graduation project 2 structural analysis and design of zayed college for nursing and optics
DESCRIPTION
An- Najah National University Faculty of Engineering Civil Engineering Department . Graduation Project 2 Structural Analysis and Design of Zayed College for Nursing and Optics. Supervisor: Dr. Mahmoud Dwaikat. Prepared By: - PowerPoint PPT PresentationTRANSCRIPT
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Graduation Project 2Structural Analysis and Design of
Zayed College for Nursing and Optics
1
An-Najah National University Faculty of Engineering
Civil Engineering Department
Supervisor: Dr. Mahmoud Dwaikat
Prepared By:Maha Sharei Amal Sabbah Jumana Khatib
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An-najah National university 2
Work Plan
3D Modeling
Seismic Load Definition
Design & Reinforcement Details
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Project Description: Zayed College for Nursing and Optics is located at An-Najah National University’s new campus- next to the faculty of medicine.
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The building has a total area of 6665m2 and consists of seven floors.
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E.g. : Ground floor
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Assumptions for Design: Analysis and design are according to ACI-318-08. ASCE 7- CODE for live load determination. UBC 97 code for seismic loads (Using response spectrum) Load combinations :
Loads are gravity and seismic loads.
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Structural Materials:
Concrete: Slabs and beams → fc` = 25 MPa Columns and footing → fc` = 30 MPa
Steel (Rebar, shrinkage mesh and stirrups): Yielding strength (Fy) = 420 MPa
Soil Properties: Allowable bearing capacity of 4 Kg/cm2
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Load Assumptions:
Dead Load:• Slab own weight: 5.54 KN/m2
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Superimposed dead load:
Total superimposed dead load = 4.8 KN/m2
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Live Load
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An-najah National university
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Slab structural system
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3D Model
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Slab modifiers
Modification Factors
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Also, for beams and columns the modifiers are as follows:
Beams:Torsional constant: 0.35Moment of inertia about 2 axis: 0.35Moment of inertia about 3 axis: 0.35
Columns:Torsional constant: 0.7Moment of inertia about 2 axis: 0.7Moment of inertia about 3 axis: 0.7
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Sap Verifications
Compatibility
Compatibility check
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Equilibrium
Total live load manually = 3306.3 KN.
Live load from sap
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Stress strain relationship check
3(from sap) should equal (manually)
3 = -203=77KN.m= = 73.55KN.m
Since 77KN.m 73.55KN.m (e=4.6%) OK.
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Dynamic Analysis (seismic loads)
UBC 97 code
Response spectrum analysis method
Response spectrum: an elastic dynamic analysis utilizing the peak dynamic response of all modes having a significant contribution to total structural response.
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Main factors according to UBC 97 code:
Seismic Zone Factor, Z
Nablus city (Zone 2B); Z= 0.2
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I= importance factor, I= 1
R = 4.2 (Dual system Concrete shear walls with OMRF) R = 6.5 (Dual systems concrete shear walls with IMRF)
Soil is rock …soil profile type SB
Ca=acceleration seismic coefficient, Ca=0.2
Cv= velocity seismic coefficient, Cv=0.2
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Definition of response spectrum function:
Response Spectrum UBC 97 Definition
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Load patterns:
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Load cases:
Earth quake –x
Scale factor= =1.83 for U1 U2: 1.83×0.3= 0.55
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Load cases:
Earth quake –y
Scale factor= =1.83 for U2 U1: 1.83×0.3= 0.55
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An-najah National university An-najah National university 25An-najah National university
Load cases:
Earth quake –z
Scale factor= =1.83 U3: 1.83×0.15= 0.275
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Load cases:
Modal case
Eigen Vectors analysis needs more than 700 modesRitz Vectors analysis needs only 21 modes to reach 90% participating mass ratios
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Period T CheckT =0.377 seconds as shown:
T= Ct (hn)3/4 .. Ct = 0.0488 … T= 0.6 seconds
…T= 0.45 seconds
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Total base shear V checkBase shear in x & y directions:
Base Reactions
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V max:
Substitute Wd = 133367 KN, Cv = 0.2, Ca= 0.2, I=1, T =0.45 s
For R =6.5 → V = 9120 KN< Vmax (10260 KN) For R =4.2 → V = 14113 KN< Vmax (15877 KN)
Vx & Vy values from the previous figure are close to manual calculations.
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Footing Design
Single Footing
Combined Footing
Wall Footing
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Single Footing
q all = 400 KN/m2 It has a dimensions of 2*2 m Thickness = 450 mm
Shear Checks : Wide Beam Shear Ø Vc ≥ Vu 507 KN > 406 KN Ok Punching Shear ≤ Ø Vc Շ 0.00141 < 1.373 Ok
Reinforcement 6 Ø16 / m in both directions.
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Combined Footing
q all = 400 KN/m2 It has a dimensions of L = 4m , B = 2.6m . Thickness = 800 mm
Shear Checks : Wide Beam Shear Ø Vc ≥ Vu 1170 KN > 419 Ok. Punching Shear Ø Vc ≥ Vu 4416.8 KN> 419 KN OK.
Check for deflection ∆SAP < ∆Max. 0.0096 < 0.01m OK. Check maximum stress σ = 363.75 KN < q all OK
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Combined Footing
Longitudinal Reinforcement As = 2056 mm² < As min = 6240 mm² Use As min 13 Ø 25
Transverse reinforcement As = 4004mm²13 Ø 20
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Wall Footing
q all = 400 KN/m² It has a dimensions of B = 2m Thickness = 550 mm
Reinforcement Longitudinal Reinforcement As min = 990 mm2/m 7 Ø 14 /m Transverse reinforcement As = 1034 mm2/m 6Ø 16 / m
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Design of columns
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Sway –Non sway Check
37
The check is done for both x and y direction for first and last floors
Columns’ Shear and Vertical loads
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Sway –Non sway Check
By applying Q index equation in x , Q index = 0.0021 <0.05 …non sway
Shear Walls’ Shear and Vertical loads
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Slenderness CheckIn project1, columns were assumed short. 3D model is used to recheck that assumption.
Alignment chart for non-sway frames
Lu = 3.7 mK conservatively assumed 1R= 0.3 h =0.3×0.4= 0.12
= = 30.83 ≤ 34- 12≤ 40M1, M2 for all columns make double curvature→ -ve → Short columns
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Columns Design
Columns Design Summary
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Design Of Beams
All beams in the building are drop and multi-span beams.
For the preliminary design , loads are calculated using tributary area method and 1-D structural model for each beam is analyzed and designed using SAP2000.
Final output was taken from 3-D model.
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Plan of beams for third basement
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Preliminary dimensions of Beams
Depth of beams was found according to ACI-318-08 (Deflection limitation) .
For beam 1 it has rectangular section with Depth = 70 cm Width = 40 cm
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Design of Beams Using Sap2000
Dimensions of beams
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Steel reinforcement
Minimum steel reinforcement
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Steel reinforcement
Main steel reinforcement
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Steel reinforcement Shear reinforcement
Concerning shear reinforcement of beams, it is recommended to unify shear reinforcement spaces for each beam near the supports due to not high shear forces in the region. In the middle we can reduce the spacing for practical and economical purposes.
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Steel reinforcement Torsion reinforcement
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Design of Transverse Beams
It has a dimensions of 30*30 cm
It has area of steel minimum : As min = p min*b*h = 0.0033*300*300 =297 mm2 Use 3 φ 12
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Design of slabs
The structural system of Zayed College is one way ribbed slab(31 cm ) with drop beams.
The ribs in the slab are analyzed using sap 2000 program ..
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Ribs distribution in the first floor
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Analysis and design of rib2
Check for shear
Shear force contours
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The maximum shear force at distance d/2 from face of support =48KN*.57 = 27.3 KN.
ØVc =29.2 KN Vu =27.3 KN
ØVc=29.2>Vu=27.3 no need for shear reinforcement , but use 1ø8/20 cm for bar fixation.
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Analysis and Design for flexure
M22 contours diagram
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Reinforcement details for rib2
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Stairway Design
Stair dimensions Going = 30 cmRiser = 16.5 cm Slab thickness = 20 cm
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Stairway Design Stair Loading Dead load = 14.1 KN/m Live Load = 7.9 KN/m
Check for deflection Deflection limitation:- - ∆L ≤ L/360 Ok - ∆ L T≤ L/480 Ok
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Stairway Design
Check for shear ØVc > Vu Ok
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Stairway Design
Design for flexure
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Stairway Design
Area of steel from sap
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Design for stair beam
Dimensions ( depth = 20 cm , Width = 40 cm ) Design of beam for flexure
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Design for stair beam
Main steel reinforcement
Minimum steel reinforcement As min = 264 mm2 Use 2 φ 14
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Design for stair beam
Design of beam for shear
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Shear wall designBending m11 modifier: 0.25, m22 modifier: 0.25 ,m12 modifier: 0.25
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•Since Pu<0.1 Aw f’c, shear walls are considered as beams in design.Where: Aw= Lw x tw , Lw is wall width , tw is wall thickness
•Since Pu< 0.1 Aw f’c and <1 ,no need for boundary zone elements .
For all shear walls Vu ≤ ØVn, so minimum reinforcement is used in horizontal direction with ρ= 0.0025
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(ρ= 0.0025)As min= 0.0025×1000×300= 750mm2
Use 5Ø10/m& 5Ø10/m in both horizontal and vertical directions
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