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PRESTRESSED CONCRETE GIRDER BRIDGE Kevin S Benítez C February 2019 Keywords: prestressed concrete; torsion; shear; box girder bridges This report will present the design of a cast-in-place, post-tensioned concrete, multi-cell box girder bridge under combined torsion, shear and flexure, based on the geometry of an example from the California Department of Transportation [1] Figure shows the elevation view of the bridge used for this example The total length of the three spans is 412’ (125.6 m): the first span is 126’ (38.4 m), the center span is 168’ (51.2 m), and the third span is 118’ (36.0 m) Figure 1––Elevation view of the bridge Figure presents the cross-section of the bridge The total width of the bridge is 58’-10’’ (17.93 m) The bridge carries three 12’ (3.7 m) traffic lanes, two 10’ (3.0 m) shoulders and two 1’-5’’ (0.45 m) concrete edge barriers Figure 2––Cross-section of the bridge The loads that act on the bridge include the self-weight, load of the asphalt concrete (A.C.) wearing surface with a thickness of in (75 mm), and the live load in accordance with AASHTO LRFD 2017 [3] HL-93 (design truck plus design lane load) Bending Moment, shear and torsional moment envelopes for bridge example Figure 3––Bending moment diagrams for separate load cases Conversion: kip-ft = 1.356 kNm, ft = 0.31 m Figure 4––Shear diagrams for separate load cases Conversion: kip = 4.45 kN, ft = 0.31 m Figure 5––Torsional moment diagrams for separate load cases Conversion: kip-ft = 1.356 kNm, ft = 0.31 m Materials Concrete Initial and 28-day concrete strength that will be used for the design of the bridge f ci '  3500 psi f c '  5000 psi (24.1 MPa) (34.5 MPa)  c = 0.15 kcf (23.6 kN/m3) Modulus of Elasticity of concrete Ec  33, 000 c1.5 f c ' Ec  4287 ksi (29.6 GPa) Reinforcing steel f y  60 ksi (414 MPa) Es  29, 000 ksi (200 GPa) Prestressing steel f pu  270 ksi (1862 MPa) f py  243 ksi (1675 MPa) f pj  202.5 ksi (1396 MPa) Design computations for flexure according to AASHTO LRFD 2017 [3] The following steps will be used for the design of the example bridge in flexure: Step Determine the tendon profile The tendon profile should be chosen so that the points of maximum eccentricities will occur at the points of maximum moments The tendon profile should include the radius of the parabolas, points of inflexion along the cross section and distances from the C.G of the cross section to points of maximum eccentricities Figure 6––Prestressing tendon profile along bridge length Step Determine the balancing forces and balancing bending moments The balancing bending moments for each one of the sections of the cable path is determined Pj f qP  l First parabola: (From ft to 110 ft) 8Pj f 1kip  3.05 ft qP1    2.02 103 kip ft l 110 ft  Second parabola: (From 110 ft to 148 ft) 8Pj f 1kip  0.72 ft qP    4.00 103 kip ft l  38 ft  Third parabola: (From 148 ft to 272 ft) 8Pj f 1kip  3.16 ft qP3    1.64 103 kip ft l 124 ft  Fourth parabola: (From 272 ft to 308 ft) 8Pj f 1kip  0.66 ft qP    4.10 103 kip ft l  36 ft  Fifth parabola: (From 308 ft to 412 ft) 8Pj f 1kip  3.11 ft qP5    2.30 103 kip ft l 104 ft  These forces should be distributed over the whole width of the bridge, for this calculation the overhanging flanges are not taken into account and a mean value of the width of 50.5 ft, between the upper and lower flanges is used, to count for the skewed exterior girders qP1  4.00 105 kip ft 5 kip qP  7.90 10 ft qP  3.25 105 kip ft 5 kip qP  8.10 10 ft qP  4.55 105 kip ft Figure 7––Balancing moments from prestressing Conversion: kip-ft = 1.356 kNm, ft = 0.31 m Step Calculate the prestressing losses according with Section 5.9.3 Calculation of Friction Losses: For the calculation of friction losses, the friction coefficient is calculated based on the following equation:  Fl  e (    kx ) µ = 0.25 and k = 0.0002 The angular friction coefficient αFl is calculated taking into account different sections of the parabolic profile of the tendon, in this example this sections are taken based on the parabolic changes of the profile Then the following equation is used for computing the angle θ  yij lij Then the cumulative coefficient is calculated adding the values of angular coefficients obtained for each section Table––1, presents the cumulative angular force coefficients for Two-End stressing Table 1––Cumulative angular force coefficients for two-end stressing LOCATION ft - 50 ft 50 ft – 110 ft 110 ft – 126 ft 126 ft – 148 ft 148 ft – 210 ft 210 ft – 272 ft 272 ft – 294 ft 294 ft – 308 ft 308 ft – 365 ft 365 ft – 412 ft LEFT END STRESSING 0,966 0,924 0.906 0.880 0.843 0.806 0.780 0.764 0.721 RIGHT END STRESSING 0.965 0.922 0.905 0.880 0.843 0.805 0.780 0.761 0.720 Losses due to Elastic shortening: The elastic shortening is computed based on a value of ΔfpES = ksi As used by the California Department of Transportation [1] It is then converted into a force coefficient: FC pES  f pES f ps  3ksi  0.015 202.5ksi Losses due to Anchor set: For the losses due to Anchor set the following value is used: FC pA  0.097 Time-dependant losses: Concrete Shrinkage and Creep To calculate the losses due to concrete shrinkage, the strain due to shrinkage is calculated using the following equation:  sh  ks khs k f ktd 0.48 103  4.54 104 And the losses due to shrinkage are calculated with: f pSH   sh  E p  kdf  13.585ksi In the case of creep, the creep coefficient is calculated as follows:  (t, ti )  1.9ks khc k f ktd ti0.118  1.213 And the losses due to creep are calculated with: f pSH  Ep Ec  0.66  (t , ti )  kdf  6.264ksi As an assumption f pR  2.4ksi The total time-dependant losses converted into a force coefficient are: fTl  f pSH  f pCR  f pR f ps  0.11 Finally the total losses of prestressing considering: friction, anchor set, elasting shortening and time dependant losses is computed with the following equation:   fi FC  1   f ps     To simplify the example, the losses will be turned into a percentage of the initial prestressing force So finally a total loss of 25% of the total load will be used for calculations Step Check stresses at Service Flexure (considering just DC and PT) Service flexure will just consider the effects of the Dead Load (DC) and prestressing force (PT), that are the force acting as soon as the bridge is terminated, it not includes the effects produced by the wearing surface (DW) and the Live Load (LL) The allowable stresses are the following: Table 2––Allowable stresses at Service Flexure Force Compression Tension Stress limit 0.60 f’ci = 2.1(ksi) 14.5(MPa) No tension allowed The stresses will be computed at different positions along the bridge length: Table 3––Calculated stresses at Service Flexure Length (ft) 50 110 126 148 210 272 294 308 365 412 Stresses at top of section (ksi) 0.643 0.468 1.675 0.892 1.214 0.632 1.155 0.939 1.752 0.388 0.643 Length (ft) 50 110 126 148 210 272 294 308 365 412 Stresses at bottom of section (ksi) 0.643 0.868 0.689 0.964 0.686 0.657 0.724 1.024 0.718 0.971 0.643 Step Check the stresses at Service Limit State (SLS III) before losses, allowing the stress limit provided in Section 5.9.2.3.1a for compression and Table 5.9.2.3.1b-1 for tension Service Limit State (SLS III) uses the following load combination DC + DW +0.8LL, considering the whole effects of the load at normal operational service Here the effects of prestressing losses will not be considered: The allowable stresses are the following: Table 4––Allowable stresses at Service III (without losses) Force Compression Tension Stress limit 0.60 f’c = 3.0(ksi) 30.7(MPa) 0.19 f 'c  0.425ksi 2.9 (MPa) Table 5––Calculated stresses at Service III (without losses) Length (ft) 50 110 126 148 210 272 294 308 365 412 Stresses at top of section (ksi) 0.643 0.725 0.833 0.752 0.750 0.951 0.708 0.691 0.820 0.620 0.643 Length (ft) 50 110 126 148 210 272 294 308 365 412 Stresses at bottom of section (ksi) 0.643 0.538 0.399 0.502 0.506 0.246 0.560 0.581 0.415 0.673 0.643 Step Check stresses at Service Limit State (SLS III) after losses allowing the stress limit provided in Table 5.9.2.3.2a-1 for compression and Table 5.9.2.3.2b-1 for tension Service Limit State (SLS III) uses the following load combination DC + DW +0.8LL, considering the whole effects of the load at normal operational service The effects of prestressing losses are considered in this case, previously a 25% losses was calculated, hence this value will be used to reduce the total jacking force Pj The allowable stresses are the following: Table 6––Allowable stresses at Service III (with Losses) Force Compression Tension Stress limit 0.60 f’c = 3.0(ksi) 30.7(MPa) 0.19 f 'c  0.425ksi 2.9 (MPa) Table 7––Allowable stresses at Service III (with Losses) Length (ft) 50 110 126 148 210 272 294 308 365 412 Stresses at top of section (ksi) 0.643 0.887 0.975 1.093 0.830 1.129 0.784 1.028 0.976 0.780 0.643 Length (ft) 50 110 126 148 210 272 294 308 365 412 Stresses at bottom of section (ksi) 0.643 0.329 0.215 0.064 0.402 0.017 0.461 0.147 0.215 0.467 0.643 Design computations according to ACI 318-14 [2] Step 1: Determine the factored bending moment, shear force, and torsional moment at the face of Bent of the box-girder based on the load combinations of Article 5.3.1 M u   56804.2kip  ft (77016kN  m) at the face of Bent M u   44485.1kip  ft (60314kN  m) at a distance d of the face of Bent Vu  3035kip (13500.5kN ) at a distance d of the face of Bent Tu  12108kip  ft (16416.5kN  m) at a distance d of the face of Bent Step 2: Determine the section properties and material properties Table 8––Section properties (ACI318-14) 44637 in2 1275 in 13684 in2 64.8 in 77in 60 in 41710 in2 1250 in 0.80 0.28 Acp pcp Ag d ds bw Aoh ph β1 γp 2.9×107 mm2 32385 mm 8.8×106 mm2 1646 mm 1956 mm 1524 mm 2.7×107mm2 31750 mm Step 3: Check the flexural resistance for the tendon layout shown in Fig 6, based on the factored bending moment produced at the face of Bent a   Aps f ps  d p    2  Pj 9300kips Aps    46in f pj 202.5ksi Mu p  f pu   p f ps  f pu 1    p    252.9ksi fc '   1 Aps Ag a  0.00336 Aps  f pu 0.85 fc ' b  5.64in 5.64in   M n  46in  252.9ksi  64.8in    59982kip  ft  81325kN  m    Mn  Mu 0.90  59982kip  ft  53983kip  ft (73193kN  m) Step 4: Compute the additional mild steel required for flexure Mu  a a    Aps f ps  d p    As f y  d   2 2   Assuming : As  30in2 19355mm2  l  As  0.00219 Ag  p   f pu d s f y f ps  f pu 1    p     l    249.9ksi   1  fc ' d p fc '    a 10 Aps  f pu  As  f y 0.85 f c ' b  6.46in For θ = 45° s Av  6.50in 165mm  Atf For θ = 35° s Av  9.25in  235mm  Atf For θ = 22° s Av  12in  305mm  Atf Design computations according to AASHTO LRFD 2017 [3] Step 1: Determine the factored bending moment, shear force, and torsional moment at the face of Bent of the box-girder based on the load combinations M u   60865.2kip  ft (82522kN  m) at the face of Bent M u   47683.3kip  ft (64650kN  m) at a distance d of the face of Bent Vu  3268.3kip (14538.2kN ) at a distance d of the face of Bent Tu  13242.1kip  ft (17954kN  m) at a distance d of the face of Bent Step 2: Determine the section properties and material properties Table 10––Section and material properties AASHTO LRFD 2017 Acp pcp Ag d ds bw Aoh ph β1 k I y Sc γ1 γ2 γ3 SECTION PROPERTIES 44637 in2 2.9×107 mm2 1275 in 32385 mm 13684 in2 8.8×106 mm2 64.8 in 1646 mm 77in 1956 mm 60 in 1524 mm 41710 in2 2.7×107mm2 1250 in 31750 mm 0.80 0.28 1.7 x 107 in4 39 in 4.4 x 105 in3 1.6 1.1 0.75 Step 3: Calculate the cracking moment M cr     f r    fcpe  Sc   54204.4kip  ft (73491.3kN  m) Mr = min(Mcr, 1.33Mu) Mr = 54204.4kip-ft(73491.3kN-m) 22 Step 4: Compute the nominal resistance of the section based on the resistance provided by the prestressing steel Pj = 9300kip (41369kN) Aps  f pu c 0.85 f 'c  1  b  k  Aps  f pu  6.85in(174mm) c  f ps  f pu  1  k   262ksi (1807 MPa) d   d  c   M n  Aps f ps  d    62234kip  ft (84378kN  m)   Step 5: Compute the nominal resistance providing mild steel Assuming As = 40in2 (25806.4mm2) Aps  f pu  As  f y  6.54in(166mm) Aps  f pu 0.85 f 'c  1  b  k  1  d a  a  f ps  f pu  1  k   260.5ksi(1796MPa) 1  d   a a   M n  Aps f pd  d p    As f y  d    76084.4kip  ft (103157kN  m) 2 2   Step 6: Calculate the net longitudinal strain Mn  63.6in(1616mm) As f y  Aps f ps α is the angle of curvature of the tendon at the section calculated, for this case α = 0.026rad d Vp  Pj    241.8kips(1075.6kN ) Veff  Vu  s  Tu  d  3440.9kip(15305.9kN ) Ao Mu  0.5 Nu  Veff  Vp  Aps  f po dv  Es As  E p Aps   0.00071 Step 7: Calculate the values of θ and β 4.8  3.13  750   s   29  3500   s  31.5  Step 8: Determine the transverse reinforcement required for shear Vc  0.0316   f ' c  bw  d  843kips(3750kN ) Vs  Vu Avs  s   Vc  V p  2546.7kip (11328.3kN )  2 Vs  0.41in 10.4 mm in mm f y  d  cot( )  23 Step Check if longitudinal reinforcement can resist the required tension Equation 1: Aps  f ps  As  f y  14362.2kips Equation 2: Mu d  0.5  Veff     Vp  0.5Vs  cot( )  12768.1kips     Nu If Equation > Equation then the provided longitudinal reinforcement is adequate for resisting the required tension Step 10 Determine the transverse reinforcement to be used on exterior girders Vu,ext = 848.70kips Veff ,ext  Vu ,ext  Mu d  s ,ext  ext  Tu  d  1021.3kips(4543kN ) Ao  0.5 N u  Veff , ext  V p  Aps f po Aps   A  Es s  E p  5    0.00105 4.8  3.13  750 s ext  29  3500 s ,ext  32.7 Vc ,ext  0.0316   f 'c  t f  d  168.6kips(750kN ) Vs ,ext  Vu ,ext  Avs ,ext   Vc ,ext  Vp ,ext  726.1kip(3230kN ) Vs ,ext f y  d cot( )  0.122 in 78.7 mm mm in Step 11 Determine the amount of transverse reinforcement required for torsion on webs Tu At   f y  Ao cot( ) cos( )  0.03 in 0.76 mm mm in Step 12 Determine the amount of transverse reinforcement required for torsion on flanges Tu Atf   f y  Ao cot( ) cos( )  0.025 in 0.64 mm mm in Step 12 Determine the amount of longitudinal reinforcement required for torsion Tu Alt    ph  Ao  f y  51.9in  33483.8mm2  24 Step 13 Determine the amount of transverse reinforcement to be used on the exterior webs Atotal ,ext  At  Av ,ext  0.151in 3.84 mm mm in Step 14 Determine the spacing to be used for stirrups Interior webs: #5 (16mm) two legged stirrups will be used with an area Av = 0.61in2 (394mm2) s Av  7.25in(185mm) Avs Exterior webs: #5 (16mm) two legged stirrups will be used with an area Av = 0.61in2 (394mm2) s Av  4in(100mm) Avs ,ext Flanges: #3 (10mm) two legged stirrups will be used with an area Av = 0.22in2 (142mm2) s Av  8.50in(215mm) Atf Design computations according to Model Code MC-2010 [5] Step 1: Determine the factored bending moment, shear force, and torsional moment at the face of Bent of the box-girder based on the load combinations M u   61102.3kip  ft (82844kN  m) at the face of Bent M u   47741.3kip  ft (64715kN  m) at a distance d of the face of Bent Vu  3162.2kip (14066.2kN ) at a distance d of the face of Bent Tu  11354kip  ft (15394kN  m) at a distance d of the face of Bent Step 2: Determine the section properties and material properties Table 10––Section and material properties MC-2010 SECTION PROPERTIES 44637 in2 2.9×107 mm2 1275 in 32385 mm 13684 in2 8.8×106 mm2 64.8 in 1646 mm 77in 1956 mm 60 in 1524 mm 41710 in2 2.7×107mm2 1250 in 31750 mm MATERIAL PROPERTIES λ γc 1.50 γs 1.15 fp,ud 235 ksi 1620.3 Mpa Acp pcp Ag d ds bw Aoh ph 25 fy,d f´c,d fp0.1k fp,d 52.2 ksi 3.33 ksi 243 ksi 211.3 ksi 360 Mpa 23 Mpa 1676 Mpa 1457 Mpa Step 3: Check the flexural resistance for the tendon layout shown in Fig 6, based on the factored bending moment produced at the face of Bent a   Aps f pd  d p   2  Pj 9300kips Aps    46in f pj 202.5ksi Mu  Aps  f pd a  fc ' b p  Aps Ag  0.00336  5.63in 5.63in   M n  46in  211.3ksi  64.8in    50127 kip  ft  67963.1kN  m    Step 4: Compute the additional mild steel required for flexure a a   M n  Aps f pd  d p    As f y  d   2    Assuming : As  40in2  25806.4mm2  a Aps  f pd  As  f y  fc ' b  6.84in M n  62433.2kip  ft (84648kN  m) This area of reinforcement will correspond to 51 #8 (25 mm) bars Step Determine the net longitudinal strain in the cross section Pj = 9300kips (41368.5kN) z (0.9  d )2  As  (0.9  d p )2  Aps (0.9  d ) As  (0.9  d p )  Aps ep = dp – 45.6in = 19.2in  63.9in(1623mm) M u  Pj cos( )  e p x    0.9d p  e p    Vu  Pj sin( )  0  Pj cos( )    z     0.000693 0.9 d  0.9d  p 2 Es  As  E p  Aps  z  z  z Step Compute the maximum shear resistance and torsional resistance For Level of Approximation ke1 = 0.55  30MPa  n fc     f 'c   0.96 kc1 = ke1(nfc) = 0.53 26 For θ = 25° Vrd ,max  kc1 f 'c bw z c Trd ,max  kc1 f 'c c sin( ) cos( )  2570.5kips(11434.2kN ) t f   Ak sin( ) cos( )  47537kip  ft (64452kN  m) For θ = 37.5° Vrd ,max  kc1 f 'c bw z c Trd ,max  kc1 f 'c c sin( ) cos( )  3241.3kips(14418kN ) t f   Ak sin( ) cos( )  59941kip  ft (81269kN  m) For θ = 45° Vrd ,max  kc1 f 'c bw z c Trd ,max  kc1 f 'c c sin( ) cos( )  3355.6kips(14927kN ) t f   Ak sin( ) cos( )  62055kip  ft (84135kN  m) For Level of Approximation II εx1 = 0.001   20  10000   x1  30 1   x1  ( x1  0.002)  cot( )   0.01 ke   0.57 1.2  551 Vrd ,max  kc f 'c bw z c Trd ,max  kc1 f 'c c kc  ke  n fc  0.55 sin   cos    3019.3kips(13430.5kN ) t f   Ak sin( ) cos( )  55835kip  ft (75702kN  m) For Level of Approximation III   20  10000   x  27 1   x  ( x  0.002)  cot( )   0.009 ke   0.59 1.2  551 Vrd ,max  kc f 'c bw z Trd ,max  kc1 c f 'c c kc  ke  n fc  0.57 sin( ) cos( )  2928.4kips(13026.2kN ) t f   Ak sin( ) cos( )  54155kip  ft (73424kN  m) 27 Step Check if the dimensions of the cross section are adequate Level of Approximation I For θ = 25°  Tu   Trd ,max   Vu      Vrd ,max    1.6  For θ = 37.5°  Tu   Trd ,max   Vu      Vrd ,max     For θ = 45°  Tu   Trd ,max   Vu      Vrd ,max    0.9  Level of Approximation II  Tu   Trd ,max   Vu      Vrd ,max    1.2  Level of Approximation III  Tu   Trd ,max   Vu      Vrd ,max    1.2  2  T   V  The dimensions of the cross section are adequate if  u    u   1.0 so then, the cross sections that T     rd ,max   Vrd ,max  not fulfill this requirement will be enlarged This process is done iterating the values until the inequality is fulfilled For θ = 25°, bw = 74in (1880mm), tf = 16in(406mm) Vrd ,max  3170.3kips(14102.2kN )  Tu   Trd ,max   Vu      Vrd ,max Trd ,max  63382.7kip  ft (85935.4kN  m)    1.0  Level of Approximation II bw = 65in (1651mm), tf = 13in (330mm) Vrd ,max  3271kips(14550.1kN )  Tu   Trd ,max   Vu      Vrd ,max Trd ,max  60488.1kip  ft (82010.8kN  m)    1.0  28 Level of Approximation III bw = 65in (1651mm), tf = 13in(330mm) Trd ,max  58668kip  ft (79543.1kN ) Vrd ,max  3172.4kips(14111.5kN )  Tu   Trd ,max   Vu      Vrd ,max    1.0  Step Compute the minimum reinforcement  w  0.08 f 'c f yk  0.0011 Level of Approximation I For θ = 25° A  b  0.084 in v ,min w,min w in  2.13 mm mm in 1.73 mm mm in 1.73 mm mm in 1.90 mm mm in 1.90 mm mm For θ = 37.5° Av ,min   w,min bw  0.068 in 2 For θ = 45° Av ,min   w,min bw  0.068 in 2 Level of Approximation II Av ,min   w,min bw  0.074 in 2 Level of Approximation III Av ,min   w,min bw  0.074 in 2 Step Compute the area of transverse reinforcement for shear Level of Approximation I For θ = 25° 2 Vu Av   0.44 in 11.2 mm s z  f  cot( ) in mm yd  For θ = 37.5° 2 Vu Av   0.73 in 18.5 mm s z  f  cot( ) in mm yd  For θ = 45° 2 Vu Av   0.95 in 24.1mm s z  f  cot( ) in mm yd       Level of Approximation II Av s  2 Vu  0.55 in 14.0 mm in mm z  f yd  cot( ) 29 Level of Approximation III Av s   2 Vu  0.48 in 12.2 mm in mm z  f yd  cot( )  Step 10 Check if longitudinal reinforcement can resist the required tension Equation Aps f pd  As f yd  11791.3kip(52450.3kN ) Equation Level of Approximation I For θ = 25° Vu cot( )  3391kips(15084kN ) For θ = 37.5° Vu cot( )  2060.5kips(9165.6kN ) For θ = 45° Vu cot( )  1581kips(7033kN ) Level of Approximation II Vu cot( )  2738.5kips(12181.5kN ) Level of Approximation III Vu cot( )  3113kips(13847.3kN ) If Equation > Equation then the longitudinal steel provided could resist the required tension Step 11 Design the transverse reinforcement for shear for exterior webs The ultimate shear load on exterior webs is Vu,ext = 765.2kips (3403.8kN) M u  Pj cos( )  e p x  5 z  Pj  0.9d p  e p   sin( )  0  cos( )    5 z     0.000654 0.9 d  0.9d  p 2 Es  As  E p  Aps  z  z   Vu  Pj 30 Level of Approximation I For θ = 25° Vu ,ext 2 Av ,ext   0.11in 2.80 mm s in mm z  f yd  cot( )   For θ = 37.5° Vu ,ext 2 Av ,ext   0.18 in 4.60 mm s in mm z  f yd  cot( )  For θ = 45° Vu ,ext 2 Av ,ext   0.23 in 5.84 mm s in mm z  f yd  cot( )    Level of Approximation II Av ,ext s  Vu ,ext z  f yd  cot( )  0.13 in in 3.30 mm mm in 3.05 mm mm 2 Level of Approximation III Av ,ext s  Vu ,ext z  f yd  cot( )  0.12 in 2 Step 12 Determine the amount of transverse reinforcement for torsion on exterior webs Level of Approximation I For θ = 25° 2 Tu At   0.020 in 0.51mm s A  f  cot( )  cos( ) in mm k yd   For θ = 37.5° 2 Tu At   0.032 in 0.813 mm s A  f  cot( )  cos( ) in mm k yd  For θ = 45° 2 Tu At   0.041in 1.04 mm s A  f  cot( )  cos( ) in mm k yd    Level of Approximation II 2 Tu At   0.024 in 0.610 mm s A  f  cot( )  cos( ) in mm k yd     Level of Approximation III At s  2 Tu  0.021in 0.533 mm in mm Ak  f yd  cot( )  cos( ) 31 Step 13 Determine the amount of transverse reinforcement for torsion on flanges Level of Approximation I For θ = 25° 2 Tu Atf   0.017 in 0.432 mm s A  f  cot( ) in mm k yd  For θ = 37.5° 2 Tu Atf   0.028 in 0.711mm s A  f  cot( ) in mm k yd   For θ = 45° 2 Tu Atf   0.037 in 0.940 mm s A  f  cot( ) in mm k yd  Level of Approximation II 2 Tu Atf   0.021in 0.533 mm s A  f  cot( ) in mm k yd     Level of Approximation III Atf s  Tu Ak  f yd  cot( )  0.019 in 0.483 mm mm in Step 14 Determine the amount longitudinal reinforcement for torsion Level of Approximation I For θ = 25°  Tu  z   Vu    cot( ) Ak   Atl   58.7in  37871mm  f yd For θ = 37.5°  Tu  z   Vu    cot( ) Ak   Atl   35.7in  23032.2mm  f yd For θ = 45°  Tu  z   Vu    cot( ) Ak   Atl   27.4in 17677.4mm  f yd Level of Approximation II  Tu  z   Vu    cot( ) Ak   Atl   47.4in  30581mm  f yd 32 Level of Approximation III  Tu  z   Vu    cot( ) Ak   Atl   54.0in  34839mm  f yd Step 15 Determine the amount of transverse reinforcement to be used on exterior webs Level of Approximation I For θ = 25° AT ,ext  Av ,ext  At  0.13 in in 3.30 mm mm in 5.33 mm mm 2 For θ = 37.5° AT ,ext  Av ,ext  At  0.21in 2 For θ = 45° AT ,ext  Av ,ext  At  0.27 in Level of Approximation II A  A  A  0.16 in T , ext v , ext t Level of Approximation III A  A  A  0.14 in T , ext v , ext t in  6.60 mm mm in  4.06 mm mm in 3.56 mm mm 2 2 Step 16 Calculate the spacing of transverse reinforcement Interior webs Use #5 (16mm) two legged stirrups with an area Av,prov of 0.61in2 (396mm2) Level of Approximation I For θ = 25° Av, prov s  6.75in Av For θ = 37.5° Av, prov s  4in Av For θ = 45° Av , prov s  3in Av Level of Approximation II Av, prov s  5.50in Av 33 Level of Approximation III Av, prov s  6.25in Av Exterior webs Use #5 (16mm) two legged stirrups with an area Av of 0.61in2 (396mm2) Level of Approximation I For θ = 25° Av , prov s  4.75in AT ,ext For θ = 37.5° Av , prov s  2.75in AT ,ext For θ = 45° Av , prov s  2in AT ,ext Level of Approximation II Av , prov s  3.75in AT ,ext Level of Approximation III Av , prov s  4.50in AT ,ext Flanges Use #5 (16mm) two legged stirrups with an area At,prov of 0.61in2 (396mm2) Level of Approximation I For θ = 25° At , prov s  12in Atf For θ = 37.5° At , prov s  7.75in Atf For θ = 45° At , prov s  5.75in Atf Level of Approximation II At , prov s  10.25in Atf 34 Level of Approximation III At , prov s  11.75in Atf Final results comparison Table 11––Results comparison of gross concrete area and required transverse reinforcement Design Code Required transverse reinforceAg ment for disregarding exterior overhanging web flanges (ft2) (shear and torsion) (in2/in) Required transverse reinforcement for interior web (shear) (in2/in) Required transverse Required Required reinforce- longitudinal longitudinal ment for reinforcement reinforcement flange for flexure for torsion (torsion) (in2) (in2) (in /in) ACI 318-14 106.3 0.195 0.111 0.035 30 83.3 AASHTO LRFD 2017 EN 1992-2004 (θ = 45°) EN 1992-2004 (θ = 35°) EN 1992-2004 (θ = 22°) 95.1 0.151 0.082 0.026 40 52.0 95.1 0.295 0.208 0.034 40 46.1 95.6 0.207 0.146 0.026 40 65.8 106.5 0.119 0.084 0.015 40 114.0 MODEL CODE 2010 –LoA (θ = 45°) 95.1 0.271 0.190 0.037 40 27.3 MODEL CODE 2010 –LoA (θ = 37.5°) 95.1 0.208 0.146 0.029 40 35.6 MODEL CODE 2010 –LoA (θ = 25°) 101.8 0.127 0.089 0.017 40 58.5 MODEL CODE 2010 –LoA 97.4 0.157 0.110 0.023 40 47.2 MODEL CODE 2010 –LoA 97.4 0.133 0.097 0.019 40 53.7 35 REFERENCES [1] California Department of Transportation 2015 Bridge Design Practice, 7-33 [2] ACI Committee 318 2014 Building Code Requirements for Structural Concrete (ACI 318-14) and Commentary, Farmington Hills, MI [3] American Association of Highway and Transportation Officials 2017 AASHTO LRFD Bridge Design Specifications, 8th Edition, Washington, D.C [4] European Committee for Standardization 2004 Eurocode 2: Design of concrete structures - Part 1-1: General rules and rules for buildings, Brussels, Belgium [5] The International Federation for Structural Concrete 2010 fib Model Code for Concrete Structures 2010, Lausanne, Switzerland 36