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

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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

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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