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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
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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.
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Introduction Project Description : Its consist of 4 stories each with 1300 m2 ,
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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.
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Soil properties Soil investigation indicated that the bearing capacity of the soil is 250 KN/m2
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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.
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Structural material
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-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
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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
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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
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-Live load:
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Load combination THE ULTIMATE LOAD COMBINATION: 1.4 × Dead Load .
1.2 × Dead Load × Live Load . 1.2 × Dead Load × Live Load × Earthquake . 0.9 × Dead Load × 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)
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PRELIMENARY DESIGN FOR SLABS, BEAMS, AND COLUMNS
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Slab preliminary design
Minimum thicknes
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Own weigh ((0.06* *0.24 )*25 + (0.2*0.24*2*12) ) /0.55 = 5.23 Kn /m2
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Check wide beam shear in the slab
Take Strep 2 : wu = 1.2( ) +1.6(4) = KN /m Wu rip = * 0.55 = KN /m Wide beam shear in slab Strip2 at the face of support.
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. From ACI Coefficient : Vu =1.15* Wu*Ln/2 Vu =1.15*10.18*4.3/2 = KN Vu = –(10.18 *0.25) = KN Ø Vc =1.1* 0.75 (24^0.5) *(1/6 ) *(150)* (0.25) = KN Vu < Ø Vc so its ok use 1Ø8 /500 for bars fixation
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Flexure design in the slab
block 1
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block 2
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Strip 1 From ACI Coefficient
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Strip 1 Positive moment : Mu = 11.55 KN.m
ƿ = , ƿ min = ,ƿ max =0.375 *β1 *0.85fc/fy =0.015 As = *150* 250 =127.5 mm = 2 Ø 10 Negative moment : Mu = KN .m Ƿ= As = *150 *250 = 180 = 2Ø12
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Strip 2 Moment from ACI coefficient
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Strip 2 Positive moment : Mu1 = 13.44 KN.m
ƿ = , ƿ min = ,ƿ max =0.375 *β1 *0.85fc/fy =0.015 As = *150* 250 =148 mm = 2 Ø 10 Mu2 = KN.m ƿ = , ƿ min = ,ƿ max =0.375 *β1 *0.85fc/fy =0.015 As = *150* 250 = mm = 2 Ø 10 Negative moment : Mu = 18.8 KN .m Ƿ= As = *150 *250 = 210 = 2Ø10
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Strip 3: Positive moment : Mu = 12.48 KN.m
ƿ = , ƿ min = ,ƿ max =0.375 *β1 *0.85fc/fy =0.015 As = *150* 250 =137 mm = 2 Ø 10
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strip 4: Positive moment : Mu = 9.07 KN.m
ƿ = , ƿ min = ,ƿ max =0.375 *β1 *0.85fc/fy =0.015 As = *150* 250 = mm = 2 Ø 10
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Design of beam block1
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block2
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Beams Dimensions
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Beams Dimensions
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beam design
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Secondary beam :
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. .
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Flexure design Bending Moment Diagram in Main beam
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Main beam Positive moment : Mu = 229 KN.m b=300mm d =540mm
ƿ = , ƿ min = ,ƿ max =0.375 *β1 *0.85fc/fy =0.015 As = *300* 540 =1011 mm = 4 Ø 20 Negative moment : Mu = 269 KN .m Ƿ= As = *300 *540 = 1449 = 5Ø20
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secondary beam Moment from ACI coefficient
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secondary beam
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THREE DIMENSIONAL MODELING AND CHECKS
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modeling of the project
Define materials
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Define section properties
Beam
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secondary beam
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Beams modifiers
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columns section properties
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columns modifiers
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define slab section
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Slab modifiers
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Model checks Check for compatibility Block1
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Check for compatibility Block2
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Equilibrium Check for block1 Area of block = (17.65 *27.6 ) = m2 For sd : sdmanual= *4.7*4 = KN = sdsap =ok
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Equilibrium Check For dead load : Dead load manual= KN Dead load sap =15697 KN %error =( ) / = 0.08 % >>> ok For live load : Live lodemanual = = KN Live lodemanual =Live load sap >> ok
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Internal Forces Check Mmanual = WL2 /8 = (21.6 *4.252 /8 ) = 48.7 Kn.m
Msap = ( m1 + m2 )/2 + m 3 = ( ) / = 48.2 KN.m %error = % <10% >>> so its ok
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FINAL DESIGN OF SLABS, BEAMS and columns.
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Final design of special structural element (Tie beam, Shear wall, Footings, and Stairs).
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Final design of Slabs .
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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 40 17.9 9.854 103 14 7.7 80 12.5 6.875 72
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Final design of Slabs
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moment Moment /RIB Ƿ As Asmin Asuse 6.2 3.4 35 131.25 2Ø10 13.7 7.535 79 22.4 12.32 130 12.7 6.985 73 15.7 8.635 90 16.3 8.965 94 12.6 6.93 72
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flexural design
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Check shear for block one
shear v13 in block one
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For block two
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: M11 Reading for strip one
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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 = mm M=23 KN/m ) ,then the reinforcement in beam is minimum 2 Ø10
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Check shear for block 2 shear v13 in block two
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shear v23 in block two
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Vmax =36KN <concrete shear capacity 44.3 KN.
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Final Design of beams
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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
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Final Design of columns
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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
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Final design of special structural element (Tie beam, Shear wall, Footings, and Stairs).
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tie beam H =40 cm , d =36cm . In the internal tie beam the area of steel found is less than the minimum. Asmin = ×300×340 = 340 mm2 Thus, it was used a minimum area of steel, 3Ø14 for both bottom and top steels.
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Design of footing for column and shear wall
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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
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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
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design for shear wall
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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
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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.
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Check for shear
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Design for flexure
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