Faculty of Engineering and Information Technology Department of Civil Engineer Graduation Project II Prepared By : Omar Daragmeh – Reg. No: 11643099 Mohammad Tammam – Reg. No: 11642560 Mohammad Odeh – Reg. No: 11610499 Under supervision of : Dr. Imad AL-Qasem Design Of Hugo Chavez Hospital 2 The Palestinian Venezuelan Ophthalmic Hospital - Hugo Chavez Hospital is a medical building , which is located in Turmousaya– Ramallah. The project consists of a three blocks with a total area of 8100 m2. Each block consists of four floors (basement , ground , first , second). Average area for each floor in block 1 is 1250 m2. Average area for each floor in block 2 is 400 m2. Average area for each floor in block 3 is 400 m2. Definition of the project 3 Definition of the project Methodology Codes: Materials: Concrete 𝑓′ 𝑐 = 28MPa (B350) ACI 318-14 Reinforcing Steel Fy= 420MPa UBC-97 Soil Bearing Capacity = 300 KN/m^2 Codes & Materials Materials Unit Weight Ξ³ kN/m3 Block 12 Tiles 26 Plastering 23 Masonry stone 26 Aggregate 18 Concrete mortal 23 Loads: (1)Dead Load: consist of the own weight of the structure. Block 1 β€’Beams weight = 11015.25 kN β€’Columns weight = 6863kN β€’Shear wall weight = 11866.61 kN β€’Slab weight = 25204.3kN Total weight due to dead loads = 54949.16 kN Block 2 β€’Beams weight = 4634.097kN β€’Columns weight = 2733.3kN β€’Shear wall weight = 5859.1 kN β€’Slab weight = 9772.5kN Total weight due to dead loads = 23003.99 kN (2)Live Load: The same for block 1&2 Total live load for Block one =30087.20 KN Total live load for Block two =10626.31 KN Usage Live Load (KN/m2) Roof 10 (3)Superimposed Load: come from the weight of all non-structural elements. Item Unit weight Ζ” (KN/m3) Thickness (mm) Calculations Weight per m2 (KN/m2) W Tile 26 30 0.03*26 0.78 W mortar 23 20 0.02*23 0.46 W fill (sand or grand) 18 10 0.1*18 1.8 W plastring 23 15 0.015*23 0.345 π‘Šπ‘π‘Žπ‘Ÿπ‘‘π‘–π‘‘π‘–π‘œπ‘›π‘ = ((H block * 𝛾block) + (H π‘π‘™π‘Žπ‘ π‘‘π‘’π‘Ÿπ‘–π‘›π‘” on both faces of the wall * π›Ύπ‘π‘™π‘Žπ‘ π‘‘π‘’π‘Ÿπ‘–π‘›π‘”))* H wall * the length of all block wall) / area of the slab π‘Šπ‘π‘Žπ‘Ÿπ‘‘π‘–π‘‘π‘–π‘œπ‘›π‘  =Weight of block walls, which has (10 cm) block thickness + weight of block walls, which has (20 cm) block thickness 9 :Distribution of super impose dead load and live load on the slabs WSD block 1 (π‘˜π‘/π‘š 2 ) WSD block 2 (π‘˜π‘/π‘š 2 ) live load (π‘˜π‘/π‘š 2 ) ground floor 5 4.5 5 first floor 5.5 5.5 5 second floor 5.5 6 5 roof 0 0 10 Load Combinations Where: U: ultimate load D: dead load L: live load W: wind load E: earthquake Load H: load due to weight or pressure of soil U = 1.4D 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 Slab For Block 1 We decide to use Two-Way Solid Slab. Largest span/Smallest span= 7.35/6.65 = 1.105 h= 16.95 cm < 9 cm h = = 16.95 cm > 9 cm ok Use h=200mm Beams: Beam in block1: we choose the beam have long span h = L/ 18.5 = 0.676 Try h = 800mm (using longer span) bw = 12.5/20 = 625 mm then the preliminary dimensions are (h = 1100 mm) and (bw = 700mm) For Block 2 We decide to use Two-Way Solid Slab. Largest span/Smallest span = 7/6.25 = 1.12 h = 16.7cm < 9cm h Use h=200mm = 16.7 cm > 9 cm ok Beams: Beam in block2: we choose the beam have long span h = L/ 18.5 = 0.676 Try h = 800 mm (using longer span) bw = 12.5/20 = 625mm then the preliminary dimensions are (h = 1100 mm) and (bw = 700mm) 3-D modal 3-D modal 3-D modal Model Checks 1-Compatibility check 2-Equilibrium check 3-Internal force check ( stress – strain check ) 4-Deflection check. Equilibrium check: This check is to compare reaction results from ETABS with hand calculations . For block 1 Dead, live, and super imposed loads from ETABS in block 1 Dead, SD, and live load from hand calculation and ETABS for block 1 All of them less than 5 %, then it’s ok For block 2 Dead, SD, and live load from hand calculation and ETABS for block 2 Dead, live, and super imposed loads from ETABS in block 2 All of them less than 5 %, then it’s ok Internal force check ( stress – strain check ): This check is to compare internal force results from ETABS with hand calculations . calculate the moment from live and dead and super dead load on the beam 46 for block 1 Shows the define load combination = 1.2D + 1.6 L For ETABS results, see the following figures: M ETABS = (522.3 +461.2) * 0.5 + 333.3 = 824.5 kN.m. %difference = ((830-824.5)/830) *100 = 0.7% < 10% β†’ OK Calculate the moment from live and dead and super dead load on the beam 16 for block 2 For ETABS results, see the following figures: M ETABS = (199 +98.5)* 0.5 + 114= 262.5kN.m. %difference = ((265.7-262.5)/265.7) *100 = 1.2% < 10% β†’ OK Block 1, Column .. Manual Wu on column from floor 1=0.2*25*1.2+1.25*5+1.6*5=20 𝐾𝑁/π‘š2 Wu on column from floor 2=0.2*0.25*1.2+1.2*5.5+1.6*5=20.6 𝐾𝑁/π‘š2 Wu on column from floor 3= 0.2*0.25*1.2+1.2*5.5+1.6*5=20.6 𝐾𝑁/π‘š2 Wu on column from floor 4=0.2*25*1.2+1.6*10=22 𝐾𝑁/π‘š2 Final Wu =83.2 𝐾𝑁/π‘š2 Pu on column from slab=83.2*7.15*6.05=3599 KN Pu on column from beam =11.8*0.5*0.6*25*1.2*4=424.8 KN Weight of column = 15.5*0.7*0.7*25*12=227.85 KN Final Pu on column=4251.7 KN From ETABS : Pu=4311.3 KN Error= 4311.3βˆ’4251.7\4311.3 βˆ— 100% = 1.38 % Block 2, Column.. Manual Wu on column from floor 1=0.2*25*1.2+1.25*4.5+1.6*5=19.4 𝐾𝑁/π‘š2 Final Wu =19.4 𝐾𝑁/π‘š2 Pu on column from slab=19.4*3.98*7.2=555.35 KN Pu on column from beam =9.7759*0.5*0.6*25*1.2=87.983 KN Weight of column = 4.1*0.7*0.7*25*1.2=60.27 KN Final Pu on column=703.6 KN From ETABS : Pu=773 KN Error= 773βˆ’703.6\773 βˆ— 100% = 9 % Seismic load calculations Seismic load will be calculated depending on UBC-97. According to the location of this project (Ramallah), soil type (SOFT ROCK) and according to Palestine seismic map that is shown in Figure (3.1) and the project location, seismic zone is 2B thus the value of zone factor is (Z= 0.2). Determination of structure period: β„Žπ‘› = 18.1 m 𝐢𝑑 = 0.0488 𝑇 = 0.0488* = 0.428 sec Determinations of base shear using static lateral force procedure: V = 0.17 * W The total design base shear need not exceed the following:- V = 0.1364 * W V = 0.17 * W > V = 0.1364 * W Then V = 0.1364 * W The total design base shear shall not be less than the following:- 𝑉 = 0.11 * πΆπ‘Ž * I * W 𝑉 = 0.033 * W 𝑉 = 0.1364 βˆ—W > 0.033 βˆ— W β†’ 𝑂𝐾 Then base shear based on UBC (97) is 𝑉 = 0.1364 * W Compatibility check and period: To ensure that the whole structure is acting as a one single unit. Long-term deflection No compression steel (ρ compression = 0). Assume 25% sustained live load. The long-term deflection is given by the following equation: βˆ† Long term = βˆ† L + λ∞ x βˆ† D + Ξ»t x βˆ† Ls Ξ»t = λ∞ , since we assumed that the live load will sustained . Ξ» = T / (1+ 50ρ′) (such that ρ β€² is the compression steel ratio, T is a multiplier for long term deflection) . Figure (4.45) shows the multipliers for long-term deflections) is used to find T from ACI 318 Check deflection for slab To ensure that the max. deflection in the structure not exceed the allowable deflection . Block 3 is a coverage for the building as spherical dome of concrete with thickness 10 cm , Ξ³concrete 25 KN/m2 , radius 11.5 m , live load 2 KN/m2 and height 2 m . NΓΈ and NΞΈ from dead load NΓΈ @ ΓΈ = 0.0= - 42.5 KN/m NΓΈ @ ΓΈ = 20= - 44 KN/m NΞΈ @ ΞΈ = 0.0 = - 42.5 KN/m NΞΈ @ ΞΈ = 20 =- 36.1 KN/m NΓΈ and NΞΈ from live load NΓΈ @ ΓΈ = 0.0 = - 34 KN/m NΓΈ @ ΓΈ = 20 = - 34 KN/m NΞΈ @ ΞΈ = 0.0 = -34 KN/m NΞΈ @ ΞΈ = 20 = -26.1 KN/m Design of dome in block 3 NΓΈ max Ultimate = 1.2* 44 + 1.6 *34 = 107.2 KN/m (compression) NΞΈ max Ultimate = 1.2* 36.1 + 1.6 *26.1 = 85.1 KN/m (compression) βˆ… Pn = βˆ… Ξ» (0 .85 f'c (Ag-As) + As (fy)) βˆ… Pn = 0.65* 0.8 *(0 .85 *28 * (Ag-As) + As (420)) Assume As = 0.0 βˆ… Pn = 0.442* f'c *Ag = (0.442*28*1000*100)/1000 = 1237.6 KN >> 107.2 KN Use minimum reinforcement As = 0.003 * 1000 *100 = 300 mm2 Use 1 βˆ… 12 mm / 250 mm Dome supported by ring beam ( 60*100 cm ) and the beam supported by 8 column (60*60 cm ) the definition as shown below : 1. Compatibility Check. By making start animation for the deformed shape of SAP2000 model, it can be noticed that the structural elements are compatible and all of them are moving together. Figure 4.76 shows the deformed shape of block 3 checks 2. Equilibrium Check. 3. Internal Stresses a) The meridian and hoop stresses at the middle of the dome from the dead load: From SAP2000: NΞΈ = 42.93 KN/m (Compression). By hand: NΞΈ = 42.5 KN/m (Compression). % Error = (42.93βˆ’42.5) / 42.5 βˆ— 100% = 1 % < 10 % OK From SAP2000: NΓΈ = 43.3 KN/m (Compression). By hand: NΓΈ = 42.5 KN/m (Compression). % Error = (43.3βˆ’42.5) /42.5 βˆ— 100% = 1.88 % < 10 % OK b) The meridian and hoop stresses at the middle of the dome from the Live load: From SAP2000: NΞΈ = 33.7 KN/m (Compression). By hand: NΞΈ = 34 KN/m (Compression). % Error = (33.7βˆ’34) / 34 βˆ— 100% = 0.88 % < 10 % OK From SAP2000: NΓΈ = 34.1 KN/m (Compression). By hand: NΓΈ = 34 KN/m (Compression). % Error = (34.1βˆ’34) / 34 βˆ— 100% = 0.3 % < 10 % OK Vu = 225 KN M+ = 180 KN.m M- = 400 KN.m Tu = 25 KN-m T = 905 KN Design of ring beam βˆ… Pn = βˆ… Ξ» (0 .85 f'c (Ag-As) + As (fy)) Design of columns Column design for block 1 Group # Critical column # Pu (kN) Ag (mm2) Dimension (mm) As (mm2 ) Reinforcement Ties (mm) 1 C30 1450 160000 400*400 1600 8 Ο† 16 1 Ο† 10 mm @ 25 cm c/c 2 C12 2812 250000 500*500 2500 10 Ο† 18 3 Ο† 10 mm @ 25 cm c/c 3 C26 4587 360000 600*600 3600 16 Ο† 18 2 Ο† 10 mm @ 25 cm c/c Column design for block 2 Group # Critical column # Pu (kN) Ag (mm2) Dimension (mm) As (mm2 ) Reinforcement Ties (mm) 1 C15 996 90000 300*300 900 8 Ο† 16 1 Ο† 10 mm / 25 cm c/c 2 C13 1966 160000 400*400 1600 8 Ο† 18 1 Ο† 10 mm / 25 cm c/c 3 C22 3036 250000 500*500 2500 10 Ο† 18 Ο† 10 mm / 25 cm c/c3 Example Shear wall βˆ… Pn = βˆ… Ξ» (0 .85 f'c (Ag-As) + As (fy)) βˆ… Pn = 0.65*0.8 (0 .85*28* (0.99Ag) + 0.01Ag (420)) = 2000 βˆ— 1000 14.44 Ag = 2000*1000 Ag = 138540 mm2 β†’ (372 x 372) mm Wall dimensions are (1000 x 250) mm Check ρ 2000 βˆ— 1000 = 0.65*0.8 (0 .85*28* (250000 - As) + 420 As ) β†’ As = - 5310 mm2 Use ρ min = 0.0018 β†’ As= 0.0018*Ag As = 0.0018 * (1000*250) = 450 mm2 Use 1 Ο† 12 mm / 25cm Design of shear wall and stairs Stairs Footings are the substructure elements that transmit the concentrated column load or wall reactions safely to the soil. The transmitted load develop a uniform stress on the soil, the applied stress should be less than or equal to the bearing capacity of the soil. The soil in the site area is mainly rocky and the bearing capacity of the soil is 300 π‘˜π‘/π‘š 2 Mat Foundation Step1 : select type of foundation The service load taken by ETABS Calculate the area of footing as single by used : Area = Pa / q all Percentage of total single area from area of mat = 270/1500 βˆ— 100% = 18 % < 50 % add the area of foundations which shear walls need the percentage will increase Taking into account the site investigation report So , select Mat foundation Design of footing Step 2 : calculate thickness of mat foundation Assume the depth of the footing = 1000 mm and the effective depth = 900 mm on the soil. Vu = 4531.5 KN βˆ… Vc punching = βˆ… (𝟏 /πŸ‘) √fβ€²c bo d bo is the perimeter of the critical section then: bo= 2(C1+d) +2(C2+d)= 2*(0.6+0.9+0.6+0.9) = 6 m = 6000 mm So, βˆ… Vc punching = (0.75 *(𝟏 /πŸ‘) √28 *6000*0.9 = 7144 KN Vu < βˆ… Vc punching So it’s okay d=0.9 m and h=1 m. Step 3 : definition 66 1- Check the Bearing Capacity: As shown above, all the mat foundation have soil pressure less than 300 KN/m2 and in compression. Step 4 : checks 2- Check the Deflection: The max deflection from ETABS = 4.966 mm nearly equal 5 mm and this displacement less than 10 mm so, is ok . 3- Punching Shear Check: As picture shows there is no Punching shear in foundation so its ok. 4- Wide Beam Shear: The max shear forces at distance d = 0.9 m from envelope combination in X and Y direction are 545 KN and 550 KN respectively from ETABS as shown below: By hand: bw= 1000 mm, d=1000-100 = 900 mm and the max Vu = 550 KN. βˆ… 𝑉𝑐 = βˆ… (𝟏/ πŸ”) √fβ€²c bw d *1.1 = 0.75 * (𝟏 /πŸ” )√28 * 1000*900*1.1/1000 = 655 KN βˆ… Vc = 655 KN > Vu = 550 KN so no need for shear reinforcement. For both negative and positive moments in X and Y direction, As min will be used. bw = 1000 mm, h=1000 mm and d= 1000-100 = 900 mm As min = ρ min * bw *h = 0.0018*1000*1000 = 1800 mm2….. use ( 6 Ø20 mm/m ) So , any zones have As more than 1800 mm2 /m the reinforcement bars will be found for it. Step 5 : Design We have zones in both X and Y direction at column need reinforcement more than the minimum. will design the max zone as one sample calculation for the zone that needs reinforcement more than the minimum. and Generalization for all zones. 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