Examples of water retaining structure design

Ir. Tu Yong Engyongengtu@yahoo.com

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

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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

Thank you

Ir. Tu Yong Engyongengtu@yahoo.com