Analysis and Redesign of Al – Tatbeqea Faculty An-Najah National University Faculty of Engineering and Information Technology Graduation Project Presentation I Analysis and Redesign of Al – Tatbeqea Faculty By : *Fayez R Abu Safaqa *Saleh M Nassasra

OUT LINE INTRODUCTION. PRELIMENARY DESIGN FOR SLABS, BEAMS, AND COLUMNS. THREE DIMENSIONAL MODELING AND CHECKS. THREE DIMENSSIONAL DYNAMIC ANALYSIS. FINAL DESIGN OF SLABS, BEAMS. DESIGN OF SPECIAL STRUTURAL ELEMENT.

Introduction Project Description : Its consist of 4 stories each with 1300 m2 ,

Object To learn about the plans and how to construct the related information from it . To analyze the structure for static and seismic load then redesign it . To Design the structural element in the building.

Soil properties Soil investigation indicated that the bearing capacity of the soil is 250 KN/m2

Design codes and design methods Design codes: (ACI) Code2008: American Concrete Institute. (UBC) 1997: Uniform Building Code . Design method: The ultimate strength method. The response spectrum method.

Structural material  

-Nonstructural materials : As shown in the following table : Density Blocks 12 Plastering 23 Mortar "plain concrete" Filling material 20 Insulation 0.4 Stone 27 Tiles 25 Marble 26

Insulation for sound & heat Structural system After the comparison between the structural system the ribbed slab was chosen Structural system economy Insulation for sound & heat Safety Solid slab expensive not isolated Safe Rib slab /waffle Isolated Safe for short building Voided In the middle

structural design loads Static (Gravity) loads are divided into three parts: Own weight dead load: calculated automatically in software. - Super imposed dead load: For the first floor stories: Sd = 4.7 KN/m2 *For the last floor (roof): Sd = 0.5 KN/m2 -External wall weight: use 20 KN/m

-Live load:

Load combination THE ULTIMATE LOAD COMBINATION: 1.4 × Dead Load . 1.2 × Dead Load + 1.6 × Live Load . 1.2 × Dead Load + 1.0 × Live Load + 1.0 × Earthquake . 0.9 × Dead Load + 1.0 × Earthquake . THE ALLWABLE LOAD COMBINARION: D D+L D+(Eh/1.4) 0.9D-/+(Eh/1.4) D+0.75L+0.75(Eh/1.4)

PRELIMENARY DESIGN FOR SLABS, BEAMS, AND COLUMNS .

Slab preliminary design Minimum thicknes

 

Own weigh ((0.06*0.55+0.15*0.24 )*25 + (0.2*0.24*2*12) ) /0.55 = 5.23 Kn /m2  

Check wide beam shear in the slab Take Strep 2 : wu = 1.2(5.23 +4.7 ) +1.6(4) =18.516 KN /m Wu rip =18.516 * 0.55 = 10.18 KN /m Wide beam shear in slab Strip2 at the face of support.

. From ACI Coefficient : Vu =1.15* Wu*Ln/2 Vu =1.15*10.18*4.3/2 = 25.17 KN Vu = 25.17 –(10.18 *0.25) = 22.63 KN Ø Vc =1.1* 0.75 (24^0.5) *(1/6 ) *(150)* (0.25) = 25.26 KN Vu < Ø Vc so its ok use 1Ø8 /500 for bars fixation

Flexure design in the slab block 1

block 2

Strip 1 From ACI Coefficient

Strip 1 Positive moment : Mu = 11.55 KN.m ƿ = 0.0034 , ƿ min =0.0035 ,ƿ max =0.375 *β1 *0.85fc/fy =0.015 As = 0.0035 *150* 250 =127.5 mm = 2 Ø 10 Negative moment : Mu = 16.17 KN .m Ƿ= 0.0048 As =0.0048 *150 *250 = 180 = 2Ø12

Strip 2 Moment from ACI coefficient

Strip 2 Positive moment : Mu1 = 13.44 KN.m ƿ = 0.0039 , ƿ min =0.0035 ,ƿ max =0.375 *β1 *0.85fc/fy =0.015 As = 0.0047 *150* 250 =148 mm = 2 Ø 10 Mu2 = 11.67 KN.m ƿ = 0.0034 , ƿ min =0.0035 ,ƿ max =0.375 *β1 *0.85fc/fy =0.015 As = 0.0035 *150* 250 =131.25 mm = 2 Ø 10 Negative moment : Mu = 18.8 KN .m Ƿ= 0.0056 As =0.0056 *150 *250 = 210 = 2Ø10

Strip 3: Positive moment : Mu = 12.48 KN.m ƿ = 0.00 37 , ƿ min =0.0035 ,ƿ max =0.375 *β1 *0.85fc/fy =0.015 As = 0.0037 *150* 250 =137 mm = 2 Ø 10

strip 4: Positive moment : Mu = 9.07 KN.m ƿ = 0.00 26 , ƿ min =0.0035 ,ƿ max =0.375 *β1 *0.85fc/fy =0.015 As = 0.0035 *150* 250 =131.25 mm = 2 Ø 10

Design of beam block1

block2

Beams Dimensions  

Beams Dimensions

beam design  

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Secondary beam :  

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Flexure design Bending Moment Diagram in Main beam

Main beam Positive moment : Mu = 229 KN.m b=300mm d =540mm ƿ = 0.0075 , ƿ min =0.0035 ,ƿ max =0.375 *β1 *0.85fc/fy =0.015 As = 0.0075 *300* 540 =1011 mm = 4 Ø 20 Negative moment : Mu = 269 KN .m Ƿ= 0.00894 As =0.00894 *300 *540 = 1449 = 5Ø20

secondary beam Moment from ACI coefficient

secondary beam  

THREE DIMENSIONAL MODELING AND CHECKS .

modeling of the project Define materials

Define section properties Beam

secondary beam

Beams modifiers

columns section properties

columns modifiers

define slab section

Slab modifiers

Model checks Check for compatibility Block1

Check for compatibility Block2

Equilibrium Check for block1 Area of block = (17.65 *27.6 ) =484.17 m2 For sd : sdmanual= 484.17*4.7*4 = 9158.232 KN = sdsap =ok

Equilibrium Check For dead load : Dead load manual=15684.1KN Dead load sap =15697 KN %error =( 15684.1-15697) /15684.1= 0.08 % >>> ok For live load : Live lodemanual = 3698.4 + 1407.6 + 586.5 = 5692.5 KN Live lodemanual =Live load sap >> ok

Internal Forces Check Mmanual = WL2 /8 = (21.6 *4.252 /8 ) = 48.7 Kn.m Msap = ( m1 + m2 )/2 + m 3 = (44.16 +44.3 ) /2 + 3.96 = 48.2 KN.m %error = 1.02 % <10% >>> so its ok

FINAL DESIGN OF SLABS, BEAMS and columns.

Final design of special structural element (Tie beam, Shear wall, Footings, and Stairs).

Final design of Slabs .

Reinforcement of strip one moment Moment /RIB Ƿ As Asmin Asuse 21.2 11.9 0.0032 125 131.25 2Ø10 13.6 7.48 0.002 78 7 3.85 0.00101 40 17.9 9.854 0.002636 103 14 7.7 0.00205 80 12.5 6.875 0.001825 72

Final design of Slabs

moment Moment /RIB Ƿ As Asmin Asuse 6.2 3.4 0.000896 35 131.25 2Ø10 13.7 7.535 0.002004 79 22.4 12.32 0.003323 130 12.7 6.985 0.001855 73 15.7 8.635 0.002304 90 16.3 8.965 0.002394 94 12.6 6.93 0.00184 72

flexural design

Check shear for block one shear v13 in block one

 

For block two

: M11 Reading for strip one

Since all the moment value is minimum moment =23 KN/m( The As min =131 Since all the moment value is minimum moment =23 KN/m( The As min =131.25 mm  M=23 KN/m ) ,then the reinforcement in beam is minimum  131.25  2 Ø10

Check shear for block 2 shear v13 in block two

shear v23 in block two

  Vmax =36KN <concrete shear capacity 44.3 KN.

Final Design of beams

beam design Beam # Width "cm" Depth, cm -As total "mm2" Right Left +As total "mm2" STIRUPS   B1 30 60 4 Ø 14 4 Ø16 2 Ø10/10cm B2 6 Ø18 4 Ø14 B3 1 Ø10/10cm B4 B5 8 Ø 14 6 Ø14 B6 6 Ø16 B7 5 Ø16 B8 2 Ø14 B9 2Ø 14 B10 4Ø 14 B11 B12 B13 B14 B15 2 Ø 14 B16

Final Design of columns

Column name Dimension (cm AS required (mm2) Number of bars Ties C1 50×30 1500 8 Ø 16 2 Ø10/25cm C2 70×30 3737 12 Ø 20 C3 30×30 900 6 Ø 14 1 Ø10/25cm C4 30*50 14 Ø 20

Final design of special structural element (Tie beam, Shear wall, Footings, and Stairs).

tie beam H =40 cm , d =36cm . In the internal tie beam the area of steel found is less than the minimum. Asmin =0.00333×300×340 = 340 mm2 Thus, it was used a minimum area of steel, 3Ø14 for both bottom and top steels.  

Design of footing for column and shear wall

Final Design for footing Footing No Ps (KN) Pu (KN) Long dimension (m) Short dimension Depth (cm) Top l.R Top T.R Bottom l.R Bottom T.R F1 1461 1911 2.6 2.4 50 7 Ø14 2 Ø14 F2 1240 1623 2.2 46 6 Ø14 F3 462 615 1.4 37 5 Ø14 F4 266 346 1.2 1 30 4 Ø14

Design of wall footing L(m) W(m) D (cm) Transverse reinforcement   L(m) W(m) D (cm) Transverse reinforcement longitudinal reinforcement W.F1 8.75 2 30 4 Ø14/m 8 Ø14 W.F2 9.8 3 12Ø14 W.F3 12.2 12 Ø14 W.E4 19 8Ø14

design for shear wall

Final design for shear wall Wall no Thickness Cm Height m Width Vertical reinforcement on each side Horizontal reinforcement on each side W1 30 12.8 6.75 4Ø14 mm/m 3Ø14 mm/m W2 W3 4.5 W4 7.5

design for stairs Stairs dimensions: *∝ = 27 o , *Going = 30 cm. *Riser = 15 cm. *Floor height = 3.2m *No. of goings = 21. *No. of risers = 22. Use solid slab with 15 cm thickness, d=120 mm.

Check for shear  

Design for flexure