# Notes 1 this clause does not apply to foundation or

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NOTES:1This clause does not apply to foundation or pavement slabs fully or partially supported onsoil.2This requirement may be waived at critical sections of a statically indeterminate member,provided it can be demonstrated this will not lead to sudden collapse of a span or a reducedcollapse load.Accessed by UNIVERSITY OF QUEENSLAND on 25 Jul 2019 (Document currency not guaranteed when printed)
AS 3600:2018218Standards Australia16.4.4Strength of beams in shear16.4.4.1Design shear strength of a beamThe design shear strength of a beam shall be taken asϕVu, where—Vu=Vuc+Vuf+Vus. . . 16.4.4.1whereVucandVusare determined from Clause 8.2.3 andVufis determined from eitherClause 16.4.4.2.1 or Clause 16.4.4.2.2.Notwithstanding Equation 16.4.4.1, the fibres component to the ultimate shear strength of abeam,Vuf, shall not exceed the greater ofVucand that determined by Clause 16.4.4.3 withVustaken as zero.16.4.4.2Contribution to shear strength by steel fibres16.4.4.2.1Design by refined calculationThe contribution of the fibres to the ultimate shear strength (Vuf) of an SFRC beam shall becalculated from the following equation:ufSgvvwvcotVK k d b f. . . 16.4.4.2.1(1)whereKS=fibre orientation casting bias factor and is taken as 0.64kg=member size factor determined by Equation 16.4.2, where area of concretewithin the tensile zone (mm2) at ultimate is calculated asActu=bvdvcotvwf= the characteristic residual tensile strength of SFRC corresponding CODwv= angle between the axis of the concrete compression strut and the longitudinalaxis of the member calculated in accordance with Clause 8.2.4.2dv= the effective shear depth of the member calculated in accordance withClause 8.2.1.9The characteristic residual tensile strengthwfshall be determined in accordance with eitherClauses 16.3.3.4, 16.3.3.5 or 16.3.3.6 where the crack width is determined as—dgvx100010.210000.125 mm1300coskdw. . . 16.4.4.2.1(2)wherexis determined from Clause 8.2.4.3.Alternatively for beams less than 1000 mm in depth,wfin Equation 16.4.4.2.1(1) may betaken as equal to1.5f.The concrete and steel components shall be determined in accordance with Clauses 8.2.3withkvdetermined from Clause 8.2.4.2.NOTE: The componentsVuc,VusandVufare coupled through the strain parameterxand theequations are solved iteratively.16.4.4.2.2Design by simplified calculationFor non-prestressed components not subjected to axial tension, and provided—(a)the specified yield strength of the longitudinal reinforcement does not exceed500 MPa;(b)the design concrete strength does not exceed 65 MPa;(c)the size of maximum aggregate particle is not less than 10 mm;(d)the length of the fibres does not exceed 70 mm; andAccessed by UNIVERSITY OF QUEENSLAND on 25 Jul 2019 (Document currency not guaranteed when printed)
219AS 3600:2018Standards Australia(e)the depth of the beam does not exceed 1000 mm.

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University of Queensland