Project analysis structure

CHAPTER 1PROJECT INFORMATION

1.0INTRODUCTIONArt is a diverse range of human activities and the products of those activities. In their most general form these activities include the production of works of art, the criticism of art, the study of the history of art, and the aesthetic dissemination of art. This article focuses primarily on the visual arts, which includes the creation of images or objects in fields including painting, sculpture, printmaking, photography, and other visual media. Architecture is often included as one of the visual arts; however, like the decorative arts, it involves the creation of objects where the practical considerations of use are essential-in a way that they usually are not in a painting, for example. Music, theatre, film, dance, and other performing arts, as well as literature and other media such as interactive media, are included in a broader definition of art or the arts. Until the 17th century, art referred to any skill or mastery and was not differentiated from crafts or sciences. In modern usage after the 17th century, where aesthetic considerations are paramount, the fine arts are separated and distinguished from acquired skills in general, such as the decorative or applied arts.An art museum or art gallery is a building or space for the exhibition of art, usually visual art. Museums can be public or private, but what distinguishes a museum is the ownership of a collection. Paintings are the most commonly displayed art objects; however, sculpture, decorative arts, furniture, textiles, costume, drawings, pastels, watercolours, collages, prints, artists books, photographs, and installation art are also regularly shown. Although primarily concerned with providing a space to show works of visual art, art galleries are sometimes used to host other artistic activities, such as performance art, music concerts, or poetry readings.An art gallery serves an important purpose by imparting the potential customer a feel of the art. Irrespective of all the advertisements and publicities, the art sales can increasing its sales when the customer gets to see or feel the art on their own. An art gallery will provide an amiable and pleasant ambiance to the customer. In addition, all the queries and information needed by the customer will be provided by the sales associates in the gallery.

1.1PROBLEM STATEMENTArt Gallery intend to build around area Parit Raja. The purpose is to attract people come to Parit Raja and make the urban people knowing more about art. The design of our Art building is more to show that we use the good materials and spent a few of budget to the building with the modern design include the long design period such as 50 years. The Art building also can be the gathering of the art of designers all Malaysian and maybe can be the place of designers of world.

1.2OBJECTIVEThe purpose of the art gallery is to attract customer and provide space for painters or artist to display and sell their artwork to people. Hence, we aim to design a modern artistic and multifunction building for this art gallery.

1.3 COMPONENTS OF ART GALLERY

1.3.1The display area

This the most important part of the gallery where people can take a closer look to the art. The display area will cover up to 60% of the gallery space. This area is where the visitors entrance.

Ensuring maximum visibility of the display area from the exteriors adds to the success of the showroom. This area is divided to a few partition to classify an artists artwork. The space is located closely associated with other spaces for visitors convenience. All other spaces in the art gallery help support this space and its function.

1.3.2 The office spaceThe office space is the area where the officials event take place. Documents, records, finances, Human Resources, meeting spaces and other related spaces are placed in this area. These spaces are kept out of bounds for the visiting customers. However there are certain office space are opened to the visitors, as these space allow the visitors to seal the deal.

1.3.3 The service area

This area has equal interest as the display area. The area is associated with the after sales service. Such as the art framing. The gallery provide the custom made frame for the artwork based on the customers preferences. Special care to be taken in ensuring proper ventilation and lighting. The service area will have small waiting area for the customers.

1.4 MATERIALS

1.4.1FlooringCommon materials used for flooring are incudes vitrified clay tiles (gloss and matte), thin set epoxy terrazzo and variations of it such as marble, granite, wood and stone. The materials should be so chosen that they match the theme of the gallery. The materials that we choose for the display area is the wood.

1.4.2 Walls There are many types of wall coverings, these includes wood panelling, fabric, stone, mirrors, tiles, rough boards, brick and in fact any material imaginable. The materials also should be chosen wisely so that it can match the theme of the gallery.

1.4.3 WindowsThere are certain things that need to be considered when choosing the type of window. Such as window orientation, climate, building design and others related. There are many window designs in use today. There are awnings type, casement, picture, horizontal slider and others. The windows are important because it provide aeration for the building.

1.4.4 DoorA door is moving structure used to block off, and allow access to, an entrance to or within an enclosed space, such as building. The doors also can the attraction to the visitors to enter the gallery. Therefore, the door type and materials should be considered carefully based on the building design and theme.

CHAPTER 2DESIGN OF ART GALLERY

2.1BUILDING OF ART GALLERY 2.1.1Front View

2.1.2Sides View

2.1.3Back View

2.1.4Structure Frame Front View

2.1.5Structure Frame Side View

2.2DESCRIPTION OF STRUCTURE

2.2.1TrussIn engineering, a truss is a structure that consists of two-force members only, where the members are organized so that the assemblage as a whole behaves as a single object. A two-force member is a structural component where force is only applied to two points. Although this rigorous definition allows the members to have any shape connected in any stable configuration, trusses typically comprise five or more triangular units constructed with straight members whose ends are connected at joints referred to as nodes. In this typical context, external forces and reactions to those forces are considered to act only at the nodes and result in forces in the members which are either tensile or compressive forces. For straight members, moments (torques) are explicitly excluded because, and only because, all the joints in a truss are treated as revolutes, as is necessary for the links to be two-force members.

2.2.2BeamA beam is a structural element that is capable of withstanding load primarily by resisting bending. The bending force induced into the material of the beam as a result of the external loads, own weight, span and external reactions to these loads is called a bending moment.

2.2.3ColumnA structural element that transmits, through compression, the weight of the structure above to other structural elements below. In other words, a column is a compression member. The term column applies especially to a large round support with a capital and base and made of stone, or appearing to be so. A small wooden or metal support is typically called a post, and supports with a rectangular or other non-round section are usually called piers.

2.2.4FrameFraming, in construction is the fitting together of pieces to give a structure support and shape and sometimes is used as a noun such as "the framing" or "framing members". Framing materials are usually wood, engineered wood, or structural steel. Building framing is divided into two broad categories, heavy-frame construction (heavy framing) if the vertical supports are few and heavy such as in timber framing, pole building framing, or steel framing or many and smaller called light-frame construction (light framing) including balloon, platform and light-steel framing.CHAPTER 3 CALCULATION AND ANALYSIS STRUCTURE

3.1TRUSS ANALYSIS

Member 1, 2, 3, 4= 2.5 metersMember 9 and 13= 1.5 metersMember 11= 3 metersMember 10 and 12= 2.92 metersMember 8,7, 6, and 5= 2.915 metersArea of zinc on the trussArea 1 = 18 x 5.83Area 2 = ( 10 x 3 )= 104.94 x 2 (both side)= 30 m2= 209.88 m2Total area zinc = 209.88 + 30= 239.88 m2Unit weight for zinc = 72 kN/m3, thickness of zinc is 0.0008 mZinc load = 239.88 x 72 x 0.0008= 13.82 / 7 ( load distributed to 7 trusses )= 1.97 KN / truss / 4 ( distributed to 5 position )= 0.39 KN / truss / position 0.4 KN / truss / positionVolume steel hollow section

0.06m 0.05 m

Area of steel = ( b x h ) ( b x h ) = - ( 0.05 ) = 0.0011 m2Total length of steel = 5.83 + 5.83 + 10 + 3 + 3+ 2.92 + 2.92 = 30.5 mUnit weight of steel is 78 kN/m3Self-weight truss=area x length x unit weight=0.0011 x 30.5 x 78=2.62 kNWind load The building excess high 8 m, so that the wind load assume that 0.67 KN/m2Area of one panel roof =3 x 5.83 = 17.49 m2Wind load per position =area x unit load/area =0.67 x 17.49=11.72 / 6 ( loading separate into 6 position )=1.95 kNWind load + zinc self-load =1,95 + 0.4=2.35 kNFor middle symmetry truss=( 1.95 x 2 ) + 0.4=4.3 kN4.3 kN

2.35 kN2.35 kN2,35 kN2.35 kNRBRA

2.62 kN

Classification M = 2j -313= 2(8) -313= 13 ( statically determine structure )Reaction MA = 02.35 ( 2.5) + 4.3(5) + 2.35(7.5) + 2.35(10) + 2.62(5)- RB(10) = 0 By(10) = 81.6 By = 8.16 kN Ay + By 16.36 = 0 Ay = 8.16 kN

Internal Forces each Members

Because of the truss consists complex calculation to balance each members. Therefore, we used staad pro to balance internal forces in the structure truss.

3.2 FRAME ANALSIS3.2.1Frame ( first floor frame )DBAC3m10m3m

Load transfer from truss 8.16 kN to first floor frame. Uniform load Span of beam is 10 metersDimension of beam ( all the roof floor beam are same sizes )wdDepth of beam, d = 0.45mWidth of beam, w = 0.30 m

Area = 0.45 x 0.3 = 0.135 m2Assumes variable load = 5.0 kN/mConcrete of beam = 25 kN/m3Dead load for beam = 25 x 0.135 = 3.38 kN/mAssume of the slab ( first floor ) dead load = 11.5 kN/mAssume of the wall dead load = 7.8 kN/mTotal dead load = 22.68 kN/m

Uniform load for beam = 1.35Gk + 1.5Qk = 1.35 ( 22.68 ) + 1.5 (5) = 38.12 39 kN/m ( from point 2 to 3 )StiffnessKCA = KAC= KAB = KBA= KBD = KDB=

DFDFCA = DFDB = 0DFAC = DFBD = = 0.77DFAB = DFBA = = 0.23

Fixed end momentFEMCA = FEMDB = 0 kNmFEMAB = = -325 kNmFEMBA = = 325 kNm kNm

Frame with non-side wayCABD

MemberCAACABBABDDB

CF00.50.50.50.50

DF00.770.230.230.770

FEM00-32532500

Dist0250.2574.75-74.75-250.250

CO125.130-37.3837.380-125.13

Dist028.788.60-8.60-28.780

Moment (kNm)125.13279.03-279.03279.03-279.03-125.13

Reaction of CA 279.03125.13

AC

MA = 0RC(3) + 125.13 + 279.03 = 0 RC(3) = - 404.16 kN RC = -134.72 kN ( ) RC + RA1 = 0 RA1 = 134.13 kN ( )Reaction of AB 8.16 kN8.16 kN

39kN/m-279.03 kNm 279.03 kNm

BA

MB1 = 0RA2(10) 279.03 + 279.03 8.16(10) 39(10)(10/2 ) = 0 RA2(10) = 2031.6 kN RA2 = 203.16 kN RB1 + RA2 8.16 8.16 390 = 0 RB1 = 203.16 kNReaction of BD-125.13 kNm-279.03 kNm

DB

MD = 0RB2(3) 125.13 279.03 = 0 RB2(3) = 402.16 kN RB2 = 134.72 kN ( ) RB2 + RD = 0 RD = - 134.13 kN ( )

Shear Force Diagram195. kN

DBAC-134.72 kN195 kN203.16. kN134.72 kN-134.72 kN

Bending Moment Diagram-279.03 kN

-279.03 kN

125.13 kN-279.03 kN208.47 kN-279.03 kN125.13 kN

3.2.2 Frame ( ground floor )AE3.5 m3.5 mD10mYC

Load transfer from frame first floor 203.16kN to ground floor frame.Self-weight for first floor column Dimension of column ( all the first and ground floor column are same ) wlength of column = 0.3 mwidth of column = 0.3 m Lhigh of column = 3 mSelf-weight of column =unit weight of concrete x volume column =25 x ( 0.3 x 0.3 x 3 ) =6.75 kNUniform load Span of beam is 10 metersDimension of beam ( all the first floor beam are same sizes )wdDepth of beam, d = 0.30 mWidth of beam, w = 0.25 m

Area = 0.3 x 0.25 = 0.075 m2Assumes variable load = 5.0 kN/mConcrete of beam = 25 kN/m3Dead load for beam = 25 x 0.075 = 1.875 kN/mAssume of the slab ( first floor ) dead load = 12 kN/mAssume of the wall dead load = 7.8 kN/mTotal dead load = 21.68 kN/mAssume variable load =5 kNUniform load for beam = 1.35Gk + 1.5Qk = 1.35 ( 21.68 ) + 1.5 (5) = 36.76 37 kN/m ( from point 2 to 3 )StiffnessKYC = KCY= KCD = KDC= KDAE = KAED=

DFDFYC = DFAED = 0DFCY = DFDAE = = 0.77DFCD = DFDC = = 0.23

Fixed end momentFEMYC = FEMCY = FEMDAE = FEMAED = 0 kNmFEMCD = = -308.33 kNmFEMDC = = 308.33 kNm kNm

Frame with non-side wayYCDAE

MemberYCCYCDDCDAEAED

CF00.50.50.50.50

DF00.740.260.260.740

FEM00-308.33308.3300

Dist0228.1680.17-80.17-228.160

CO114.080-40.0940.090-114.08

Dist029.6710.42-10.42-29.670

Moment (kNm)114.08257.83-257.83257.83-257.83-114.08

Reaction of YC 257.83114.08

CY

MC = 0RY(3) + 114.08 + 257.83 = 0 RY(3.5) = - 371.91 kN RY = -106.26 kN ( ) RY + RC1 = 0 RC1 = 106.26 kN ( )Reaction of CD 209.91 kN209.91 kN

37kN/m-257.83 kNm 257.83 kNm

DC

MD1 = 0RC2(10) 257.83 + 257.83 209.91(10) 37(10)(10/2 ) = 0 RC2(10) = 3949.1 kN RC2 = 394.91 kN RD1 + RC2 209.91 209.91 370 = 0 RD1 = 394.91 kNReaction of DAE-114.08 kNm-257.83 kNm

AED

MAE = 0RD2(3.5) 257.83 114.08 = 0 RD2(3.5) = 371.91 kN RD2 = 106.26 kN ( ) RD2 + RAE = 0 RAE = - 106.26 kN ( ) Shear force diagram .185 kNC

AE-106.26 kN106.26 kND185. kN-106.26 kN106.26kNY

Bending Moment Diagram-257.83 kN-257.83 kN

114.08 kN-257.83 kN204.67 kN114.08 kN-257.83 kN

3.3BEAM ANALYSIS3.3.1Steel beam section I to T JMI

Cross section area = 0.06 x 0.05 = 0.0011 m2 0.06 m

0.05 mSelf-weight of steel beamVolume of beam = 0.0011 x 18 = 0.0198 m3Self- weight = unit weight x volume = 78 x 18 = 1.54 kNLoading from trusses is 8.16 kN transfer to the beam, and middle of the beam add with self-weight steel beam is 9.7. StiffnessKIM = KMI = KMJ = KJM = DFDFIM= DFJM = 0 DFMI = DFMJ = = 0.5FEMFEMIM = FEMMJ = - = - 16.32 kNmFEMMI = FEMLM = + = 16.32 kNmTableIMJ

MEMBERIMMIMJJM

CF00.50.50

DF00.50.50

FEM-16.3216.32-16.3216.32

DIST0000

MOMENT (kNm)-16.3216.32-16.3216.32

Reaction for IM9.78.168.16

-16.32 kNm16.32 kNm

I

M

MI = 0-16.32 + 16.32 + 8.16(3) + 8.16 ( 6) + 9.7(9) My (9) = 0 My = 17.86 kNIy + My = 26.02 kN Iy = 8.16 kN Reaction for MJ8.168.169.7

16.32 kNm

-16.32 kNm

JM

MJ = 0-16.32 + 16.32 - 8.16(3) - 8.16 ( 6) - 9.7(9) + My (9) = 0 My = 17.86 kNJy + My = 26.02 kN Jy = 8.16 kN Shear force diagram 8.16 kN8.16 kN

-8.16 kN-8.16 kN-16.32-16.32

Bending Moment diagram -16.32

8.168.16

3.3.2Beam concrete H to D3m3m6m3m3mAGAFAHAEHD

Point load at point AE and AH is loading from roof floor = 132.75 kNPoint load at point AF and AG is loading from roof floor= 132.30 kNPoint load at between AF and AG is loading from roof floor = 218.59 kNSelf-weight for first floor beam Dead load Area for beam = 0.075 m2Assume loading from slab = 12 kN/mAssume of wall= 10 kN/mDead load for beam= 25x (0.075 ) = 1.875 kN/mTotal dead load= 23.88 kN/mAssume load = 5 kN/mUniform load acting on beam = 1.35Gk + 1.5Qk= 1.35(23.88) + 1.5(5)= 39.74 40 kN/mStiffness KHAE = KAEH = KAEAF = KAFAE = KAGAH = KAHAG = KAHD = KDAH = KAFAG = KAGAF = KAFAG = KAGAF =

DFDFHAE= DFDAH = 0DFAEH= DFAEAF = DFAHAG = DFAHD = 0.5 DFAFAG = DFAGAH = 0.67DFAFAG= DFAGAF = 0.33FEMFEMHAE = FEMAEAF = FEMAGAH = FEMAHAD = = -30 kNmFEMAEH = FEMAFAE = FEMAHAG = FEMDAH = = 30 kNmFEMAFAG = - = - 285.3 kNmFEMAGAF = + = 285.3 kNmHAEAFAGAHD

MEMBERHAEAEHAEAFAFAEAFAGAGAFAGAHAHAGAHDDAH

CF00.50.50.50.50.50.50.50.50

DF00.50.50.670.330.330.670.50.50

FEM-3030-3030-285.3285.3-3030-3030

DIST000171.0584.25-84.25-171.05000

CO0085.530-42.1342.130-85.5300

DIST0-42.77-42.7728.2313.9-13.9-28.2342.7742.770

MOMENT ( kNm)-30-12.7712.77229.28-229.28229.28-229.28-12.7612.7730

Reaction HAE132.75 kN

-30 kNm40kN/mH

-12.77 kNm

3mAE

MA = 0-12.77 - 30 + 40(3)(1.5) + 132.75 ( 3) AEy (3) = 0 AEy = 178.49 kNAEy + Hy = 252.75 kN Hy = 74.26 kN Reaction for AEAF132.75 kN132.3 kN40kN/m

229.28 kNm12.77 kNm

3m

AEAF

MAE = 012.77 + 229.28 + 40(3)(1.5) - AF( 3) + 132.3 (3) = 0 AFy = 272.98 kNAEy + AFy = 385.05 kN AEy = 112.07 kN Reaction AFAG218.59 kNAGAF132.3 kN132.3 kN

40kN/m

229.28 kNm-229.28 kNm

3m3m

-229.28 + 229.28 + 218.59(3) 40(6)( 3) + 132.3 (6) AGy(6)= 0 AGy = 361.60 kNAGy + AFy =723.19 kN AFy = 361.60 kN

Shear force diagram 272.98361.60

178.49112.07

74.26

-178.49-112.07-74.26

-272.98-361.60

Bending moment diagram

-30-12.77-229.28-229.28-12.77-30

20.83

20.7820.58.94.96278.62

CHAPTER 4INFLUENCE LINE FOR RAMP7 m19.6 m

Separated to two span of beam;1. 19.6 m span2. 7 m spanMoving direction

B1A

7 mCB219.6 m15 kN10 kN

2.7 m

4.1INFLUENCE ANALYSIS BEAM FOR 19.6 METERSFor span A B1 (19.6 m)

1. At centre of the beam

0 x 9.8RA9.8 - xx

F = 0RA = 1 + vv = 1 - 1v = -

M = 0RA (9.8) 1 (9.8 x) M M = (1 - ) (9.8) 9.8 + x M = 9.8 x 19.69.8

RA

F = 0RA = vv = 1 -

M = 0RA (9.8) M = 0M = (1 - ) (9.8) M = 9.8 -

So, shear and bending moment Shear Bending moment

1st trial1510

-0.362.77.10.5

v = 10 (-0.36) + 15 (0.5)v = 3.9 kN1st trial1015

3.552.77.14.9

M = 10 (3.55) + 15 (4.9)v = 109 kNm (MAX)

2nd trial15

0.369.812.5 0.510

v = 10 (0.5) + 15 (0.36)v = 10.4 kN (MAX)2st trial10

15

4.92.77.1

3.55

M = 10 (4.9) + 15 (3.55)v = 102.25 kNm

2. At 5m from A

0 x 55 x RAx

F = 0RA = 1 + vv = 1 - 1v = -

M = 0RA (5) 1 (5 x) M M = (1 - ) (5) 5 + x M = 5 x 19.69.8

RA

F = 0RA = vv = 1 -

M = 0RA (9.8) M = 0M = (1 - ) (5) M = 5 -

Shear Bending moment

1st trial1510

0.74-0.122.72.3

v = 10 (-0.12) + 15 (0.74)v = 9.9 kN1st trial1015

3.721.712.72.3

M = 10 (1.71) + 15 (3.72)v = 72.9 kNm

2nd trial15

0.6110

0.74

5

7.7

v = 10 (0.74) + 15 (0.61)v = 16.55 kN (MAX)2nd trial10

15

3.72

3.04

57.7

M = 10 (3.72) + 15 (3.04)v = 82.8 kNm (MAX)

3. At 15m from A

0 x 1515 x RAx

F = 0RA = 1 + vv = 1 - 1v = -

M = 0RA (15) 1 (15 x) M M = (1 - ) (15) 15 + x M = 15 x 19.615

RA

F = 0RA = vv = 1 -

M = 0RA (15) M = 0M = (1 - ) (15) M = 15 -

Shear Bending moment

1st trial1510

0.23-0.63

12.712.3

v = 10 (-0.63) + 15 (0.23)v = -2.85 kN1st trial1015

3.522.89

12.712.3

M = 10 (2.89) + 15 (3.52)v = 81.7 kNm (MAX)

2nd trial15

0.6110

0.23

15

17.7

v = 10 (0.23) + 15 (0.10)v = 3.8 kN (MAX)2nd trial10

15

2.89

1.45

17.715

M = 10 (2.89)+ 15 (1.45)v = 50.65 kNm

4.2INFLUENCE ANALYSIS BEAM FOR 7 METERSFor span B2 C (7 m)

1. At centre of beam0 x 3.5RA3.5 - xx

F = 0RA = 1 + vv = 1 - 1v = -

M = 0RA (3.5) 1 (3.5 x) M M = (1 - ) (3.5) 3.5 + x M = 3.5 x 73.5

RA

F = 0RA = vv = 1 -

M = 0RA (3.5) M = 0M = (1 - ) (3.5) M = 3.5 -

Shear Bending moment

1st trial1510

0.5-0.11

2.70.8

v = 10 (-0.11) + 15 (0.5)v = -6.4 kN1st trial1015

1.750.4

2.70.8

M = 10 (0.4) + 15 (1.75)v = 30.25 kNm (MAX)

2nd trial15

0.1110

0.5

3.5

6.2

v = 10 (0.5) + 15 (0.11)v = 6.65 kN (MAX)2nd trial10

15

1.75

0.4

6.23.5

M = 10 (1.75)+ 15 (0.4)v = 23.5 kNm

2. At 1.75m from B20 x 1.751.75 - xRAx

F = 0RA = 1 + vv = 1 - 1v = -

M = 0RA (1.75) 1 (1.75 x) M M = (1 - ) (1.75) 1.75 + x M = 1.75 x 71.75

RA

F = 0RA = vv = 1 -

M = 0RA (3.5) M = 0M = (1 - ) (1.75) M = 1.75 -

Shear Bending moment

1st trial1015

0.86

0.95

v = 15 (0.86)v = 12.9 kN1st trial1015

1.31

0.95

M = 15 (1.31) v = 19.65 kNm

2nd trial15

0.3610

0.86

3.5

4.45

v = 10 (0.86) + 15 (0.36)v = 14 kN (MAX)2nd trial10

15

1.31

0.64

4.45

M = 10 (1.31)+ 15 (0.64)v = 22.7 kNm (MAX)

3. At 5.25 from B20 x 5.255.25 - xRAx

F = 0RA = 1 + vv = 1 - 1v = -

M = 0RA (5.25) 1 (5.25 x) M M = (1 - ) (5.25) 5.25 + x M = 5.25 x 75.25

RA

F = 0RA = vv = 1 -

M = 0RA (3.5) M = 0M = (1 - ) (5.25) M = 5.25 -

Shear Bending moment

1st trial1510

0.25-0.36

2.72.55

v = 10 (-0.36) + 15 (0.25)v = 0.5 kN1st trial1015

1.310.64

2.552.7

M = 10 (0.64) + 15 (1.31)v = 26.65 kNm (MAX)

2nd trial15

10

0.25

5.25

1.75

v = 10 (0.25) v = 2.5 kN (MAX)2nd trial1510

1.31

1.75

M = 10 (1.31)v = 13.1kNm (MAX)

4.3CONCLUSION

As the conclusion, a successful in designing this art gallery is to attract the public to the visit the gallery. Looking into the future, with space, energy will become a big constraint. Thus, our role are provide an environmental friendly and an energy saving buildings design. Energy efficiency and operational efficiency will contribute towards the better market. An attempt has been made in this dissertation to understand the various design aspects involved in an art gallery. The design of the building will affect the art gallerys visitor number. Thus it will affecting the art sales. There are also some factors to the success in art sales; sales representatives manner and customers trust.

ATTACHMENT Dimension YCBJADACABAAZXIEG

WVUTSMRQNPO

AEYDCBALJ

Building information High between truss and roof ( B to C )= 3 mHigh between first floor to roof floor ( C to B )= 3 mHigh between ground to first floor ( Y to C ) = 3.5 mThe the length building = 18 mThe width of building = 10 mThe length between column to column = 3 m

The displacement truss for every point

Beam for section EB42.47 kN

BRQNPOE

EOPNQRB

MEMBEREOOEOPPOPNNPNQQNQRRQRBBR

CF00.50.50.50.50.50.50.50.50.50.50

DF00.50.50.50.50.50.50.50.50.50.50

FEM-31.531.5-31.531.5-31.531.5-31.531.5-31.531.5-31.531.5

DIST000000000000

MOMENT-31.531.5-31.531.5-31.531.5-31.531.5-31.531.5-31.531.5

Reaction E = 63 kNReaction O = 126 kNReaction P = 125.55 kNReaction N = 211.84 kNReaction Q = 125.55 kNReaction R = 126 kNReaction B = 63 kNBeam for section GC

132.75 kN132.75 kN218.59 kN132.3 kN132.3 kN

CWVUTSG

GSTUVWC

MEMBERGSSGSTTSTUUTUVVUVWWVWCCW

CF00.50.50.50.50.50.50.50.50.50.50

DF00.50.50.50.50.50.50.50.50.50.50

FEM-3030-3030-3030-3030-3030-3030

DIST000000000000

MOMENT-3030-3030-3030-3030-3030-3030

Reaction G = 60.75 kNReaction S = 384.74 kNReaction T = 384.6 kNReaction U =557.18 kNReaction V = 384.6 kNReaction W = 384.74 kNReaction C= 60.75 kN

Wind Load Charateristic

PROJECT STRUCTURE ANALYSIS |BUILDING ART GALLERY 44