A one way slab is to be designed for an effective span 3.3 m . The super imposed load including finishing is 4 KN/m2 .Taking modification factor 1.2.Design the slab. Sketch c/s of slab showing reinforcement details. Use concrete M20 and steel Fe 415 .
Given:
- Span (l): 3.3 m
- Live Load (L.L.) and Floor Finish (F.F.): 4 KN/m²
- Modification Factor (M.F.): 1.2
1. Design Constants:
- Yield strength of steel (fy): 415 N/mm²
- Characteristic compressive strength of concrete (fck): 20 N/mm²
- Maximum depth of neutral axis (Xu max): 0.48d
- Ultimate moment of resistance (Mu lim): 0.138 fck b d²
2. Estimation of Slab Thickness:
d=span20×M.F.d = frac{text{span}}{20 times text{M.F.}}
d=330020×1.2d = frac{3300}{20 times 1.2}
d=137.5 mmd = 137.5 text{ mm}
Assuming 10 mm Ø main bars and nominal cover as 20 mm:
D=d+dc+ϕ2D = d + d_c + frac{phi}{2}
D=137.5+20+102D = 137.5 + 20 + frac{10}{2}
D=162.5 mmD = 162.5 text{ mm}
Say D=165 mmD = 165 text{ mm}
Therefore:
davail=D−dc−ϕ2d_{avail} = D – d_c – frac{phi}{2}
davail=165−20−102d_{avail} = 165 – 20 – frac{10}{2}
davail=140 mmd_{avail} = 140 text{ mm}
3. Effective Span:
Le=3.3 m (given)
L_e = 3.3 text{ m (given)}
4. Load Calculations (considering 1 m wide strip):
Self weight of slab=1×1×0.165×25=4.125 KN/mtext{Self weight of slab} = 1 times 1 times 0.165 times 25 = 4.125 text{ KN/m}
Weight of F.F & L.L.=1×1×4=4 KN/mtext{Weight of F.F & L.L.} = 1 times 1 times 4 = 4 text{ KN/m}
Total load (w)=8.125 KN/mtext{Total load (w)} = 8.125 text{ KN/m}
Factored load:
wd=1.5×8.125w_d = 1.5 times 8.125
wd=12.1875 KN/m
w_d = 12.1875 text{ KN/m}
5. Factored (Design) Maximum Bending Moment:
Md=(wd×Le28)M_d = left(frac{w_d times L_e^2}{8}right)
Md=(12.1875×3.328)M_d = left(frac{12.1875 times 3.3^2}{8}right)
Md=16.59 KNmM_d = 16.59 text{ KNm}
6. Required Overall Depth and Effective Depth:
Equating MulimM_u text{lim} to MdM_d:
0.138fckbd2=Md0.138 f_{ck} b d^2 = M_d
0.138×20×1000×d2=16.59×1060.138 times 20 times 1000 times d^2 = 16.59 times 10^6
d=77.53 mmd = 77.53 text{ mm}
davail=140 mm>drequiredd_{avail} = 140 text{ mm} > d_{required}:
Hence, OK.text{Hence, OK.}
Provide D=165 mmD = 165 text{ mm} and davail=140 mmd_{avail} = 140 text{ mm}
7. Area of Main Steel:
Ast=0.5fckfy[1−1−4.6Mdfckbd2]bdA_{st} = frac{0.5 f_{ck}}{f_y} left[ 1 – sqrt{1 – frac{4.6 M_d}{f_{ck} b d^2}} right] b d
Ast=0.5×20415[1−1−4.6×16.59×10620×1000×1402]×1000×140A_{st} = frac{0.5 times 20}{415} left[ 1 – sqrt{1 – frac{4.6 times 16.59 times 10^6}{20 times 1000 times 140^2}} right] times 1000 times 140
Ast=346.13 mm2
Spacing of Main Reinforcement
a) Calculating Spacing (S10S_{10})
S10=1000×(π4×102)346.13S_{10} = frac{1000 times left(frac{pi}{4} times 10^2right)}{346.13}
S10=1000×78.54346.13S_{10} = frac{1000 times 78.54}{346.13}
S10=226.91 mmS_{10} = 226.91 , text{mm}
Spacing = 225 mm c/c (rounded to nearest standard size)
b) 3d Calculation
3d=3×140=420 mm3d = 3 times 140 = 420 , text{mm}
c) Minimum Spacing
Minimum Spacing=300 mmtext{Minimum Spacing} = 300 , text{mm}
Ast provided
Ast provided=1000×(π4×102)225text{Ast provided} = frac{1000 times left(frac{pi}{4} times 10^2right)}{225}
Ast provided=1000×78.54225text{Ast provided} = frac{1000 times 78.54}{225}
Ast provided=349.06 mm2
8. Area and Spacing of Distribution Steel:
Step 1: Calculate the Minimum Area of Distribution Steel (Astmin)
Astmin=0.15%×b×Dtext{Ast}_{text{min}} = 0.15 % times b times D
=0.15100×1000 mm×165 mm= frac{0.15}{100} times 1000 , text{mm} times 165 , text{mm}
=247.5 mm2= 247.5 , text{mm}^2
Step 2: Determine the Spacing of 6 mm Diameter Mild Steel Distribution Bars
Sd=(Area of one barAstmin)×1000S_d = left( frac{text{Area of one bar}}{text{Ast}_{text{min}}} right) times 1000
Area of one bar=π×(62)2text{Area of one bar} = pi times left(frac{6}{2}right)^2
=π×32= pi times 3^2
=28.27 mm2= 28.27 , text{mm}^2
Calculate Spacing (S)
Sd=(28.27 mm2247.5 mm2)×1000 mmS_d = left( frac{28.27 , text{mm}^2}{247.5 , text{mm}^2} right) times 1000 , text{mm}
=114.22 mm= 114.22 , text{mm}
≈110 mmapprox 110 , text{mm}
Step 3: Check Maximum Allowable Spacing
Smax=5×DepthS_{text{max}} = 5 times text{Depth}
=5×140 mm= 5 times 140 , text{mm}
=700 mm= 700 , text{mm}
Given Maximum Spacing
=450 mm= 450 , text{mm}
Therefore, the minimum spacing calculated is 114.22 mm, approximated to 110 mm.
Conclusion:
Provide 6 mm diameter distribution bars at 110 mm center-to-center.
9. Reinforcement Details:
