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7/27/2019 SPAN DEFLECTION (DOUBLE INTEGRATION METHOD)
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FACULTY OF CIVIL AND ENVIRONMENTAL
ENGINEERING
DEPARTMENT OF STRUCTURE AND MATERIAL
ENGINEERING
LAB MATERIAL
REPORT Subject Code BFC 21201
Code & Experiment Title SPAN DEFLECTION (DOUBLE INTERGRATION METHOD)
Course Code 2 BFF/1
Date 03/10/2011
Section / Group 2
Name MUHAMAD ASYRAF BIN AB MALIK (DF100108)
Members of Group 1.MUHAMMAD IKHWAN BIN ZAINUDDIN (DF100018)2.AHMAD FARHAN BIN RAKAWI (DF100142)
3.IDAMAZLIZA BINTI ISA (DF100128)
4.AINUN NAZHIRIN BINTI ABD JALIL (DF100076)
Lecturer/Instructor/Tutor EN MOHAMAD HAIRI BIN OSMAN
Received Date 10 OCTOBER 2011
Comment by examiner Received
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STUDENT CODE OF ETHIC
(SCE)DEPARTMENT OF STRUCTURE AND MATERIAL
ENGINEERING
FACULTY OF CIVIL & ENVIRONMENTAL ENGINEERING
UTHM
We, hereby confess that we have prepared this report on our effort. We also admit not to receive
or give any help during the preparation of this report and pledge that everything mentioned in the
report is true.
___________________________
Student Signature
Name : MUHAMAD ASYRAF AB MALIK
Matric No. : DF100108
Date : 03/10/2011
_______________________
Student Signature
Name : MUHAMMAD IKHWAN ZAINUDDIN
Matric No. : DF100018
Date : 03/10/2011
___________________________
Student Signature
Name : AHMAD FARHAN BIN RAKAWI
Matric No. : DF100142
Date : 03/10/2011
___________________________
Student Signature
Name : AINUN NAZHIRIN ABD JALIL
Matric No. : DF100076
Date : 03/10/2011
___________________________
Student Signature
Name : IDAMAZLIZA ISA
Matric No. : DF100128
Date : 03/10/2011
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1.0 OBJECTIVE
Main propose of our experiment is to determine the relationship between span and
deflection.
2.0 INTRODUCTIONA beam must possess sufficient stiffness so that excessive deflections do not have
an adverse effect on adjacent structural members. In many cases, maximum allowable
deflections are specified by Code of Practice in terms of the dimensions of the beam,
particularly the span. The actual deflections of a beam must be limited to the elastic range
of the beam, otherwise permanent distortion result. Thus in determining the deflections of
beam under load, elastic theory is used.
In this experiment double integrations method is used to give the complete
deflected shape of the beam.
3.0 THEORY
L/2-x
A x c B
X
L/2 L/2
Beam with point load at mid span
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bdI
backboardthefromobtainedbecanEwhere
support)(at16EI
PLvL/2;x
c)span;(mid48EI
PL
Y0;When x
48
PLB
B96
PL
32
PL0;yL/2;When x
0A,0dyo;When x
BAx12Px
8PLxEIyy
A4
Px
4
PLx
dx
dyEIV
2
L
2
P
dx
ydEIM
3
2
mak
3
mak
3
33
32
x-x
2
xx
2
2
x-x
b
d
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4.0 APPARATUS
Figure 1: Apparatus for Span Deflection Experiment ( Double Integration Method )
Figure 2 : Digital Dial Test Indicator Figure 3 : Hanger And Masses
Figure 2 : Specimen Beam ( Steel )
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5.0 PROSEDURE1. The moveable knifeedge supports had been positioned so that they are 400mm.2. The chosen beam had placed on the support.3. The hanger and the digital dial test indicator had placed at mid span. The Digital
Reading had been zeroed.
4. Incremental load had been applied and the deflection recorded for each increment inthe table below.
5. The above step repeated using span of 300mm and 500mm.
6.0 RESULTSpecimen Beam : Steel
Youngs Modulus, E Steel = 207 GN/m2
= 207
x 109Nm
-2
Second moment of area, Irectangle
b = 18.97 mm = bd3
d = 3.15 mm 12
=(18.97 x 10-3
)(3.15 x 10-3
)3
12
= 4.941 x 10-11
mm4
EI for rectangular Steel = (207 x 109 )( 4.941 x 10-11 )
= 10.25 Nm2
= 10.25 x 106
Nmm2
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Experiment 1 : Span = 500mm : 0.5m
No. Mass*(N) Deflection (experimental)
(mm)
Theoretical Def.
( )MakY )
% Different
1. 0.981 0.26 - 0.249 4.417 %
2. 1.962 0.52 - 0.498 4.417 %
3. 2.943 0.81 0.748 8.289 %
Experiment 2 : Span = 400mm : 0.4m
No. Mass*(N) Deflection (experimental)
( Ymax) (mm)
Theoretical Def.
(Ymax) (mm)
% Different
1. 0.981 0.17 - 0.128 32.813 %
2. 1.962 0.29 - 0.255 13.37 %
3. 2.943 0.43 - 0.383 12.272 %
Experiment 3 : Span = 300mm : 0.3m
No. Mass*(N) Deflection (experimental)
(mm)
Theoretical Def.
( )MakY )
% Different
1. 0.981 0.08 -0.054 44.82 %
2. 1.962 0.14 -0.108 29.62 %
3. 2.943 0.18 -0.162 11.11 %
Ymax
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EI
PLY
48
3
max
100
max
max
Y
DeflectionY
EI
PLY
48
3
max
EI
PLY
48
3
max
100max
max
Y
DeflectionY
100max
max
Y
DeflectionY
7.0 ANALYSISExperiment 1 : Span = 500mm : 0.5m
MASS (N) DEFLECTION (( )MakY ) DIFFERENT (%)
0.981
= - 0.981 (500)3
48( 10.25 x 106)
= - 0.249 mm
0.249 mm
= 0.2490.26) x 100
0.249
= 4.417 %
1.962
= - 1.962 (500)3
48( 10.25 x 106)
= - 0.498 mm
= (0.4980.52) x 100
0.498
= 4.417 %
2.943
= - 2.943 (500)3
48( 10.25 x 106)
= - 0.748 mm
= (0.7480.81) x 100
0.748
= 8.289%
-0.498mm
-0.748mm
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Experiment 2 : Span = 400mm : 0.4m
MASS (N) DEFLECTION (( )MakY ) DIFFERENT (%)
0.981
= - 0.981 (400)
48( 10.25 x 106)
= - 0.128 mm
= ( 0.1280.17) x 100
0.128
= 32.813 %
1.962
= - 1.962 (400)
48( 10.25 x 106)
= - 0.255 mm
= ( 0.255 0.29) x 100
0.255
= 13.37 %
2.943
= - 2.943 (400)
48( 10.25 x 106)
= - 0.383 mm
= ( 0.3830.43) x 100
0.383
= 12.272 %
-0.128mm
-0.255mm
-0.383mm
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Experiment 3 : Span = 300mm : 0.3m
MASS (N) DEFLECTION (( )MakY ) DIFFERENT (%)
0.981
= - 0.981 (300)
48( 10.25 x 106
)
= - 0.054 mm
= (0.054 0.08) x 100
0.054
= 44.82 %
1.962
= - 1.962 (300)
48( 10.25 x 106)
= - 0.108 mm
= (0.108 0.14) x 100
0.108
= 29.62 %
2.943
= - 2.943 (300)
48( 10.25 x 106)
= -0.162 mm
= (0.162 0.18) x 100
0.162
= 11.11 %
-0.054mm
-0.108mm
-0.162mm
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8.0 DISCUSSIONComment on the different between the theatrical and experimental result.
From experiment 1 and the span is 500mm we get the different between the
theoretical and experiment 1 result for 0.981N=4.417%, 1.962N=4.417% and
2.943N=8.289%. Then, for experiment 2 with the span is 400mm we get for
0.981N=32..813%, 1.962N=13.37% and 2.943N=12.272%.. Finally, for experiment 3
with the span is 300mm we get for 0.981N=44.82%, 1.962N=29.62% and
2.943N=11.11%.
Based on this different show that our experiment is accurate and success for
experiment 1 because our different value is quite small. It can be because we followed the
procedure without any error while doing it. But experiment 2 and 3 not accurate and
both has a big different of theory and experimental. This can be some errors due to
equipment experiment or environmental interference.
9.0 EXTRA QUESTION
9.1 Calculate the deflection when x = L/3 (experiment 1, no 3). Check the result byplacing the digital dial at this position.
P
L/3-x
A X x C B
X
L/3 2L/3
MB = 0 MX = 0
= RA (L)P( 2L/3 ) = RA (L/3x ) - Mx-x
:- RA = P(2L/3) Mx-x = RA (L/3RA( x )
L = 2P/3(L/3) - 2P/3(x)
= 2P/3 = 2PL/92Px/3
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Mx-x = EI d2y = 2PL - 2Px
dx2
9 3
Vx-x = EI dy = 2PLx - 2Px2
+ A
dx 9 6
Mx-x = EI Ymax = 2PLx2
- 2Px3
+Ax + B
18 18
= PLx2
- Px3
+Ax + B
9 9
X=0, dy = 0 A=0
dx
X=L/3 ,
Ymax = PL3
- PL3
+ B
81 243
B = -2PL3
243EI
= - 2(2.943) (500)3
243( 10.25 x 106)
= - 0.295mm
Experimental Value = 0.65 mm
% diffrent = ( 0.2950.65) x 100
0.295
= 120.338 %
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9.2 Calculate max.V in experiment 2, no. 2.Experiment 2 : Span = 400mm : 0.4m
Specimen Beam : Steel
Vmax = EI dy = PL3
+ A
dx 16
Vmax = PL2
16 EI
= 1.962 (400)2
16( 10.25 x 106)
= 1.914 x 10-3 mm
10.0 CONCLUSION
We can conclude that the experimental value and the theoretical value are not
exactly same. We can see that there are small and big different values. It means that, ourexperiment (span deflection) is not success. From the result, the value for theoretical
deflection is negative. This is because our experiment is in tension condition.
Besides that, we are able to know how much the span can support the load and
have a maximum deflection level until it reached to failure mood. Although wise, we can
design the safety factor from this action.
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11.0 REFERENCES
Mechanics of materials / Ferdinand P. Beer, John T. DeWolf Mechanics of materials / Madhukar Vable Mechanics of materials / James M. Gere, Barry J. Goodno Mechanics of materials / Ansel C. Ugural