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SumitSaha80142

Seismic

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An-Najah National University Faculty of Engineering Civil Engineering Department Terra Santa School Structural Design and Analysis Prepared By: Bara Shawahna Khaled Malhis Nadeem AL-Masri Supervised By : Dr. Mahmud Dwaikat
OUTLINE  Introduction & general description of the project  3D modeling  Shear walls design  Design of columns  Design of beams  Slabs design  Foundation design
INTRODUCTION ο‚„ Our Graduation project is the design of a school in Jericho named as Terra Santa School. This school was designed by Al-Diyar Consultant and we will check on their design. ο‚„ The school consists of three floors with total plan area of (3866.5m2).
Scope of work
ο‚„ SAP 2000 program will be used as main analysis tool. ο‚„ ACI 318-08 for design. ο‚„ Live loads are taken from ASCE 7 -05 code. ο‚„ UBC 97 for seismic design. Methodology
MATERIALS USED IN THE PROJECT: ο‚„ Reinforced concrete ο‚„ The unit weight of concrete (𝛾) = 25 kN/m3. ο‚„ The required compressive strength after 28 days ο‚„ For slabs & beams fc is 25 MPa. ο‚„ For Columns ,shear walls & footing fc is 30 MPa. ο‚„ The yield steel bars required Fy = 420 MPa.
STRUCTURAL SYSTEMS ο‚„ For each block we chose the following floor system: 1. Block A ( one way ribs slab ) 2. Block B ( one way ribs slab ) 3. Block C ( two way ribs slab ) Blocks A & B. It’s very clear to see that block A , B has a uniform grid for columns with clear path of loading (one-way), therefore, we took the architectural layout for the columns. Ribbed slabs are known for their economic efficiency. The thickness calculation follows the ACI-318 code Block C. Block C has a different shape and dimensions and according to ACI – code requirements the slab should be designed as two-way ribbed slab for economic and deflection requirements. This is because the spans of each panel in Block C have approximately equal lengths.
LOADS Dead load: ο‚„ Own weight for one way slabs = 3.54 kN/m2 . ο‚„ Own weight for two way slabs = 4.86 kN/m2 . ο‚„ Super imposed load = 3 kN/m2 . Live load: ο‚„ for all the class room = 2 KN/m2 . ο‚„ for corridors = 5 KN/m2.
ANALYSIS AND DESIGN AGAINST SEISMIC LOADS ο‚„ We will design the seismic load by using SAP2000. Several methods is used in SAP for seismic which is: ο‚„ Dynamic analysis: 1. Response spectrum. 2. Time history. ο‚„ Equivalent static force Equivalent static method will be used for comparison [Block A only] and as a cross-check on the results of response spectrum analysis. Because response spectrum is more realistic and covers the modal shapes of the building, we will use it as a main tool for seismic design.
ο‚„ The seismic force effect on the structure can be translated to equivalent lateral force at the base of the structure and then this force will be distributed to the different stories and then to the vertical structural elements (frames and/ or shear walls). ο‚„ This method is best applied to Regular Structure only.
DESIGN FOR EARTHQUAKE BY EQUIVALENT LATERAL FORCE METHOD (STATIC METHOD) FOR BLOCK A ο‚„ We will use the UBC97 method for analysis, because of available data and factors in our region. ο‚„ The total design base shear in a given direction shall be determined from the following formula: ο‚„ Where 𝑉 = 𝐢𝑣 𝐼 𝑅 𝑇 𝑀 1. Z= seismic zone factor. 2. I= importance factor. 3. R= Response Modification Factor 4. CV= velocity seismic coefficient. 5. W= the total dead load.
SEISMIC ZONE FACTOR Z From this map the project in Jericho Z = 0.3
CV AND CA TABLE Soil type D CV=0.54 Soil type D Ca=0.36
IMPORTANCE FACTOR TABLE I = 1.25
RESPONSE MODIFICATION FACTOR β€œR” TABLE The overall system is dual system R for building between (4.2-6) =5.6
T CALCULATION ο‚„ 𝑇 = 𝐢𝑑(β„Žπ‘›) 3 4 ο‚„ Hn= height of structure in meters = 15m ο‚„ Ct = factor for this case =0.0488 ο‚„ T= is the basic natural period of a simple one degree of freedom system which is the time required to complete one whole cycle during dynamic loading. ο‚„ And after calculation T=0.3719 sec, T from sap for block A =0.25 sec. Difference can be due to presence of shear walls that increase the stiffness of the building block.
ο‚„ Then calculate 𝑉 = 0.54𝑋1.25 5.5𝑋0.3719 Γ— w = 0.33W ο‚„ π‘‰π‘šπ‘–π‘› = 0.11 𝑋 πΆπ‘Ž 𝑋 𝐼 𝑋 π‘Š ο‚„ Vmin. = 0.11X0.36 X1.25X w=0.0396W ο‚„ Vmax= 2.5π‘‹πΆπ‘Žπ‘‹πΌ 𝑅 π‘Š  0.2W ο‚„ Vmax= 0.20 W use it because it Vcalculated>Vmax ο‚„ For block A the weigh is = 12206 kN ο‚„ V=0.20 X12206 =2441.2kN ο‚„ From SAP Vtotal equivalent static force equal =2125kN. ο‚„ From SAP Vtotal Response spectrum equal =1950kN.
RESPONSE SPECTRUM METHOD ο‚„ We use sap to design and analyze the project the design response spectrum is shown in
ο‚„ We find CA=0.36,CV=0.54 Then we have this curve
ο‚„ Then we should define a load cases in the X-direction, Y-direction, Z- direction ο‚„ When we define X-direction for example: ο‚„ We scale factor is for X direction = 𝑔𝐼 𝑅 = 9.81𝑋1.25 5.6 = 2.227 ο‚„ And 33% of it for to the other direction Y = 0.33X2.227=0.7329 ( as per UBC97 and ASCE requirements) .
3D MODELING STRUCTURE OVERVIEW
BLOCK B
BLOCK C
COMPATIBILITY CHECK
EQUILIBRIUM CHECK FOR BLOCK A ο‚„ Dead load: 1. Slab dead load = area X slab own weight per square meter =1185.84 X 3.54=4136.4 kN 2. Shear wall load = walls volume X concrete weight per volume = 28.7 X0.3 X15 X 25=3288.75 kN 3. Columns dead load = column volume X concrete weight per volume = 30 X 15 X0.4 X 0.4 X 25=1800 kN 4. Beams dead load = beams volume X concrete weight per volume = 3691.4 kN 5. Total dead load = 12917.6 kN 6. Total dead load form SAP = 12956.699 7. % Error = 0.3% which is acceptable
ο‚„ Live load: Structure area X live load per square meter =1185.84 X 5 =5929.2 kN Live load from SAP = 5929.2 kN % Error = 0 ο‚„ Superimposed dead load: Structure area X superimposed load per square meter = 1185.84 X 4.5 = 5336.28 kN Superimposed load from SAP = 5336.2 kN % Error = 0
MOMENT EQUILIBRIUM CHECK ο‚„ We take block A as example to do this check. In this check we take the moment from 3D modeling and find the weight and then comparing it with the hand calculated weight ο‚„ Moment resulted from dead load in block A in beam B1 is:
ο‚„ Calculating the moment = 39+44.75 2 + 28.9 = 70.775 π‘˜π‘. π‘š ο‚„ 𝑀×𝑙2 8 = 𝑀 , 70.775Γ—8 6.82 = 12.244 π‘˜π‘ π‘š ο‚„ To convert it to kilo newton per square meter divide it over tributary area which equal 3 meter ο‚„ W = 4.08 kN/m2 ο‚„ Calculated dead load = 3.54 kN/m2. ο‚„ % of error = 4.08βˆ’3.54 3.54 Γ— 100% = 15% ο‚„ Check for sway and non-sway: ο‚„ βˆ‘π‘ƒπ‘’ 𝑋 π·π‘’π‘“π‘™π‘–π‘π‘‘π‘–π‘œπ‘› βˆ‘π‘‰π‘’ 𝑋 π‘™π‘–π‘›π‘”π‘‘β„Ž π‘œπ‘“ π‘π‘œπ‘™π‘’π‘šπ‘› < 0.05 ο‚„ 361300 6068820 = 0.059 ο‚„ Its acceptable value.  non-sway
DESIGN OF SHEAR WALLS ο‚„ Design B2 in block B ο‚„ axial force value Pu < 0.1 Ag fc ο‚„ The maximum moment in the wall is 6824 kN.m ο‚„ Assume singly reinforced section with reinforcement at d= 0.8h: ο‚„ 𝜌 = 0.85 𝑓𝑐 𝑓𝑦 1 βˆ’ 1 βˆ’ 2𝑀 0.85ᴓ𝑏𝑑2𝑓𝑐 ο‚„ = 0.85Γ—30 420 1 βˆ’ 1 βˆ’ 2Γ—6824Γ—106 0.85Γ—0.9Γ—300Γ—45002Γ—30 = 3 Γ— 10βˆ’3 ο‚„ 𝐴𝑆 = 0.003 Γ— 1000 Γ— 300 = 900 π‘šπ‘š2 π‘š 5βˆ…16/π‘š ο‚„ 𝑉 𝑐 = 1 6 Γ— 𝑓𝑐 β€² Γ— 𝑏𝑀 Γ— 𝑑=0.1666π‘₯ 30π‘₯300π‘₯4500/1000 = 1233π‘˜π‘ ο‚„ Vu from sap =756 kN ο‚„ 𝑉 𝑛 = 𝑉𝑒 βˆ… , π‘€β„Žπ‘’π‘Ÿπ‘’ βˆ… = 0.75 ο‚„ 𝑉 𝑛 = 756 0.75 = 1008 π‘˜π‘ ο‚„ 𝑉 𝑐 > 𝑉 𝑛 𝑒𝑠𝑒 π‘šπ‘–π‘›π‘–π‘šπ‘’π‘š 𝑠𝑑𝑒𝑒𝑙 =(1 βˆ…12/20π‘π‘š)
DESIGN OF COLUMNS
DESIGN OF THE COLUMNS AGAINST SEISMIC LOAD ο‚„ Columns in Category C4(110X50)cm ο‚„ We take the moment value from 3D –modal of 𝑀1 𝑀2 we have 𝑀1 𝑀2 = 312 298 = 1.04. ο‚„ ο‚„ For the effective length factor (K) we take the building non sway so we take k from Ο†major= Ο† a= 𝐸𝑐𝐼𝑐/𝑙𝑐 𝐸𝑏𝐼𝑏/𝑙𝑏 = 12.4 , Ο† b=50 and Ο† minor= 𝐸𝑐𝐼𝑐/𝑙𝑐 𝐸𝑏𝐼𝑏/𝑙𝑏 = 11.99 and Ο† b=50the which is between equal 0.98 take it 1. ο‚„ 𝐾𝐿 𝑅 = 34 βˆ’ 12 𝑀1 𝑀2 ο‚„ Check slenderness limit 1𝑋3000 0.3 β„Ž < 34 βˆ’ 12 312 298 οƒ  non-slender column ο‚„ Using strength calculation, Pu<𝝫Pn for the same column, we got Ag = 5500000mm2. So the other dimension is5500000/500= 110mm. This dimension should be larger or equal the least dimension so let it equal 500mm, so we took it equal to 500 mm ο‚„ The dimension of column 110X50 cm so Ag= 5500cm2 ο‚„ Strength calculations for this column are done as follows:
ο‚„ From SAP2000 and 3D-model we take Pu=8287 and it equal 𝝫Pn then we go to the interaction diagram and take the steel ratio value= 0.012 ο‚„ 𝝫Pn/bh=15 =2.15 ksi ο‚„ Β₯=h-2couver/h = 0.9 ο‚„ Mu/bh2=1.44 = 0.2 ksi
𝑉 𝑐 = 1 6 Γ— 𝑓𝑐 β€² Γ— 𝑏𝑀 Γ— 𝑑 =β‡’ 𝑉 𝑐 = 1 6 Γ— 30 Γ— 1100 Γ— 460/1000 = 461.9 π‘˜π‘ 𝑉 𝑛 = 𝑉 𝑒 βˆ… , π‘€β„Žπ‘’π‘Ÿπ‘’ βˆ… = 0.75 𝑉 𝑛 = 88 0.75 = 117 π‘˜π‘ From SAP2000 the shear value on the column is equal 117 kN <463.9 it’s Ok Since the column is not subjected to shear, we use minimum steel for shear reinforcement so as to hold vertical bars and confine concrete. The stirrup spacing must be the minimum of the following β€’48 ds (diameter of stirrups)………..From code ACI 318-08 β€’16 db(diameter of bar) )………..From code ACI 318-08 β€’Least diminution of the section )………..From code ACI 318-08 β€’Not more than d/2……..From ACI 315-99 Therefore, use α΄“10 β€’48 * 10 = 480 mm β€’16 * 20= 320 mm β€’500 mm β€’460/2= 230 mm Therefore use 4α΄“10/150mm for stirrups along the column
Column Dimension Main Steel Stirrups C1 40X40=1600 8α΄“20 1α΄“10/150mm C2 50X50=2500 12α΄“18 3α΄“10/150mm C3 85X50=4250 18α΄“18 3α΄“10/150mm C4 110X50=5500 28α΄“18 4α΄“10/150mm
DESIGN OF BEAMS ο‚„ Beam distribution and categories
FOR BEAM B1 AT BLOCK A UNDER SEISMIC LOAD (50X35 CM)
ο‚„ πœŒπ‘šπ‘–π‘› = π‘šπ‘Žπ‘₯ 1.4 𝑓𝑦 , 0.25βˆ— 𝑓𝑐 𝑓𝑦 = π‘šπ‘Žπ‘₯ 0.00333, 0.00297 ο‚„ ο‚„ 𝑒𝑠𝑒 πœŒπ‘šπ‘–π‘› = 0.00333 ο‚„ For left span ο‚„ Negative moment need 1284 mm2 (5𝝫20 mm). ο‚„ For positive moment need 945 mm2 (4𝝫18 mm). ο‚„ ο‚„ For the right span ο‚„ Use (5𝝫20 mm) for top steel ο‚„ And (4𝝫18 mm) for bottom steel
DESIGN FOR SHEAR
ο‚„ Near the column face ο‚„ 𝐴𝑣 𝑠 = 0.774 ο‚„ 𝑠 = 157 0.774 = 200 π‘šπ‘š ο‚„ Use 1 𝝫 10 /100 mm ο‚„ At the middle ο‚„ 𝐴𝑣 𝑠 = 0.292 ο‚„ S=500> S max ο‚„ So use S max 230 mm ο‚„ Use 1 𝝫 10 /150 mm ο‚„ Right span ο‚„ 𝐴𝑣 𝑠 = 0.596 ο‚„ S= 263 mm > S max ο‚„ Use 1 𝝫 10 /100mm
SLAB DESIGN ο‚„ The maximum span length is about 2.7m as shown in Block A,B Map, and therefore the thickness of the slab (assumed one-way ribbed) according to the table will be L /18.5 =15 cm and we used 20 cm for block A& B(S1).
ο‚„ The maximum span length is about 7.55m as shown in Block C Map, and therefore the thickness of the slab two-way ribbed according to the table will be Ln /30 =25.1 cm and we used 30 cm.
ο‚„ For Slab S1 at block A (20 cm): ο‚„ Slab S1 is one way ribbed slab with 9 spans 2.7m for each. ο‚„ Dead and super imposed load = 8.04 kN/m2 ο‚„ Live load= 5kN/m2 due to corridor live load. ο‚„ All 𝜌 value is less than πœŒπ‘šπ‘–π‘› so use πœŒπ‘šπ‘–π‘› for design. ο‚„ Asmin = πœŒπ‘šπ‘–π‘› X bw X d= 0.0033X120X180=72mm .Used 2αΆ²10.(As bottom steel
ο‚„ Top rib steel All 𝜌 value is less than πœŒπ‘šπ‘–π‘› so use πœŒπ‘šπ‘–π‘› for design. ο‚„ Asmin = πœŒπ‘šπ‘–π‘› X bw X d= 0.0033X120X180=72mm .Used 2αΆ²10. ο‚„ For Shrinkage net steel ο‚„ Asmin = πœŒπ‘šπ‘–π‘› X b X d= 0.0018X1000X60=108mm2 Used 2αΆ²8
DESIGN FOR TWO WAY RIBBED SLAB IN BLOCK C m11 (x-direction moment)
ο‚„ For left span the positive moment is ο‚„ Mu = 86.4 (18.7βˆ’15.8) = 29.8 π‘˜π‘. π‘š ο‚„ 𝜌 = 0.85 𝑓𝑐 𝑓𝑦 1 βˆ’ 1 βˆ’ 2𝑀𝑒 0.85ᴓ𝑏𝑑2𝑓𝑐 = 0.0092 ο‚„ As = 𝜌 X bw X d= 0.0092X120X280=310 mm ο‚„ Use 2𝝫16 ο‚„ For right span the positive moment is ο‚„ Mu = 30 16.5βˆ’15.5 = 30 π‘˜π‘. π‘š ο‚„ 𝜌 = 0.85 𝑓𝑐 𝑓𝑦 1 βˆ’ 1 βˆ’ 2𝑀𝑒 0.85ᴓ𝑏𝑑2𝑓𝑐 = 0.00928 ο‚„ As = 𝜌 X bw X d= 0.00928 X 120 X280= 310 mm ο‚„ Use 2βˆ…16
ο‚„ Design for negative moment (A-B section in x-direction) Negative moment at the middle near the beam face. ο‚„ Mu = 78 (16βˆ’14) = 39 π‘˜π‘. π‘š ο‚„ 𝜌 = 0.85 𝑓𝑐 𝑓𝑦 1 βˆ’ 1 βˆ’ 2𝑀𝑒 0.85ᴓ𝑏𝑑2𝑓𝑐 = 0.0125 ο‚„ As = 𝜌 X bw X d = 0.0125 X 120 X 280= 420 mm ο‚„ Use 2𝝫20 (as top steel) Negative moment at right edge ο‚„ Mu = 60 (18βˆ’16) = 30 π‘˜π‘. π‘š ο‚„ 𝜌 = 0.85 𝑓𝑐 𝑓𝑦 1 βˆ’ 1 βˆ’ 2𝑀𝑒 0.85ᴓ𝑏𝑑2𝑓𝑐 = 0.00928 ο‚„ As = 𝜌 X bw X d = 0.00928 X 120 X 280= 312 mm ο‚„ Use 2𝝫16 for both left and right edge (as top steel).
Foundation
Design of F1 for C1 ο‚„ Pservice = 1200kN. ο‚„ Pultimate=1500kN. ο‚„ Bearing capacity= 250 kN/m2 ο‚„ Area of footing = π‘†π‘’π‘Ÿπ‘£π‘–π‘π‘’ π‘™π‘œπ‘Žπ‘‘ π‘π‘’π‘Žπ‘Ÿπ‘–π‘›π‘” π‘π‘Žπ‘π‘Žπ‘π‘–π‘‘π‘¦ = 1200 250 = 4.8 m2 ο‚„ So we will use B= 2.2, L= 2.2 ο‚„ Footing long side (l)= πΏβˆ’πΏ π‘π‘œπ‘™π‘’π‘šπ‘› 2 = 2.2βˆ’0.4 2 =0.9m
Wide beam shear ο‚„ q Ultimate = 𝑃𝑒𝑙𝑑𝑖 𝐡𝑋𝐿 = 1500 2.2𝑋2.2 =310kN ο‚„ Assuming effective depth (d) of 400mm ο‚„ 𝝫VC = 0.75 Γ— 1 6 Γ— 𝑓𝑐 β€² 𝑏𝑀𝑑 = 273.8 π‘˜π‘ ο‚„ 𝑉 𝑒 = π‘žπ‘’π‘™π‘‘. Γ— 𝑏𝑀 2 βˆ’ π‘π‘œπ‘™π‘’π‘šπ‘› π‘ β„Žπ‘œπ‘Ÿπ‘‘ π‘‘π‘–π‘šπ‘–π‘›π‘ π‘–π‘œπ‘› 2 βˆ’ 𝑑 1000 ο‚„ 𝑉 𝑒 = 310 Γ— 2.2 2 βˆ’ 0.4 2 βˆ’ 400 1000 = 155 π‘˜π‘ ο‚„ 𝝫VC>𝑉 𝑒 , so d is accepted. Check for punching shear ο‚„ VC in punching shear is: (𝝫=0.75) ο‚„ 𝝫VC = 1 6 1 + 2 𝛽 𝑓𝑐 β€² 𝑏0𝑑 = 0.75 Γ— 0.17 Γ— 1 + 2 1 Γ— 30 Γ— 3200 Γ— 400 1000 = 2629π‘˜π‘ ο‚„ 𝑉 𝑒,𝑝 = Pultimate – π‘žπ‘’π‘™π‘‘. Γ— 0.4 + 𝑑 Γ— 0.4 + 𝑑 π‘€β„Žπ‘’π‘Ÿπ‘’ 𝑑 𝑖𝑛 π‘šπ‘’π‘‘π‘’π‘Ÿ ο‚„ 𝑉 𝑒,𝑝 =1301.65 kN ο‚„ βˆ…π‘‰πΆ,𝑝 = 2629.6 π‘˜π‘ > 𝑉 𝑒,𝑝 𝑖𝑑′ 𝑠 𝑂. 𝐾 , π‘‘β„Žπ‘–π‘π‘˜π‘›π‘’π‘ π‘  𝑖𝑠 π‘’π‘›π‘œπ‘’π‘”β„Ž.
ο‚„ Design for flexure ο‚„ M ultimate= π‘žπ‘’π‘™π‘‘ Γ— π‘™π‘œ 2 2 = 125.5 kN/m2 ο‚„ Where π‘™π‘œis length of cantilever in x or y direction ο‚„ 𝜌 = 0.85 𝑓𝑐 𝑓𝑦 1 βˆ’ 1 βˆ’ 2𝑀 0.85ᴓ𝑏𝑑2𝑓𝑐 = 0.2108% ο‚„ πœŒπ‘šπ‘Žπ‘₯. = πœŒπ›·=0.9,βˆˆπ‘‘=0.005 = 0.375𝛽1 0.85×𝑓𝑐 β€² 𝑓𝑦 , π‘€β„Žπ‘’π‘Ÿπ‘’ 𝛽1 = 0.85 ο‚„ πœŒπ‘šπ‘Žπ‘₯. = 0.0161 πœŒπ‘šπ‘Žπ‘₯. > 𝜌 ο‚„ As = 0.002108 X 1000 X 400=843.1 mm2 ο‚„ Asmin. = 0.0018 X 1000 X 400=864 mm2 ο‚„ Asmin>As ο‚„ So use As =Asmin=864mm2 footing number p service p ultimate footing area b l d As As,min F1 1200 1500 4.8 2.2 2.2 400 843 864 F2 2860 3720 11.44 3.5 3.5 500 1862 1044 F3 4000 5227 16 4 4 600 2273 1224 F4 6200 8245 24.8 5 5 900 2507 1764
DESIGN OF MAT (1)IN BLOCK C Mat 1 plan Mat 1 displacement
moment (m11) in x-axis
the maximum moment is 238kN.m. in x-axis 𝜌 = 0.85 𝑓𝑐 𝑓𝑦 1 βˆ’ 1 βˆ’ 2𝑀 0.85ᴓ𝑏𝑑2𝑓𝑐 = 0.1% 𝜌 < πœŒπ‘šπ‘–π‘› Use 𝜌 = 0.0018 As = 0.0018Γ—1000Γ—800=1440 mm/m the maximum moment is 414 kN.m in y-axis T hen calculate 𝜌 = 0.85 𝑓𝑐 𝑓𝑦 1 βˆ’ 1 βˆ’ 2𝑀 0.85ᴓ𝑏𝑑2𝑓𝑐 = 0.173% 𝜌 < πœŒπ‘šπ‘–π‘› Use 𝜌 = 0.0018 As = 0.0018Γ—1000Γ—800=1440 mm/m For shrinkage steel As = 0.5Γ—0.0018Γ—1000Γ—800 =720 mm/m (5 𝝫14 mm / m).
C2 C2 C1 C1
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presentation1_28.pptx

  • 1. An-Najah National University Faculty of Engineering Civil Engineering Department Terra Santa School Structural Design and Analysis Prepared By: Bara Shawahna Khaled Malhis Nadeem AL-Masri Supervised By : Dr. Mahmud Dwaikat
  • 2. OUTLINE  Introduction & general description of the project  3D modeling  Shear walls design  Design of columns  Design of beams  Slabs design  Foundation design
  • 3. INTRODUCTION ο‚„ Our Graduation project is the design of a school in Jericho named as Terra Santa School. This school was designed by Al-Diyar Consultant and we will check on their design. ο‚„ The school consists of three floors with total plan area of (3866.5m2).
  • 5. ο‚„ SAP 2000 program will be used as main analysis tool. ο‚„ ACI 318-08 for design. ο‚„ Live loads are taken from ASCE 7 -05 code. ο‚„ UBC 97 for seismic design. Methodology
  • 6. MATERIALS USED IN THE PROJECT: ο‚„ Reinforced concrete ο‚„ The unit weight of concrete (𝛾) = 25 kN/m3. ο‚„ The required compressive strength after 28 days ο‚„ For slabs & beams fc is 25 MPa. ο‚„ For Columns ,shear walls & footing fc is 30 MPa. ο‚„ The yield steel bars required Fy = 420 MPa.
  • 7. STRUCTURAL SYSTEMS ο‚„ For each block we chose the following floor system: 1. Block A ( one way ribs slab ) 2. Block B ( one way ribs slab ) 3. Block C ( two way ribs slab ) Blocks A & B. It’s very clear to see that block A , B has a uniform grid for columns with clear path of loading (one-way), therefore, we took the architectural layout for the columns. Ribbed slabs are known for their economic efficiency. The thickness calculation follows the ACI-318 code Block C. Block C has a different shape and dimensions and according to ACI – code requirements the slab should be designed as two-way ribbed slab for economic and deflection requirements. This is because the spans of each panel in Block C have approximately equal lengths.
  • 8.
  • 9. LOADS Dead load: ο‚„ Own weight for one way slabs = 3.54 kN/m2 . ο‚„ Own weight for two way slabs = 4.86 kN/m2 . ο‚„ Super imposed load = 3 kN/m2 . Live load: ο‚„ for all the class room = 2 KN/m2 . ο‚„ for corridors = 5 KN/m2.
  • 10. ANALYSIS AND DESIGN AGAINST SEISMIC LOADS ο‚„ We will design the seismic load by using SAP2000. Several methods is used in SAP for seismic which is: ο‚„ Dynamic analysis: 1. Response spectrum. 2. Time history. ο‚„ Equivalent static force Equivalent static method will be used for comparison [Block A only] and as a cross-check on the results of response spectrum analysis. Because response spectrum is more realistic and covers the modal shapes of the building, we will use it as a main tool for seismic design.
  • 11. ο‚„ The seismic force effect on the structure can be translated to equivalent lateral force at the base of the structure and then this force will be distributed to the different stories and then to the vertical structural elements (frames and/ or shear walls). ο‚„ This method is best applied to Regular Structure only.
  • 12. DESIGN FOR EARTHQUAKE BY EQUIVALENT LATERAL FORCE METHOD (STATIC METHOD) FOR BLOCK A ο‚„ We will use the UBC97 method for analysis, because of available data and factors in our region. ο‚„ The total design base shear in a given direction shall be determined from the following formula: ο‚„ Where 𝑉 = 𝐢𝑣 𝐼 𝑅 𝑇 𝑀 1. Z= seismic zone factor. 2. I= importance factor. 3. R= Response Modification Factor 4. CV= velocity seismic coefficient. 5. W= the total dead load.
  • 13. SEISMIC ZONE FACTOR Z From this map the project in Jericho Z = 0.3
  • 14. CV AND CA TABLE Soil type D CV=0.54 Soil type D Ca=0.36
  • 16. RESPONSE MODIFICATION FACTOR β€œR” TABLE The overall system is dual system R for building between (4.2-6) =5.6
  • 17. T CALCULATION ο‚„ 𝑇 = 𝐢𝑑(β„Žπ‘›) 3 4 ο‚„ Hn= height of structure in meters = 15m ο‚„ Ct = factor for this case =0.0488 ο‚„ T= is the basic natural period of a simple one degree of freedom system which is the time required to complete one whole cycle during dynamic loading. ο‚„ And after calculation T=0.3719 sec, T from sap for block A =0.25 sec. Difference can be due to presence of shear walls that increase the stiffness of the building block.
  • 18. ο‚„ Then calculate 𝑉 = 0.54𝑋1.25 5.5𝑋0.3719 Γ— w = 0.33W ο‚„ π‘‰π‘šπ‘–π‘› = 0.11 𝑋 πΆπ‘Ž 𝑋 𝐼 𝑋 π‘Š ο‚„ Vmin. = 0.11X0.36 X1.25X w=0.0396W ο‚„ Vmax= 2.5π‘‹πΆπ‘Žπ‘‹πΌ 𝑅 π‘Š  0.2W ο‚„ Vmax= 0.20 W use it because it Vcalculated>Vmax ο‚„ For block A the weigh is = 12206 kN ο‚„ V=0.20 X12206 =2441.2kN ο‚„ From SAP Vtotal equivalent static force equal =2125kN. ο‚„ From SAP Vtotal Response spectrum equal =1950kN.
  • 19. RESPONSE SPECTRUM METHOD ο‚„ We use sap to design and analyze the project the design response spectrum is shown in
  • 20. ο‚„ We find CA=0.36,CV=0.54 Then we have this curve
  • 21. ο‚„ Then we should define a load cases in the X-direction, Y-direction, Z- direction ο‚„ When we define X-direction for example: ο‚„ We scale factor is for X direction = 𝑔𝐼 𝑅 = 9.81𝑋1.25 5.6 = 2.227 ο‚„ And 33% of it for to the other direction Y = 0.33X2.227=0.7329 ( as per UBC97 and ASCE requirements) .
  • 26. EQUILIBRIUM CHECK FOR BLOCK A ο‚„ Dead load: 1. Slab dead load = area X slab own weight per square meter =1185.84 X 3.54=4136.4 kN 2. Shear wall load = walls volume X concrete weight per volume = 28.7 X0.3 X15 X 25=3288.75 kN 3. Columns dead load = column volume X concrete weight per volume = 30 X 15 X0.4 X 0.4 X 25=1800 kN 4. Beams dead load = beams volume X concrete weight per volume = 3691.4 kN 5. Total dead load = 12917.6 kN 6. Total dead load form SAP = 12956.699 7. % Error = 0.3% which is acceptable
  • 27. ο‚„ Live load: Structure area X live load per square meter =1185.84 X 5 =5929.2 kN Live load from SAP = 5929.2 kN % Error = 0 ο‚„ Superimposed dead load: Structure area X superimposed load per square meter = 1185.84 X 4.5 = 5336.28 kN Superimposed load from SAP = 5336.2 kN % Error = 0
  • 28. MOMENT EQUILIBRIUM CHECK ο‚„ We take block A as example to do this check. In this check we take the moment from 3D modeling and find the weight and then comparing it with the hand calculated weight ο‚„ Moment resulted from dead load in block A in beam B1 is:
  • 29. ο‚„ Calculating the moment = 39+44.75 2 + 28.9 = 70.775 π‘˜π‘. π‘š ο‚„ 𝑀×𝑙2 8 = 𝑀 , 70.775Γ—8 6.82 = 12.244 π‘˜π‘ π‘š ο‚„ To convert it to kilo newton per square meter divide it over tributary area which equal 3 meter ο‚„ W = 4.08 kN/m2 ο‚„ Calculated dead load = 3.54 kN/m2. ο‚„ % of error = 4.08βˆ’3.54 3.54 Γ— 100% = 15% ο‚„ Check for sway and non-sway: ο‚„ βˆ‘π‘ƒπ‘’ 𝑋 π·π‘’π‘“π‘™π‘–π‘π‘‘π‘–π‘œπ‘› βˆ‘π‘‰π‘’ 𝑋 π‘™π‘–π‘›π‘”π‘‘β„Ž π‘œπ‘“ π‘π‘œπ‘™π‘’π‘šπ‘› < 0.05 ο‚„ 361300 6068820 = 0.059 ο‚„ Its acceptable value.  non-sway
  • 30. DESIGN OF SHEAR WALLS ο‚„ Design B2 in block B ο‚„ axial force value Pu < 0.1 Ag fc ο‚„ The maximum moment in the wall is 6824 kN.m ο‚„ Assume singly reinforced section with reinforcement at d= 0.8h: ο‚„ 𝜌 = 0.85 𝑓𝑐 𝑓𝑦 1 βˆ’ 1 βˆ’ 2𝑀 0.85ᴓ𝑏𝑑2𝑓𝑐 ο‚„ = 0.85Γ—30 420 1 βˆ’ 1 βˆ’ 2Γ—6824Γ—106 0.85Γ—0.9Γ—300Γ—45002Γ—30 = 3 Γ— 10βˆ’3 ο‚„ 𝐴𝑆 = 0.003 Γ— 1000 Γ— 300 = 900 π‘šπ‘š2 π‘š 5βˆ…16/π‘š ο‚„ 𝑉 𝑐 = 1 6 Γ— 𝑓𝑐 β€² Γ— 𝑏𝑀 Γ— 𝑑=0.1666π‘₯ 30π‘₯300π‘₯4500/1000 = 1233π‘˜π‘ ο‚„ Vu from sap =756 kN ο‚„ 𝑉 𝑛 = 𝑉𝑒 βˆ… , π‘€β„Žπ‘’π‘Ÿπ‘’ βˆ… = 0.75 ο‚„ 𝑉 𝑛 = 756 0.75 = 1008 π‘˜π‘ ο‚„ 𝑉 𝑐 > 𝑉 𝑛 𝑒𝑠𝑒 π‘šπ‘–π‘›π‘–π‘šπ‘’π‘š 𝑠𝑑𝑒𝑒𝑙 =(1 βˆ…12/20π‘π‘š)
  • 32. DESIGN OF THE COLUMNS AGAINST SEISMIC LOAD ο‚„ Columns in Category C4(110X50)cm ο‚„ We take the moment value from 3D –modal of 𝑀1 𝑀2 we have 𝑀1 𝑀2 = 312 298 = 1.04. ο‚„ ο‚„ For the effective length factor (K) we take the building non sway so we take k from Ο†major= Ο† a= 𝐸𝑐𝐼𝑐/𝑙𝑐 𝐸𝑏𝐼𝑏/𝑙𝑏 = 12.4 , Ο† b=50 and Ο† minor= 𝐸𝑐𝐼𝑐/𝑙𝑐 𝐸𝑏𝐼𝑏/𝑙𝑏 = 11.99 and Ο† b=50the which is between equal 0.98 take it 1. ο‚„ 𝐾𝐿 𝑅 = 34 βˆ’ 12 𝑀1 𝑀2 ο‚„ Check slenderness limit 1𝑋3000 0.3 β„Ž < 34 βˆ’ 12 312 298 οƒ  non-slender column ο‚„ Using strength calculation, Pu<𝝫Pn for the same column, we got Ag = 5500000mm2. So the other dimension is5500000/500= 110mm. This dimension should be larger or equal the least dimension so let it equal 500mm, so we took it equal to 500 mm ο‚„ The dimension of column 110X50 cm so Ag= 5500cm2 ο‚„ Strength calculations for this column are done as follows:
  • 33. ο‚„ From SAP2000 and 3D-model we take Pu=8287 and it equal 𝝫Pn then we go to the interaction diagram and take the steel ratio value= 0.012 ο‚„ 𝝫Pn/bh=15 =2.15 ksi ο‚„ Β₯=h-2couver/h = 0.9 ο‚„ Mu/bh2=1.44 = 0.2 ksi
  • 34. 𝑉 𝑐 = 1 6 Γ— 𝑓𝑐 β€² Γ— 𝑏𝑀 Γ— 𝑑 =β‡’ 𝑉 𝑐 = 1 6 Γ— 30 Γ— 1100 Γ— 460/1000 = 461.9 π‘˜π‘ 𝑉 𝑛 = 𝑉 𝑒 βˆ… , π‘€β„Žπ‘’π‘Ÿπ‘’ βˆ… = 0.75 𝑉 𝑛 = 88 0.75 = 117 π‘˜π‘ From SAP2000 the shear value on the column is equal 117 kN <463.9 it’s Ok Since the column is not subjected to shear, we use minimum steel for shear reinforcement so as to hold vertical bars and confine concrete. The stirrup spacing must be the minimum of the following β€’48 ds (diameter of stirrups)………..From code ACI 318-08 β€’16 db(diameter of bar) )………..From code ACI 318-08 β€’Least diminution of the section )………..From code ACI 318-08 β€’Not more than d/2……..From ACI 315-99 Therefore, use α΄“10 β€’48 * 10 = 480 mm β€’16 * 20= 320 mm β€’500 mm β€’460/2= 230 mm Therefore use 4α΄“10/150mm for stirrups along the column
  • 35. Column Dimension Main Steel Stirrups C1 40X40=1600 8α΄“20 1α΄“10/150mm C2 50X50=2500 12α΄“18 3α΄“10/150mm C3 85X50=4250 18α΄“18 3α΄“10/150mm C4 110X50=5500 28α΄“18 4α΄“10/150mm
  • 36. DESIGN OF BEAMS ο‚„ Beam distribution and categories
  • 37.
  • 38. FOR BEAM B1 AT BLOCK A UNDER SEISMIC LOAD (50X35 CM)
  • 39. ο‚„ πœŒπ‘šπ‘–π‘› = π‘šπ‘Žπ‘₯ 1.4 𝑓𝑦 , 0.25βˆ— 𝑓𝑐 𝑓𝑦 = π‘šπ‘Žπ‘₯ 0.00333, 0.00297 ο‚„ ο‚„ 𝑒𝑠𝑒 πœŒπ‘šπ‘–π‘› = 0.00333 ο‚„ For left span ο‚„ Negative moment need 1284 mm2 (5𝝫20 mm). ο‚„ For positive moment need 945 mm2 (4𝝫18 mm). ο‚„ ο‚„ For the right span ο‚„ Use (5𝝫20 mm) for top steel ο‚„ And (4𝝫18 mm) for bottom steel
  • 41. ο‚„ Near the column face ο‚„ 𝐴𝑣 𝑠 = 0.774 ο‚„ 𝑠 = 157 0.774 = 200 π‘šπ‘š ο‚„ Use 1 𝝫 10 /100 mm ο‚„ At the middle ο‚„ 𝐴𝑣 𝑠 = 0.292 ο‚„ S=500> S max ο‚„ So use S max 230 mm ο‚„ Use 1 𝝫 10 /150 mm ο‚„ Right span ο‚„ 𝐴𝑣 𝑠 = 0.596 ο‚„ S= 263 mm > S max ο‚„ Use 1 𝝫 10 /100mm
  • 42.
  • 43. SLAB DESIGN ο‚„ The maximum span length is about 2.7m as shown in Block A,B Map, and therefore the thickness of the slab (assumed one-way ribbed) according to the table will be L /18.5 =15 cm and we used 20 cm for block A& B(S1).
  • 44. ο‚„ The maximum span length is about 7.55m as shown in Block C Map, and therefore the thickness of the slab two-way ribbed according to the table will be Ln /30 =25.1 cm and we used 30 cm.
  • 45. ο‚„ For Slab S1 at block A (20 cm): ο‚„ Slab S1 is one way ribbed slab with 9 spans 2.7m for each. ο‚„ Dead and super imposed load = 8.04 kN/m2 ο‚„ Live load= 5kN/m2 due to corridor live load. ο‚„ All 𝜌 value is less than πœŒπ‘šπ‘–π‘› so use πœŒπ‘šπ‘–π‘› for design. ο‚„ Asmin = πœŒπ‘šπ‘–π‘› X bw X d= 0.0033X120X180=72mm .Used 2αΆ²10.(As bottom steel
  • 46. ο‚„ Top rib steel All 𝜌 value is less than πœŒπ‘šπ‘–π‘› so use πœŒπ‘šπ‘–π‘› for design. ο‚„ Asmin = πœŒπ‘šπ‘–π‘› X bw X d= 0.0033X120X180=72mm .Used 2αΆ²10. ο‚„ For Shrinkage net steel ο‚„ Asmin = πœŒπ‘šπ‘–π‘› X b X d= 0.0018X1000X60=108mm2 Used 2αΆ²8
  • 47. DESIGN FOR TWO WAY RIBBED SLAB IN BLOCK C m11 (x-direction moment)
  • 48. ο‚„ For left span the positive moment is ο‚„ Mu = 86.4 (18.7βˆ’15.8) = 29.8 π‘˜π‘. π‘š ο‚„ 𝜌 = 0.85 𝑓𝑐 𝑓𝑦 1 βˆ’ 1 βˆ’ 2𝑀𝑒 0.85ᴓ𝑏𝑑2𝑓𝑐 = 0.0092 ο‚„ As = 𝜌 X bw X d= 0.0092X120X280=310 mm ο‚„ Use 2𝝫16 ο‚„ For right span the positive moment is ο‚„ Mu = 30 16.5βˆ’15.5 = 30 π‘˜π‘. π‘š ο‚„ 𝜌 = 0.85 𝑓𝑐 𝑓𝑦 1 βˆ’ 1 βˆ’ 2𝑀𝑒 0.85ᴓ𝑏𝑑2𝑓𝑐 = 0.00928 ο‚„ As = 𝜌 X bw X d= 0.00928 X 120 X280= 310 mm ο‚„ Use 2βˆ…16
  • 49. ο‚„ Design for negative moment (A-B section in x-direction) Negative moment at the middle near the beam face. ο‚„ Mu = 78 (16βˆ’14) = 39 π‘˜π‘. π‘š ο‚„ 𝜌 = 0.85 𝑓𝑐 𝑓𝑦 1 βˆ’ 1 βˆ’ 2𝑀𝑒 0.85ᴓ𝑏𝑑2𝑓𝑐 = 0.0125 ο‚„ As = 𝜌 X bw X d = 0.0125 X 120 X 280= 420 mm ο‚„ Use 2𝝫20 (as top steel) Negative moment at right edge ο‚„ Mu = 60 (18βˆ’16) = 30 π‘˜π‘. π‘š ο‚„ 𝜌 = 0.85 𝑓𝑐 𝑓𝑦 1 βˆ’ 1 βˆ’ 2𝑀𝑒 0.85ᴓ𝑏𝑑2𝑓𝑐 = 0.00928 ο‚„ As = 𝜌 X bw X d = 0.00928 X 120 X 280= 312 mm ο‚„ Use 2𝝫16 for both left and right edge (as top steel).
  • 51. Design of F1 for C1 ο‚„ Pservice = 1200kN. ο‚„ Pultimate=1500kN. ο‚„ Bearing capacity= 250 kN/m2 ο‚„ Area of footing = π‘†π‘’π‘Ÿπ‘£π‘–π‘π‘’ π‘™π‘œπ‘Žπ‘‘ π‘π‘’π‘Žπ‘Ÿπ‘–π‘›π‘” π‘π‘Žπ‘π‘Žπ‘π‘–π‘‘π‘¦ = 1200 250 = 4.8 m2 ο‚„ So we will use B= 2.2, L= 2.2 ο‚„ Footing long side (l)= πΏβˆ’πΏ π‘π‘œπ‘™π‘’π‘šπ‘› 2 = 2.2βˆ’0.4 2 =0.9m
  • 52. Wide beam shear ο‚„ q Ultimate = 𝑃𝑒𝑙𝑑𝑖 𝐡𝑋𝐿 = 1500 2.2𝑋2.2 =310kN ο‚„ Assuming effective depth (d) of 400mm ο‚„ 𝝫VC = 0.75 Γ— 1 6 Γ— 𝑓𝑐 β€² 𝑏𝑀𝑑 = 273.8 π‘˜π‘ ο‚„ 𝑉 𝑒 = π‘žπ‘’π‘™π‘‘. Γ— 𝑏𝑀 2 βˆ’ π‘π‘œπ‘™π‘’π‘šπ‘› π‘ β„Žπ‘œπ‘Ÿπ‘‘ π‘‘π‘–π‘šπ‘–π‘›π‘ π‘–π‘œπ‘› 2 βˆ’ 𝑑 1000 ο‚„ 𝑉 𝑒 = 310 Γ— 2.2 2 βˆ’ 0.4 2 βˆ’ 400 1000 = 155 π‘˜π‘ ο‚„ 𝝫VC>𝑉 𝑒 , so d is accepted. Check for punching shear ο‚„ VC in punching shear is: (𝝫=0.75) ο‚„ 𝝫VC = 1 6 1 + 2 𝛽 𝑓𝑐 β€² 𝑏0𝑑 = 0.75 Γ— 0.17 Γ— 1 + 2 1 Γ— 30 Γ— 3200 Γ— 400 1000 = 2629π‘˜π‘ ο‚„ 𝑉 𝑒,𝑝 = Pultimate – π‘žπ‘’π‘™π‘‘. Γ— 0.4 + 𝑑 Γ— 0.4 + 𝑑 π‘€β„Žπ‘’π‘Ÿπ‘’ 𝑑 𝑖𝑛 π‘šπ‘’π‘‘π‘’π‘Ÿ ο‚„ 𝑉 𝑒,𝑝 =1301.65 kN ο‚„ βˆ…π‘‰πΆ,𝑝 = 2629.6 π‘˜π‘ > 𝑉 𝑒,𝑝 𝑖𝑑′ 𝑠 𝑂. 𝐾 , π‘‘β„Žπ‘–π‘π‘˜π‘›π‘’π‘ π‘  𝑖𝑠 π‘’π‘›π‘œπ‘’π‘”β„Ž.
  • 53. ο‚„ Design for flexure ο‚„ M ultimate= π‘žπ‘’π‘™π‘‘ Γ— π‘™π‘œ 2 2 = 125.5 kN/m2 ο‚„ Where π‘™π‘œis length of cantilever in x or y direction ο‚„ 𝜌 = 0.85 𝑓𝑐 𝑓𝑦 1 βˆ’ 1 βˆ’ 2𝑀 0.85ᴓ𝑏𝑑2𝑓𝑐 = 0.2108% ο‚„ πœŒπ‘šπ‘Žπ‘₯. = πœŒπ›·=0.9,βˆˆπ‘‘=0.005 = 0.375𝛽1 0.85×𝑓𝑐 β€² 𝑓𝑦 , π‘€β„Žπ‘’π‘Ÿπ‘’ 𝛽1 = 0.85 ο‚„ πœŒπ‘šπ‘Žπ‘₯. = 0.0161 πœŒπ‘šπ‘Žπ‘₯. > 𝜌 ο‚„ As = 0.002108 X 1000 X 400=843.1 mm2 ο‚„ Asmin. = 0.0018 X 1000 X 400=864 mm2 ο‚„ Asmin>As ο‚„ So use As =Asmin=864mm2 footing number p service p ultimate footing area b l d As As,min F1 1200 1500 4.8 2.2 2.2 400 843 864 F2 2860 3720 11.44 3.5 3.5 500 1862 1044 F3 4000 5227 16 4 4 600 2273 1224 F4 6200 8245 24.8 5 5 900 2507 1764
  • 54.
  • 55.
  • 56. DESIGN OF MAT (1)IN BLOCK C Mat 1 plan Mat 1 displacement
  • 57. moment (m11) in x-axis
  • 58. the maximum moment is 238kN.m. in x-axis 𝜌 = 0.85 𝑓𝑐 𝑓𝑦 1 βˆ’ 1 βˆ’ 2𝑀 0.85ᴓ𝑏𝑑2𝑓𝑐 = 0.1% 𝜌 < πœŒπ‘šπ‘–π‘› Use 𝜌 = 0.0018 As = 0.0018Γ—1000Γ—800=1440 mm/m the maximum moment is 414 kN.m in y-axis T hen calculate 𝜌 = 0.85 𝑓𝑐 𝑓𝑦 1 βˆ’ 1 βˆ’ 2𝑀 0.85ᴓ𝑏𝑑2𝑓𝑐 = 0.173% 𝜌 < πœŒπ‘šπ‘–π‘› Use 𝜌 = 0.0018 As = 0.0018Γ—1000Γ—800=1440 mm/m For shrinkage steel As = 0.5Γ—0.0018Γ—1000Γ—800 =720 mm/m (5 𝝫14 mm / m).
  • 60.