection as indicated in Fig. III-1(a). For most beams of
practical interest, this deflection will be small, so that the slope
dy/dx will be very
small compared to 1. Hence, a very good approximation is
But, since σm = Ec/r = Mc/I, there results the differential
equation of the elastic curve:
(III-7)
To obtain the elastic curve of the beam, y(x), and the maximum
deflection, ym, it is necessary to integrate Eq. (III-7) using the
moment function M(x) in Fig. III-1(c). Thus, using M(x) = Px/2
for
0 ≤ x ≤ a and M(x) = Pa/2 for a ≤ x ≤ a + b, it is found that
and that the maximum deflection at x = a + b/2 is
(III-8)
In particular, for a = b = L/3,
(III-9)
PROCEDURES
This experiment will be applied on a 1018 steel beam
,which has E = 30x10*6 psi, using the MTS testing machine.
Because the beam is not perfectly symmetric we had to measure
the cross-sectional dimension carefully and the location of the
loading points. After that, we placed the beam in the testing
machine and aligned the 12-inch black marks on the beam with
roller supports of the lower fixture. We had to make sure the
beam is centered on the lower support with strain gauge facing
down. Then we went to the computer and entered the TestWork
4 software. When prompted, we had to make sure the name field
under login says “306A_Lab” then clicked OK to login then
under the Open Method dialog, we had to select“exp-3 4 Point
Flex Mod X” after that we had to select the Motor Reset button
right corner by clicking on it. We zeroed the load by right
clicking on the load cell icon and selecting zero channel. Then,
we had position the upper bending fixture over the beam using
the handset after enabling the handset by pressing the unlock
button, then slowly lowering the crosshead using the down
arrow until the fixture is touching the beam. After that, we used
the thumb wheel of the handset to lower the fixture onto the
beam. We watched for the load reading increase when the upper
fixture makes contact. Then, we slowly lifted the fixture until a
very slight preload of approximately 0.2 lb is applied. Then we
locked the handset. Then we connected the strain gauge wire to
the #1 strain channel of the grey DAQ box, data acquisition is
conducted by the LabVIEW software. We took the magnetic
base holding the dial indicator, and positioned it with the dial
indicator in the center of the beam on the bottom side. We made
sure the dial indicator is not touching the strain gauge, after that
we locked the MTS frame by activating the magnetic base then
zeroed the dial indicator. Next, we started the LabVIEW
software by double click the icon on the desktop, selecting
“Open” and double clicking “exp2&3-Strain Mod-15 LV7.1”
then we pressed the white arrow to start the strain gauge
acquisition then we zeroed the strain. After that we pressed the
green arrow on the TestWorks 4 GUI. We loaded the beam up to
1000 lb for each in 100 lb and recorded the data. After reaching
1000 lb we repeated the experiment after making sure it zeroed
out just to assure accuracy of our work.
SUMMARY OF IMPORTANT RESULTS
I
In graph I. we can notice that it’s a deflection vs load diagram
as titled. we have the theoretical values and the experimental
ones. As we can see from the graph the is a noticeable
difference between the two values as the two lines get different
slops.
II
As we can see we, graph II is about the strees vs load diagram,
where the load is the x-axis and the stress is the y-axis which is
represented in a straight line as shown.
SAMPLE CALCULATION AND ERROR ANALYSIS
To find the use the following equation:
Using the position points, calculate the neutral axis, but it is
important to remember that this value isn’t the maximum
neutral axis, which is what is needed. In that case, simply
subtract total distance by the calculated neutral axis.
The moment of Inertia is calculated by the following formula:
Manually determined moment of Inertia
To find the section modulus we use the formula:
Where I is the calculated moment of Inertia and c is the distance
from the neutral axis
To find the maximum bending stress we start with the following
equation:
Where,
P (maximum load) = 997.3 lb
a (distance away from reference) = 4 in
Providing the applied moment of,
Now the maximum bending stress can be found using Eq. (1)
The deflection is determined by the following formula:
Or
Now, Eq. (9) is used to find the deflection of the crossbeam
given the maximum pressure, moment of inertia, and distance
between loads on the beam. Using the previous maximum load,
initial length of 12 in, initial modulus elasticity and calculated
moment of Inertia, the maximum deflection comes out to be…
The Stress and Deflection percent error were calculated using
the results obtained by the load applied by the MTS machine
and the dial indicator and a thermometer.
Stress () % Error = x 100% = x 100% = 1.20%
Deflection % Error = x 100% = x 100% = .0083 %
Ai
Yi
Ay
D
AD^2
I
1
0.261
0.1295
0.034
0.3875
0.039
0.00146
2
0.116
0.524
0.061
0.007
5.7x10^-6
2.7x10-3
3
0.253
0.63
0.915
0.231
0.326
0.398
0.4
1.36x10^-6
D1= 0.517 – 0.1295 = 0.3875 in Y=∑yA/∑A = 0.326/0.63
= 0.517
D2= 0.524 – 0.517 = 0.007 in
D3= 0.915 – 0.517= 0.398 in
I1= (bh^3)/12 = (1.007(0.259)^3)/12= 1.46x10^-3 in ^4, I2=
2.714x10^-3 in^4, I3= 1.36x10^-3 in^4
Moment = 0.039 + 5.7x10^-6 + 0.04 + 1.4x10^-3 + 2.714x10^-3
+ 1.36x10^-3 = 0.0845 in^4
Error Analysis
Human error is a common one that could happen in any
experiments, in this example, human error might happened in
the setup of the equipment. Another possibility is that we didn’t
get the write readings. Also, the machine could be old and not
accurate enough. Last one is that the sample we tested might not
manufactured properly. So these reasons explain why we got
1.20% error for stress and 0.0083% for deflection.
DISCUSSION AND CONCLUSION
After finishing the experiment and calculating the data and
graph them,we figured out many features of the beam.
Whenever the load increases, the deflection and the stress
increase too. We can use the strain to find the theoretical stress
and we can use moment, moment of inertia, and the neutral axis
to find the experimental stress. And about the deflection, we
used the dial indicator to find it directly from the experiment
for the experimental and we used load, length, modulus of
elasticity, and the moment of inertia to find the theoretical data.
The graph clearly shows that the load and deflection and load
and stress have a directly proportional relationship.
REFRENCES
[1] CSUF EGME 306A Lab Manual.
Appendix
deflection(in)
Max Stress at B
0.0009
1225.223179
0.002
2450.446358
0.0033
3663.417306
0.0046
4888.640485
0.0059
6113.863664
0.0073
7351.339075
0.0086
8539.805559
0.0099
9814.037665
0.0103
11049.06263
0.0107
12217.92554
Distance of Neutral Axis
Moment of Inertia
Applied Moment
Modulus of Elasticity
0.5179
0.0845397
200
30000000
400
598
798
998
1200
1394
1602
1803.6
1994.4
Deflection Vs. Load
Theoretical Calcuations 100 200 299 399 499 600 697
801 901.8 997.2 1.2091623488399204E-3
2.4183246976798409E-3 3.6153954230313621E-3
4.8245577718712823E-3 6.0337201207112034E-3
7.2549740930395222E-3 8.4278615714142458E-3
9.6853904142077622E-3 1.0904226061838402E-2
1.2057766942631687E-2 Experimental Values 100
200 299 399 499 600 697 801 901.8 997.2
8.9999999999999998E-4 2E-3 3.3E-3
4.5999999999999999E-3 5.8999999999999999E-3
7.3000000000000001E-3 8.6E-3
9.9000000000000008E-3 1.03E-2
1.0699999999999999E-2
Load (lbs)
Deflection(in)
Stress vs. Load
100 200 299 399 499 600 697 801 901.8 997.2
1225.223179169077 2450.4463583381539
3663.4173057155399 4888.6404848846169
6113.8636640536934 7351.3390750144608
8539.8055588084662 9814.0376651443075
11049.062629746735 12217.925542674035 100
200 299 399 499 600 697 801 901.8 997.2
Load (lbs)
Stress (PSI)

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EGME 306A The Beam Page 1 of 18 Group 2 EXPER.docx

  • 1. EGME 306A The Beam Page 1 of 18 Group 2 EXPERIMENT 3:The Beam Group 2 Members: Ahmed Shehab Marvin Penaranda Edwin Estrada Chris May Bader Alrwili Paola Barcenas
  • 2. Deadline Date: 10/23/2015 Submission Date: 10/23/2015 EGME 306A – UNIFIED LABORATORY EGME 306A The Beam Page 2 of 18 Group 2 Abstract (Bader): The main objective for this experiment was to determine the stress, deflection, and strain of a supported beam under loading, and to experimentally verify the beam stress and flexure formulas. Additionally, maximum bending stress and maximum deflection were determined. To accomplish this, a 1018 steel I-beam with a strain gage bonded to the underside was utilized in conjunction with a dial indicator to monitor beam deflection. In order to determine the values for strain and deflection, the beam underwent testing utilizing the MTS Tensile Testing machine, which applied a controlled, incrementally increasing load to the beam. This data was then utilized along with calculations for the beams neutral axis, moment of inertia, and section modulus to determine
  • 3. the required objective values. Final values of 12,150 psi for the maximum actual stress (vs. 12,784.8 psi for theoretical stress), and 0.0138 in for the maximum actual deflection (vs. .0130 in for theoretical deflection) correlated closely with each other, and successfully verify established beam stress and flexure formulas. EGME 306A The Beam Page 3 of 18 Group 2
  • 4. Table of Contents: List of Symbols and Units 4 Theory 5 Procedure and Experimental Set-up 8 Results 9 Sample Calculations and Error Analysis 12 Discussion and Conclusion 15 Bibliography 16 Appendix 17
  • 5. EGME 306A The Beam Page 4 of 18 Group 2 List of Symbols and Units (Chris): List of Symbols and Units Name of variables (units) Units � Stress psi � Applied load lbf � Moment of Inertia in.4 � Strain in/in � Length of the bar in Z Section Modulus of Beam in3 � Distance to Beam Neutral Axis in
  • 6. � Modulus of Elasticity psi EGME 306A The Beam Page 5 of 18 Group 2 Theory (Edwin): There are two main objectives for this experiment: to determine maximum bending stress values in the beam and to determine the deflection in the beam. To help visualize this phenomena, imagine cutting a section of a symmetrically loaded beam: Now, examine diagrams of this section before (Fig. A) and after bending (Fig. B): (Fig. A)
  • 7. (Fig. B) The main points to take away from the above diagrams are as follows: When the moment, M is applied as shown in Fig. A, forces will be in compression near the top (positive moment) and in tension near the bottom (negative moment). The effects from this moment are seen in Fig. B. For determining max stress values, one concept to note is that our bending moment M can help calculate bending stress. First, we recall our basic definition of normal strain: � = �′ − � � EGME 306A The Beam Page 6 of 18 Group 2 Since the beam is bending, however, we will need to alter the above formula by taking into account
  • 8. the radius of curvature, r, and the differential angle, �. Since there are locations where forces are compressive and locations where forces are tensile, there must be a location where forces are neither compressive nor tensile. This axis is called the Neutral Axis and is labeled in Fig. B as N-N. We can now alter the above formula by measuring the strain at distance +� from the neutral axis: � = �′ − � � = (� + �)∆� − �∆� �∆� = � � The strain equation from above can be converted to stress by using Hooke’s Law: � = � � � Now, if we allow c to be the max distance from the neutral axis,
  • 9. we can build upon the above expression as follows: � = �� � � To obtain the beam stress formula, we still need to define where the neutral axis is located. We can do this by relating the radius of curvature and the bending moment, which will be determined by applying a moment equilibrium equation about the neutral axis: �� = ∫ � � �� = �� � ∫ � � �� = 0 Since moment due to all forces is summed up by the product of their forces and the moment arm about the neutral axis, we deduce: � = ∫ ��� = ∫ ���� =
  • 10. �� � ∫ �2�� We will then let I be defined as the second moment of area about the neutral axis, commonly called the moment of inertia: � = ∫ �2�� Finally, we deduce the moment equation as: �� = �� � = � � Our final objective involved determining beam deflection. Using similar methods from bending stress, we will examine the relationship between the radius of curvature, r and the moment, M, at any given point on a beam. We begin with the radius of curvature for any point of a function from calculus: EGME 306A The Beam
  • 11. Page 7 of 18 Group 2 � � = �2� ��2 (1 + � 2� ��2 ) 3 2 We will note that for small deflections, the first derivative or slope is small and becomes even smaller when it is squared and can therefore be neglected. This will greatly simplify the radius of curvature to: 1 � = �2� ��2
  • 12. By combing the above equation with the bending stress equation, we develop the standard moment curvature equation: �� �2� ��2 = �(�) The differential equation will only be useful if we apply it to the beam with specific boundaries, or boundaries of integration. Integrating with M(x)=Px/2 for 0≤x≤a andM(x)=Pa/2 for a≤x≤a+b, we get: �(�) = � 2�� ( �3 6 − ��(� + �) 2 ) , ��� 0 ≤ � ≤ � �(�) = �
  • 13. 2�� ( �3 6 − ��(2� + �) 2 + ��2 2 ) , ��� � ≤ � ≤ � + � The max deflection at x = a + b/2 is: −�� = �� 2�� ( �2 3 + �� 2 + �2
  • 14. 8 ) Lastly, by allowing a=b=L/3: −�� = 23��3 48�� = 23��3 1296�� EGME 306A The Beam Page 8 of 18 Group 2 Procedure and Experimental Set-Up (Marvin): This experiment requires the use of MTS Insight Tensile Testing Machine, Beam Deflection Dial indicator, digital caliper, a 1018 steel beam and two separate data acquisition software. The experiment begins with the
  • 15. measurement of the length and the cross-sectional dimensions of the 1018 steel beam with the use of the digital caliper. The beam is then carefully placed into the MTS machine while making sure that the beam's black marks are aligned with the roller supports of the bending fixture. Once the beam is squarely nested on the roller supports, TestWorks 4 software is accessed in order to zero the load reading. Now that the parameters have been set up, the MTS' handset is used to position the upper bending fixture over the beam. While the crosshead lowers, the digital load readout from the software is observed. When the load increases as the upper loading component of the fixture makes contact with the beam, the fixture is raised until a preload of 0.2 lb is applied. The Beam Deflection Dial indicator is positioned in the center of the beam on the bottom side. Do note, the dial indicator needs to be set to zero. Now, the LabVIEW software is accessed. On the software, the strain indicator is zeroed by pressing “zero strain”. The software also requires the inputs dialog for the thickness and width, a value of 0.125 is used for thickness and a value of 0.500 is used for width. The MTS machine loads the beam up to 1000 lb in increments of 100 lb. When the digital load readout reaches approximately 100 lb, the “Pause”
  • 16. button is pressed since the data readings need to be taken for every 100 lbs. Pausing for every 100 lb increments are continued up until 1000 lbs. The procedure is repeated again since the average of the 2 sets of readings will be used for experimental calculations. Keep in mind, the dial indicator and the strain indicator needs to be re- zeroed again for the next set of experiment. Note: for a detailed procedure, see Appendix A.1 EGME 306A The Beam Page 9 of 18 Group 2
  • 17. Results (Chris): -TEST 1 / TEST 2 - - AVERAGE OF TEST 1 & 2 - LOAD ( lbf ) STRAIN GAGE ELONGATIO N ( ε ) INDICATOR DEFLECTIO N (in.) LOA D ( lbf ) STRAIN GAGE ELONGATIO N ( ε ) INDICATOR DEFLECTIO N (in.) 100/99 47/46 .0011/.0016 99.5 46.5 .00135
  • 18. 202/19 8 91/88 .0023/.0029 200 89.5 .0026 316/29 5 144/131 .004/.0041 305.5 137.5 .00405 396/39 5 179/174 .005/.0059 395.5 176.5 .00545 497/50 0 222/224 .0064/.0071 498.5 223 .00675 596/59 8 268/268 .008/.0087 597 268 .00835 695/69 6 312/310 .0093/.010 695.5 311 .00965
  • 19. 797/80 0 354/355 .0109/.0114 798.5 354.5 .01115 897/89 8 318/400 .0122/.0129 897.5 359 .01255 996/99 6 365/445 .0134/.0142 996 405 .0138 TABLE 1. Strain gauge, indicator, and load data (reference Appendix for calculations) EGME 306A The Beam Page 10 of 18 Group 2 LOAD ( lbf )
  • 20. ACTUAL VALUES / DEFLECTION ( in. ) THEORETICAL VALUES / DEFLECTION ( in. ) 99.5 0.00135 0.0013 200 0.0026 0.0026 305.5 0.0041 0.0039 395.5 0.0055 0.0051 498.5 0.00675 0.0065 597 0.00835 0.0077 695.5 0.00965 0.0091 798.5 0.01115 0.0104 897.5 0.01255 0.0117 996 0.0138 0.013 TABLE 2. Actual and Theoretical Values for Beam Deflection (reference Appendix for calculations) Figure 2-1. Actual vs. Theoretical Deflection using dial indicator and calculated values. The Load vs. Deflection curve consists of the actual experimental values obtained during the
  • 21. experiment compared with theoretical calculated values to measure deflection in relation to applied load. 0 200 400 600 800 1000 1200 0 0.005 0.01 0.015 Lo ad ( lb f ) Deflection ( in. ) Actual vs. Theoretical Deflection Actual Deflection
  • 22. Theoretical Deflection Linear (Actual Deflection) Linear (Theoretical Deflection) EGME 306A The Beam Page 11 of 18 Group 2 LOAD ( lbf ) ACTUAL STRESS ( psi ) THEORETICAL STRESS ( psi ) 99.5 1,410 1,277.20 200 2,685 2,567.23 305.5 4,125 3,921.44 395.5 5,295 5,076.70
  • 23. 498.5 6,690 6,398.82 597 8,040 7,663.18 695.5 9,330 8,927.54 798.5 10,635 10,249.66 897.5 10,770 11,520.44 996 12,150 12,784.80 TABLE 3. Actual and Theoretical Values for Beam Stress (reference Appendix for calculations) Figure 2-1. Actual vs. Theoretical Stress curve using experimental and calculated values. The Load vs. Stress curve consists of the actual experimental values obtained during the experiment compared with theoretical calculated values to measure stress in relation to applied load. 0 200 400
  • 24. 600 800 1000 1200 0 5,000 10,000 15,000 Lo ad ( lb f ) Stress ( psi ) Actual vs. Theoretical Stress Actual Stress Theoretical Stress Linear (Actual Stress) Linear (Theoretical Stress) EGME 306A The Beam Page 12 of 18
  • 25. Group 2 Sample Calculations and Analysis (Paola): E=30x106 psi Load =P= 996 lb Dimensions: a = b= 4” Bending Moment: � = � × � � = 996 �� × 4" � = 3984 lb-in Shear force: � = � 2 � = 996 �� 2 � = 498 �� Height of the beam: 0.278+0.438+0.290= 1.006” Neutral axis (c):
  • 27. - 498 lb 3984 lb-in 1 2 3 C 996 �� 2 498 �� 996 �� 2 498 �� EGME 306A The Beam Page 13 of 18 Group 2 � = 1.006
  • 28. 2 = 0.503" Moment of inertia: Ai xi Ai xi Ici di Ix 1 0.275776 0.867 0.2390977 0.00177608 0.367397 0.0389556913 2 0.083658 0.509 0.04258192 0.001337 0.009694 0.001345 3 0.28768 0.145 0.0417136 0.002016 0.354603 0.038189 ∑ 0.647114 0.32339322 0.0784896913 � = ∑ �� �� ∑ �� = 0.32339322 0.647114 = .499746 ��. Moment of inertia: �� = ∑(�� + �� �� 2) = ∑ �� Ic= 0.07837 in4 Section modulus of the beam cross-section:
  • 29. � = � � = 0.07837 ��4 0.503" = 0.15581 ��4 Maximum Bending Stress: ���� = �� � = (3984 �� − ��)(0.503 ��) 0.07837 ��4 = 25570.4 ��� Maximum Deflection: −�� = 23 ��3 1296 �� = 23(996��)(123) 1296(30 × 106���)(0.07837��4)
  • 30. = 0.01299 �� Error Analysis: A) Determining the maximum bending stress values in the beam experimentally and theoretically B) Determining the deflection in the beam by experimental and theoretical methods. A-1) Experimental Method: ���� = �� � = (3984 �� − ��)(0.503 ��) 0.07837 ��4 = 25570.4 ��� ���� = ����� = (30 × 106���)���� = 25570.4 ��� ���� = 8.523 × 10−4 �� �� Bending stress uncertainty: Δ� � =
  • 31. Δ� � = 10−6 8.523 × 10−4 �� �� = 0.0011732 EGME 306A The Beam Page 14 of 18 Group 2 Bending stress uncertainty: 0.0011732 A-2) Theoretical Method: Bending stress uncertainty: Δ� � = Δ� � + Δℎ ℎ
  • 32. + ℎ3∆� + 3�ℎ2∆ℎ + 2�3∆� + 6��2∆� �ℎ3 − 6��3 Δ� � = 0.0001 996 + 0.001 in 0.438 + 0.4383 (0.001) + 3(0.191)(0.4382 )0.001 + 2(0.2783 )0.001 + 6(.992)(0.2782 )0.001 (0.191)(0.4383) − 6(0.992)(0.278)3 Δ� � = -0.023679 B-1) Experimental Method for finding the deflection: The deflection value, which is directly the uncertainty of the dial indicator, is 10-4in. B-2) Theoretical Method:
  • 33. Δ� � = Δ� � + 3Δ� � + ℎ3∆� + 3�ℎ2∆ℎ + 2�3∆� + 6��2∆� �ℎ3 − 6��3 Δ� � = 0.0001 996 + 3(0.001) 12 + 0.4383 (0.001) + 3(0.191)(0.4382 )0.001 + 2(0.2783 )0.001 + 6(.992)(0.2782)0.001 (0.191)(0.4383) − 6(0.992)(0.278)3
  • 34. Δ� � = −0.0054985 EGME 306A The Beam Page 15 of 18 Group 2 Discussion and Conclusion (Ahmed): In this experiment we examined stress, deflection and strain of the load applied to a 1018 steel I-beam. The beam was placed horizontally and a force P was applied to it by the MTS Insight Tensile Testing Machine. A load was applied and readings were taken incrementally every 100 lbf until the test was concluded at 1000 lbf. This process was repeated twice and the average of the two tests was taken. This data was then utilized along with calculations for the beams neutral axis, moment of inertia, and section modulus to determine the required objective values. Final values of 12,150 psi for the maximum
  • 35. actual stress (vs. 12,784.8 psi for theoretical stress), and 0.0138 in for the maximum actual deflection (vs. .0130 in for theoretical deflection) correlated closely with each other, and successfully verified established beam stress and flexure formulas. EGME 306A The Beam Page 16 of 18 Group 2
  • 36. BIBLIOGRAPHY (Edwin): 1) Beckwith, T. G., Buck, N. L. and Marangoni, R. D., Mechanical Measurements, Addison-Wesley. 2) Popov, E.P. Mechanics of Materials, Prentice-Hall Inc. 3) Sharma, P. The Beam (EGME 306A lab manual). CSU- Fullerton. 4) Thomas, G.B., and Finney R.L., Calculus and Analytical Geometry, Addison-Wesley,
  • 37. EGME 306A The Beam Page 17 of 18 Group 2 Appendix (All): A.1 Procedures 1. Measure the cross-sectional dimensions of the beam by using the caliper and measure the location of the loading and support pins. 2. Place the beam into the MTS machine while making sure that beam's black marks aligns with the roller supports of the lower bending fixture. 3. Access the TestWorks 4 software by double clicking its icon on the desktop 4. If prompted, make sure that the name field under the User Login says "306A_lab" then click OK to login 5. Under the Open Method dialog, select "exp-3 4 Point Flex Mod X'
  • 38. 6. Select the Motor Reset button in the bottom right corner by clicking on it 7. Zero the "load" readout by right clicking on the "Load cell" icon and selecting "zero channel" 8. Do not close the TestWorks 4 software as it will be used again in the later parts of the experiment. Leave the software running. 9. Next, use the MTS' handset to position the upper bending fixture over the beam 10. Enable the handset by pressing "unlock" button at the top right of the handset 11. Slowly lower the crosshead using the down arrow until fixture is nearly touching the beam. Make sure not to pinch the strain gauge lead wires between the fixture and the beam while lowering. 12. Once the crosshead is near the beam, use the thumb wheel of the handset to lower/raise the fixture onto the beam. 13. Observe the digital load readout from the screen while lowering/raising the fixture and aim for a pre- load value of 0.2 lb 14. Once finished, return control to the computer software by locking the handset by pressing the
  • 39. "unlock" button. 15. Take the magnetic base holding the dial indicator and position it with the dial indicator in the center of the beam on the bottom side. Make sure the dial indicator is not touching the strain gauge. 16. Once the dial indicator is in position, lock it to the MTS frame by activating the magnetic base. 17. Zero the dial indicator by turning the dial of the indicator. 18. Start the LabVIEW software by double clicking its icon on the desktop. NOTE: LabVIEW and TestWorks 4 should be open side-by-side. 19. In LabVIEW, select "Open" and double click "exp2&3- Strain Mod 9-15 LV7.1" 20. Press the white arrow near the upper left corner of the screen to start the strain gauge acquisition EGME 306A The Beam Page 18 of 18 Group 2 21. Press the "Zero Strain" button to zero-out the strain indicator 22. Press the Green Arrow on the TestWorks 4 window
  • 40. 23. Name a sample ID for the test 24. Under the required inputs dialog, input 0.125 for thickness and 0.500 for width. 25. Press OK when ready to start 26. When the digital load readout reaches approximately 100 lb, press the "Pause" button to pause the test. 27. Record the actual load from the digital readout, also record the strain reading from the LabVIEW window, and record the deflection of the beam from the dial indicator. 28. Continue taking readings in increments of 100 lb until reaching approximately 1000 lb (steps 26-27 will be repeated 10 times) 29. Steps 18-29 will be repeated again for the next set of experiment. III-1
  • 41. Experiment III The Beam OBJECTIVES The objectives of this experiment are (a) to determine the stress, deflection and strain of a simply supported beam under load, and (b) to experimentally verify the beam stress and flexure formulas. THEORY Structural members are usually designed to carry tensile, compressive, or transverse loads. A member which carries load transversely to its length is called a beam. In this experiment, a beam will be symmetrically loaded as shown in Fig. III-1(a), where P is the applied load. Note that at any cross section of the beam there will be a shear force V (Fig. III-1(b)) and moment M (Fig. III-1c). Also, in the central part of the beam (between the loads P/2) V is zero and M has its maximum constant value. Notice the sign convention of a positive moment, M, causing a negative (downward) deflection, y. If in this part a small slice EFGH of the beam is imagined to be cut out, as shown, then it is clear that the external applied moment, M, must be balanced by internal forces (stresses) at the sections (faces) EF and GH. For M applied as shown in Fig. III- 2(a), these forces would be
  • 42. compressive near the top, EG, and tensile near the bottom, FH. Since the beam material is considered elastic, these forces would deform the beam such that the length EG would tend to become shorter, and FH would tend to become longer. The first fundamental assumption of the beam theory can be stated as follows: “Sections, or cuts, which are plane (flat) before deformation remain plane after deformation.” Thus, under this assumption, the parallel and plane sections EF and GH will deform into plane sections E F′ ′ and G H′ ′ which will intersect at point O, as shown in Fig. III-2(b). Since E F′ ′ and G H′ ′ are no longer parallel, they can be thought of as being sections of a circle at some radial distance from O. Convince yourself of this by drawing a square on an eraser and observe its shape when you bend the eraser. Since the forces near E G′ ′ are compressive, and those near F H′ ′ are tensile, there must be some radial distance r where the forces are neither compressive nor tensile, but zero. This axis, N-N, is called the neutral axis. Notice that N-N is not assumed to lie in the center of the beam. Consider an arc of distance +η, from the neutral axis, or distance r + η from O (Fig. III-2(b)). At this radius, the length of arc is l′ =(r + η) Δθ. As shown in Fig. III-2(a), the length of the arc was l before the deformation. This length is also equal to rΔθ (because at N-N there are no forces to change the length). Thus, the strain at distance +η from the neutral axis can be found by:
  • 43. ε η θ θ θ η = l - l l = (r + ) - r r = r ′ Δ Δ Δ (III-1) III-2
  • 45. ym b - Figure III-1. Symmetrically Loaded Beam (a), with Shear Force Diagram (b) and Bending Moment Diagram (c) L b c a 2 P 2 P V x M x 2 Pa
  • 46. 2 P + III-3 Figure III-2. Stresses and Strains of a Beam III-4 In other words, the axial strain is proportional to the distance from the neutral axis. It is remarked that this strain is positive, because positive η was taken on the tensile side of N-N in Fig. III- 2(b). Had η been taken in the opposite direction, then the strain would have been negative, as appropriate for the compressive side. The second fundamental assumption is that Hooke’s Law applies both in tension and compression with the same modulus of Elasticity. Thus, from Eqs. (I-3) and (III-1), σ η = E r
  • 47. (III-2) If c is the maximum distance from the neutral axis (largest positive or negative value of η), then the maximum stress (compressive or tensile) is given by σm = Ec/r, and Eq. (III-2) can also be written as σ σ η = cm (III-3) That is, the stress at a section EF or GH, due to applied moment M, varies linearly from zero at the neutral axis to some maximum value σm (positive or negative) when η = c. To obtain the beam stress formula, it remains to define where the neutral axis is located, and to relate σm to M. To locate the neutral axis, it is observed that the tensile and compressive forces on a section are equal to the stress times a differential element of area, as shown in Fig. III-2(c). For static equilibrium, the sum (or integral) of all these internal forces must be zero. That is, m A A dF = dA = dA = 0 c
  • 48. σσ η∫ ∫ where, the integrals are over the whole cross-sectional area. Thus, it is seen that the neutral axis is located such that the first moment of area about it is zero; that is, the neutral axis passes through the centroid of the cross-sectional area. In Fig. III-2(c), a rectangular area was used for illustration; however, any shape of vertically symmetric cross- sectional area is valid for the area integral. In a similar fashion, the moment due to all the forces is the sum (or integral) of the forces times their moment arms about the neutral axis, and this must be equal to the external applied moment. Thus, M = dF = dA = c dAm 2∫ ∫ ∫η ησ σ η (III-4) If I is defined as the second moment of area about the neutral axis, commonly called the moment of inertia, I = dA2∫η (III-5) then Eq. (III-4) can be written as: m =
  • 49. Mc I = M Z σ (III-6) III-5 where Z = I/c is the section modulus, which depends only on the cross-sectional geometry of the beam. Equation (III-6) is the beam stress equation which relates the maximum (compressive or tensile) stress to the applied moment. Notice its similarity to Equation (I-1), the stress equation for uniaxial tension. It is understood, of course, that σm is the maximum bending stress at a particular location, x, along the beam. In general, both σm and M are functions of x, and are related by Eq. (III-6). The remaining question about the beam concerns its degree of deformation, or flexure. That is, how is the radius of curvature, r, related to the moment M (or load P)? From calculus, it can be shown that the curvature of a function y(x) is given by 1 r =
  • 50. d y dx (1 + d y dx ) 2 2 2 2 3 2 Thus, if x is the distance along the beam, y will be the deflection as indicated in Fig. III-1(a). For most beams of practical interest, this deflection will be small, so that the slope dy/dx will be very small compared to 1. Hence, a very good approximation is 1 r = d y dx 2
  • 51. 2 But, since σm = Ec/r = Mc/I, there results the differential equation of the elastic curve: EI d y dx = M(x) 2 2 (III-7) To obtain the elastic curve of the beam, y(x), and the maximum deflection, ym, it is necessary to integrate Eq. (III-7) using the moment function M(x) in Fig. III- 1(c). Thus, using M(x) = Px/2 for 0 ≤ x ≤ a and M(x) = Pa/2 for a ≤ x ≤ a + b, it is found that for 3P ax(a + b)xy(x) = - , 0 x a 2EI 6 2 ⎛ ⎞ ≤ ≤⎜ ⎟ ⎝ ⎠ ( ) for 3 2P ax(2a + b)a ax y x = - + , a x a + b 2EI 6 2 2 ⎛ ⎞
  • 52. ≤ ≤⎜ ⎟ ⎝ ⎠ and that the maximum deflection at x = a + b/2 is )( 8 b + 2 ab + 3 a 2EI Pa = y- 22 m (III-8) In particular, for a = b = L/3, - y = 23 Pa 48EI = 23 PL 1296EIm
  • 53. 3 3 (III-9) III-6 Although the above stress and flexure formulas are quite simple, it took some of the best minds of the 17th and 18th centuries to derive them correctly. Part of the difficulty in obtaining the correct results at that time was that there were no methods of verifying the results experimentally. Today, with the advent of sensitive displacement dial gauges, the verification is more convenient. Figure III-3. MTS Insight Tensile Testing Machine PROCEDURE 1. The test will be conducted on a 1018 steel beam (E = 30x106 psi) using the MTS testing machine. The position of the beam in the testing machine is shown in Fig. III-3. 2. Observe that the cross-section of the I-beam is not a symmetric; one flange of the beam is thicker than the other. Measure carefully the cross-sectional dimensions and the location of the loading and support points.
  • 54. 3. Carefully place the beam into the MTS machine if it is not already installed. Align the 12- inch black marks on the beam with the roller supports of the lower bending fixture. Make sure the beam is squarely resting and centered on the lower support with the strain gauge facing down. Beam Deflection Dial indicator Beam to Be Tested III-7 4. Enter the TestWorks 4 software by double clicking on the icon on the desktop. a. When prompted, make sure the Name field under the User Login says “306A_lab”. b. Click OK to login. c. Under the Open Method dialog, select “exp-3 4 Point Flex Mod X”. d. Now, select the Motor Reset button in the bottom right corner by clicking on it. 5. Zero the “load” readout by right clicking on the “Load cell” icon and selecting “zero channel”.
  • 55. 6. Next we will use the handset to position the upper bending fixture over the beam. a. Enable the handset by pressing the “unlock” button at the top right of the handset b. Slowly lower the crosshead using the down arrow until the fixture is NEARLY touching the beam. c. Make sure not to pinch the strain gauge lead wires between the fixture and the beam. d. While observing the digital load readout on the screen, use the thumb wheel of the handset to lower the fixture onto the beam. Watch for the load reading to increase when the upper loading component of the fixture makes contact with the beam. e. Now, slowly raise the fixture with the thumb wheel until only a very slight pre-load of approximately 0.2 lb is applied. f. When finished, return control to the computer software by locking the handset using the same button as before. Note that the beam has a strain gauge bonded to its surface. See Figure III-4. This strain gauge has a strain gauge factor of 2.14 ± 0.5% and 350 ohm resistance. Connect the strain gauge lead wires to the “#1 Strain” channel of the grey DAQ box. Data acquisition is conducted by the LabVIEW software.
  • 56. 7. Take the magnetic base holding the dial indicator, and position it with the dial indicator in the center of the beam on the bottom side. Make sure the dial indicator is not touching the strain gauge. Also, be VERY CAREFUL not to damage the dial indicator while doing this! When the dial indicator is in position, lock it to the MTS frame by activating the magnetic base. Finally, zero the dial indicator by turning the dial of the indicator. Note the divisions on the dial indicator and the accuracy with which it can be read. 8. Start the LabVIEW software. a. Double click the LabVIEW icon on the desktop, or follow the procedure suggested for Experiment II. b. Select “Open” and double click “exp2&3-Strain Mod 9-15 LV7.1”. c. Press the white arrow near the upper left corner of the screen to start the strain gauge acquisition. d. Press the “Zero Strain” button to zero-out the strain indicator. 9. You are now ready to run the experiment. a. Press the Green Arrow on the TestWorks 4 GUI. b. Give the sample ID for your group.
  • 57. III-8 c. Under the Required Inputs dialog, enter 0.125 for the thickness and 0.500 for the width. DO NOT press OK yet! 10. You will load the beam up to 1000 lb in increments of 100 lb. To do this: a. “OK” the Required Inputs dialog from above and the test will begin. b. When the digital load readout reaches approximately 100 lb, press the “Pause” button to temporarily stop the test. c. Record the actual load on your data sheet from the digital readout. d. Toggle over to the LabVIEW screen and record the strain reading from the software. e. Record the deflection of the beam from the dial indicator. f. Return to the TestWorks 4 screen and press the “Pause” button again to resume the test. g. Continue pausing and taking readings approximately every 100 lb from 0 to 1000 lb. 11. At 1000 lb the test will stop. Collect your last data point before pressing “OK” at the last dialog. 12. Before repeating the experiment, make sure your load readout, strain indicator, and dial indicator are all re-zeroed.
  • 58. 13. Repeat steps 7 through 9. Figure III-4. Beam with Bonded Strain Gauge REPORT REQUIREMENTS 1. Determine the neutral axis, moment of inertia, and section modulus of the beam cross- section. 2. Draw shear force and bending moment diagrams for the beam for the maximum load. 3. Compute the maximum bending stress and deflection of the beam for the maximum load. 4. Plot deflection versus load, and also stress versus load. Each graph should contain a curve based completely on theoretical calculations, and another using the experimental data points. III-9 5. By error calculation, determine if the beam theory is vindicated within the precision of the instruments. REFERENCES [1] Beckwith, T. G., Buck, N. L. and Marangoni, R. D.,
  • 59. Mechanical Measurements, Addison- Wesley. [2] Popov, E.P. Mechanics of Materials, Prentice-Hall Inc. [3] Thomas, G.B., and Finney R.L., Calculus and Analytical Geometry, Addison-Wesley, 19 Experiment 3 The Beam Ali Almoslim EGME-306A Mo-8:30a.m ABSTRACT The main objectives of this experiment are to determine the stress, deflection, and the strain of a supported beam under load. Also to experimentally verify the beam stress and flexure formulas. The experiment was accomplished by using the machine which applies a load to a supported beam and measure the deflection and the strain of it. The moment of inertia was 0.0845 and deflection of 0.0083%. TABLE OF CONTENTS Abstract Table of Contents Introduction and Theory Procedures Summary of Important Results
  • 60. Sample Calculations and Error Analysis Discussion and Conclusion References Appendix INTRODUCTION AND THEORY Structural members are usually designed to carry tensile, compressive, or transverse loads. A member which carries load transversely to its length is called a beam. In this experiment, a beam will be symmetrically loaded as shown in Fig. III-1(a), where P is the applied load. Note that at any cross section of the beam there will be a shear force V (Fig. III-1(b)) and moment M (Fig. III-1c). Also, in the central part of the beam (between the loads P/2) V is zero and M has its maximum constant value. Notice the sign convention of a positive moment, M, causing a negative (downward) deflection, y. If in this part a small slice EFGH of the beam is imagined to be cut out, as shown, then it is clear that the external applied moment, M, must be balanced by internal forces (stresses) at the sections (faces) EF and GH. For M applied as shown in Fig. III- 2(a), these forces would be compressive near the top, EG, and tensile near the bottom, FH. Since the beam material is considered elastic, these forces would deform the beam such that the length EG would tend to become shorter, and FH would tend to become longer. The first fundamental assumption of the beam theory can be stated as follows: “Sections, or cuts, which are plane (flat) before deformation, remain plane after deformation.” Thus, under this assumption, the parallel and plane section EF and GH will deform into plane sections E’F’ and G’H’ which will intersect at point O, as shown in Fig. III-2(b). Since E’F’ and G’H’ are no longer parallel, they can be thought of as being sections of a circle at some radial distance from O. Convince yourself of this by drawing a square on an eraser and observe its shape when you bend the eraser. Since the forces near E’G’ are
  • 61. compresiive, and those near F’H’ are tensile, there must be some radial distance r where the forces are neither compressive nor tensile, but zero. This axis, N-N, is called the neutral axis. Notice that N-N is not assumed to lie in the center of the beam. Consider an arc of distance +η, from the neutral axis, or distance r + η from O (Fig. III-2(b)). At this radius, the length of arc is l’=(r + η) Δθ. As shown in Fig. III-2(a), the length of the arc was l before the deformation. This length is also equal to rΔθ (because at N-N there are no forces to change the length). Thus, the strain at distance +η from the neutral axis can be found by: (III-1) In other words, the axial strain is proportional to the distance from the neutral axis. It is remarked that this strain is positive, because positive η was taken on the tensile side of N-N in Fig. III- 2(b). Had η been taken in the opposite direction, then the strain would have been negative, as appropriate for the compressive side. The second fundamental assumption is that Hooke’s Law applies both in tension and compression with the same modulus of Elasticity. Thus, from Eqs. (I-3) and (III-1), (III-2) If c is the maximum distance from the neutral axis (largest positive or negative value of η), then the maximum stress (compressive or tensile) is given by σm = Ec/r, and Eq. (III-2) can also be written as (III-3) That is, the stress at a section EF or GH, due to applied moment M, varies linearly from zero at the neutral axis to some maximum value σm (positive or negative) when η = c. To obtain the beam stress formula, it remains to define where the neutral axis is located, and to relate σm to M. To locate the neutral axis, it is observed that the tensile and
  • 62. compressive forces on a section are equal to the stress times a differential element of area, as shown in Fig. III-2(c). For static equilibrium, the sum (or integral) of all these internal forces must be zero. That is, where, the integrals are over the whole cross-sectional area. Thus, it is seen that the neutral axis is located such that the first moment of area about it is zero; that is, the neutral axis passes through the centroid of the cross-sectional area. In Fig. III-2(c), a rectangular area was used for illustration; however, any shape of vertically symmetric cross-sectional area is valid for the area integral. In a similar fashion, the moment due to all the forces is the sum (or integral) of the forces times their moment arms about the neutral axis, and this must be equal to the external applied moment. Thus, (III-4) If I is defined as the second moment of area about the neutral axis, commonly called the moment of inertia, (III-5) then Eq. (III-4) can be written as: (III-6) where Z = I/c is the section modulus, which depends only on the cross-sectional geometry of the beam. Equation (III-6) is the beam stress equation which relates the maximum (compressive or tensile) stress to the applied moment. Notice its similarity to Equation (I-1), the stress equation for uniaxial tension. It is understood, of course, that σm is the maximum bending stress at a particular location, x, along the beam. In general, both σm and M are functions of x, and are related by Eq. (III-6). The remaining question about the beam concerns its degree of
  • 63. deformation, or flexure. That is, how is the radius of curvature, r, related to the moment M (or load P)? From calculus, it can be shown that the curvature of a function y(x) is given by Thus, if x is the distance along the beam, y will be the deflection as indicated in Fig. III-1(a). For most beams of practical interest, this deflection will be small, so that the slope dy/dx will be very small compared to 1. Hence, a very good approximation is But, since σm = Ec/r = Mc/I, there results the differential equation of the elastic curve: (III-7) To obtain the elastic curve of the beam, y(x), and the maximum deflection, ym, it is necessary to integrate Eq. (III-7) using the moment function M(x) in Fig. III-1(c). Thus, using M(x) = Px/2 for 0 ≤ x ≤ a and M(x) = Pa/2 for a ≤ x ≤ a + b, it is found that and that the maximum deflection at x = a + b/2 is (III-8) In particular, for a = b = L/3, (III-9) PROCEDURES This experiment will be applied on a 1018 steel beam ,which has E = 30x10*6 psi, using the MTS testing machine. Because the beam is not perfectly symmetric we had to measure the cross-sectional dimension carefully and the location of the loading points. After that, we placed the beam in the testing machine and aligned the 12-inch black marks on the beam with roller supports of the lower fixture. We had to make sure the beam is centered on the lower support with strain gauge facing
  • 64. down. Then we went to the computer and entered the TestWork 4 software. When prompted, we had to make sure the name field under login says “306A_Lab” then clicked OK to login then under the Open Method dialog, we had to select“exp-3 4 Point Flex Mod X” after that we had to select the Motor Reset button right corner by clicking on it. We zeroed the load by right clicking on the load cell icon and selecting zero channel. Then, we had position the upper bending fixture over the beam using the handset after enabling the handset by pressing the unlock button, then slowly lowering the crosshead using the down arrow until the fixture is touching the beam. After that, we used the thumb wheel of the handset to lower the fixture onto the beam. We watched for the load reading increase when the upper fixture makes contact. Then, we slowly lifted the fixture until a very slight preload of approximately 0.2 lb is applied. Then we locked the handset. Then we connected the strain gauge wire to the #1 strain channel of the grey DAQ box, data acquisition is conducted by the LabVIEW software. We took the magnetic base holding the dial indicator, and positioned it with the dial indicator in the center of the beam on the bottom side. We made sure the dial indicator is not touching the strain gauge, after that we locked the MTS frame by activating the magnetic base then zeroed the dial indicator. Next, we started the LabVIEW software by double click the icon on the desktop, selecting “Open” and double clicking “exp2&3-Strain Mod-15 LV7.1” then we pressed the white arrow to start the strain gauge acquisition then we zeroed the strain. After that we pressed the green arrow on the TestWorks 4 GUI. We loaded the beam up to 1000 lb for each in 100 lb and recorded the data. After reaching 1000 lb we repeated the experiment after making sure it zeroed out just to assure accuracy of our work. SUMMARY OF IMPORTANT RESULTS I In graph I. we can notice that it’s a deflection vs load diagram
  • 65. as titled. we have the theoretical values and the experimental ones. As we can see from the graph the is a noticeable difference between the two values as the two lines get different slops. II As we can see we, graph II is about the strees vs load diagram, where the load is the x-axis and the stress is the y-axis which is represented in a straight line as shown. SAMPLE CALCULATION AND ERROR ANALYSIS To find the use the following equation: Using the position points, calculate the neutral axis, but it is important to remember that this value isn’t the maximum neutral axis, which is what is needed. In that case, simply
  • 66. subtract total distance by the calculated neutral axis. The moment of Inertia is calculated by the following formula: Manually determined moment of Inertia To find the section modulus we use the formula: Where I is the calculated moment of Inertia and c is the distance from the neutral axis To find the maximum bending stress we start with the following equation: Where, P (maximum load) = 997.3 lb a (distance away from reference) = 4 in Providing the applied moment of, Now the maximum bending stress can be found using Eq. (1) The deflection is determined by the following formula: Or Now, Eq. (9) is used to find the deflection of the crossbeam given the maximum pressure, moment of inertia, and distance between loads on the beam. Using the previous maximum load, initial length of 12 in, initial modulus elasticity and calculated moment of Inertia, the maximum deflection comes out to be… The Stress and Deflection percent error were calculated using the results obtained by the load applied by the MTS machine and the dial indicator and a thermometer. Stress () % Error = x 100% = x 100% = 1.20% Deflection % Error = x 100% = x 100% = .0083 % Ai Yi Ay D AD^2 I
  • 67. 1 0.261 0.1295 0.034 0.3875 0.039 0.00146 2 0.116 0.524 0.061 0.007 5.7x10^-6 2.7x10-3 3 0.253 0.63 0.915 0.231 0.326 0.398 0.4 1.36x10^-6 D1= 0.517 – 0.1295 = 0.3875 in Y=∑yA/∑A = 0.326/0.63 = 0.517 D2= 0.524 – 0.517 = 0.007 in D3= 0.915 – 0.517= 0.398 in I1= (bh^3)/12 = (1.007(0.259)^3)/12= 1.46x10^-3 in ^4, I2= 2.714x10^-3 in^4, I3= 1.36x10^-3 in^4 Moment = 0.039 + 5.7x10^-6 + 0.04 + 1.4x10^-3 + 2.714x10^-3 + 1.36x10^-3 = 0.0845 in^4
  • 68. Error Analysis Human error is a common one that could happen in any experiments, in this example, human error might happened in the setup of the equipment. Another possibility is that we didn’t get the write readings. Also, the machine could be old and not accurate enough. Last one is that the sample we tested might not manufactured properly. So these reasons explain why we got 1.20% error for stress and 0.0083% for deflection. DISCUSSION AND CONCLUSION After finishing the experiment and calculating the data and graph them,we figured out many features of the beam. Whenever the load increases, the deflection and the stress increase too. We can use the strain to find the theoretical stress and we can use moment, moment of inertia, and the neutral axis to find the experimental stress. And about the deflection, we used the dial indicator to find it directly from the experiment for the experimental and we used load, length, modulus of elasticity, and the moment of inertia to find the theoretical data. The graph clearly shows that the load and deflection and load and stress have a directly proportional relationship. REFRENCES [1] CSUF EGME 306A Lab Manual. Appendix
  • 69. deflection(in) Max Stress at B 0.0009 1225.223179 0.002 2450.446358 0.0033 3663.417306 0.0046 4888.640485 0.0059 6113.863664 0.0073 7351.339075 0.0086 8539.805559 0.0099 9814.037665 0.0103 11049.06263 0.0107 12217.92554 Distance of Neutral Axis Moment of Inertia Applied Moment Modulus of Elasticity 0.5179 0.0845397 200 30000000
  • 71. 1803.6 1994.4 Deflection Vs. Load Theoretical Calcuations 100 200 299 399 499 600 697 801 901.8 997.2 1.2091623488399204E-3 2.4183246976798409E-3 3.6153954230313621E-3 4.8245577718712823E-3 6.0337201207112034E-3 7.2549740930395222E-3 8.4278615714142458E-3 9.6853904142077622E-3 1.0904226061838402E-2 1.2057766942631687E-2 Experimental Values 100 200 299 399 499 600 697 801 901.8 997.2 8.9999999999999998E-4 2E-3 3.3E-3 4.5999999999999999E-3 5.8999999999999999E-3 7.3000000000000001E-3 8.6E-3 9.9000000000000008E-3 1.03E-2 1.0699999999999999E-2 Load (lbs) Deflection(in) Stress vs. Load 100 200 299 399 499 600 697 801 901.8 997.2 1225.223179169077 2450.4463583381539 3663.4173057155399 4888.6404848846169 6113.8636640536934 7351.3390750144608 8539.8055588084662 9814.0376651443075 11049.062629746735 12217.925542674035 100 200 299 399 499 600 697 801 901.8 997.2 Load (lbs)