From: Subject: ACES PSC Module - Shear Design Date: Sun, 12 Jul 2009 22:04:15 +0930 MIME-Version: 1.0 Content-Type: text/html; charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable Content-Location: file://C:\ACES6\Tempdata\PSC Module-Page-Data.htm X-MimeOLE: Produced By Microsoft MimeOLE V6.00.2900.3028 ACES PSC Module - Shear Design

ACES - PSC Design Module V1.000:   Run=20 date:  23-JUN-09
-------------------------------------------= ------------------------------------------------------
Heading: &= nbsp; Test=20 module
Job Name: AS5100 SUPER-T CLOSED TOP : = SUPERT-CLSD3
Designer:=20  GS

Comments: No comments

Units: =    mm, kN,=20 kN.m,=20 MPa
------------------------------------------------------------------= -------------------------------

DESIGN=20 CODE: AS5100.5-2004

  Test shear loading


SHEAR CHECK

  Ultimate design shear force:
            
  Calculate the default ultimate design shear = force=20 (V*) if the user has not provided one:
     (V* =3D Vult =3D LFsw*Vdsw + = LFslab*Vdslab +=20 LFsdl*Vdsdl + LFll*Vdll)
           
  Design parameters:      
           
  Distance of section from support node (x)

=3D

1200

  mm
           
  Ultimate design shear force (V* =3D Vult)

=3D

1264.0

  kN
  Corresponding ultimate design moment (M* =3D = Mult)

=3D

300.0

  kN.m
           
  28 day girder concrete strength (f'cg)

=3D

50.00

  MPa
  Allowable principal tensile strength (st = =3D=20 0.33*f'cg^0.5)

=3D

2.33

  MPa (AS5100.5-2004 Clause = 8.2.7.2(b))
  Characteristic flexural tensile strength = (f'cf =3D=20 0.6*f'cg^0.5)   

=3D

4.24

  MPa (AS5100.5-2004 Clause = 6.1.1(b))
           
  Overall depth of composite section (D)

=3D

1400

  mm
  Location of girder centroid (Yb)

=3D

605

  mm
  Location of composite centroid (Yc)

=3D

941

  mm
  Width of a single web (Tw)

=3D

100

  mm
  Width of bottom flange (Wbf)

=3D

814

  mm
  Moment of inertia of girder (Ig)

=3D

   1.1710E+11

  mm^4
  Section modulus of girder at bottom flange = (Zb)

=3D

1.9360E+08

  mm^3
  Q for composite girder at centroid (Qna)

=3D

2.3670E+08

  mm^3
           
  Modulus of elasticity of prestressing steel = (Ep)

=3D

200000

  MPa
  Total area of pretensioned bonded strand = (Apt)

=3D

0

  mm^2
  Area of pretensioned girder (Ag)

=3D

585200

  mm^2
           
  Section cracking moment:        
           
  Final prestress force (P)

=3D

4896.1

  kN
  Eccentricity CG girder to CG strand group = (e)

=3D

111.7

  mm
  Ultimate shrinkage strain (us)

=3D

300.0

  microstrain (AS5100.5-2004 Figure = 6.1.7)
           
  Ratio of total strand area to girder area = (r =3D=20 (Ast+Ap)/Ag)   

=3D

0.018

   
           
  Shrinkage tensile stress at extreme fibre - = uncracked=20 section (fcs):
       fcs =3D = 1.5*r*Ep*us*10^-6/(1 +=20 50*r)

=3D

0.859

  MPa
           
  Section cracking moment (Mcr):      
       Mcr =3D Zgb*10^-6(f'cf - = fcs +=20 1000*P/Ag) + P*e/1000

=3D

3739

  kN.m
           
  Flexure shear cracking:=20   (AS5100.5-2004 8.2.7.2(a))
           
  Compression stress due to PS at girder = bottom=20 (scpf)        
       scpf =3D 1000*P/Ag +=20 1000*P*e*Yb/Ig

=3D

11.19

  MPa
  Decompression moment (Mo =3D scpf*Zgb/10^6)

=3D

3040.1

  kN.m
           
  Shear force corresponding to the decompression = moment:        
       Vo =3D Mo*Vult/Mult

=3D

12808.9

  kN
           
  Distance from extreme compression fibre to = the=20 centroid of the outermost layer of tensile R/F or tendons = (do):
       (Default value for do =3D = D -=20 Ybar(min))

=3D

1350.0

  mm
  Minimum allowable value of 'do' (doMin =3D = 0.8D)

=3D

1120.0

  mm
  Sum of widths of both webs in section (Bw =3D = 2*Tw)

=3D

200.0

  mm
           
  Shear coefficients:      
       =DF1 =3D 1.1*(1.6 -=20 do/1000):   If =DF1<1.1 then =DF1 =3D 1.1

=3D

1.10

   
  Shear coefficient =DF2:

=3D

1.00

  (For Super-Ts: AS5100.5-2004 Clause=20 8.2.7.1) 
  Shear coefficient =DF3:        
    =20   =DF3=3D2*do*1000/x:  If =DF3<1 then = =DF3=3D1:  =20 If =DF3>2 then =DF3=3D2   

=3D

1.00

   
           
  Flexural shear capacity (V'uc):        
     V'uc =3D=20 =DF1*=DF2*=DF3*Tw*do*((Ast+Ap)*f'cg/(Tw*do))^1/3/1000 =   

=3D

235

  kN
  Ultimate flexural shear capacity of concrete alone = (Vucs):        
     Vucs =3D V'uc + Vo

=3D

13044

  kN
           
  Web shear cracking at the centroid of the = composite=20 section:   (AS5100.5-2004 8.2.7.2(b))
           
  Since there is no bending stress at the = centroid the=20 ultimate shear capacity of the concrete
  can be determined directly.
           
  Flexural stress at composite centroid, uncracked = sectn=20 (sc):   

=3D

-7.65

  MPa
    sc =3D -P*1000/Ag + P*e*1000*(Yc - = Yb)/Ic        
           
  First moment of area of composite section above = composite=20 centroid & about composite centroid:
       Qlbwc =3D = Qna/(Ic*Bw)

=3D

4.6280E-06

  mm^-2
  Shear stress 'tau' (tc =3D (st^2 - st*sc)^0.5 =3D = 0 if sc <=20 st)

=3D

4.83

  MPa
           
  Principal tensile stress due to design prestress = (s1c):        
       s1c =3D ((0.5*sc)^2 + = tc*tc)^0.5 +=20 0.5*sc

=3D

2.33

  MPa
  Allowable principal tensile strength (st = =3D=20 0.33*f'cg^0.5)

=3D

2.33

  MPa
           
  Concrete ultimate shear capacity assuming = web-cracking=20 (Vtc)        
       Vtc =3D = tc/(Qlbwc*1000)

=3D

1043

  kN
           
  Corresponding moment capacity (Mtc =3D Mult*Vtc = /Vult)

=3D

247

  kN.m
  Section cracking moment (Mcr)

=3D

3739

  kN.m
           
  If Mtc > Mcr then section is cracked; otherwise = it is=20 ------------>  

OK

   
    and the ultimate shear capacity of = concrete=20 alone (Vucc =3D Vtc)   

=3D

1043

  kN
           
  Web shear cracking at the bottom of the = web-flange=20 interface:    
           
  The solution for the ultimate concrete shear = capacity=20 assuming web-cracking (Vtf) is based on
  the method proposed by RF Warner, BV Rangan, = AS Hall=20 and KA Faulkes in their text book
  "Concrete Structures", (Addison Wesley = Longman, Australia, page 331, ISBN 0582 802474)
           
  Given that:        
           
       e =3D Eccentricity (CG = girder - CG=20 strand group)

=3D

111.7

  mm
       y2 =3D Dist from = web/flange=20 joint to comp centroid (Yc-Dwf)   

=3D

618

  mm
       y3 =3D Dist from = web/flange=20 join to centroid of gird (Yb-Dwf)  

=3D

282

  mm
       Dwf =3D Dist from = bottom of=20 girder to web/flange interface

=3D

323

  mm
       Bw =3D Sum of widths of = both=20 webs

=3D

200

  mm
       Ag =3D Area of precast = girder

=3D

585200

  mm^2
       Ig =3D Moment of Inertia = of precast=20 girder

=3D

1.1710E+11

  mm^4
       Ic =3D Composite moment = of=20 inertia

=3D

2.5570E+11

  mm^4
       Qfw =3D Shear flow = constant at=20 flange/web junction

=3D

1.9840E+08

  mm^3
           
       P =3D Final prestress = force

=3D

4896

  kN
       Mcf =3D Corresponding ult = moment=20 factor (=3D Mult/Vult)

=3D

0.24

   
           
  and if:        
       sf =3D Flexural stress = web-flange=20 interface (uncracked sectn)          
       st =3D Allowable = principal=20 tensile strength        
       tf =3D Ult shear = stress=20 capacity of concrete on its own (tau)        
           
  The relationships in the solution process = are:        
   
       sf =3D (-P/Ag) - = (P*e*y3/Ig) +=20 Vtf*Mcf*y2/Ic ..................... (1) Flexural stress at = web-flange=20 interface
       st =3D = 0.33SQRT(f'cg) =3D=20 SQRT((sf/2)^2 + tf^2) + sf/2 ....... (2) Tensile stress = capacity
       Vtf =3D tf*Ic*Bw/Qfw = ......................................................... (3) = Shear force=20 capacity
           
  Assuming:        
       s1 =3D - P*1000/Ag -=20 P*e*y3*1000/Ig

=3D

-9.68

  MPa
       s2 =3D y2*10^6/Ic

=3D

2.4180E-03

 
       st =3D = 0.33SQRT(f'cg)

=3D

2.33

  MPa
       Qlbwf =3D = Qfw/(Ic*Bw)

=3D

3.8800E-06

  mm^-2
           
  Equations 1 - 3 can be combined to form a = quadratic=20 equation where the only unknown is Vtf viz:
       (Qlbwf*Vtf)^2 +=20 (st*Mcf*s2)*Vtf + (st*s1 - st^2) =3D 0 ........ (4)
           
  Setting:        
       a =3D = Qlbwf*Qlbwf*10^6

=3D

1.5060E-05

   
       b =3D st*Mcf*s2

=3D

1.3390E-03

 
       c =3D st*s1 - st^2

=3D

-28.04

   
           
  Expression (4) reduces to:        
       a*Vtf^2 + b*Vtf + c =3D=20 0     and        
       Vtf =3D (-b +- SQRT(b*b - = 4*a*c))/2*a ................... (5)        
           
  The determinant of equation (5) is:      
       f =3D b*b - 4*a*c = =3D=20 (st*Mcf*s2)^2 - 4*Qlbwf^2*(st*s1 - st^2)  

=3D

1.6910E-03

   
   
  If f > 0 then a solution exists and:  

 
       Vtf =3D (-b + = SQRT(f))/2*a

=3D

1320.9

  kN
       Corresponding moment=20 capacity  (Mtf =3D Mult*Vtf/Vult)

=3D

313.5

  kN.m
           
  NOTE:   If f < 0 then Mtf is set to = 999999=20 and Vtf to 0.0          
           
  If f > 0 flexural stress at web-flange=20 interface, uncracked sectn:        
       sf =3D=20 -P*1000/Ag-P*e*1000*y3/Ig+Vtf*Mcf*y2*10^6/Ic

=3D

-8.93

  MPa
           
  If f > 0 ultimate shear stress capacity of = concrete on=20 its own (tau):        
       tf =3D = 1000*Vtf*Qlbwf

=3D

5.13

  MPa
           
  If f > 0 principal tension stress due to design = prestress (s1f):        
       s1f =3D ((0.5*sf)^2 + = tf*tf)^0.5 +=20 0.5*sf

=3D

2.33

  MPa
  Allowable principal tensile strength (st = =3D=20 0.33*f'cg^0.5)

=3D

2.33

  MPa
           
  If Mtf > Mcr then the section is cracked; = otherwise it=20 is --------->  

   
      and the ultimate shear capacity = of=20 concrete alone (Vucf =3D Vtf)

=3D

1321

  kN
           
  Design of web shear reinforcement:=20   (AS5100.5-2004 Section 8.2.8 - 8.2.10)
           
  Ultimate shear capacity of concrete alone, = ignoring=20 shear reinforcement (Vuc) is given by:
      Vuc =3D Minimum of = Vucs,=20 Vucc, Vucf   

=3D

1043

  kN
  Ultimate design shear force (V* =3D Vult)

=3D

1264

  kN
           
  Ultimate shear strength of girder provided = with=20 minimum amount of shear R/F (=D8Vmin):
      =D8Vumin =3D 0.7*(Vuc +=20 0.6*Bw*do/1000)

=3D

843

  kN
           
  Ultimate shear strength of girder limited by = web=20 crushing failure (=D8Vmax):
      =D8Vumax =3D=20 0.7*0.2*f'cg*Bw*do/1000

=3D

1890

  kN
           
  Yield strength of shear reinforcement (fsy.f)

=3D

  MPa
  Diameter of shear reinforcement (Dsr)

=3D

16

  mm
  Area of both legs of shear stirrup (Asv =3D=20 2*3.14*Dsr^2/4)

=3D

401.9

  mm^2
           
  OVumin > Vult: Concrete web shear = capacity is=20 sufficient  

   
    Shear steel required

=3D

N/A

   
    Angle between inclined conc = comp strut=20 & long. axis (=D8v)

=3D

0.0

  Degrees (prescribed minimum)
    Required area of shear = reinforcmt=20 (Asvsr1 =3D 0.35*Bw/fsyf)

=3D

0.0

  mm^2 per mm
    Actual shear reinforcement = spacing (=3D=20 maximum allowed)   

=3D

300.0

  mm
           
  OVumin < Vult: Concrete web shear = capacity is=20 insufficient:  

  Asv Reqd

   
    Required contribution of the = steel=20 reiforcement to the shear capacity of the section (Vus):
        Vus =3D (Vult = -=20 0.7*Vuc)/0.7

=3D

763

  kN
    Angle between inclined conc = comp strut=20 & long. axis (=D8v1)   

=3D

36.0

  Degrees
        =D8v2 = =3D 30 +=20 15*(Vult - =D8Vumin)/ (=D8Vumax - =D8Vumin)

=3D

36.0

   
        If =D8v1 < = 30 then=20 =D8v1 =3D 30        
        If =D8v1 > = 45 then=20 =D8v1 =3D 45 otherwise =D8v1 =3D =D8v2        
    Required area of of shear = reinforcement=20 per unit length along beam (Asvsr2):
        Asvsr2 = =3D=20 ABS(Vus*1000*TAN(=D8v1)/(do*fsyf)))

=3D

8.9

  mm^2 per mm
    Required shear reinforcement = spacing=20 (Svr2 =3D Asv/Asvsr2)  

=3D

45

  mm
    Maximum allowable shear = reinforcement=20 spacing   

=3D

300

  mm
           
  Shear Reinforcement:        
    Actual shear stirrup spacing (Sv)

=3D

300

  mm
    Actual steel area per unit length = (Asvs)

=3D

8.86

  mm^2 per mm