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

Comments: {COMMENT1}

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$}