Lecture 8 RC Water Retaining Example1
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Transcript of Lecture 8 RC Water Retaining Example1
Circular tank
• Hydrostatic loading– Horizontally: Circumferential tension– Vertically: Moment depends on the end condition– Depends on radius and height ratio.
Rectangular tank
• Hydrostatic pressure– Horizontally – moment, shear and direct tension– Vertically – moment and shear– Sensitive to the end condition (pin, free or fixed)– Depends on the width, length and height ratio.
EX 12.2
• Information given:– Wall fully restraint at both end– Direct tension 265 kN/m– Maximum Crack width 0.15mm– fyk = 500 Mpa– C30/37
• Determine sectional thickness & reinforcement detail
ULS
• T = 265 x 1.2 = 318 kN/m• As required = T/(0.87 x fyk) = 731 mm2/m
Thickness calculation
• As,minσs = kc k fct,eff Act• fct,eff is the mean value of the tensile strength of
the concrete effective at the time when the cracks may first be expected to occur:
• fct,eff = fctm or lower, (fctm(t)), if cracking is expected earlier than 28 days
• C30/37 fctm = 2.9 Mpa• Remark: For pure tension, concrete is not
contributing to the resistance, hence, thinner section will have less crack.
Thickness calculation
• k is the coefficient which allows for the effect of non-uniform self-equilibrating
• stresses, which lead to a reduction of restraint forces
• = 1,0 for webs with h ≤ 300 mm or flanges with widths less than 300 mm
• = 0,65 for webs with h ≥ 800 mm or flanges with widths greater than 800 mm (0.75 C660)
• intermediate values may be interpolated
Capacity• βcc(t) = exp{s[1-(28/t)(1/2) }• βcc(t) is a coefficient which depends on the age of the
concrete t• t is the age of the concrete in days• s is a coefficient which depends on the type of cement:• = 0,20 for cement of strength Classes CEM 42,5 R, CEM
52,5 N and CEM 52,5 R (Class R)• = 0,25 for cement of strength Classes CEM 32,5 R, CEM
42,5 N (Class N)• = 0,38 for cement of strength Classes CEM 32,5 N
(Class S)
EX 12.2
• fctm = 2.9 MPa (C30/37)• fct, eff = 1.73 MPa (3 days)• Check the minimum steel content based on 3
day strength (text book) or 28 days (more critical)?
• A s,min = 0.00346 Act (3 days)• A s,min = 0.0058 Act (28 days)• Try thickness of 150 mm
Minimum steel content
• Minimum steel content– 519 mm2/m (based on 3 days strength)– 817 mm2/m (based on 28 days strength)
Thermal and Shrinkage crack
• Properties• Modular ratio αe = 7• Steel Provided H25 – 100 (4908 mm2/m)• Act = 150 x 1000 = 150 000• ρ = 0.0327• Es = 200 GPa• c = 63mm
EX 12.2
• Restraint both end (Annex M, expression M1)
= 0.00027Crack spacing = 474mmCrack width = 0.13mm
EX 12.2
• If thickness changes to 100mm• Steel Provided H20 – 100 (3141 mm2/m)• Act = 100 x 1000 = 100 000• ρ = 0.0314• c = 40 mm• Strain = 0.000282• Crack spacing = 352• Crack width = 0.099mm
Crack width calculation
• sr,max = k3c + k1k2k4φ /ρp,eff (7.11)• where:• φ is the bar diameter (refer to 7.12)• c is the cover to the longitudinal reinforcement• k1 is a coefficient which takes account of the bond
properties of the bonded reinforcement:• = 0,8 for high bond bars• = 1,6 for bars with an effectively plain surface
(e.g. prestressing tendons)
Crack width calculation• k2 is a coefficient which takes account of the distribution of strain:• = 0,5 for bending• = 1,0 for pure tension• For cases of eccentric tension or for local areas, intermediate values of k2
should be used which may be calculated from the relation:• k2 = (ε1 + ε2)/2ε1 (7.13)• Where ε1 is the greater and ε2 is the lesser tensile strain at the boundaries
of the section considered, assessed on the basis of a cracked section.• The recommended values for k3 and k4 (Both NDP) are 3,4 and 0,425
respectively.• Where the spacing of the bonded reinforcement exceeds 5(c+φ/2) or
where there is no bonded reinforcement within the tension zone, an upper bound to the crack width may be found by assuming a maximum crack spacing:
• sr,max = 1,3 (h - x) (7.14)
Direct tension
• Steel stress = 54 MPa• εsm -εcm may be calculated from the
expression:
σs is the stress in the tension reinforcement assuming a cracked section.αe is the ratio Es/Ecm
EX 12.2
• Ap’ and Ac,eff are as defined in 7.3.2 (3)• ξ1 according to Expression (7.5)• kt is a factor dependent on the duration of the
load• kt = 0,6 for short term loading• kt = 0,4 for long term loading
Ac,eff
• Ac,eff is the effective area of concrete in tension surrounding the reinforcement or prestressingtendons of depth, hc,ef , where hc,ef is the lesser of 2,5(h-d), (h-x)/3 or h/2
Ex 12.2
• εsm –εcm = 5.7 x 10 -5
• wk = 474 x 5.7 x 10 -5 = 0.03mm
Ex. 12.3
• Information given:– Rectangular tank supported by beams– H = 2.3m– Slab T1 = 18 K, T2 = 10 K– Width = 6.5m– M (-ve) = 42.26kNm/m– M(+ve) = 42.96 kNm/m– Direct tension = 23.54 kN/m
Thermal and Shrinkage
• Minimum steel content– As = 0.00346 x 1000 x 300 = 1038 mm2/m – (As = 0.0058 x 1000 x 300 = 1740 mm2/m)– 12mm – 200 (566 mm2/m per face)
• Early age strain– Restraint factor = 0.3 – ε imp = RT1α = 65 microstrain (α = 12 με)– Capacity = 75 με– No crack
Ex. 12.3
• Long term• S r, max = 1217 mm • wk = 0.18mm • check the NA location• µ = 0.027 (allow for creep, long term E = 0.5
Ec)• x/d = 0.21• x = 50 mm < 0.2 h (60 mm)
Ex. 12.3
• Crack width allowed = 0.19 mm (hydraulic gradient = 6.7)
• Design crack width• h eff = 83 mm• Ρp,eff = 0.0068• S r, max = 436 mm • wk = 0.23mm • Revised rebar to H12-175 (646mm2/m per face)
CIRIA C660
• Retaining wall on a rigid foundation• Information given:
– Retaining wall 4m high 0.5m thick x 12m long– Rigid foundation 0.8m thick 2.85m wide– Concrete C30/37 with 30% fly ash– Cover 40mm– 18mm thk plywood formwork– Α = 12 με
CIRIA C660
• Pressure gradient = 8 (crack width limitation = 0.18 mm)
• Binder content 365 kg/m3 with 30% fly ash• Estimated Temperature rise T1 = 27 K• Autogeneous shrinkage = 15 με• Restraint at the joint = = 0.62
• Restraint at the top = 0.18 oo
nn
EAEA
1
1
CIRIA C660
• Creep coefficient K1= 0.65• Early age strain = k1 (αT1+ εca) R = 136 με• Tensile capacity = 76 με• Early age crack inducing strain = 136 – 0.5 x 76• = 99 με• Min steel = kckAct fcteff/fky = 780 mm2/m• Kc = 0.9• Reinforcement T16 - 225
CIRIA C660
• S r, max = 1177 mm • wk = 0.12mm • Long term• T2 = 20 K• Autogeneous shrinkage = 33 με (28 days)• Drying shrinkage = 100 με (humidity = 90%)• Tensile strain capacity = 109 με
CIRIA C660
• Total crack inducing strain – = k1 {(αT1+ εca) R1 + αT1 R2+ εcd R3} – 0.5 εctu = 227
με– (R1 = R2 = R3 = 0.62, k1 = 0.65)
• wk = 0.27mm
Expansion joint BS 8007
Complete Contraction joint BS 8007
Complete Contraction joint BS 8007
Partial Contraction joint BS 8007
Partial Contraction joint BS 8007
