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One Way Ribbed Slab
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Step 1: Get Wu
Given: F'c = 25 Mpa cc = 20 mm
Fy = 400 Mpa 8 mm
Fyt = 400 Mpa 16 mm
Slab Type= 1 (Ribed=1 , Solid=0)
Dimensions: Beam : h = 26 cmAvg. Beam Width = 15 cmHordis Width = 42 cmSupported Width = 57 cmLn = 3.2 m Flange Th = 6 cmTiling Th = 15 cm
Loads: SLL = 3
SDL = 4Weight Reduction Factor For The Slab = 0.77
Weight Of Masonry Block = 0
Wu = 11.220883 KN/m
Step 2: Check a Mu = 14.36 KN-m d = 224 mm
1- Ru = 0.5578803 m = 18.823529
0.0014135
3- Ast = 180.47638
0.0035 0.00350.003125
117.6
5- Ast = 180.47638
14 mm 153.86
6- #Bars = 2 1.1729909 Ast = 307.72
7- Check a a = 10.162064
ψStr =
ψBar =
KN/m2
KN/m2
KN/m2
2- ρ =
mm2
ρMIN =
4- AstMIN = mm2
mm2
Used φBars =
mm2
It can be Designed as Rectangular Section
Step 3: Design For Shear 11.781
1- Vu = 12.3 KN
23.562 KN Use Minimum Shear Reinforcement
14.7 112 mm
6 mm 14.7 14.713.02
Area = 28.26
#Bars = 1 Use 2 φ 6
Step 4: Temperature & Shrinkage Steel
In direction Perpendicular to the ribs Flange Thickness = 60 mm 120
108
Ast = 120 6 28.26
Per 1000 mm Widh : 5 Bars are Used
Spacing = 200 mm
300 mm Adequate
In direction Parallel to the ribs Flange Thickness = 60 mm 68.4
61.56
Ast = 68.4 10 78.5
Per 1000 mm Widh : 1 Bars are Used
Spacing = 1000 mm
300 mm Not Adequate
Vu = 0.5*Wu*Ln If Regular BeamVu = Wu*Ln If Cantiliever
2- φVc =
3- AvMIN = mm2 SMAX =
4- Try φ =
mm2
mm2 Try φ =
SMAX =
mm2 Try φ =
SMAX =
Since this 1 φ 10 is already provided on top of every ribthis means that a φ 10 is sufficient but then the spacing
between each two φ 10 will > 30 cm, so we must provide a φ 6 bar on top of each Hordis block
Step 1: Get Wu
Given: F'c = 25 Mpa cc = 20 mm
Fy = 400 Mpa 8 mm
Fyt = 400 Mpa 16 mm
Slab Type= 1 (Ribed=1 , Solid=0)
Dimensions: Beam : h = 20 cmAvg. Beam Width = 16 cmHordis Width = 40 cmSupported Width = 56 cmLn = 3.2 m Flange Th = 6 cmTiling Th = 15 cm
Loads: LL = 3
Weight Reduction Factor For The Slab = 0.8
Weight Of Masonry Block = 0
Wu = 7.68768 KN/m
Step 2: Check a Mu = 9.84023 d = 164 mm
1- Ru = 0.725917 m = 18.82353
0.001847
3- Ast = 169.619
0.0035 0.00350.003125
91.84
5- Ast = 169.619
14 mm 153.86
6- #Bars = 2 1.102424 Ast = 307.72
7- Check a a = 10.34353
It can be Designed as Rectangular Section
ψStr =
ψBar =
KN/m2
KN/m2
2- ρ =
mm2
ρMIN =
4- AstMIN = mm2
mm2
Used φBars =
mm2
Step 3: Design For Shear 9.2004
1- Vu = 12.3 KN
18.4008 KN Use Minimum Shear Reinforcement
11.48 82 mm
6 mm 11.48 11.4810.168
Area = 28.26
#Bars = 1 Use 2 φ 6
Step 4: Temperature & Shrinkage Steel
In direction Perpendicular to the ribs Flange Thickness = 60 mm 120
108
Ast = 120 6 28.26
Per 1000 mm Widh : 5 Bars are Used
Spacing = 200 mm
300 mm Adequate
In direction Parallel to the ribs Flange Thickness = 60 mm 67.2
60.48
Ast = 67.2 10 78.5
Per 1000 mm Widh : 1 Bars are Used
Spacing = 1000 mm
300 mm Not Adequate
Vu = 0.5*Wu*Ln If Regular BeamVu = Wu*Ln If Cantiliever
2- φVc =
3- AvMIN = mm2 SMAX =
4- Try φ =
mm2
mm2 Try φ =
SMAX =
mm2 Try φ =
SMAX =
Since this 1 φ 10 is already provided on top of every ribthis means that a φ 10 is sufficient but then the spacingbetween each two φ 10 will > 30 cm, so we must provide a φ 6 bar on top of each Hordis block
Step 1: Get Wu
Given: F'c = 25 Mpa cc = 20 mm
Fy = 400 Mpa 8 mm
Fyt = 400 Mpa 16 mm
Slab Type= 1 (Ribed=1 , Solid=0)
Dimensions: Beam : h = 20 cmAvg. Beam Width = 16 cmHordis Width = 40 cmSupported Width = 56 cmLn = 3.2 m Flange Th = 6 cmTiling Th = 15 cm
Loads: LL = 3
Weight Reduction Factor For The Slab = 0.8
Weight Of Masonry Block = 0
Wu = 7.68768 KN/m
Step 2: Check a Mu = 9.84023 d = 164 mm
1- Ru = 0.725917 m = 18.82353
0.001847
3- Ast = 169.619
0.0035 0.00350.003125
91.84
5- Ast = 169.619
14 mm 153.86
6- #Bars = 2 1.102424 Ast = 307.72
7- Check a a = 10.34353
It can be Designed as Rectangular Section
ψStr =
ψBar =
KN/m2
KN/m2
2- ρ =
mm2
ρMIN =
4- AstMIN = mm2
mm2
Used φBars =
mm2
Step 3: Design For Shear 9.2004
1- Vu = 12.3 KN
18.4008 KN Use Minimum Shear Reinforcement
11.48 82 mm
6 mm 11.48 11.4810.168
Area = 28.26
#Bars = 1 Use 2 φ 6
Step 4: Temperature & Shrinkage Steel
In direction Perpendicular to the ribs Flange Thickness = 60 mm 120
108
Ast = 120 6 28.26
Per 1000 mm Widh : 5 Bars are Used
Spacing = 200 mm
300 mm Adequate
In direction Parallel to the ribs Flange Thickness = 60 mm 67.2
60.48
Ast = 67.2 10 78.5
Per 1000 mm Widh : 1 Bars are Used
Spacing = 1000 mm
300 mm Not Adequate
Vu = 0.5*Wu*Ln If Regular BeamVu = Wu*Ln If Cantiliever
2- φVc =
3- AvMIN = mm2 SMAX =
4- Try φ =
mm2
mm2 Try φ =
SMAX =
mm2 Try φ =
SMAX =
Since this 1 φ 10 is already provided on top of every ribthis means that a φ 10 is sufficient but then the spacingbetween each two φ 10 will > 30 cm, so we must provide a φ 6 bar on top of each Hordis block
