UNIVERSITY OF SALAHADDIN COLLEGE OF ENGINEERING CIVIL DEPARTMENT

1. UNIVERSITY OF SALAHADDINCOLLEGE OF ENGINEERING CIVIL DEPARTMENT STRUCTURAL ANALYSIS AND DESIGN OF MULTI-STORY REINFORCED CONCRETE COMMERCIAL BUILDING Supervised by Prepared by Dr. Dilshad K. Jaf Ameen Younis Muhammed Ismat ShakewMusleh

2. The mechanism of load transformation from Slabs to beams: The load is transferred from slab to beams by distributing the load over the beam. The slab load (Dead and Live), expressed in units of weight per area, is converted into weight per length of the beam. The slab should rest on the beam that carries its weight. The area weight is distributed along the beam by three methods depending on the reinforcement direction and the geometrical dimension of the slab. It involves two mechanisms : • Two-way • One-way

3. Function of the building: 1- Basement: for parking cars. 2- Ground floor: for reception and providing a comfortable waiting for people, and accessibility to other floors. 3 - 1st to 7th floor: used as office service rooms. 4 - Roof floor: provides flexible accessibility.

5. Example of analysis and design • Slabs : The slab thickness is calculated by the following formula: t min = ; where β = . • Sample of calculation: Calculating for slab S2 (8500×7500) Ln (long) = 8500 - 300 = 8200 mm Ln (short) = 7500 – 300 = 7200 mm t min = = 195 mm use h min equal to 210 mm for all slabs of all floors.

6. Beams The beam depth is calculated using the following formula: h min = for one end continuous. h min = for both end continuous. h min = = 404.76 mm ; based on B2. h min = = 247 mm ; based on B1. Use (h min) equal to 510 mm for practical purposes and (bw) equal to 510/1.7 = 300 mm

7. load calculation: • Slabs: • Dead load: 1)Basement slab: - Self-weight of slab = 0.21 * 24 = 5.04 kN/m2 - Tiles and mortar (5.0cm thick) = 1.15 kN/m2 - Cement plastering (2.0cm thick) = 0.46kN/m2 - HVAC = 0.5 kN/m2 Total dead load = 5.04 + 1.15 + 0.46 + 0.5 = 7.15 kN/m2 2)Ground floor, 1st to 7th floor slabs: - Self-weight of slab = 0.21 * 24 = 5.04 kN/m2 - Tiles and mortar (50cm thick) = 1.15 kN/m2 - Gypsum plastering (2.0cm thick) = 0.26 kN/m2 - HVAC = 0.5 kN/m2 - False ceiling = 0.5 kN/m2 - Light-weight partition wall ( thermo stone ) = 1.8 kN/m2 Total dead load = 5.04 + 1.15 + 0.26 + 1 + 1.8 = 9.25 kN/m2

8. 3)Roof floor: - Self-weight of slab = 0.21 * 24 = 5.04 kN/m2 -Roof tiles + roof sand + water proofing (2.0cm thick) = 0.92+1.92+0.27 = 3.11 kN/m2 - Gypsum plastering (2.0cm thick) = 0.26 kN/m2 - HVAC + false ceiling = 0.5 + 0.5 = 1 kN/m2 - Total dead load = 5.04 + 3.11 + 0.26 + 1 = 9.41 kN/m2 Live load: Basement slab: Total live load = 5 kN/m2 Ground floor, 1st to 7th floor slabs: Total live load = 2 kN/m2 Roof floor slab: Total live load = 1.5 kN/m2

9. Columns: • - Assume preliminary section size 500×500 mm. • Roof: • Exterior column ( C6 ): • load from slabs = = 465.069 kN • weight of beam stem = 0.3*0.3*24*( = 21.978 kN • weight of column = 0.5*0.5*3.6*24 = 21.6 kN • load of req. floor col. = 465.069+21.978+21.6 = 508.647 kN • area req. = = 42387.23 mm2 • Interior column ( C10 ): • Slab loads = • = 867.552 kN • weight of beam stem = 0.3*0.3*24*( = 29.02 kN • weight of column = 0.5*0.5*3.6*24 = 21.6 kN • load of req. floor col. = 867.552+29.02+21.6 = 918.172 kN • area req. = = 76514.33 mm2

11. Analysis of frames : For basement moments : Mu @ face = Cm*Wu*Ln2 Mab = * 40 * 3.652 = 33.3 kN.m Mba= * 40 * 3.652 = 48.45 kN.m Mbc= * 70.6413 * 7.62 = 408 kN.m Mposext. = * 40 * 3.652 = 38.1 kN.m Mposint. = * 70.6413 * 7.62 = 255 kN.m

12. Design of slabs Asmin= 0.0018 *b*t = 0.0018 *1000*210 = 378 mm2/m Spacing of Ø10 = 71*1000/378 = 188 mm Use Ø10@175 mm c/c As provided = 71*1000/175 = 405.71 mm2/m d = 210 – 20 -10/2 = 185 mm a = = = 6.682 mm ; c = a/0.85 = 6.682/0.85 = 7.86 mm Ɛt = ok ; use Ø = 0.9 Mu = ØAs.fy (d-a/2) = 0.9 *420 * 405.71 * (185 - 6.682/2 ) = 27.86 kN.m

13. Design of columns: For interior column C10: Mu y = 93.84 kN.m , Mu x = 294.5 kN.m , Pu = 8556.775 kN. Mmin= Pu (15 +0.03 h ) = 8556.775 *( 15 + 0.03*900 ) = 359.4 kN.m So Mu y and Mu x will not be considered ; use ρ =1 % and b = h =900 mm As = 0.01 * 900 * 900 = 8100 mm2 Pu = Ø*0.8(0.85*fc`*(Ag – As)+As fy) = 0.65 * 0.8 *(0.85*30*(900*900– 8100) + 8100*420 = 12402.2 kN > 8556.775 kN So use ρ =1 % No. of bars required = 8100π* /4= 16.5 , use 18Ø25 mm • Ties : - 16 * Dia. of bar = 16 *25 = 400 mm  controls • - 48 * Dia. of tie = 48 *10 = 480 mm • - least column dimension = 900 mm • Use Ø10@380 mm .

14. Beams results: Reinforcement details of beam B3 and B4 is shown in the figure below:

15. Beams results: cross section of the beam at mid span cross section of the beam at face of the support.

16. Column results: • Interior column reinforcement details at ground floor are shown in the figure below 18Ø25

17. Discussion: • The results of the analysis and design are satisfied according to the codes, the slab is reinforced by Ø10@175 mm c/c in both long and short direction at top and bottom . • Beam B1 for the basement was analyzed and designed using 2 Ø 20 mm as topreinforcement and 2 Ø 20 mm as bottom reinforcement. • An interior column (C10) at basement was designed and design illustrated that we need to use 18Ø25 mm as longitudinal bars and 10 Ø380 mm as tie reinforcement.