Analysis and Design of Sahel Hotel

8 -Story Steel Frame With Concrete Shear Wall

Design:Mehrdad Modirkhazeni

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Analysis and Design of Sahel Hotel

Chapter 1

General & technicalDescription

Project Description:

Name: Sahel Hotel Dimension: 40m x 30m Number of stories: 8(3 under ground, 5 above) Total Height: 19.25 m Location: Iran, Mashhad

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Analysis and Design of Sahel Hotel

Occupation category: Hotel Type of Structure: Steel Frame Seismic force-resisting system: intermediate concrete shear wall Type of decks: Composit Metak Deck Importance factor: Intermediate Irregularity: Horizental Irregular and Vertical Regular Type of Foundation: Strip in two directions Site Classification: type 3

References :

1. Iranian Building Code, Mabhas10, Steel structures2. Iranian Building Code, Mabhas 9, Concrete structures3. AISC- ASD 894. ACI 318-995. 2800 Standard of Iran, Seismic Design for Buildings

Chapter 2

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Analysis and Design of Sahel Hotel

Preminarly Modeling

1. Concrete characteristic strength:

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Analysis and Design of Sahel Hotel

2. Steel Yeild Strength:

3. Frame sections :

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Analysis and Design of Sahel Hotel

4. Deck Section:

5. Shear wall section:

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Analysis and Design of Sahel Hotel

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Analysis and Design of Sahel Hotel

Chapter 3

Loading

Dead load

Weight of Decks Materials:

Weight per Weight per ThicknessMateriallayerNo

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Analysis and Design of Sahel Hotel

AreaKg/m^2

volumeKg/m^3)m(

7021000.033Tile+plasterFLOORING1408000.05Light ConcreteLEVELING2

AUTOMATICConcreteSLAB43216000.02Gypsum BoardCEILING5100----------------------------PARTITIONS6242TOTAL

Weight of Roof Materials:

Weight per Area

Kg/m^2

Weight per volumeKg/m^3

Thickness)m(MateriallayerNo

4522500.02MusaicFLOORING152.521000.025CEMENTPLASTER2

808000.10Light ConcreteLEVELING2

15.522000.007INSULATION4AUTOMATICCONCRTESLAB5

3216000.02Gypsum BoardCEILING6

225TOTAL

Weight of Exterior walls in Face of the building:

Weight per Area Kg/m^2

Weight per volumeKg/m^3

Thickness(m) Material layer No

50 2500 0.02 Luxtury Stone Exterior layer 1

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Analysis and Design of Sahel Hotel

52.5 2100 0.025 Cement Plaster 2

85 850 0.10Light

Concrete block

Structure layer 3

32 1600 0.02 Gypsum Board Interior layer 4

220 Total with 35% opening

Weight of Exterior walls other sides of Building:

Weight per Area Kg/m^2

Weight per volumeKg/m^3

Thickness

)m(MateriallayerNo

52.521000.025Cement plasterExterior layer1

858500.10Light Concrete blockStructure layer2

3216000.02Gypsum BoardInterior layer3

169.5Total

Weigth of Partitions:

Weight per Area Kg/m^2

Weight per volumeKg/m^3

Thickness) m(

MateriallayerNo

648000.08Light Concrete block

Structure layer1

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Analysis and Design of Sahel Hotel

1616000.01Gypsum BoardExterior21616000.01Gypsum BoardInterior396Total

Live Load

SpacesLoads1RoomsLL=200 kg /m2

2Stares and CoridorsLL=350 kg /m2

3RoofLL=150 kg /m2

4ParkingsLL=500 kg /m2

Equivalent Lateral Analysis

V (Seismic Base Shear) V= C.WC= A*B*I/R

A (Base Acceleration) regarding to hazard zoning :

A=0 .3 g

B (Response coefficient)

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Analysis and Design of Sahel Hotel

0<T <T 0⇒B=(s+1)( TT 0

)

T 0<T <T s⇒B=(s+1)

T>T s⇒B=(s+1)(

T sT

)23

Regarding with 2800 standard for site classification type III:

S=1.75 T s=0 .7

T 0=0 .15

Experimental Period of structure:

T x=0 . 05H34 sec

T x=0 . 05×19 .334=0.46 sec

T>T s⇒B=(s+1)(

T sT

)23=(1 .75+1)( 0 .7

0. 46)0.67=2. 75

Occupancy Category: Hotel →I(importance factor)=1

Type of force-resistance system: intermediate shear wall

Rx=Ry=7

Cx=ABIRx

=0 .3×2 .75×17

=0 .1179>0 .1 AI=0 .025⇒C x=C y=0 . 1179

V x=V y=0 .1179×3187 . 5ton=375 . 8 ton

Response Spectrum Analysis

Remarks: Using 2800 Standard Specrum Using SRSS and CQC for combination of results Scaling the base shear of ELA metod with Response Spectral

Analysis base shear

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Analysis and Design of Sahel Hotel

Load Cases Terms

Dead load DL Live load LL Roof Live Load Lr Snow S Static Earthquake load case X EX Static Earthquake load case Y EY Spectrum case X SPX

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Analysis and Design of Sahel Hotel

Spectrum case Y SPY Spectrum case X with eccentricity SPXT Spectrum case Y with eccentricity SPYT

Load Combinations for Concrete Shear Wall:

COMB1= 1.25DL+1.5LLCOMB2= 1.25DL+1.5LL+ 1.5(Lr or S)COMB3= DL+1.2LL+1. 2SPXCOMB4= DL+1.2LL+1.2SPX+ 0.36SPYCOMB5= DL+1.2LL+1.2SPXTCOMB6= DL+1.2LL+1.2SPXT+0.36SPYCOMB7= DL+1.2LL+1.2SPY

COMB8= DL+1.2LL+1.2SPY+ 0.36SPXCOMB9= DL+1.2LL+1.2SPYT

COMB10= DL+1.2LL+1.2SPYT+0.36SPXCOMB11= 0.85DL+1.2SPXCOMB12= 0.85DL+1.2SPX+ 0.36 SPYCOMB13= 0.85DL+1.2SPXTCOMB14= 0.85DL+1.2SPXT+ 0.36 SPYCOMB15= 0.85DL+1.2SPYCOMB16= 0.85DL+1.2SPY+ 0.36 SPXCOMB17= 0.85DL+1.2SPYTCOMB18= 0.85DL+1.2SPYT+0.36 SPX

Load Combinations for Steel Elements:

COMB1= DLCOMB2= DL+LLCOMB3= DL+ (Lr or S)COMB4= DL+0.75 LL+0.75 EXCOMB5= DL+0.75 LL – 0.75 EXCOMB5= DL+0.75 LL+0.75 EYCOMB6= DL+0.75 LL – 0.75 EYCOMB7= 0.6 DL+ 0.7 EX

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Analysis and Design of Sahel Hotel

COMB8= 0.6 DL - 0.7 EXCOMB9= 0.6 DL+ 0.7 EYCOMB10= 0.6 DL - 0.7 EY

Soil pressure

Kg/cm2 2.5 = σγ = 1.6 g/cm3= 1.6 ton/m3

k0 = 1-sin φ = 1-sin (27.63) = 0.536Ph = k0 γ h = (0.536) (1.6) h = (o.858 h) ton/m2

Chapter 4

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Analysis and Design of Sahel Hotel

Analysis & Control

1. Building separation distance: dMax=0 .005H=0. 005×19 .25=9 .6 cm

2. Control distance between Center of Mass & Center of Rigidity

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Analysis and Design of Sahel Hotel

yrotSmgarhpaiD MCX MCY RCX RCY %ye %xe9RTS 3D 708.32 275.62 492.32 643.82 3.1 9.5-8RTS 3D 108.22 260.61 188.91 373.6 3.7 3.237RTS 3D 787.22 817.51 357.91 997.6 6.7 7.926RTS 3D 628.22 888.51 545.91 605.7 2.8 9.725RTS 3D 818.22 698.51 593.91 918.8 6.8 6.324RTS 3D 987.22 528.51 540.02 576.11 9.6 8.313RTS 2D 91.02 581.02 592.62 556.22 3.51- 2.8-2RTS 1D 194.91 282.12 458.42 114.32 4.31- 1.7-1RTS 1D 535.91 172.12 700.42 470.42 2.11- 3.9-

ey= XCM−XCRLX

×100

ex=YCM−YCRLy

×100

STORY MAXIMUM AND AVERAGE LATERAL DISPLACEMENTS STORY LOAD DIR MAXIMUM AVERAGE RATIO PENT EX X 0.0546 0.0512 1.068 STORY8 EX X 0.0479 0.0306 1.564 STORY7 EX X 0.0369 0.0238 1.549 STORY6 EX X 0.0260 0.0170 1.526 STORY5 EX X 0.0158 0.0107 1.487 STORY4 EX X 0.0073 0.0052 1.398 STORY3 EX X 0.0008 0.0008 1.008 STORY2 EX X 0.0004 0.0004 1.020 STORY1 EX X 0.0002 0.0002 1.027 PENT EY Y 0.0704 0.0692 1.017 STORY8 EY Y 0.0661 0.0592 1.116 STORY7 EY Y 0.0515 0.0461 1.118 STORY6 EY Y 0.0368 0.0329 1.120 STORY5 EY Y 0.0228 0.0204 1.118 STORY4 EY Y 0.0106 0.0096 1.100 STORY3 EY Y 0.0019 0.0015 1.217 STORY2 EY Y 0.0008 0.0007 1.255 STORY1 EY Y 0.0004 0.0003 1.288

Ax = (∆max/1.2∆ave) 2 = [(0.0479/ (1.2*0.0306)] 2=1.7Ay = ((∆max/1.2∆ave) 2 = [(0.0661/ (1.2*0.0592)] 2=1

3. Story Drifts Control:

∆M: Maximum lateral displacement of structure during the Earthquake∆W: lateral displacement of structure in linear static analysis

T=0.46sec≤0.7 →∆M=0.025h

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Analysis and Design of Sahel Hotel

∆W*0.7R≤∆M=0.025h

Check Drift (X):

Story DiaphragmLoad UX X Y Z ∆W ∆W*0.7R ∆M*h RESULTPENT D3 SPX 0.0867 23.807 26.572 32.4STR8 D3 SPX 0.0307 22.801 16.062 30.1 0.0075 0.03675 0.09375 OKSTR7 D3 SPX 0.0232 22.787 15.718 26.35 0.0067 0.03283 0.09375 OKSTR6 D3 SPX 0.0165 22.826 15.888 22.6 0.0064 0.03136 0.09375 OKSTR5 D3 SPX 0.0101 22.818 15.896 18.85 0.0053 0.02597 0.09375 OKSTR4 D3 SPX 0.0048 22.789 15.825 15.1 0.0041 0.02009 0.10625 OKSTR3 D2 SPX 0.0007 20.19 20.185 10.85 0.0003 0.00147 0.10625 OKSTR2 D1 SPX 0.0004 19.491 21.282 6.6 0.0002 0.00098 0.08125 OKSTR1 D1 SPX 0.0002 19.535 21.271 3.35 0.0002 0.00098 0.08375 OK

Check Drift (Y):

yrotSmgarhpaiD daoL YU X Y Z W∆ R7.0*W∆ h*M∆ TLUSERTNEP 3D YPS 1960.0 708.32 275.62 4.238RTS 3D YPS 5940.0 108.22 260.61 1.03 1110.0 93450.0 57390.0 KO7RTS 3D YPS 4830.0 787.22 817.51 53.62 110.0 9350.0 57390.0 KO6RTS 3D YPS 4720.0 628.22 888.51 6.22 4010.0 69050.0 57390.0 KO5RTS 3D YPS 710.0 818.22 698.51 58.81 900.0 1440.0 57390.0 KO4RTS 3D YPS 800.0 987.22 528.51 1.51 7600.0 38230.0 52601.0 KO3RTS 2D YPS 3100.0 91.02 581.02 58.01 7000.0 34300.0 52601.0 KO2RTS 1D YPS 6000.0 194.91 282.12 6.6 3000.0 74100.0 52180.0 KO1RTS 1D YPS 3000.0 535.91 172.12 53.3 3000.0 74100.0 57380.0 KO

4. Overturning control:

MEx = 7917 ton.mMRx = W.YCM = 19.5 x 3187.6 = 62158 ton.mS.F= MRx/ MEx = 62158/7917=7.85 ok

MEy =7825 ton.mMRy = W.XCM = 21.3 x 3187.6 = 67895 ton.mS.F= MRy/ MEy = 67895/7825 = 8.67 ok

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