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An-Najah National University
Faculty of Engineering
Civil Engineering Department
Structural analysis and design of Tareq Al-Alool building - Nablus
Prepared By:
Yazan saif
Motaz othman
Mohammad masalha
Supervisor: : Dr. Riyad Abd Alkareem
Outline :
16/5/2018
Graduation project 11
 Background information
 Analysis and design inputs
 Conceptual design
 ETABs modeling & checks.
 Design.
16/5/2018
Graduation project 11
Location:
The project addresses is the analysis and design of Tareq Al-Alool building (5500 m2) located
in Nablus.
16/5/2018
Graduation project 11
Basement floor
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Graduation project 11
Ground floor
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Graduation project 11
1st floor
Floor number Function Summation of
Areas (m2)
Height of floors(m)
Basement (B3) Parking,
660
3.25
Basement(B2,B1) Stores
1115 B2=3.25
B1=3.5
floors(GF,F1,F2,F3,F4) Offices 2240
GF = 4.5
Floors=3
floors (F5,F6) Apartments 810 3
Sixth floor roof(F6) Apartment 55 3
Floors, functions areas and heights
16/5/2018
Graduation project 11
16/5/2018
Graduation project 11
What we do in GP 11?
Dead Superimposed
Gravity Load
Live
 Concrete:
Unit weight of reinforced concrete = 25 kN/m³
Structural element Fc` (MPs) Concrete type Modulus of elasticity(MPs)
Beams, slabs , stairs 24 B300 23000
Mat, columns, shear walls 28 B350 24900
 Reinforcement steel:
Yielding strength (Fy) = 420MPa.
Modulus of elasticity (Es) = 200GP
Materials:
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Graduation project 11
.
 Soil properties:
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Graduation project 11
• Soil investigation indicated that the
• bearing capacity of the soil is 250 KN/m2.
 Loads :
Gravity Lateral
Loads
Dead Live Earthquake
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Graduation project 11
 Load:
 Dead Load:
13
Live load:
In this project, a live load of 2.5 kN/m2 will be used for Garage and 5
kN/m2 for corridors and floors according to IBC 2015
 Slab one way ribbed = 3.63 KN/ m2
 Slab two way ribbed =4.82 KN/m2
 Superimposed dead load= 4 KN/m2
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Graduation project 11
Load combinations:
According to the "ACI 318-2011 , the load factors that are used in the analysis and
design are:
Load combinations:
• U = 1.4D
• U = 1.2D + 1.6L
• U = 1.2D + 1.6L + 0.5(Lr or S or R)
• U = 1.2D + 1.6(Lr or S or R) + (1.0L or 0.5W)
• U = 1.2D + 1.0W + 1.0L + 0.5(Lr or S or R)
• U = 1.2D + 1.0E + 1.0L + 0.2S
• U = 0.9D + 1.0W
• U = 0.9D + 1.0E
16/5/2018
Graduation project 11
Computer programs:
1) ETABS 2016: is used to analyze and design the structural elements.
2) Sap 2000 for design and checks some of structural elements .
3) AutoCAD: is used for plans and draw structural detailing.
4) Other programs such as Excel and word.
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Graduation project 11
 Preliminary design for building:
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Graduation project 11
1. Slab :
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For two end continuous ribbed slab thickness of slab h=span length/21
Ribbed slab thickness = 6.75 m/21 = 32 cm
2. Beam Dimensions :
Beam Dimensions(mm2)
main beams B2 320×1000
secondary beams B1 320×350
Graduation project 11
After making checks, the final beam dimensions are shown in this
table:
Graduation project 11
3. Column Dimensions :
Critical column is shown below in figure below:
Graduation project 11
By using tributary area method,
Graduation project 11
Tributary area = (6.6 *5.76 ) = 38 m2
ØPn₀= 9239.8kN
Assume ρ =0.01
As= ρ *Ag
ØPn₀= ‫×ג‬Ø (0.85𝑓𝑐 ′Ac + As × fy)
=0.65*0.8*(0.85*28*(Ag -0.01(Ag)) +420*0.01*Ag)/1000
Ag=460000 mm2 Assume b=h
b=h=700 mm these dimensions consider no moment .
Consider moment use column 800 *800 mm
Graduation project 11
Columns dimensions are shown in table:
Original dimensions
(mm)
C1 800*800
C2 600*600
C3 1000*1000
Shear Wall
Graduation project 11
Wall Thickness
(mm)
Wall 1 200
Wall 2 250
Wall 3 300
Graduation project 11
Modifiers for area structural element
Modifiers/Section Slab Shear walls
Membrane f11 Direction 1 1
Membrane f22 Direction 1 1
Membrane f12 Direction 1 1
Bending m11 Direction 0.25 0.7
Bending m22 Direction 0.25 0.7
Bending m12 Direction 0.25 0.7
Shear v13 Direction 1 1
Shear v23 Direction 1 1
Mass 1 1
Weight 1 1
.
Graduation project 11
Modifiers for frame structural element
Modifiers/Section beams Columns
Cross Section(axial) Area 1 1
Sheer Area in 2 direction 1 1
Sheer Area in 3 direction 1 1
Torsional Constant 0.35 0.7
Moment of Inertia about 2
axis
0.35 0.7
Moment of Inertia about 3
axis
0.35 0.7
Mass (Depth-200)/200 1
Weight (Depth-200)/200 1
DYNAMIC ANALYSIS
Response Spectrum Analysis
Modes and Period
Drift Check
Definition of mass resource
Seismic Parameters
Seismic Zone Factor Z
Site Classification Sd
Seismic Coefficient Ca
Seismic Coefficient Cv
Seismic Importance Factor I
Define Of Response Spectrum Function
Response Spectrum In X-Direction
Response Spectrum In Y-Direction
Check Base Shear
• Cs =
0.4×1
5.5×0.77
=0.0945
• W= 82044 kN
• V= 82044 × 0.0945 = 7753 kN
• Not ok
After increase scale factor
the value become large than V
TABLE: Base Reactions
Load Case/Combo FX FY FZ MX MY MZ
Unit kN kN kN kN-m kN-m kN-m
EQ x Max 9811.4 2980.2 3079.4 51365.9 220549.0 74680.1
EQ y Max 3226.3 9926.7 3796.3 239332.6 48945.5 73167.4
Check period
• By method A
• The period, T is given by: T = Ct(Hn)
3
4
• So, T = 0.0488(35.50)
3
4= 0.77sec
• And from Etabs
• Tetab < 1.2 Tcal
Period in mode 1 → T= 0.74 sec.
Checks
Model
check
Equilibrium
check
Deflection
check
Structural
period T
check
Stress
strain
check
Model check
compatibility check
Equilibrium check
Type
ETABS results
(KN)
Hand calculation
(KN)
Error%
Live 22200 22182 0
Superimposed 21588 21574 0
Dead 55643 54811 0
 M Hand =115.4KN.m

𝑀1+𝑀3
2
+ 𝑀2 =
47.3+44.7
2
+ 30.5 = 77 KN.m
 % Difference =33%
 We made the internal force check from live load for this beam as following in the figure:
Stress strain check
Stress strain check
• We made the internal force check from live load for this coulmn as following
in the figure:
D= (-From hand:
D= (37.46*2.5*3)+(37.46*5*7)=1592KN
1564 KN
Error 0 < 10 % OK
Stress strain check
• We made the internal force check from SD load for slab as following in the figure:
• -From hand:
• SD =𝑊𝑢×𝑙2/8=(4×6.45^2)/8=20KN.𝑚
• From Etab
• -(M1 + M2)/ 2 + M3 =(13+12.2)/2 + 8.6= 21.5KN.m.
• Error 5 % < 10 OK
•
 ACI code limitations on immediate and long term deflections
 Long term deflection
ΔLT = ΔL + λ∞ ΔD + λ t ΔLS
Deflection check
immediate term deflection for ribbed slab :
Deflection check
Maximum deflection (Uz) = 22.8-(AVG OF (2.5, 3.56,2.8, 3.9))
=19.61 mm
𝐴𝑙𝑙𝑜𝑤𝑎𝑏𝑙𝑒 𝑑𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛 = 𝐿 /240
=(7.25∗ 1000)/ 240
= 30.2 𝑚𝑚.
19.61 mm < 30.2 → Ok
Deflection For Beam
•
For beam:
•
Allowable deflection=L / 240= (7.25 *1000)/240= 30.2
mm
•
Deflection from sap = 9.7- ((3.9+2.3)/2) = 6.6 mm
Check for shear in slab :
• Hand ΦVc= 0.75X
1
6
X 1X 24X 150X 280/1000= 25 kN/Rib
• = 46.7 KN/m
From above Figures it was noticed that ɸVc ≫ Vu so
no need for shear reinforcement in slab.
DESIGN Design of element
for flexure
Design for slab
Asmin = ρ b h
= 0.0033 X 150 X 320 = 140 mm2 (use 2 Φ
10/Ribbed)
So, As = 140 mm2  0.55 m
ΦMn= 0.9 X 140X 420 X (280-(19.3/2)) X 10-6
= 14.6 kN.m /0.55m
=26 KN.mm
Moment (11)→ bottom and top
2φ10mm/ribbed mm for top steel
Design for coulmn
• From ETABS we found:
• The axial force in column =548 kN
• M11=113kN.m
• M22=195kN.m
• At first we will check the slenderness effect (KL/r):
• KL/r = 1 X 3.25 / 0.24 = 13.54 should be less than 34−12
𝑀11
𝑀22
•
• 34−12
𝑀11
𝑀22
= 27 > 13.54 , so is a short column
•
1
Ø𝑃𝑛
=
1
Ø𝑃𝑛𝑥
+
1
Ø𝑃𝑛𝑦
−
1
Ø𝑃𝑛₀
• Ø𝑃𝑛𝑥:
• M2=113kN.m
• Pu=548.5kN.
•
𝑒
ℎ
=
360
800
X
1
0.7
= 0.64
• ϒ=
800−80
800
= 0.9
• ρ = 0.01
• From interaction diagram
• Ø𝑃𝑛𝑥 =0.2X 800 X 800 X 7/1000=896kN.
• Ø𝑃𝑛𝑦:
• =1254
• Pn>Pu
• As=0.01×800×800=6400mm2(17Ø22)
Design for matt foundation
• Checks:
• Bearing capacity of soil
maximum spring load = 40
• Bearing capacity = 40/0.5*0.5 = 160 < 250 K.n/m2 ok
Shear capacity check(punching):
• Vu = pu – q (critical section )
𝑉
𝑐 =
1
3
𝑓′𝑐 bd
• Vu = 4954
• Vc = 6773
VC>VU Check is ok
Slab Moment and Area of steel
• As min = ρ b h
• = 0.0018 × 1000 × 1600 = 2880 mm2 (use 10 Φ 20mm)
• Φ Mn= 0.9×1880×420× 1500 - (33.1/2)) ×10-6 =1747 kN.m /m
All value is less than 1747 K.n/m in both m11 & m22
The reinforcement is 10 Φ 20mm/m
Design of stairs:
Deflection shape
• Based on the long term deflection, max. immediate deflection from SAP2000,
∆ =2.7 mm
• Deflection limitation:
• 𝑎𝑙𝑙𝑎𝑤𝑎𝑏𝑙𝑒 𝐿𝑜𝑛𝑔 𝑡𝑒𝑟𝑚, ∆𝐿 ≤ 𝑙 /240 = 2700/240= 11.25 mm
• The long term deflection from SAP found approximately 2.7 mm, which
means that is smaller than the allowable deflection → OK
• Check for shear:
• Vu13 and Vu23 from SAP analysis = 67 kN and 62 kN respectively.
• ɸ VC= 0.75×
1
6
×λ× Fc
;
×b×d
• = 0.75×
1
6
× 28×1000×110/1000 =72.8 kN
• Vu= 67 kN
• Vu < ɸVc The slab thickness is OK
Design for flexural:
• Mu11 and Mu22 from landing analysis by SAP2000 = 9 , 10.6 kN.m
respectively.
• Mu max =8 kN.m
• ρ =
0.85×28
420
( 1 − 1 −
2.61×10×106
28×1000×1102 ) =0.00236
• As= 0.00236 × 1000×110 = 260mm2
• Asmin =0.0018×1000×150=270 mm2 (Consider it as slab)
•
• As > Asmin, so Use As= 270mm2
(4ɸ10/m in two direction)
• ρ =
0.85×28
420
(1 − 1 −
2.61×8.6×106
28×1000×1102 ) = 0.00191
• As=0.00191× 1000×110 =210 mm2
• Asmin =0.0018×1000×150=270 mm2
• As < Asmin so Use As= 270 mm2
(4ɸ10/m)
•
• As we defined the stair as one-way solid slab, so the transverse moment is too small,
so we used:
• As shrinkage = As min = 0.0018 × 150 × 1000
= 270 mm2 in both face (Use 4 ∅10 /m for each layer)
For Flight:
Mu22 from flight analysis by SAP2000= 8.6kN.m
Design For Interior Panel (B12 ):
:
1-Parameters:
• 2-Steel Reinforcement:
• Mu,max Positive = 120 kN.m >>> 6Ø16 mm.
• Mu,max Nevative = 225 kN.m >>> 8Ø20 mm.
• Vu = 242 kN >>> 2Φ10 / 300 mm.
•
b (mm) h (mm) d (mm) 𝑓𝑠 (Mpa) fy (Mpa)
1000 320 240 24 420
AutoCAD Details :
Design For Shear Wall (SW-3 ):
Notes:
• 5 internal ultimate forces can be computed; namely, P, Vx, Vy, Mx and My .
• Design for V-y in a way similar to beams .
• Design for M-y can be carried out by assuming there acts as Couple.
• Design for P and Mx in a way similar to columns.
• The stresses in the SW-3 are less than 0.1 F`c, there is no need for interaction
diagram check.
Reinforcement For SW-3 :
• AS min=0.0025×b×h=0.0025*1000*300 >>> (1Ø12/200mm) on each side.
• According to ACI315-04 section 2.10 these rules are following:
• Transverse and Longitudinal rebar ratio shall be at least 0.0025
• Spacing of reinforcement shall not exceed:
450
3ℎ
𝑙𝑤
5
mm.
• h: the thickness of shear wall (300mm).
• So the maximum spacing is 450 mm which is larger than 300 mm, so it is OK.
Reinforcement For SW-3 :
• First : Add Two column to the shear wall , To minimize the moment on shear wall
and make Minimum Reinforcement.
• Second : Use General Reinforcement At Etabs Pier Design.
Shear Wall Results :
General Reinforcement:
Define General Reinforcement for Etabs using minimum Reinforcement ,But many Etabs
alerts for moment and shear so we increase the Reinforcement.
Etabs Reinforcement Check Result
:
Shear Wall Reinforcement Results
:
Vertical
reinforcement/m
As/2
Reinforcement
For wall
Horizontal
reinforcement/m
Reinforcement
For wall
Shear
Reinforcement
For wall
Vertical
reinforcement
For Column
Shear
Reinforcement
for Column
SW-3 1600 800 1φ12/130mm 850
1φ18/200mm
1φ12/200mm
Use
1φ18/200mm
1φ25/150mm
2φ10/200mm
AutoCAD Details :
Thank you

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ANU Civil Engineering Structural Analysis of Tareq Al-Alool Building

  • 1. An-Najah National University Faculty of Engineering Civil Engineering Department Structural analysis and design of Tareq Al-Alool building - Nablus Prepared By: Yazan saif Motaz othman Mohammad masalha Supervisor: : Dr. Riyad Abd Alkareem
  • 2. Outline : 16/5/2018 Graduation project 11  Background information  Analysis and design inputs  Conceptual design  ETABs modeling & checks.  Design.
  • 4. Location: The project addresses is the analysis and design of Tareq Al-Alool building (5500 m2) located in Nablus.
  • 8. Floor number Function Summation of Areas (m2) Height of floors(m) Basement (B3) Parking, 660 3.25 Basement(B2,B1) Stores 1115 B2=3.25 B1=3.5 floors(GF,F1,F2,F3,F4) Offices 2240 GF = 4.5 Floors=3 floors (F5,F6) Apartments 810 3 Sixth floor roof(F6) Apartment 55 3 Floors, functions areas and heights 16/5/2018 Graduation project 11
  • 9. 16/5/2018 Graduation project 11 What we do in GP 11? Dead Superimposed Gravity Load Live
  • 10.  Concrete: Unit weight of reinforced concrete = 25 kN/m³ Structural element Fc` (MPs) Concrete type Modulus of elasticity(MPs) Beams, slabs , stairs 24 B300 23000 Mat, columns, shear walls 28 B350 24900  Reinforcement steel: Yielding strength (Fy) = 420MPa. Modulus of elasticity (Es) = 200GP Materials: 16/5/2018 Graduation project 11
  • 11. .  Soil properties: 16/5/2018 Graduation project 11 • Soil investigation indicated that the • bearing capacity of the soil is 250 KN/m2.
  • 12.  Loads : Gravity Lateral Loads Dead Live Earthquake 16/ 5/2018 Graduation project 11
  • 13.  Load:  Dead Load: 13 Live load: In this project, a live load of 2.5 kN/m2 will be used for Garage and 5 kN/m2 for corridors and floors according to IBC 2015  Slab one way ribbed = 3.63 KN/ m2  Slab two way ribbed =4.82 KN/m2  Superimposed dead load= 4 KN/m2 16/5/2018 Graduation project 11
  • 14. Load combinations: According to the "ACI 318-2011 , the load factors that are used in the analysis and design are: Load combinations: • U = 1.4D • U = 1.2D + 1.6L • U = 1.2D + 1.6L + 0.5(Lr or S or R) • U = 1.2D + 1.6(Lr or S or R) + (1.0L or 0.5W) • U = 1.2D + 1.0W + 1.0L + 0.5(Lr or S or R) • U = 1.2D + 1.0E + 1.0L + 0.2S • U = 0.9D + 1.0W • U = 0.9D + 1.0E 16/5/2018 Graduation project 11
  • 15. Computer programs: 1) ETABS 2016: is used to analyze and design the structural elements. 2) Sap 2000 for design and checks some of structural elements . 3) AutoCAD: is used for plans and draw structural detailing. 4) Other programs such as Excel and word. 16/5/2018 Graduation project 11
  • 16.  Preliminary design for building: 25/5/2017 Graduation project 11
  • 18. 16/5/2018 Graduation project 11 For two end continuous ribbed slab thickness of slab h=span length/21 Ribbed slab thickness = 6.75 m/21 = 32 cm
  • 19. 2. Beam Dimensions : Beam Dimensions(mm2) main beams B2 320×1000 secondary beams B1 320×350 Graduation project 11 After making checks, the final beam dimensions are shown in this table:
  • 21.
  • 22. 3. Column Dimensions : Critical column is shown below in figure below: Graduation project 11
  • 23. By using tributary area method, Graduation project 11 Tributary area = (6.6 *5.76 ) = 38 m2 ØPn₀= 9239.8kN Assume ρ =0.01 As= ρ *Ag ØPn₀= ‫×ג‬Ø (0.85𝑓𝑐 ′Ac + As × fy) =0.65*0.8*(0.85*28*(Ag -0.01(Ag)) +420*0.01*Ag)/1000 Ag=460000 mm2 Assume b=h b=h=700 mm these dimensions consider no moment . Consider moment use column 800 *800 mm
  • 24. Graduation project 11 Columns dimensions are shown in table: Original dimensions (mm) C1 800*800 C2 600*600 C3 1000*1000
  • 25. Shear Wall Graduation project 11 Wall Thickness (mm) Wall 1 200 Wall 2 250 Wall 3 300
  • 26. Graduation project 11 Modifiers for area structural element Modifiers/Section Slab Shear walls Membrane f11 Direction 1 1 Membrane f22 Direction 1 1 Membrane f12 Direction 1 1 Bending m11 Direction 0.25 0.7 Bending m22 Direction 0.25 0.7 Bending m12 Direction 0.25 0.7 Shear v13 Direction 1 1 Shear v23 Direction 1 1 Mass 1 1 Weight 1 1
  • 27. . Graduation project 11 Modifiers for frame structural element Modifiers/Section beams Columns Cross Section(axial) Area 1 1 Sheer Area in 2 direction 1 1 Sheer Area in 3 direction 1 1 Torsional Constant 0.35 0.7 Moment of Inertia about 2 axis 0.35 0.7 Moment of Inertia about 3 axis 0.35 0.7 Mass (Depth-200)/200 1 Weight (Depth-200)/200 1
  • 28. DYNAMIC ANALYSIS Response Spectrum Analysis Modes and Period Drift Check
  • 29. Definition of mass resource
  • 36. Define Of Response Spectrum Function
  • 37. Response Spectrum In X-Direction
  • 38. Response Spectrum In Y-Direction
  • 40. • Cs = 0.4×1 5.5×0.77 =0.0945 • W= 82044 kN • V= 82044 × 0.0945 = 7753 kN • Not ok
  • 41. After increase scale factor the value become large than V TABLE: Base Reactions Load Case/Combo FX FY FZ MX MY MZ Unit kN kN kN kN-m kN-m kN-m EQ x Max 9811.4 2980.2 3079.4 51365.9 220549.0 74680.1 EQ y Max 3226.3 9926.7 3796.3 239332.6 48945.5 73167.4
  • 42. Check period • By method A • The period, T is given by: T = Ct(Hn) 3 4 • So, T = 0.0488(35.50) 3 4= 0.77sec • And from Etabs • Tetab < 1.2 Tcal Period in mode 1 → T= 0.74 sec.
  • 46. Equilibrium check Type ETABS results (KN) Hand calculation (KN) Error% Live 22200 22182 0 Superimposed 21588 21574 0 Dead 55643 54811 0
  • 47.
  • 48.  M Hand =115.4KN.m  𝑀1+𝑀3 2 + 𝑀2 = 47.3+44.7 2 + 30.5 = 77 KN.m  % Difference =33%  We made the internal force check from live load for this beam as following in the figure: Stress strain check
  • 49. Stress strain check • We made the internal force check from live load for this coulmn as following in the figure: D= (-From hand: D= (37.46*2.5*3)+(37.46*5*7)=1592KN 1564 KN Error 0 < 10 % OK
  • 50. Stress strain check • We made the internal force check from SD load for slab as following in the figure: • -From hand: • SD =𝑊𝑢×𝑙2/8=(4×6.45^2)/8=20KN.𝑚 • From Etab • -(M1 + M2)/ 2 + M3 =(13+12.2)/2 + 8.6= 21.5KN.m. • Error 5 % < 10 OK •
  • 51.  ACI code limitations on immediate and long term deflections  Long term deflection ΔLT = ΔL + λ∞ ΔD + λ t ΔLS Deflection check
  • 52. immediate term deflection for ribbed slab : Deflection check Maximum deflection (Uz) = 22.8-(AVG OF (2.5, 3.56,2.8, 3.9)) =19.61 mm 𝐴𝑙𝑙𝑜𝑤𝑎𝑏𝑙𝑒 𝑑𝑒𝑓𝑙𝑒𝑐𝑡𝑖𝑜𝑛 = 𝐿 /240 =(7.25∗ 1000)/ 240 = 30.2 𝑚𝑚. 19.61 mm < 30.2 → Ok
  • 54. • For beam: • Allowable deflection=L / 240= (7.25 *1000)/240= 30.2 mm • Deflection from sap = 9.7- ((3.9+2.3)/2) = 6.6 mm
  • 55. Check for shear in slab : • Hand ΦVc= 0.75X 1 6 X 1X 24X 150X 280/1000= 25 kN/Rib • = 46.7 KN/m From above Figures it was noticed that ɸVc ≫ Vu so no need for shear reinforcement in slab.
  • 56. DESIGN Design of element for flexure
  • 57. Design for slab Asmin = ρ b h = 0.0033 X 150 X 320 = 140 mm2 (use 2 Φ 10/Ribbed) So, As = 140 mm2 0.55 m ΦMn= 0.9 X 140X 420 X (280-(19.3/2)) X 10-6 = 14.6 kN.m /0.55m =26 KN.mm
  • 58. Moment (11)→ bottom and top 2φ10mm/ribbed mm for top steel
  • 59. Design for coulmn • From ETABS we found: • The axial force in column =548 kN • M11=113kN.m • M22=195kN.m • At first we will check the slenderness effect (KL/r):
  • 60. • KL/r = 1 X 3.25 / 0.24 = 13.54 should be less than 34−12 𝑀11 𝑀22 • • 34−12 𝑀11 𝑀22 = 27 > 13.54 , so is a short column
  • 61. • 1 Ø𝑃𝑛 = 1 Ø𝑃𝑛𝑥 + 1 Ø𝑃𝑛𝑦 − 1 Ø𝑃𝑛₀ • Ø𝑃𝑛𝑥: • M2=113kN.m • Pu=548.5kN. • 𝑒 ℎ = 360 800 X 1 0.7 = 0.64 • ϒ= 800−80 800 = 0.9 • ρ = 0.01 • From interaction diagram • Ø𝑃𝑛𝑥 =0.2X 800 X 800 X 7/1000=896kN.
  • 62. • Ø𝑃𝑛𝑦: • =1254 • Pn>Pu • As=0.01×800×800=6400mm2(17Ø22)
  • 63. Design for matt foundation
  • 64. • Checks: • Bearing capacity of soil maximum spring load = 40 • Bearing capacity = 40/0.5*0.5 = 160 < 250 K.n/m2 ok
  • 65. Shear capacity check(punching): • Vu = pu – q (critical section ) 𝑉 𝑐 = 1 3 𝑓′𝑐 bd • Vu = 4954 • Vc = 6773 VC>VU Check is ok
  • 66. Slab Moment and Area of steel • As min = ρ b h • = 0.0018 × 1000 × 1600 = 2880 mm2 (use 10 Φ 20mm) • Φ Mn= 0.9×1880×420× 1500 - (33.1/2)) ×10-6 =1747 kN.m /m
  • 67. All value is less than 1747 K.n/m in both m11 & m22 The reinforcement is 10 Φ 20mm/m
  • 70. • Based on the long term deflection, max. immediate deflection from SAP2000, ∆ =2.7 mm • Deflection limitation: • 𝑎𝑙𝑙𝑎𝑤𝑎𝑏𝑙𝑒 𝐿𝑜𝑛𝑔 𝑡𝑒𝑟𝑚, ∆𝐿 ≤ 𝑙 /240 = 2700/240= 11.25 mm • The long term deflection from SAP found approximately 2.7 mm, which means that is smaller than the allowable deflection → OK
  • 71. • Check for shear: • Vu13 and Vu23 from SAP analysis = 67 kN and 62 kN respectively.
  • 72. • ɸ VC= 0.75× 1 6 ×λ× Fc ; ×b×d • = 0.75× 1 6 × 28×1000×110/1000 =72.8 kN • Vu= 67 kN • Vu < ɸVc The slab thickness is OK
  • 73. Design for flexural: • Mu11 and Mu22 from landing analysis by SAP2000 = 9 , 10.6 kN.m respectively.
  • 74. • Mu max =8 kN.m • ρ = 0.85×28 420 ( 1 − 1 − 2.61×10×106 28×1000×1102 ) =0.00236 • As= 0.00236 × 1000×110 = 260mm2 • Asmin =0.0018×1000×150=270 mm2 (Consider it as slab) • • As > Asmin, so Use As= 270mm2 (4ɸ10/m in two direction)
  • 75. • ρ = 0.85×28 420 (1 − 1 − 2.61×8.6×106 28×1000×1102 ) = 0.00191 • As=0.00191× 1000×110 =210 mm2 • Asmin =0.0018×1000×150=270 mm2 • As < Asmin so Use As= 270 mm2 (4ɸ10/m) • • As we defined the stair as one-way solid slab, so the transverse moment is too small, so we used: • As shrinkage = As min = 0.0018 × 150 × 1000 = 270 mm2 in both face (Use 4 ∅10 /m for each layer) For Flight: Mu22 from flight analysis by SAP2000= 8.6kN.m
  • 76. Design For Interior Panel (B12 ): : 1-Parameters: • 2-Steel Reinforcement: • Mu,max Positive = 120 kN.m >>> 6Ø16 mm. • Mu,max Nevative = 225 kN.m >>> 8Ø20 mm. • Vu = 242 kN >>> 2Φ10 / 300 mm. • b (mm) h (mm) d (mm) 𝑓𝑠 (Mpa) fy (Mpa) 1000 320 240 24 420
  • 78. Design For Shear Wall (SW-3 ):
  • 79. Notes: • 5 internal ultimate forces can be computed; namely, P, Vx, Vy, Mx and My . • Design for V-y in a way similar to beams . • Design for M-y can be carried out by assuming there acts as Couple. • Design for P and Mx in a way similar to columns. • The stresses in the SW-3 are less than 0.1 F`c, there is no need for interaction diagram check.
  • 80. Reinforcement For SW-3 : • AS min=0.0025×b×h=0.0025*1000*300 >>> (1Ø12/200mm) on each side. • According to ACI315-04 section 2.10 these rules are following: • Transverse and Longitudinal rebar ratio shall be at least 0.0025 • Spacing of reinforcement shall not exceed: 450 3ℎ 𝑙𝑤 5 mm. • h: the thickness of shear wall (300mm). • So the maximum spacing is 450 mm which is larger than 300 mm, so it is OK.
  • 81. Reinforcement For SW-3 : • First : Add Two column to the shear wall , To minimize the moment on shear wall and make Minimum Reinforcement. • Second : Use General Reinforcement At Etabs Pier Design.
  • 82. Shear Wall Results : General Reinforcement: Define General Reinforcement for Etabs using minimum Reinforcement ,But many Etabs alerts for moment and shear so we increase the Reinforcement.
  • 84. Shear Wall Reinforcement Results : Vertical reinforcement/m As/2 Reinforcement For wall Horizontal reinforcement/m Reinforcement For wall Shear Reinforcement For wall Vertical reinforcement For Column Shear Reinforcement for Column SW-3 1600 800 1φ12/130mm 850 1φ18/200mm 1φ12/200mm Use 1φ18/200mm 1φ25/150mm 2φ10/200mm