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 "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. 153-1F, The University of Texas at Austin, August 1974 54' Type TX54 Girders (0.851 k / ft ) 6 Girders Spaced @ 8.00' with 3' overhangs 2" Haunch Span 2 112' Type TX54 Girders (0.851 k / ft ) 6 Girders Spaced @ 8.00' with 3' overhangs 3.75" Haunch Span 3 54' Type TX54 Girders (0.851 k / ft ) 6 Girders Spaced @ 8.00' with 3' overhangs 2" Haunch All Spans The basic bridge geometry can be found on the Bridge Layout located in the Appendices. Deck is 46ft wide Type T551 Rail (0.382k/ft) 8" Thick Slab (0.100 ksf) Assume 2" Overlay @ 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 (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
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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) asprescribed by TxDOT Bridge Design Manual - LRFD (May 2009).
"BDM-LRFD" refers to the TxDOTBridge Design Manual - LRFD (May2009)
"TxSP" refers to TxDOT guidance,recommendations, and standardpractice.
"Furlong & Mirza" refers to "Strengthand Servicability of Inverted T-BeamBent Caps Subject to CombinedFlexure, Shear, and Torsion", Centerfor Highway Research ResearchReport No. 153-1F, The University ofTexas at Austin, August 1974
54' Type TX54 Girders (0.851k/ft)6 Girders Spaced @ 8.00' with 3' overhangs2" Haunch
Span 2112' Type TX54 Girders (0.851k/ft)6 Girders Spaced @ 8.00' with 3' overhangs3.75" Haunch
Span 354' Type TX54 Girders (0.851k/ft)6 Girders Spaced @ 8.00' with 3' overhangs2" Haunch
All SpansThe basic bridge geometry can befound on the Bridge Layout located inthe Appendices.
BentsUse 36" Diameter Columns (Typical for Type TX54 Girders)
LRFD Inverted Tee Bent Cap Design Example 1 June 2010
Define Variables
Back Span Forward Span
Span1 54ft� Span2 112ft� Span Length
GdrSpa1 8ft� GdrSpa2 8ft� Girder Spacing
GdrNo1 6� GdrNo2 6� Number of Girders in Span
GdrWt1 0.851klf� GdrWt2 0.851klf� Weight of Girder
Haunch1 2in� Haunch2 3.75in� Size of Haunch
Bridge
Skew 0deg� Skew of Bents
BridgeW 46ft� Width of Bridge Deck
RdwyW 44ft� Width of Roadway
GirderD 54in� Depth of Type TX54 Girder
BrgSeat 1.5in� Bearing Seat Buildup
BrgPad 2.75in� Bearing Pad Thickness
SlabThk 8in� Thickness of Bridge Slab
OverlayThk 2in� Thickness of Overlay
RailWt 0.382klf� Weight of Rail
wc 0.150kcf� Unit Weight of Concrete for Loads
WOlay 0.140kcf� Unit Weight of Overlay
Bents
fc 3.60ksi� Concrete Strength
wcE 0.145kcf� Unit Weight of Concrete for Ec
Design Parameters (Con't)
Ec = 33 000 wcE1.5
� fc��� Ec 3457 ksi�� Modulus of Elasticity of Concrete,(AASHTO LRFD Eq. 5.4.2.4-1)
fy 60ksi� Yield Strength of reinforcement
Es 29000ksi� Modulus of Elasticity of Steel
Dcolumn 36in� Diameter of Columns
Other VariablesDynamic load allowance,(AASHTO LRFD Table 3.6.2.1-1)IM 33%�
LRFD Inverted Tee Bent Cap Design Example 2 June 2010
Determine Cap Dimensions
Stem Width The stem is typically at least 3" widerthan the Diameter of the Column(36") to allow for the extension of thecolumn reinforcement into the Cap.(TxSP)
bstem Dcolumn 3in�� bstem 39 in��
Stem Height
Distance From Top of Slab to Top of Ledge
DSlab_to_Ledge SlabThk Haunch2� GirderD� BrgPad� BrgSeat�� Haunch 2 is the larger of the twohaunches.
DSlab_to_Ledge 70.00 in��
StemHaunch 3.75 in�� 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 stemmust have at least 2" of haunch on it, but the haunch shouldnot be less than either of the haunches of the adjacent spans.
dstem DSlab_to_Ledge SlabThk� StemHaunch� 0.5in�� The stem must accommodate 1/2" ofbituminous fiber.
dstem 57.75 in��
Use: dstem 57 in�� Round the Stem Height down to thenearest 1". (TxSP)
LRFD Inverted Tee Bent Cap Design Example 3 June 2010
Determine Cap Dimensions (Con't)Ledge Width
The Ledge Width must be adequatefor Bar M to develop fully. 26" istypically used for TX54 girders, as it isadequate to develop a # 6 bar withthe typical 2.5" cover. If the cover isincreased to 3", allowing for amodification factor of 0.7, a 24" Ledgeis adequate to develop a # 7 bar.
I-Beams have 9.5" from the face ofthe cap to the CL of Bearing. Thetypical ledge width for these bridgesis 23".
"Ldh,prov" must be greater than orequal to "Ldh,req" for bar M.
"cover" is measured from the centerof the transverse bars.cover 2.5in�
"L" is the length of the Bearing Padalong the girder. A typical type TX54bearing pad is 8"x21" as shown in theIGEB standard.
L 8in�
Determine the Required Development Length of Bar M:
Try # 6 Bar for Bar M.
dbar_M 0.750in�
Abar_M 0.44in2�
Basic Development Length
=38.0 dbar_M�
fcLdh 15.02 in�� (AASHTO LRFD Eq. 5.11.2.4.1-1)
Modification Factors for Ldh: (AASHTO LRFD 5.11.2.4.2)
Is Top Cover greater than or equal to 2.5", and Side Cover greater than or equal to 2"?
"Side Cover" and "Top Cover" arethe clear cover on the side and topof the hook respectively. Thedimension "cover" is measuredfrom the center of Bar M.
SideCover coverdbar_M
2�� SideCover 2.13 in��
TopCover coverdbar_M
2�� TopCover 2.13 in��
No, Factor = 1.0
Ldh
LRFD Inverted Tee Bent Cap Design Example 4 June 2010
Determine Cap Dimensions (Con't)The Required Development Length is the larger of the following: (AASHTO LRFD 5.11.2.4.1)
Ldh Factor� 15.02 in��
8 dbar_M� 6.00 in��
6in
Therefore,
Ldh_req 15.02 in��
hcap 85 in��
bledge_min Ldh_req cover� 12in�L2
�� bledge_min 25.52 in��
Use:bledge 26 in��
Width of Bottom Flange
bf 2 bledge� bstem�� bf 91 in��
Ledge Depth
As a general rule of thumb LedgeDepth is greater than or equal to 2'-3".This is the depth at which a bent froma typical bridge will pass thepunching shear check (calculationsfound on Pg. 21).
Use a Ledge Depth of 28"
dledge 28in�
Total Depth of Cap
hcap dstem dledge�� hcap 85 in��
Summary of Cross Sectional Dimensions
bstem 39 in�� From Pg. 3
dstem 57 in�� From Pg. 3
bledge 26 in��
dledge 28 in��
The distance from the face ofthe stem to the center of bearingis 12" for TxGirders. (IGEB)
LRFD Inverted Tee Bent Cap Design Example 5 June 2010
Determine Cap Dimensions (Con't)Length of Cap
First define Girder Spacing and End Distance:
S 8ft� Girder Spacing
"c" is the distance from the CenterLine of the Exterior Girder to theEdge of the Cap measured alongthe Cap.
c 2ft�
LCap S GdrNo1 1�( )� 2 c��� LCap 44 ft� Length of Cap
TxDOT policy is as follows, "The edge distance between the exterior bearing pad and the end of the invertedT-beam shall not be less than 12in." (BDM-LRFD, Ch. 4, Sect. 5, Design Criteria) replacing the statement inAASHTO LRFD 5.13.2.5.5 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 (IGEB standard)
L 8 in�� Length of Bearing Pad
W 21in� Width of Bearing Pad
Cross Sectional Properties of Cap
Ag dledge bf� dstem bstem��� Ag 4771 in2��
Distance from bottom of cap to thecenter of gravity of the capybar
dledge bf�12
dledge��
��
� dstem bstem� dledge12
dstem���
��
��
Ag�
ybar 33.80 in��
Igbf dledge
3�
12bf dledge� ybar
12
dledge���
��
2��
bstem dstem3
�
12bstem dstem� ybar dledge
12
dstem���
��
����
���
2���
����
Ig 2.91 106� in4
��
LRFD Inverted Tee Bent Cap Design Example 6 June 2010
Cap AnalysisCap Model
Assume:
4 Columns Spaced @ 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 ModelStation = 0.5'
The circled numbers are the stations that will be used in the CAP18 input file. One station is 0.5ft in the direction perpendicular tothe pgl, not parallel to the bent.
station 0.5ft� Station increment for CAP18
Recall:
Ec 3457.14 ksi�� (Pg. 2) Ig 2.91 106� in4
�� (Pg.6)
Ec Ig� 1.01 1010� kip in2
��� / 12inft
�
��
2= Ec Ig� 6.99 107
� kip ft2���
LRFD Inverted Tee Bent Cap Design Example 7 June 2010
Cap Analysis (Con't)Dead Load
Values used in the followingequations can be found on Pg. 2.SPAN 1
Rail weight is distributed evenlyamong stringers, up to 3 stringersper rail. (TxSP)
Rail12 RailWt�
Span12
�
min GdrNo1 6��( )� Rail1 3.44
kipgirder
��
Increase slab DL by 10% toaccount for haunch and thickenedslab ends.
Slab1 wc GdrSpa1� SlabThk�Span1
2� 1.10�� Slab1 23.76
kipgirder
��
Girder1 GdrWt1Span1
2�� Girder1 22.98
kipgirder
��
Overlay is calculated separately,because it has a different loadfactor than the rest of the deadloads.
DLRxn1 Rail1 Slab1� Girder1�� DLRxn1 50.17kip
girder��
Overlay1 WOlay GdrSpa1� OverlayThk�Span1
2�� Overlay1 5.04
kipgirder
�� Design for future overlay.
SPAN 2
Rail22 RailWt�
Span22
�
min GdrNo2 6��( )� Rail2 7.13
kipgirder
��
Slab2 wc GdrSpa2� SlabThk�Span2
2� 1.10�� Slab2 49.28
kipgirder
��
Girder2 GdrWt2Span2
2�� Girder2 47.66
kipgirder
��
DLRxn2 Rail2 Slab2� Girder2�� DLRxn2 104.07kip
girder��
Overlay2 WOlay GdrSpa2� OverlayThk�Span2
2�� Overlay2 10.45
kipgirder
��
CAP
Cap wc Ag� 4.970kipft
��� * 0.5ftstation
= Cap 2.485kip
station��
LRFD Inverted Tee Bent Cap Design Example 8 June 2010
Cap Analysis (Con't)Live Load (AASHTO LRFD 3.6.1.2.2 and 3.6.1.2.4)
LongSpan max Span1 Span2��( )�
ShortSpan min Span1 Span2��( )�
LongSpan 112.00 ft�
ShortSpan 54.00 ft�
IM 0.33�
Use HL-93 Live Load. For maximumreaction at interior bents, "DesignTruck" will always govern over"Design Tandem". For the maximumreaction when the long span is morethan twice as long as the short span,place the rear (32 kip) axle over thesupport and the middle (32 kip) andfront (8 kip) axles on the long span.For the maximum reaction when thelong span is less than twice as longas the short span, place the middle(32 kip) axle over the support, thefront (8 kip) axle on the short spanand the rear (32 kip) axle on the longspan.
P 16.0kip 1 IM�( )�� The Live Load is applied to the slabby two 16 kip wheel loads increasedby the dynamic load allowance withthe remainder of the live loaddistributed over a 10 ft (AASHTOLRFD 3.6.1.2.1) design lane width.(TxSP)
P 21.28 kip��
wLLRxn 2 P�( )�
10ft�
w 9.83kipft
�� * 0.5ftstation
The Live Load applied to the slab isdistributed to the beams assumingthe slab is hinged at each beamexcept the outside beam.(BDM-LRFD, Ch. 4, Sect. 5, StructuralAnalysis)
w 4.92kip
station��
LRFD Inverted Tee Bent Cap Design Example 9 June 2010
Cap 18 InputMultiple Presence Factors, m (AASHTO LRFD Table 3.6.1.1.2-1) Input "Multiple Presence Factors" into
Cap18 as "Load Reduction Factors".No. of Lanes Factor "m"
1 2 3>3
1.20 1.000.850.65
Limit States (AASHTO LRFD 3.4.1) The cap design need only considerStrength I, Service I, and Service Iwith DL. (TxSP)Strength I
Live Load and Dynamic Load Allowance LL + IM = 1.75
Dead Load Components DC = 1.25
Dead Load Wearing Surface (Overlay) DW = 1.50 TxDOT allows the Overlay Factor tobe reduced to 1.25 (TxSP), sinceoverlay is typically used in designonly to increase the safety factor, butin this example we will use DW = 1.50.
Cap Analysis (Con't)
Live Load and Dynamic Load Allowance LL + IM = 1.00
Dead Load and Wearing Surface DC & DW = 1.00
Dead Load
TxDOT considers Service level Dead Load only with a limitreinforcement stress of 22 ksi to minimize cracking. (BDM-LRFD, Chapter 4, Section 5, Design Criteria)
Cap 18 OutputMax +M Max -M
Dead Load: MposDL 250.1kip ft�� MnegDL 379.4� kip ft�� These loads are the maximum loadsfrom the Cap 18 Output File located inthe Appendices.Service Load: MposServ 492.5kip ft�� MnegServ 590.9� kip ft��
LRFD Inverted Tee Bent Cap Design Example 10 June 2010
Cap Analysis (Con't)Girder Reactions on Ledge:
Dead LoadFor calculations of these loadssee Pg. 8.DLSpan1 Rail1 Slab1� Girder1�� DLSpan1 50.17
kipgirder
��
Overlay1 5.04kip
girder��
DLSpan2 Rail2 Slab2� Girder2�� DLSpan2 104.07kip
girder��
Overlay2 10.45kip
girder��
Live Load (AASHTO LRFD 3.6.1.2.2 and 3.6.1.2.4)
Loads per Lane:Use HL-93 Live Load. For maximumreaction at interior bents, "DesignTruck" will always govern over"Design Tandem" for Spans greaterthan 26ft. For the maximum reaction,place the back (32 kip) axle over thesupport.
LaneSpan1 0.64klfSpan1
2�
��
�� LaneSpan1 17.28kip
lane��
LaneSpan2 0.64klfSpan2
2�
��
�� LaneSpan2 35.84kip
lane��
TruckSpan1 32.0kip 32.0kipSpan1 14ft�
Span1�
��
�� 8.0kipSpan1 28ft�
Span1�
��
��� TruckSpan1 59.56kip
lane��
TruckSpan2 32.0kip 32.0kipSpan2 14ft�
Span2�
��
�� 8.0kipSpan2 28ft�
Span2�
��
��� TruckSpan2 66.00kip
lane��
LRFD Inverted Tee Bent Cap Design Example 11 June 2010
Cap Analysis (Con't)Girder Reactions on Ledge (Con't) :
Live Load (Con't)
Loads per Lane (Con't) :
IM 0.33�
Combine "Design Truck" and "DesignLane" loadings. (AASHTO LRFD 3.6.1.3)
gVSpan1_Int 0.814� The Live Load Reactions areassumed to be the Shear Live LoadDistribution Factor multiplied by theLive Load Reaction per Lane. TheShear Live Load Distribution Factorwas calculated using the "LRFD LiveLoad Distribution Factors"Spreadsheet found in theAppendices.
The Exterior Girders must have a LiveLoad Distribution Factor equal to orgreater than the Interior Girders. Thisis to accommodate a possible futurebridge widening. Widening thebridge would cause the exteriorgirders to become interior girders.
LRFD Inverted Tee Bent Cap Design Example 13 June 2010
Cap Analysis (Con't)Torsional Loads Strength I Limit State, (AASHTO LRFD 3.4.1)
To maximize the torsion, the liveload only acts on the longer span inthe configuration shown.
The loads are applied to the cap as depicted in the following picture:
av 12in� "av" is the value for the distance from the face of the stem to the center of bearing forthe girders. 12" is the typical value for TxGirders on Inverted Tee Bents (IGEB). 9" isthe typical value for I-Beams (IBEB). The lever arm for the torsional loads is thedistance from the center line of bearing to the centerline of the cap (1/2bstem + av).
bstem 39 in�� From Pg. 3
LeverArm av12
bstem��� LeverArm 31.5 in��
LRFD Inverted Tee Bent Cap Design Example 14 June 2010
Analyzed assuming Bents are torsionally rigid at Effective Face of Columns.
Tu 660kip ft�� Maximum Torsion on Cap
LRFD Inverted Tee Bent Cap Design Example 15 June 2010
Cap Analysis (Con't)Load Summary
Ledge LoadsInterior Girder
Service Load
Vs_Int max Vs_Span1Int Vs_Span2Int��� �� Vs_Int 215.15 kip��
Factored Load
Vu_Int max Vu_Span1Int Vu_Span2Int��� �� Vu_Int 321.86 kip��
Exterior GirderService Load
Vs_Ext max Vs_Span1Ext Vs_Span2Ext��� �� Vs_Ext 215.15 kip��
Factored Load
Vu_Ext max Vu_Span1Ext Vu_Span2Ext��� �� Vu_Ext 321.86 kip��
Cap LoadsPositive Moment (From CAP 18)
Dead Load: MposDL 250.10 kip ft���
Service Load: MposServ 492.50 kip ft���
Factored Load: MposUlt 741.70 kip ft���
Negative Moment (From CAP 18)
Dead Load: MnegDL 379.40� kip ft���
Service Load: MnegServ 590.90� kip ft���
Factored Load: MnegUlt 852.10� kip ft���
Maximum Torsion and Concurrent Shear and Moment (Strength I)
Tu 660 kip ft��� Located two stations awayfrom centerline of column.
Vu 448.1kip� Vu and Mu values arefrom CAP 18Mu 335.6kip ft��
In this example the maximum Torsion and the maximum Shear are concurrent with each other. If they are not, itbecomes necessary to check the location of the maximum Torsion with its concurrent Shear and the location ofthe maximum Shear with its concurrent Torsion.
LRFD Inverted Tee Bent Cap Design Example 16 June 2010
Locate and Describe Reinforcement
Recall:
bstem 39 in�� From Pg. 3
dstem 57 in�� From Pg. 3
bledge 26 in�� From Pg. 5
dledge 28 in�� From Pg. 5
bf 91 in�� From Pg. 5
hcap 85 in�� From Pg. 5
cover 2.50 in�� From Pg. 4Measured from Center of bar
LRFD Inverted Tee Bent Cap Design Example 17 June 2010
Locate and Describe Reinforcement (Con't)Describe Reinforcing Bars
Use # 11 bars for Bar A
Abar_A 1.56in2� dbar_A 1.410in�
Use # 11 bars for Bar B
Abar_B 1.56in2� dbar_B 1.410in�
Use # 6 bars for Bar MBar M must be a # 6 bar or smaller toallow it to fully develop, as stated onPg 4.
Abar_M 0.44in2� dbar_M 0.75in�
Use # 6 bars for Bar N
To prevent confusion, use the samebar size for Bar N as Bar M.Abar_N 0.44in2
� dbar_N 0.75in�
Use # 6 bars for Bar S
Abar_S 0.44in2� dbar_S 0.75in�
Use # 6 bars for Bar T
Abar_T 0.44in2� dbar_T 0.750in�
Calculate Dimensions
ds_neg hcap cover�12
dbar_S��12
dbar_A���
ds_neg 81.42 in��
ds_pos hcap cover�12
max dbar_S dbar_M��� ���12
dbar_B���
ds_pos 81.42 in��
av 12 in�� Typical for TX Girders on InvertedTee Bent Caps (IGEB standard)
af av cover�� af 14.50 in��
de dledge cover�� de 25.50 in��
df dledge cover�12
dbar_M��12
dbar_B��� df 24.42 in��
h dledge BrgSeat�� h 29.50 in�� "BrgSeat" is the height of the BearingSeat Buildup. This value is definedon Pg. 2.
LRFD Inverted Tee Bent Cap Design Example 18 June 2010
Locate and Describe Reinforcement (Con't)Calculate Dimensions (Con't)
α 90deg� Angle of Bars S
Recall:
L 8 in�� From Pg. 4
W 21 in�� From Pg. 6
LRFD Inverted Tee Bent Cap Design Example 19 June 2010
Check Bearing (AASHTO LRFD 5.7.5)
The load on the bearing pad propagates along atruncated pyramid whose top has the area A1 andwhose base has the area A2. A1 is the loadedarea (the bearing pad area: LxW ). A2 is the areaof the lowest rectangle contained wholly within thesupport (the Inverted Tee Cap). A2 must notoverlap the truncated pyramid of another load ineither direction, nor can it extend beyond theedges of the cap in any direction. Plan View
ϕ 0.7� (AASHTO LRFD 5.5.4.2.1)
A1 W L�� A1 168 in2�� Area under Bearing Pad
Interior Girders"B" is the distance from theperimeter of A1 to theperimeter of A2, as seen inthe above figures.
LRFD Inverted Tee Bent Cap Design Example 20 June 2010
Check Punching Shear
Vu_Ext From Pg. 16
ϕ 0.9� (AASHTO LRFD 5.5.4.2.1)
Determine if the Shear Cones Intersect
Is12
S�12
W�� df� ? Yes. Therefore Shear Cones do not intersect in the longitudinaldirection of the Cap.
TxDOT uses "df" instead of "de" forPunching Shear (BDM-LRFD, Ch. 4,Sect. 5, Design Criteria). This isbecause "df" has traditionally beenused for inverted tee bents and wasused in the Inverted Tee Research(Furlong & Mirza pg. 58).
12
S�12
W�� 37.50 in��
df 24.42 in��
Is12
bstem av�12
L�� df� ? Yes. Therefore Shear Cones do not intersect in the transversedirection of the Cap.
d ds_neg� d 81.42 in�� See Pg. 18 for the calculation of"ds_neg,"
b bf� b 91 in�� See Pg. 5 for the calculation of "bf."
fc 3.60 ksi�� Compressive Strength of Concrete
fy 60ksi� Yield Strength of Rebar
β1 = 0.85 0.05 fc 4ksi�� ��� (AASHTO LRFD 5.7.2.2)
Bounded by: 0.65 β1� 0.85� β1 0.85�
cAs fy�
0.85 fc� β1� b�� c 1.98 in�� Depth of Cross Section under
Compression under Ultimate Load(AASHTO LRFD Eq. 5.7.3.1.2-4)
This "c" is the distance from the extreme compression fiber to the neutral axis, not thedistance from the center of bearing of the last girder to the end of the cap.
a c β1�� a 1.68 in�� Depth of Equivalent Stress Block(AASHTO LRFD 5.7.2.2)
Note: "a" is less than "dledge" therefore the equivalent stress block acts over arectangular area. If "a" was greater than "dledge" it would act over a Tee shaped area.
Check Dead Load BDM-LRFD, Chapter 4, Section 5, Design CriteriaTxDOT limits dead load stress to 22ksi. This is due to observed crackingunder dead load.
Check allowable Mdl: fdl 22ksi�
Ma As d� j� fdl�1ft
12in�� Ma 1116.04 kip ft��� Allowable Dead Load Moment
(AASHTO LRFD 5.7.3.4)
Mdl 379.40 kip ft��� < Ma DeadLoadMom "OK!"�
sActual
bstem 2 cover12
dstirrup��12
dbar_A���
��
��
BarANo 1��
LRFD Inverted Tee Bent Cap Design Example 25 June 2010
Flexural Reinforcement for Positive Bending (Bars B) (Tension in Bottom)
Mdl MposDL� Mdl 250.1 kip ft���
Ms MposServ� Ms 492.5 kip ft��� From Cap 18 Output. See Pg. 10
Factored Flexural Resistance, Mr, must be greater than or equal to the lesser of 1.2 Mcr (Cracking Moment)or 1.33 Mu (Ultimate Moment)
yt ybar� yt 33.80 in�� Distance to the Center of Gravity ofthe Cap from the top of the CapSee Pg. 6 for calculations of "ybar"
SIgyt
� S 8.62 104� in3
�� Section Modulus for the extremetension fiber
Mcr S fr�1ft
12in�� Mcr 3269.9 kip ft��� Cracking Moment
(AASHTO LRFD Eq. 5.7.3.3.2-1)
Mf = minimum of: Design for the lesser of 1.2Mcr or1.33Mu when determining minimumarea of steel required. 1.2 Mcr� 3923.9 kip ft���
1.33 Mu� 986.5 kip ft���
Thus, Mr must be greater than Mf 986.5 kip ft���
Moment Capacity Design (AASHTO LRFD 5.7.3.2)
Try, 11 ~ #11's Bottom
BarBNo 11� Number of bars in tension
dbar_B 1.41in� Diameter of main reinforcing bars
Abar_B 1.56in2� Area of one main reinforcing bar
As BarBNo( ) Abar_B�� As 17.16 in2�� Area of steel in tension
d ds_pos� d 81.42 in�� See Pg. 18 for the calculation of"ds_pos."
b bstem� b 39 in�� See Pg. 3 for the calculation of "bstem."
cAs fy�
0.85 fc� β1� b�� c 10.15 in�� Depth of Cross Section under
Compression under Ultimate Load(AASHTO LRFD Eq. 5.7.3.1.2-4)
This "c" is the distance from the extreme compression fiber to the neutral axis, not thedistance from the center of bearing of the last girder to the end of the cap.
a c β1�� a 8.63 in�� Depth of Equivalent Stress Block(AASHTO LRFD 5.7.2.2)
Note: "a" is less than "dstem" therefore the equivalent stress block acts over arectangular area. If "a" was greater than "dstem" it would act over a Tee shaped area.
Check Dead Load BDM-LRFD, Chapter 4, Section 5, Design Criteria
TxDOT limits dead load stress to 22ksi. This is due to observed crackingunder dead load.
Check allowable Mdl: fdl 22ksi�
Ma As d� j� fdl�1ft
12in�� Ma 2340.19 kip ft��� Allowable Dead Load Moment
Flexural Reinforcement for Positive Bending (Con't) (Bars B)
Mdl 250.10 kip ft��� < Ma DeadLoadMom "OK!"�
Flexural Steel Summary: Use 5~#11 Bars on Top& 11~#11 Bars on Bottom
LRFD Inverted Tee Bent Cap Design Example 28 June 2010
Ledge Reinforcement (Bars M&N)Use trial and error to determine thespacing needed for the ledgereinforcing. Bars M @ 5" and Bars N@ 10" failed one of the checks by asmall margin (1%). 4.95" spacingwas chosen because it shoud passall the checks.
Try Bars M at a 4.95" spacing, and Bars N at a 9.90" spacing.
sbar_M 4.95in�
sbar_N 9.90in�
It is typical for Bars N to be at everyother Bar M.
Determine Distribution WidthsThese distribution widths will be used on the following three pages to determine the required ledgereinforcement per foot of cap.
Distribution Width for Shear (AASHTO LRFD 5.13.2.5.2) Note: These are the same distributionwidths used for the Shear Frictioncheck on Pg. 22.Interior Girders
bs_Int = minimum of:
W 4 av� 69.00 in��
S 96.00 in�� "S" is the girder spacing.(From Pg. 6)
bs_Int 69.00 in��
Exterior Girdersbs_Ext = minimum of:
W 4 av� 69.00 in��
S 96.00 in��
2 c 48.00 in�� "c" is the distance from the centerof bearing of the outside beam tothe end of the ledge. (From Pg. 6)bs_Ext 48.00 in��
Distribution Width for Bending and Axial Loads (AASHTO LRFD 5.13.2.5.3)
Interior Girdersbm_Int = minimum of:
W 5 af� 93.50 in��
S 96.00 in��
bm_Int 93.50 in��
Exterior Girdersbm_Ext = minimum of:
W 5 af� 93.50 in��
S 96.00 in��
2 c 48.00 in��
bm_Ext 48.00 in��
LRFD Inverted Tee Bent Cap Design Example 29 June 2010
Ledge Reinforcement (Con't) (Bars M&N)The reinforcing required for shear friction, flexure, and axial tension will be calculated on Pgs. 30-32. The reinforcingprovided by Bars M and Bars N must exceed the checks that are found on the bottom of Pg. 32 (These checkedcan be found on pages 33 and 34). These checks combine the reinforcing requirements for shear friction, flexure,and axial tension, as per AASHTO LRFD 5.13.2.4.2.
Reinforcing Required for Shear Friction (AASHTO LRFD 5.8.4.1)
ϕ 0.9� (AASHTO LRFD 5.5.4.2.1)
μ 1.4� c1 0ksi� Pc 0kip� "�" is 1.4 for monolithically placedconcrete. (AASHTO LRFD 5.8.4.3)
Recall: de 25.5 in�� (From Pg. 18)For clarity, the cohesion factor islabeled "c1". This is to preventconfusion with "c", the distance fromthe last girder to the edge of the cap.c1 is 0ksi for corbels and ledges.(AASHTO LRFD 5.8.4.3)
(AASHTO LRFD Eq. 5.7.3.1.2-4)This "c" is the distance from theextreme compression fiber to theneutral axis, not the distance from thecenter of bearing of the last girder tothe end of the cap.
(AASHTO LRFD Eq. 5.7.3.1.2-4)This "c" is the distance from theextreme compression fiber to theneutral axis, not the distance from thecenter of bearing of the last girder tothe end of the cap.
The following checks bars M and N. These are the checks found on Pg. 32, modified to check the steel perunit length instead of the total steel.
Check Interior GirdersBar M:
Check if: as as_min� (AASHTO LRFD 5.13.2.4.1)
as af_Int an_Int�� (AASHTO LRFD 5.13.2.4.2)
as2 avf_Int�
3an_Int�� (AASHTO LRFD Eq. 5.13.2.4.2-5)
as 1.07in2
ft��
as_min 0.73in2
ft�� < as
af_Int an_Int� 0.54in2
ft�� < as
2 avf_Int�
3an_Int� 0.65
in2
ft�� < as
BarMCheck "OK!"�
Bar N:Check if: ah 0.5 as an_Int�� ��� (AASHTO LRFD Eq. 5.13.2.4.2-6)
"as" in this equation is the steelrequired for Bar M, based on therequirements for Bar M in AASHTOLRFD 5.13.2.4.2.This is derived from the suggestionthat Ah should not be less than Af/2nor less than Avf/3 (Furlong & Mirzapg. 73 & 74)
as = The maximum of:
af_Int an_Int�
2 avf_Int�
3an_Int�
as 0.54in2
ft��
ah 0.53in2
ft��
0.5 as an_Int�� �� 0.19in2
ft�� < ah
BarNCheck "OK!"�
LRFD Inverted Tee Bent Cap Design Example 33 June 2010
Check Required Reinforcing (Con't)
Check Exterior GirdersBar M
Check if: as as_min� (AASHTO LRFD 5.13.2.4.1)
as af_Ext an_Ext�� (AASHTO LRFD 5.13.2.4.2)
as2 avf_Ext�
3an_Ext�� (AASHTO LRFD Eq. 5.13.2.4.2-5)
as 1.07in2
ft��
as_min 0.73in2
ft�� < as
af_Ext an_Ext� 1.06in2
ft�� < as
2 avf_Ext�
3an_Ext� 1.01
in2
ft�� < as
BarMCheck "OK!"�
Bar NCheck if: ah 0.5 as an_Int�� ��� (AASHTO LRFD Eq. 5.13.2.4.2-6)
as = The maximum of: "as" in this equation is the steelrequired for Bar M, based on therequirements for Bar M in AASHTOLRFD 5.13.2.4.2.This is derived from the suggestionthat Ah should not be less than Af/2nor less than Avf/3 (Furlong & Mirzapg. 73 & 74)
af_Int an_Int�
2 avf_Int�
3an_Int�
as 1.06in2
ft��
ah 0.53in2
ft��
Ledge Reinforcement (Con't) (Bars M&N)
0.5 as an_Int�� �� 0.46in2
ft�� < ah
BarNCheck "OK!"�
Ledge Reinforcement Summary: Use #6 primary ledge reinforcing @ 4.95" maximum spacing& #6 auxiliary ledge reinforcing @ 9.90" maximum spacing
LRFD Inverted Tee Bent Cap Design Example 34 June 2010
Try Double # 6 Stirrups at a 8" spacing. Use trial and error to determine thespacing needed for the hangerreinforcing.sbar_S 8.5in�
Ahr 2stirrups Abar_S�� Ahr 0.88 in2�� Ahr is the area of one leg of hanger
reinforcement. In this example thearea of the bar is multiplied by 2because the Inverted Tee hasdouble stirrups.
Criteria ~ Modified to limit thedistribution width to the girderspacing. This will prevent distributionwidths from overlapping)
Vall
Vall 236 kip��
Vs_Int 215 kip�� < Vall ServiceCheck "OK!"�
Exterior GirdersVall = minimum of:
Vall for the interior girder
(BDM-LRFD Ch.4, Sect. 5, DesignCriteria ~ Modified to limit thedistribution width to the edge of thecap. This will prevent distributionwidths from extending over the edgeof the cap.)
Ahr23
fy��
��
�
sbar_S
W 3 av��
2c�
�
��
� 217 kip��
Ahr23
fy��
��
�
sbar_S
S2
c��
��
� 298 kip�� (BDM-LRFD Ch.4, Sect. 5, DesignCriteria ~ Modified to limit thedistribution width to half the girderspacing and the distance to the edgeof the cap. This will preventdistribution widths from overlappingor extending over the edge of thecap.)
Hanger Reinforcement (Bars S)
Vall 217 kip��
Vs_Ext 215 kip�� < Vall ServiceCheck "OK!"�
LRFD Inverted Tee Bent Cap Design Example 35 June 2010
� 447 kip�� (AASHTO LRFD Eq. 5.13.2.5.5-2)Modified to limit the distribution widthto the edge of the cap.
0.063 fc� bf� df�� �Ahr fy�
sbar_S
W 2 df��
2c�
�
��
�� 632 kip�� (AASHTO LRFD Eq. 5.13.2.5.5-3)Modified to limit the distribution widthto the edge of the cap.
Hanger Reinforcement (Con't) (Bars S)
Vn 447 kip��
Check Strength Limit State
ϕVn 403 kip��
Vu_Ext 322 kip�� < ϕVn UltimateCheck "OK!"�
LRFD Inverted Tee Bent Cap Design Example 36 June 2010
Combined Shear and TorsionThe following calculations are for Station 36. All critical locations must be checked. See the the Concrete SectionShear Capacity spreadsheet in the appendices for calculations at other locations. Shear and Moments werecalculated using the CAP 18 program.
Mu 335.6kip ft�� Vu 448.1 kip�� Nu 0kip� Tu 660 kip ft��� These loads can be found on Pg. 16
Recall: β1 0.85� fy 60 ksi��
fc 3.6 ksi�� Es 29000 ksi��
bf 91 in�� hcap 85 in�� bstem 39 in��
bv bstem� bv 39.00 in��
Find dv:
The Moment calculations to the leftreiterate those on Pg. 24.As Abar_A BarANo�� As 7.80 in2
��
Shears are maximum near thecolumn faces. In these regions thecap is in negative bending withtension in the top of the cap.Therefore, the calculations are basedon the steel in the top of the bent cap.
cAs fy�
0.85 fc� β1� bf�� c 1.98 in��
a c β1�� a 1.68 in��
ds ds_neg� d 81.42 in��
Mn As fy� da2
��
��
�� Mn 3142.6 kip ft���
Hanger Reinforcement (Con't) (Bars S)
Aps 0in2�
deAps fps� dp� As fy� ds��
Aps fps� As fy��� de 81.42 in�� (AASHTO LRFD Eq. 5.8.2.9-2)
dv need not be less than the greater of 0.9de and 0.72h: (AASHTO LRFD 5.8.2.9)
dv = maximum of:
MnAs fy� Aps fps��
80.58 in�� (AASHTO LRFD Eq. C5.8.2.9-1)
0.9 de� 73.28 in��
0.72 h� 21.24 in��
dv 80.58 in��
The method for calculating �and � used in this design example are from AASHTO LRFD AppendixB5. The method from AASHTO LRFD 5.8.3.4.2 may be used instead. The method from 5.8.3.4.2 isbased on the method from Appendix B5; however, it is less accurate and more conservative (oftenexcessively conservative). The method from Apendix B5 is preferred because it is more accurate,but it requires iterating to a solution. The method from 5.8.3.4.2 can be used when doingcalculations by hand.
LRFD Inverted Tee Bent Cap Design Example 37 June 2010
Check Combined Shear and Torsion (Con't)Determine � and �:
ϕv 0.9� (AASHTO LRFD 5.5.4.2.1)
vuVu ϕv Vp�� ��
ϕv bv� dv�� vu 0.16 ksi�� Shear Stress on the Concrete
(AASHTO LRFD Eq. 5.8.2.9-1)
vufc
0.04�
Using Table B5.2-1 withvufc
0.04� and εx 0.001� Determining � and � is an iterativeprocess, therefore, assume initialshear strain value �x of 0.001per LRFD B5.2 and then verifythat the assumption was valid.
θ 36.4 deg� and β 2.23�
εx
Mudv
0.5 Nu� 0.5 Vu Vp� cot θ( )� Aps fpo���
���
2 Es As� Ep Aps��� �� Strain halfway between thecompressive and tensile resultants(AASHTO LRFD Eq. B5.2-1)If �x < 0, then use equationB5.2-3 and re-solve for �x.
Hanger Reinforcement (Con't) (Bars S)
where, Mu 335.60 kip ft��� Must be > Vu Vp� dv� 3008.98 kip ft���
εx 1.66 10 3� inin
�� > 1.00 10 3��
inin
use εx 1.00 10 3��
inin
� For values of �x greater than 0.001,the tensile strain in the reinforcing,�t is greater than 0.002. (�t = 2 �x - �c , where �c is < 0)Grade 60 steel yields at a strain of60 ksi / 29,000 ksi = 0.002. By limiting the tensile strain in thesteel to the yield strain and usingthe Modulus of Elasticity of thesteel prior to yield, this limits thetensile stress of the steel to theyield stress. �x has not changedfrom the assumed value, therefore noiterations are required.
Vp 0kip� "Vp" is zero as there is noprestressing.
Ac bstemhcap
2�� Ac 1657.5 in2
�� (AASHTO LRFD B5.2)"Ac" is the area of concrete on theflexural tension side of the cap, fromthe extreme tension fiber to one halfthe cap depth."Ac" is needed if AASHTO LRFD Eq.B5.2-1 is negative.
s sbar_S� s 8.50 in��
LRFD Inverted Tee Bent Cap Design Example 38 June 2010
Hanger Reinforcement (Con't) (Bars S)Check Combined Shear and Torsion (Con't)
Av 2 legs 2� stirrups Abar_S�� Av 1.76 in2�� The transverse reinforcement, "Av", is
double closed stirrups. The failuresurface intersects four stirrup legs,therefore the area of the shear steel isfour times the stirrup bar's area(0.44in2). See the sketch of thefailure plane to the left.
At 1leg Abar_S�� At 0.44 in2�� "At" is the area of the outer stirrup. At
is not independent of Av, but it is oneleg of the outer stirrup of Av.
Aoh dstem� � bstem 2 cover��� �� dledge 2 cover��� � bf 2 cover��� ���� "Aoh" is the area inside the centerlineof the exterior stirrup.
Aoh 3916 in2��
Ao 0.85 Aoh�� Ao 3329 in2�� (AASHTO LRFD C5.8.2.1)
Ao = 0.85 * the area inside thecenterline of the exterior stirrup.
Total Required Transverse Steel (AASHTO LRFD 5.8.3.6.1)
areq av_req 2 sides at_req��� areq 0.53in2
ft�� The transverse reinforcement is
designed for the side of the sectionwhere the effects of shear and torsionare additive. (AASHTO LRFDC5.8.3.6.1) Both sides of the sectionneed this reinforcing because thetorsion can act on either side of thesection.
aprovAv
sbar_S� aprov 2.48
in2
ft��
Hanger Reinforcement (Con't) (Bars S)
aprov > areq TransverseSteelCheck "OK"�
LRFD Inverted Tee Bent Cap Design Example 40 June 2010
Maximum Spacing of Transverse Reinforcement (AASHTO LRFD 5.8.2.7)
Shear Stress
vuVu ϕv Vp��
ϕv bv� dv�� vu 0.158 ksi�� (AASHTO LRFD Eq. 5.8.2.9-1)
0.125 fc� 0.450 ksi��
if vu 0.125 fc�� , smax = minimum of: (AASHTO LRFD Eq. 5.8.2.7-1)
0.8 dv� 64.46 in��
& 24in
if vu 0.125 fc�� , smax = minimum of: (AASHTO LRFD Eq. 5.8.2.7-2)
0.4 dv� 32.23 in��
& 12in
Since vu < 0.125*fc, smax 24.00 in��
Hanger Reinforcement (Con't) (Bars S)
TxDOT limits the maximum transverse reinforcementspacing to 12", therefore:
(BDM-LRFD, Ch. 4, Sect. 4, Detailing)
smax 12.00 in��
sbar_S 8.50 in�� < smax SpacingCheck "OK!"�
Hanger Reinforcement Summary: Use double #6 stirrups @ 8.5" maximum spacing
Vertical End Reinforcement (Bars C) (BDM-LRFD, Ch. 4, Sect. 5, Detailing)
Extra vertical reinforcing across end surfaces of the stem is provided to resist cracking which has beenobserved in existing bridges. (BDM-LRFD, Ch. 4, Sect. 5, Detailing) We will place #5 bars atapproximately 6 in spacing. (TxSP) See the Bent Cap Details in the Appendices for detail of Bar C.
Use 6~#5 bars evenly spaced across each end of the stem
LRFD Inverted Tee Bent Cap Design Example 41 June 2010
Skin Reinforcement (Bars T)
Abar_T 0.44 in2��
NoTBarsStem 7�
NoTBarsLedge 3�
"a" must be within 2/3 of de
(AASHTO LRFD 5.13.2.4.1)
23
de� 17.00 in��
TxDOT typically uses: a 6in� (TxSP)
Required Area of Skin Reinforcement(AASHTO LRFDEq. 5.7.3.4-2)Ask_Req = 0.012 d 30�( )�
Ask_Req 0.62in2
ft��
Ask need not be greater than one quarter of the main reinforcing (As/4) per side face within d/2 of the mainreinforcing. (AASHTO LRFD Eq. 5.7.3.4)
Ask_max = maximum of:
Abar_A BarANo�
4
ds_neg2
� 0.57in2
ft��
Abar_B BarBNo�
4
ds_pos2
� 1.261ft
in2��
Ask_max 1.26in2
ft��
AskReq = minimum of:
Ask_Req 0.62in2
ft��
Ask_max 1.261ft
in2��
AskReq 0.62in2
ft��
Try 7~#6 bars in stem and 3~#6 bars in Ledge on each side
LRFD Inverted Tee Bent Cap Design Example 42 June 2010
Skin Reinforcement (Con't) (Bars T)
SkinSpacing "OK!"�
Required Spacing of Skin Reinforcement
Skin Reinforcement Summary:
sreq = minimum of:
Abar_TAskReq
8.56 in��
ds_neg6
13.57 in��
Use 7~#6 bars in stem and 3~#6 bars in Ledge on each side
See the Bent Cap Detail Sheet in the Appendices for the resulting design of all the preceding calculations.
sreq 8.56 in��
Actual Spacing of Skin ReinforcementCheck T bars Spacing in Stem:
"cover" is measured to centerof shear reinforcement.
htop 56.67 in��
sskStemhtop
NoTBarsStem 1�� sskStem 7.08 in��
sreq 8.56 in�� > sskStem SkinSpacing "OK!"�
Check T bars Spacing in Ledge:
hbot dledge coverdbar_M
2�
dbar_T2
��
��
� coverdbar_S
2�
dbar_B2
��
��
�� "cover" is measured to centerof shear reinforcement.
hbot 21.17 in��
sskLedgehbot a�
NoTBarsLedge 1�� sskLedge 7.58 in��
sreq 8.56 in�� > sskLedge SkinSpacing "OK!"�
Check if "a" dimension is less than or equal to s req:
a 6.00 in�� < sreq 8.56 in��
ds_pos6
13.57 in��
12in&
htop dstem coverdbar_S
2�
dbar_A2
��
��
� coverdbar_M
2�
dbar_T2
��
��
��
(AASHTO LRFD 5.7.3.4)
LRFD Inverted Tee Bent Cap Design Example 43 June 2010
LRFD Inverted Tee Bent Cap Design Example 46 June 2010
APR 22, 2010 TEXAS DEPARTMENT OF TRANSPORTATION (TxDOT) PAGE 1 CAP18 BENT CAP ANALYSIS Ver 6.1
PSF HIGHWAY PD- CONTROL- CODED NO COUNTY NO IPE SECTION-JOB BY DATE 00001 ___County____ Highwy Pro# XXXX-XX-XXX BRG APR 22, 2010 Comment
CAP18 Version 6.10 Inverted Tee Cap Design Example, Skew = 0.00 PROB 1 (Spans 54'-112'-54', Type TX54 Girder @ 8.0', 8" Slab, 2" O'lay)
ENGLISH SYSTEM UNITS
TABLE 1. CONTROL DATA ENVELOPES TABLE NUMBER OF MAXIMUMS 2 3 4 KEEP FROM PRECEDING PROBLEM (1=YES) 0 0 0 0 CARDS INPUT THIS PROBLEM 16
OPTION TO CLEAR ENVELOPES BEFORE LANE LOADINGS (1=YES) 0
OPTION TO OMIT PRINT (-1=TABLE 4A, -2=TABLE 5, -3=BOTH) 0
SKEW ANGLE, DEGREES 0.000
TABLE 2. CONSTANTS
NUMBER OF INCREMENTS FOR SLAB AND CAP 92 INCREMENT LENGTH, FT 0.500 NUMBER OF INCREMENTS FOR MOVABLE LOAD 20 START POSITION OF MOVABLE-LOAD STA ZERO 2 STOP POSITION OF MOVABLE-LOAD STA ZERO 70 NUMBER OF INCREMENTS BETWEEN EACH POSITION OF MOVABLE LOAD 1
TABLE 5. MULTI-LANE LOADING SUMMARY ( WORKING STRESS ) ( *--CRITICAL NUMBER OF LANE LOADS)
MOMENT ( FT-K ) ------------------------------------------------------------------------ AT DEAD LD LANE POSITIVE LOAD AT LANE NEGATIVE LOAD AT STA EFFECT ORDER MAXIMUM LANE STA ORDER MAXIMUM LANE STA ------------------------------------------------------------------------
MOMENT ( FT-K ) ------------------------------------------------------------------------ AT DEAD LD LANE POSITIVE LOAD AT LANE NEGATIVE LOAD AT STA EFFECT ORDER MAXIMUM LANE STA ORDER MAXIMUM LANE STA ------------------------------------------------------------------------
SHEAR ( K ) ------------------------------------------------------------------------ AT DEAD LD LANE POSITIVE LOAD AT LANE NEGATIVE LOAD AT STA EFFECT ORDER MAXIMUM LANE STA ORDER MAXIMUM LANE STA ------------------------------------------------------------------------
SHEAR ( K ) ------------------------------------------------------------------------ AT DEAD LD LANE POSITIVE LOAD AT LANE NEGATIVE LOAD AT STA EFFECT ORDER MAXIMUM LANE STA ORDER MAXIMUM LANE STA ------------------------------------------------------------------------
REACTION ( K ) ------------------------------------------------------------------------ AT DEAD LD LANE POSITIVE LOAD AT LANE NEGATIVE LOAD AT STA EFFECT ORDER MAXIMUM LANE STA ORDER MAXIMUM LANE STA ------------------------------------------------------------------------
TABLE 5. MULTI-LANE LOADING SUMMARY ( LOAD FACTOR) ( *--CRITICAL NUMBER OF LANE LOADS)
MOMENT ( FT-K ) ------------------------------------------------------------------------ AT DEAD LD LANE POSITIVE LOAD AT LANE NEGATIVE LOAD AT STA EFFECT ORDER MAXIMUM LANE STA ORDER MAXIMUM LANE STA ------------------------------------------------------------------------
MOMENT ( FT-K ) ------------------------------------------------------------------------ AT DEAD LD LANE POSITIVE LOAD AT LANE NEGATIVE LOAD AT STA EFFECT ORDER MAXIMUM LANE STA ORDER MAXIMUM LANE STA ------------------------------------------------------------------------
SHEAR ( K ) ------------------------------------------------------------------------ AT DEAD LD LANE POSITIVE LOAD AT LANE NEGATIVE LOAD AT STA EFFECT ORDER MAXIMUM LANE STA ORDER MAXIMUM LANE STA ------------------------------------------------------------------------
SHEAR ( K ) ------------------------------------------------------------------------ AT DEAD LD LANE POSITIVE LOAD AT LANE NEGATIVE LOAD AT STA EFFECT ORDER MAXIMUM LANE STA ORDER MAXIMUM LANE STA ------------------------------------------------------------------------
REACTION ( K ) ------------------------------------------------------------------------ AT DEAD LD LANE POSITIVE LOAD AT LANE NEGATIVE LOAD AT STA EFFECT ORDER MAXIMUM LANE STA ORDER MAXIMUM LANE STA ------------------------------------------------------------------------
Moment LLDF Correction for Skew (Table 4.6.2.2.2e-1) Check �� 0° < 30° Set � = 0°Corr. = 1 - c1(tan �)^1.5
= 1 - 0(tan 0)^1.5 where: c1 =
Corr. = 1.000 c1 = 0.000 because � < 30°
XXX-XX-XXXXAnyXXXXBRIDGE
TXDOT ANY
Inverted Tee Design Example, Spans 1 & 3
The Final LLDF may be modified according to the following TxDOT policies: � Exterior beams use the interior LLDF when OH � S/2. � When OH > S/2 the exterior beam LLDF is determined by the lever rule for a single lane with a multiple presence factor of 1.0. �In no case shall the LLDF for the exterior beams be less than the LLDFs for the interior beams. �When the Roadway width is less than 20ft, all beams are designed for one lane loaded only. �In no case shall the LLDF be less than m·NL÷Nb.
DIVISION distribution_factors_i.xls
LRFD Live Load Distribution Factors*
Live Load Distribution Factors are calculated according to AASHTO LRFD Bridge Design Specifications, 4th Edition (2007 with no interim revisions) as prescribed by TxDOT policies (LRFD Design Manual July 2008) and practices. The Lever Rule is used when outside the Range of Applicability. The Range of Applicability for the Skew Correction Factors is ignored.
�tan0.12
20.00.13.03
��
�
��
g
s
KLt
5.025.0
30.1225.0 �
�
�
��
�
�
LS
LtK
s
g
LRFD Inverted Tee Bent Cap Design Example 70 June 2010
2 of 8
County: Highway: Design: BRG Date: 2/2/10 2007 LRFD SpecsC-S-J: ID #: Ck Dsn: BRG Date: Rev. 8/08 - (No Interim)Descrip: File: Sheet:
XXX-XX-XXXXAnyXXXXBRIDGE
TXDOT ANY
Inverted Tee Design Example, Spans 1 & 3DIVISION distribution_factors_i.xls
INTERIOR BEAM: Shear LL Distribution Per Lane (Table 4.6.2.2.3a-1):
Equationde = dist. b/w CL web to curbde = OH - Rail Widthde = 3ft - 1ft = 2.0 ft
e =
e = 0.6 + (2.0/10) = 0.800
g = e*gVint2+Eq
g = 0.800 * 0.814 = 0.651Skew Correction is included in gV(interior).
Range of Applicability (ROA) Checks Interior ROA is implicitly applied to the exterior beam.Check Interior Beam ROA: OKCheck de: -1.0' � 2.0' � 5.5' OKCheck Nb: 6 � 3 OK
Use Equation from Table 4.6.2.2.3b-1 because all criteria is OK.gVext2+ = 0.651
TxDOT Policy states gVExterior must be � gVinterior
gVinterior = 0.814TxDOT Policy states gVExterior must be � m·NL÷Nb
>> TxDOT Policy states that if OH � S/2, gVExterior is gVinterior.TxDOT Policy states that if OH > S/2 and W < 20ft, gV Exterior is the Maximum of: gVext1, gVinterior, and
m·NL÷Nb.TxDOT Policy states that if OH > S/2 ans W � 20ft, gVExterior is the Maximum of: gVext1, gVext2+, gVinterior,
and m·NL÷Nb.
0.814gVexterior =
TxDOT uses a multiple presence factor of 1.0 for one lane loaded on the exterior beam.
� �
��10
6.0 ed
LRFD Inverted Tee Bent Cap Design Example 73 June 2010
5 of 8
County: Highway: Design: BRG Date: 2/2/10 2007 LRFD SpecsC-S-J: ID #: Ck Dsn: BRG Date: Rev. 8/08 - (No Interim)Descrip: File: Sheet:
XXX-XX-XXXXAnyXXXXBRIDGE
TXDOT ANY
Inverted Tee Design Example, Spans 1 & 3DIVISION distribution_factors_i.xls
EXTERIOR BEAM:Moment LL Distribution Per Lane (Table 4.6.2.2.2d-1):
One Lane LoadedLever Rule
mg = 0.625 * 1.0 = 0.625Modify for Skew:
skew correction = 1.000mg = 0.625 * 1.000 = 0.625
Use Lever Rule as per AASHTO LRFD Table 4.6.2.2.2d-1.gMext1 = 0.625
Two or More Lanes LoadedLever Rule (Table 3.6.1.1.2)
g = 0.99 * 0.794 = 0.786Skew Correction included in gM(interior).
Range of Applicability (ROA) Checks Interior ROA is implicitly applied to the exterior beam.Check Interior Beam ROA: OKCheck de: -1.0' � 2.0' � 5.5' OKCheck Nb: 6 � 3 OK
Use Equation from Table 4.6.2.2.2d-1 because all criteria is OK.gMext2+ = 0.786
TxDOT Policy states gMExterior must be � gMinterior
gMinterior = 0.794TxDOT Policy states gMExterior must be � m·NL÷Nb
>> TxDOT Policy states that if OH � S/2, gMExterior is gMinterior.TxDOT Policy states that if OH > S/2 and W < 20ft, gMExterior is the Maximum of: gMext1, gMinterior, and
m·NL÷Nb.TxDOT Policy states that if OH > S/2 ans W � 20ft, gMExterior is the Maximum of: gMext1, gMext2+, gMinterior,
and m·NL÷Nb.
0.794gMexterior =
0.990
TxDOT uses a multiple presence factor of 1.0 for one lane loaded on the exterior beam.
e = � �
��1.9
77.0 ed
LRFD Inverted Tee Bent Cap Design Example 74 June 2010
6 of 8
County: Highway: Design: BRG Date: 2/2/10 2007 LRFD SpecsC-S-J: ID #: Ck Dsn: BRG Date: Rev. 8/08 - (No Interim)Descrip: File: Sheet:
XXX-XX-XXXXAnyXXXXBRIDGE
TXDOT ANY
Inverted Tee Design Example, Spans 1 & 3DIVISION distribution_factors_i.xls
LEVER RULE S = 8.0 ft
INTERIOR
For S < 4:One Lane = = 0.500
For 4 � S < 6:One Lane = = 0.500
Two Lanes = = 0.750
>>For 6 � S < 10:One Lane = = 0.625
Two Lanes = = 0.875
For 10 � S < 12:One Lane = = 0.625
Two Lanes = = 0.750
For 12 � S < 16:One Lane = = 0.625
Two Lanes = = 0.750
Three Lanes = = 0.500
For 16 � S < 18:One Lane = = 0.625
Two Lanes = = 0.750
Three Lanes = = 0.500
Four Lanes = = 0.000
3216
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LRFD Inverted Tee Bent Cap Design Example 75 June 2010
7 of 8
County: Highway: Design: BRG Date: 2/2/10 2007 LRFD SpecsC-S-J: ID #: Ck Dsn: BRG Date: Rev. 8/08 - (No Interim)Descrip: File: Sheet:
XXX-XX-XXXXAnyXXXXBRIDGE
TXDOT ANY
Inverted Tee Design Example, Spans 1 & 3DIVISION distribution_factors_i.xls
LEVER RULE S = 8.0 ft
INTERIOR (con't)
For 18 � S < 22:One Lane = = 0.625
Two Lanes = = 0.750
Three Lanes = = -0.125
Four Lanes = = -0.625
For 22 � S � 24:One Lane = = 0.625
Two Lanes = = 0.750
Three Lanes = = -0.125
Four Lanes = = -1.500
EXTERIOR
S = OH =
Rail Width = RW = X = S+OH-RW-2ft =
For X < 6:One Lane = = 0.500
>>For 6 � X < 12:One Lane = = 0.625
For 12 � X < 18:One Lane = = 0.625
Two Lanes = = 0.375
8.0 ft3.0 ft1.0 ft8.0 ft
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LRFD Inverted Tee Bent Cap Design Example 76 June 2010
8 of 8
County: Highway: Design: BRG Date: 2/2/10 2007 LRFD SpecsC-S-J: ID #: Ck Dsn: BRG Date: Rev. 8/08 - (No Interim)Descrip: File: Sheet:
XXX-XX-XXXXAnyXXXXBRIDGE
TXDOT ANY
Inverted Tee Design Example, Spans 1 & 3DIVISION distribution_factors_i.xls
LEVER RULE
EXTERIOR (con't) S = 8.0 ft OH = 3.0 ftRW = 1.0 ft X = S+OH-RW-2ft = 8.0 ft
For 18 � X < 24:One Lane = = 0.625
Two Lanes = = -0.250
For 24 � X < 30:One Lane = = 0.625
Two Lanes = = -0.250
Three Lanes = = -1.250
For 30 � X < 36:One Lane = = 0.625
Two Lanes = = -0.250
Three Lanes = = -2.625
For 36 � X < 42:One Lane = = 0.625
Two Lanes = = -0.250
Three Lanes = = -2.625
Four Lanes = = -4.375
For 42 � X � 48:One Lane = = 0.625
Two Lanes = = -0.250
Three Lanes = = -2.625
Four Lanes = = -6.500
INTERIOR EXTERIOR
One Lane Loaded = 0.625 One Lane Loaded = 0.625
Two Lanes Loaded = 0.875 Two Lanes Loaded = 0.625
Three Lanes Loaded = 0.875 Three Lanes Loaded = 0.625
Four Lanes Loaded = 0.875 Four Lanes Loaded = 0.625
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LRFD Inverted Tee Bent Cap Design Example 77 June 2010
1 of 8
County: Highway: Design: BRG Date: 2/2/10 2007 LRFD SpecsC-S-J: ID #: Ck Dsn: BRG Date: Rev. 8/08 - (No Interim)Descrip: File: Sheet:
Moment LLDF Correction for Skew (Table 4.6.2.2.2e-1) Check �� 0° < 30° Set � = 0°Corr. = 1 - c1(tan �)^1.5
= 1 - 0(tan 0)^1.5 where: c1 =
Corr. = 1.000 c1 = 0.000 because � < 30°
XXX-XX-XXXXAnyXXXXBRIDGE
TXDOT ANY
Inverted Tee Design Example, Span 2
The Final LLDF may be modified according to the following TxDOT policies: � Exterior beams use the interior LLDF when OH � S/2. � When OH > S/2 the exterior beam LLDF is determined by the lever rule for a single lane with a multiple presence factor of 1.0. �In no case shall the LLDF for the exterior beams be less than the LLDFs for the interior beams. �When the Roadway width is less than 20ft, all beams are designed for one lane loaded only. �In no case shall the LLDF be less than m·NL÷Nb.
DIVISION distribution_factors_i.xls
LRFD Live Load Distribution Factors*
Live Load Distribution Factors are calculated according to AASHTO LRFD Bridge Design Specifications, 4th Edition (2007 with no interim revisions) as prescribed by TxDOT policies (LRFD Design Manual July 2008) and practices. The Lever Rule is used when outside the Range of Applicability. The Range of Applicability for the Skew Correction Factors is ignored.
�tan0.12
20.00.13.03
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LRFD Inverted Tee Bent Cap Design Example 78 June 2010
2 of 8
County: Highway: Design: BRG Date: 2/2/10 2007 LRFD SpecsC-S-J: ID #: Ck Dsn: BRG Date: Rev. 8/08 - (No Interim)Descrip: File: Sheet:
XXX-XX-XXXXAnyXXXXBRIDGE
TXDOT ANY
Inverted Tee Design Example, Span 2DIVISION distribution_factors_i.xls
INTERIOR BEAM: Shear LL Distribution Per Lane (Table 4.6.2.2.3a-1):
Equationde = dist. b/w CL web to curbde = OH - Rail Widthde = 3ft - 1ft = 2.0 ft
e =
e = 0.6 + (2.0/10) = 0.800
g = e*gVint2+Eq
g = 0.800 * 0.814 = 0.651Skew Correction is included in gV(interior).
Range of Applicability (ROA) Checks Interior ROA is implicitly applied to the exterior beam.Check Interior Beam ROA: OKCheck de: -1.0' � 2.0' � 5.5' OKCheck Nb: 6 � 3 OK
Use Equation from Table 4.6.2.2.3b-1 because all criteria is OK.gVext2+ = 0.651
TxDOT Policy states gVExterior must be � gVinterior
gVinterior = 0.814TxDOT Policy states gVExterior must be � m·NL÷Nb
>> TxDOT Policy states that if OH � S/2, gVExterior is gVinterior.TxDOT Policy states that if OH > S/2 and W < 20ft, gV Exterior is the Maximum of: gVext1, gVinterior, and
m·NL÷Nb.TxDOT Policy states that if OH > S/2 ans W � 20ft, gVExterior is the Maximum of: gVext1, gVext2+, gVinterior,
and m·NL÷Nb.
0.814gVexterior =
TxDOT uses a multiple presence factor of 1.0 for one lane loaded on the exterior beam.
� �
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6.0 ed
LRFD Inverted Tee Bent Cap Design Example 81 June 2010
5 of 8
County: Highway: Design: BRG Date: 2/2/10 2007 LRFD SpecsC-S-J: ID #: Ck Dsn: BRG Date: Rev. 8/08 - (No Interim)Descrip: File: Sheet:
XXX-XX-XXXXAnyXXXXBRIDGE
TXDOT ANY
Inverted Tee Design Example, Span 2DIVISION distribution_factors_i.xls
EXTERIOR BEAM:Moment LL Distribution Per Lane (Table 4.6.2.2.2d-1):
One Lane LoadedLever Rule
mg = 0.625 * 1.0 = 0.625Modify for Skew:
skew correction = 1.000mg = 0.625 * 1.000 = 0.625
Use Lever Rule as per AASHTO LRFD Table 4.6.2.2.2d-1.gMext1 = 0.625
Two or More Lanes LoadedLever Rule (Table 3.6.1.1.2)
g = 0.99 * 0.649 = 0.643Skew Correction included in gM(interior).
Range of Applicability (ROA) Checks Interior ROA is implicitly applied to the exterior beam.Check Interior Beam ROA: OKCheck de: -1.0' � 2.0' � 5.5' OKCheck Nb: 6 � 3 OK
Use Equation from Table 4.6.2.2.2d-1 because all criteria is OK.gMext2+ = 0.643
TxDOT Policy states gMExterior must be � gMinterior
gMinterior = 0.649TxDOT Policy states gMExterior must be � m·NL÷Nb
>> TxDOT Policy states that if OH � S/2, gMExterior is gMinterior.TxDOT Policy states that if OH > S/2 and W < 20ft, gMExterior is the Maximum of: gMext1, gMinterior, and
m·NL÷Nb.TxDOT Policy states that if OH > S/2 ans W � 20ft, gMExterior is the Maximum of: gMext1, gMext2+, gMinterior,
and m·NL÷Nb.
0.649gMexterior =
0.990
TxDOT uses a multiple presence factor of 1.0 for one lane loaded on the exterior beam.
e = � �
��1.9
77.0 ed
LRFD Inverted Tee Bent Cap Design Example 82 June 2010
6 of 8
County: Highway: Design: BRG Date: 2/2/10 2007 LRFD SpecsC-S-J: ID #: Ck Dsn: BRG Date: Rev. 8/08 - (No Interim)Descrip: File: Sheet:
XXX-XX-XXXXAnyXXXXBRIDGE
TXDOT ANY
Inverted Tee Design Example, Span 2DIVISION distribution_factors_i.xls
LEVER RULE S = 8.0 ft
INTERIOR
For S < 4:One Lane = = 0.500
For 4 � S < 6:One Lane = = 0.500
Two Lanes = = 0.750
>>For 6 � S < 10:One Lane = = 0.625
Two Lanes = = 0.875
For 10 � S < 12:One Lane = = 0.625
Two Lanes = = 0.750
For 12 � S < 16:One Lane = = 0.625
Two Lanes = = 0.750
Three Lanes = = 0.500
For 16 � S < 18:One Lane = = 0.625
Two Lanes = = 0.750
Three Lanes = = 0.500
Four Lanes = = 0.000
3216
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LRFD Inverted Tee Bent Cap Design Example 83 June 2010
7 of 8
County: Highway: Design: BRG Date: 2/2/10 2007 LRFD SpecsC-S-J: ID #: Ck Dsn: BRG Date: Rev. 8/08 - (No Interim)Descrip: File: Sheet:
XXX-XX-XXXXAnyXXXXBRIDGE
TXDOT ANY
Inverted Tee Design Example, Span 2DIVISION distribution_factors_i.xls
LEVER RULE S = 8.0 ft
INTERIOR (con't)
For 18 � S < 22:One Lane = = 0.625
Two Lanes = = 0.750
Three Lanes = = -0.125
Four Lanes = = -0.625
For 22 � S � 24:One Lane = = 0.625
Two Lanes = = 0.750
Three Lanes = = -0.125
Four Lanes = = -1.500
EXTERIOR
S = OH =
Rail Width = RW = X = S+OH-RW-2ft =
For X < 6:One Lane = = 0.500
>>For 6 � X < 12:One Lane = = 0.625
For 12 � X < 18:One Lane = = 0.625
Two Lanes = = 0.375
8.0 ft3.0 ft1.0 ft8.0 ft
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LRFD Inverted Tee Bent Cap Design Example 84 June 2010
8 of 8
County: Highway: Design: BRG Date: 2/2/10 2007 LRFD SpecsC-S-J: ID #: Ck Dsn: BRG Date: Rev. 8/08 - (No Interim)Descrip: File: Sheet:
XXX-XX-XXXXAnyXXXXBRIDGE
TXDOT ANY
Inverted Tee Design Example, Span 2DIVISION distribution_factors_i.xls
LEVER RULE
EXTERIOR (con't) S = 8.0 ft OH = 3.0 ftRW = 1.0 ft X = S+OH-RW-2ft = 8.0 ft
For 18 � X < 24:One Lane = = 0.625
Two Lanes = = -0.250
For 24 � X < 30:One Lane = = 0.625
Two Lanes = = -0.250
Three Lanes = = -1.250
For 30 � X < 36:One Lane = = 0.625
Two Lanes = = -0.250
Three Lanes = = -2.625
For 36 � X < 42:One Lane = = 0.625
Two Lanes = = -0.250
Three Lanes = = -2.625
Four Lanes = = -4.375
For 42 � X � 48:One Lane = = 0.625
Two Lanes = = -0.250
Three Lanes = = -2.625
Four Lanes = = -6.500
INTERIOR EXTERIOR
One Lane Loaded = 0.625 One Lane Loaded = 0.625
Two Lanes Loaded = 0.875 Two Lanes Loaded = 0.625
Three Lanes Loaded = 0.875 Three Lanes Loaded = 0.625
Four Lanes Loaded = 0.875 Four Lanes Loaded = 0.625
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LRFD Inverted Tee Bent Cap Design Example 85 June 2010
Maximum stirrup spacing, Smax in 24.0 24.0 24.0 22.9 22.9 24.0 24.0 24.0
Conclusion
Shear Reinforcing OK OK OK OK OK OK OK OKLongitudinal Reinforcing FAILED OK OK FAILED FAILED OK OK FAILED
Note: Longitudinal Reinforcing check can be ignored for typical multi-column bent caps. For straddle bents with no overhangs, this check must be considered. Refer to LRFD 5.8.3.5 for further information.If torsion is not being considered, leave last five rows of input data blank.