November1998 63

igh-performance high-strength concrete (HPHSC)is defined as concrete that meets special perfor-mance and uniformity requirements which cannotalways be achieved by using only the conventional

materials and normal mixing, placing, and curing practices.1

The performance requirements may include ease of place-ment and compaction without segregation, enhanced short-term and long-term mechanical properties, high-early-agestrength and long life in severe environments.

HPHSC is usually proportioned with a low water-to-cementitiousmaterials ratio and has a high compressivestrength in the range of 50 to 100 MPa (7 to 15 ksi). The sup-plementary cementitious materials may include blast furnaceslag, fly ash, or silica fume, which are used either as cementreplacement or as additives to the concrete mixture.

Considerable information is already available in the litera-ture regarding HPHSC.2-9 Several codes have design propos-als to include HPHSC.10-13 The aim of this paper is to reviewthe existing information and to suggest certain proposals forthe design of HPHSC beams, columns, and walls. These pro-posals cover concrete grades in the range of 20 to 100 MPa(3 to 15 ksi).

Material properties

Mix proportions and curingThe mix proportions of HPHSC vary depending on the avail-ability of local materials. General guidelines on proportion-ing of HPHSC are available.2,3 It is, however, necessary thatthe designers should work closely with the local ready-mixplants to produce the optimum mix required for their project.A great deal of cooperation and team effort between the en-gineer, owner, contractor, and concrete producer are essen-tial for the successful application of HPHSC.

Curing conditions have a significant influence on the prop-erties of HPHSC. If HPHSC is allowed to dry out beforecompletion of curing there is a degradation of its properties.An adequate curing regime for at least seven days is there-fore recommended.

Modulus of elasticity and Poisson’s ratioExtensive data on the modulus of elasticity Ec of HPHSC arereported elsewhere.1-5 Measured values show good correla-tion with the expression:

Ec = (3320√√ f'c + 6900) (ρρ/2300)1.5 Eq. 1

where f 'c is the cylinder compressive strength in MPa and ρis the density of concrete in kg/m3 . Eq. 1 is incorporated inthe Canadian and New Zealand Standards.11,12 In the absenceof measured values Eq. 1 is therefore recommended. Limiteddata on Poisson’s ratio of HPHSC indicate that a suitablevalue may be taken as 0.2.

Tensile strengthThe tensile strength test data usually show a large scatter. Anumber of equations have been proposed in the literature.The following expressions given in the codes14,15 may beused to obtain lower bound estimates:

Principal tensile strength = 0.4 √√ f'c

Flexural tensile strength = 0.6 √√ f'c Eq. 2

where f 'c is expressed in terms of MPa.To allow for restrained shrinkage and temperature effects

in members with large surface area such as slabs and walls,the flexural tensile strength may be taken as 0.3√f 'c as rec-ommended by the Canadian Standard.11

Shrinkage and creepLimited data available in the literature show that shrinkageof HPHSC is similar to other concretes. Therefore, the de-sign data given in the codes may be used to estimate the finalshrinkage strain of HPHSC.13,14

Limited data indicate that the creep coefficient (i.e., the ra-tio of creep strain to elastic strain) of HPHSC is significantlysmaller than that of other concretes. More data are urgentlyneeded to formulate design rules.

Durability and specificationA primary consideration in the utilization of HPHSC is thedurability of concrete to resist aggressive environments. Theminimum compressive strength and minimum cover require-ments specified in codes alone are not sufficient to producedurable HPHSC mixes. Many other requirements, such asminimum and maximum cement content, maximum water-cement ratio, aggregate quality, shrinkage limit, curing re-quirements, water permeability, and resistance againstchemical attack, should be included in the formulation of aspecification for HPHSC mixes. Field measurement of spec-ified properties is essential to ensure that the required perfor-mance criteria are met.

H

High-Performance High-Strength Concrete: Design Recommendations

by B. Vijaya Rangan

64 ConcreteInternational

Fire performanceTest data show that the fire performance of HPHSC is signif-icantly different to that of other concretes especially betweenroom temperature and about 450 C (840 F).16,17 The currentcode provisions regarding fire protection may not be appli-cable in the case of HPHSC.

It has been pointed out that HPHSC is more susceptible toexplosive spalling failure and compressive strength losswhen exposed to temperatures above 300 C (570 F) than oth-er concretes. Insufficient information is available and moreresearch is urgently needed.

HPHSC beams

Flexural strengthThe flexural strength of a beam is customarily calculated byassuming a linear strain distribution over the depth of thesection and considering the equilibrium of forces and mo-ment.18 To apply this procedure, two factors require atten-tion. First, the ultimate concrete compressive strain εcu atwhich extreme compression face of the member reaches fail-ure should be known. Secondly, the actual distribution ofcompressive stresses in the concrete should be defined.

Although the ultimate compressive strain varies with con-crete strengths, a value equal to 0.003 represents the test re-sults satisfactorily. The scatter of test data does not justify avariation of the strain with the concrete compressivestrength.2 This value is specified in several codes.12,14,15 Inthe Canadian Standard,11 εcu is taken as 0.0035.

In codes and standards, the actual distribution of compres-sive stresses in the concrete is replaced by an equivalent rect-angular stress block. The stress block has a uniform stress of0.85 f 'c and a depth less than the neutral axis depth. It is gen-erally accepted that the uniform stress should be smaller than0.85 f 'c for HPHSC. In the New Zealand Standard,12 the depthof equivalent rectangular stress block is taken as γ times thedepth of neutral axis and the uniform stress is taken as α f 'cwhere:

γγ = 0.85 −− 0.008 ( f'c −− 30) Eq. 3

within the limits 0.65 ≤ γ ≤ 0.85 and

αα = 0.85 −− 0.004 ( f 'c −− 55) Eq. 4

within the limits 0.75 ≤ α ≤ 0.85.Note that α = 0.85 when f 'c ≤ 55 MPa (8 ksi) and α = 0.75

when f 'c ≥ 80 MPa (12 ksi).In the Canadian Standard,11 α and γ are given by the fol-

lowing:

αα = (0.85 −− 0.0015 f 'c ) ≥ ≥ 0.67 Eq. 5

γ γ = (0.97 −− 0.0025 f 'c ) ≥ ≥ 0.67 Eq. 6

The minimum values of 0.67 do not apply until f 'c > 125 MPa(18 ksi). In Eq. 3 to 6, f 'c must be substituted in MPa.

Beams designed in practice are under-reinforced and theirflexural strength is controlled by the yield force in the tensilesteel. The values of rectangular stress block parameters willtherefore have insignificant effect on the design calculations.From this viewpoint, any of the above proposals will be suit-able.

Eq. 3 is similar to the expression given in the currentcodes14,15 and Eq. 4 reflects the observed test trend. There-fore Eq. 3 and 4 together with εcu = 0.003 are recommended.

Maximum tensile steel ratioIf we take εcu= 0.003, the depth of neutral axis at balancedfailure dnb is given by:

dnb = [0.003/(0.003 + εεy)]d Eq. 7

where εy is the yield strain of reinforcing steel and d is theeffective depth. When fy = 400 MPa (58 ksi), dnb = 0.60 d andfy = 500 MPa (73 ksi), dnb = 0.55d. To ensure a ductile failure,codes and standards limit the depth of neutral axis dn to a val-ue smaller than dnb. In the Australian Standard,14 dn is limitedto a maximum of 0.4d.

If we take dn = 0.4d, for a reinforced concrete rectangularsection with tensile steel only, the maximum tensile steel ra-tio ρmax is given by:

ρρmax = 0.4 α γα γ f 'c /fy Eq. 8

When f 'c = 80 MPa (12 ksi), from Eq. 3 and 4, γ = 0.65 andα = 0.75. For fy = 400 MPa (58 ksi), from Eq. 8 we get ρmax

= 0.039, or 3.9 percent. In other words, for HPHSC large val-ues of ρmax are possible. In the New Zealand Standard,12 ρmaxis taken as 0.025 when designing for seismic effects.

Minimum tensile steel ratioTo prevent a brittle failure at first cracking, the tensile steelratio should not be less than a minimum. In the CanadianStandard,11 the minimum area of tensile steel Ast min is givenby:

Ast min = 0.2 √√f 'c bt D/fy Eq. 9

where f 'c is the cylinder compressive strength in MPa, bt isthe width of the tension zone of the section under consider-ation and D is the overall depth of beam.

In the New Zealand Standard,12 Ast min is given by:

Ast min/bw d = ρρmin = (0.25 √√f 'c )/fy Eq. 10

where bw is the width of the web.Both Eq. 9 and 10 are similar and acceptable. The format

of Eq. 10 follows the current practice14,15 and this expressionis therefore recommended. According to Eq. 10, when f 'c =30 MPa (4 ksi), ρmin = 1.4/fy and when f 'c = 80 MPa (12 ksi),ρmin = 2.2/fy.

Shear strengthThe shear design provisions contained in the AustralianStandard14 can be used for HPHSC beams. Accordingly, theultimate shear strength Vn of a reinforced concrete beam withvertical stirrups and carrying no axial force is given by:

Vn = (Vc + Vs) ≤≤ Vmax Eq. 11

where:

Vc = ββ1 bvdo (Ast f 'c /bv do)1/3 Eq. 12

β β1 = 1.1 (1.6 −− do /1000) ≥≥ 1.1 Eq. 13

here, f'c is the cylinder compressive strength in MPa, bv is theeffective width of web, do is the distance from the extremecompression fiber of the concrete to the centroid of the out-ermost layer of tensile reinforcement, and Ast is the cross-sectional area of longitudinal reinforcement in the tensionzone and fully anchored at the cross section under consider-ation. Eq. 12 may be suitably modified for prestressed con-crete beams or when there is axial force.14

The shear strength is limited to a maximum value:

Vmax = 0.2 f 'c bv do Eq. 14

November1998 65

Also, the shear force Vs resisted by vertical stirrups is giv-en by:

Vs = Asv fy (do /s) cot θθ Eq. 15

where Asv is the area of vertical stirrups, s is the spacing ofstirrups, and θ is the angle between the axis of the concretecompression strut and the longitudinal axis of the beam, tak-en to vary linearly from 30 deg when Vu = φVmin to 45 degwhen Vu = φVmax .

The area of minimum shear reinforcement may be takenas: 11

Asv.min = 0.06√√ f'c sbv /fy Eq. 16

Then, the shear strength of a beam that contains this mini-mum shear reinforcement, designated as Vmin , is given by:

Vmin = Vc + 0.10√√ f'c bv do Eq. 17

where f 'c is expressed in terms of MPa. The design require-ment is:

Vu ≤≤ φφVn Eq. 18

where Vu is the design (factored) shear force and φ is thestrength reduction factor.

The shear strength calculated by the above design provi-sions has shown good correlation with the test strength of147 HPHSC beams.19 The mean value of test/calculatedstrength was 1.22 with a coefficient of variation of 36 per-cent. The large coefficient of variation (see Fig. 1) was dueto the considerable scatter in measured shear strength of testbeams.19

HPHSC columns

Pure axial load capacityThe axial load capacity Po under concentric compres-sion is usually given by:

Po = 0.85 f 'c (Ag −− As) + fy As Eq. 19

where Ag is the gross concrete area and As is the area oflongitudinal steel.

Well-confined HPHSC columns may reach strengthin excess of that predicted by Eq. 19. Extensive testdata reported by Cusson and Paultre20 show that Eq. 19may be acceptable provided that the column cross sec-tion contains at least eight evenly distributed longitu-dinal bars.

In the New Zealand Standard,12 Po is taken as:

Po = αα f'c (Ag −− As) + fy As Eq. 20

where α is given by Eq. 4.In Eq. 20, the factor α replaces 0.85 in Eq. 19 to ac-

count for the observed reduction in the strength ofHPHSC columns provided with nominal quantity oftransverse reinforcement. A similar but more conser-vative approach is adopted by the Canadian Stan-dard.11 Eq. 20 together with Eq. 4 is recommended.Note that when f 'c ≤ 55 MPa (8 ksi), α = 0.85 and Eq.20 becomes Eq. 19.

Reinforcement detailsTest results available in the literature2,5,18 show that the nom-inal spacing of closed ties as specified in codes and standardsmay not be adequate for HPHSC columns. In the CanadianStandard,11 the tie spacing in HPHSC columns is reduced by25 percent when f 'c > 50 MPa (7 ksi). Accordingly, the spac-ing of closed ties or the pitch of a helix used as lateral rein-forcement in HPHSC columns should not exceed the smallerof 0.75 times the least lateral dimension of the cross sectionof the member or 12 times the smallest longitudinal bar di-ameter. The closed ties must be provided with 135 deghooks.

Other reinforcement details of HPHSC columns may be inaccordance with the current practice given in the codes andstandards.

Columns in combined bending and compressionThe strength of a HPHSC column subjected to combined ax-ial compression and bending moment when there is tensionover part of the cross section is calculated by assuming a lin-ear strain distribution over the depth of the section and con-sidering the equilibrium of forces and moment. Similar to thecase of beams, the failure concrete strain εcu is taken as 0.003and the compression zone concrete is represented by theequivalent rectangular stress block as defined by the param-eters γ and α given by Eq. 3 and 4. The strength interactiondiagram for a HPHSC column is then obtained in the usualmanner.18

The equivalent rectangular stress block concept may bevalid up to the point on the strength interaction diagramwhen the depth of neutral axis dn is equal to the depth of ex-treme layer of tensile steel do measured from the compres-sion face. For dn > do, the rectangular stress block concept isnot applicable. This part of the interaction diagram is usuallyapproximated by a straight line between pure axial load ca-pacity (0, Po) and the point corresponding to dn = do.18

Fig. 1 — Correlation of shear strength of HPHSC beams predicted by the Australian Standard.19

Predicted shear capacity (kN)

Test

she

ar c

apac

ity

(kN

)

66 ConcreteInternational

Slender columnsSlender HPHSC columns may be proportioned using the mo-ment magnifier method given in the codes and standards.However, the empirical expressions given in the codes andstandards invariably result in very conservative designs es-pecially for HPHSC columns. The procedure presented else-where is therefore recommended.9 This procedure has showngood correlation with 143 test column results, the mean val-ue of test/calculated failure load was 1.08 with a coefficientof variation of 12 percent.21

According to this procedure, if Pu is the factored axial loadat an equivalent eccentricity e then the co-existing magnifiedfactored moment Me is given by:

Me = Pu(e + ∆∆y + ∆∆cp) Eq. 21

In Eq. 21, the deflection ∆y at failure may be approximatedby the following:• For Pu ≥ φPb

∆ ∆y = ∆∆yb ( φφ Po −− Pu)/( φφPo −− φφPb) Eq. 22

• For Pu ≤ φ Pb

∆ ∆y = ∆∆yo + (∆∆yb −− ∆∆yo) (Pu /φφPb) Eq. 23

where:

∆∆yb = (0.003 + εεy) Le2 /ππ2do Eq. 24

∆∆yo = 1.6 εε y Le2 /ππ2do Eq. 25

Pb is the particular axial load strength at balanced failureconditions, φ is the strength reduction factor, Po is the axialload capacity under concentric compression given by Eq. 20,Le is the column effective length, εy and fy are, respectively,the yield strain and the yield strength of reinforcing steel, anddo is the depth of extreme layer of tensile steel measuredfrom the compression face. In Eq. 21, the equivalent eccen-tricity e may be taken as:

e = km M2 /Pu Eq. 26

where M2 is the value of the larger factored end moment, km

is given by:

km = (0.6 −− 0.4 M1/M2) ≥≥ 0.4 Eq. 27

and M 1 is the smaller factored end moment. The ratio M1/M2

is less than or equal to unity and is taken as negative whenthe column is bent in single curvature and positive for doublecurvature.14,15

Also, the creep deflection ∆cp of the column that is treatedas an additional eccentricity in Eq. 21, may be calculated by:

∆∆ cp = ∆∆tot −− ∆∆e Eq. 28

where ∆tot is the total deflection of the column due to sus-tained load and ∆e is its elastic component. These deflectionsare given by the following expressions:22

∆∆tot = e/[(Pc /Pφφ) −− 1] Eq. 29

where: Pc = ππ2 ( EI/L e2) Eq. 30

EI = λλEcIg/(1 + 0.8 φφ c c) Eq. 31

λλ = [0.6 + (eb /8e)] ≤≤ 1.0 Eq. 32

Pφ is the axial thrust due to sustained loads, φcc is the creepcoefficient, and eb is the value of e corresponding to balancedfailure in combined axial compression and bending.

Also, ∆∆ e = e/[(Pc o /P φφ) −− 1] Eq. 33

where Pco = λπλπ2 ΕΕc ΙΙg //Le2 Eq. 34

That is, ∆e is the particular value of ∆tot when φcc = 0.Based on the preceding information, the following steps

are proposed for the design of HPHSC slender columns inbraced frames:• Select a trial cross section for the column. Calculate theeffective length Le of the column using the methods given inthe codes.11-15

• Calculate the eccentricity e by Eq. 26.• Calculate the design strength interaction diagram for thecolumn cross section using the equivalent rectangular stressblock as defined by Eq. 3 and 4.• Calculate ∆cp by Eq. 28, 29, and 33 and ∆y by Eq. 22 or 23.For these values of e, ∆y, and ∆cp, and given value of Pu , cal-culate Me by Eq. 21. Check whether the design strength ofthe column cross section is adequate to resist the combinedeffect of the factored actions Pu and Me.

The preceding design method is illustrated below.

Design exampleFig. 2 is a cross section of a reinforced rectangular HPHSCcolumn. All are Y28 bars (area of each bar = 620 mm2

[1in.2]) used in the Australian practice. The columnis bent about the major axis and must carry an axialcompressive (factored) force Pu = 5000 kN(1124kips) at an equivalent eccentricity e = 95 mm(3.74 in.). The effective length Le = 10 m (33 ft),sustained axial load Pφ = 3000 kN (675 kips), Ec =36,500 MPa (5.3 × 106 psi), and Es = 200 × 103 MPa(29 × 106 psi). Take f 'c = 80 MPa (12 ksi), fy =400MPa (58 ksi), εy = 0.002, creep coefficient φcc =1.5 and strength reduction factor φ = 0.6.

Check the adequacy of the cross section to carrythe design loads.

Eq. 4: α = 0.85 − 0.004(80 − 55) = 0.75Eq. 3: γ = 0.85 − 0.008(80 − 30)

= 0.45 < 0.65, take γ = 0.65Fig. 2 — Example column.

November1998 67

Eq. 20: Po = 0.75 × 80(450 × 600 − 14 × 620) + 14 × 620 × 400

= 19,151 kN (4309 kips)φPo = 0.6 × 19,151 = 11,491 kN (2585 kips)

Balanced failureFrom Fig. 2, the depth of extreme layer of steel from com-pression face do = 540 mm (21.3 in.). For balanced failure,take that the steel in this layer just yields, i.e., ε5 = εy = 0.002.From strain diagram, dn = [0.003/(0.003 + 0.002)](540)

= 324 mm (12.8 in.)In these calculations, compression is taken as positive.ε1 = (0.003/324)(324 − 60) = 0.0024 > εy,

σ1 = 400 MPa (58ksi)ε2 = (0.003/324)(324 − 180) = 0.0013,

σ2 = 267 MPa (39ksi)ε3 = (0.003/324)(324 − 300), σ3 = 44 MPa (6.4 ksi)ε4 = (0.003/324)(324 − 420), σ4 = −178 MPa (−26 ksi)ε5 = −0.002, σ5 = −400 MPa (−58 ksi)From Fig. 2:Cc = 0.75 × 80 × 0.65 × 324 × 450 = 5686 kN (1278 kips)F1 = 4 × 620 × 400 = 992 kN (223 kips)F2 = 2 × 620 × 267 = 331 kN (74 kips)F3 = 2 × 620 × 44 = 55 kN (12 kips)F4 = 2 × 620 × (−178) = −221 kN (−50 kips)F5 = 4 × 620 × (−400) = −992 kN (−223 kips)Equilibrium of forces gives:Pb = 5686 + 992 + 331 + 55 − 221 − 992

= 5851 kN (1316 kips)φPb = 0.6 × 5851 = 3511 kN (790 kips)By summing moments about the plastic centroid:Mb = 5686(300 − 0.65 × 324/2) + 992(300 − 60) +

331(300 − 180) + 55(300 − 300) + (−221)(300 − 420)+ (−992)(300 − 540)

= 1107 + 238 + 40 + 0 + 27 + 238 = 1650 kNm (1218 ft-kips)

φMb = 0.6 × 1650 = 990 kNm (730 ft-kips)eb = 1650/5851 = 282 mm (11.1 in.)

Magnified moment MeEq. 24: ∆yb = (0.003 + 0.002)(10 × 103)2/(π2 × 540)

= 94 mm (3.7 in.)Since Pu > φPb:Eq. 22: ∆y = 94(11,491 − 5000)/(11,491 − 3511)

= 76 mm (3.0 in.)Eq. 32: λ = 0.6 + (282/8 × 95) = 0.97Eq. 31: EI = 0.97 × 36,500 × 1⁄12 × 450 × 6003/1 + (0.8 × 1.5)

= 130,355 × 109 mm4 (313 × 106 in.4)Eq. 30: Pc = π2 × 130,355 × 109/(10,000)2

=12,866 kN (2895 kips)Eq. 29: ∆tot = 95/[(12,866/3000) − 1] = 29 mm (1.1 in.)Eq. 34: Pco = 0.97 × π2 × 36,500 × 1⁄12 × 450 × 6003/(10,000)2

= 28,305 kN (6369 kips)Eq. 33: ∆e = 95/[(28,305/3000) − 1] = 11 mm (0.4 in.)Eq. 28: ∆cp = 29 − 11 = 18 mm (0.7 in.)Eq. 21: Me = 5000 (95 + 76 + 18) = 945 kNm (697 ft-kips)

Adequacy of cross sectionIt is necessary to check whether the cross section is adequateto carry the combined effect of axial thrust Pu = 5000 kN(1125 kips) and magnified moment Me = 945 kNm (697 ft-kips). Note that Me/Pu = 945/5000 = 189 mm (7.4 in.) < eb;therefore the type of failure will be primary compression and

the neutral axis depth dn will be greater than the value calcu-lated for balanced failure.

Another point on the strength interaction diagram is calcu-lated when dn = do = 540 mm (21.2 in.). For this value of dn ,from the strain diagram (Fig. 2):

ε1 = (0.003/540)(540 − 60) = 0.0027, σ1 = 400 MPa (58 ksi)ε2 = (0.003/540)(540 − 180) = 0.002, σ2 = 400 MPa (58 ksi)ε3 = (0.003/540)(540 − 300) = 0.00133, σ3 = 267 MPa (39 ksi)ε4 = (0.003/540)(540 − 420) = 0.00067, σ4 = 133 MPa

(19 ksi)ε5 = 0, σ5 = 0 From Fig. 2:Cc = 0.75 × 80 × 0.65 × 540 × 450 = 9477 kN (2132 kips)F1 = 4 × 620 × 400 = 992 kN (223 kips)F2 = 2 × 620 × 400 = 496 kN (112 kips)F3 = 2 × 620 × 267 = 331 kN (74 kips)F4 = 2 × 620 × 133 = 165 kN (37 kips)F5 = 0Equilibrium of forces gives:Pn = 9477 + 992 + 496 + 331 + 165 + 0 = 11,461 kN (2579 kips)φPn = 0.6 × 11,461 = 6877 kN (1547 kips)By summing moments about the plastic centroid: Mn = 9477(300 − 0.65 × 540/2) + 992 (300 − 60)

+ 496 (300 − 180) + 331(300 − 300) + 165 (300 − 420) + 0 (300 − 540)

= 1180 + 238 + 60 + 0 − 20 − 0 = 1458 kNm (1076 ft-kips)φMn = 0.6 × 1458 = 875 kNm (646 ft-kips)The strength interaction diagram between this point (φMn

= 875, φPn = 6877) and the balanced failure point (φMb =990, φPb = 3511) may be conservatively approximated as astraight line.

The equation of this straight line is given by: (φMn − 875) / (990 − 875) = (φPn − 6877) / (3511 − 6877) or φMn = 875 + (6877 − φPn )/ 29.3For this example column, if we substitute φPn = Pu =

5000kN (1125 kips) in the above expression φMn =939 kNm (693ft-kips) which is close enough to Me =945 kNm (697ft-kips). Therefore the column cross sectionshown in Fig. 2 is acceptable.

HPHSC walls

Flexure and shear strengthsThe flexural strength of HPHSC walls may be calculated bythe usual theory of reinforced concrete sections subjected tocombined bending moment and axial compression (see pre-vious section on HPHSC columns).

In an earlier paper, strength equations for the shear designof structural walls made of all grades of concrete were devel-oped.23 The predictions from the equations correlated wellwith test results; the mean of test/calculated shear strengthswas 1.09 with a coefficient of variation of 12 percent.

Accordingly, the shear strength Vn of a wall is given by:

Vn = twdw ( ρρ l fly + Nu /Ag) tan θ ≤θ ≤ Vmax Eq. 35

where tw is the wall thickness, dw is the horizontal length ofwall between centers of end elements, Lw is the length ofwall, ρ l = Al /tw Lw , Al is the area of vertical steel in wall onboth faces in length Lw , fly is the yield strength of verticalsteel, Nu is the design ultimate compressive load on wall, andAg is the gross concrete area of wall cross section. In Eq. 35,θ is the inclination of concrete strut (failure plane) to the lon-gitudinal axis given by:

68 ConcreteInternational

tan θθ = dw /Hw Eq. 36

within the limits 30 deg ≤ θ ≤ 60 deg, where Hw is the heightof wall.

To ensure yielding of vertical steel, the shear strength islimited to a maximum of Vmax given by:

Vmax = k3 f 'c tw dw sin θθ cos θθ /(1.14 + 0.68 cot2 θθ) Eq. 37

where k3 is the reduction factor to relate cylinder strength toin situ concrete strength and is given by:8

k3 = 0.6 + (10/f 'c ) ≤ ≤ 0.85 Eq. 38

In addition to vertical steel, the wall must also contain hor-izontal steel. For adequate control of cracking due to re-strained shrinkage and temperature effects, the minimumvalue of horizontal steel ratio is taken as 1.4/fsy, where fsy isthe yield strength of horizontal steel. This value is recom-mended by the Australian Standard14 for restrained slabswhen moderate degree of control over cracking is required.

It is also necessary to ensure that the vertical steel ratio ρl

is not less than 0.0025 for crack control purposes. Further-more, the spacing of bars should not exceed the lesser of2.5tw or 500 mm (20 in.).14,15

The preceding design equations were illustrated by an ex-ample in Ref. 23.

ConclusionsBased on this brief state-of-the-art report of the design ofhigh-performance high-strength concrete structural mem-bers, the following suggestions for HPHSC with compres-sive strength in the range of 20 to 100 MPa (3 to 15 ksi) aremade:

1. The modulus of elasticity of HPHSC may be estimatedusing Eq. 1.

2. Eq. 2 provides a lower bound of tensile strength ofHPHSC.

3. The final shrinkage strain of HPHSC may be estimatedusing the design data given in the codes.13,14

4. The final creep coefficient of HPHSC is significantlysmaller than that of other concretes.

5. The flexural strength of HPHSC beams and columnsmay be calculated by taking the ultimate concrete compres-sive strain εcu as 0.003 and by using an equivalent rectangu-lar stress block for the compression zone concrete. The depthof the stress block is taken as γ times the neutral axis depthand the uniform stress is α f 'c , where γ and α are given by Eq.3 and 4.

6. The maximum and the minimum tensile steel ratios ofHPHSC beams may be taken as given by Eq. 8 and 10.

7. The shear design provisions given in the AustralianStandard14 may be used for HPHSC beams provided that theminimum area of shear reinforcement is taken as given byEq. 16.

8. The pure axial load capacity of HPHSC columns may becalculated by Eq. 20 where α is given by Eq. 4.

9. Slender HPHSC columns may be designed using theprocedure described in the paper.

10. The shear strength of HPHSC walls may be calculatedby Eq. 35.

References1. Russell, H. G., “High-Performance Concrete — From Buildings to

Bridges,” Concrete International, V. 19, No. 8, August 1997, pp. 62-63.2. ACI Committee 363, “State-of-the-Art Report on High-Strength Con-

crete,” ACI Journal, V. 81, No. 4, July-August 1984, pp. 364 - 411.3. FIP-CEB, “High-Strength Concrete: State of the Report,” FIP, London,

1990.4. Choy, R. S., “High-Strength Concrete,” Technical Report No. TR/F112,

Cement & Concrete Association of Australia, Sydney, May 1988.5. “High-Strength Concrete,” Cement & Concrete Association of Austra-

lia, June 1992, 31 pp.6. Lloyd, N. A., and Rangan, B. V., “High-Strength Concrete: A Review,”

Research Report No. 1/93, School of Civil Engineering, Curtin Universityof Technology, Perth, January 1993, 132 pp.

7. ACI-ASCE Committee 441, “High-Strength Concrete Columns: Stateof the Art,” ACI Structural Journal, V. 94, No. 3, May-June 1997, pp. 323-335.

8. Collins, M. P.; Mitchell, D.; and MacGregor, J. G., “Structural DesignConsiderations for High-Strength Concrete,” Concrete International, V. 15,No. 5, May 1993, pp. 27 - 34.

9. Rangan, B. V., “Applications of High-Strength Concrete (HSC),” Chap-ter 7, Large Concrete Buildings, Longman, 1996, pp. 158-182.

10. Norwegian Standards, “Concrete Structures, Design Rules,” NS3473,1989.

11. Standards Council of Canada, “Design of Concrete Structures forBuildings,” CAN3-A23.3-M94, Canadian Standards Association, Rexdale(Toronto), Canada, December 1994, 199 pp.

12. Standards Association of New Zealand, “Concrete Structures NZS3101 - Part 1: Design,” 1995.

13. Eurocode No. 2, “Design of Concrete Structures. Part 1: GeneralRules and Rules for Buildings,” Commission of the European Communi-ties, ENV 1992 - 1.1, December 1991, 253 pp.

14. “Australian Standard for Concrete Structures, AS 3600,” StandardsAustralia, Sydney, 1994, 155 pp.

15. ACI Committee 318, “Building Code Requirements for StructuralConcrete (ACI 318-95) and Commentary (ACI 318R-95),” American Con-crete Institute, Farmington Hills, 1995, 369 pp.

16. Phan, L. T., “Fire Performance of High-Strength Concrete: A Reportof the State-of-the-Art,” Report No. NISTIR-5934, National Institute ofStandards and Technology, December 1996, 105 pp.

17. Phan, L. T.; Carino, N. J.; Duthinh, D.; and Garboczi, E. (Editors), In-ternational Workshop on Fire Performance of High-Strength Concrete,NIST Special Publication 919, National Institute of Standards and Technol-ogy, September 1997, 164 pp.

18. Warner, R. F.; Rangan, B. V.; Hall, A. S.; and Faulkes, K. A., ConcreteStructures, Addison Wesley Longman, Melbourne, 1998, 975 pp.

19. Kong, P. Y. L., and Rangan, B. V., “Reinforced High-Strength Con-crete (HSC) Beams in Shear,” Australian Civil/Structural EngineeringTransactions, V. CE 39, No. 1, 1997, pp. 43-50.

20. Cusson, D., and Paultre, P., “High-Strength Concrete Columns Con-fined by Rectangular Ties,” ASCE Journal of Structural Engineering, V.120, No. 3, March 1994, pp. 783-804.

21. Basappa Setty, R. H., and Rangan, B. V., “Failure Load of High-Strength Concrete (HSC) Columns under Eccentric Compression,” Austra-lian Civil/Structural Engineering Transactions, V. CE39, No. 1, September1996, pp. 19-30.

22. Rangan, B. V., “Strength of Reinforced Concrete Slender Columns,”ACI Structural Journal, V. 87, No. 1, January-February 1990, pp. 32-38.

23. Rangan, B. V., “Rational Design of Structural Walls,” Concrete Inter-national , V. 19, No. 11, November 1997, pp. 29-33.

Received and reviewed under Institute publication policies.

ACI Fellow B. Vijaya Rangan is Professor and Head of Civil Engineering, Curtin University of Technology, Perth, Western Australia. He is an associate member of several ACI Committees, including 435, Deflection of Concrete Building Structures; and Joint ACI-ASCE Committees 441, Reinforced Concrete

Columns, and 445, Shear and Torsion. He is also a member of Standards Australia Committee on Concrete Structures.