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8/13/2019 Copy of Rcc Beam Design Task-01
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REFERENCE
Analysis
Length L 4.2
Width B 0.225
Dead load g 13.62 kN/m
Imposed load i 12 kN/m
Ultimate Design Load w 38.27 kN/m
Single span fixed end condition
BM @ fixed end MA 56.25 kNm
BM @ mid span MC 28.13
SF V 80.36 kN
Design Data
Material Properties
Concrete grade fcu 20 N/mm2
Characteristic steel strength fy 460 N/mm2
Youngs modulus of concrete EC 26000 N/mm2
Youngs modulus of steel ES 200000 N/mm2
Density of concrete pct 24 kN/m3
Durabilty Requirements
Exposure condition - Severe
Beam
Design of Reinforcement
Sizes
Top r/f = 12 mm
Bottom r/f = 16 mm
Links = 6 mm
Cover = 25 mm
Width of the beam bt = 225 mm
overall depth of the beam hb = 350 mm
Design ofTopReinforcement
Effective depth d = 311 mm d = 311
Design bending moment M = 56.25 kNm
Cl 3.4.4.4 K = M/[bd2fcu]
BS 8110; 1985 = 0.103 k = 0.103
Since K < 0.156 no compresiion r/f required
Z = [0.5+(0.25-K/0.9)0.5
] * d
= 0.87 d
Take Z = 0.83d Z = 270.57 mm Z = 270.5
Required area of steel to carry ultimate BM
CALCULATION OU
8/13/2019 Copy of Rcc Beam Design Task-01
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Effective depth d = 311 mm d =
Design bending moment M = 28.13 kNm
K = M/[bd2fcu]
= 0.052 k =
Since K < 0.156 no compresiion r/f required
Z = [0.5+(0.25-K/0.9)0.5
] * d
= 0.94 d
Take Z = 0.92d Z = 292.34 mm Z =
Required area of steel to carry ultimate BM
As req,M = M/[0.95*fy*Z]
= 220 mm2/m
Area of steel provided As prov 270 mm2/m
Provide T 12 1 Nos T
and T 10 2 Nos T
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Ckeck for Shear Resistance
Ultimate shear force at support Vu = 80.36 kN
Shear stress = Vu/[bv*d]
= 1.15 N/mm2 = 1
Maximum shear = 3.6 N/mm2
100 As/[bv*d] = 0.80
400/d = 1.29
c = 0.62
c + 0.4 = 1.02
Since c+0.4 < < max shear, design links to be provided
Link size = 6 mm
Shear link spacing Sv < 113.73
Max spacing < 0.75d < 233.25
Provide R6 @ 150 mm c/c upto 1.5 m from both ends
Middle area spacing shall be 200 mm c/c
Deflection Checking
SS condition assumed to check deflection
Table 3.10 BS
8110; 1985Basic Span/Effective depth ratio = 20
Service stress in the tension r/f fs = 5*fy*As req/(8*As prov*b)
= 245 N/mm2
Modification factor for tension r/f MF = 0.55+ [477-fs]/[120*(0.9+M/bd2)]
= 1.1 ( < 2.0 )
Hence allowable Span/Effective depth ratio = MF * Basic Span/Eff. Depth
= 22
Actual Span/Effective Depth ratio = 6.225/382
= 14.23
Since [Span/Eff.Depth]actual < [Span/Eff.Depth]allowabledeflection is within the limit Deflecti
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Design ofBottomReinforcement
Effective depth d = 382 mm d = 382
Design bending moment M = 79.18 kNm
K = M/[bd2fcu]
= 0.096 k = 0.09
Since K < 0.156 no compresiion r/f required
Z = [0.5+(0.25-K/0.9)0.5
] * d
= 0.88 d
Take Z = 0.89d Z = 336.16 mm Z = 336
Required area of steel to carry ultimate BM
As req,M = M/[0.95*fy*Z]
= 539 mm2/m
Area of steel provided As prov 629 mm2/m
Provide T 16 0 Nos T 16
and T 20 2 Nos T 20
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Ckeck for Shear Resistance
Ultimate shear force at support Vu = 121.82 kN
Shear stress = Vu/[bv*d]
= 1.42 N/mm2 = 1
Maximum shear = 4.0 N/mm2
100 As/[bv*d] = 0.83
400/d = 1.05
c = 0.60
c + 0.4 = 1.00
Since c+0.4 < < max shear, design links to be provided
Link size = 10 mm
Shear link spacing Sv < 202.83
Max spacing < 0.75d < 286.5
Provide R10 @ 150 mm c/c upto 1.5 m from both ends
Middle area spacing shall be 200 mm c/c
Deflection Checking
SS condition assumed to check deflection
Table 3.10 BS
8110; 1985Basic Span/Effective depth ratio = 20
Service stress in the tension r/f fs = 5*fy*As req/(8*As prov*b)
= 300 N/mm2
Modification factor for tension r/f MF = 0.55+ [477-fs]/[120*(0.9+M/bd2)]
= 0.91 ( < 2.0 )
Hence allowable Span/Effective depth ratio = MF * Basic Span/Eff. Depth
= 18.2
Actual Span/Effective Depth ratio = 6.225/382
= 16.30
Since [Span/Eff.Depth]actual < [Span/Eff.Depth]allowabledeflection is within the limit Deflecti
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REFERENCE
Analysis SLAB THICK 150 MM
Length L 3
Width B 0.225
Dead load [slab+finishes] g 14.88 kN/m
Imposed load i 12 kN/m
Ultimate Design Load w 40.03 kN/m
Total Design Load F 120.10 kN
Single span fixed end condition
BM @ C MC 39.63 kNm
BM @ B MB 32.43
SF @ A V 54.04 kN
Design Data
Material Properties
Concrete grade fcu 25 N/mm2
Characteristic steel strength fy 460 N/mm2
Youngs modulus of concrete EC 26000 N/mm2
Youngs modulus of steel ES 200000 N/mm2
Density of concrete pct 24 kN/m3
Durabilty Requirements
Exposure condition - Severe
Beam
Design of Reinforcement
Sizes
Top r/f = 12 mm
Bottom r/f = 12 mm
Links = 6 mm
Cover = 25 mm
Width of the beam bt = 225 mm
overall depth of the beam hb = 350 mm
Design ofTopReinforcement @ C
Effective depth d = 313 mm d = 313
Design bending moment M = 39.63 kNm
Cl 3.4.4.4 K = M/[bd2fcu]
BS 8110; 1985 = 0.072 k = 0.07
Since K < 0.156 no compresiion r/f required
Z = [0.5+(0.25-K/0.9)0.5
] * d
CALCULATION O
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and T 16 0 Nos T 16
Design ofBottomReinforcement @ B
Effective depth d = 313 mm d = 313
Design bending moment M = 32.43 kNm
K = M/[bd2fcu]
= 0.059 k = 0.05
Since K < 0.156 no compresiion r/f required
Z = [0.5+(0.25-K/0.9)0.5
] * d
= 0.93 d
Take Z = 0.92d Z = 291.09 mm Z = 291.
Required area of steel to carry ultimate BM
As req,M = M/[0.95*fy*Z]
= 255 mm2/m
Area of steel provided As prov 339 mm2/m
Provide T 12 3 Nos T 12
and T 0 Nos T 0
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Ckeck for Shear Resistance
Ultimate shear force at support Vu = 54.04 kN
Shear stress = Vu/[bv*d]
= 0.77 N/mm2 = 0
Maximum shear = 4.0 N/mm2
100 As/[bv*d] = 0.48
400/d = 1.28
c = 0.53
c + 0.4 = 0.93
Since c+0.4 > > 0.5 c max shear, minimum links required
Link size = 6 mm
Shear link spacing Sv < 149.29
Max spacing < 0.75d < 234.75
Provide R6 @ 150 mm c/c upto 1.5 m throughout
Deflection Checking
SS condition assumed to check deflection
Table 3.10 BS
8110; 1985Basic Span/Effective depth ratio = 20
Service stress in the tension r/f fs = 5*fy*As req/(8*As prov*b)
= 270 N/mm2
Modification factor for tension r/f MF = 0.55+ [477-fs]/[120*(0.9+M/bd2)]
= 1.28 ( < 2.0 )
Hence allowable Span/Effective depth ratio = MF * Basic Span/Eff. Depth
= 25.6
Actual Span/Effective Depth ratio = 6.225/382
= 10.30
Since [Span/Eff.Depth]actual< [Span/Eff.Depth]allowabledeflection is within the limit Deflecti
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REFERENCE
Analysis SLAB THICK 150 MM
Length L 3
Width B 0.225
Dead load [slab+finishes] g 14.88 kN/m
Imposed load i 12 kN/m
Ultimate Design Load w 40.03 kN/m
Total Design Load F 120.10 kN
Single span fixed end condition
BM @ E ME 28.82 kNm
BM @ D MD 25.22 kNm
SF @ C V 66.05 kN
Design Data
Material Properties
Concrete grade fcu 25 N/mm2
Characteristic steel strength fy 460 N/mm2
Youngs modulus of concrete EC 26000 N/mm2
Youngs modulus of steel ES 200000 N/mm2
Density of concrete pct 24 kN/m3
Durabilty Requirements
Exposure condition - Severe
Beam
Design of Reinforcement
Sizes
Top r/f = 12 mm
Bottom r/f = 12 mm
Links = 6 mm
Cover = 25 mm
Width of the beam bt = 225 mm
overall depth of the beam hb = 350 mm
Design ofTopReinforcement @ C
Effective depth d = 313 mm d = 313
Design bending moment M = 28.82 kNm
Cl 3.4.4.4 K = M/[bd2fcu]
BS 8110; 1985 = 0.052 k = 0.05
Since K < 0.156 no compresiion r/f required
Z = [0.5+(0.25-K/0.9)0.5
] * d
CALCULATION O
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and T 16 0 Nos T 16
Design ofBottomReinforcement @ B
Effective depth d = 313 mm d = 313
Design bending moment M = 25.22 kNm
K = M/[bd2fcu]
= 0.046 k = 0.04
Since K < 0.156 no compresiion r/f required
Z = [0.5+(0.25-K/0.9)0.5
] * d
= 0.95 d
Take Z = 0.94d Z = 297.35 mm Z = 297.
Required area of steel to carry ultimate BM
As req,M = M/[0.95*fy*Z]
= 194 mm2/m
Area of steel provided As prov 226 mm2/m
Provide T 12 2 Nos T 12
and T 0 Nos T 0
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Ckeck for Shear Resistance
Ultimate shear force at support Vu = 66.05 kN
Shear stress = Vu/[bv*d]
= 0.94 N/mm2 = 0
Maximum shear = 4.0 N/mm2
100 As/[bv*d] = 0.48
400/d = 1.28
c = 0.53
c + 0.4 = 0.93
Since c+0.4 < < max shear, design links to be provided
Link size = 6 mm
Shear link spacing Sv < 144.91
Max spacing < 0.75d < 234.75
Provide R6 @ 125 mm c/c upto 1.5 m from support
Deflection Checking
SS condition assumed to check deflection
Table 3.10 BS
8110; 1985Basic Span/Effective depth ratio = 20
Service stress in the tension r/f fs = 5*fy*As req/(8*As prov*b)
= 190 N/mm2
Modification factor for tension r/f MF = 0.55+ [477-fs]/[120*(0.9+M/bd2)]
= 1.72 ( < 2.0 )
Hence allowable Span/Effective depth ratio = MF * Basic Span/Eff. Depth
= 34.4
Actual Span/Effective Depth ratio = 6.225/382
= 10.30
Since [Span/Eff.Depth]actual< [Span/Eff.Depth]allowabledeflection is within the limit Deflecti
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REFERENCE
Analysis SLAB THICK 125 MM
Length L 3
Width B 1
Dead load g 4.6 kN/m
Imposed load i 4 kN/m
Ultimate Design Load w 12.84 kN/m
Single span fixed end condition
BM @ fixed end MA 10.60 kNm
BM @ mid span MC 8.70 kNm
SF V 19.26 kN
Design Data
Material Properties
Concrete grade fcu 25 N/mm2
Characteristic steel strength fy 460 N/mm2
Youngs modulus of concrete EC 26000 N/mm2
Youngs modulus of steel ES 200000 N/mm2
Density of concrete pct 24 kN/m3
Durabilty Requirements
Exposure condition - Severe
Beam
Design of Reinforcement
Sizes
Top r/f = 0 mm
Bottom r/f = 10 mm
Links = mm
Cover = 25 mm
Width of the strip bt
= 1000 mm
overall depth of the slab hb = 125 mm
Design ofTopReinforcement
Effective depth d = 95 mm d = 95
Design bending moment M = 10.60 kNm
Cl 3.4.4.4 K = M/[bd2fcu]
BS 8110; 1985 = 0.047 k = 0.0
Since K < 0.156 no compresiion r/f required
Z = [0.5+(0.25-K/0.9)0.5] * d
= 0.94 d
CALCULATION O
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Design ofBottomReinforcement
Effective depth d = 95 mm d = 95
Design bending moment M = 8.70 kNm
K = M/[bd2fcu]
= 0.039 k = 0.0
Since K < 0.156 no compresiion r/f required
Z = [0.5+(0.25-K/0.9)0.5
] * d
= 0.95 d
Take Z = 0.92d Z = 90.25 mm Z = 90
Required area of steel to carry ultimate BM
As req,M = M/[0.95*fy*Z]
= 221 mm2/m
Area of steel provided As prov 314 mm2/m
Provide T 10 4 Nos T 10
T10 @ 250 mm spacing
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Ckeck for Shear Resistance
Ultimate shear force at support Vu = 19.26 kN
Shear stress = Vu/[bv*d]
= 0.2 N/mm2 =
Maximum shear = 4.0 N/mm2
100 As/[bv*d] = 0.33
400/d = 4.21
c = 0.63
c + 0.4 = 1.03
Since < 0.5 c
Deflection Checking
Continous condition assumed to check deflection
Table 3.10 BS
8110; 1985Basic Span/Effective depth ratio = 27
Service stress in the tension r/f fs = 5*fy*As req/(8*As prov*b)
= 202 N/mm2
Modification factor for tension r/f MF = 0.55+ [477-fs]/[120*(0.9+M/bd2)]
= 1.78 ( < 2.0 )
Hence allowable Span/Effective depth ratio = MF * Basic Span/Eff. Depth
= 48.06
Actual Span/Effective Depth ratio = 3000/85
= 31.58
Since [Span/Eff.Depth]actual < [Span/Eff.Depth]allowabledeflection is within the limit Deflect
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REFERENCE
Analysis SLAB THICK 125 MM
Length L 6
Width B 1
Dead load g 4.6 kN/m
Imposed load i 4 kN/m
Ultimate Design Load w 12.84 kN/m
Single span fixed end condition
BM @ fixed end MA 3.46 kNm
BM @ mid span MC 4.76 kNm
SF V 38.52 kN
Design Data
Material Properties
Concrete grade fcu 25 N/mm2
Characteristic steel strength fy 460 N/mm2
Youngs modulus of concrete EC 26000 N/mm2
Youngs modulus of steel ES 200000 N/mm2
Density of concrete pct 24 kN/m3
Durabilty Requirements
Exposure condition - Severe
Beam
Design of Reinforcement
Sizes
Top r/f = 0 mm
Bottom r/f = 10 mm
Links = 10 mm
Cover = 25 mm
Width of the strip bt
= 1000 mm
overall depth of the slab hb = 125 mm
Design ofTopReinforcement
Effective depth d = 85 mm d = 85
Design bending moment M = 3.46 kNm
Cl 3.4.4.4 K = M/[bd2fcu]
BS 8110; 1985 = 0.019 k = 0.0
Since K < 0.156 no compresiion r/f required
Z = [0.5+(0.25-K/0.9) 0.5] * d
= 0.98 d
CALCULATION O
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Design ofBottomReinforcement
Effective depth d = 85 mm d = 85
Design bending moment M = 4.76 kNm
K = M/[bd2fcu]
= 0.026 k = 0.0
Since K < 0.156 no compresiion r/f required
Z = [0.5+(0.25-K/0.9)0.5
] * d
= 0.97 d
Take Z = 0.92d Z = 82.45 mm Z = 82.4
Required area of steel to carry ultimate BM
As req,M = M/[0.95*fy*Z]
= 132 mm2/m
Area of steel provided As prov 314 mm2/m
Provide T 10 4 Nos T 10
T10 @ 250 mm spacing
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Ckeck for Shear Resistance
Ultimate shear force at support Vu = 38.52 kN
Shear stress = Vu/[bv*d]
= 0.45 N/mm2 = 0.4
Maximum shear = 4.0 N/mm2
100 As/[bv*d] = 0.37
400/d = 4.71
c = 0.67
c + 0.4 = 1.07
Since < 0.5 c
Deflection Checking [Not necessary in long span direction]
SS condition assumed to check deflection
Table 3.10 BS
8110; 1985Basic Span/Effective depth ratio = 20
Service stress in the tension r/f fs = 5*fy*As req/(8*As prov*b)
= 121 N/mm2
Modification factor for tension r/f MF = 0.55+ [477-fs]/[120*(0.9+M/bd2)]
= 2.45 ( < 2.0 )
Hence allowable Span/Effective depth ratio = MF * Basic Span/Eff. Depth
= 49
Actual Span/Effective Depth ratio = 3000/120
= 35.29
Since [Span/Eff.Depth]actual < [Span/Eff.Depth]allowabledeflection is within the limit Deflection
8/13/2019 Copy of Rcc Beam Design Task-01
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REFERENCE
Analysis SLAB THICK 150 MM
Length L 7
Width B 0.225
Dead load [slab+finishes] g 15.42 kN/m
Imposed load i 12 kN/m
Ultimate Design Load w 40.79 kN/m
Single span fixed end condition
BM @ fixed end MA 166.55 kNm
BM @ mid span MC 83.28
SF V 142.76 kN
Design Data
Material Properties
Concrete grade fcu 25 N/mm2
Characteristic steel strength fy 460 N/mm2
Youngs modulus of concrete EC 26000 N/mm2
Youngs modulus of steel ES 200000 N/mm2
Density of concrete pct 24 kN/m3
Durabilty Requirements
Exposure condition - Severe
Beam
Design of Reinforcement
Sizes
Top r/f = 20 mm
Bottom r/f = 16 mm
Links = 10 mm
Cover = 25 mm
Width of the beam bt = 225 mm
overall depth of the beam hb = 475 mm
Design ofTopReinforcement
Effective depth d = 432 mm d = 432
Design bending moment M = 149.90 kNm
Cl 3.4.4.4 K = M/[bd2fcu]
BS 8110; 1985 = 0.143 k = 0.14
Since K < 0.156 no compresiion r/f required
Z = [0.5+(0.25-K/0.9)0.5] * d
= 0.8 d
CALCULATION O
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Design ofBottomReinforcement
Effective depth d = 432 mm d = 432
Design bending moment M = 104.09 kNm
K = M/[bd2fcu]
= 0.099 k = 0.09
Since K < 0.156 no compresiion r/f required
Z = [0.5+(0.25-K/0.9)0.5
] * d
= 0.87 d
Take Z = 0.89d Z = 375.84 mm Z = 375
Required area of steel to carry ultimate BM
As req,M = M/[0.95*fy*Z]
= 634 mm2/m
Area of steel provided As prov 742 mm2/m
Provide T 12 1 Nos T 12
and T 20 2 Nos T 20
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Ckeck for Shear Resistance
Ultimate shear force at support Vu = 142.76 kN
Shear stress = Vu/[bv*d]
= 1.47 N/mm2 = 1
Maximum shear = 4.0 N/mm2
100 As/[bv*d] = 0.97
400/d = 0.93
c = 0.61
c + 0.4 = 1.01
Since c+0.4 < < max shear, design links to be provided
Link size = 10 mm
Shear link spacing Sv < 193.73
Max spacing < 0.75d < 324
Provide R10 @ 150 mm c/c upto 1.5 m from both ends
Middle area spacing shall be 200 mm c/c
Deflection Checking
SS condition assumed to check deflection
Table 3.10 BS
8110; 1985Basic Span/Effective depth ratio = 20
Service stress in the tension r/f fs = 5*fy*As req/(8*As prov*b)
= 303 N/mm2
Modification factor for tension r/f MF = 0.55+ [477-fs]/[120*(0.9+M/bd2)]
= 0.87 ( < 2.0 )
Hence allowable Span/Effective depth ratio = MF * Basic Span/Eff. Depth
= 17.4
Actual Span/Effective Depth ratio = 6.225/382
= 16.72
Since [Span/Eff.Depth]actual < [Span/Eff.Depth]allowabledeflection is within the limit Deflecti
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REFERENCE
Analysis SLAB THICK 150 MM
Length L 3
Width B 0.225
Dead load [slab+finishes] g 17.18 kN/m
Imposed load i 14 kN/m
Ultimate Design Load w 46.45 kN/m
Total Design Load F 139.36 kN
Single span fixed end condition
BM @ C MC 45.99 kNm
BM @ B MB 37.63
SF @ A V 62.71 kN
Design Data
Material Properties
Concrete grade fcu 25 N/mm2
Characteristic steel strength fy 460 N/mm2
Youngs modulus of concrete EC 26000 N/mm2
Youngs modulus of steel ES 200000 N/mm2
Density of concrete pct 24 kN/m3
Durabilty Requirements
Exposure condition - Severe
Beam
Design of Reinforcement
Sizes
Top r/f = 12 mm
Bottom r/f = 12 mm
Links = 6 mm
Cover = 25 mm
Width of the beam bt = 225 mm
overall depth of the beam hb = 350 mm
Design ofTopReinforcement @ C
Effective depth d = 313 mm d = 313
Design bending moment M = 45.99 kNm
Cl 3.4.4.4 K = M/[bd2fcu]
BS 8110; 1985 = 0.083 k = 0.08
Since K < 0.156 no compresiion r/f required
Z = [0.5+(0.25-K/0.9)0.5
] * d
CALCULATION O
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and T 16 2 Nos T 16
Design ofBottomReinforcement @ B
Effective depth d = 313 mm d = 313
Design bending moment M = 37.63 kNm
K = M/[bd2fcu]
= 0.068 k = 0.06
Since K < 0.156 no compresiion r/f required
Z = [0.5+(0.25-K/0.9)0.5
] * d
= 0.92 d
Take Z = 0.92d Z = 287.96 mm Z = 287.
Required area of steel to carry ultimate BM
As req,M = M/[0.95*fy*Z]
= 299 mm2/m
Area of steel provided As prov 339 mm2/m
Provide T 12 3 Nos T 12
and T 0 Nos T 0
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Ckeck for Shear Resistance
Ultimate shear force at support Vu = 62.71 kN
Shear stress = Vu/[bv*d]
= 0.89 N/mm2 = 0
Maximum shear = 4.0 N/mm2
100 As/[bv*d] = 0.57
400/d = 1.28
c = 0.56
c + 0.4 = 0.96
Since c+0.4 > > 0.5 c max shear, minimum links required
Link size = 6 mm
Shear link spacing Sv < 149.29
Max spacing < 0.75d < 234.75
Provide R6 @ 150 mm c/c upto 1.5 m throughout
Deflection Checking
SS condition assumed to check deflection
Table 3.10 BS
8110; 1985Basic Span/Effective depth ratio = 20
Service stress in the tension r/f fs = 5*fy*As req/(8*As prov*b)
= 267 N/mm2
Modification factor for tension r/f MF = 0.55+ [477-fs]/[120*(0.9+M/bd2)]
= 1.22 ( < 2.0 )
Hence allowable Span/Effective depth ratio = MF * Basic Span/Eff. Depth
= 24.4
Actual Span/Effective Depth ratio = 6.225/382
= 10.30
Since [Span/Eff.Depth]actual< [Span/Eff.Depth]allowabledeflection is within the limit Deflecti
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REFERENCE
Analysis SLAB THICK 150 MM
Length L 3
Width B 0.225
Dead load [slab+finishes] g 17.18 kN/m
Imposed load i 14 kN/m
Ultimate Design Load w 46.45 kN/m
Total Design Load F 139.36 kN
Single span fixed end condition
BM @ E ME 33.45 kNm
BM @ D MD 29.26 kNm
SF @ C V 76.65 kN
Design Data
Material Properties
Concrete grade fcu 25 N/mm2
Characteristic steel strength fy 460 N/mm2
Youngs modulus of concrete EC 26000 N/mm2
Youngs modulus of steel ES 200000 N/mm2
Density of concrete pct 24 kN/m3
Durabilty Requirements
Exposure condition - Severe
Beam
Design of Reinforcement
Sizes
Top r/f = 12 mm
Bottom r/f = 12 mm
Links = 6 mm
Cover = 25 mm
Width of the beam bt = 225 mm
overall depth of the beam hb = 350 mm
Design ofTopReinforcement @ C
Effective depth d = 313 mm d = 313
Design bending moment M = 33.45 kNm
Cl 3.4.4.4 K = M/[bd2fcu]
BS 8110; 1985 = 0.061 k = 0.06
Since K < 0.156 no compresiion r/f required
Z = [0.5+(0.25-K/0.9)0.5
] * d
CALCULATION O
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and T 16 0 Nos T 16
Design ofBottomReinforcement @ B
Effective depth d = 313 mm d = 313
Design bending moment M = 29.26 kNm
K = M/[bd2fcu]
= 0.053 k = 0.05
Since K < 0.156 no compresiion r/f required
Z = [0.5+(0.25-K/0.9)0.5
] * d
= 0.94 d
Take Z = 0.94d Z = 294.22 mm Z = 294.
Required area of steel to carry ultimate BM
As req,M = M/[0.95*fy*Z]
= 228 mm2/m
Area of steel provided As prov 226 mm2/m
Provide T 12 2 Nos T 12
and T 0 Nos T 0
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Ckeck for Shear Resistance
Ultimate shear force at support Vu = 76.65 kN
Shear stress = Vu/[bv*d]
= 1.09 N/mm2 = 1
Maximum shear = 4.0 N/mm2
100 As/[bv*d] = 0.48
400/d = 1.28
c = 0.53
c + 0.4 = 0.93
Since c+0.4 < < max shear, design links to be provided
Link size = 6 mm
Shear link spacing Sv < 106.24
Max spacing < 0.75d < 234.75
Provide R6 @ 125 mm c/c upto 1.5 m from support
Deflection Checking
SS condition assumed to check deflection
Table 3.10 BS
8110; 1985Basic Span/Effective depth ratio = 20
Service stress in the tension r/f fs = 5*fy*As req/(8*As prov*b)
= 223 N/mm2
Modification factor for tension r/f MF = 0.55+ [477-fs]/[120*(0.9+M/bd2)]
= 1.5 ( < 2.0 )
Hence allowable Span/Effective depth ratio = MF * Basic Span/Eff. Depth
= 30
Actual Span/Effective Depth ratio = 6.225/382
= 10.30
Since [Span/Eff.Depth]actual< [Span/Eff.Depth]allowabledeflection is within the limit Deflecti
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Design of Pad Footing TO BE EDITED
F2 - 600mm x 300mm plate resting on footing
Load F 193
Weight of footing 18.432
Service load Pser 211.432
Ultimate load Pult 317.148
Bearing capacity of soil 125 kN/m2
Footing area 1.69
Say, Length 1.6
Width 1.6
Actual area 2.56 > 1.69 m2
Overall depth "h" 300 mm > 178.09 mm
Design for Bending
Ultimate bearing pressure 123.89 < 125.00 kN/m2
Critical BM at colum face BM 41.873 kNm
Bar diameter assumed 12 mm
Cover 50 mm
Effective depth 232
Average depth 238
M/bd2
0.486
100As/bd 0.1
Reinforcement required As 371.2 mm2
Minimum r/f required 624 mm2
Provide T12 @ 300 c/c
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r/f provided 678.24 mm2
> 624.00 mm2
Check for Punching Shear
Max shear v 1.110 N/mm2
< 4.00 N/mm2
Critical perimeter 4056 mm
Area outside the perimeter 1.532 m2
Punching shear V 189.77 kN
Stress v 0.197 N/mm2
< 4.00 N/mm2
Shear r/f not required
Detail
Foundation depth (to bottom of footing) : 1200 mm
Size : 1600 x 1600 mm
Footing thickness : 300 mm
Reinforcement : T12 @ 300 c/c both direction