span deflection (double integration method)

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


Date : 03/10/2011

___________________________

Student Signature

Name : AHMAD FARHAN BIN RAKAWI


Date : 03/10/2011

___________________________

Student Signature

Name : AINUN NAZHIRIN ABD JALIL


Date : 03/10/2011

___________________________

Student Signature

Name : IDAMAZLIZA ISA


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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12

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 %



( 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 %



(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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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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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

span deflection (double integration method)

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