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Home Design Slab,Beam,Stairs

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Home Design Slab,Beam,Stairs

  1. 1. ENGR.HAMMAD BASHIR WEBSITE http://hammadjatt.weebly.com/ 1 | P a g e
  2. 2. ENGR.HAMMAD BASHIR : Waqas Rahmat B.Sc. Electrical Engineer : This residential building structural component is designed according to ACI 14 Code. Hammad Bashir BSC Civil ENGINEER Pakistan, Islamabad Email: hammad.bashir96@gmail.com Website: www.hammadjatt.weebly.com Cell: +92343-0817733 : Engr. Sajjid rasheed khokar M.SC CIVIL ENGINEER PEC. NO CIVIL/7998 Engr.Naveed Rashid Pec no civil /5984 Structural engineer Naveed associates WEBSITE http://hammadjatt.weebly.com/ 2 | P a g e
  3. 3. ENGR.HAMMAD BASHIR PLOT LOCATION: GROUND FLOOR PLAN: WEBSITE http://hammadjatt.weebly.com/ 3 | P a g e DESIGN 1) SLAB 2) BEAM 3) STAIRS
  4. 4. ENGR.HAMMAD BASHIR FIRST FLOOR..YOU CAN MAKE IT ON YOUR CHOICE..HOWEVER YOU CAN PUT SAME STRUCTURE ON FIRST FLOOR. WEBSITE http://hammadjatt.weebly.com/ 4 | P a g e
  5. 5. ENGR.HAMMAD BASHIR SLAB THICKNESS: According to ACI 318 the minimum thickness of slab should be 5 inch. Load calculation: Service dead loads: Material THICKNESS (INCH) THICKNESS (FEET) DENSITY LOAD CALCULATION KSF SLAB 5” 5”/12 0.15 0.0625 MUD 4” 4”/12 0.12 0.04 TILE 2” 2”/12 0.12 0.02 TOTAL DEAD LOADS= 0.1225 KSF WEBSITE http://hammadjatt.weebly.com/ 5 | P a g e
  6. 6. ENGR.HAMMAD BASHIR FACTORED DEAD LOAD 1.2 0.1225 0.147 Ksf Service dead loads: FOR BUILDING LIVE LOAD WILL BE ACCORDING TO LOADING CRITERIA…… Sr.no Occupancy or use Live load Kgs/m2 Pascal N/m2 lb/ft2 1 Private rooms, school class rooms. 200 1900 40 2 Offices. 250 to 425 2400 to 4000 50 to 85 3 Fixed-seats, assembly halls, library reading rooms. 300 2900 60 4 Corridors in public building 400 3800 80 5 Movable seats assembly hall 500 4800 100 6 Wholesales stores, light storage warehouses. 610 6000 125 7 Library stack rooms 730 7200 150 8 Heavy manufacturing, heavy storage warehouses, side walks and driveways subject to truckling 1200 12000 250 9 Stairs, general 500 4800 100 10 Stairs, upto two-family residences, 50% more than specifications. 300 2900 60 WE USE 40 Psf = 0.040 Ksf FACTORED LIVE LOAD 1.6 0.04 0.064 Ksf TOTAL FACTORED LOAD= FACTORED LIVE LOAD + FACTORED DEAD LOAD = 0.1225 + 0.064 = 0.211 Ksf WEBSITE http://hammadjatt.weebly.com/ 6 | P a g e
  7. 7. ENGR.HAMMAD BASHIR BENDING MOMENT CALCULATION ASPECT RATIO: m= la/lb la= shorter length…. lb= longer length BENDING MOMENT Co-efficients WEBSITE http://hammadjatt.weebly.com/ 7 | P a g e
  8. 8. ENGR.HAMMAD BASHIR Calculating moments using ACI Coefficients: Ma, neg = Ca, neg wula 2 WEBSITE http://hammadjatt.weebly.com/ 8 | P a g e
  9. 9. ENGR.HAMMAD BASHIR Mb, neg = Cb, neg wulb 2 Ma, pos, (dl + ll) = M a, pos, dl + M a, pos, ll = Ca, pos, dl × wu, dl × la 2 + Ca, pos, ll × wu, ll × la 2 Mb, pos, (dl + ll) = Mb, pos, dl + Mb, pos, ll = Cb, pos, dl × wu, dl × lb 2 + Cb, pos, ll × wu, ll × lb 2 Design of Two-Way Slab First determining capacity of min. reinforcement: As,min = 0.002bhf = 0.12 in2 Using #3 bars: Spacing for As,min = 0.12 in2 = (0.11/0.12) × 12 = 11″ c/c However ACI max spacing for two way slab = 2h = 2(5) = 10″ or 18″ = 10″ c/c Hence using #3 bars @ 10″ c/c For #3 bars @ 10″ c/c: As,min = (0.11/10) × 12 = 0.132 in2 Capacity for As,min: a = (0.132 × 40)/(0.85 × 3 × 12) = 0.17″ ΦMn = ΦAsminfy(d – a/2) = 0.9 × 0.132 × 40(4 – (0.17/2)) = 18.60 in-kip Therefore, for Mu values ≤ 18.60 in-k/ft, use As,min (#3 @ 10″ c/c) & for Mu values > 18.6 in-kip/ft, calculate steel area using trial & error procedure. Shrinkage Reinforcement IF NEEDED: Ast = 0.002bhf = 0.12 in2 (#3 @ 11″ c/c) WEBSITE http://hammadjatt.weebly.com/ 9 | P a g e
  10. 10. ENGR.HAMMAD BASHIR However, for facilitating field work, we will use #3 @ 10″ c/c WEBSITE http://hammadjatt.weebly.com/ 10 | P a g e
  11. 11. ENGR.HAMMAD BASHIR WEBSITE http://hammadjatt.weebly.com/ 11 | P a g e
  12. 12. ENGR.HAMMAD BASHIR WEBSITE http://hammadjatt.weebly.com/ 12 | P a g e SLAB STEEL DRAWING
  13. 13. ENGR.HAMMAD BASHIR Beam Design Step 01: Sizes Let depth of beam = 18″ ln + depth of beam = 15.875′ + (18/12) = 17.375′ c/c distance between beam supports = 16.375 + (4.5/12) = 16.75′ Therefore l = 16.75′ Depth (h) = (16.75/18.5) × (0.4 + 40000/100000) × 12 = 8.69″ (Minimum requirement of ACI 9.5.2.2). Take h = 1.5′ = 18″ d = h – 3 = 15″ b = 12″ Step 02: Loads Load on beam will be equal to Factored load on beam from slab + factored self weight of beam web Factored load on slab = 0. 211 ksf Load on beam from slab = 0. 211 ksf x 5 = 1.055 k/f Factored Self load of beam web = = 1.2 x (13 × 12/144) × 0.15 = 0.195 k/f Total load on beam = 1.055 + 0.195 = 1.25 kip/ft WEBSITE http://hammadjatt.weebly.com/ 13 | P a g e
  14. 14. ENGR.HAMMAD BASHIR WEBSITE http://hammadjatt.weebly.com/ 14 | P a g e
  15. 15. ENGR.HAMMAD BASHIR STAIR DRAWING: RISER= 175 mm Tread = 300mm Landing: 1.2m WEBSITE http://hammadjatt.weebly.com/ 15 | P a g e
  16. 16. ENGR.HAMMAD BASHIR Reference I. Design of concrete structure by NIlson II. Notes of PROF. ZIAAUDDIN MIAN UET LAHORE PAKISTAN III. Design of concrete structures by z.a siddique UET lahore WEBSITE http://hammadjatt.weebly.com/ 16 | P a g e

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