Space Frame

FACULTY OF CIVIL AND ENVIRONMENTAL ENGINEERING DEPARTMENT OF STRUCTURE AND MATERIAL ENGINEERING LAB MATERIAL

REPORT Subject Code Code & Experiment Title Course Code Date Section / Group Name Members of Group

Lecturer/Instructor/Tutor Received Date

Comment by examiner

BFC 31901 U5 – SPACE FRAME 2 BFF 21/03/2012 SECTION 9 / GROUP 7 MUHAMMAD IKHWAN BIN ZAINUDDIN 1.NUR EZRYNNA BINTI MOHD ZAINAL 2.MUHAMMAD NUH BIN AHMAD ZAIRI 3.NUR EEZRA ATHIRLIA BINTI GHAZALI 4.MUHAMMAD HUZAIR BIN ZULKIFLI 5.ZIRWATUL FAUZANA BINTI CHE JEMANI EN. MOHD KHAIRY BIN BURHANUDIN 4/04/2012

Received

(DF100018) (DF100118) (DF100093) (DF100147) (DF100040) (DF100027)

STUDENT CODE OF ETHIC (SCE) DEPARTMENT OF STRUCTURE AND MATERIAL ENGINEERING FACULTY OF CIVIL & ENVIRONMENTAL ENGINEERING UNIVERSITI TUN HUSSEIN ONN MALAYSIA

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.

Name

___________________________

___________________________

Student Signature

Student Signature

: MUHAMMAD IKHWAN BIN ZAINUDDIN

Name

: NUR EZRYNNA BINTI MOHD ZAINAL

Matric No. : DF100018


Date

Date

Name

: 04/04/2012

: 04/04/2012

___________________________

___________________________

Student Signature

Student Signature

: NUR EEZRA ATHIRLIA BINTI GHAZALI

Name

: MUHAMMAD HUZAIR BIN ZULKIFLI



Date

Date

Name

: 04/04/2012

: 04/04/2011

_______________________

_______________________

Student Signature

Student Signature

: MUHAMMAD NUH BIN AHMAD ZAIRI

Name

: ZIRWATUL FAUZANA BINTI CHE JEMANI



Date

Date

: 04/04/2012

: 04/04/2012

1.0

INTRODUCTION In architecture and structural engineering, a space frame or space structure is a truss-like,

lightweight rigid structure constructed from interlocking struts in a geometric pattern. Space frames can be used to span large areas with few interior supports. Like the truss, a space frame is strong because of the inherent rigidity of the triangle; flexing loads (bending moments) are transmitted as tension and compression loads along the length of each structure.

Figure 1.0 - The roof of this industrial building is supported by a space frame structure

Space frames were independently developed by Alexander Graham Bell around 1900 and Buckminster Fuller in the 1950s. Bell's interest was primarily in using them to make rigid frames for nautical and aeronautical engineering. Few of his designs were realized. Buckminster Fuller's focus was architectural structures; his work had greater influence.

Figure 2.0 - Simplified space frame roof with the half-octahedron highlighted in blue

2.0

OBJECTIVE To verify member forces obtain from experiment with tension coefficient method

3.0

LEARNING OUTCOME 2.1

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 problem, solving and finding out appropriate solution through laboratory application.

4.0

THEORY If the 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 nodeA (xA,yA) and B (xB,yB).

Figure 3.0 – Space truss component Assume the 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: TABcosӨ = tABLABcosӨ = tABLAB (xB-XA) LAB = tAB (xB-xA)

Used 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 t. Convert it into force using:

5.0

PROCEDURES Part 1: 1. Select any weight between 10 to 50 N. 2. Ensure distance a = 500mm and place load hanger on D. 3. Measure the distance b,c dan d and record it in Table 1. 4. Record the dynamometer readings for members S1, S2 dan S3. 5. Put the selected load on the hanger at D and record the 6. Repeat step (2) to (4) with different value of a. 7. Calculate the theoretical member forces and record it in Table 1. Part 2: 1. For part 2, use a distance of 350 mm for a. 2. Place the hanger on D. 3. Measure the distance b, c and d. Record the dynamometer readings for member S1, S2 and S3 in Table 2. 4. Put a load of 5N on the hanger and record the dynamometer readings.. 5. Repeat step 2 to 4 using different load. 6. Complete Table 2 by calculating the theoretical member value. 7. Plot the graph of force against load for the theoretical and experimental results.

Figure 4.0 - Space Frame Equipment

Figure 5.0 – Space Frame Dimension

6.0

RESULT TABLE 1

Dimension (mm)

Dynamometer Reading S1

a

b

c

Force (N)

S2

S3

Experiment

Theory

d Unloaded

Loaded

Unloaded

Loaded

Unloaded

Loaded

S1

S2

S3

S1

S2

S3

500 460 295 360

0.3

0.53

0.3

0.57

5

13

0.23

0.27

8

5.35

5.35

-10.93

400 490 247 360

0.6

7

0.7

6.8

6

17

6.4

6.1

11

5.44

5.44

-16.46

300 515 186 360

1.4

10

1.4

9.7

6

22

0.4

8.3

16

11.15

11.15

-16.73

200 540

2.5

16

2.7

16.1

8

31

13.5

13.4

23

17.55

17.55

-27.20

LX=b, LY=b/2 LZ=a-c F=L x t

65

360

TABLE 2 Dimension (mm)

LOAD

a

b

c

Dynamometer Reading S1

d

Force (N)

S2

S3

Unloaded Loaded Unloaded Loaded Unloaded

(N)

Experiment

Theory

Loaded

S1

S2

S3

S1

S2

S3

5

350 495 257

360

1

4.4

1.2

4.7

6

12

3.4

3.5

6

5.35

5.35

-5.56

10

350 505 221

360

1

8

1.2

8.4

6

19

7

7.2

13

5.51

5.51

-5.51

15

350 513 187

360

1

12

1.2

12.3

6

25

11

11.1

19

11.26

11.26

-21.42

20

350 518 160

360

1

16

1.2

16.5

6

31

15

15.3

25

17.41

17.41

-32.53

25

350 523 115

360

1

20.5

1.2

20.8

6

37

19.5

19.6

31

24.04

24.04

-37.48

LX=b, LY=b/2 LZ=a-c F=L x t

7.0

DATA ANALYSIS

7.1

PART 1

Force (N) = Theory 

Load = 10 N



Dimension a = 500 mm



Dimension b = 460 mm



Dimension c = 295 mm



Dimension d = 360 mm

Member

Lx (mm)

Ly (mm)

Lz (mm)

S1

460

-180

205

S2

460

180

S3

460 0

Load L=

t

F

534.81

0.01

5.35

205

534.81

0.01

5.35

0

-295

546.47

- 0.02

-10.93

0

-10

Lx2 + Ly2 + Lz2

L=

L (mm)

Lx2 + Ly2 + Lz2

= 4602 + (-180)2 + 2052

= 4602 + (-295)2

= 534.81 mm

= 546.47mm

FOR F : t * L ∑ Fx = 0 460 ts1 + 460 ts2 + 460 ts3 = 0

(1)

∑ Fy = 0 -180 ts1 + 180 ts2 + 0 + 0 = 0 180 ts2 = 180 ts1 ts2 = ts1

(2)

∑ Fz = 0 205ts1 + 205 ts2 - 295 ts3 = 10

(3)

Substitute ts2= ts1 into (1) and (3) 460 ts1 + 460 ts2 + 460 ts3 = 0

(1)

920 ts2 + 460 ts3 = 0

(4)

205ts1 + 205 ts2 - 295 ts3 = 10

(3)

410 ts2 - 295 ts3 = 10

(5)

From equation ( 4 ) 920 ts2 + 460 ts3

=0

ts3

= - 920 ts2 460 = -2 ts2

ts3 Substitute into ( 5)

410 ts2 - 295 ts3 = 10 410 ts2 -295 ( -2 ts2 ) = 10 410 ts2 + 590 ts2 = 10 1000 ts2 = 10 ts2 =

10 1000

ts2 = ts1 = 0.01 Substitute into ( 4 ) ts3 = -2 ts2 = - 2 (0.01) = - 0.02

*Do the same step for other calculation 

Load = 10 N













Member

Lx (mm)

Ly (mm)

Lz (mm)

L (mm)

t

F

S1

490

-180

153

543.98

0.01

5.44

S2

490

180

153

543.98

0.01

5.44

S3

490

0

-247

548.73

-0.03

-16.46

0

0

-10

Load

L=

Lx2 + Ly2 + Lz2

L=

Lx2 + Ly2 + Lz2

= 4902 + (-180)2 +1532

= 4902 +(-247)2

= 543.98 mm

= 548.73mm

FOR F : t * L ∑ Fx = 0 490 ts1 + 490 ts2 + 490 ts3 = 0

(1)

∑ Fy = 0 -180 ts1 + 180 ts2 + 0 + 0 = 0 180 ts2 = 180 ts1 ts2 = ts1

(2)

∑ Fz = 0 153ts1 + 153 ts2 - 247 ts3 = 10

(3)



Load = 10 N













Member

Lx (mm)

Ly (mm)

Lz (mm)

L (mm)

t

F

S1

515

-180

114

557.33

0.02

11.15

S2

515

180

114

557.33

0.02

11.15

S3

515

0

-186

557.56

-0.03

-16.73

-10

Load L=

Lx2 + Ly2 + Lz2

L=

Lx2 + Ly2 + Lz2

= 5152 + (-180)2 +1142

= 5152 + (-186)2

= 557.33 mm

= 557.56mm

FOR F : t * L ∑ Fx = 0 515 ts1 + 515 ts2 + 515 ts3 = 0

(1)

∑ Fy = 0 -180 ts1 + 180 ts2 + 0 + 0 = 0 180 ts2 = 180 ts1 ts2 = ts1

(2)

∑ Fz = 0 114ts1 +114 ts2 - 186 ts3 = 10

(3)



Load = 10 N






Dimension b = 540mm



Dimension c = 65 mm




Member

Lx (mm)

Ly (mm)

Lz (mm)

L (mm)

t

F

S1

540

-180

135

585.00

0.03

17.55

S2

540

180

135

585.00

0.03

17.55

S3

540

0

-65

543.90

-0.05

-27.20

0

0

-10

0

0

0

Load L=

Lx2 + Ly2 + Lz2

L=

Lx2 + Ly2 + Lz2

= 5402 + (-1802) +1352

= 5402 + (-652)

= 585.00 mm

= 543.90mm

FOR F : t * L ∑ Fx = 0 540 ts1 + 540 ts2 + 540 ts3 = 0

(1)

∑ Fy = 0 -180 ts1 + 180 ts2 + 0 + 0 = 0 180 ts2 = 180 ts1 ts2 = ts1

(2)

∑ Fz = 0 135ts1 +135 ts2 - 65 ts3 = 10

(3)

7.2

PART 2

Force (N) = Theory 

LOAD = 5 N













 Member

Lx (mm)

Ly (mm)

Lz (mm)

L (mm)

t

F

S1

495

-180

93

534.86

0.01

5.35

S2

495

180

93

534.86

0.01

5.35

S3

495

0

-257

557.74

-0.01

-5.56

-5

Load L=

Lx2 + Ly2 + Lz2

L=

Lx2 + Ly2 + Lz2

= 4952 + 1802 + 932

= 4952 + (-257)2

= 534.86mm

= 557.74mm

FOR F : t * L ∑ Fx = 0 495 ts1 + 495 ts2 + 495 ts3 = 0

(1)

∑ Fy = 0 -180 ts1 + 180 ts2 + 0 + 0 = 0 180 ts2 = 180 ts1 ts2 = ts1

(2)

∑ Fz = 0 93 ts1 + 93 ts2 -257 ts3 = 5

(3)



LOAD = 10 KN













Member

Lx (mm)

Ly (mm)

Lz (mm)

L (mm)

t

F

S1

505

-180

129

551.42

0.01

5.51

S2

505

180

129

551.42

0.01

5.51

S3

505

0

-221

551.24

-0.03

-5.51

-10

Load L=

Lx2 + Ly2 + Lz2

L=

Lx2 + Ly2 + Lz2

= 5052 + 1802 + 1292

= 5052 + (-221) 2

= 551.42 mm

= 551.24 mm

FOR F : t * L ∑ Fx = 0 505 ts1 + 505ts2 + 505 ts3 = 0

(1)

∑ Fy = 0 -180 ts1 + 180 ts2 + 0 + 0 = 0 180 ts2 = 180 ts1 ts2 = ts1

(2)

∑ Fz = 0 129 ts1 +129 ts2 - 221 ts3 = 10

(3)



LOAD = 15 KN













Member

Lx (mm)

Ly (mm)

Lz (mm)

L (mm)

t

F

S1

513

-180

172

563.04

0.02

11.26

S2

513

180

172

563.04

0.02

11.26

S3

513

0

-187

535.45

-0.04

-21.42

-15

Load L=

Lx2 + Ly2 + Lz2

L=

Lx2 + Ly2 + Lz2

= 5132 + 1802 + 1722

= 5132 + (-187)2

= 563.04 mm

= 535.45mm

FOR F: t * L ∑ Fx = 0 513 ts1 + 513 ts2 + 513 ts3 = 0

(1)

∑ Fy = 0 -180 ts1 + 180 ts2 + 0 + 0 = 0 180 ts2 = 180 ts1 ts2 = ts1

(2)

∑ Fz = 0 172ts1 +172 ts2 - 187 ts3 = 15

(3)



LOAD = 20 KN













Member

Lx (mm)

Ly (mm)

Lz (mm)

L (mm)

t

F

S1

518

-180

190

580.37

0.03

17.41

S2

518

180

190

580.37

0.03

17.41

S3

518

0

-160

542.15

-0.06

-32.53

-20

Load L=

Lx2 + Ly2 + Lz2

L=

= 5182 + (-1802) + 1902

Lx2 + Ly2 + Lz2

= 5182 + (-1602)

= 580.37 mm

= 542.15 mm

FOR F : t * L ∑ Fx = 0 518 ts1 + 518ts2 + 518 ts3 = 0

(1)

∑ Fy = 0 -180 ts1 + 180 ts2 + 0 + 0 = 0 180 ts2 = 180 ts1 ts2 = ts1

(2)

∑ Fz = 0 190ts1 +190 ts2 - 160 ts3 = 20

(3)



LOAD = 25 KN













Member

Lx (mm)

Ly (mm)

Lz (mm)

L (mm)

t

F

S1

523

-180

235

600.96

0.04

24.04

S2

523

180

235

600.96

0.04

24.04

S3

523

0

-115

535.49

-0.07

-37.48

-25

Load L=

Lx2 + Ly2 + Lz2

L=

Lx2 + Ly2 + Lz2

= 5232 +(- 1802) + 2352

= 5232 +(- 1152)

= 600.96 mm

= 535.49 mm

FOR F : t * L ∑ Fx = 0 523 ts1 + 523ts2 + 523 ts3 = 0

(1)

∑ Fy = 0 -180 ts1 + 180 ts2 + 0 + 0 = 0 180 ts2 = 180 ts1 ts2 = ts1

(2)

∑ Fz = 0 235ts1 +235 ts2 - 115 ts3 = 25

(3)

8.0

RESULT

1.

Compare the graph of theoretical and experimental results. Comment on the results. From the graph, the theoretical and experimental result is almost the same. When we compare the both graph, it’s a linear. For the third graph, there are negative gradient. It is because of the some error while doing experiment. There are may be errors in calibrating the scale of the apparatus and all the reading taken consist of errors. Zero error occurs. Where the scale does not start with zero or the zero ends is spoilt

2.

Gives reasons for any discrepancy in the results. Several reading of the value are taken, their values may differ from one another. However these reading and closed to the real value. The errors may be due to the charges in condition of the experiment or surrounding which can be controlled. We should record the average of this value as the best reading for this quantity

9.0

CONCLUSION From the experiment, we had verified member forces obtain from experiment with tension

coefficient method. From this experiment, the value of S1 and S2 was almost the same compared to the value of S3 caused by distance and angle influences. The values of theoretical and experimental graph are just a little bit different. From S1 and S2 graph, it shown the gradient are almost the same while S3 graph had negative gradient

10.0

REFERENCES

1. Structure Analysis ; UTHM 2. Teori Struktur ; Yusuf Ahmad ; UTM 3. Fundamentals of Structural Analysis; Tung Au; Carnegie; Mellon University

Space Frame

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