One way slab design by Sabuj Chowdhury, Lecturer, Department of CIvil Engineering, Ahsanullah University Of Science and Technology AUST
7 likes1,734 views
The document outlines the design principles for one-way slabs, detailing various support configurations and load transfer mechanisms. It includes steps for calculating slab thickness, moment, and reinforcement requirements based on specific design data and loading conditions. Additionally, it emphasizes students' responsibility to perform their own analyses and design tasks using provided data.
One way slab design by Sabuj Chowdhury, Lecturer, Department of CIvil Engineering, Ahsanullah University Of Science and Technology AUST
- 1. DESIGN OF ONE-WAY SLAB Sabuj Chowdhury Lecturer Department of Civil Engineering Ahsanullah University of Science and Technology
- 2. One Way Slab Design Sabuj Chowdhury, Lecturer, Department of CE, AUST Plan view of one-way slab may be--- 1. Supported on two opposite edges (Shown in Fig 01) 2. Supported on all edges (L/B ≥ 2) (Shown in Fig 02) Load Transfer LoadTransfer Beam Beam Beam Beam Beam Fig 01 : One-way Slab In Fig 01 slab is supported on two opposite sides only. In this case the structural action of the slab is essentially one way.
- 3. One Way Slab Design Sabuj Chowdhury, Lecturer, Department of CE, AUST Fig 02 : One-way Slab In Fig 02 there are beams on all four sides with a intermediate beam. Now if L & B ratio is 2 or greater, slab is one way even though supports are provided on all sides. L/B ≥ 2
- 4. One Way Slab Design Sabuj Chowdhury, Lecturer, Department of CE, AUST One way slab can be 1. Solid 2. Hollow or 3. Ribbed One-way Slab HollowSolid Ribbed
- 5. Design Drawing Sabuj Chowdhury, Lecturer, Department of CE, AUST 12‘ 12‘ 6‘ 27‘27‘ Load Transfer in Shorter Direction L/B=2.25 L/B=2.25L/B=2.25 L/B=2.25 Support in one direction only 1 1
- 6. Design Data Sabuj Chowdhury, Lecturer, Department of CE, AUST Note: Here it is the Example Data. Students have to analyse and design the slab by their own data, which are supplied in class where only A is unknown where, A = (Student No.+8) ft Design Data Dimension, A 12 ft (Student No. :13.02.03.004) Strength of Concrete, 3 ksi Strength of Steel, 𝑓𝑦 40 ksi Live Load on Floor 60 psf Floor Finish(FF) 25 psf 𝑓𝑐 ‘
- 7. One Way Slab Design Step 01 : Determination of Thickness (From consideration of Deflection) To control deflection, ACI Code 9.5.2.1 specifies minimum thickness values for one-way solid slabs. Here, l is the clear span Multiplying Factor = 0.4 + 𝑓𝑦 100 ; 𝑤𝑒𝑟𝑒, 𝑓𝑦 in ksi If , Thickness < 6 inch then upper rounding is 0.25 Thickness ≥ 6 inch then upper rounding is 0.50 t = 5.2 in = 5.25 in t = 5.3 in = 5.50 in t = 5.6 in = 5.75 in t = 5.8 in = 6.00 in t = 6.2 in = 6.50 in t = 6.3 in = 6.50 in t = 6.6 in = 7.00 in t = 6.8 in = 7.00 in Sabuj Chowdhury, Lecturer, Department of CE, AUST
- 8. Example 10” Wall 10” Wall 12” Beam 12 ft 12 ft 6 ft One end Continuous Both end Continuous Cantilever 𝑙 = 12 × 12 − 10 2 − 12 2 𝑙 = 133 in 𝑙 = 12 × 12 − 12 2 − 10 2 𝑙 = 133 in 𝑙 = 6 × 12 − 10 2 𝑙 = 67 in M F = 0.4 + 𝑓𝑦 100 = 0.8As, 𝑓𝑦 = 40 ksi; MF ≠ 1 𝑡 𝑚𝑎𝑥 = 5.36 in = 5.5 in (upper-rounding in 0.25 as 𝑡 𝑚𝑎𝑥< 6 in) t = 𝑙 24 × 𝑀𝐹 = 4.433𝑖𝑛 t = 𝑙 28 × 𝑀𝐹 = 3.80𝑖𝑛 t = 𝑙 10 × 𝑀𝐹 = 5.36𝑖𝑛 Sabuj Chowdhury, Lecturer, Department of CE, AUST Section 1-1
- 9. One Way Slab Design Step 02 : Load Calculation Dead Load Self Weight Super-imposed Dead Load FF (Floor) LC (Roof) PW (Floor) No PW (Roof) Live Load Total Load, 𝑤 𝑢 = 1.2 × 𝐷𝐿 + 1.6 × 𝐿𝐿 Sabuj Chowdhury, Lecturer, Department of CE, AUST
- 10. Example Dead Load = Self Wt.+ FF(Floor) or LC(Roof)+PW(if Given) Self Wt. = 𝑡 12 × 150 psf = 5.5 12 × 150 𝑝𝑠𝑓 = 𝟔𝟖. 𝟕𝟓 𝒑𝒔𝒇 t is found in Step 01 FF = 25 psf (Given) PW = 0 psf (As not given) LL = 60 psf (Given for floor) For Floor, DL = Self Wt.+ FF(Floor) +PW(if Given) = (68.75 + 25 + 0) psf = 93.75 psf LL = 𝟔𝟎 𝒑𝒔𝒇 Total Load, 𝑤 𝑢 = 1.2 × 𝐷𝐿 + 1.6 × 𝐿𝐿 = (1.2 × 93.75 + 1.6 × 60) psf = 208.5 𝑝𝑠𝑓 = 0.2085 𝑘𝑠𝑓 ≈ 𝟎. 𝟐𝟏 𝒌𝒔𝒇 Sabuj Chowdhury, Lecturer, Department of CE, AUST
- 11. One Way Slab Design Step 03 : Moment Calculation Sabuj Chowdhury, Lecturer, Department of CE, AUST
- 12. 12 ft 12 ft 6 ft Moment Coefficients are given below : 0 1 11 1 12 1 12 1 11 1 11 1 2 𝑙 = 133 in = 11.083 ft 𝑙 = 67 in = 5.583 ft 𝑙 = 133 in = 11.083 ft From Step 01 Example Sabuj Chowdhury, Lecturer, Department of CE, AUST
- 13. 1 11 1 12 1 2 0 0 + - - + - - 𝑤 𝑢 = 0.21 𝑘𝑠𝑓 (𝐹𝑟𝑜𝑚 𝑺𝒕𝒆𝒑 𝟎𝟐) Example 𝑤 𝑢 𝑙2 𝑤 𝑢 𝑙2 1 12 𝑤 𝑢 𝑙2 1 11 𝑤 𝑢 𝑙2 1 11 𝑤 𝑢 𝑙2 𝑤 𝑢 𝑙2 2.35 kip-ft/ft 2.35 kip-ft/ft 2.35 kip-ft/ft2.15 kip-ft/ft 2.15 kip-ft/ft 3. 27 kip-ft/ft 𝑀 𝑚𝑎𝑥 = 2.15 kip-ft/ft 𝑀 𝑚𝑎𝑥 = 3.27 kip-ft/ft Sabuj Chowdhury, Lecturer, Department of CE, AUST
- 14. 12 ft 12 ft 6 ft 2.35 kip-ft/ft 2.35 kip-ft/ft −2.15 kip-ft/ft −3.27 kip-ft/ft Example Sabuj Chowdhury, Lecturer, Department of CE, AUST
- 15. One Way Slab Design Step 04 : Effective depth, d check 𝜌 = 𝜌0.005 = 0.85𝛽1 𝑓𝑐 𝑓𝑦 𝜖 𝑢 𝜖 𝑢 + 0.005 ‘ 𝑑 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑈𝑛𝑖𝑡 𝑜𝑓 𝑀 𝑢 𝑖𝑠 𝒌𝒊𝒑 − 𝒊𝒏/𝒇𝒕 𝑎𝑠 𝑎𝑙𝑙 𝑜𝑡𝑒𝑟 𝑢𝑛𝑖𝑡𝑠 𝑎𝑟𝑒 𝑖𝑛 𝑘𝑖𝑝 𝑎𝑛𝑑 𝑖𝑛𝑐 𝑑 𝑝𝑟𝑜𝑣𝑖𝑑𝑒𝑑 = 𝑡𝑖𝑐𝑘𝑛𝑒𝑠𝑠 𝑡 − 1 𝑖𝑛𝑐 𝒅 𝒑𝒓𝒐𝒗𝒊𝒅𝒆𝒅 ≥ 𝒅 𝒓𝒆𝒒𝒖𝒊𝒓𝒆𝒅 Sabuj Chowdhury, Lecturer, Department of CE, AUST
- 16. Example 𝜌 = 𝜌0.005 = 0.85𝛽1 𝑓𝑐 𝑓𝑦 𝜖 𝑢 𝜖 𝑢 + 0.005 = 0.85 × 0.85 × 3 40 × 3 8 = 0.0203 ‘ ∅ = 0.4833 + 83.33𝜖 𝑡 ∅ = 0.4833 + 83.33 × 0.005 𝜖 𝑡 = 0.005𝐴𝑡, = 0.9 Here, 𝑀 𝑢 𝑖𝑠 𝑡𝑒 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑚𝑜𝑚𝑒𝑛𝑡 𝑓𝑜𝑢𝑛𝑑 𝑖𝑛 𝑺𝒕𝒆𝒑 𝟎𝟑 𝑀 𝑢 = 3.27 kip − ft/ft = 39.24 kip − in/ft 𝑑 𝑟𝑒𝑞 = 2.31 𝑖𝑛𝑐 𝑑 𝑝𝑟𝑜𝑣𝑖𝑑𝑒𝑑 = (𝑡 − 1) 𝑖𝑛𝑐 t is found in Step 01 = (5.5 − 1) 𝑖𝑛𝑐 = 4.5 𝑖𝑛𝑐 𝒅 𝒑𝒓𝒐𝒗𝒊𝒅𝒆𝒅 > 𝒅 𝒓𝒆𝒒𝒖𝒊𝒓𝒆𝒅 ok Sabuj Chowdhury, Lecturer, Department of CE, AUST
- 17. One Way Slab Design Step 05 : Reinforcement Area Determination & Minimum Reinforcement area for temperature and shrinkage reinforcement in slabs based on gross concrete area = 0.002𝑏𝑡 = 0.018𝑏𝑡 = 0.018 × 60 𝑓𝑦(𝑖𝑛 𝑘𝑠𝑖) 𝑏𝑡 𝐴 𝑠,𝑚𝑖𝑛 (𝑊𝑒𝑛, 𝑓𝑦 = 40 𝑘𝑠𝑖 𝑜𝑟 50 𝑘𝑠𝑖) (𝑊𝑒𝑛, 𝑓𝑦 = 60 𝑘𝑠𝑖) (𝑊𝑒𝑛, 𝑓𝑦 > 60 𝑘𝑠𝑖) H𝑒𝑟𝑒, 𝑏 = 12 𝑖𝑛𝑐 𝑎𝑛𝑑 𝑡 𝑖𝑠 𝑓𝑜𝑢𝑛𝑑 𝑖𝑛 𝑺𝒕𝒆𝒑 𝟎𝟏 If, 𝐴 𝑠 < 𝐴 𝑠,𝑚𝑖𝑛 then 𝐴 𝑠 = 𝐴 𝑠,𝑚𝑖𝑛 (𝐴 𝑠,𝑚𝑖𝑛 𝑎𝑣𝑒 𝑡𝑜 𝑏𝑒 𝑝𝑟𝑜𝑣𝑖𝑑𝑒𝑑) Sabuj Chowdhury, Lecturer, Department of CE, AUST
- 18. 12 ft 12 ft 6 ft 2.35 kip-ft/ft 2.35 kip-ft/ft −2.15 kip-ft/ft −3.27 kip-ft/ft Example 28.2 kip-in/ft −25.8 kip-in/ft −39.24 kip-in/ft 28.2 kip-in/ft 0.179 𝑖𝑛2 /𝑓𝑡 > 𝐴 𝑠,𝑚𝑖𝑛 −0.163 𝑖𝑛2 /𝑓𝑡 > 𝐴 𝑠,𝑚𝑖𝑛 −0.251 𝑖𝑛2 /𝑓𝑡 > 𝐴 𝑠,𝑚𝑖𝑛 𝐴 𝑠,𝑚𝑖𝑛 = 0.002 𝑏𝑡 𝐴𝑠 𝑓𝑦 = 40 𝑘𝑠𝑖 = 0.132 𝑖𝑛2 /𝑓𝑡 0.179 𝑖𝑛2 /𝑓𝑡 > 𝐴 𝑠,𝑚𝑖𝑛 𝐻𝑒𝑟𝑒, 𝑏 = 12 𝑖𝑛 & 𝑡 = 5.5 𝑖𝑛 (𝐹𝑟𝑜𝑚 𝑆𝑡𝑒𝑝 01) ok ok ok ok Sabuj Chowdhury, Lecturer, Department of CE, AUST
- 19. Step 06 : Reinforcement Spacing Determination One Way Slab Design If, we use #3 bar then area = 0.11 𝑖𝑛2 If, we use #4 bar then area = 0.20 𝑖𝑛2 𝑆𝑝𝑎𝑐𝑖𝑛𝑔, 𝑠 ≤ 3𝑡 𝑜𝑟 18 𝑖𝑛𝑐 𝑤𝑖𝑐𝑒𝑣𝑒𝑟 𝑖𝑠 𝑙𝑒𝑠𝑠𝐹𝑜𝑟 𝑆𝑜𝑟𝑡𝑒𝑟 𝐷𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 (Main Reinforcement) 𝑆𝑝𝑎𝑐𝑖𝑛𝑔, 𝑠 ≤ 5𝑡 𝑜𝑟 18 𝑖𝑛𝑐 𝑤𝑖𝑐𝑒𝑣𝑒𝑟 𝑖𝑠 𝑙𝑒𝑠𝑠𝐹𝑜𝑟 𝐿𝑜𝑛𝑔𝑒𝑟 𝐷𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 𝑆𝑝𝑎𝑐𝑖𝑛𝑔, 𝑠 = 12 × 0.11 𝐴 𝑠 12 × 0.20 𝐴 𝑠 𝑖𝑓 #3 𝑏𝑎𝑟 𝑖𝑓 #4 𝑏𝑎𝑟 Sabuj Chowdhury, Lecturer, Department of CE, AUST
- 20. Example 12 ft 12 ft 6 ft 0.179 𝑖𝑛2 /𝑓𝑡 −0.163 𝑖𝑛2 /𝑓𝑡 −0.251 𝑖𝑛2 /𝑓𝑡 0.179 𝑖𝑛2 /𝑓𝑡 0.179 𝑖𝑛2 /𝑓𝑡 : 𝑖𝑓 𝑤𝑒 𝑢𝑠𝑒 #3 𝑏𝑎𝑟 0.179 𝑖𝑛2 𝑎𝑟𝑒𝑎 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑖𝑛 12𝑖𝑛𝑐 1 𝑖𝑛2 𝑎𝑟𝑒𝑎 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑖𝑛 𝑖𝑛𝑐 0.11 𝑖𝑛2 𝑎𝑟𝑒𝑎 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑖𝑛 𝑖𝑛𝑐 12 0.179 12 × 0.11 0.179 = 7.37 𝑖𝑛𝑐 ≈ 7.0 𝑖𝑛𝑐(0.5" 𝑙𝑜𝑤𝑒𝑟 𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔) 𝑠 = 12 × 0.11 𝐴 𝑠 (#3 𝑏𝑎𝑟 𝑖𝑠 𝑢𝑠𝑒𝑑) 𝑂𝑅 = 12 × 0.11 0.179 = 7.37 𝑖𝑛𝑐 ≈ 7.0 𝑖𝑛𝑐(0.5" 𝑙𝑜𝑤𝑒𝑟 𝑟𝑜𝑢𝑛𝑑𝑖𝑛𝑔) ∴ #𝟑 @ 𝟕" 𝒄/𝒄 𝒂𝒍𝒕. 𝒄𝒌𝒅 Sabuj Chowdhury, Lecturer, Department of CE, AUST 𝐹𝑜𝑟 𝑆𝑜𝑟𝑡𝑒𝑟 𝐷𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 ∶ 𝑠 < 16.5“ (For Shorter Direction)
- 21. Example 12 ft 12 ft 6 ft ckd bar remain at top when negative moment ckd bar remain at bottom when positive moment Sabuj Chowdhury, Lecturer, Department of CE, AUST
- 22. Example 7” 7” 7” 7” 7” 7” 14” 14” Distance between straight bar & ckd bar = 7” Distance between ckd bar & ckd bar = 14” Sabuj Chowdhury, Lecturer, Department of CE, AUST
- 23. Example Sabuj Chowdhury, Lecturer, Department of CE, AUST 0.163 𝑖𝑛2 /𝑓𝑡 : (For Negative Moment) ∴ #3 @ 14" 𝑐/𝑐 𝑎𝑙𝑟𝑒𝑎𝑑𝑦 𝑝𝑟𝑜𝑣𝑖𝑑𝑒𝑑 𝐼𝑛 14" 𝑝𝑟𝑜𝑣𝑖𝑑𝑒𝑑 𝑟𝑒𝑖𝑛𝑓𝑜𝑟𝑐𝑒𝑚𝑒𝑛𝑡 𝑎𝑟𝑒𝑎 = 0.11 𝑖𝑛2 𝐼𝑛 12" 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑟𝑒𝑖𝑛𝑓𝑜𝑟𝑐𝑒𝑚𝑒𝑛𝑡 𝑎𝑟𝑒𝑎 = 0.163 𝑖𝑛2 𝐼𝑛 1" 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑟𝑒𝑖𝑛𝑓𝑜𝑟𝑐𝑒𝑚𝑒𝑛𝑡 𝑎𝑟𝑒𝑎 = 𝑖𝑛2 𝐼𝑛 14" 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑟𝑒𝑖𝑛𝑓𝑜𝑟𝑐𝑒𝑚𝑒𝑛𝑡 𝑎𝑟𝑒𝑎 = 𝑖𝑛2 0.163 12 0.163 × 14 12 = 0.19 𝑖𝑛2 (𝑇𝑜𝑡𝑎𝑙 𝑅𝑒𝑞𝑢𝑖𝑟𝑒𝑚𝑒𝑛𝑡) 𝐴𝑙𝑟𝑒𝑎𝑑𝑦 𝑃𝑟𝑜𝑣𝑖𝑑𝑒𝑑 𝑅𝑒𝑖𝑛𝑓𝑜𝑟𝑐𝑒𝑚𝑒𝑛𝑡 𝐴𝑟𝑒𝑎 𝑖𝑛 14" = 0.11 𝑖𝑛2 𝐸𝑥𝑡𝑟𝑎 𝑛𝑒𝑒𝑑𝑒𝑑 𝑟𝑒𝑖𝑛𝑓𝑜𝑟𝑐𝑒𝑚𝑒𝑛𝑡 𝑎𝑟𝑒𝑎 𝑡𝑜 𝑓𝑢𝑙𝑓𝑖𝑙 𝑡𝑜𝑡𝑎𝑙 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑚𝑒𝑛𝑡 = (0.19 − 0.11) 𝑖𝑛2 = 0.08 𝑖𝑛2 𝐼𝑓 #3 𝑖𝑠 𝑝𝑟𝑜𝑣𝑖𝑑𝑒𝑑 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑒𝑥𝑡𝑟𝑎 𝑡𝑜𝑝 𝑏𝑎𝑟 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 = 0.08 0.11 = 0.73 ≈ 1 1 #3 extra top between ckd. bar
- 24. Example 1 #3 extra top between ckd. bar Sabuj Chowdhury, Lecturer, Department of CE, AUST
- 25. Example Sabuj Chowdhury, Lecturer, Department of CE, AUST 0.251 𝑖𝑛2 /𝑓𝑡 : (For Negative Moment) ∴ #3 @ 14" 𝑐/𝑐 𝑎𝑙𝑟𝑒𝑎𝑑𝑦 𝑝𝑟𝑜𝑣𝑖𝑑𝑒𝑑 𝐼𝑛 14" 𝑝𝑟𝑜𝑣𝑖𝑑𝑒𝑑 𝑟𝑒𝑖𝑛𝑓𝑜𝑟𝑐𝑒𝑚𝑒𝑛𝑡 𝑎𝑟𝑒𝑎 = 0.11 𝑖𝑛2 𝐼𝑛 12" 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑟𝑒𝑖𝑛𝑓𝑜𝑟𝑐𝑒𝑚𝑒𝑛𝑡 𝑎𝑟𝑒𝑎 = 0.251 𝑖𝑛2 𝐼𝑛 1" 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑟𝑒𝑖𝑛𝑓𝑜𝑟𝑐𝑒𝑚𝑒𝑛𝑡 𝑎𝑟𝑒𝑎 = 𝑖𝑛2 𝐼𝑛 14" 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑟𝑒𝑖𝑛𝑓𝑜𝑟𝑐𝑒𝑚𝑒𝑛𝑡 𝑎𝑟𝑒𝑎 = 𝑖𝑛2 0.251 12 0.251 × 14 12 = 0.293 𝑖𝑛2 (𝑇𝑜𝑡𝑎𝑙 𝑅𝑒𝑞𝑢𝑖𝑟𝑒𝑚𝑒𝑛𝑡) 𝐴𝑙𝑟𝑒𝑎𝑑𝑦 𝑃𝑟𝑜𝑣𝑖𝑑𝑒𝑑 𝑅𝑒𝑖𝑛𝑓𝑜𝑟𝑐𝑒𝑚𝑒𝑛𝑡 𝐴𝑟𝑒𝑎 𝑖𝑛 14" = 0.11 𝑖𝑛2 𝐸𝑥𝑡𝑟𝑎 𝑛𝑒𝑒𝑑𝑒𝑑 𝑟𝑒𝑖𝑛𝑓𝑜𝑟𝑐𝑒𝑚𝑒𝑛𝑡 𝑎𝑟𝑒𝑎 𝑡𝑜 𝑓𝑢𝑙𝑓𝑖𝑙 𝑡𝑜𝑡𝑎𝑙 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑚𝑒𝑛𝑡 = (0.293 − 0.11) 𝑖𝑛2 = 0. 183 𝑖𝑛2 𝐼𝑓 #4 𝑖𝑠 𝑝𝑟𝑜𝑣𝑖𝑑𝑒𝑑 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑒𝑥𝑡𝑟𝑎 𝑡𝑜𝑝 𝑏𝑎𝑟 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 = 0.183 0.20 = 0.73 ≈ 1 1 #4 extra top between ckd. bar
- 26. Example 1 #3 extra top between ckd. bar 1 #4 extra top between ckd. bar Sabuj Chowdhury, Lecturer, Department of CE, AUST
- 27. Example 𝐹𝑜𝑟 𝐿𝑜𝑛𝑔𝑒𝑟 𝐷𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛 ∶ 𝐴 𝑠,𝑚𝑖𝑛 = 0.132 𝑖𝑛2 /𝑓𝑡 𝐼𝑓 𝑤𝑒 𝑢𝑠𝑒 #3 𝑏𝑎𝑟 𝑡𝑒𝑛 𝑠 = 12 × 0.11 0.132 = 10" #3 @ 10" 𝑐/𝑐 Sabuj Chowdhury, Lecturer, Department of CE, AUST 𝑠 < 18“ (For Longer Direction) 𝑠 < 18“ (OK)
- 28. Step 07 : Detailing One Way Slab Design Sabuj Chowdhury, Lecturer, Department of CE, AUST
- 29. Example Detailing in Plan View/Top View a) #3 @ 7” c/c alt. ckd. b) 1 #3 extra top between ckd. bar c) 1 #4 extra top between ckd. bar d) #3 @ 10” c/c a b c d t = 5.5 inch Sabuj Chowdhury, Lecturer, Department of CE, AUST 1 1 2 2
- 30. #3 @ 10” c/c 12 ft 6 ft #3 @ 7” c/c alt. ckd. 1 #3 extra top 1 #4 extra top 12 ft 𝑙 = 133 in = 11.083 ft 𝑙 = 67 in = 5.583 ft 𝑙 = 133 in = 11.083 ft 1’-7” 2’-10” 2’-10” 2’-10” 3’-9” 3’-9” 3’-9” 𝑆𝑒𝑐𝑡𝑖𝑜𝑛 1 − 1 Example Sabuj Chowdhury, Lecturer, Department of CE, AUST
- 31. Example Sabuj Chowdhury, Lecturer, Department of CE, AUST 𝑆𝑒𝑐𝑡𝑖𝑜𝑛 2 − 2 27 ft 27 ft #3 @ 10” c/c #3 @ 7” c/c alt. ckd.