ACES - PSC Design Module V{VERSION}:   Run date:  {DATE}
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Job Name: {JOBNAME}
Designer:  {DESIGNER}

Units:    mm, kN, kN.m, MPa

Design Code:   {CODE\$} (Section 8.2) {DEC 0}
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SECTION:   {Sectnum}

 Distance (x) of section from the first node  = {x}  mm Strand segment number:  {SectSSeg} Passive R/F segment number:  {SectPSeg}

SHEAR & TORSION DESIGN

 Design parameters: Design torsion (T*) = {Tdesign} kN.m Design shear force  (V*) = {Vdesign} kN Design moment  (M*) = {Mdesign} kN.m Design axial force (N*) = {Ndesign} kN.m Final design prestress force  (P) = {P} kN 28 day girder concrete strength (f'c) = {f'cg} MPa Yield stress of shear & torsion reinforcement (fsy.f) = {fsysr} MPa Young’s modulus of concrete  (Ec) = {Eg} MPa Young’s modulus of prestressing strand  (Ep) = {Ep} MPa Young’s modulus of steel reinforcement  (Es) = {Ep} MPa {DEC 1} Capacity reduction factor for shear (Øs) = {Os} Capacity reduction factor for torsion (Øt) = {Ot} {DEC 0} Check if torsional effects are to be considered:   (Clause 8.2.1.2) Torsional effects are to be considered if  T* > 0.25ØtTcr Eq 8.2.1.2(1) where the torsional cracking moment (Tcr) is given by: Tcr = 0.33*(f`c^0.5)*(Acp^2/uc)*SQR((1+scp/(.33f’c^2))) = {Tcr} kN.m  (Eq 8.2.1.2(2)) perimeter of outside cross-section resisting torsion  (uc) = {uc} mm total area enclosed by outer perimeter of concrete section  (Acp) = {Acp} mm^2  {DEC 1} effective prestress at the centroid of the concrete cross-section  (scp) = {Scp} MPa  {DEC 0} therefore  Tcrmin = 0.25ØtTcr = {Tcrmin} kN.m and the design torsion (T*) = {Tdesign} kN.m {TorChck\$} Check torsional requirements for cellular structures: This check requires that  Acp^2/pc  <= 2Ao*bv (Eq 8.2.1.2(3)) where area enclosed by outermost line of closed stirrups  (Aoh) = {Aoh} mm^2 area enclosed by the shear flow path  (Ao = 0.85Aoh) = {Ao} mm^2  (Cl. 8.2.5.6) outside perimeter of the concrete cross-section (pc = uc) = {uc} mm effective width of both webs for shear (bv) = {bv} mm Set  a1 = Acp^2/pc = {a1} and a2 = 2Ao*bv = {a2} {TrCheck\$} {DEC 0} Equivalent shear (V*eq) due to combined shear + torsion: Equivalent shear is given by  V*eq = V* + T*ds/2Ao = {Veq} kN  (Eq 8.2.1.2(5)) Area enclosed by the shear flow path  (Ao) = {Ao} mm^2 Dist. from extreme compn fibre to centroid of non-PS tensile RF (Dcfcnp) = {Dcfcnp} mm Longitudinal strain εx for shear at mid-depth of section:   (Eqn 8.2.4.4(1)) εx = [(M*/dv)+SQR(V*^2+(0.9T*uh/2Ao)^2)+0.5N* – Apt*fpo]/2(EsAst+EpApt) Effective shear depth dv is the greater of .72D and .9d    (dv) = {dv} mm Dist. from extreme compn fibre to centroid of all steel in tension  (d) = {dcts} mm Overall depth of composite section                                   (D) = {D} mm Stress in PS strands when surrounding conc. stress = 0   (fpo) = {fpo} MPa Perimeter of centreline of closed transverse torsion RF  (uh) = {uh} mm Area of tendons in tensile zone at ultimate load              (Apt) = {Atens} mm^2 Area of reinforcement in tensile zone at ultimate load    (Ast) = {Ast} mm^2 ({PasNote\$}) {DEC 6} Set  ex1 = εx  = ex = {ex1} {DEC 0} If εx < 0 then recalculate εx with Ec*Act in denominator where: (Eq 8.2.4.4(2)) Act = area of concrete from mid-depth on flexural tensile side = {Act} mm^2  {DEC 6} Set  ex2 = εx    (If ex2 > 0 set ex2 = 0) = {ex2} and εx = ex = minimum of ex1 and ex2 = {ex} {DEC 0} Check if  M* >= dv*SQR(V*^2 + (0.9T*uh/2Ao)^2) (Eq 8.2.4.4(3)) Where the design moment  M* = {Mdesign} kN.m and  dv*SQR(V*^2 + (0.9T*uh/2Ao)^2)   =  Mexmin = {Mexmin} kN.m   {DEC 0} {MexChck\$} Contribution of concrete (Vuc) to the ultimate shear strength:   (Eqn 8.2.4.1) Vuc = kv*SQR(f’c)*bv*dv = {Vuc} kN   {DEC 3} where  kv = 0.4/(1+1500*εx) = {kv} (Eq 8.2.4.2(4)) {DEC 1} Angle of inclination of concrete compression strut  (θv) (Eq 8.2.4.2(5)) θv = 29 + 7000*εx = {Ov} Deg. {DEC 0} Calculate requirements for transverse shear reinforcement: Reinforcement must be provided if  V* > Vucap (Eq 8.2.1.6) where the design shear force  V* = {Vdesign} kN and  Vucap = 0.5*Øs*(Vuc + Pv)      (Note: for straight strands  Pv = 0) = {Vucap} kN   {DEC 2} Øs = capacity reduction factor for shear = {Os} {DEC 0} {ShearRF\$} Shear force (Vus) carried by transverse reinforcement:   (Clause 8.2.5.2) where Vus = (Asv*fsy.f*dv/s)*cot(θv) = {Vus} kN  (Eq 8.2.5.2(1) and dv = effective shear depth = {dv} mm s = centre to centre spacing of shear reinforcement = {s} mm Asv = area of steel stirrups provided by {nVbars} / {Dsr}  diam. bars = {Asv} mm^2  {DEC 1} θv = angle of inclination of concrete compression strut = {Ov} Deg.     {DEC 0} fsy.f  = yield stress of shear & torsion reinforcement = {fsysr} MPa Minimum required transverse shear  reinforcement:   (Clause 8.2.1.7) Asvmin = 0.08*SQR(f’c)*bv*s / fsy.f = {Asvmin} mm^2 where bv = combined width of both webs = {bv} mm f'c  = 28 day girder concrete strength = {f'cg} MPa Asv = area of steel stirrups provided = {Asv} mm^2 {MinShRF\$} Design shear strength (Vu) of section:   (Clause 8.2.3.1) Design shear strength is given by  Vu = Vuc + Vus + Pv  (and Pv=0) = {Vu} kN Factored shear strength  =   Øt*Vu = {phiVu} kN Equivalent shear due to combined shear & torsion (V*eq) = {Veq} kN {VuNote1\$} Check if ultimate shear strength (Vu.max) is limited by web crushing: Web crushing should not occur before the ultimate shear strength (Vu.max) is reached: where Vu.max = 0.55*f’c*bv*dv[(cot(θv)+cot(av))/(1+cot(θv)^2)] + Pv = {Vumax} kN  (Eq 8.2.3.3(1)) av = angle of inclined transverse RF to the longitudinal RF = 90° (Note:  Pv = 0) and the design shear strength Vu = {Vu} kN {VuNote2\$} Check web crushing due to combined shear and torsion:    (Clause 8.2.4.5) Minimum wall thickness (twmin) required to avoid crushing from combined shear & torsion twmin = Aoh/uh = {twmin} mm  (Eq 8.2.4.5(1)) where Aoh = area enclosed by outermost line of closed stirrups = {Aoh} mm^2 uh = perimeter of centreline of closed transverse torsion RF = {uh} mm and  tw = thickness of walls = 2bw = {tw} mm  {DEC 2} {WebCrsh\$} (Eq 8.2.4.5(2)) If twmin > tw then  V*/(bv*dv) + T*/(1.7*tw*Aoh)  <=  Øt*Vumax/(bv*dv) (Eqn 8.2.4.5(2)) where  V*/(bv*dv) + T*/(1.7*tw*Aoh) = Svt = {Svt} N/mm^2 and      Øt*Vumax/(bv*dv)  =  Svmax = {Svmax} N/mm^2 {Crush\$} {DEC 0} Torsional resistance (Tus) of section: Tus = 2*Ao*(Asw*fsy.f/s)*cot(θv) = {Tus} kN.m  (Eq 8.2.5.6) where Asw = cross sectional area of bars forming the closed fitment = {Asw} mm^2 Ao = area enclosed by the shear flow path = {Ao} mm^2 and the design torsion (T*) = {Tdesign} kN.m {TusNote\$} Minimum required torsional reinforcement:    (Clause 8.2.5.5) Torsional strength of section Tus must be > than the cracking torque (Eq 8.2.5.5(1) where Tus = {Tus} and the torsional cracking moment (Tcr) = {Tcr} kN.m  {DEC 2} {TcrNote\$} Must also satisfy the requirement that:   Asw/s  >=  0.2*y1/fsy.f (Eq 8.2.5.5(2) where  Asw/s = {ReoRatio} 0.2*y1/fsy.f = fsyRatio = {fsyRatio} {DEC 0} Asw = cross sectional area of bars forming the closed fitment = {Asw} mm^2 s = centre to centre spacing of shear reinforcement = {s} mm y1 = larger overall dimension of the closed filament = {y1} mm {ReoRati\$} Required transverse reinforcement for combined shear and torsion:   (Clause 8.2.5.3) Area of steel required for shear  (rearrange Eqn 8.2.5.2(1)) Asv =  (V* - ØsVuc)*s / (fsy.f*dv*cot(θv))  (for two webs) = {Aswsv} mm^2 Area of steel required for torsion  (rearrange Eqn 8.2.5.6) Asw =  T*s / (2*Ao*fsy.f*cot(θv)) = {Aswst} mm^2 Total area of transverse steel required Asvtotl= 0.5Asvr + Aswr = {Asvtotl} mm^2 Minimum required area of transverse steel =  Asvmin = {Asvmin} mm^2 {Asvmin\$} Determine required longitudinal reinforcement on flexural tension side   (Clause 8.2.7) Additional longitudinal force due to shear and tension is given by: ΔFtd = SQR[((V* - ypPv) – (0.5ØsVus))^2 +(0.45uhT*/2Ao)^2] cot(θv) = {deltaFtd} kN  (Pv = 0) Factored additional longitudinal design force = ΔFtd/Øs = ReqdFt = {ReqdFt} kN Area of additional longitudinal bars needed to resist this force = AbarsFt = {AbarsFt} Additional longitudinal force produced by these bars = ΔAsfsy = ActualFt = {ActualFt} kN Factored longitudinal design force = ΔFtd/Øs = ReqdFt must be < ActualFt {AsNote1\$} Determine required longitudinal reinforcement on flexural compression side   (Clause 8.2.8) Additional longitudinal force due to shear and torsion given by: ΔFcd = ΔFtd – F*c =  incFcd = {incFcd} kN  (Pv = 0) where F*c = design force in comp. zone due to flexure & axial actions = {Fc} kN and   ΔFcd = {Fclong} kN {AsNote2\$} Area of additional longitudinal bars required to resist ΔFcd = AbarsFc = {AbarsFc} Additional longitudinal force produced by these bars = As*fsy = {ActualFc} kN {AsNote3\$}