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 Communicate effectively in group.

2.4 To identify the problem, solving and finding out the appropriate solution

through laboratory application.

3.0 THEORY

If a members of a truss system is situated not in a two dimensional plane, then the

truss is defined as a space frame truss. In other words, space truss has components

in three axis i.e. x, y and z.

Consider a member with node A (xA, yA) and B (xB,yB)

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Assume te force in the member is TAB (+ve tension) and length LAB

Definition of tension coefficient (t), tAB = TAB

LAB

At A, the horizontal component TAB is :

TAB cos θ = tAB LAB cosθ = tAB LAB (xB – xA)

LAB

= tAB (xB – xA)

Use the same method, the vertical component at A is :

= tAB (yB – yA)

At B, the horizontal component TAB = tAB (xA – xB)

Vertical component TAB = tAB (yA – yB)

Using statics, write the equation for each joint using the coordinate value and

solve for it. Convert it into force using:

TAB = tAB LAB = √ (xB – xA)2 + (yB – yA)2

A space frame or space structure is a truss-like, lightweight rigid structure

constructed from interlocking struts in a geometric pattern. Space frames usually

utilize 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 frame; flexing loads (bending moments) are transmitted as tension and

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

throughout the world have expressed their design freedoms with Space Frame

Systems. The simplicity of these systems provides a natural link between

yesterday and today. For this reason, designers have specified Space Frame

Systems for an increasing variety of renovation and remodeling applications. The

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

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prices. Space frame systems give you the architectural beauty you desire within

the budget you set.

Some space frame applications include:

1) Parking canopy’sH

2) Hotel/Hospital/commercial building entrances

3) Commercial building lobbies/atriums

Some advantages of space frame systems over conventional systems are:

1) Random column placement

2) Column-free spaces

3) Minimal perimeter support

4) Controlled load distribution

5) Redundant integrity

6) Design freedom

7) Supports all types of roofing

8) Exposing building exterior to view adds color, texture and style

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Figure 1 : Space frame for construction

4.0 APPARATUS

Space frame apparatus

PROSEDURES

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Part 1

1. First of all, we select any weight between 10 to 50 N.

2. Then, ensure that the distance a = 500 mm and place load hanger on D.

3. Measure the distance b, c and d, and then record it in table 1.

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4. We record the dynamometer readings for members S1, S2 and S3.

5. After that, we put the selected load on hanger at D and record it.

6. Then, we repeat step (2) to (4) with the different value of a.

7. Finally, calculate the theoretical member forces and record it in table one.

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Part 2

1. For part 2, we use a distance of 350 mm for a.

2. Then, we place the hanger on D.

3. After that, measure the distance b, c, and d. Then we record the dynamometer

reading for member S1, S2, S3 in table 2.

4. The next step is we put a load of 5 N on the hanger and record the

dynamometer readings.

5. We repeat step 2 to 4 using the different load.

6. We complete the table 2 by calculating the theoretical member value.

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

experimental results.

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5.0 RESULT

Table 1

Dimension (mm) Dynamometer Reading Force (N)

a b c dS1 S2 S3 Experiment Theory

Loaded Unloaded Loaded Unloaded Loaded Unloaded S1 S2 S3 S1 S2 S3

500 483 260 365 60 40 55 0 80 1 20 55 80 142.35 142.35 -274.27

400 503 215 365 78 70 74 40 110 0 8 34 110 141.54 141.54 -273.51

300 532 170 365 109 15 104 11 150 10 94 93 140 72.16 72.16 -139.63

200 548 110 365 181 31 175 28 240 40 150 147 200 146.14 146.14 -279.47

Table 2

Load

(N)

Dynamometer Reading Force (N)

S1 S2 S3 Experiment Theory

Loaded Unloaded Loaded Unloaded Loaded Unloaded S1 S2 S3 S1 S2 S3

5 50 10 45 7 70 20 40 38 50 53.48 53.48 -98.97

10 90 10 85 7 120 20 80 78 100 106.96 106.96 -197.95

15 134 10 130 7 180 20 124 123 160 160.43 160.43 -296.92

20 178 10 175 7 245 20 168 168 225 213.91 213.91 -395.89

25 230 10 225 7 305 20 220 218 285 267.39 267.39 -494.86

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

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Data Analysis

Part 1:

Example Of Experiment Calculation

a = 500 mm

1. S1 : loaded = 60NUnloaded = 40N

So S1 = 60N – 40N= 20N

2. S2 : loaded = 55NUnloaded = 0N

So S2 = 55N – 0N = 55N

3. S3 : loaded = 80NUnloaded = 0N

So S3 = 80N –0N = 80 N

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Part 1:

Table 1 theoretical calculationLoad F = 10 N

1. a = 500mm, b = 483mm, c = 260mm, d = 365mm.

Lx = bLy = d/2 (for S1 and S2)Lz = a – c (for S1 and S2)F = L x tL = √ (Lx² + Ly² + Lz²)

Member Lx

(mm)Ly

(mm)Lz

(mm)L (mm) t F (N) Remarks

S1 483 -182.5 240 569.38 0.25 142.35 TensionS2 483 182.5 240 569.38 0.25 142.35 TensionS3 483 0.00 260 548.53 -0.50 -274.27 CompressionForce (N) 0 0 -10 - - - -

∑ Fx = 0; 483s1 + 483s2 + 483s3 = 0

∑ FY = 0;-182.5s1 + 182.5s2 + 0s3 = 0

∑ Fz = 0; 240s1 + 240s2 + 260s3 = -10

By using calculator

So ts1 = 0.25 ts2 = 0.25 ts3 = -0.50

And S1 = 142.35 S2 = 142.35 S3 = -274.27

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2. a = 400mm, b = 503mm, c = 215mm, d = 365mm.

Lx = bLy = d/2 (for S1 and S2)Lz = a – c (for S1 and S2)F = L x tL = √ (Lx² + Ly² + Lz²)

Member Lx

(mm)Ly

(mm)Lz

(mm)L (mm) t F (N) Remarks

S1 503 -182.5 185 566.16 0.25 141.54 TensionS2 503 182.5 185 566.16 0.25 141.54 TensionS3 503 0.00 215 547.02 -0.50 -273.51 CompressionForce (N) 0 0 -10 - - - -

∑ Fx = 0; 503s1 + 503s2 + 503s3 = 0

∑ FY = 0;-182.5s1 + 182.5s2 + 0s3 = 0

∑ Fz = 0; 185s1 + 185s2 + 215s3 = -10

By using calculator

So ts1 = 0.25 ts2 = 0.25

ts3 = -0.50

And S1 = 141.54 S2 = 141.54 S3 = -273.51

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3. a = 300mm, b = 532mm, c = 170mm, d = 365mm.

Lx = bLy = d/2 (for S1 and S2)Lz = a – c (for S1 and S2)F = L x tL = √ (Lx² + Ly² + Lz²)

Member Lx

(mm)Ly

(mm)Lz

(mm)L (mm) t F (N) Remarks

S1 532 -182.5 130 577.26 0.125 72.16 TensionS2 532 182.5 130 577.26 0.125 72.16 TensionS3 532 0.00 170 558.50 -0.25 -139.63 CompressionForce (N) 0 0 -10 - - - -

∑ Fx = 0; 532s1 + 532s2 + 532s3 = 0

∑ FY = 0; -182.5s1 + 182.5s2 + 0s3 = 0

∑ Fz = 0; 130s1 + 130s2 + 170s3 = -10

By using calculator

So ts1 = 0.125ts2 = 0.125

ts3 = -0.25

And S1 = 72.16 S2 = 72.16 S3 = -139.63

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4. a = 200mm, b = 548mm, c = 110mm, d = 365mm.

Lx = bLy = d/2 (for S1 and S2)Lz = a – c (for S1 and S2)F = L x tL = √ (Lx² + Ly² + Lz²)

Member Lx

(mm)Ly

(mm)Lz

(mm)L (mm) t F (N) Remarks

S1 548 -182.5 90 584.56 0.25 146.14 TensionS2 548 182.5 90 584.56 0.25 146.14 TensionS3 548 0.00 110 558.93 -0.50 -279.47 CompressionForce (N) 0 0 -10 - - - -

∑ Fx = 0; 548s1 + 548s2 + 548s3 = 0

∑ FY = 0; -182.5s1 + 182.5s2 + 0s3 = 0

∑ Fz = 0; 90s1 + 90s2 + 110s3 = -10

By using calculator

So ts1 = 0.25ts2 = 0.25

ts3 = -0.50

And S1 = 146.14 S2 = 146.14 S3 = -279.47

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Part 2 :

Table 2 theoretical calculation:

1. Load F = 5 Na = 350mm, b = 521mm, c = 185mm, d = 365mm.

Lx = bLy = d/2 (for S1 and S2)Lz = a – c (for S1 and S2)F = L x tL = √ (Lx² + Ly² + Lz²)

Member Lx

(mm)Ly

(mm)Lz

(mm)L (mm) t F (N) Remarks

S1 350 -182.5 165 427.82 0.125 53.48 TensionS2 350 182.5 165 427.82 0.125 53.48 TensionS3 350 0.00 185 395.89 -0.25 -98.97 CompressionForce (N)

0 0 -5 - - - -

∑ Fx = 0; 350s1 + 350s2 + 350s3 = 0

∑ FY = 0; -182.5s1 + 182.5s2 + 0s3 = 0

∑ Fz = 0; 165s1 + 165s2 + 185s3 = -5

By using calculator

So ts1 = 0.125ts2 = 0.125

ts3 = -0.25

And S1 = 53.48S2 = 53.48S3 = -98.97

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2. Load F = 10 N

a = 350mm, b = 521mm, c = 185mm, d = 365mm.

Lx = bLy = d/2 (for S1 and S2)Lz = a – c (for S1 and S2)F = L x tL = √ (Lx² + Ly² + Lz²)

Member Lx

(mm)Ly

(mm)Lz

(mm)L (mm) t F (N) Remarks

S1 350 -182.5 165 427.82 0.25 106.96 TensionS2 350 182.5 165 427.82 0.25 106.96 TensionS3 350 0.00 185 395.89 -0.50 -197.95 CompressionForce (N)

0 0 -10 - - - -

∑ Fx = 0; 350s1 + 350s2 + 350s3 = 0

∑ FY = 0; -182.5s1 + 182.5s2 + 0s3 = 0

∑ Fz = 0; 165s1 + 165s2 + 185s3 = -10

By using calculator

So ts1 = 0.25ts2 = 0.25

ts3 = -0.50

And S1 = 106.96S2 = 106.96S3 = -197.95

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3. Load F = 15 N

a = 350mm, b = 521mm, c = 185mm, d = 365mm.

Lx = bLy = d/2 (for S1 and S2)Lz = a – c (for S1 and S2)F = L x tL = √ (Lx² + Ly² + Lz²)

Member Lx

(mm)Ly

(mm)Lz

(mm)L (mm) t F (N) Remarks

S1 350 -182.5 165 427.82 0.375 160.43 TensionS2 350 182.5 165 427.82 0.375 160.43 TensionS3 350 0.00 185 395.89 -0.75 -296.92 CompressionForce (N)

0 0 -15 - - - -

∑ Fx = 0; 350s1 + 350s2 + 350s3 = 0

∑ FY = 0; -182.5s1 + 182.5s2 + 0s3 = 0

∑ Fz = 0; 165s1 + 165s2 + 185s3 = -15

By using calculator

So ts1 = 0.375ts2 = 0.375

ts3 = -0.75

And S1 = 160.43S2 = 160.43S3 = -296.92

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4. Load F = 20 N

a = 350mm, b = 521mm, c = 185mm, d = 365mm.

Lx = bLy = d/2 (for S1 and S2)Lz = a – c (for S1 and S2)F = L x tL = √ (Lx² + Ly² + Lz²)

Member Lx (mm) Ly (mm) Lz

(mm)L (mm) t F (N) Remarks

S1 350 -182.5 165 427.82 0.5 213.91 TensionS2 350 182.5 165 427.82 0.5 213.91 TensionS3 350 0.00 185 395.89 -1 -395.89 CompressionForce (N)

0 0 -20 - - - -

∑ Fx = 0; 350s1 + 350s2 + 350s3 = 0

∑ FY = 0; -182.5s1 + 182.5s2 + 0s3 = 0

∑ Fz = 0; 165s1 + 165s2 + 185s3 = -20

By using calculator

So ts1 = 0.5ts2 = 0.5

ts3 = -1

And S1 = 213.91S2 = 213.91S3 = -395.89

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5. Load F = 25 N

a = 350mm, b = 521mm, c = 185mm, d = 365mm.

Lx = bLy = d/2 (for S1 and S2)Lz = a – c (for S1 and S2)F = L x tL = √ (Lx² + Ly² + Lz²)

Member Lx

(mm)Ly

(mm)Lz

(mm)L (mm) t F (N) Remarks

S1 350 -182.5 165 427.82 0.625 267.39 TensionS2 350 182.5 165 427.82 0.625 267.39 TensionS3 350 0.00 185 395.89 -1.25 -494.86 CompressionForce (N)

0 0 -25 - - - -

∑ Fx = 0; 350s1 + 350s2 + 350s3 = 0

∑ FY = 0; -182.5s1 + 182.5s2 + 0s3 = 0

∑ Fz = 0; 165s1 + 165s2 + 185s3 = -25

By using calculator

So ts1 = 0.625ts2 = 0.625

ts3 = -1.25

And S1 = 267.39S2 = 267.39S3 = -494.86

GRAPH

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Graph of Force versus Load

0

50

100

150

200

250

300

5 10 15 20 25

Load (N)

Fo

rce

(N)

S1 Exp.

S1 Theory

Graph of Force versus Load 2

0

50

100

150

200

250

300

5 10 15 20 25

Load (N)

Fo

rce

(N)

S2 Exp.

S2 Theory

6.0 DISSCUSSION

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Based 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. When more load were applied, the value of force was also

increase.

It 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.

But for the graph3, the results of the theoretical and the experimental is totally

difference because for the experiment, the results is in range 0 to 300 while for the results

of the theoretical is around range -0 to -300. For 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. Beside that, it maybe cause by the error of the

apparatus which is not in good condition.

7.0 CONCLUSION

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

value. Space frames usually utilize 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 frame; flexing loads (bending moments) are transmitted as

tension and compression loads along the length of each structure.

In many ways this looks like the horizontal jib of a tower crane repeated many

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

pyramids in which all the struts have unit length. More technically this is referred to as an

isotropic vector matrix or in a single unit width an octet truss.

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