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Moment Diagram

FRAME ANALYSIS

MAX = 278kN.m

MAX = -42.6kN.m MAX = -42.6kN.m

MAX = 278kN.m

MAX = 254kN.m

MAX = 583kN.m

MAX = 0kN.m MAX = 0kN.m

SLAB COVER ANALYSIS

Moment Diagram

MAX = -583kN.m

ELEMENT DESIGN to BS 8110:1997 SOLID SLABS

INPUT Location Deck Mid SpanDesign moment, M 583.0 kNm/m fcu 40 N/mm c = 1.50

b 1.00 fy 460 N/mm s = 1.05span 3300 mm

Height, h 350 mm Section location SIMPLY SUPPORTED SPANBar 25 mmcover 30 mm to this reinforcement

OUTPUT Deck Mid Span Compression steel = Noned = 350 - 30 - 25/2 = 307.5 mm .

(3.4.4.4) K' = 0.156 > K = 0.154 ok .(3.4.4.4) z = 307.5 [0.5 + (0.25 - 0.154 /0.9)^ = 240.0 > 0.95d = 292.1 mm(3.4.4.1) As = 583.00E6 /460 /240.0 x 1.05 = 5544 > min As = 455 mm/m

PROVIDE T25 @ 75 = 6545 mm/m .(Eqn 8) fs = 2/3 x 460 x 5544 /6545 /1.00 = 259.8 N/mm(Eqn 7) Tens mod factor = 0.55 + (477 - 259.8) /120 /(0.9 + 6.166) = 0.806(3.4.6.3) Permissible L/d = 20.0 x 0.806 = 16.124. Actual L/d = 3300 /307.5 = 10.732 ok .

. .

ELEMENT DESIGN to BS 8110:1997 SOLID SLABS

INPUT Location Upper Slab SupportDesign moment, M 0.0 kNm/m fcu 40 N/mm c = 1.50

b 1.00 fy 460 N/mm s = 1.05span 3300 mm

Height, h 350 mm Section location SUPPORTBar 25 mmcover 30 mm to this reinforcement

OUTPUT Upper Slab Support Compression steel = Noned = 350 - 30 - 25/2 = 307.5 mm FAILS on Max Spacing.

(3.4.4.4) K' = 0.156 > K = 0.000 ok .(3.4.4.4) z = 307.5 [0.5 + (0.25 - 0.000 /0.9)^ = 307.5 > 0.95d = 292.1 mm(3.4.4.1) As = 0.00E6 /460 /292.1 x 1.05 = 0 < min As = 455 mm/m

PROVIDE T25 @ 1100 = 446 mm/m .

Project

CRACK WIDTH CALCULATIONS - FLEXURE -

INPUTfcu= 40 N/mm2fy= 460 N/mm2

Area of reinforcement " As " = 44179 mm2b = 1000 mmh = 350 mmd = 283 mm

Minimum cover to tension reinforcement " CO " = 30 mmMaxmum bar spacing " S " = 100 mm

Bar dia " DIA " = 75 mm " acr " =(((S/2)^2+(CO+DIA/2)^2)^(1/2)-DIA/2) as default or enter other value = 46.5 mm "acr " is distance from the point considered to the surface of the nearest longitudinal bar

Applied service moment " Ms "= 388.7 KNmCALCULATIONS

moduli of elasticity of concrete " Ec" = (1/2)*(20+0.2*fcu) = 14.0 KN/mm2moduli of elasticity of steel " Es " = 200.0 KN/mm2

Modular ratio "a " = (Es/Ec) = 14.29 " r " = As/bd = 0.156depth to neutral axis, "x" = (- . +(( .)2 + 2. .)0.5.d = 238 mm

" Z " = d-(x/3) = 203Reinforcement stress " fs " = Ms/(As*Z) = 43 N/mm2Concrete stress " fc " = (fs*As)/(0.5*b*x) = 16.09 N/mm2

Strain at soffit of concrete beam/slab " e1" = (fs/Es)*(h-x)/(d-x) = 0.000543Strain due to stiffening effect of concrete between cracks " e2 " =

e2 = b.(h-x)2/(3.Es.As.(d-x)) for crack widths of 0.2 mm Usede2 = 1.5.b.(h-x)2/(3.Es.As.(d-x)) for crack widths of 0.1 mm n/a

e2 = 0.000011Average strain for calculation of crack width "em "=e1-e2 = 0.000532Calculated crack width, " w " = 3.acr.m/(1+2.(acr-c)/(h-x))

CALCULATED CRACK WIDTH, 'w' = 0.06 mm

Client Made by Date PageBox Culvert

Location Upper Slab 1Crack Width Calculations to BS8110: 1997/ BS8007:1987 Checked Revision Job No

REINFORCED CONCRETE

Moment Diagram

WALL ANALYSIS

MAX = 278kN.m

MAX = -42.6kN.m

ELEMENT DESIGN to BS 8110:1997 SOLID SLABS

INPUT Location WallDesign moment, M 278.0 kNm/m fcu 40 N/mm c = 1.50

b 1.00 fy 460 N/mm s = 1.05span 3000 mm

Height, h 300 mm Section location SIMPLY SUPPORTED SPANBar 25 mmcover 30 mm to this reinforcement

OUTPUT Wall Compression steel = Noned = 300 - 30 - 25/2 = 257.5 mm .

(3.4.4.4) K' = 0.156 > K = 0.105 ok .(3.4.4.4) z = 257.5 [0.5 + (0.25 - 0.105 /0.9)^ = 222.8 > 0.95d = 244.6 mm(3.4.4.1) As = 278.00E6 /460 /222.8 x 1.05 = 2848 > min As = 390 mm/m

PROVIDE T25 @ 175 = 2805 mm/m .(Eqn 8) fs = 2/3 x 460 x 2848 /2805 /1.00 = 311.3 N/mm(Eqn 7) Tens mod factor = 0.55 + (477 - 311.3) /120 /(0.9 + 4.193) = 0.821(3.4.6.3) Permissible L/d = 20.0 x 0.821 = 16.422. Actual L/d = 3000 /257.5 = 11.650 ok .

. .

Project

CRACK WIDTH CALCULATIONS - FLEXURE -

INPUTfcu= 40 N/mm2fy= 460 N/mm2

Area of reinforcement " As " = 2805 mm2b = 1000 mmh = 300 mmd = 258 mm

Minimum cover to tension reinforcement " CO " = 30 mmMaxmum bar spacing " S " = 175 mm

Bar dia " DIA " = 25 mm " acr " =(((S/2)^2+(CO+DIA/2)^2)^(1/2)-DIA/2) as default or enter other value = 84.8 mm "acr " is distance from the point considered to the surface of the nearest longitudinal bar

Applied service moment " Ms "= 185.3 KNmCALCULATIONS

moduli of elasticity of concrete " Ec" = (1/2)*(20+0.2*fcu) = 14.0 KN/mm2moduli of elasticity of steel " Es " = 200.0 KN/mm2

Modular ratio "a " = (Es/Ec) = 14.29 " r " = As/bd = 0.011depth to neutral axis, "x" = (- . +(( .)2 + 2. .)0.5.d = 109 mm

" Z " = d-(x/3) = 221Reinforcement stress " fs " = Ms/(As*Z) = 299 N/mm2Concrete stress " fc " = (fs*As)/(0.5*b*x) = 15.37 N/mm2

Strain at soffit of concrete beam/slab " e1" = (fs/Es)*(h-x)/(d-x) = 0.001922Strain due to stiffening effect of concrete between cracks " e2 " =

e2 = b.(h-x)2/(3.Es.As.(d-x)) for crack widths of 0.2 mm Usede2 = 1.5.b.(h-x)2/(3.Es.As.(d-x)) for crack widths of 0.1 mm n/a

e2 = 0.000146Average strain for calculation of crack width "em "=e1-e2 = 0.001776Calculated crack width, " w " = 3.acr.m/(1+2.(acr-c)/(h-x))

CALCULATED CRACK WIDTH, 'w' = 0.29 mm

Client Made by Date Page Location WALL CRACK WIDTH

Crack Width Calculations to BS8110: 1997/ BS8007:1987 Checked Revision Job No

REINFORCED CONCRETEBox Culvert

Moment Diagram

BASE SLAB ANALYSIS

MAX = 278kN.m MAX = 278kN.m

MAX = 254kN.m

ELEMENT DESIGN to BS 8110:1997 SOLID SLABS

INPUT Location Base SlabDesign moment, M 254.0 kNm/m fcu 40 N/mm c = 1.50

b 1.00 fy 460 N/mm s = 1.05span 3300 mm

Height, h 300 mm Section location SIMPLY SUPPORTED SPANBar 25 mmcover 30 mm to this reinforcement

OUTPUT Base Slab Compression steel = Noned = 300 - 30 - 25/2 = 257.5 mm .

(3.4.4.4) K' = 0.156 > K = 0.096 ok .(3.4.4.4) z = 257.5 [0.5 + (0.25 - 0.096 /0.9)^ = 226.3 > 0.95d = 244.6 mm(3.4.4.1) As = 254.00E6 /460 /226.3 x 1.05 = 2562 > min As = 390 mm/m

PROVIDE T25 @ 175 = 2805 mm/m .(Eqn 8) fs = 2/3 x 460 x 2562 /2805 /1.00 = 280.1 N/mm(Eqn 7) Tens mod factor = 0.55 + (477 - 280.1) /120 /(0.9 + 3.831) = 0.897(3.4.6.3) Permissible L/d = 20.0 x 0.897 = 17.938. Actual L/d = 3300 /257.5 = 12.816 ok .

Project

CRACK WIDTH CALCULATIONS - FLEXURE -

INPUTfcu= 40 N/mm2fy= 460 N/mm2

Area of reinforcement " As " = 2805 mm2b = 1000 mmh = 300 mmd = 258 mm

Minimum cover to tension reinforcement " CO " = 30 mmMaxmum bar spacing " S " = 175 mm

Bar dia " DIA " = 25 mm " acr " =(((S/2)^2+(CO+DIA/2)^2)^(1/2)-DIA/2) as default or enter other value = 84.8 mm "acr " is distance from the point considered to the surface of the nearest longitudinal bar

Applied service moment " Ms "= 169.3 KNmCALCULATIONS

moduli of elasticity of concrete " Ec" = (1/2)*(20+0.2*fcu) = 14.0 KN/mm2moduli of elasticity of steel " Es " = 200.0 KN/mm2

Modular ratio "a " = (Es/Ec) = 14.29 " r " = As/bd = 0.011depth to neutral axis, "x" = (- . +(( .)2 + 2. .)0.5.d = 109 mm

" Z " = d-(x/3) = 221Reinforcement stress " fs " = Ms/(As*Z) = 273 N/mm2Concrete stress " fc " = (fs*As)/(0.5*b*x) = 14.04 N/mm2

Strain at soffit of concrete beam/slab " e1" = (fs/Es)*(h-x)/(d-x) = 0.001756Strain due to stiffening effect of concrete between cracks " e2 " =

e2 = b.(h-x)2/(3.Es.As.(d-x)) for crack widths of 0.2 mm Usede2 = 1.5.b.(h-x)2/(3.Es.As.(d-x)) for crack widths of 0.1 mm n/a

e2 = 0.000146Average strain for calculation of crack width "em "=e1-e2 = 0.001610Calculated crack width, " w " = 3.acr.m/(1+2.(acr-c)/(h-x))

CALCULATED CRACK WIDTH, 'w' = 0.26 mm

Location BASE CRACK WIDTHCrack Width Calculations to BS8110: 1997/ BS8007:1987 Checked Revision Job No

REINFORCED CONCRETEBox Culvert Client Made by Date Page

1-Mid Span Slab Calc.pdf (p.1)2-Upper Slab Support.pdf (p.2)3-Upper Slab Crack Width.pdf (p.3)4-Wall Calc.pdf (p.4)5-Wall Crack Width.pdf (p.5)6-Base Slab Calc.pdf (p.6)7-Base Slab Crack Width.pdf (p.7)