27381847 Space Full Report

8/14/2019 27381847 Space Full Report

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1.0 OBJECTIVE

The objective is to verify member forces obtain from experiment with tension

coefficient method.

2.0 LEARNING OUTCOME

There are many learning outcome that we can get from this laboratory test:

2.1 The application of theoretical engineering knowledge through practical

application.

2.2 To enhance the technical competency in structural engineering through

laboratory application.

2.3 ommunicate effectively in group.

2.! To identify the problem" solving and finding out the appropriate solution

through laboratory application.

3.0 THEORY

#f a members of a truss system is situated not in a two dimensional plane" then thetruss is defined as a space frame truss. #n other words" space truss has components

in three axis i.e. x" y and $.

onsider a member with node % &x%" y%' and ( &x("y('

%ssume te force in the member is T%( &)ve tension' and length *%(

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+efinition of tension coefficient &t'" t%( , T%(

*%(

%t %" the hori$ontal component T%(is :

T%(cos - , t%( *%( cos- , t%( *%( &x( x%'

*%(

, t%( &x( x%'

/se the same method" the vertical component at % is :

, t%( &y( y%'

%t (" the hori$ontal component T%( , t%( &x% x('

0ertical component T%( , t%( &y% y('

/sing statics" write the euation for each joint using the coordinate value and

solve for it. onvert it into force using:

T%( , t%( *%( , &x( x%'2 ) &y( y%'2

% space frame or space structure is a trusslike" lightweight rigid structure

constructed from interlocking struts in a geometric pattern. 4pace frames usuallyutili$e a multidirectional span" and are often used to accomplish long spans with

few supports. They derive their strength from the inherent rigidity of the

triangular frame5 flexing loads &bending moments' are transmitted as tension and

compression loads along the length of each strut. 6any architects and engineers

throughout the world have expressed their design freedoms with 4pace 7rame

4ystems. The simplicity of these systems provides a natural link between

yesterday and today. 7or this reason" designers have specified 4pace 7rame

4ystems for an increasing variety of renovation and remodeling applications. The

modular systems allow fast track delivery and job site assembly at affordable

prices. 4pace frame systems give you the architectural beauty you desire within

the budget you set.

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4ome space frame applications include:

1' 8arking canopy9s

2' otel;ospital;commercial building entrances

3' ommercial building lobbies;atriums

4ome advantages of space frame systems over conventional systems are:

1'


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S!"e #r!$e !!r!(u'

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3. 6easure the distance b" c and d" and then record it in table 1.

!. Ee record the dynamometer readings for members 41" 42 and 43.

=. %fter that" we put the selected load on hanger at + and record it.

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>. Then" we repeat step &2' to &!' with the different value of a.

?. 7inally" calculate the theoretical member forces and record it in table one.

8art 2

1. 7or part 2" we use a distance of 3=C mm for a.

2. Then" we place the hanger on +.3. %fter that" measure the distance b" c" and d. Then we record the dynamometer

reading for member 41" 42" 43 in table 2.

!. The next step is we put a load of = D on the hanger and record the

dynamometer readings.

=. Ee repeat step 2 to ! using the different load.

>. Ee complete the table 2 by calculating the theoretical member value.

?. The last one is we plot the graf of force against load for the theoretical and

experimental results.

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+.0 RESULT

T!,-e 1

i$e&'i%& /$$ &!$%$e(er Re!i&g F%r"e /N

! , "

S1 S2 S3 Eeri$e&( T4e%r

L%!e U&-%!e L%!e U&-%!e L%!e U&-%!e S1 S2 S3 S1 S2 S3

=CC !@3 2>C 3>= >C !C == C @C 1 2C == @C 1!2.3= 1!2.3= 2?

!CC =C3 21= 3>= ?@ ?C ?! !C 11C C @ 3! 11C 1!1.=! 1!1.=! 2?

3CC =32 1?C 3>= 1CF 1= 1C! 11 1=C 1C F! F3 1!C ?2.1> ?2.1> 13

2CC =!@ 11C 3>= 1@1 31 1?= 2@ 2!C !C 1=C 1!? 2CC 1!>.1! 1!>.1! 2?

T!,-e 2

L%!

/N

&!$%$e(er Re!i&g F%r"e /N

S1 S2 S3 Eeri$e&( T4e%r

L%!e U&-%!e L%!e U&-%!e L%!e U&-%!e S1 S2 S3 S1 S2 S3

= =C 1C != ? ?C 2C !C 3@ =C =3.!@ =3.!@ F@

1C FC 1C @= ? 12C 2C @C ?@ 1CC 1C>.F> 1C>.F> 1F1= 13! 1C 13C ? 1@C 2C 12! 123 1>C 1>C.!3 1>C.!3 2F

2C 1?@ 1C 1?= ? 2!= 2C 1>@ 1>@ 22= 213.F1 213.F1 3F

2= 23C 1C 22= ? 3C= 2C 22C 21@ 2@= 2>?.3F 2>?.3F !F

Dimension a = 350mm Dimension b = 521mm Dimension c = 185mm Dimension d = 365mm

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!(! A&!-'i'

8art 1:

Axample Bf Axperiment alculation

a = 500 mm

1. 41 : loaded , >CD/nloaded , !CD

4o S1 = 60N 40N=20N

2. 42 : loaded , ==D/nloaded , CD

4o 42 , 55N 0N= 55N

3. 43 : loaded , @CD/nloaded , CD

4o S3 = 80N 0N= 80 N

*!r( 1:

F


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Table 1 theoretical calculation*oad 7 , 1C D

1. !, =CCmm" ,, !@3mm" ", 2>Cmm" , 3>=mm.

*x , b*y , d;2 &for 41 and 42'

*$ , a c &for 41 and 42'7 , * x t* , &*xG ) *yG ) *$G'

H 7x , C5 !@3s1 ) !@3s2 ) !@3s3 , C

H 7I , C5 1@2.=s1 ) 1@2.=s2 ) Cs3 , C

H 7$ , C5 2!Cs1 ) 2!Cs2 ) 2>Cs3 , 1C

(y using calculator

4o ts1 , C.2= ts2 , C.2= ts3 , C.=C

%ndS1 = 142.35 S2 = 142.35

S3 = -274.27

2. !, !CCmm" ,, =C3mm" ", 21=mm" , 3>=mm.

1C

Me$,er L/$$

L/$$

L5/$$

L /$$ ( F /N Re$!r6'

S1 !@3 1@2.= 2!C =>F.3@ C.2= 1!2.3= TensionS2 !@3 1@2.= 2!C =>F.3@ C.2= 1!2.3= TensionS3 !@3 C.CC 2>C =!@.=3 C.=C 2?!.2? ompression7orce &D' C C 1C


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*x , b*y , d;2 &for 41 and 42'


H 7x , C5 =C3s1 ) =C3s2 ) =C3s3 , C

H 7I , C51@2.=s1 ) 1@2.=s2 ) Cs3 , C

H 7$ , C5 1@=s1 ) 1@=s2 ) 21=s3 , 1C

(y using calculator

4o ts1 , C.2= ts2 , C.2=

ts3 , C.=C

%nd S1 = 141.54 S2 = 141.54

S3 = -273.51

3. !, 3CCmm" ,, =32mm" ", 1?Cmm" , 3>=mm.

*x , b*y , d;2 &for 41 and 42'

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Me$,er L/$$

L/$$

L5/$$

L /$$ ( F /N Re$!r6'

S1 =C3 1@2.= 1@= =>>.1> C.2= 1!1.=! TensionS2 =C3 1@2.= 1@= =>>.1> C.2= 1!1.=! TensionS3 =C3 C.CC 21= =!?.C2 C.=C 2?3.=1 ompression7orce &D' C C 1C


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H 7x , C5 =32s1 ) =32s2 ) =32s3 , C

H 7I , C5 1@2.=s1 ) 1@2.=s2 ) Cs3 , C

H 7$ , C5 13Cs1 ) 13Cs2 ) 1?Cs3 , 1C

(y using calculator

4o ts1 , C.12=ts2 , C.12=

ts3 , C.2=

%nd S1 = 72.16 S2 = 72.16

S3 = -139.63

!. !, 2CCmm" ,, =!@mm" ", 11Cmm" , 3>=mm.

*x , b*y , d;2 &for 41 and 42'*$ , a c &for 41 and 42'

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Me$,er L/$$

L/$$

L5/$$

L /$$ ( F /N Re$!r6'

S1 =32 1@2.= 13C =??.2> C.12= ?2.1> TensionS2 =32 1@2.= 13C =??.2> C.12= ?2.1> TensionS3 =32 C.CC 1?C ==@.=C C.2= 13F.>3 ompression7orce &D' C C 1C


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7 , * x t* , &*xG ) *yG ) *$G'

H 7x , C5 =!@s1 ) =!@s2 ) =!@s3 , C

H 7I , C5 1@2.=s1 ) 1@2.=s2 ) Cs3 , C

H 7$ , C5 FCs1 ) FCs2 ) 11Cs3 , 1C

(y using calculator

4o ts1 , C.2=ts2 , C.2=

ts3 , C.=C

%ndS1 = 146.14S2 = 146.14

S3 = -279.47

*!r( 2 :

Table 2 theoretical calculation:

1. Load F = 5 N

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Me$,er L/$$

L/$$

L5/$$

L /$$ ( F /N Re$!r6'

S1 =!@ 1@2.= FC =@!.=> C.2= 1!>.1! TensionS2 =!@ 1@2.= FC =@!.=> C.2= 1!>.1! TensionS3 =!@ C.CC 11C [email protected] C.=C 2?F.!? ompression

7orce &D' C C 1C


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!, 3=Cmm" ,, =21mm" ", 1@=mm" , 3>=mm.

*x , b*y , d;2 &for 41 and 42'


H 7x , C5 3=Cs1 ) 3=Cs2 ) 3=Cs3 , C

H 7I , C5 1@2.=s1 ) 1@2.=s2 ) Cs3 , C

H 7$ , C5 1>=s1 ) 1>=s2 ) 1@=s3 , =

(y using calculator

4o ts1 , C.12=ts2 , C.12=

ts3 , C.2=

%nd S1 = 53.48S2 = 53.48

S3 = -98.97

1!

Me$,er L/$$ L /$$ L5/$$ L /$$ ( F /N Re$!r6'

S1 3=C 1@2.= 1>= !2?.@2 C.12= =3.!@ TensionS2 3=C 1@2.= 1>= !2?.@2 C.12= =3.!@ TensionS3 3=C C.CC 1@= 3F=.@F C.2= [email protected]? ompression7orce&D'

C C =


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2. Load F = 10 N

!, 3=Cmm" ,, =21mm" ", 1@=mm" , 3>=mm.

*x , b*y , d;2 &for 41 and 42'

*$ , a c &for 41 and 42'7 , * x t

* , &*xG ) *yG ) *$G'

H 7x , C5 3=Cs1 ) 3=Cs2 ) 3=Cs3 , C

H 7I , C5 1@2.=s1 ) 1@2.=s2 ) Cs3 , C

H 7$ , C5 1>=s1 ) 1>=s2 ) 1@=s3 , 1C

(y using calculator

4o ts1 , C.2=ts2 , C.2=

ts3 , C.=C

%nd S1 = 106.96S2 = 106.96

S3 = -197.95

1=

Me$,er L/$$

L /$$ L5/$$

L /$$ ( F /N Re$!r6'

S1 3=C 1@2.= 1>= !2?.@2 C.2= 1C>.F> TensionS2 3=C 1@2.= 1>= !2?.@2 C.2= 1C>.F> TensionS3 3=C C.CC 1@= 3F=.@F C.=C 1F?.F= ompression7orce&D'

C C 1C


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3. Load F = 15 N

!, 3=Cmm" ,, =21mm" ", 1@=mm" , 3>=mm.

*x , b*y , d;2 &for 41 and 42'

*$ , a c &for 41 and 42'7 , * x t

* , &*xG ) *yG ) *$G'

H 7x , C5 3=Cs1 ) 3=Cs2 ) 3=Cs3 , C

H 7I , C5 1@2.=s1 ) 1@2.=s2 ) Cs3 , C

H 7$ , C5 1>=s1 ) 1>=s2 ) 1@=s3 , 1=

(y using calculator

4o ts1 , C.3?=ts2 , C.3?=

ts3 , C.?=

%nd S1 = 160.43S2 = 160.43

S3 = -296.92

1>

Me$,er L/$$

L/$$

L5/$$

L /$$ ( F /N Re$!r6'

S1 3=C 1@2.= 1>= !2?.@2 C.3?= 1>C.!3 TensionS2 3=C 1@2.= 1>= !2?.@2 C.3?= 1>C.!3 TensionS3 3=C C.CC 1@= 3F=.@F C.?= 2F>.F2 ompression7orce&D'

C C 1=


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4. Load F = 20 N

!, 3=Cmm" ,, =21mm" ", 1@=mm" , 3>=mm.

*x , b*y , d;2 &for 41 and 42'

*$ , a c &for 41 and 42'7 , * x t

* , &*xG ) *yG ) *$G'

H 7x , C5 3=Cs1 ) 3=Cs2 ) 3=Cs3 , C

H 7I , C5 1@2.=s1 ) 1@2.=s2 ) Cs3 , C

H 7$ , C5 1>=s1 ) 1>=s2 ) 1@=s3 , 2C

(y using calculator

4o ts1 , C.=ts2 , C.=

ts3 , 1

%nd S1 = 213.91S2 = 213.91

S3 = -395.89

1?

Me$,er L /$$ L /$$ L5/$$

L /$$ ( F /N Re$!r6'

S1 3=C 1@2.= 1>= !2?.@2 C.= 213.F1 TensionS2 3=C 1@2.= 1>= !2?.@2 C.= 213.F1 TensionS3 3=C C.CC 1@= 3F=.@F 1 3F=.@F ompression7orce&D'

C C 2C


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5. Load F = 25 N

!, 3=Cmm" ,, =21mm" ", 1@=mm" , 3>=mm.

*x , b*y , d;2 &for 41 and 42'

*$ , a c &for 41 and 42'7 , * x t

* , &*xG ) *yG ) *$G'

H 7x , C5 3=Cs1 ) 3=Cs2 ) 3=Cs3 , C

H 7I , C5 1@2.=s1 ) 1@2.=s2 ) Cs3 , C

H 7$ , C5 1>=s1 ) 1>=s2 ) 1@=s3 , 2=

(y using calculator

4o ts1 , C.>2=ts2 , C.>2=

ts3 , 1.2=

%nd S1 = 267.39S2 = 267.39

S3 = -494.86

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Graph of Force versus Loa

0

50

100

150

200

250

300

5 10 15 20 25

Load (N)

Force

(N)

S1 Exp

S1 The

Graph of Force versus Load

0

50

100

150

200

250

300

5 10 15 20 25

Load (N)

Force

(N)

S2 Exp

S2 The

7.0 ISSCUSSION

1F


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(ased on the graph that have been plotted" we can see that for the graph1" the

comparison between the theoretical and the experimental results is there is not much

different for the two lines. Ehen more load were applied" the value of force was also

increase.#t is same like the graph2" which is there is a little difference between the

theoretical and the experimental results. The value of force in increase due to the

increasing of load.

(ut for the graph3" the results of the theoretical and the experimental is totally

difference because for the experiment" the results is in range C to 3CC while for the results

of the theoretical is around range C to 3CC. 7or the theoretical" when more load were

applied" the value of force were decrease but for the experimental" when more load were

applied" the force will increase.

The reason of discrepancy in the results maybe cause by the spring that used was

not elastic anymore after being stretched for many time of doing experiment" it might

have a mistake during taking the results. (eside that" it maybe cause by the error of the

apparatus which is not in good condition.

8.0 CONCLUSION

The experiment is to prove experimental and theoretical have a small relative

value. 4pace frames usually utili$e a multidirectional span" and are often used to

accomplish long spans with few supports. They derive their strength from the inherent

rigidity of the triangular frame5 flexing loads &bending moments' are transmitted as

tension and compression loads along the length of each structure.

#n many ways this looks like the hori$ontal jib of a tower crane repeated many

times to make it wider. % stronger purer form is composed of interlocking tetrahedral

pyramids in which all the struts have unit length. 6ore technically this is referred to as anisotropic vector matrix or in a single unit width an octet truss.

2C

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27381847 Space Full Report