Kuwait Oil Company Facility Upgrade and Relocation of Under Ground Process Project: JI-180 Doc No. Design of Cantilever retaining wall Rev No. AS PER BS 8110 Calculation Of Crack Width Crack width for drying shrinkage /thermal movement : Maximum allowable crack width = 0.3 mm (Per BS 8110-2 : 1985 clause 3.2. fck = Characteristic strength of reinforced concrete = 30 fy = Characteristic strength of reinforcing steel as per table 3.1 of BS = 414 ( Refer equation 14 of clause 3.8.4.2 of BS 8110-2) R = 0.6 ( Per Table 3.3 of BS 8110-2) = coefficient of thermal expansion of mature concrete = 0.000012 Table 7.3 of BS 8110 ( Part 2 ) = Fall in temperatue between hydration peak and ambient ( per Table 3. = 20 0.0001152 Design surface crack width, W1 where acr = Dist from point considered to the surface of the nearest long bar acr acr = 103.3578405 mm f = Size of each reinforcing bar = 20 mm D = Depth of wall = B = 400 mm S = Spacing of reinforcement = 150 mm c/c Ast = Area of steel = 2093.3333 W1 = 0.030728432 mm ( 1 ) Crack width in flexure : Maximum allowable crack width = 0.3 mm (for severe exposure) Crack width, ( w ) = Where, acr = Dist from point considered to the surface of the nearest l acr = 103.3578 mm = average strain at the level where the cracking is being co = = 0.001039 = bt*(h-x)*(a'-x)/3*Es*As*(d-x) for crack width of 0.3mm = 0.000367 Moment in section, M = 129 kNm ( Unfactored moment) Stress in steel, fs = M/(As*(d-S'/2)) N/mm 2 N/mm 2 Thermal strain er = 0.8*Dt*a*R a DT Thermal strain er = 3*acr*er/(1+2*((acr-cmin)/(h-x))) =sqrt((S/2)^2+(c+f/2)^2)-(f/2) mm 2 3*acr*em/(1+2*((acr-cmin)/(h-x))) =sqrt((S/2)^2+(c+f/2)^2)-(f/2) em e1 -e2 e2 acr
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Sheet1Kuwait Oil CompanyFacility Upgrade and Relocation of Under
Ground ProcessProject:JI-180Doc No.Design of Cantilever retaining
wallRev No.AS PER BS 8110Calculation Of Crack WidthCrack width for
drying shrinkage /thermal movement :Maximum allowable crack
width=0.3mm (Per BS 8110-2 : 1985 clause 3.2.4)fck=Characteristic
strength of reinforced concrete=30N/mm2fy=Characteristic strength
of reinforcing steel as per table 3.1 of BS 8110=414N/mm2Thermal
strain er =0.8*Dt*a*R( Refer equation 14 of clause 3.8.4.2 of BS
8110-2)R =0.6( Per Table 3.3 of BS 8110-2)a=coefficient of thermal
expansion of mature concrete=0.000012Table 7.3 of BS 8110 ( Part 2
)DT=Fall in temperatue between hydration peak and ambient ( per
Table 3.2 of BS 8110-2)=20Thermal strain er =0.0001152Design
surface crack width, W1
=3*acr*er/(1+2*((acr-cmin)/(h-x)))whereacr=Dist from point
considered to the surface of the nearest long
baracr=sqrt((S/2)^2+(c+f/2)^2)-(f/2)acr =103.3578404875mmf=Size of
each reinforcing bar=20mmD=Depth of wall = B=400mmS=Spacing of
reinforcement=150mm c/cAst=Area of steel=2093.3333333333mm2W1
=0.0307284324mm( 1 )Crack width in flexure :Maximum allowable crack
width=0.3mm (for severe exposure)Crack width, ( w
)=3*acr*em/(1+2*((acr-cmin)/(h-x)))Where,acr=Dist from point
considered to the surface of the nearest long
baracr=sqrt((S/2)^2+(c+f/2)^2)-(f/2)=103.3578404875mmem=average
strain at the level where the cracking is being considered=e1
-e2=0.001039169e2=bt*(h-x)*(a'-x)/3*Es*As*(d-x)for crack width of
0.3mm=0.0003674112Moment in section, M=129kNm( Unfactored
moment)Stress in steel, fs=M/(As*(d-S'/2))Stress in steel,
fs=212.8228106159N/mm2OKe1=fs/Es*((h-x)/(d-x))=0.0014065802h=Overall
depth of member = D=400mmbt=Width of section at the centroid of
tensile steel=1000mmEs=Modulus of elasticity of
reinforcement=200000N/mm2As=Area of tensile
reinforcement=2093.3333333333mm2a'=distance from the compression
face to the point at which thecrack width is being
calculated=Overall depth ( D )=400mmd=Effective depth=D - cmin -
f=315mmcmin=Minimum cover to tensile steel=75mmNow as per clause
3.4.4.4 of BS 8110 ( Part - I )k=Mu / fck bd20.060670194
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