Design Parameters. Span 1. Span 2. Span 3. All Spans. Bents

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1 County: Any Hwy: Any Design: BRG Date: 6/2010 Inverted Tee Bent Cap Design Example Design example is in accordance with the AASHTO LRFD Bridge Design Specifications, 5th Ed. (2010) as prescribed by TxDOT Bridge Design Manual - LRFD (May 2009). Design Parameters Span 1 54' Type TX54 Girders (0.851 k / ft ) 6 Girders 8.00' with 3' overhangs 2" Haunch Span 2 112' Type TX54 Girders (0.851 k / ft ) 6 Girders 8.00' with 3' overhangs 3.75" Haunch Span 3 54' Type TX54 Girders (0.851 k / ft ) 6 Girders 8.00' with 3' overhangs 2" Haunch All Spans Deck is 46ft wide Type T551 Rail (0.382k/ft) 8" Thick Slab (0.100 ksf) Assume 2" 140 pcf (0.023 ksf) Use Class "C" Concrete f' c =3.60 ksi w c =150 pcf (for weight) w c =145 pcf (for Modulus of Elasticity calculation) Grade 60 Reinforcing F y =60 ksi "AASHTO LRFD" refers to the AASHTO LRFD Bridge Design Specification, 5th Ed. (2010) "BDM-LRFD" refers to the TxDOT Bridge Design Manual - LRFD (May 2009) "TxSP" refers to TxDOT guidance, recommendations, and standard practice. "Furlong & Mirza" refers to "Strength and Servicability of Inverted T-Beam Bent Caps Subject to Combined Flexure, Shear, and Torsion", Center for Highway Research Research Report No F, The University of Texas at Austin, August 1974 The basic bridge geometry can be found on the Bridge Layout located in the Appendices. (TxSP) (BDM-LRFD, Ch. 4, Sect. 5, Materials) (BDM-LRFD, Ch. 4, Sect. 5, Materials) Bents Use 36" Diameter Columns (Typical for Type TX54 Girders) LRFD Inverted Tee Bent Cap Design Example 1 June 2010

2 Design Parameters Define Variables (Con't) Back Span Span1 GdrSpa1 Forward Span 54ft Span2 112ft 8ft GdrSpa2 8ft GdrNo1 6 GdrNo2 6 GdrWt1 Haunch klf GdrWt klf 2in Haunch2 3.75in Span Length Girder Spacing Number of Girders in Span Weight of Girder Size of Haunch Bridge Skew 0deg BridgeW 46ft RdwyW 44ft GirderD 54in BrgSeat 1.5in BrgPad 2.75in SlabThk 8in OverlayThk 2in RailWt 0.382klf w c 0.150kcf Skew of Bents Width of Bridge Deck Width of Roadway Depth of Type TX54 Girder Bearing Seat Buildup Bearing Pad Thickness Thickness of Bridge Slab Thickness of Overlay Weight of Rail Unit Weight of Concrete for Loads W Olay 0.140kcf Unit Weight of Overlay Bents f c 3.60ksi Concrete Strength w ce 0.145kcf Unit Weight of Concrete for E c 1.5 E c = w ce f c E c 3457ksi Modulus of Elasticity of Concrete, (AASHTO LRFD Eq ) f y 60ksi Yield Strength of reinforcement E s 29000ksi Modulus of Elasticity of Steel D column 36in Diameter of Columns Other Variables IM 33% Dynamic load allowance, (AASHTO LRFD Table ) LRFD Inverted Tee Bent Cap Design Example 2 June 2010

3 Determine Cap Dimensions Stem Width b stem D column 3in b stem 39in Stem Height The stem is typically at least 3" wider than the Diameter of the Column (36") to allow for the extension of the column reinforcement into the Cap. (TxSP) Distance From Top of Slab to Top of Ledge D Slab_to_Ledge SlabThk Haunch2 GirderD BrgPad BrgSeat Haunch 2 is the larger of the two haunches. D Slab_to_Ledge 70.00in StemHaunch 3.75in The top of the stem must be 2.5" below the bottom of the slab. (BDM-LRFD, Ch. 4, Sect. 5, Geometric Constraints) Accounting for the 1/2" of bituminus fiber, the top of the stem must have at least 2" of haunch on it, but the haunch should not be less than either of the haunches of the adjacent spans. d stem D Slab_to_Ledge SlabThk StemHaunch 0.5in Use: d stem 57in d stem 57.75in The stem must accommodate 1/2" of bituminous fiber. Round the Stem Height down to the nearest 1". (TxSP) LRFD Inverted Tee Bent Cap Design Example 3 June 2010

4 Determine Cap Dimensions Ledge Width (Con't) The Ledge Width must be adequate for Bar M to develop fully. 26" is typically used for TX54 girders, as it is adequate to develop a # 6 bar with the typical 2.5" cover. If the cover is increased to 3", allowing for a modification factor of 0.7, a 24" Ledge is adequate to develop a # 7 bar. I-Beams have 9.5" from the face of the cap to the CL of Bearing. The typical ledge width for these bridges is 23". "L dh,prov " must be greater than or equal to "L dh,req " for bar M. cover 2.5in "cover" is measured from the center of the transverse bars. L 8in Determine the Required Development Length of Bar M: "L" is the length of the Bearing Pad along the girder. A typical type TX54 bearing pad is 8"x21" as shown in the IGEB standard. Try # 6 Bar for Bar M. d bar_m 0.750in A bar_m 0.44in 2 Basic Development Length L dh = 38.0d bar_m L dh 15.02in (AASHTO LRFD Eq ) f c Modification Factors for L dh: (AASHTO LRFD ) Is Top Cover greater than or equal to 2.5", and Side Cover greater than or equal to 2"? SideCover cover d bar_m 2 SideCover 2.13in TopCover cover d bar_m 2 TopCover 2.13in "Side Cover" and "Top Cover" are the clear cover on the side and top of the hook respectively. The dimension "cover" is measured from the center of Bar M. No, Factor = 1.0 LRFD Inverted Tee Bent Cap Design Example 4 June 2010

5 Determine Cap Dimensions (Con't) The Required Development Length is the larger of the following: (AASHTO LRFD ) L dh Factor 8d bar_m 6in Therefore, L dh_req 15.02in 6.00in 15.02in L b ledge_min L dh_req cover 12in b 2 ledge_min 25.52in The distance from the face of the stem to the center of bearing is 12" for TxGirders. (IGEB) Use: b ledge 26in Width of Bottom Flange b f 2b ledge b stem b f 91in Ledge Depth Use a Ledge Depth of 28" d ledge 28in As a general rule of thumb Ledge Depth is greater than or equal to 2'-3". This is the depth at which a bent from a typical bridge will pass the punching shear check (calculations found on Pg. 21). Total Depth of Cap h cap d stem d ledge h cap 85in Summary of Cross Sectional Dimensions b stem 39in From Pg. 3 d stem 57in From Pg. 3 b ledge 26in d ledge 28in h cap 85in LRFD Inverted Tee Bent Cap Design Example 5 June 2010

6 Determine Cap Dimensions (Con't) Length of Cap First define Girder Spacing and End Distance: S 8ft c 2ft L Cap S( GdrNo1 1) 2c L Cap 44 ft Girder Spacing "c" is the distance from the Center Line of the Exterior Girder to the Edge of the Cap measured along the Cap. Length of Cap TxDOT policy is as follows, "The edge distance between the exterior bearing pad and the end of the inverted T-beam shall not be less than 12in." (BDM-LRFD, Ch. 4, Sect. 5, Design Criteria) replacing the statement in AASHTO LRFD stating it shall not be less than d f. Preferably, the stem should extend at least 3" beyond the edge of the bearing seat. Bearing Pad Dimensions L 8in (IGEB standard) Length of Bearing Pad W 21in Width of Bearing Pad Cross Sectional Properties of Cap A g d ledge b f d stem b stem A g 4771in 2 ybar d ledge b f 1 2 d ledge 1 d stem b stem d ledge 2 d stem A g Distance from bottom of cap to the center of gravity of the cap ybar 33.80in 3 b f d ledge I g 12 3 b stem d stem 12 b f d ledge 2 1 ybar 2 d ledge I g in 4 1 b stem d stem ybar d ledge 2 d stem 2 LRFD Inverted Tee Bent Cap Design Example 6 June 2010

7 Cap Analysis Cap Model Assume: 4 Columns 12'-0" The cap will be modeled as a continuous beam with simple supports using TxDOT's CAP18 program. TxDOT does not consider frame action for typical multi-column bents. (BDM-LRFD, Ch. 4, Sect. 5, Structural Analysis) Cap 18 Model Station = 0.5' The circled numbers are the stations that will be used in the CAP 18 input file. One station is 0.5ft in the direction perpendicular to the pgl, not parallel to the bent. station 0.5ft Station increment for CAP18 Recall: E c ksi (Pg. 2) I g in 4 (Pg.6) E c I g kipin 2 / 12 in ft 2 = E c I g kipft 2 LRFD Inverted Tee Bent Cap Design Example 7 June 2010

8 Cap Analysis Dead Load SPAN 1 Rail1 (Con't) Span1 2RailWt 2 kip Rail min( GdrNo1 6) girder Values used in the following equations can be found on Pg. 2. Rail weight is distributed evenly among stringers, up to 3 stringers per rail. (TxSP) Slab1 Span1 kip w c GdrSpa1SlabThk 1.10 Slab girder Increase slab DL by 10% to account for haunch and thickened slab ends. Girder1 GdrWt1 Span1 kip Girder girder kip DLRxn1 Rail1 Slab1 Girder1 DLRxn girder Overlay is calculated separately, because it has a different load factor than the rest of the dead loads. Overlay1 Span1 kip W Olay GdrSpa1OverlayThk Overlay girder Design for future overlay. SPAN 2 Rail2 Slab2 Span2 2RailWt 2 kip Rail min( GdrNo2 6) girder Span2 kip w c GdrSpa2SlabThk 1.10 Slab girder Girder2 GdrWt2 Span2 kip Girder girder kip DLRxn2 Rail2 Slab2 Girder2 DLRxn girder Overlay2 Span2 kip W Olay GdrSpa2OverlayThk Overlay girder CAP Cap w c A g kip * ft 0.5ft station kip = Cap station LRFD Inverted Tee Bent Cap Design Example 8 June 2010

9 Cap Analysis (Con't) Live Load (AASHTO LRFD and ) LongSpan ShortSpan max( Span1 Span2) min( Span1 Span2) LongSpan ft ShortSpan ft IM 0.33 Lane 0.64klf Lane kip lane LongSpan ShortSpan LongSpan 14ft Truck 32kip 32kip 8kip LongSpan Truck kip lane LLRxn Lane Truck( 1 IM) LLRxn kip lane P 16.0kip( 1 IM) P 21.28kip LLRxn ( 2P) w 10ft w 9.83 kip * ft 0.5ft station kip w 4.92 station 2 LongSpan 28ft LongSpan Use HL-93 Live Load. For maximum reaction at interior bents, "Design Truck" will always govern over "Design Tandem". For the maximum reaction when the long span is more than twice as long as the short span, place the rear (32 kip) axle over the support and the middle (32 kip) and front (8 kip) axles on the long span. For the maximum reaction when the long span is less than twice as long as the short span, place the middle (32 kip) axle over the support, the front (8 kip) axle on the short span and the rear (32 kip) axle on the long span. Combine "Design Truck" and "Design Lane" loadings. (AASHTO LRFD ) Dynamic load allowance, IM, does not apply to "Design Lane." (AASHTO LRFD ) The Live Load is applied to the slab by two 16 kip wheel loads increased by the dynamic load allowance with the remainder of the live load distributed over a 10 ft (AASHTO LRFD ) design lane width. (TxSP) The Live Load applied to the slab is distributed to the beams assuming the slab is hinged at each beam except the outside beam. (BDM-LRFD, Ch. 4, Sect. 5, Structural Analysis) LRFD Inverted Tee Bent Cap Design Example 9 June 2010

10 Cap Analysis Cap 18 Input Multiple Presence Factors, m (AASHTO LRFD Table ) Input "Multiple Presence Factors" into Cap18 as "Load Reduction Factors". No. of Lanes Factor "m" >3 (Con't) Limit States (AASHTO LRFD 3.4.1) The cap design need only consider Strength I Live Load and Dynamic Load Allowance LL + IM = 1.75 Dead Load Components DC = 1.25 Strength I, Service I, and Service I with DL. (TxSP) Dead Load Wearing Surface (Overlay) Service I Live Load and Dynamic Load Allowance DW = 1.50 LL + IM = 1.00 TxDOT allows the Overlay Factor to be reduced to 1.25 (TxSP), since overlay is typically used in design only to increase the safety factor, but in this example we will use DW = Dead Load and Wearing Surface DC & DW = 1.00 Dead Load TxDOT considers Service level Dead Load only with a limit reinforcement stress of 22 ksi to minimize cracking. (BDM-LRFD, Chapter 4, Section 5, Design Criteria) Cap 18 Output Max +M Max -M Dead Load: M posdl 250.1kipft M negdl 379.4kipft Service Load: M posserv 492.5kipft M negserv 590.9kipft These loads are the maximum loads from the Cap 18 Output File located in the Appendices. Factored Load: M posult 741.7kipft M negult 852.1kipft LRFD Inverted Tee Bent Cap Design Example 10 June 2010

11 Cap Analysis (Con't) Girder Reactions on Ledge: Dead Load kip DLSpan1 Rail1 Slab1 Girder1 DLSpan girder kip Overlay girder kip DLSpan2 Rail2 Slab2 Girder2 DLSpan girder kip Overlay girder For calculations of these loads see Pg. 8. Live Load (AASHTO LRFD and ) Loads per Lane: Use HL-93 Live Load. For maximum reaction at interior bents, "Design Truck" will always govern over "Design Tandem" for Spans greater than 26ft. For the maximum reaction, place the back (32 kip) axle over the support. LaneSpan1 LaneSpan2 Span1 0.64klf LaneSpan kip 2 lane Span2 0.64klf LaneSpan kip 2 lane Span1 14ft TruckSpan1 32.0kip 32.0kip Span1 28ft 8.0kip TruckSpan kip Span1 Span1 lane Span2 14ft TruckSpan2 32.0kip 32.0kip Span2 28ft 8.0kip TruckSpan kip Span2 Span2 lane LRFD Inverted Tee Bent Cap Design Example 11 June 2010

12 Cap Analysis (Con't) Girder Reactions on Ledge (Con't) : Live Load (Con't) Loads per Lane (Con't) : IM 0.33 LLRxnSpan1 LaneSpan1 TruckSpan1 ( 1 IM) LLRxnSpan kip lane LLRxnSpan2 LaneSpan2 TruckSpan2 ( 1 IM) LLRxnSpan kip lane gv Span1_Int gv Span1_Ext gv Span2_Int gv Span2_Ext Combine "Design Truck" and "Design Lane" loadings. (AASHTO LRFD ) Dynamic load allowance, "IM", does not apply to "Design Lane." (AASHTO LRFD ) The Live Load Reactions are assumed to be the Shear Live Load Distribution Factor multiplied by the Live Load Reaction per Lane. The Shear Live Load Distribution Factor was calculated using the "LRFD Live Load Distribution Factors" Spreadsheet found in the Appendices. The Exterior Girders must have a Live Load Distribution Factor equal to or greater than the Interior Girders. This is to accommodate a possible future bridge widening. Widening the bridge would cause the exterior girders to become interior girders. LLSpan1Int gv Span1_Int LLRxnSpan1 kip LLSpan1Int girder LLSpan1Ext gv Span1_Ext LLRxnSpan1 kip LLSpan1Ext girder LLSpan2Int gv Span2_Int LLRxnSpan2 kip LLSpan2Int girder LLSpan2Ext gv Span2_Ext LLRxnSpan2 kip LLSpan2Ext girder LRFD Inverted Tee Bent Cap Design Example 12 June 2010

13 Cap Analysis (Con't) Girder Reactions on Ledge (Con't) : Span 1 Interior Girder Service Load Service I Limit State, (AASHTO LRFD 3.4.1) V s_span1int DLSpan1 Overlay1 LLSpan1Int V s_span1int 134kip Factored Load Strength I Limit State, (AASHTO LRFD 3.4.1) V u_span1int 1.25DLSpan1 1.5Overlay1 1.75LLSpan1Int V u_span1int 208kip Exterior Girder Service Load Service I Limit State, (AASHTO LRFD 3.4.1) V s_span1ext DLSpan1 Overlay1 LLSpan1Ext V s_span1ext 134kip Factored Load Strength I Limit State, (AASHTO LRFD 3.4.1) V u_span1ext 1.25DLSpan1 1.5Overlay1 1.75LLSpan1Ext V u_span1ext 208kip Span 2 Interior Girder Service Load Service I Limit State, (AASHTO LRFD 3.4.1) V s_span2int DLSpan2 Overlay2 LLSpan2Int V s_span2int 215kip Factored Load Strength I Limit State, (AASHTO LRFD 3.4.1) V u_span2int 1.25DLSpan2 1.5Overlay2 1.75LLSpan2Int V u_span2int 322kip Exterior Girder Service Load Service I Limit State, (AASHTO LRFD 3.4.1) V s_span2ext DLSpan2 Overlay2 LLSpan2Ext V s_span2ext 215kip Factored Load Strength I Limit State, (AASHTO LRFD 3.4.1) V u_span2ext 1.25DLSpan2 1.5Overlay2 1.75LLSpan2Ext V u_span2ext 322kip LRFD Inverted Tee Bent Cap Design Example 13 June 2010

14 Cap Analysis (Con't) Torsional Loads Strength I Limit State, (AASHTO LRFD 3.4.1) To maximize the torsion, the live load only acts on the longer span in the configuration shown. The loads are applied to the cap as depicted in the following picture: a v 12in "a v " is the value for the distance from the face of the stem to the center of bearing for the girders. 12" is the typical value for TxGirders on Inverted Tee Bents (IGEB). 9" is the typical value for I-Beams (IBEB). The lever arm for the torsional loads is the distance from the center line of bearing to the centerline of the cap ( 1 / 2 b stem + a v ). b stem 39in From Pg. 3 1 LeverArm a v 2 b stem LeverArm 31.5in LRFD Inverted Tee Bent Cap Design Example 14 June 2010

15 Cap Analysis (Con't) Torsional Loads (Con't) Interior Girders Girder Reactions R u_span1 1.25DLSpan1 1.5Overlay1 R u_span1 70kip R u_span2 Torsional Load 1.25DLSpan2 1.5Overlay2 1.75gV Span2_Int [ LaneSpan2 TruckSpan2 ( 1 IM) ] R u_span2 322kip T u_int R u_span1 R u_span2 LeverArm T u_int 660kipft Exterior Girders Girder Reactions R u_span1 1.25DLSpan1 1.5Overlay1 R u_span1 70kip R u_span2 Torsional Load 1.25DLSpan2 1.5Overlay2 1.75gV Span2_Ext [ LaneSpan2 TruckSpan2 ( 1 IM) ] R u_span2 322kip T u_ext R u_span1 R u_span2 LeverArm T u_ext 660kipft Torsion on Cap Analyzed assuming Bents are torsionally rigid at Effective Face of Columns. T u 660kipft Maximum Torsion on Cap LRFD Inverted Tee Bent Cap Design Example 15 June 2010

16 Cap Analysis (Con't) Load Summary Ledge Loads Interior Girder Service Load V s_int max V s_span1int V s_span2int V s_int kip Factored Load V u_int max V u_span1int V u_span2int V u_int kip Exterior Girder Service Load V s_ext Factored Load max V s_span1ext V s_span2ext V s_ext kip V u_ext max V u_span1ext V u_span2ext V u_ext kip Cap Loads Positive Moment (From CAP 18) Dead Load: M posdl kipft Service Load: M posserv kipft Factored Load: M posult kipft Negative Moment (From CAP 18) Dead Load: M negdl kipft Service Load: M negserv kipft Factored Load: M negult kipft Maximum Torsion and Concurrent Shear and Moment (Strength I) T u 660kipft V u 448.1kip M u 335.6kipft Located two stations away from centerline of column. V u and M u values are from CAP 18 In this example the maximum Torsion and the maximum Shear are concurrent with each other. If they are not, it becomes necessary to check the location of the maximum Torsion with its concurrent Shear and the location of the maximum Shear with its concurrent Torsion. LRFD Inverted Tee Bent Cap Design Example 16 June 2010

17 Locate and Describe Reinforcement Recall: b stem 39in From Pg. 3 d stem 57in From Pg. 3 b ledge 26in From Pg. 5 d ledge 28in From Pg. 5 b f 91in From Pg. 5 h cap 85in From Pg. 5 cover 2.50in From Pg. 4 Measured from Center of bar LRFD Inverted Tee Bent Cap Design Example 17 June 2010

18 Locate and Describe Reinforcement Describe Reinforcing Bars (Con't) Use # 11 bars for Bar A A bar_a 1.56in 2 d bar_a 1.410in Use # 11 bars for Bar B A bar_b 1.56in 2 d bar_b 1.410in Use # 6 bars for Bar M A bar_m 0.44in 2 d bar_m 0.75in Bar M must be a # 6 bar or smaller to allow it to fully develop, as stated on Pg 4. Use # 6 bars for Bar N A bar_n 0.44in 2 d bar_n 0.75in To prevent confusion, use the same bar size for Bar N as Bar M. Use # 6 bars for Bar S A bar_s 0.44in 2 d bar_s 0.75in Use # 6 bars for Bar T A bar_t 0.44in 2 d bar_t 0.750in Calculate Dimensions d s_neg h cap cover 1 2 d bar_s 1 2 d bar_a d s_neg 81.42in d s_pos h cap cover 1 2 max d bar_s d bar_m 1 2 d bar_b d s_pos 81.42in a v 12in Typical for TX Girders on Inverted Tee Bent Caps (IGEB standard) a f a v cover a f 14.50in d e d ledge cover d e 25.50in 1 d f d ledge cover 2 d 1 bar_m 2 d bar_b d f 24.42in h d ledge BrgSeat h 29.50in "BrgSeat" is the height of the Bearing Seat Buildup. This value is defined on Pg. 2. LRFD Inverted Tee Bent Cap Design Example 18 June 2010

19 Locate and Describe Reinforcement Calculate Dimensions (Con't) (Con't) α 90deg Angle of Bars S Recall: L 8in From Pg. 4 W 21in From Pg. 6 LRFD Inverted Tee Bent Cap Design Example 19 June 2010

20 Check Bearing (AASHTO LRFD 5.7.5) The load on the bearing pad propagates along a truncated pyramid whose top has the area A 1 and whose base has the area A 2. A 1 is the loaded area (the bearing pad area: LxW ). A 2 is the area of the lowest rectangle contained wholly within the support (the Inverted Tee Cap). A 2 must not overlap the truncated pyramid of another load in either direction, nor can it extend beyond the edges of the cap in any direction. Elevation View Plan View ϕ 0.7 A 1 WL A 1 168in 2 (AASHTO LRFD ) Area under Bearing Pad Interior Girders 1 B min b ledge a v 2 L a 1 v 2 b stem 1 2 L 2d 1 ledge 2 S 1 2 W B 10.00in L 2 L 2B L in "B" is the distance from the perimeter of A 1 to the perimeter of A 2, as seen in the above figures. W 2 W 2B W in A 2 L 2 W 2 A in 2 m = the minimum of: Modification Factor A & 2 m 2.00 (AASHTO LRFD Eq ) A 1 ϕv n ϕ0.85f c A 1 m ϕv n 720kip (AASHTO LRFD Eq & AASHTO LRFD Eq ) V u_int 322kip < ϕv n BearingChk "OK!" V u_int From Pg. 16 Exterior Girders B min b ledge a v 1 2 L a v 1 2 b stem 1 2 L 2d 1 ledge 2 S 1 2 W c 1 2 W B 10.00in L 2 L 2B L in W 2 W 2B W in "B" is the distance from the perimeter of A 1 to the perimeter of A 2. A 2 L 2 W 2 A in 2 m = the minimum of: Modification Factor A A 1 & 2 m 2.00 (AASHTO LRFD Eq ) ϕv n ϕ0.85f c A 1 m ϕv n 720kip V u_ext 322kip < ϕv n BearingChk "OK!" (AASHTO LRFD Eq & AASHTO LRFD Eq ) V u_ext From Pg. 16 LRFD Inverted Tee Bent Cap Design Example 20 June 2010

21 Check Punching Shear (AASHTO LRFD with modifications from BDM-LRFD, Ch. 4, Sect. 5, Design Criteria) ϕ 0.9 (AASHTO LRFD ) Determine if the Shear Cones Intersect Is 1 2 S 1 2 W d f? Yes. Therefore Shear Cones do not intersect in the longitudinal direction of the Cap. Is 1 2 S 1 d f 2 W 24.42in 1 2 b stem a v in 1 2 L TxDOT uses "d f " instead of "d e " for Punching Shear (BDM-LRFD, Ch. 4, Sect. 5, Design Criteria). This is because "d f " has traditionally been used for inverted tee bents and was used in the Inverted Tee Research (Furlong & Mirza pg. 58). d f? Yes. Therefore Shear Cones do not intersect in the transverse direction of the Cap. 1 2 b stem a v d f 24.42in 1 2 L in Interior Girders V n f c W 2L 2d f d f V n 497kip (BDM-LRFD, Ch. 4, Sect. 5, Design Criteria) ϕv n 447kip V u_int 322kip < ϕv n PunchingShearChk "OK!" V u_int From Pg. 16 Exterior Girders V n = minimum of: (BDM-LRFD, Ch. 4, Sect. 5, Design Criteria) f c 2 W L d f c d f 388kip V n f c W 2L 2d f 388kip d f 497kip ϕv n 349kip V u_ext 322kip < ϕv n PunchingShearChk "OK!" V u_ext From Pg. 16 LRFD Inverted Tee Bent Cap Design Example 21 June 2010

22 Check Shear Friction (AASHTO LRFD ) Checks are for concrete only see ϕ 0.9 (AASHTO LRFD ) Determine the Distribution Width (AASHTO LRFD ) Interior Girders b s_int = minimum of: W 4 a v 69.00in "Ledge Reinforcement" for reinforcement checks for Bars M and N. b s_int S 69.00in 96.00in "S" is the girder spacing. Exterior Girders b s_ext = minimum of: W 4 a v 69.00in b s_ext S 96.00in 2 c 48.00in 48.00in "S" is the girder spacing. Interior Girders A cv d e b s_int A cv 1759in 2 V n = minimum of: 0.2 f c A cv 1267kip (AASHTO LRFD Eq ) 0.8 ksia cv 1408kip (AASHTO LRFD Eq ) V n ϕv n V u_int 1267kip 1140kip 322kip < ϕv n ShearFrictionChk "OK!" V u_int From Pg. 16 Exterior Girders A cv d e b s_ext A cv 1224in 2 V n = minimum of: 0.2 f c A cv 0.8 ksia cv 881kip 979kip (AASHTO LRFD Eq ) (AASHTO LRFD Eq ) V n ϕv n V u_ext 881kip 793kip 322kip < ϕv n ShearFrictionChk "OK!" V u_ext From Pg. 16 LRFD Inverted Tee Bent Cap Design Example 22 June 2010

23 Flexural Reinforcement for Negative Bending (Bars A) (Tension in Top) M dl M negdl M dl 379.4kipft M s M negserv M s 590.9kipft From Cap 18 Output. See Pg. 10 M u M negult M u 852.1kipft Minimum Flexural Reinforcement (AASHTO LRFD ) Factored Flexural Resistance, M r, must be greater than or equal to the lesser of 1.2 M cr (Cracking Moment) or 1.33 Mu (Ultimate Moment) I g in 4 Gross Moment of Inertia (From Pg. 6) h cap 85in Depth of Cap (From Pg. 5) ybar 33.80in f r = 0.24 f c f r 0.455ksi y t h cap ybar y t 51.20in Distance to the Center of Gravity of the Cap from the bottom of the Cap (From Pg. 6) Modulus of Rupture (BDM-LRFD, Ch. 4, Sect. 5, Design Criteria) Distance from Center of Gravity to extreme tension fiber S I g S in 3 y t Section Modulus for the extreme tension fiber M cr 1ft Sf r M 12in cr kipft Cracking Moment (AASHTO LRFD Eq ) M f = minimum of: 1.2M cr 1.33M u kipft kipft Design for the lesser of 1.2M cr or 1.33Mu when determining minimum area of steel required. Thus, M r must be greater than M f kipft LRFD Inverted Tee Bent Cap Design Example 23 June 2010

24 Flexural Reinforcement for Negative Bending (Con't) (Bars A) Moment Capacity Design (AASHTO LRFD ) Try, 5 ~ #11's Top BarANo 5 d bar_a 1.41in Number of bars in tension Diameter of main reinforcing bars A bar_a 1.56in 2 Area of one main reinforcing bar A s ( BarANo) A bar_a A s 7.80in 2 Area of steel in tension d stirrup d bar_s d stirrup 0.75in Diameter of shear reinforcing bars From Pg. 18 d d s_neg d 81.42in See Pg. 18 for the calculation of "d s_neg, " b b f b 91in See Pg. 5 for the calculation of "b f." f c 3.60ksi Compressive Strength of Concrete f y 60ksi Yield Strength of Rebar β 1 = f c 4ksi (AASHTO LRFD ) Bounded by: 0.65 β β c A s f y c 1.98in 0.85f c β 1 b Depth of Cross Section under Compression under Ultimate Load (AASHTO LRFD Eq ) This "c" is the distance from the extreme compression fiber to the neutral axis, not the distance from the center of bearing of the last girder to the end of the cap. a cβ 1 a 1.68in Depth of Equivalent Stress Block (AASHTO LRFD ) Note: "a" is less than "d ledge " therefore the equivalent stress block acts over a rectangular area. If "a" was greater than "d ledge " it would act over a Tee shaped area. a 1ft M n A s f y d M 2 12in n kipft d c ε s c ε s Nominal Flexural Resistance (AASHTO LRFD Eq ) Strain in Reinforcing at Ultimate ε s > FlexureBehavior "Tension Controlled" (AASHTO LRFD ) ϕ M 0.90 (AASHTO LRFD ) M r ϕ M M n M r kipft Factored Flexural Resistance (AASHTO LRFD Eq ) M f kipft < M r MinReinfChk "OK!" M u 852.1kipft < M r UltimateMom "OK!" LRFD Inverted Tee Bent Cap Design Example 24 June 2010

25 Flexural Reinforcement for Negative Bending (Con't) (Bars A) Check Serviceability (AASHTO LRFD ) To find s max : Modular Ratio: E s n n 8.39 E c For service loads, the stress on the cross-section is located as drawn: Tension Reinforcement Ratio: A s ρ bd ρ k dk ( 2ρn) ( ρn) 2 ( ρn) k in < d ledge 28.00in k j 1 j Therefore, the compression force acts over a rectangular area. If the compression force does not act over a rectangular area, j will not be 1-k/3. f ss M s 12in f A s j d 1ft ss 11.65ksi f a 0.6f y f a 36.00ksi f ss < f a ServiceStress "OK!" 1 d c cover 2 d 1 stirrup 2 d bar_a d c 3.58in Exposure Condition Factor: γ e 1.00 d c β s 1 β 0.7 h cap d s 1.06 c s max = minimum of: Service Load Bending Stress in outer layer of the reinforcing Allowable Bending Stress in the outer layer of the reinforcing (BDM-LRFD Ch. 4, Sect. 5, Design Criteria) "cover" is measured to center of shear reinforcement. For class 1 exposure conditions. For areas where deicing chenicals are frequently used, design for Class 2 Exposure ( e = 0.75). (BDM-LRFD, Ch. 4, Sect. 5, Design Criteria) 700γ e β s f ss 2d c 49.38in (AASHTO LRFD Eq ) & 12in A good practice is to place a bar every 12in along each surface of the s max 12.00in bent. (TxSP) 1 b stem 2 cover 2 d 1 stirrup 2 d bar_a s Actual BarANo 1 s Actual 7.96in < s max ServiceabilityCheck "OK!" Check Dead Load Check allowable M dl : f dl 22ksi BDM-LRFD, Chapter 4, Section 5, Design Criteria 1ft M a A s dj f dl M 12in a kipft TxDOT limits dead load stress to 22 ksi. This is due to observed cracking under dead load. Allowable Dead Load Moment M dl kipft < M a DeadLoadMom "OK!" LRFD Inverted Tee Bent Cap Design Example 25 June 2010

26 Flexural Reinforcement for Positive Bending (Bars B) (Tension in Bottom) M dl M posdl M dl 250.1kipft M s M posserv M s 492.5kipft From Cap 18 Output. See Pg. 10 M u M posult M u 741.7kipft Minimum Flexural Reinforcement (AASHTO LRFD ) Factored Flexural Resistance, M r, must be greater than or equal to the lesser of 1.2 M cr (Cracking Moment) or 1.33 Mu (Ultimate Moment) y t S M cr ybar y t 33.80in I g S in 3 y t 1ft Sf r M 12in cr kipft M f = minimum of: 1.2M cr 1.33M u kipft 986.5kipft Distance to the Center of Gravity of the Cap from the top of the Cap See Pg. 6 for calculations of "ybar" Section Modulus for the extreme tension fiber Cracking Moment (AASHTO LRFD Eq ) Design for the lesser of 1.2M cr or 1.33Mu when determining minimum area of steel required. Thus, M r must be greater than M f 986.5kipft Moment Capacity Design (AASHTO LRFD ) Try, 11 ~ #11's Bottom BarBNo 11 d bar_b 1.41in A bar_b 1.56in 2 Number of bars in tension Diameter of main reinforcing bars Area of one main reinforcing bar A s ( BarBNo) A bar_b A s 17.16in 2 Area of steel in tension d d s_pos d 81.42in See Pg. 18 for the calculation of "d s_pos." b b stem b 39in See Pg. 3 for the calculation of "b stem." c A s f y c 10.15in 0.85f c β 1 b Depth of Cross Section under Compression under Ultimate Load (AASHTO LRFD Eq ) This "c" is the distance from the extreme compression fiber to the neutral axis, not the distance from the center of bearing of the last girder to the end of the cap. a cβ 1 a 8.63in Depth of Equivalent Stress Block (AASHTO LRFD ) Note: "a" is less than "d stem " therefore the equivalent stress block acts over a rectangular area. If "a" was greater than "d stem " it would act over a Tee shaped area. M n a 1ft A s f y d M 2 12in n kipft Nominal Flexural Resistance (AASHTO LRFD Eq ) LRFD Inverted Tee Bent Cap Design Example 26 June 2010

27 Flexural Reinforcement for Positive Bending (Con't) (Bars B) Moment Capacity Design (Con't) d c ε s c ε s > FlexureBehavior ε s Strain in Reinforcing at Ultimate "Tension Controlled" (AASHTO LRFD ) ϕ M 0.90 (AASHTO LRFD ) M r ϕ M M n M r kipft Factored Flexural Resistance (AASHTO LRFD Eq ) M u 741.7kipft < M r MinReinfChk "OK!" M f 986.5kipft < M r UltimateMom "OK!" Check Serviceability (AASHTO LRFD ) To find s max : 1 d c cover 2 d 1 stirrup 2 d bar_b d c 3.58in "cover" is measured to center of shear reinforcement. Tension Reinforcement Ratio: A s ρ ρ bd For service loads, the stress on the cross-section is located as drawn: k dk ( 2ρn) ( ρn) 2 ( ρn) k in < d stem 57.00in Therefore, the compression force acts over a rectangular area. k j 1 j f ss M s 12in f A s j d 1ft ss 4.63ksi f a 0.6f y f a 36.00ksi f ss < f a ServiceStress "OK!" Exposure Condition Factor: γ e 1.00 d c β s 1 β 0.7 h cap d s 1.06 c s max = minimum of: & 12in s max 12.00in 700γ e 2d β s f c ss in Service Load Bending Stress in outer layer of the reinforcing Allowable Bending Stress in the outer layer of the reinforcing (BDM-LRFD Ch. 4, Sect. 5, Design Criteria) For class 1 exposure conditions. For areas where deicing chenicals are frequently used, design for Class 2 Exposure ( e = 0.75). (BDM-LRFD, Ch. 4, Sect. 5, Design Criteria) (AASHTO LRFD Eq ) A good practice is to place a bar every 12in along each surface of the bent. (TxSP) LRFD Inverted Tee Bent Cap Design Example 27 June 2010