Strut-and-Tie Modeling (STM) for Concrete Structures: Design Examples

Publication No FHWANHI17071 NHI Course No 130126 October 2017 StrutandTie Modeling STM for Concrete Structures Design Examples 108 in 108 in 64 in 54 in 54 in 600 k A B C D E F 498 498 307 600 k 600 k 600 k P1 2005 k P2 926 k 543 1601 697 A B C E D E A 850 1330 C 1328 psi C 685 067 031 217 D T 057 ft 057 ft 368 ft 862 ft 862 ft 842 ft 842 ft 499 ft 360 ft G A A B H C I J K L L F F D E 1620 kips 51735 kips 43879 kips 44934 kips 3217 kips 1620 kips 2033 kips 2071 kips 1650 kips 2596 kips G 512 ft T C T C z x y A D E H C B F G I J FA 17636 kips FD 17636 kips FC 3391 kips FB 3391 kips R2 2595 kips R3 2595 kips R4 11650 kips R1 11650 kips FOREWORD This Manual provides four design examples illustrating the application of the strutandtie method for a variety of structural configurations including a simplysupported deep beam a cantilever bent cap an invertedtee moment frame straddle bent cap and a drilled shaft footing Each design example is based on the 8th Edition of the AASHTO LRFD Bridge Design Specifications This Manual is intended for state DOT bridge and structures engineers and practicing bridge engineers who are responsible for concrete bridge design and evaluation This Manual will serve as a reference and a guide for engineers of all levels including designers consultants reviewers maintenance engineers management engineers and load rating engineers This document is part of a training program that also includes a oneandahalfday instructorled training ILT course Notice This document is disseminated under the sponsorship of the US Department of Transportation USDOT in the interest of information exchange The US Government assumes no liability for the use of the information contained in this document The US Government does not endorse products or manufacturers Trademarks or manufacturers names appear in this report only because they are considered essential to the objective of the document Quality Assurance Statement The Federal Highway Administration FHWA provides highquality information to serve Government industry and the public in a manner that promotes public understanding Standards and policies are used to ensure and maximize the quality objectivity utility and integrity of its information FHWA periodically reviews quality issues and adjusts its programs and processes to ensure continuous quality improvement TECHNICAL REPORT DOCUMENTATION PAGE 1 Report No FHWANHI17071 2 Government Accession No 3 Recipients Catalog No FHWANHI17071 4 Title and Subtitle StrutandTie Modeling STM for Concrete Structures Design Examples 5 Report Date October 2017 6 Performing Organization Code 7 Authors Aaron B Colorito PE Kenneth E Wilson PE SE Oguzhan Bayrak PhD PE and Francesco M Russo PhD PE 8 Performing Organization Report No MBIDE155714 9 Performing Organization Name and Address Michael Baker International 100 Airside Drive Moon Township PA 15108 10 Work Unit No 11 Contract or Grant No DTFH6111D000465011 12 Sponsoring Agency Name and Address Federal Highway Administration National Highway Institute 1310 North Courthouse Road Arlington VA 22201 13 Type of Report and Period Final Document August 2016 October 2017 14 Sponsoring Agency FHWA 15 Supplementary Notes Michael Baker Project Manager Scott D Vannoy PE FHWA Contracting Officer Representative Melonie Barrington PE PMP FHWA Technical Team Leader Reggie H Holt PE 16 Abstract This Manual provides four design examples illustrating the application of the strutandtie method for a variety of structural configurations including a simplysupported deep beam a cantilever bent cap an invertedtee moment frame straddle bent cap and a drilled shaft footing Each design example is based on the 8th Edition of the AASHTO LRFD Bridge Design Specifications This Manual is intended for state DOT bridge and structures engineers and practicing bridge engineers who are responsible for concrete bridge design and evaluation This Manual will serve as a reference and a guide for engineers of all levels including designers consultants reviewers maintenance engineers management engineers and load rating engineers This document is part of a training program that also includes a oneanda halfday instructorled training ILT course 17 Key Words Concrete reinforcement strutandtie method strutandtie model strut tie node DRegion BRegion nodal zone concrete efficiency factor confinement modification factor CCC node CCT node CTT node crack control reinforcement smeared node 18 Distribution Statement This manual can be obtained from the National Highway Institute 1310 North Courthouse Road Arlington VA 22201 19 Security Classif of this report Unclassified 20 Security Classif of this page Unclassified 21 No of Pages 196 22 Price Form DOT F 17007 872 Reproduction of completed page authorized ii SI MODERN METRIC CONVERSION FACTORS APPROXIMATE CONVERSIONS TO SI UNITS Symbol When You Know Multiply By To Find Symbol LENGTH in inches 254 millimeters mm ft feet 0305 meters m yd yards 0914 meters m mi miles 161 kilometers km AREA in 2 square inches 6452 square millimeters mm 2 ft 2 square feet 0093 square meters m 2 yd 2 square yard 0836 square meters m 2 ac acres 0405 hectares ha mi 2 square miles 259 square kilometers km 2 VOLUME fl oz fluid ounces 2957 milliliters mL gal gallons 3785 liters L ft 3 cubic feet 0028 cubic meters m 3 yd 3 cubic yards 0765 cubic meters m 3 NOTE volumes greater than 1000 L shall be shown in m 3 MASS oz ounces 2835 grams g lb pounds 0454 kilograms kg T short tons 2000 lb 0907 megagrams or metric ton Mg or t TEMPERATURE exact degrees oF Fahrenheit 5 F329 Celsius oC or F3218 ILLUMINATION fc footcandles 1076 lux lx fl footLamberts 3426 candelam 2 cdm 2 FORCE and PRESSURE or STRESS lbf poundforce 445 newtons N lbfin 2 poundforce per square inch 689 kilopascals kPa APPROXIMATE CONVERSIONS FROM SI UNITS Symbol When You Know Multiply By To Find Symbol LENGTH mm millimeters 0039 inches in m meters 328 feet ft m meters 109 yards yd km kilometers 0621 miles mi AREA mm 2 square millimeters 00016 square inches in 2 m 2 square meters 10764 square feet ft 2 m 2 square meters 1195 square yards yd 2 ha hectares 247 acres ac km 2 square kilometers 0386 square miles mi 2 VOLUME mL milliliters 0034 fluid ounces fl oz L liters 0264 gallons gal m 3 cubic meters 35314 cubic feet ft 3 m 3 cubic meters 1307 cubic yards yd 3 MASS g grams 0035 ounces oz kg kilograms 2202 pounds lb Mg or t megagrams or metric ton 1103 short tons 2000 lb T TEMPERATURE exact degrees oC Celsius 18C32 Fahrenheit oF ILLUMINATION lx lux 00929 footcandles fc cdm 2 candelam 2 02919 footLamberts fl FORCE and PRESSURE or STRESS N newtons 0225 poundforce lbf kPa kilopascals 0145 poundforce per square inch lbfin 2 SI is the symbol for the International System of Units Appropriate rounding should be made to comply with Section 4 of ASTM E380 Revised March 2003 Visit httpwwwfhwadotgovpublicationsconvtablcfm for a 508 compliant version of this table FHWANHI130126 Design Examples StrutandTie Modeling STM for Concrete Structures This page intentionally left blank FHWANHI130126 Design Examples StrutandTie Modeling STM for Concrete Structures Table of Contents Acknowledgments i Glossary iii How to Use These Design Examples v STM Handout for Each Design Example vii Design Example 1 SimplySupported Deep Beam 11 Design Example 2 Cantilever Bent Cap 21 Design Example 3 InvertedTee Moment Frame Straddle Bent Cap 31 Design Example 4 Drilled Shaft Footing 41 FHWANHI130126 Design Examples StrutandTie Modeling STM for Concrete Structures This page intentionally left blank FHWANHI130126 Design Examples StrutandTie Modeling STM for Concrete Structures i Acknowledgments The examples contained in this document greatly benefited from the example problems previously developed in an implementation project sponsored by the Texas Department of Transportation TxDOT In this context the contributions of Christopher Williams and Dean Deschenes have been invaluable and are gratefully acknowledged Further the support of Gregg Freeby TxDOTs Bridge Division Director is also gratefully acknowledged FHWANHI130126 Design Examples StrutandTie Modeling STM for Concrete Structures ii This page intentionally left blank FHWANHI130126 Design Examples StrutandTie Modeling STM for Concrete Structures iii Glossary Available Length The tie width for a CCT or CTT node over which the stirrups considered to carry the force in the tie can be spread Back Face The face of a nodal zone at which neither a load reaction nor strut is located Beam or Bernoulli Region BRegion Regions of concrete members in which Bernoullis hypothesis of straightline strain profiles linear for bending and uniform for shear applies Bearing Face The face of a nodal zone at which a load or reaction is applied Bottleshaped Strut A strut that is wider at midlength than at its ends CCC Node A node where only struts intersect CCT Node A node where a tie intersects the node in only one direction Concrete Efficiency Factor A factor based on the node type CCC CCT or CTT and the node face bearing face back face or struttonode interface that is used to compute the limiting compressive stress at a node face Confinement Modification Factor A factor based on the relative proportion of the supporting surface to the loaded area that is used to compute the limiting compressive stress at a node face Crack Control Reinforcement Reinforcement based on 0003 times the effective area of the strut intended to control the width of cracks and to ensure a minimum ductility for the member so that if required significant redistribution of internal stresses is possible CTT Node A node where ties intersect in two different directions Curved Bar Node A node resulting from bending a large reinforcing bar such as No 11 14 or 18 Development Length The distance required to develop the specified strength of a reinforcing bar or prestressing strand Direct Strut Model A strutandtie model in which a single direct strut is used to connect the nodes at two bearing faces FHWANHI130126 Design Examples StrutandTie Modeling STM for Concrete Structures iv Disturbed or Discontinuity Region DRegion Regions of concrete members encompassing abrupt changes in geometry or concentrated forces in which strain profiles more complex than straight lines exist Interior Node A node that is not located at the end reactions of the member Load and Resistance Factor Design A reliabilitybased design methodology in which force effects caused by factored loads are not permitted to exceed the factored resistance of the components LRFD Load and Resistance Factor Design Nodal Zone The volume of concrete around a node that is assumed to transfer strut andtie forces through the node Node A point in a strutandtie model where the axes of the struts ties and concentrated forces acting on the joint intersect Singular Node A node with a clearly defined geometry Smeared Node An interior node that is not bounded by a bearing plate STM Strutandtie model strutandtie modeling strutandtie method Strut A compression member in a strutandtie model representing the resultant of a parallel or a fanshaped compression field StrutandTie Method A procedure used principally in regions of concentrated forces and geometric discontinuities to determine concrete proportions and reinforcement quantities and patterns based on an analytic model consisting of compression struts in the concrete tensile ties in the reinforcement and the geometry of nodes at their points of intersection StrutandTie Model A truss model of a member or of a DRegion in such a member made up of struts and ties connected at nodes and capable of transferring the factored loads to the supports or to adjacent BRegions StruttoNode Interface The face of a nodal zone at which a strut is located Tie A tension element in a strutandtie model Two Panel Model A strutandtie model in which an intermediate vertical tie is introduced between the nodes at two bearing faces such that there are two panels between the nodes at the two bearing faces FHWANHI130126 Design Examples StrutandTie Modeling STM for Concrete Structures v How to Use These Design Examples This document provides four design examples illustrating the application of the strut andtie method for a variety of structural configurations including the following Design Example 1 SimplySupported Deep Beam Design Example 2 Cantilever Bent Cap Design Example 3 InvertedTee Moment Frame Straddle Bent Cap Design Example 4 Drilled Shaft Footing There are several characteristics that are common to all four design examples that are intended to benefit the designer as they use this document At the beginning of each design example is a table of contents specific to that example as well as a flowchart of the various design steps Each design step in the table of contents and in the flowchart is then clearly identified within the document Each design example contains a wealth of figures to illustrate and supplement the concepts being presented in the narrative In addition most of the design examples also contain several tables Each design example is based on Load and Resistance Factor Design LRFD As used in this document AASHTO LRFD is used as an abbreviation of AASHTO LRFD Bridge Design Specifications In addition STM is used as an abbreviation for strutandtie model strutandtie modeling or strutandtie method It is generally clear which is meant References to AASHTO LRFD articles figures tables and equations are presented throughout the design examples The designer can refer to those portions of AASHTO LRFD for clarification or for more information about the information being presented in the design examples In addition tipparagraphs are presented throughout the design examples in the following format Tipparagraphs are additional supplemental information that may not necessarily be required to complete the design example but that is useful information for the designer to know as they seek to apply the strutandtie method to other structural configurations Tip paragraphs are set apart from all other paragraphs in two ways 1 they are indented on the left and right and 2 they are presented in narrow font as illustrated in this paragraph Each design example is based on the 8th Edition of the AASHTO LRFD Bridge Design Specifications FHWANHI130126 Design Examples StrutandTie Modeling STM for Concrete Structures vi This page intentionally left blank FHWANHI130126 Design Examples StrutandTie Modeling STM for Concrete Structures vii STM Handout for Each Design Example The following pages contain the basic strutandtie model layout for each of the four design examples FHWANHI130126 Design Examples StrutandTie Modeling STM for Concrete Structures viii This page intentionally left blank FHWANHI130126 Design Examples StrutandTie Modeling STM for Concrete Structures ix Design Example 1 SimplySupported Deep Beam 108 in 108 in 64 in 54 in 54 in 600 k A B C D E F 498 498 307 600 k 600 k 600 k FHWANHI130126 Design Examples StrutandTie Modeling STM for Concrete Structures x This page intentionally left blank FHWANHI130126 Design Examples StrutandTie Modeling STM for Concrete Structures xi Design Example 2 Cantilever Bent Cap P1 2005 k P2 926 k 543 1602 697 A B C E D E A 043 019 1570 k 1664 k 3751 k 845 k 1330 D 283 29 504 C 1328 psi C 381 555 064 685 067 031 217 T 840 FHWANHI130126 Design Examples StrutandTie Modeling STM for Concrete Structures xii This page intentionally left blank FHWANHI130126 Design Examples StrutandTie Modeling STM for Concrete Structures xiii Design Example 3 InvertedTee Moment Frame Straddle Bent Cap 057 ft 057 ft 364 ft 862 ft 862 ft 842 ft 842 ft 505 ft 354 ft G A A B H C I J K L F D E 162 kips 5174 kips 4388 kips 4493 kips 322 kips 162 kips 204 kips 207 kips 165 kips 260 kips G 038 ft 050 ft 032 ft 105 ft 6539 8647 4643 4639 12970 5056 4289 1950 10639 10639 10281 2367 5323 5174 4605 9305 Values shown without units are element forces in kips indicates tension indicates compression 2689 kips 2841 kips 8647 kips 2689 L 115 ft 032 ft 6208 12138 12138 kips 2408 kips 2689 F 2689 kips FHWANHI130126 Design Examples StrutandTie Modeling STM for Concrete Structures xiv This page intentionally left blank FHWANHI130126 Design Examples StrutandTie Modeling STM for Concrete Structures xv Design Example 4 Drilled Shaft Footing For Load Case 1 z x y A D E H C B F G I J FA 17636 kips FD 17636 kips FC 3391 kips FB 3391 kips R2 2595 kips R3 2595 kips R4 11650 kips R1 11650 kips 12953 kips 9484 kips 11829 kips 10607 kips FHWANHI130126 Design Examples StrutandTie Modeling STM for Concrete Structures xvi For Load Case 2 z x y A D E H C B F G J FA 10268 kips FD 10268 kips FC 4718 kips FB 4718 kips R2 1007 kips R3 1007 kips R4 6557 kips R1 6557 kips 543 kips 10610 kips 9690 kips 902 kips 9958 kips 6337 kips 697 kips 4699 kips 4718 kips 4699 kips 437 kips 6337 kips I 697 kips 437 kips 6362 kips 9958 kips 5838 kips 1007 kips 10770 kips 10770 kips 1420 kips 1420 kips 4718 kips K L M N 1007 kips FHWANHI130126 StrutandTie Modeling STM for Concrete Structures Design Example 1 SimplySupported Deep Beam 11 Design Example 1 SimplySupported Deep Beam Table of Contents Page Design Step 1 Define StrutandTie Model Input 13 Design Step 2 Determine the Locations of the B and DRegions 13 Design Step 3 Define Load Cases 14 Design Step 4 Analyze Structural Components 14 Design Step 5 Size Structural Components Using the Shear Serviceability Check 15 Design Step 6 Develop a StrutandTie Model 18 Design Step 7 Proportion Ties 113 Design Step 8 Perform Nodal Strength Checks 117 Design Step 9 Proportion Crack Control Reinforcement 128 Design Step 10 Provide Necessary Anchorage for Ties 132 Design Step 11 Draw Reinforcement Layout 134 FHWANHI130126 StrutandTie Modeling STM for Concrete Structures Design Example 1 SimplySupported Deep Beam 12 Design Example 1 is one of the classic problems used to demonstrate the application of the strutandtie method STM to the analysis and design of concrete members This example demonstrates the sizing analysis and designcode checking of a deep beam supporting two concentrated loads Although this does not represent a specific member in a bridge a real world analogy to this example would be a straddle bent cap spanning another roadway at a skewed crossing It also has features similar to the design of a deep pile cap supporting drilled shafts The example features the elements of strutand tie design of concrete members listed below Design Step 1 Define StrutandTie Model Input Design Step 3 Define Load Cases Design Step 4 Analyze Structural Components Design Step 5 Size Structural Components Using the Shear Serviceability Check Design Step 6 Develop a StrutandTie Model Design Step 7 Proportion Ties Design Step 8 Perform Nodal Strength Checks Design Step 9 Proportion Crack Control Reinforcement Design Step 10 Provide Necessary Anchorage for Ties Design Step 2 Determine the Locations of the B and DRegions Design Step 11 Draw Reinforcement Layout FHWANHI130126 StrutandTie Modeling STM for Concrete Structures Design Example 1 SimplySupported Deep Beam 13 Design Step 1 Define StrutandTie Model Input PSTRENGTH 600 k PSERVICE 400 k PSTRENGTH 600 k PSERVICE 400 k 90 90 90 60 h 15 h 15 h 15 h 10 Overhang Typ Figure 11 Beam Defined for Design Example 1 The beam to be used in this design example is a simplysupported beam which resists two equal concentrated loads on its top face and is supported by bearings at each end refer to Figure 11 The compressive strength for design fc is taken to be 50 ksi and the yield strength of the steel reinforcing fy is taken as 60 ksi The overall depth of the beam h is assumed to be 6 ft 72 in and the beam width is assumed to be 4 ft 48 in The two loads are located at 15 times the overall depth of the beam from the centerline of bearing 15h The overall beam span is 27 ft Design Step 2 Determine the Locations of the B and DRegions The definitions of B and DRegions are given in AASHTO LRFD Article 5512 Because it is assumed that all regions within d of a concentrated load qualify as D Regions the entire length of this beam is governed by the strutandtie design method To reinforce the concept of the designation of DRegions refer to AASHTO LRFD Figure 551211 reproduced on the following page as Figure 12 FHWANHI130126 StrutandTie Modeling STM for Concrete Structures Design Example 1 SimplySupported Deep Beam 14 R1 P d BRegion d d d d R2 D Region DRegion Figure 12 Definition of B and DRegions In Figure 12 d is taken as the effective depth of the member or the distance from the extreme compression fiber to the centroid of the tension steel Note that the BRegion of this member is the area more than d away from a disturbance load reaction etc In this example for a member 6 ft deep the distance d will be nearly the full height of the member ie d and h are essentially the same Design Step 3 Define Load Cases Two concentrated loads are applied to the top surface of the beam They are assumed to be a combination of dead load and live load and include the selfweight of the beam The total service load such as could come from the AASHTO LRFD Service I load combination is 400 kips The factored strength load such as from the Strength I load combination is 600 kips The loads are assumed to be point loads distributed to the top of the beam by bearing plates These are assumed equal to the width of the beam 48 in and 12 in long parallel to the member Design Step 4 Analyze Structural Components Starting first with the statics of the problem it is shown that each reaction is equal to one of the applied point loads The service and strength load combination forces and reactions are as follows FHWANHI130126 StrutandTie Modeling STM for Concrete Structures Design Example 1 SimplySupported Deep Beam 15 PSTRENGTH 600 k PSERVICE 400 k PSTRENGTH 600 k PSERVICE 400 k RSTRENGTH LEFT 600 k RSERVICE LEFT 400 k RSTRENGTH RIGHT 600 k RSERVICE RIGHT 400 k Figure 13 Beam Applied Loads and Reactions In strutandtie modeling traditional or beam theory flexure and shear are not the mechanisms of internal load distribution However in Design Step 5 the moments determined through traditional beam theory equations at the strength limit state will be used as a tool in establishing the strutandtie truss geometry Design Step 5 Size Structural Components Using the Shear Serviceability Check Design of the simplysupported beam begins with the selection of member dimensions that can be reasonably reinforced Because the intent of the example is to demonstrate the use of the strutandtie method to design DRegions the geometry of the example beam applied loads and supports is chosen such that the entire beam is governed by DRegions ie there are no locations along the beam where Bernoulli beam theory applies Recall that the beam height is 72 in and the beam width is 48 in AASHTO LRFD Equation C58221 limits the applied shear to the following value Vcr with corresponding minimum and maximum values limited as follows where bw width of the members web in d effective depth of the member in FHWANHI130126 StrutandTie Modeling STM for Concrete Structures Design Example 1 SimplySupported Deep Beam 16 This equation estimates the shear at which diagonal cracks form in DRegions Where the applied service load shears are less than Vcr reasonable assurance is provided that diagonal shear cracks will not form Since this check is performed at the service limit state the calculated cracking shear will be compared against the service load shear force of 400 kips A value for d must be assumed at this point Although it is technically not correct the LRFD equations for beamtheory flexural strength will be used to determine a trial value of d in order to determine Vcr The required flexural strength is a function of the estimated maximum moment This is found using simple beam statics and the geometry loads and reactions shown in Figures 11 and 13 To begin it is assumed that a 2 in bottom cover is provided No 5 stirrups are used and two layers of reinforcement will be required equally divided between the two layers and a 2 in clear space is provided between layers If No 10 reinforcing bars are used the trial value of d is Next determine the required area of reinforcing As using AASHTO LRFD Equation 563221 modified by omitting the prestressed reinforcement terms which is approximated as In the equation above the term jd replaces the term d a2 As a firstpass approximation take j 09 Using longitudinal reinforcing steel with a yield strength of 60 ksi ϕf 09 and the assumed values for d and j the required area of reinforcing steel is computed as The area of steel for the tie is intentionally shown as an approximate or rounded number at this point since it is based on a series of approximations We must now verify if the required area is consistent with the several assumptions of the values of j and d FHWANHI130126 StrutandTie Modeling STM for Concrete Structures Design Example 1 SimplySupported Deep Beam 17 The required area may be provided by 16 No 10 bars with 8 in each layer A sketch of the trial reinforcement layout is provided in Figure 14 The sketch allows verification that the width of the beam is adequate and that the reinforcing may be arranged consistently with the initial assumptions of the effective depth d 49 to Tie CL d 7 spa 5 ⅞ in Figure 14 Assumed Tie Location This layout is consistent with prior assumptions for the effective depth d namely the reinforcing bar size and clear spacing between the layers are as assumed The distance to the center of the two layers is Checking against the initial assumption of d 671 in The value of d is rounded to 67 in for simplicity The distance a is known as the shear span It is the distance from the centerline of the concentrated load to the centerline of bearing In this example a is equal to 15h or 108 in Now that the values of a and d are known the value of is found Since the calculated value of 00388 is less than the lower bound of 00632 the lower bound value is used The value of Vcr may now be calculated FHWANHI130126 StrutandTie Modeling STM for Concrete Structures Design Example 1 SimplySupported Deep Beam 18 Thus diagonal cracking should not be expected at the service load shear of 400 kips Therefore the following parameters will be used throughout the rest of this design example Total beam height h 6 ft 72 in Beam width bw 48 in Effective depth d 67 in rounded Design Step 6 Develop a StrutandTie Model For this design example two likely load paths may be chosen to distribute the point loads to the bearings Each is described below and will be used for onehalf of the beam design The first load path termed the direct strut model is shown in Figure 15 AASHTO LRFD Figure 58222 The flow of forces is via a direct strut between the applied load on the top surface and the bearing This model is valid as long as the angle between the inclined strut and the tie is greater than or equal to 25 degrees Bearing Face Interface Back Face CCC Nodal Zone CCT Nodal Zone Back Face Interface Bearing Face Figure 15 Direct StrutandTie Model of a Deep Beam If by geometry the angle between the strut and tie cannot be greater than 25 degrees one or more intermediate ties may be introduced such that the angle is kept greater than or equal to 25 degrees This is known as the two panel model The number of vertical ties required is dependent on the ratio of ad and will increase as the ratio ad increases AASHTO LRFD Article C5822 states the following regarding the layout of an STM model FHWANHI130126 StrutandTie Modeling STM for Concrete Structures Design Example 1 SimplySupported Deep Beam 19 Minimize the number of vertical ties between a load and a support using the least number of truss panels possible while still satisfying the 25 degree minimum as shown in Figure C58223 The 25degree Limit The goal of the 25degree limit in AASHTO LRFD Article C5822 is to preclude excessive strains in the tensile reinforcement in the member Limiting the tensile strains in the reinforcement limits crack sizes When introducing vertical ties the designer should try to use only the number of ties needed to comply with the 25degree limit Using only the minimum required number of ties required results in an efficient beam design This is illustrated by AASHTO LRFD Figure C58223 reproduced here as Figure 16 Studying Figure 16 one will note that on the left side only a single vertical tie is used On the right side three vertical ties are used Both are statically admissible and valid strutandtie models However by examination the truss system using only one vertical tie is more efficient than the three tie system L L 2 L 2 P 05P 05P 05P 05P 25 25 EFFICIENT INEFFICIENT 05P 05P 25 25 25 25 Figure 16 Efficient and Inefficient STM Models By simple statics it is found that the force in all of the vertical ties is 05P In the right hand truss the additional ties serve no structural purpose because the resultant force in each tie is not reduced from a singletie truss Therefore the additional vertical ties serve only to increase the quantity of reinforcing steel required because every tie must be reinforced to support 05P FHWANHI130126 StrutandTie Modeling STM for Concrete Structures Design Example 1 SimplySupported Deep Beam 110 15 h 15 h h 075 h 075 h PSTRENGTH 600 k PSERVICE 400 k PSTRENGTH 600 k PSERVICE 400 k hSTM Figure 17 Example Beam and Trial StrutandTie Layout Both the direct strut model and the two panel model will be used in this design example each to model onehalf of the example beam The primary purpose of this is to illustrate the differences between the two models as well as to give example calculations for both methods The left side point load will be distributed using the two panel model and the right point load will be distributed using the direct strut model The example beam and trial strutandtie layout are shown in Figure 17 DirectStrut and TwoPanel Models The direct strut model is the most efficient way to model the flow of forces in a strutandtie model Examining Figure 17 the reader will note that the lefthand side of the beam may also be modeled using a direct strut It is modeled as a twopanel model so that calculations for both models may be presented in this example The angles of the struts for the example beam are shown in Figure 18 The introduction of the vertical tie requires that it must carry the entire vertical force in the truss panel This may be proven by calculation using the method of sections to solve for the vertical tie force In the direct strut model the vertical force is transferred purely by diagonal compression No stirrups ie vertical ties would be required other than the crack control reinforcement which is covered in Design Step 9 Developing a StrutandTie Model Any staticallyadmissible truss that is in external and internal equilibrium may be used for a strutandtie model However the reader would be correct in pointing out that the direct strut model used in the right half of the beam is an unstable truss This type of truss is acceptable to use in strutandtie design Refer to AASHTO LRFD Article C5822 for additional information FHWANHI130126 StrutandTie Modeling STM for Concrete Structures Design Example 1 SimplySupported Deep Beam 111 In order to establish the geometry of the truss nodes must be located and the vertical distance between the truss chords hSTM must be determined The distance hSTM is shown in Figure 17 Nodes are located at each point load and reaction location Additionally for the two panel model nodes are introduced at each end of a vertical tie There are no fixed rules for establishing the height of the truss hSTM as shown in AASHTO LRFD Figure C58222 It is accepted practice to use a flexural model to determine an approximate height of the truss Since the effective depth d was determined in Design Step 5 it will be used in this design step to find the height of the truss By assuming the center of the tie is located at the centroid of the bottom reinforcing the depth of the flexural compressive stress block can be found and used to estimate the location of the top horizontal compressive strut The StrutandTie Model Truss The height of the strutandtie model truss may also be found by trial and error by varying the height of the top strut the top chord of the truss and checking for equilibrium The use of beam theory to estimate the height of the strut is an expeditious way to estimate hSTM However remember that the results of the beam theory calculations for the beams internal forces are not valid within the DRegions of the beam and should not be used for any other calculations The height of the compression chord is estimated using the traditional Whitney stress block approximation of the depth of the compression zone of a flexural member Recall that 16 No 10 bars were assumed in Design Step 5 The area of reinforcing is then The other design variables have already been defined fy 60 ksi fc 50 ksi b 48 in Solving for a which is rounded up to 6 in FHWANHI130126 StrutandTie Modeling STM for Concrete Structures Design Example 1 SimplySupported Deep Beam 112 The dimension a is the assumed height of the top strut The top nodes are located at the midthickness of this dimension ie 3 in from the top surface of the beam Therefore the resulting height of the truss hSTM is The resulting truss is shown in Figure 18 Node designations are given in circles 108 in 108 in 64 in 54 in 54 in 600 k A B C D E F 498 498 307 600 k 600 k 600 k Figure 18 Design Example 1 STM Truss Using the forces at the Strength I load combination two loads of 600 kips each the forces in the truss members are found Calculations for the individual truss element forces are not given here however they may be calculated simply by the method of sections or a similar method A negative sign indicates compression and a positive sign indicates tension In the sketch of the truss it is common practice that struts be drawn as dashed lines and ties be drawn as solid lines This graphic convention is adhered to throughout this example FHWANHI130126 StrutandTie Modeling STM for Concrete Structures Design Example 1 SimplySupported Deep Beam 113 Table 11 StrutandTie Model Forces Node Member Force kips A AB 506 A AD 785 B AB 506 B BC 1013 B BD 600 B BE 785 C BC 1013 C CF 1177 D AD 785 D BD 600 D DE 506 E BE 785 E DE 506 E EF 1013 F CF 1177 F EF 1013 Design Step 7 Proportion Ties Now that the forces in all of the truss members have been determined the verification of the ties struts and nodes will begin This example begins with the verification of the bottom tie and then the vertical ties or stirrups Proportion the Bottom Tie From the truss analysis the maximum force in the bottom chord tie is in member BC a force of 1013 kips The assumed reinforcing pattern developed previously includes two rows of 8 No 10 reinforcing bars AASHTO LRFD Article 58241 provides the tie strength requirements The nominal resistance of a tie is given by AASHTO LRFD Equation 582411 FHWANHI130126 StrutandTie Modeling STM for Concrete Structures Design Example 1 SimplySupported Deep Beam 114 Strength of Ties For strutandtie models containing no prestressed reinforcement the prestressed reinforcement terms in Equation 582411 may be taken as zero Where there is no non prestressed reinforcement the term fy may be taken as 60 ksi in the second term of Equation 582411 but the sum of fpe and fy shall not be greater than the yield strength of the prestressing steel The goal of this equation is to limit the stress in the prestressing steel to a value less than or equal to its yield strength which aids in limiting cracking For a strutandtie model the resistance factor ϕ is given by AASHTO LRFD Article 5542 as follows For compression in strutandtie models 070 For tension in strutandtie models o Reinforced concrete 090 o Prestressed concrete 100 Thus for a tie ϕ 090 Check the factored strength of the tie using the area of reinforcement determined previously The factored load to be resisted is 1013 kips Since ϕPn Pu the bottom tie design is acceptable Proportion the Vertical Tie In order to proportion the vertical tie the amount and placement of the reinforcing that comprise the tie must be determined The tie connects two interior nodes that is nodes not acted on by direct loads or bounded by bearing plates AASHTO LRFD Article C5822 indicates that a check of the stresses at such nodes is unnecessary but the tie must still be proportioned and detailed accordingly Figure 19 suggests a method of locating the vertical steel corresponding to the fan shaped strut Although the strutandtie truss model implies that struts and ties occupy a specific location in reality the stresses spread out over portions of the member as shown in AASHTO LRFD Figure C58222 reproduced here as Figure 19 AASHTO LRFD suggests that an available length be determined over which the stirrups can be distributed It is both inefficient and unwise to concentrate all of the reinforcing steel exactly where the theoretical tie is located By spreading the vertical steel out the individual bars better resist the overall distribution of forces as the concentrated load from the reaction and applied load fan out across the web depth and length FHWANHI130126 StrutandTie Modeling STM for Concrete Structures Design Example 1 SimplySupported Deep Beam 115 hSTM x tan25 Available Length la hSTM x tan25 Shear Span a 25 P R Determine the Tie Width P R 25 hSTM FanShaped Strut Stirrups Comprising Tie Figure 19 FanShaped Struts Engaging Reinforcement Forming a Tie The reader is encouraged to refer back to the discussion in Design Step 6 to review the 25degree limit from which Figure 19 is developed Using Figure 19 as a guide Figure 110 is generated to determine the available length over which to distribute the vertical steel la FHWANHI130126 StrutandTie Modeling STM for Concrete Structures Design Example 1 SimplySupported Deep Beam 116 hSTM x tan25 la hSTM x tan25 108 in 25 A B D E Figure 110 Design Example FanShaped Strut For this design example the following variables have been calculated previously and are used to determine the available length Shear span a 108 in hSTM 64 in Therefore which is rounded up to 30 in Therefore the available length la is given by In the strutandtie truss model the force in the vertical tie is in member BD From Table 11 the force in the member BD is found to be 600 kips The required area of reinforcing steel may be found by setting AASHTO LRFD Equation 582411 equal to 600 kips and solving for Ast Thus a minimum of 1111 in2 of reinforcing steel must be placed within the available length la Try 9 sets of 2 No 5 stirrups a total of 4 legs per stirrup FHWANHI130126 StrutandTie Modeling STM for Concrete Structures Design Example 1 SimplySupported Deep Beam 117 Check the centertocenter spacing of the individual stirrups against the minimum spacing of 15 in given by AASHTO LRFD Article 51031 The vertical tie reinforcement layout is shown in Figure 111 48 in 108 in 30 in 30 in 8 Spa 6 in 48 in A A ELEVATION SECTION AA 2 No 5 Stirrups 2 No 5 Stirrups Typ 48 in 72 in Figure 111 Vertical Tie Reinforcing Layout Design Step 8 Perform Nodal Strength Checks Next the various nodes must be proportioned and checked for adequate strength This step begins by examining the node at the right bearing Node C Types of Nodes Nodes may be characterized as CCC CCT or CTT nodes CCC nodes Compression CompressionCompression are nodes where only struts intersect CCT nodes CompressionCompressionTension are nodes where a tie intersects the node in only one direction CTT CompressionTensionTension nodes are nodes where ties intersect a node in two different directions Check Node C The forces acting at Node C are shown in Figure 112 There are three forces intersecting at this node two compressive forces and one tensile force therefore this is a CCT node Two rows of reinforcing steel and an assumed 12 in by 48 in bearing plate which supports the 600 kip reaction force are shown FHWANHI130126 StrutandTie Modeling STM for Concrete Structures Design Example 1 SimplySupported Deep Beam 118 600 k 1177 kips 6 in 1013 kips 307 6 in Figure 112 Forces Acting on Node C The geometry of a CCT node is given in AASHTO LRFD Figure 58221b reproduced here as Figure 113 Bearing Face lb ha θs Back Face 05ha ha x cosθs lb x sinθs StruttoNode Interface Figure 113 Geometry of a CCT Node The variables in Figure 113 are defined below Height of the back face of the CCT node ha Length of the bearing face lb Angle between a strut and longitudinal axis of the member θs These variables may be defined for this example using previouslycalculated values and the geometries given in Figure 112 and Figure 113 The dimension of the bearing FHWANHI130126 StrutandTie Modeling STM for Concrete Structures Design Example 1 SimplySupported Deep Beam 119 plate is assumed at this point to be 12 in therefore lb 12 in The height of the back face of the CCT node is taken as the height of the tie This is assumed to be twice the depth from the bottom fiber of the beam to the centroid of the tie reinforcing per AASHTO LRFD Article 58252 Consequently ha is taken as a rounded dimension of 10 in The width of the struttonode interface herein called w is given by The resistance of each node face must be checked against the factored loads on each face The nominal resistance of a node face is given by AASHTO LRFD Equation 582511 where Pn nominal resistance of a node face kips Acn effective crosssectional area of the node face in2 fcu limiting compressive stress at the node face ksi The value of Acn is determined according to AASHTO LRFD Article 58252 The depth of an individual node face is determined according to AASHTO LRFD Figure 58221 The outofplane dimension may be determined by the bearing device dimensions or the width of member as appropriate The value of fcu is determined according to AASHTO LRFD Article 58253 The value of fcu is given by Equation 58253a1 where fc compressive strength of concrete used in design ksi m confinement modification factor defined below v concrete efficiency factor The confinement modification factor m is defined by FHWANHI130126 StrutandTie Modeling STM for Concrete Structures Design Example 1 SimplySupported Deep Beam 120 where A1 area under the bearing device in2 A2 notional area defined by AASHTO LRFD Article 565 in2 Notional Area A2 The stress on the face of a node is assumed to be uniformly distributed Where a supporting surface is larger than the loaded area area A2 is calculated assuming that the load spreads out from the loaded area at a rate of 2H1V until the edge of a member is met Otherwise if the loaded area is equal to the total member area m is taken as 10 The areas A1 and A2 are illustrated in AASHTO LRFD Figure 5651 reproduced below as Figure 114 45 45 Loaded Area A1 PLAN Loaded Area A1 A2 is measured on this plane 2 1 ELEVATION Figure 114 Determining the Areas A1 and A2 The concrete efficiency factor v is dependent on the presence of crack control reinforcement in the member under consideration If crack control reinforcement is not present as specified by AASHTO LRFD Article 5826 v shall be taken as 045 For structures that do contain crack control reinforcement defined by Article 5826 v is determined from AASHTO LRFD Table 58253a1 reproduced below as Table 12 Calculations for satisfying the crack control reinforcement requirement will be shown in Design Step 9 FHWANHI130126 StrutandTie Modeling STM for Concrete Structures Design Example 1 SimplySupported Deep Beam 121 Table 12 Concrete Efficiency Factors v Face CCC Node CCT Node CTT Node Bearing Face 085 070 045 ν 065 Back Face 085 070 045 ν 065 StruttoNode Interface 045 ν 065 045 ν 065 045 ν 065 Recall that Node C is a CCT node Therefore for the bearing face And for the struttonode interface For both node faces since the strut is the full widththickness of the member the confinement modification factor m equals 10 The resistance factor ϕ for compression in a strutandtie model is found in AASHTO LRFD Article 5542 which is equal to 070 Thus the factored resistance of a nodal face in a strutandtie model is calculated by AASHTO LRFD Equation 582511 modified as shown For the bearing face FHWANHI130126 StrutandTie Modeling STM for Concrete Structures Design Example 1 SimplySupported Deep Beam 122 For the struttonode interface A check of the back face is not required Bond stresses from reinforcing steel development need not be applied to the back face of a CCT node per AASHTO LRFD Article 58253b If the bars were anchored with headed reinforcing or an anchor plate the stresses on the back face of the node would be checked The Back Face of a CCT Node AASHTO LRFD Article C58253b states that there were no experimental cases reviewed where the back face stress of a CCT node controlled the strength of that node when the tie was composed of deformed reinforcing steel However this is not the case if the stress on the back face is applied by a bearinganchor plate or a headed bar If the tie is composed of a headed bar or is anchored by a bearing plate the back face of the node should be checked assuming that the bar or tendon is unbonded and that all of the tie force is transferred through the anchor plate or bar head Check Node A Because the reactions at Nodes A and C are the same and the bearings are assumed to be the same size the check of the bearing surface is satisfied automatically The diagonal compressive load in member AD is significantly less than member CF 785 kips vs 1177 kips and by geometry the width of the struttonode interface is larger at Node A than at Node C Therefore Node A is adequate by inspection Check Node F The geometry of Node F and loads acting on the node faces are given in Figure 115 The upper bearing plate is assumed to be 12 in long and 48 in wide similar to the lower bearing plates The height of the horizontal strut loading the left side of the node ha is the assumed strut depth found in Design Step 6 using the flexural analogy which is 6 in This strut is in equilibrium with the tie in the bottom of the beam FHWANHI130126 StrutandTie Modeling STM for Concrete Structures Design Example 1 SimplySupported Deep Beam 123 lb sin θ ha cos θ lb 12 in ha 6 in 600 k 1177 kips 1013 kips θ θ Figure 115 Forces Acting on Node F Because all of the forces acting on this node are compressive this is a CCC node The geometry of a CCC node is given in AASHTO LRFD Figure 58221a reproduced here as Figure 116 α x lb sinθs ha x cosθs α x lb θs θs ha 1α x lb lb Back Face StruttoNode Interface Centerline of Faces Consistent with Model Geometry Bearing Face Figure 116 Geometry of a CCC Node FHWANHI130126 StrutandTie Modeling STM for Concrete Structures Design Example 1 SimplySupported Deep Beam 124 The width of the strut at the node interface may be found as for Node C CCC Nodes and the Factor α In CCC nodes where diagonal struts enter a node from both sides and an external load P is applied it is sometimes beneficial to separate the applied load into two statically equivalent loads P1 and P2 such that P1 P2 P These two loads are then assumed to act in the center of the tributary area of the bearing plate The factor α denotes the portion of the load supported by the right diagonal strut and 1 α denotes the load carried by the left diagonal strut This approximation is not used in this example as there is only one diagonal strut therefore referring to Figure 116 it is assumed that α 1 and the full length of the bearing plate lb is used Refer to AASHTO LRFD Article C5822 and Figure C58224 for additional information Next calculate the effective crosssectional area of each node face Acn and determine the concrete efficiency factors v Recall that the concrete efficiency factors are found in AASHTO LRFD Table 58253a1 and in Table 12 of this design example The concrete efficiency factors for CCC nodes are illustrated in AASHTO LRFD Figure C58253a1a reproduced below as Figure 117 085 C 085 fc 20 ksi 085 C C Figure 117 Concrete Efficiency Factors for CCC Nodes FHWANHI130126 StrutandTie Modeling STM for Concrete Structures Design Example 1 SimplySupported Deep Beam 125 For the bearing face For the back left face For the inclined strut face Next using these calculated values determine the factored resistance of each node face using AASHTO LRFD Equations 582511 and 58253a1 For both node faces since the width of the strut is equal to the full widththickness of the members the confinement modification factor m equals 10 Recall that ϕ 07 for compression in strutandtie models The strength is then calculated by AASHTO LRFD Equation 582511 For the bearing face For the back left face For the inclined strut face Two of the checks at this node do not meet the requirements of the specification First in order to increase the strength for the inclined strut the bearing plate length is increased to 14 in This increases the struttonode interface width w to 123 in resulting in a new effective crosssectional area of 590 in2 This results in a factored FHWANHI130126 StrutandTie Modeling STM for Concrete Structures Design Example 1 SimplySupported Deep Beam 126 compressive resistance of 1239 kips which is greater than 1177 kips Consequently this node is now sufficient For consistency all of the bearing plates will be increased to 14 inches in length For the horizontal top strut originally sized to meet the assumed flexural analogy from Bernoulli beam theory the strut does not have sufficient resistance In this example it is decided to reinforce the top strut Since some longitudinal mild steel is required anyway for crack control and to anchor the vertical stirrups it is quite simple to add steel reinforcement to increase the node faces factored resistance Increasing the Strength of a Nodes Face The four options for increasing the strength of the deficient node face presented below vary in difficulty and time required to implement They are listed in decreasing order of difficulty Change the truss geometry Changing the geometry of the STM truss results in a new truss analysis All of the truss forces must then be recalculated and all tie and nodalstrength checks must be reperformed which can be very timeconsuming for all but the simplest models Increase the concrete strength Increasing the concrete strength requires recalculation of the concrete efficiency factors v and subsequent verification that all calculated node strengths are still sufficient Increase the beam width Similar to increasing the concrete strength increasing the beam width requires another iteration of STM design checks It will also require recalculating the beam selfweight loads Reinforce the struts Reinforcing the strut only requires calculation of the area of reinforcing steel required to make up the deficiency in resistance This option is often the simplest The top strut is deficient in resistance by Because the strut is a compressive member the provisions of AASHTO LRFD Article 5644 may be applied The area of reinforcing steel required is found by applying AASHTO LRFD Equation 56443 modified by including only the mild steel reinforcement terms since the resistance of the concrete strut has already been determined and there is no prestressed reinforcement Rearranging and solving for Ast FHWANHI130126 StrutandTie Modeling STM for Concrete Structures Design Example 1 SimplySupported Deep Beam 127 The area of reinforcing steel required to satisfy the strength requirement at this node is 467 in2 Try providing 6 No 8 bars in the strut Therefore provide 6 No 8 reinforcing bars near the top of the beam Check Node E At Node E multiple forces intersect at the same location there is a load applied at a bearing and three internal truss member loads at this node The strutandtie design method as defined by experiment and adopted by AASHTO was developed by assuming three forces intersect each node Therefore nodes where there are more than three intersecting loads must have some forces combined into resultant forces This also results in somewhat simpler computations 600 k E 506 k 1013 k 785 k 600 k 1013 k 600 k 1013 k E 600 k 1013 k 600 k 1013 k E R θ Figure 118 Combining Nodal Forces into Resultants The topleft portion of Figure 118 depicts the loads at Node E as they are drawn in the strutandtie model Depicted is the 600kip applied load and the resultant strut forces The topright portion of Figure 118 shows the forces at Node E with the 785kip diagonal load separated into its horizontal and vertical components Note that all of the forces acting on Node E remain in equilibrium The bottom portion of Figure 118 shows a resultant force R determined by combining the force in member DE and the horizontal and vertical components of force in member BE The resultant force R and the angle of its line of action θ are found by FHWANHI130126 StrutandTie Modeling STM for Concrete Structures Design Example 1 SimplySupported Deep Beam 128 The resulting force of R 1177 kips is the same force as exists by symmetry in member CF acting at the same angle θ and was previously analyzed in the prior check of Node F Because Node F was eventually shown to be satisfactory Node E is therefore also satisfactory since it is statically equivalent to the forces at Node F Static Equivalency Recall that the left side of the strutandtie model was chosen to be a two panel model only to demonstrate the design of a beam with a vertical tie If it were replaced with a direct strut model it would be identical to the righthand side of the strutandtie model Nodes B and D Nodes B and D are not checked for compressive resistance As discussed in Design Step 7 when the vertical tie was designed the tension force within the tie is actually spread out over the available length la Hence no check is required Smeared Nodes Refer to Figure 19 which was referenced while designing the vertical tie between Nodes B and D Recall that a vertical tie in a strutandtie model does not exist in a distinct location rather forces spread out over the shear span a the distance between the applied load and reaction The nodes at the ends of the vertical tie are referred to as smeared nodes Smeared nodes do not have a geometry that can be clearly defined by a bearing plate or the boundaries of the member Therefore the limits of a nodal region cannot be determined with any degree of certainty The forces in a smeared node are able to disperse over a large area and thus are less critical than singular nodes or nodes with clearly defined geometries and checking the concrete stresses at smeared nodes is typically unnecessary Design Step 9 Proportion Crack Control Reinforcement Crack control reinforcement provided as orthogonal grids of reinforcing bars is provided both to limit the width of cracks and to provide a minimum level of ductility so that if required inelastic redistribution of stresses can occur Because the nodal zones of the example beam have been designed using the concrete efficiency factors of AASHTO LRFD Table 58253a1 crack control reinforcement is required The spacing of the crack control reinforcing may not exceed the smaller of 𝑑𝑑 4 or 120 in in both directions FHWANHI130126 StrutandTie Modeling STM for Concrete Structures Design Example 1 SimplySupported Deep Beam 129 Controlling Cracks Placing crack control reinforcement is a requirement for satisfactory performance of a concrete member under service loads This reinforcement aids in controlling the width of cracks under service loads and it also restrains the compressive stress within struts Restraining the compressive stress helps prevent sidebursting failure of the concrete in the struts The crack control reinforcement in the vertical direction shall satisfy AASHTO LRFD Equation 58261 The crack control reinforcement in the horizontal direction shall satisfy AASHTO LRFD Equation 58262 where Av area of vertical reinforcement within spacing sv in2 Ah area of horizontal reinforcement within spacing sh in2 bw width of the members web in sv spacing of vertical crack control reinforcement in sh spacing of horizontal crack control reinforcement in Thick and Thin Members For thinner members the crack control reinforcement will consist of two mats of reinforcing one placed near each face of the member For thicker members multiple mats of reinforcing placed throughout the width of the member may be required in order to satisfy the 03 requirement of Equations 58261 and 2 AASHTO LRFD Figure C58261 reproduced below as Figure 119 illustrates the variables of Equations 58261 and 2 Note the difference in placement of reinforcement of a thin and thick member The orthogonal grid of reinforcement restrains the concrete in the strut shown in gray from failure by bursting outward from the face of the member FHWANHI130126 StrutandTie Modeling STM for Concrete Structures Design Example 1 SimplySupported Deep Beam 130 A A sv d SECTION AA bw sh Av Ah Thick Member bw Thin Member h hatop hatop hatop habot habot Figure 119 Distribution of Crack Control Reinforcement Solve for the area of reinforcing required in the horizontal direction Try providing 4 No 6 bars per foot FHWANHI130126 StrutandTie Modeling STM for Concrete Structures Design Example 1 SimplySupported Deep Beam 131 Solve for the area of reinforcing required in the vertical direction Try providing 2 No 5 closed stirrups 4 legs total spaced at 8 in centertocenter For the right half of the beam where the direct strut model is used this is the only reinforcing steel required For the left half of the beam recall that the vertical tie had its own reinforcing steel requirements That reinforcing requirement was met by 2 No 5 closed stirrups 4 legs total at 6 in centers a tighter spacing than necessary for the direct strut model A comparison of the direct strut and two panel models reinforcement requirements is given in Table 13 Table 13 Reinforcement Requirements Comparison Location Direct Strut Model Two Panel Model Vertical Reinforcing 4 Legs of No 5 Stirrups at 8 in Centers 4 Legs of No 5 Stirrups at 8 in Centers Typ Except 4 Legs of No 5 Stirrups at 6 in Centers Centered Around the Vertical Tie Bottom Chord Tie Reinforcing 16 No 10 Bars 16 No 10 Bars Note that the bottom tie reinforcing is independent of the direct strut or two panel truss model In the region between the point loads the region of constant moment the moment in the pseudosimple beam is independent of the truss configuration and would be the same no matter what discretization is used between the point loads and the reactions The comparison in Table 13 demonstrates the efficiency of the direct strut model versus the two panel model Recall that the two panel model was used in this example only to demonstrate its application and in this example it can be replaced by a direct strut model The addition of the vertical tie requires providing additional reinforcing to carry the same forces carried by the direct strut model FHWANHI130126 StrutandTie Modeling STM for Concrete Structures Design Example 1 SimplySupported Deep Beam 132 Design Step 10 Provide Necessary Anchorage for Ties The ends of the hooked No 10 longitudinal bars are checked for required development Unlike a flexural model where the reinforcing steel is checked for development at the points of maximum moment the tie of an STM truss model has a constant force between its end nodes Consequently the anchorage of reinforcing steel particularly near the ends of beams and faces of members is critical Refer to Figure 120 below AASHTO LRFD Figure C582421 The reinforcing bars of the ties must be properly anchored to guarantee that the tie force is fully developed and the structure can achieve the resistance calculated in the strutandtie model In order for a tie to be considered properly anchored the full yield resistance of the tie should be developed at the point where the centroid of the reinforcing steel exits the extended nodal zone as shown Available Length Assume Strut is Prismatic Nodal Zone Extended Nodal Zone Critical Section for Development of Tie Figure 120 Available Development Length for Ties Critical Section for Development of the Tie The location of the critical section for the development of the tie reinforcing is shown graphically in Figure 120 This point occurs where the centroid of the reinforcing steel in the tie passes through the edge of strut that intersects the tie Its location may be determined by assuming that the strut is prismatic where its edges are parallel to its centerline The demand in the bottom chord is a maximum in member BC and was found to be 1013 kips In Design Step 7 the tie capacity was found to be 1097 kips Because FHWANHI130126 StrutandTie Modeling STM for Concrete Structures Design Example 1 SimplySupported Deep Beam 133 some excess capacity is present it may be used to reduce the required development length The required development length for a hooked reinforcing bar in tension is given by AASHTO LRFD Equation 510824a1 where lhb is given by AASHTO LRFD Equation 510824a2 where λ concrete density modification factor λrc reinforcement confinement factor λcw reinforcement coating factor λer excess reinforcement factor The following values are assumed for calculation λ 10 Normalweight concrete λrc 10 Conservative assumption λcw 10 Uncoated reinforcement ie black bars The excess reinforcement factor λer is calculated by Therefore which is rounded up to 20 in The available development length is determined graphically in Figure 121 Recall that the top bearing plates were lengthened to 14 in during the analysis of Node F and this change was carried through to the bottom bearing plates as well FHWANHI130126 StrutandTie Modeling STM for Concrete Structures Design Example 1 SimplySupported Deep Beam 134 Anchorage of the tie reinforcing is checked at the critical section near Node C The distance from the end of the bearing plate to the critical section x is determined by geometry lb 14 in ha 10 in θ 5 in 5 in 7 in x Strut Nodal Zone Tie Reinforcing Figure 121 Geometry of Tie End Hook Development near Node C The available development length is determined by Therefore the development length is adequate and the tie is capable of developing its required strength Design Step 11 Draw Reinforcement Layout At this point the strutandtie analysis for the example simplysupported deep beam is complete Sketches of the final beam dimensions and reinforcing layouts follow FHWANHI130126 StrutandTie Modeling STM for Concrete Structures Design Example 1 SimplySupported Deep Beam 135 8 Spa 6 in 48 in A A LEFTSIDE ELEVATION 2 No 5 Stirrups Typ Spa 8 in Max 4 Spa 8 in Max 16 No 10 8 ea Row 5 Equal Spa 4 No 6 6 No 8 Figure 122 Reinforcing Layout Left Side A A RIGHTSIDE ELEVATION 2 No 5 Stirrups Typ Spa 8 in Max 4 No 6 16 No 10 8 ea Row 5 Equal Spa 6 No 8 Figure 123 Reinforcing Layout Right Side FHWANHI130126 StrutandTie Modeling STM for Concrete Structures Design Example 1 SimplySupported Deep Beam 136 SECTION AA 6 No 8 8 No 10 8 No 10 4 No 6 Typ No 5 Stirrup 72 in 48 in Figure 124 Section through Example Beam FHWANHI130126 Design Example 2 Cantilever Bent Cap StrutandTie Modeling STM for Concrete Structures 21 Design Example 2 Cantilever Bent Cap Table of Contents Page Design Step 1 Define StrutandTie Model Input 23 Design Step 2 Determine the Locations of the B and DRegions 28 Design Step 3 Define Load Cases 29 Design Step 4 Analyze Structural Components 212 Design Step 5 Size Structural Components Using the Shear Serviceability Check 213 Design Step 6 Develop a StrutandTie Model 215 Design Step 7 Proportion Ties 221 Design Step 8 Perform Nodal Strength Checks 226 Design Step 9 Proportion Crack Control Reinforcement 233 Design Step 10 Provide Necessary Anchorage for Ties 233 Design Step 11 Draw Reinforcement Layout 234 FHWANHI130126 Design Example 2 Cantilever Bent Cap StrutandTie Modeling STM for Concrete Structures 22 Design Example 2 presents the design of a cantilever bent cap utilizing the strutand tie modeling STM design procedure Stepbystep STM procedures are presented and a complete design is demonstrated for one of the load cases that must be considered This design example includes developing a strutandtie model for a slightly sloped structure The cantilever bent cap is sloped to accommodate the banked grade of the roadway supported by the bent and the applied loads are therefore not perpendicular to the primary longitudinal chord of the STM A flowchart of the various design steps is presented below Design Step 1 Define StrutandTie Model Input Design Step 2 Determine the Locations of the B and DRegions Design Step 3 Define Load Cases Design Step 4 Analyze Structural Components Design Step 5 Size Structural Components Using the Shear Serviceability Check Design Step 6 Develop a StrutandTie Model Design Step 7 Proportion Ties Design Step 8 Perform Nodal Strength Checks Design Step 9 Proportion Crack Control Reinforcement Design Step 10 Provide Necessary Anchorage for Ties Design Step 11 Draw Reinforcement Layout FHWANHI130126 Design Example 2 Cantilever Bent Cap StrutandTie Modeling STM for Concrete Structures 23 It should be noted that the STM provisions in the 8th edition of AASHTO LRFD 2017 are based primarily on research conducted at the University of Texas at Austin Birrcher et al 2009 This design example is based primarily on one of the example problems included in an implementation project sponsored by TxDOT 5525301 Williams et al 2011 In addition figures in this design example have been adapted from Williams et al 2011 The example problem originally prepared by Williams et al 2011 has been revised to provide additional explanation and to be fully compliant with the STM provisions of the 8th edition of AASHTO LRFD as appropriate Design Step 1 Define StrutandTie Model Input Elevation and plan views of the cantilever bent cap are presented in Figures 21 and 2 2 For clarity a simplified view excluding bearing pads and bearing seats is shown in Figure 21 However a more detailed geometry of the cap is presented in Figure 22 For this design example the cantilever bent cap supports two prestressed concrete U beams from one direction and two steel girders from the opposite direction Each of the Ubeams rests on two neoprene bearing pads while each of the steel girders is supported by a single pot bearing The bearing conditions of each girder are shown in Figure 22 FHWANHI130126 Design Example 2 Cantilever Bent Cap StrutandTie Modeling STM for Concrete Structures 24 500 L Column C 1842 1000 Column L Column C L Bent C L Column C 850 00516 500 800 Column 2842 800 Cap 400 400 PLAN ELEVATION Figure 21 Plan and Elevation Views of Cantilever Bent Cap Simplified Geometry FHWANHI130126 Design Example 2 Cantilever Bent Cap StrutandTie Modeling STM for Concrete Structures 25 175 L Bent L Column C C L Column C L Column C 850 00516 275 800 2842 400 400 PLAN ELEVATION 1650 500 225 225 517 125 079 L Girder 1 C L Girder 2 C 1500 225 225 L Beam 1 C L Beam 2 C 342 Beam Spacing Girder Spacing L Bearing Girders C L Bearing Beams C Figure 22 Plan and Elevation Views of Cantilever Bent Cap Detailed Geometry Define Material Properties Material properties for this design example are as follows Concrete strength Reinforcement strength Concrete unit weight FHWANHI130126 Design Example 2 Cantilever Bent Cap StrutandTie Modeling STM for Concrete Structures 26 Since normal weight concrete is being used and since fc is less than 15 ksi and fy is less than 75 ksi the AASHTO provisions for STM are applicable AASHTO LRFD Article 5821 Design Iterations Using STM When using STM design iterations may be necessary to determine both the concrete strength and bent cap width to provide adequate strength to the critical node Since the geometry of the strutandtie model is dependent on the value of fc and the cap width the geometry of the STM must be updated for every iteration that is performed This design example presents the development of final strutandtie models for the last iteration that was performed for this problem Determine the Bearing Areas For this design example each of the bearing pads supporting the prestressed concrete Ubeams is 16 inches by 9 inches The steel girders are supported by pot bearings with masonry plates that rest on the bearing seats The sizes of the masonry plates for Girder 1 and Girder 2 are 42 by 295 inches and 24 by 24 inches respectively Each bearing pad or plate is placed on a bearing seat that allows the applied force to spread over an area of the cap surface that is larger than the pad or plate itself The longitudinal dimensions ie effective lengths of the effective areas are measured at the top surface of the bent cap and labeled in Figure 23 A plan view of the bearings is presented in Figure 24 The transverse dimensions ie effective widths shown in Figure 24 are measured at the centerline of each bearing pad or plate AASHTO LRFD Article 565 specifies a slope of 1 vertical to 2 horizontal for computing the effective areas However for simplification a slope of 1 vertical to 1 horizontal was used in this design example The effective width of the bearing area of Girder 1 has been limited to prevent overlap with the effective bearing area of Beam 1 The dimensions of the bearing areas are summarized in Table 21 along with the size of the effective bearing area for each beam or girder FHWANHI130126 Design Example 2 Cantilever Bent Cap StrutandTie Modeling STM for Concrete Structures 27 420 270 270 540 4819 1 1 240 270 270 540 3012 300 300 Girder 1 Girder 2 270 135 135 540 160 160 1 1 292 432 2193 2473 Beam 1 Beam 2 270 135 135 540 160 160 199 339 2007 2287 Figure 23 Elevation View Showing Effective Bearing Areas Considering Effect of Bearing Seats 1 1 L Beam 1 Bearings C 16 x 9 Bearing Pads 42 x 295 Plate 325 L Girder 1 Bearing C 1484 1764 L Girder 2 Bearing C 1578 1298 1 1 L Beam 2 Bearings C 16 x 9 Bearing Pads 300 24 x 24 Plate Figure 24 Plan View Showing Effective Bearing Areas Considering Effect of Bearing Seats FHWANHI130126 Design Example 2 Cantilever Bent Cap StrutandTie Modeling STM for Concrete Structures 28 Table 21 Bearing Sizes and Effective Bearing Areas for Each Beam or Girder Characteristic Girder 1 Beam 1 Pad 1 Beam 1 Pad 2 Girder 2 Beam 2 Pad 1 Beam 2 Pad 2 Bearing Size 42x295 16x9 16x9 24x24 16x9 16x9 Effective Length 4819 2193 2473 3012 2007 2287 Effective Width 325 1484 1764 300 1298 1578 Effective Area 1566 in2 326 in2 436 in2 904 in2 260 in2 361 in2 Nodes Directly Below Applied Loads A simplification is provided to facilitate definition of the geometry of the nodes that are located directly below the applied superstructure loads Specifically the bearing areas are assumed to be square and located concentrically with the longitudinal axis of the bent cap Design Step 2 Determine the Locations of the B and DRegions The entire cantilever bent cap is a DRegion due to the applied superstructure loads ie load discontinuities and the geometric discontinuity of the frame corner The behavior of the bent cap is therefore dominated by a nonlinear distribution of strains Transition from DRegion to BRegion The transition from a DRegion to a BRegion occurs approximately one member depth away from a load or geometric discontinuity AASHTO LRFD Article 55121 Considering the bent in this design example the DRegionBRegion interface is assumed to be located at a distance of one column width ie 10 feet from the bottom of the bent cap The limit of the DRegion is shown in Figure 25 FHWANHI130126 Design Example 2 Cantilever Bent Cap StrutandTie Modeling STM for Concrete Structures 29 1000 500 500 DRegion BRegion Figure 25 Limit of DRegion in Bent Cap Design Step 3 Define Load Cases The factored superstructure loads from the two steel girders and two concrete Ubeams are shown in Figure 26a Loads for both the strength and service limit states are presented These loads correspond to one particular load case that must be considered during the design process In this design example the Girder 1 reaction is significantly greater than the Girder 2 reaction due to various additional loads applied to the fascia girder that are not applied to the interior girder The final design of the bent cap must satisfy the design requirements for all governing load cases The superstructure design loads are assumed to act at the point where the longitudinal centerline of a beam or girder coincides with the transverse centerline of the respective bearing pads Resolving Point Loads Point loads in close proximity to one another can be resolved together to simplify the load case and facilitate development of a practical strutandtie model The factored loads on the left and right can be resolved into single loads through superposition as shown in Figure 26b The locations of the resolved loads are determined by the calculations presented below In these calculations x1 is the horizontal distance from the centerline of the column to the left resolved load P1 Similarly x2 is the horizontal distance from the centerline of the column to the right resolved load P2 as shown in the plan view of Figure 26b FHWANHI130126 Design Example 2 Cantilever Bent Cap StrutandTie Modeling STM for Concrete Structures 210 The dimensions in the above calculations are illustrated in Figure 26a and the resolved loads are assumed to act at the longitudinal centerline of the top of the bent cap as illustrated in Figure 26b 500 L Bearing C 175 1650 517 2842 L Beam 1 C L Beam 2 C Load on Bent Cap L Column C 500 275 1500 567 L Bent L Column C C x1 197 L Column C 543 1602 x2 1799 L Column C L Bent C L Girder 1 C L Girder 2 C 404 k 1396 k 404 k 367 k L Column C L Column C P2 771 k P1 1800 k 500 500 500 500 263 k 907 k 263 k 238 k L Column C L Column C P2 501 k P1 1170 k 500 500 500 500 Strength Limit State Loads Service Limit State Loads a Loads from Each Beam or Girder b Resolved Loads Figure 26 Factored Superstructure Loads Acting on Bent Cap FHWANHI130126 Design Example 2 Cantilever Bent Cap StrutandTie Modeling STM for Concrete Structures 211 After the design loads from the superstructure acting on the bent cap have been computed as shown in Figure 26b the selfweight of the cantilever bent cap must also be considered The factored selfweight based on tributary volumes is added to each load as presented in Figure 27 As previously defined the unit weight of the reinforced concrete is 150 pcf The magnitude of each load acting on the strutandtie model including the selfweight of the bent cap is computed as follows The first value in each calculation is the factored superstructure load and the second value is the tributary selfweight of the cantilever bent cap factored by 125 for the Strength I load combination AASHTO LRFD Tables 3411 and 3412 These calculations result in the final design loads for the strength limit state acting on the bent cap as shown in Figure 27 Tributary Volumes The tributary volumes in this design example include concrete details eg bearing seats and bent cap end details not shown for clarity Taking these details into account the location of the selfweight dividing line between P1 and P2 is approximately 122 feet from the right end of the cantilever bent cap as shown in Figure 27 An alternate method is to define the selfweight dividing line as the midpoint between the two applied loads Selfweight Included in P1 P1 2005 k P2 926 k Selfweight Included in P2 122 ft Figure 27 Total Factored Loads Based on Superstructure Loads and Bent Cap Selfweight FHWANHI130126 Design Example 2 Cantilever Bent Cap StrutandTie Modeling STM for Concrete Structures 212 Referring to Table 21 in Design Step 1 the effective bearing area for the load P1 acting on the bent cap in Figure 27 is the combination of the effective bearing areas for Beam 1 and Girder 1 or 2328 in2 and it is assumed to be a 482inch by 482inch square ie 2328 𝑖𝑖𝑖𝑖2 482 𝑖𝑖𝑖𝑖 Similarly the effective bearing area for the load P2 acting on the cap is assumed to be a 391inch by 391inch square Both loads are assumed to act at the center of these effective bearing areas Design Step 4 Analyze Structural Components Assuming a linear distribution of stress at the interface of the B and DRegion based on St Venants principle as per AASHTO LRFD Article C55121 the linear stress distribution is as shown in Figure 28 and the extreme fiber stress for the right side of the column is computed as follows where AColumn crosssectional area of the column in2 IColumn moment of inertia of the column in4 M moment at the centerline of the column due to P1 and P2 kin c distance from the extreme fiber to the centerline of the column in It should be noted that at the strength limit state the concrete stress distribution will not be linear but rather the concrete would be cracked The designer can therefore treat the boundary section as a cracked section Doing so however would be overly conservative For this design example the linear stress distribution at the interface of the B and D Regions is computed based on St Venants principle A primary purpose for calculating the bending stresses in this manner is to define the locations of the struts within the column and linear stress distribution is used to optimize conservatism More specifically if we assumed that the column is at its flexural capacity and recognizing that the column section has symmetrically distributed reinforcement the depth of the compression zone would be very small Although this analysis technique would be FHWANHI130126 Design Example 2 Cantilever Bent Cap StrutandTie Modeling STM for Concrete Structures 213 acceptable it would be overly conservative By justifying a greater compression zone our solution is sufficiently but not overly conservative DRegion P1 2005 k P2 926 k C x1 197 543 x2 1799 500 381 619 1328 psi ƒRight 1000 500 500 T Figure 28 Linear Stress Distribution at the Boundary of the B and DRegions Design Step 5 Size Structural Components Using the Shear Serviceability Check The likelihood of the formation of diagonal cracks in the cantilevered portion of the bent cap should be considered To limit diagonal cracking the service level shear force should be less than the estimated diagonal cracking strength of the member Using the AASHTO LRFD Service I load combination the service level shear force is computed at the inside face of the column Based on Figure 26 the service limit state value of P2 is 501 kips and the selfweight of the cantilever portion of the bent cap is 188 kips Therefore the total service level shear force at the inside face of the column is 689 kips FHWANHI130126 Design Example 2 Cantilever Bent Cap StrutandTie Modeling STM for Concrete Structures 214 A value for d must be assumed at this point It is assumed that a cover of 2 inches is provided No 6 stirrups are used No 11 longitudinal rebar is used in two layers and a clear space of 2 inches is provided between layers Thus the value for d is found as The distance a is known as the shear span In this design example the shear span is the distance from the applied load P2 to the right face of the column Therefore as illustrated in Figure 28 a is equal to 1299 feet or 1559 inches As specified in AASHTO LRFD Equation C58221 the estimated resistance at which diagonal cracks begin to form Vcr for the cantilever portion of the bent cap is computed as follows where a shear span 1559 in as previously described and computed d effective depth of the member 968 in as previously described and computed In addition to the equation presented above Vcr must not be greater than nor less than AASHTO LRFD Article C5822 The estimated diagonal cracking resistance is considerably greater than the service level shear force Therefore diagonal cracks are not expected to form under the service loads considered in this design example If this check did not produce a favorable result the designer could resize the cap andor increase the compressive strength of concrete to satisfy this design check The equation for Vcr presented in AASHTO LRFD Article C5822 is based on shear resistance and does not consider torsional effects If significant torsion is present the Vcr expression can be modified by taking torsion into account not currently defined in AASHTO LRFD For this design example since torsion is not significant its effects need not be considered FHWANHI130126 Design Example 2 Cantilever Bent Cap StrutandTie Modeling STM for Concrete Structures 215 Design Step 6 Develop a StrutandTie Model For this design example two STMs are used to model the flow of forces within the cantilevered portion of the bent The first model a direct strut model shown in Figure 2 9 features one truss panel in the cantilevered portion and models a direct flow of forces to the column As defined in AASHTO LRFD Article 5822 the angle between the axes of a strut and tie should be limited to angles greater than 25 degrees As presented in Figure 29 the angle between the strut and tie for the direct strut model is 283 degrees Therefore the direct strut model for this design example satisfies the AASHTO requirements In addition for this design example the direct strut model is a more efficient and correct model and it better represents the actual flow of forces within the cantilever bent cap However simply as a learning exercise a two panel model is presented in Figure 210 This second model features two truss panels with an intermediate vertical tie and was developed to investigate the requirements of the vertical tie within the cantilever when using a two panel model All other characteristics of the STM geometry are the same for both models The description of the STM development presented in this section applies primarily to the first model shown in Figure 29 unless otherwise noted As specified in AASHTO LRFD Article C5822 the designer should minimize the number of vertical ties between a load and a support using the least number of truss panels possible while still satisfying the 25 degree minimum Since the direct strut model satisfies the 25 degree minimum requirement the two panel model is not necessary and is not recommended by AASHTO In addition for this design example the two panel model is a less efficient and less correct model and it does not accurately represent the actual flow of forces within the cantilever bent cap It is included in this design example solely as a learning exercise As previously explained the geometry of the STM is dependent on the value of fc and the cap geometry and the STM must correspond to the applied loads and chosen geometry The following explanation applies to the development of the final STMs for the last iteration that was performed In Figures 29 and 210 the width of the trapezoid defining the location of Strut EE is determined by setting the trapezoidal stress volume equal to the value of P2 as follows The locations of Struts EE and DD can then be determined using basic centroid equations found in geometry books The location of Strut EE determined by computing the centroid of the trapezoidal stress volume defined above is 031 feet from the right side of the column The location of Strut DD determined by computing the FHWANHI130126 Design Example 2 Cantilever Bent Cap StrutandTie Modeling STM for Concrete Structures 216 centroid of the remaining triangular compressive stress volume is 217 feet from Strut EE These values are illustrated in Figures 29 and 210 Figure 29 StrutandTie Model Featuring One Truss Panel for the Cantilever Bent Cap P1 2005 k P2 926 k 543 1602 697 A B C E D E A 043 019 1570 k 1664 k 3751 k 845 k 1330 D 283 29 504 C 1328 psi C 381 555 064 685 067 031 217 T 840 FHWANHI130126 Design Example 2 Cantilever Bent Cap StrutandTie Modeling STM for Concrete Structures 217 P1 2005 k P2 926 k 937 A B C E D E A 1437 k 1573 k 043 1183 k 2254 k 2017 k 1570 k 1664 k 3751 k 845 k 850 665 019 786 k 926 k 786 k F G 665 D 665 1183 k 543 697 00516 485 29 504 C 1328 psi C 381 555 064 685 067 031 217 T Figure 210 StrutandTie Model Featuring Two Truss Panels for the Cantilever Bent Cap Placement of Struts The first step in developing the STMs is to determine the locations of the vertical struts within the column identified as Struts DD and EE in Figures 29 and 210 As previously described the struts are located to correspond with the resultants of the compressive portion of the linear stress diagram at the boundary of the DRegion If a single strut is used to model the forces within the column as illustrated in Figure 2 11a it should be positioned to correspond with the resultant of the compressive portion of the linear stress diagram For this design example if only P2 acted on the FHWANHI130126 Design Example 2 Cantilever Bent Cap StrutandTie Modeling STM for Concrete Structures 218 bent cap one vertical strut within the column would be sufficient However since there are two applied loads a second vertical strut is needed to model the direct transfer of load P1 into the column as shown in Figure 211b Placement of Struts Correct strutandtie models are developed by making a series of rational assumptions Further there is generally not one unique solution or strutandtie model for each case There is much flexibility associated with STM design For this design example the use of a single strut coupled with the cracked column section at flexural ultimate gives a very conservative size to the flexural compression zone Therefore the designer has two options 1 use an oversized column to satisfy a perceived problem triggered by one STM or 2 look for other load paths a different STM that demonstrates the fact that the actual column size is in fact acceptable For this design example the second option is applied and two vertical struts in the column are used to model the direct transfer of the two applied loads P1 P2 P1 P2 Becomes Subdivide this node a b Figure 211 Modeling Compressive Forces within the Column a Using a Single Strut and b Using Two Struts In order to position the two vertical struts within the column the compressive portion of the stress diagram is subdivided into two parts a trapezoidal shape and a triangular shape as shown in Figures 29 and 210 The geometry of each subdivision is determined by setting its resultant force equal to the corresponding force within the structure The resultant of the trapezoidal shape at the right is generally equal to the magnitude of P2 and the resultant of the triangular shape is generally equal to P1 plus the resultant of the tensile portion of the stress diagram For this design example however the resulting forces in the vertical struts within the column Struts DD and EE do not equal the resultants of the stress diagram subdivisions that were previously determined This is to be expected since Tie AA within the column must coincide with the column reinforcement and therefore does not coincide with the resultant of the tensile portion of the stress diagram The slight angle of Ties AB and BC also contributes to the difference in forces The combined effect of the forces in Strut DD Strut EE and Tie AA however is equivalent to the axial force and moment within the FHWANHI130126 Design Example 2 Cantilever Bent Cap StrutandTie Modeling STM for Concrete Structures 219 column at the DRegionBRegion interface The strutandtie models therefore satisfy the STM requirements for design Placement of Ties The next step in developing the STMs is to determine the placement of Ties AB BC and AA in Figure 29 The locations of the ties must correspond with the centroids of the longitudinal tension steel that will be provided within the structure AASHTO LRFD Article C5822 Design iterations are generally needed to achieve this level of accuracy When using the STM procedure the designer should compare the final reinforcement details ie the centroids of the longitudinal reinforcement with the locations of the longitudinal ties of the STM to decide whether another iteration would affect the final design As previously explained it is assumed that a cover of 2 inches is provided No 6 stirrups are used No 11 longitudinal rebar is used in two layers and a clear space of 2 inches is provided between layers The distance from the top surface of the bent cap to the centroid of the reinforcement along the top of the bent cap is therefore computed as follows 2 075 141 1 516 in The centroid of the main tension steel within the column is assumed to be located 80 inches from the left face of the column Considering the final reinforcement layout presented in Figures 219 and 220 following the STM design the locations of Ties AB and BC described above correspond with the centroids of the main longitudinal reinforcement within the bent cap Placement of the Corner Node Node E Before the remaining members of the STM are positioned the location of Node E should be determined The horizontal position of Node E is defined by the location of the vertical strut near the right face of the column Strut EE Only the vertical position of the node therefore needs to be determined In contrast to the placement of the column struts a linear distribution of stress cannot be used to position the node since no DRegionBRegion interface exists within the cap ie the entire cap is a DRegion The vertical position of Node E is therefore defined by optimizing the height of the STM ie the moment arm jd of the bent cap to achieve efficient use of the bent cap depth Node E is placed so that the factored force acting on the back face will be approximately equal to its design strength In other words the moment arm jd is as large as possible while still ensuring that the back face of Node E has adequate strength The calculation necessary to determine the vertical location of Node E is shown below and is illustrated in Figure 212 The moment at the right face of the column due to load P2 neglecting the slight angle of the bent cap is set equal to the factored resistance of the back face of Node E times the moment arm jd FHWANHI130126 Design Example 2 Cantilever Bent Cap StrutandTie Modeling STM for Concrete Structures 220 The resistance factor ϕ in the calculation is the AASHTO LRFD factor of 07 for compression in strutandtie models AASHTO LRFD Article 5542 The concrete efficiency factor ν is taken as the factor for the back face of Node E 085 for a CCC node as presented in AASHTO LRFD Table 58253a1 The term left of the equal sign is the moment at the right face of the column The vertical location of Node E is taken as 223 inches from the bottom face of the bent a distance equal to a2 The exact location of Node E is clearly shown in Figure 214 in the section presenting the nodal strength checks for Node E P2 926 k 1299 a2 516 Take moment about this point M 12029 kft CCC Node d 968 E C Figure 212 Determining the Vertical Position of Node E Placement of the Remaining Nodes The remaining nodes within the strutandtie model shown in Figure 29 can now be positioned Node D is located vertically to align with Node E and it is located horizontally to align with Node D Strut DE connects the two nodes Nodes B and C are located vertically below the applied superstructure loads Struts AD BD and CE are then added to model the elastic flow of forces within the bent cap These struts connect the nodes that have already been defined FHWANHI130126 Design Example 2 Cantilever Bent Cap StrutandTie Modeling STM for Concrete Structures 221 Similarly the remaining nodes within the strutandtie model shown in Figure 210 can also be positioned The vertical Tie FG is located midway between Strut EE and Node C Strut EG is parallel to the bottom face of the bent cap at a distance of 223 inches from the face Compute the Member Forces After the geometry of the STMs has been determined as described above the member forces of the struts and ties are computed by enforcing equilibrium Computing the Member Forces Since both models are statically determinate systems all member forces can be calculated by satisfying equilibrium at the joints of the truss This can be accomplished using the method of joints Given the small number of joints the forces can easily be determined using hand calculations It should be noted that these are not real trusses in that they do not satisfy the requirements of stability If this truss was modeled in a computer program it would not generate results due to the presence of incomplete panels and triangles Dummy members would need to be added for the computer to generate results For this STM example this truss is superimposed on a rigid element While it provides a convenient way to visualize the flow of forces it does not necessarily need to be stable in and of itself Design Step 7 Proportion Ties The only significant difference between the two models presented in Figures 29 and 2 10 is the additional vertical Tie FG in Figure 210 Since a vertical tie is not provided within the cantilevered portion of the bent cap in Figure 29 the STM in Figure 210 was developed as a learning exercise to determine the amount of stirrups required to resist the force in Tie FG However since we previously determined that the direct strut model in Figure 29 satisfies AASHTO requirements the design results from that model will be used for final design Design Requirement for Tie FG Using Two Truss Panels The nodes at the ends of the vertical tie Nodes F and G are both smeared nodes Although the strutandtie truss model implies that struts and ties occupy a specific location in reality the stresses spread out over portions of the member as shown in Figure 213 AASHTO LRFD suggests that an available length be determined over which the stirrups can be distributed AASHTO LRFD Figure C58222 By spreading the vertical steel out the individual bars better resist the overall distribution of forces as the concentrated load from the reaction and applied load fan out across the cantilever bent cap depth and length The 25degree angle in Figure 213 comes from the fact that FHWANHI130126 Design Example 2 Cantilever Bent Cap StrutandTie Modeling STM for Concrete Structures 222 compression is assumed to spread at 21 vertical to horizontal within the member This corresponds to an angle of 266 degrees which is rounded to 25 degrees in AASHTO LRFD Article 5822 Available Length la hSTM x tan25 Shear Span a 25 25 hSTM FanShaped Strut Stirrups Comprising Tie Determine the Tie Width hSTM x tan25 Figure 213 FanShaped Struts Engaging Reinforcement Forming a Tie The available length over which the reinforcement comprising Tie FG can be distributed is therefore computed as follows FHWANHI130126 Design Example 2 Cantilever Bent Cap StrutandTie Modeling STM for Concrete Structures 223 where 1299 feet is the shear span a as previously described and 946 inches is the vertical distance between Nodes F and G or the length of Tie FG The value of la is slightly conservative because the cross slope of the bent cap is ignored in its calculation The resistance factor ϕ for tension in strutandtie models for reinforced concrete is 09 AASHTO LRFD Article 5542 The strength of the ties must satisfy the following equation AASHTO LRFD Equations 58231 and 582411 Distributing fourlegged 6 stirrups over the available length the required spacing necessary to carry the force in Tie FG is determined as follows Factored load Tie capacity The number of 6 stirrups with 4 legs each required and its corresponding stirrup spacing are then computed as follows Therefore the spacing of fourlegged 6 stirrups should be no greater than 69 inches to satisfy the requirements for Tie FG for the STM with two truss panels as depicted in Figure 210 Design Requirement for Crack Control Reinforcement The stirrup requirements previously computed for the STM with two truss panels will now be compared to the minimum crack control reinforcement requirement The crack control reinforcement must satisfy the following two equations AASHTO LRFD Equations 58261 and 58262 FHWANHI130126 Design Example 2 Cantilever Bent Cap StrutandTie Modeling STM for Concrete Structures 224 where Av total area of vertical crack control reinforcement within spacing sv in2 Ah total area of horizontal crack control reinforcement within spacing sh in2 sv spacing of vertical crack control reinforcement in sh spacing of horizontal crack control reinforcement in bw width of member web in Using fourlegged 6 stirrups the required spacing of the vertical crack control reinforcement is computed as follows It should be noted that the stirrup spacing necessary for Tie FG as previously computed is greater than the stirrup spacing necessary for crack control reinforcement Therefore if the two panel model in Figure 210 were being used to design the cantilever bent cap the crack control reinforcement would be sufficient to resist the force in Tie FG However since the direct strut model in Figure 29 is acceptable for this design example this comparison is simply a learning exercise The vertical crack control reinforcement detailed above ie fourlegged 6 stirrups will be used throughout the bent cap with the single exception of the region directly above the column In the region above the column twolegged 8 stirrups will be used to alleviate congestion and enhance constructability The required spacing of the vertical crack control reinforcement above the column is computed as follows Finally the required spacing of 8 bars provided as horizontal crack control reinforcement is computed as follows FHWANHI130126 Design Example 2 Cantilever Bent Cap StrutandTie Modeling STM for Concrete Structures 225 The required crack control reinforcement is used along the entire length of the bent cap Summary Based on the above design computations the following reinforcement will be used in the bent cap Use 4 legs of 6 stirrups with spacing less than 61 inches within the cantilevered portion of the bent cap Use 2 legs of 8 stirrups with spacing less than 55 inches above the column Use 8 bars with spacing less than 55 inches as horizontal crack control reinforcement Final reinforcement details are provided in Figures 218 219 and 220 Proportion Longitudinal Ties AB and BC Since the forces in Ties AA AB and BC are all similar a constant amount of reinforcement will be provided along the top of the bent cap and then down the tension face of the column For the longitudinal reinforcement along the top of the bent cap the force in Tie BC controls Two layers of 11 bars will be provided The reinforcement is proportioned as follows Factored load Tie capacity The number of required 11 bars is then computed as follows Therefore use 20 11 bars in two layers for the longitudinal reinforcement along the top of the bent cap Proportion Column Tie AA Similarly for the reinforcement in the column comprising Tie AA two layers of 11 bars will be provided as the main tension steel The reinforcement is proportioned as follows Factored load Tie capacity FHWANHI130126 Design Example 2 Cantilever Bent Cap StrutandTie Modeling STM for Concrete Structures 226 The number of required 11 bars is then computed as follows Therefore use 20 11 bars in two layers for the tension reinforcement in the column Requirements for Final Reinforcement Details The calculated amount of main column tension reinforcement is only satisfactory for the load case under consideration and the STM analysis that was performed The final reinforcement details for the column are dependent on the complete design that considers all governing load cases and applicable articles in AASHTO LRFD Design Step 8 Perform Nodal Strength Checks The strength of each node of the STM is now checked to ensure that it is sufficient to resist the applied forces The limiting compressive stress at the node face fcu is computed as follows AASHTO LRFD Equation 58253a1 where m confinement modification factor described later in this design example v concrete efficiency factor AASHTO LRFD Table 58253a1 fc compressive strength of concrete ksi Node E CCC Due to the limited geometry of Node E and the high forces it resists it is identified as the most critical node of the STM The geometry of Node E is detailed in Figure 214 Referring back to Figure 29 the lateral spread of Strut EE at Node E will be limited by the right face of the column The bottom bearing face of Node E and the width of Strut EE is therefore taken as twice the distance from the centroid of Strut EE to the right face of the column or 2376 inches 75 inches The length of the back face or vertical face of Node E is double the vertical distance from the center of Node E ie the point where the centroids of the struts meet to its bottom bearing face This length can be calculated as follows FHWANHI130126 Design Example 2 Cantilever Bent Cap StrutandTie Modeling STM for Concrete Structures 227 where 29 is the angle between the longitudinal axis of the cap and the horizontal ie the cross slope of the cap The other dimensions are shown in Figures 29 and 214 As illustrated in Figure 214 the length of the struttonode interface ws where Strut CE enters Node E is computed as follows The use of a computeraided design program can facilitate determination of the geometry of such a node A designer can calculate the nodal geometry as has been done in this design example or in more complicated cases the designer can draw it in a CAD platform and let the CAD program perform the descriptive geometry work 49 Strut DE 1570 k 845 k 1783 k Strut EE 75 376 376 Column Surface 283 Measured from Horizontal 79 223 245 Strut CE Bent Cap Surface 29 Figure 214 Geometry and Forces at Node E FHWANHI130126 Design Example 2 Cantilever Bent Cap StrutandTie Modeling STM for Concrete Structures 228 Node E is a CCC node with concrete efficiency factors of 085 for the bearing and back faces and 055 for the struttonode interface as specified in AASHTO LRFD Table 58253a1 see calculation below The confinement modification factor m is 1 since the column and the bent cap have the same width The faces of Node E are checked as follows Confinement modification factor Captocolumn bearing face Factored load Concrete efficiency factor Concrete capacity Back face Factored load Concrete efficiency factor Concrete capacity Struttonode interface Factored load Concrete efficiency factor Concrete capacity Although the struttonode interface does not have enough capacity to resist the applied stress according to the calculation above the percent difference between the applied force and the resistance is less than 2 percent as computed below This difference is relatively insignificant and it is reasonably safe to assume that fc will exceed the specified value by 17 only 102 psi Therefore the struttonode interface is considered to have adequate strength Therefore the strength of each face of Node E is sufficient to resist the applied forces FHWANHI130126 Design Example 2 Cantilever Bent Cap StrutandTie Modeling STM for Concrete Structures 229 Node B CCT The geometry and forces at Node B are shown in Figure 215 Its geometry is defined by the effective square bearing area previously calculated in this design example the location of the tie along the top of the bent cap and the angle of Strut BD The length of the bearing face of the node is equal to the dimension of the effective square bearing area or 482 inches As specified in AASHTO LRFD Article 58252 the length of the back face is taken as double the distance from the centroid of the longitudinal reinforcement or Tie AB to the top face of the bent cap measured perpendicular to the top face The length of the struttonode interface is computed as follows where 832 is the angle of Strut BD relative to the top surface of the cap 482 491 2017 k 832 Parallel to Bent Cap Surface 103 2005 k Strut BD Bent Cap Surface 1437 k Tie AB 1573 k Tie BC Figure 215 Geometry and Forces at Node B The strength of each bearing area at Node B ie those supporting Beam 1 and Girder 1 should be checked for adequacy The size of each bearing pad or plate is summarized in Table 21 and the factored load corresponding to each beam or girder is presented in Figure 26a Since Node B is a CCT node ie ties intersect the node in only one direction a concrete efficiency factor ν of 070 is applied to the strengths of the bearings AASHTO LRFD Table 58253a1 In the following computations the actual bearing areas are used to be conservative As previously stated the bearings sit on seats that allow the force to be spread over a larger area If the following calculations revealed that the node was insufficient to resist FHWANHI130126 Design Example 2 Cantilever Bent Cap StrutandTie Modeling STM for Concrete Structures 230 the applied forces the confinement modification factor m could be applied to the concrete capacity taking into account the larger spread areas Bearing for Beam 1 Bearing area Factored load Concrete efficiency factor Concrete capacity Bearing for Girder 1 Bearing area Factored load Concrete efficiency factor Concrete capacity The tie forces at Node B result from the anchorage of the reinforcing bars and do not concentrate at the back face In cases where the back face does not resist a direct force no back face check is necessary The strength of the struttonode interface of Node B is checked as shown below As explained with previous computations the use of the confinement modification factor m can be used as needed It is included here for illustrative purposes A2 is taken as the width of the cap beam 96 inches see Figure 21 and A1 is taken as the dimension of the effective square bearing area 482 inches see Design Step 3 As specified in AASHTO LRFD Article 58253a the confinement modification factor m is computed as follows Struttonode interface Factored load Concrete efficiency factor FHWANHI130126 Design Example 2 Cantilever Bent Cap StrutandTie Modeling STM for Concrete Structures 231 Concrete capacity Therefore the strength of Node B is sufficient to resist the applied forces Node C CCT Node C is shown in Figure 216 The geometry of the node is determined in a manner similar to that of Node B The length of the bearing face of the node 391 inches was previously calculated in this design example The following set of checks is similar to that performed for Node B since both nodes are CCT nodes The length of the strutto node interface is computed as follows where 312 is the angle of Strut CE relative to the top surface of the cap 291 103 926 k 391 312 Parallel to Bent Cap Surface Bent Cap Surface Strut CE 1783 k 1573 k Tie BC Figure 216 Geometry and Forces at Node C Bearing for Beam 2 Bearing area Factored load FHWANHI130126 Design Example 2 Cantilever Bent Cap StrutandTie Modeling STM for Concrete Structures 232 The bearing check for Beam 2 is the same as that for Beam 1 and therefore satisfies the requirements Bearing for Girder 2 Bearing area Factored load Concrete efficiency factor Concrete capacity Confinement modification factor AASHTO LRFD Article 58253a Struttonode interface Factored load Concrete efficiency factor Concrete capacity Therefore the strength of Node C is sufficient to resist the applied forces Node A CTT CurvedBar Node Node A is a curvedbar node located in the top left corner of the cantilever bent cap As specified in AASHTO LRFD Article C5822 curvedbar node checks are not required Therefore a check of Node A is not presented in this design example Node D CCC Node D is an interior node with no bearing plate or geometrical boundaries to clearly define its geometry It is therefore a smeared node As specified in AASHTO LRFD Article C5822 a check of concrete stresses in smeared nodes is unnecessary Therefore a check of Node D is not presented in this design example FHWANHI130126 Design Example 2 Cantilever Bent Cap StrutandTie Modeling STM for Concrete Structures 233 Design Step 9 Proportion Crack Control Reinforcement Crack control reinforcement design provisions are presented in AASHTO LRFD Article 5826 Since crack control reinforcement was compared with proportioning of the ties it was computed as part of Design Step 7 Based on those computations the following reinforcement will be used in the bent cap Use 4 legs of 6 stirrups with spacing less than 61 inches within the cantilevered portion of the bent cap Use 2 legs of 8 stirrups with spacing less than 55 inches above the column Use 8 bars with spacing less than 55 inches as horizontal crack control reinforcement Final reinforcement details are provided in Figures 218 219 and 220 Design Step 10 Provide Necessary Anchorage for Ties Anchorage at Node C As specified in AASHTO LRFD Articles 58242 and C58242 the primary longitudinal reinforcement of the cantilever must be properly developed at Node C The available length for the development of the tie bars is measured from the point where the centroid of the reinforcement enters the extended nodal zone assuming the diagonal strut is prismatic to the tip of the cantilever leaving the required clear cover as illustrated in Figure 217 Available Length 652 391 Assume Prismatic Strut Extended Nodal Zone Nodal Zone 312 Node C 2 min 52 Critical Section Figure 217 Anchorage of Longitudinal Bars at Node C FHWANHI130126 Design Example 2 Cantilever Bent Cap StrutandTie Modeling STM for Concrete Structures 234 Providing 2 inches of clear cover the available length for the primary longitudinal reinforcement of the cantilever measured at the centroid of the bars is computed as follows Each of the dimensional values in the above calculation is shown in Figure 217 As specified in AASHTO LRFD Article 510821a the straight development length is then computed as follows The 13 factor in the above equation is a modification factor to account for more than 12 inches of fresh concrete cast below the reinforcement AASHTO LRFD Article 510821b In addition AASHTO LRFD Article 510821c presents modification factors that decrease the development length Based on these computations sufficient straightbar anchorage is not provided for the tie at Node C and hooks are used to provide sufficient anchorage Design requirements for standard hooks in tension are provided in AASHTO LRFD Article 510824 and are not presented here Splice between Cantilever and Column Reinforcement In addition to ensuring adequate anchorage of the tie bars a splice is designed between the primary longitudinal reinforcement of the cantilever and the main column tension reinforcement All 20 longitudinal reinforcing bars will be spliced and the ratio of the area of the steel provided to the area required is less than 2 The splice is therefore a Class B splice with a required length of 13ld AASHTO LRFD Article 510843a calculated as follows The required splice length must be provided within the depth of the cap and the top portion of the column Design Step 11 Draw Reinforcement Layout The reinforcement details for the load case considered in this design example are presented in Figures 218 219 and 220 Any reinforcement details shown in these FHWANHI130126 Design Example 2 Cantilever Bent Cap StrutandTie Modeling STM for Concrete Structures 235 figures that were not previously described within this design example can be adjusted based on the specific state or agency policies and practices 850 025 A A B B 22 Equal Spaces at 55 Max 1000 No 8 Stirrups 2 Legs 37 Equal Spaces at 6 Max 1817 No 6 Stirrups 4 Legs Figure 218 Elevation View of Reinforcement Details Based on STM Specifications SECTION AA No 6 Stirrups at 6 Max oc 86 80 11 No 8 Bars 20 No 11 Bars 17 Eq Spa No 8 Bars s 5 Max Figure 219 Section AA Showing Reinforcement Details Based on STM Specifications FHWANHI130126 Design Example 2 Cantilever Bent Cap StrutandTie Modeling STM for Concrete Structures 236 SECTION BB No 8 Stirrups at 55 Max oc 86 80 11 No 8 Bars 20 No 11 Bars 17 Eq Spa No 8 Bars s 5 Max Cap Bars No 11 Bars Column Bars No 11 Bars Figure 220 Section BB Showing Reinforcement Details Based on STM Specifications References 1 Birrcher DB Tuchscherer RG Huizinga MR Bayrak O Wood SL and Jirsa JO Strength and Serviceability Design of Reinforced Concrete Deep Beams Technical Report 052531 Center for Transportation Research Bureau of Engineering Research University of Texas at Austin April 2009 376 pp 2 Williams CS Deschenes DJ and Bayrak O StrutandTie Model Design Examples for Bridges Implementation Report 5525301 Center for Transportation Research Bureau of Engineering Research University of Texas at Austin October 2011 272 pp FHWANHI130126 Design Example 3 InvertedTee Moment Frame StrutandTie Modeling STM for Concrete Structures Straddle Bent Cap 31 Design Example 3 InvertedTee Moment Frame Straddle Bent Cap Table of Contents Page Design Step 1 Define StrutandTie Model Input 33 Design Step 2 Determine the Locations of the B and DRegions 35 Design Step 3 Define Load Cases 35 Design Step 4 Analyze Structural Components and Develop Global StrutandTie Model 38 Design Step 5 Size Structural Components Using the Shear Serviceability Check 316 Design Step 6 Develop Local StrutandTie Models 317 Design Step 7 Proportion Ties 321 Design Step 8 Proportion Ledge Reinforcement 328 Design Step 9 Perform Nodal Strength Checks 330 Design Step 10 Proportion Crack Control Reinforcement 344 Design Step 11 Provide Necessary Anchorage for Ties 345 Design Step 12 Draw Reinforcement Layout 349 FHWANHI130126 Design Example 3 InvertedTee Moment Frame StrutandTie Modeling STM for Concrete Structures Straddle Bent Cap 32 Design Example 3 presents the application of the strutandtie method STM to the analysis and design of an invertedtee bent cap beam A complete stepbystep design is presented for one of multiple load cases that must be considered in the design of a structure of this type The invertedtee bent cap beam is part of a moment frame straddle bent which will carry a flyover ramp over a highway below This design example requires the use of global and local strutandtie models to fully model the flow of forces within the cap beam The example features the elements of strutandtie design of concrete members listed below Design Step 1 Define StrutandTie Model Input Design Step 3 Define Load Cases Design Step 4 Analyze Structural Components and Develop Global StrutandTie Model Design Step 5 Size Structural Components Using the Shear Serviceability Check Design Step 6 Develop Local StrutandTie Models Design Step 7 Proportion Ties Design Step 9 Perform Nodal Strength Checks Design Step 10 Proportion Crack Control Reinforcement Design Step 11 Provide Necessary Anchorage for Ties Design Step 2 Determine the Locations of the B and DRegions Design Step 12 Draw Reinforcement Layout Design Step 8 Proportion Ledge Reinforcement FHWANHI130126 Design Example 3 InvertedTee Moment Frame StrutandTie Modeling STM for Concrete Structures Straddle Bent Cap 33 Please note that this example is based on an example problem developed in an implementation project sponsored by TxDOT Report No 5525301 Williams et al 2011 Figures included in this design example are adapted from this report The example has been revised herein to provide additional explanations and to provide compliance with the STM provisions of the 8th edition of AASHTO LRFD as appropriate Design Step 1 Define StrutandTie Model Input Elevation and plan views of the moment frame straddle bent are shown in Figure 32 The straddle bent supports three trapezoidal box beams of a flyover ramp using neoprene bearing pads which rest on the ledges of the invertedtee straddle bent cap beam The bent cap is 4750 ft long A depth of 600 ft is used for this design example The stem of the cap beam is 334 ft wide with 133 ft wide ledges projecting from each side resulting in a total beam width of 600 ft at the ledges The cap beam is supported by two 500 ft by 300 ft rectangular columns The cap beam has a crossslope to accommodate the superelevation of the curved flyover ramp In this example the crossslope is deemed insignificant to the design of the beam therefore a simplified orthogonal model will be used for design Designing the cap beam as sloped or orthogonal may be valid dependent on the crossslope of the beam therefore the designer should use hisher discretion to decide which approach is most appropriate 071 ft Typ 233 ft 334 ft 600 ft 133 ft Typ 600 ft Bearing Pad Typ 334 ft 600 ft CAP BEAM SECTION CAP BEAM SECTION AT BEAM LEDGE Figure 31 Typical Sections of Straddle Bent Cap Beam FHWANHI130126 Design Example 3 InvertedTee Moment Frame StrutandTie Modeling STM for Concrete Structures Straddle Bent Cap 34 600 ft 4750 ft 4200 ft 275 ft 275 ft 916 ft 842 ft 842 ft 2150 ft C Column A L C Column B L C Beam 3 L C Beam 2 L C Beam 1 L 133 ft 258 ft 133 ft 258 ft C Bent Cap Beam L C Bearings L 32 in x 8 in Bearing Pad Typ PLAN ELEVATION 600 ft 133 ft 133 ft 167 ft 167 ft Elev 8141 CJ Elev 7635 CJ Elev 5633 CJ Elev 5533 CJ Elev 7547 Elev 8041 500 ft x 300 ft Column Typ 233 ft 50 ft Typ CJ Construction Joint Figure 32 InvertedTee Beam Straddle Bent for Design Example 3 FHWANHI130126 Design Example 3 InvertedTee Moment Frame StrutandTie Modeling STM for Concrete Structures Straddle Bent Cap 35 The design compressive strength of the concrete fc is taken as 60 ksi and the yield strength of the steel reinforcing fy is taken as 600 ksi Six trapezoidal box beams each rest on a rectangular 340 in by 80 in neoprene bearing pad on the cap beam ledges The bearing stresses from the bearing pads may be assumed to spread laterally through the bearing seats however for simplicity this effect will be ignored in this design example Design Step 2 Determine the Locations of the B and DRegions The subject straddle bent cap beam in this design example contains numerous disturbances including the superstructure loads the beam ledges and the corners of the moment frame itself Referring to Figure 33 although a small region of the cap beam may be considered a BRegion the entire cap beam is more appropriately designed using the STM procedures 600 ft 500 ft 500 ft 600 ft 600 ft 1234 ft 034 ft BRegions BRegion DRegion DRegion Column A Column B Figure 33 Bent Divided into B and DRegions Design Step 3 Define Load Cases Factored beam loads applied to the cap beam are shown in Figure 34 The loads are applied symmetrically about the centerline of the cap beam This load case is only one load case that must be considered during final design Additional load cases should be evaluated to determine if other load cases govern the design of certain strutandtie members FHWANHI130126 Design Example 3 InvertedTee Moment Frame StrutandTie Modeling STM for Concrete Structures Straddle Bent Cap 36 C Column A L C Column B Beam Line 1 C Bent Cap Beam L L Beam Line 2 Beam Line 3 2485 kips 2485 kips 2090 kips 2164 kips 2090 kips 2164 kips Total Factored Loads Per Beam Line 4970 kips 4181 kips 4328 kips Figure 34 Factored Beam Loads per Beam Line In addition to the factored beam loads the factored selfweight load of the beam cap must be determined The selfweight loads are determined using a unit weight of concrete of 150 lbft3 A load factor of 125 is applied to the selfweight in accordance with the AASHTO LRFD Strength I load combination Since the selfweight of the cap beam must eventually be distributed to each of the nodes in the STM truss model the magnitude of each of the selfweight nodal loads is dependent on the actual geometry of the STM A diagram of the selfweight loads acting on the bent cap beam is given in Figure 35 on the following page The uniform dead loads of the basic rectangular beam and the beam ledge will be resolved into point loads for application to the strutandtie truss FHWANHI130126 Design Example 3 InvertedTee Moment Frame StrutandTie Modeling STM for Concrete Structures Straddle Bent Cap 37 Beam Reaction Typ Uniform Dead Load of Beam Ledge Uniform Dead Load of Basic Rectangular Beam Figure 35 Loads Acting on the Global StrutandTie Model FHWANHI130126 Design Example 3 InvertedTee Moment Frame StrutandTie Modeling STM for Concrete Structures Straddle Bent Cap 38 Design Step 4 Analyze Structural Components and Develop Global StrutandTie Model Design of the straddle bent cap beam requires the use of global and local strutandtie models to effectively model the flow of forces within the cap beam The global model is used to model the flow of forces from the beam bearing pads to the supporting columns The local strutandtie models illustrate the flow of forces around the cross section of the cap beam which are used to design the beam ledges The global and local models together form a threedimensional strutandtie model of the cap beam Before determining the geometry of the global strutandtie model an analysis of the moment frame bent itself must be performed to determine the structure reactions to the externallyapplied loads Each frame member in this analysis is located at the center of gravity of its respective crosssection A constant flexural stiffness is assumed for the cap beam and both columns are modeled as 5 ft by 3 ft rectangular sections Modeling the Cap Beam With the widespread availability of finite element analysis software today it is relatively easy to model the cap beam using the actual sections of the cap beam Correctly modeling the cap beam stiffness will improve the strutandtie model and is recommended The simplified model of the cap beam used in this example allows the example moment frame analysis to be performed by hand as a check The internal forces in the moment frame determined in Figure 36 are used to estimate the locations of the struts and ties in the bent columns The uniform cap beam load and uniform beam ledge load are resolved into point loads and applied to the moment frame The loads are applied where the estimated STM truss nodes will be located Single Analysis vs Multiple Analyses The approach shown in this design example uses a single analysis of the moment frame assuming the locations and magnitudes of the cap beam selfweight loads and analysis of the moment frame are performed in a single step This approach gives reasonable estimates of the cap beam forces The moment frame analysis may be improved by performing the analysis in multiple steps First only the external beam loads are applied to the moment frame and the frame reactions and internal forces are determined The locations of the vertical struts and ties in the frame columns are then located based on the results of this analysis The global strut andtie model of the straddle bent cap beam is then defined The factored beam self weights are then calculated by tributary volumes and distributed to each of the nodes in the STM truss The moment frame analysis is then performed again to eliminate discrepancies between the moment frame internal forces and the member forces in the STM truss The strut and tie locations may then be adjusted and the truss reanalyzed Iterating in this manner will improve the results of the analysis FHWANHI130126 Design Example 3 InvertedTee Moment Frame StrutandTie Modeling STM for Concrete Structures Straddle Bent Cap 39 4200 ft 842 ft 842 ft 499 ft 5377 kips 4595 kips 4658 kips 2302 ft 2314 ft 2689 kips 9730 kips 5806 kips 21653 kipft 19069 kipft 150 ft 862 ft 862 ft 142 ft 260 kips 322 kips 324 kips 800 ft Location of the DRegion BRegion Interface 2689 kips 2689 kips 9730 kips 5806 kips 21653 19069 Location of the DRegion BRegion Interface 2689 kips Values in brackets are frame moments in kipfeet 40239 40239 31531 15739 60270 58444 17928 18731 29329 43146 43146 21637 MOMENT DIAGRAM MOMENT FRAME ANALYSIS Figure 36 Moment Frame Analysis and Resulting Moment Diagram FHWANHI130126 Design Example 3 InvertedTee Moment Frame StrutandTie Modeling STM for Concrete Structures Straddle Bent Cap 310 The locations of the column struts are based on the results of the moment frame analysis in Figure 36 The distribution of stress within the column is assumed to be linear at a point equal to one member depth below the bottom of the cap beam ie at the BRegionDRegion interface The internal forces at the interface are shown below in Figure 37 9730 kips 5806 kips 18731 kipft 21637 kipft 2689 kips 2689 kips Figure 37 Moment Frame Forces at BRegionDRegion Interface Next the stress distributions at the interface must be determined in order to locate the column struts and ties These stress distributions will be assumed to be linear note that in reality the stress will probably not be linear because of cracking of the concrete The area and moment of inertia of the columns are first calculated Next determine the stresses at each face of each column Compressive stress will be taken as negative Beginning with the left column Column A FHWANHI130126 Design Example 3 InvertedTee Moment Frame StrutandTie Modeling STM for Concrete Structures Straddle Bent Cap 311 The location of the compression force resultant may now be obtained by using similar triangles 500 ft 077 ksi 131 ksi 315 ft 105 ft Tension Compression Figure 38 Stress Distribution in Column A By similar triangles the neutral axis is located at 315 ft from the right face of Column A The compression strut in the column will then be located at the centroid of the triangular compression area Therefore the compression strut will be placed at 105 ft from the right face of Column A Similarly determine the stresses in the right column Column B 500 ft 075 ksi 165 ksi 344 ft Tension Compression 115 ft Figure 39 Stress Distribution in Column B FHWANHI130126 Design Example 3 InvertedTee Moment Frame StrutandTie Modeling STM for Concrete Structures Straddle Bent Cap 312 By similar triangles the neutral axis is located at 344 ft from the left face of Column B The compression strut will be located at 115 ft from the left face of Column B approximately onethird of 344 ft The following discussion details the development of the global STM of the moment frame For the arrangement of the STM truss refer to Figure 310 Figure 310 also includes the individual truss member forces calculated by a truss analysis Now that the locations of the column struts have been determined the ties may be located The ties will be located at the center of gravity of the longitudinal column reinforcement This location is assumed to be 38 in 032 ft from the outer face of the columns which allows space for the column stirrupsties and clear cover Next the locations of the chords of the global truss model must be located Positive and negative moment regions will exist within the straddle bent beam requiring ties in both the top and bottom chords Therefore the truss chords will be located at the centers of gravity of the longitudinal reinforcement in the straddle bent cap beam In Figure 310 the top chord is located at 46 in 038 ft from the top of the cap beam and the bottom chord is located 60 in 050 ft from the bottom of the cap beam Note that this is a departure from the previous design examples where the depth of the compression block a of an analogous Bernoulli beam was used to locate the struts The resulting truss depth in the cap beam is 512 ft Next vertical Ties CI DJ and EK are placed at the positions of the applied superstructure loads These ties represent the reinforcement required to hang the loads applied to the ledges of the invertedtee beam and transfer the stress from the beam ledges to the components of the global strutandtie model truss Recall that the angle between a tie and an adjacent strut should not be less than 25 degrees as stipulated in AASHTO LRFD Article 5822 In order to meet this requirement Tie BH is placed halfway between Nodes G and I Note that all of the struts are oriented such that they will be in compression This is also the location of the assumed cap beam selfweight applied in the moment frame analysis Continuing the total factored loads from each beam line are applied to bottom chord Nodes I J and K The factored selfweight loads from the moment frame analysis are distributed to each of the nodes in the STM truss except Nodes A and F Because of their location at the upper corners of the truss applying any selfweight at these nodes would be unreasonable To account for the shears in the frame at the DRegionBRegion interfaces Ties AG and LF are added at the base of the STM truss to accept the horizontal shear loads and Struts AG and FL are added to anchor Ties AG and LF FHWANHI130126 Design Example 3 InvertedTee Moment Frame StrutandTie Modeling STM for Concrete Structures Straddle Bent Cap 313 Subdividing Nodes and Struts Keep in mind that nodes with multiple struts entering the nodal zone ie Nodes C G and L must be subdivided per AASHTO LRFD Commentary C58224 Subdivision of the nodes will addressed during the node design checks Additionally recall that in Design Example 2 Cantilever Bent Cap two vertical struts were used to carry the compressive force in the bent column In this design example the column struts will similarly be subdivided Using only one strut in each column simplifies the truss analysis However an STM truss with 2 struts in each column may be solved reasonably quickly using a structural analysis software package The column reactions that are applied to the strutandtie model at the BRegionD Region interface are found using the results of the moment frame analysis in Figure 37 This is done such that the forces in Ties AA and FF and Struts GG and LL are in equilibrium with the internal forces in each column The bending moments at the D RegionBRegion interface are found to be 18731 kipft and 21637 kipft in Columns A and B respectively The axial compressions in Columns A and B are found to be 5806 kips and 9730 kips respectively In order to determine the strut and tie forces in each column two systems of two simultaneous equations are solved for the strut and tie forces The first equation in each system ensures static equilibrium with respect to the axial force in each column The second equation in each system equates the moments about the centerline of each column to the column bending moment from the moment frame analysis Taking forces that are vertical up as positive the system of simultaneous equations for Column A the left column is Note that the distances 218 ft and 145 ft are the distances from the tie and strut to the centerline of the column respectively Solving Similarly the system of simultaneous equations for Column B the right column is Solving FHWANHI130126 Design Example 3 InvertedTee Moment Frame StrutandTie Modeling STM for Concrete Structures Straddle Bent Cap 314 Now that the forces in each of the column struts and ties are known the remaining forces in the strutandtie model may be found using simple statics or by using a structural analysis software package Solving the STM Truss The strutandtie model truss may be solved manually by applying the equations of statics using either the method of joints or the method of sections If the truss will be solved using a structural analysis software the forces in the column struts and ties should be imposed on those members for the analysis This forces equilibrium between the moment frame analysis beam theory results and the assumed truss shape FHWANHI130126 Design Example 3 InvertedTee Moment Frame StrutandTie Modeling STM for Concrete Structures Straddle Bent Cap 315 057 ft 057 ft 364 ft 862 ft 862 ft 842 ft 842 ft 505 ft 354 ft G A A B H C I J K L F D E 162 kips 5174 kips 4388 kips 4493 kips 322 kips 162 kips 204 kips 207 kips 165 kips 260 kips G 038 ft 050 ft 032 ft 105 ft 6539 8647 4643 4639 12970 5056 4289 1950 10639 10639 10281 2367 5323 5174 4605 9305 Values shown without units are element forces in kips indicates tension indicates compression 2689 kips 2841 kips 8647 kips 2689 L 115 ft 032 ft 6208 12138 12138 kips 2408 kips 2689 F 2689 kips Figure 310 Global StrutandTie Model for the InvertedTee Cap Beam FHWANHI130126 Design Example 3 InvertedTee Moment Frame StrutandTie Modeling STM for Concrete Structures Straddle Bent Cap 316 Design Step 5 Size Structural Components Using the Shear Serviceability Check In Design Step 1 the depth of the cap beam was chosen as 600 ft It was chosen to apply the strutandtie design method to the entire cap beam even though there is a small region within the beam that may be considered a BRegion The beam stem width must now be verified AASHTO LRFD Equation C58221 limits the applied shear to the following value Vcr with corresponding minimum and maximum values limited as follows As in previous examples this equation is used to estimate the shear at which diagonal cracks form in DRegions Where the applied service load shears are less than Vcr reasonable assurance is provided that diagonal shear cracks will not form Recall that this check is performed at the AASHTO LRFD Service I load combination After performing an elastic analysis the maximum shear force is found near the right end of the bent cap beam The Service I shear at Column B the right column is found to be Therefore the shear serviceability check will determine the risk of crack formation in the shear span between Beam Line 3 and the centerline of Strut LL The shear span a is taken as the horizontal distance between Nodes K and L which is 606 in or 505 ft Because the moment in this area of the cap beam is negative the distance to the tension reinforcement d is calculated from the bottom face of the cap beam to the centroid of the top reinforcing steel Now that all quantities are known Vcr may now be calculated First check the limits on the term This value is less than the upper limit of 0158 and greater than the lower limit of 00632 therefore use the value of 0110 FHWANHI130126 Design Example 3 InvertedTee Moment Frame StrutandTie Modeling STM for Concrete Structures Straddle Bent Cap 317 In addition it is wise to check the left end of the cap beam because of the longer shear span The shear span in this region is The maximum service load shear in this shear span is found to be First check the limits on the term Since 0047 00632 use the lower bound value of 00632 Therefore both checks are satisfactory and the designer should not expect diagonal cracking at service loads The values of Vcr should also be checked using other load cases and if the value of d varies significantly from the assumed value of 674 in Design Step 6 Develop Local StrutandTie Models Since the flow of forces in the invertedtee bent cap beam is very complex separate strutandtie models should be developed at each location where a beam load is supported by the bent cap beam ledge The strutandtie model for a section cut at Beam Line 1 is shown in Figure 311 on the next page Ties AsGs and BsHs are placed to coincide with the locations of the vertical stirrup legs also known as hanger reinforcement which serves as the transverse reinforcement in the stem of the bent cap beam In the same way Tie CsFs is located at the top horizontal leg of the stirrups provided in the beam ledges The position of Strut GsHs coincides with the location of the bottom chord of the global strutandtie model FHWANHI130126 Design Example 3 InvertedTee Moment Frame StrutandTie Modeling STM for Concrete Structures Straddle Bent Cap 318 85 in 2485 kips 85 in 2485 kips 24 in 196 in 60 in 24 in 24 in 352 in 400 in 1382 1382 2587 2587 102 kips 102 kips 102 kips 102 kips Gs Cs Ds As Bs Es Fs Hs Values shown without units are element forces in kips indicates tension indicates compression Edge of Cap Beam Figure 311 Local StrutandTie Model at Beam Line 1 Recall that the reason behind developing the local strutandtie models is to design the ledge reinforcement It is important to note that the development of the local strutand tie models is dependent only the applied the beam loads at the beam line in question the forces in the global strutandtie model have no influence on the design of the ledge reinforcement The area of reinforcement required for Tie CsFs in Figure 311 is dependent only on the applied beam selfweights and the applied beam reactions InvertedTee Beam Terminology Additional terminology is introduced here to completely describe the reinforcement used in the design of an invertedtee beam Refer to Figure 312 on the next page Hanger reinforcement or hanger ties refers to the vertical reinforcement in the beam stem within a specified distance from an applied ledge load The hanger reinforcement carries the beam load upward toward the compression face of the beam Ledge reinforcement is the horizontal reinforcement provided in the beam ledges which carries the tensile forces created by the applied ledge loads FHWANHI130126 Design Example 3 InvertedTee Moment Frame StrutandTie Modeling STM for Concrete Structures Straddle Bent Cap 319 TYPICAL SECTION OF AN INVERTEDTEE BEAM Ledge Reinforcement Hanger Reinforcement Figure 312 InvertedTee Beam Terminology Before continuing it is worth stating that the designer should keep in mind that the flow of forces within an invertedtee beam may be visualized as a single threedimensional strutandtie model Such a visualization can make it easier to determine if the chords of the truss members are placed correctly It is therefore reasonable to place Strut GsHs such that it would intersect the bottom chord of the global strutandtie model The applied loads in the local strutandtie model of Figure 311 are the applied factored beam loads 2485 kips each and the tributary selfweight of the cap beam which is evenly distributed to Nodes As Bs Gs and Hs recall that these selfweight loads are factored loads The individual member forces are then found by satisfying equilibrium at each node Local strutandtie models must also be developed at Beam Lines 2 and 3 These models are given in Figure 313 and Figure 314 Each local strutandtie model is geometrically identical but is subject to a different set of external forces By comparing each of the three local models design of the ledge reinforcement Tie CsFs and the nodal strength checks will be governed by the model at Beam Line 1 the location of the largest applied beam loads In order to simplify detailing and construction the spacing of the ledge reinforcement required at Beam Line 1 will be provided along the entire ledge All other reinforcement details will be based on the global strutandtie model FHWANHI130126 Design Example 3 InvertedTee Moment Frame StrutandTie Modeling STM for Concrete Structures Straddle Bent Cap 320 85 in 2090 kips 85 in 2090 kips 24 in 196 in 60 in 24 in 24 in 352 in 400 in 1158 1158 2194 2194 104 kips 104 kips 104 kips 104 kips Gs Cs Ds As Bs Es Fs Hs Values shown without units are element forces in kips indicates tension indicates compression Figure 313 Local StrutandTie Model at Beam Line 2 85 in 2164 kips 85 in 2164 kips 24 in 196 in 60 in 24 in 24 in 352 in 400 in 1204 1204 2247 2247 83 kips 83 kips 83 kips 83 kips Gs Cs Ds As Bs Es Fs Hs Values shown without units are element forces in kips indicates tension indicates compression Figure 314 Local StrutandTie Model at Beam Line 3 FHWANHI130126 Design Example 3 InvertedTee Moment Frame StrutandTie Modeling STM for Concrete Structures Straddle Bent Cap 321 Design Step 7 Proportion Ties The forces from the global strutandtie model will be used to determine the longitudinal reinforcement in the top and bottom chords of the cap beam as well as for the exterior faces of the columns A constant amount of longitudinal steel will be provided along the cap beam for ease of detailing and construction Bottom Chord The force in Ties HI and IJ controls the design of the bottom chord of the global strut andtie model The amount of reinforcing required is found by applying AASHTO LRFD Equations 58231 and 582411 Using No 11 reinforcing bars Therefore use 13 No 11 bars Top Chord The force in tie AB controls the design of the top chord of the global strutandtie model The reinforcing required is determined as for the bottom chord Using No 11 reinforcing bars FHWANHI130126 Design Example 3 InvertedTee Moment Frame StrutandTie Modeling STM for Concrete Structures Straddle Bent Cap 322 Therefore use 6 No 11 bars Note that this is the minimum amount of reinforcing required Additional reinforcing may be required to supplement the strength of the nodes This will be explored in Design Step 9 Column Vertical Ties For simplicity of detailing and construction identical reinforcing will be provided in both columns The amount of reinforcing required will be controlled by column tie AA Using No 11 reinforcing bars Use 8 No 11 bars in each column Recall that this reinforcing is determined only for one load case Final reinforcement details should be determined by a complete design that considers all governing load cases Hanger Reinforcement Vertical Ties The geometries of the nodes above Beam Lines 1 2 and 3 are dependent on the distribution of the vertical tie reinforcing at Nodes C D and E respectively As opposed to a strutandtie model with loads applied to its top chord a strutandtie model loaded by its bottom chord requires hanger reinforcement to transfer the applied superstructure loads to its compression chord Referring to AASHTO LRFD Figure 584352 reproduced on the next page as Figure 315 the length over which the hanger reinforcement may be distributed ie the width of hanger tie is W 2df where W width of the bearing pad measured along the length of the cap beam in df distance from the top face of the ledge to the centroid of the bottom horizontal leg of the ledge stirrups in FHWANHI130126 Design Example 3 InvertedTee Moment Frame StrutandTie Modeling STM for Concrete Structures Straddle Bent Cap 323 bf Vu Vu Ahr df Vu W 2df W Figure 315 InvertedTee Beam Hanger Reinforcement In addition the effective tie widths for the ledge reinforcement are limited per AASHTO LRFD Article 58435 The distributed width for the interior beam is taken as W 2df for interior beams and 2c for exterior beams where c distance from the centerline of bearing to end of the beam ledge in Any effects of the tapered ends of the beam ledges are conservatively neglected Hanger and Ledge Reinforcement AASHTO LRFD Article 58435 covers empirical design of hanger and ledge reinforcing The empirical guidelines given in this article are followed in this example even though they are not specifically referenced by the AASHTO LRFD strutandtie provisions The available widths for Ties CI and EK are determined first Referring to Figure 316 on the next page the available width is determined thus FHWANHI130126 Design Example 3 InvertedTee Moment Frame StrutandTie Modeling STM for Concrete Structures Straddle Bent Cap 324 df 2c 2c W 2df 620 in 620 in 8525 in df 2563 in 842 ft 842 ft c c Figure 316 Available Hanger Reinforcement Widths The distance df is determined by assuming a 20 in clear cover and a No 6 stirrup giving Now the available width of Tie DJ is found The hanger reinforcement along the ledge will be determined first then the required stirrup spacing for Tie BH will be determined Tie EK Tie EK is the most critical hanger tie in the bent cap beam because it must carry the largest tensile load with a relatively narrow band of reinforcement Because of this limitation bundled No 6 stirrups with two legs will be used Alternatively the designer has the option to use fourlegged No 6 stirrups instead The required spacing of the stirrups is found in the same manner as for the longitudinal beam ties above using AASHTO LRFD Equations 58232 and 582411 FHWANHI130126 Design Example 3 InvertedTee Moment Frame StrutandTie Modeling STM for Concrete Structures Straddle Bent Cap 325 Using the No 6 bundled stirrups Recalling that the available width for the stirrups is 620 in the required stirrup spacing is Thus use 2 bundled No 6 stirrups 4 legs total with a spacing of less than 63 in Ties CI and DJ Ties CI and DJ are proportioned next The reinforcement detailed for Tie CI will be used along the entire length of the beam ledge except at the Tie EK region detailed previously Here No 6 stirrups with 2 legs will be used The required reinforcing is The available length is again 620 in giving a required stirrup spacing of Thus use No 6 stirrups 2 legs total with a spacing of less than 57 in Tie BH In contrast to Nodes C D and E Nodes B and H are smeared nodes with undefined geometries Hence Tie BH is contained within a fanshaped strut which connects Nodes C and G and the reinforcement for Tie BH will be determined using the method explained in Design Example 1 Design Step 7 The shear span a is the distance between Nodes C and G FHWANHI130126 Design Example 3 InvertedTee Moment Frame StrutandTie Modeling STM for Concrete Structures Straddle Bent Cap 326 Referring back to Figure 310 the height of the global strutandtie model truss is 512 ft or 614 in The available width for the tie is then hSTM x tan25 Available Length la hSTM x tan25 Shear Span a 25 P R 25 hSTM FanShaped Strut Stirrups Comprising Tie Figure 317 Geometry of a FanShaped Strut Excerpt of AASHTO LRFD Figure C58222 However in reality this available length la will be partially occupied by the reinforcement for Tie CI Therefore the reinforcement must be spread over a smaller distance chosen to be equal to the average spacing between Nodes G H and I which is 862 ft or 1035 in Using No 6 stirrups with 2 legs the stirrups will be centered on the tie and spaced evenly over the available length Applying AASHTO LRFD Equations 58231 and 582411 FHWANHI130126 Design Example 3 InvertedTee Moment Frame StrutandTie Modeling STM for Concrete Structures Straddle Bent Cap 327 Thus use No 6 stirrups 2 legs total with a spacing of less than 92 in It will be demonstrated that the minimum crack control reinforcement required will ultimately control the detailing in this region of bent cap beam Crack control reinforcement is detailed in Design Step 10 Ties AG and LF Ties AG and LF distribute the shear that results from the moment frame analysis into the strutandtie model of the cap beam These ties are required because the presence of these ties influences the forces in the global strutandtie model These ties are also anchored by smeared nodes therefore they will be proportioned in the same manner as Tie BH The available shear span a is taken as the distance between Nodes G and G or 600 in The height of the STM in these locations will be taken as the smaller of the distances between Nodes A and G and Nodes L and F This is found to be 354 ft or 425 in between Nodes L and F Thus This reinforcement will be provided on either side of Ties AG and LF as No 6 closed column ties The column ties will be centered on the each tie and spaced evenly over the available length Applying AASHTO LRFD Equations 58231 and 582411 Thus use No 6 closed column ties 2 legs total with a spacing of less than 36 in FHWANHI130126 Design Example 3 InvertedTee Moment Frame StrutandTie Modeling STM for Concrete Structures Straddle Bent Cap 328 Design Step 8 Proportion Ledge Reinforcement As was stated earlier Tie CsFs of the local strutandtie model at Beam Line 1 refer to Figure 311 was found to control the design of the ledge reinforcement According to AASHTO LRFD Article 58433 the reinforcement encompassing this tie should be uniformly spaced over a distance of W 5af or 2c whichever is less where af distance from centerline of girder reaction to vertical reinforcement in backwall or stem of inverted tee in subject to the limitation that the widths of these regions shall not overlap The distribution of this reinforcement is given in AASHTO LRFD Figure 584331 reproduced below as Figure 318 The limit of W 5af may be applied to the interior beam and the limit of 2c may be applied for the exterior beams Vu As de Vu W 5af W h af 2c s c Figure 318 Flexural Reinforcement for Ledges AASHTO LRFD Figure 584331 ThreeDimensional Judgement Consider again the threedimensional flow of forces within the invertedtee bent cap beam The ledge reinforcement and hanger reinforcement must work together to carry the applied beam forces around the cross section of the cap beam In this design example recall that the available width of Tie CI is 620 in The job of Tie CI is to hang up the ledge reinforcement which is represented by Tie CsFs in the local STM at Beam Line 1 Therefore instead of applying the provisions of AASHTO LRFD Article 58433 the width of Tie CsFs will be limited to the width of Tie CI or 620 in In this design example the width of Tie CI just so happens to match the result obtained by applying the 2c limitation for an exterior beam FHWANHI130126 Design Example 3 InvertedTee Moment Frame StrutandTie Modeling STM for Concrete Structures Straddle Bent Cap 329 Since the force in Tie CsFs of the local strutandtie model at Beam Line 1 controls and the available width of Tie CsFs is smaller at the exterior beams than the interior beam using the spacing determined using the forces and available width at Tie CsFs to determine the required reinforcement will result in a conservative design for the entire beam ledge Assuming No 6 reinforcing bars will be used for the ledge and applying AASHTO LRFD Equations 58231 and 582411 Thus use No 6 bars with a spacing of less than 107 in Providing stirrups within the ledge satisfies this requirement see Figure 319 on the following page To simplify construction each of the ledge stirrups will be paired with the stirrups in the cap beam stem Since the required stirrup spacings for the cap beam are all less than that required for the ledge reinforcing ie they are all less than 107 in pairing the stirrups this way ensures sufficient reinforcement is provided over the entire ledge length FHWANHI130126 Design Example 3 InvertedTee Moment Frame StrutandTie Modeling STM for Concrete Structures Straddle Bent Cap 330 Top portion of stirrups carries force in Tie CsFs Figure 319 Stirrups Carrying Tie Force CsFs Design Step 9 Perform Nodal Strength Checks Figure 320 is a representation of how the struts and nodes fit within the global strut andtie model of the invertedtee beam An arbitrary size is given to smeared Nodes B and H as they are only drawn for illustrative purposes The nodes with multiple intersecting struts may be resolved to simplify the nodal geometries G A A B H C I J K L L F F D E G Figure 320 Struts and Nodes within the InvertedTee Cap Beam Within Design Step 9 the nodes of the global strutandtie model will be evaluated first The most critical nodes will be identified and the corresponding strength checks will be carried out Some of the remaining nodes may be deemed to have adequate strength by inspection Nodes A and F are curved bar nodes which will be explained later in this design step The nodes in the local strutandtie model at Beam Line 1 will then be evaluated FHWANHI130126 Design Example 3 InvertedTee Moment Frame StrutandTie Modeling STM for Concrete Structures Straddle Bent Cap 331 Check Node G CCCCCT Nodes G and L are located near the inside faces of the frame corners Due to the tight geometric constraints and large forces acting on the nodes these nodes are among the most highlystressed regions in the bent cap beam The geometry of Node G is given in Figure 321 The total width of the bearing bottom face of the node is taken as twice the distance from the face of Column A to the centerline of Strut GG 210 ft or 252 in The height of the back face of the node is taken as twice the distance from the bottom surface of the bent cap beam to the centroid of the bottom chord reinforcement or 120 in Because multiple struts intersect at Node G the node will be subdivided and struts will be resolved to ensure that no more than three forces intersect at a single node The two diagonal Struts AG and AG will be resolved into a single strut called Strut AAG The force acting on the bearing face of the left portion of the node equilibrates the vertical component of the strut acting on the left node face Strut AAG and a portion of the applied selfweight Equilibrium is satisfied for the right portion of the node in the same manner Note that the inclinations of the struts must also be revised to account for the subdivision of the node The new inclination angles are shown with the original inclination angles in Figure 321 252 in 859 in 1661 in 60 in 120 in Strut AAG Resolved Struts AG and AG Strut GG Right Face of Column A Bent Cap Beam Face 110 k 212 k SelfWeight 2948 k 5699 k 1950 k 2114 3179 3069 Per Global STM Figure 321 Geometry of Node G The dimension of the bearing face for each nodal subdivision is determined based on the magnitude of the vertical component of each diagonal strut in relation to the net vertical force from Strut GG and the applied selfweight at the node Uniform bearing pressure will be maintained over the total 252 in width of Strut GG to maintain equilibrium FHWANHI130126 Design Example 3 InvertedTee Moment Frame StrutandTie Modeling STM for Concrete Structures Straddle Bent Cap 332 First resolve Struts AG and AG into a single strut Strut AAG The angle of inclination of the resolved strut is found next The length of each bearing face is determined thus where 8647 kips is the force in Strut GG 322 kips is the selfweight applied at Node G 7872 kips is the force in Strut AAG 2114 is the angle of inclination of Strut AAG The revised angle of inclination of Strut BG is now calculated where 6144 in is the height of the strutandtie model hstm 4416 in is the horizontal distance from Node G to Tie AA 10344 in is the horizontal distance from Node G to Node H 1661 in is the subdivided bearing face width for Strut BG The widths of the struttonode interfaces ws are now calculated FHWANHI130126 Design Example 3 InvertedTee Moment Frame StrutandTie Modeling STM for Concrete Structures Straddle Bent Cap 333 Only compression forces act on the left portion of Node G while one tensile force acts on the right half of Node G Therefore the left portion will be treated as a CCC node and the right half will be treated as a CCT node Node GRight CCT Because the bent cap beam is wider than both columns the confinement modification factor m may be applied to the strength of Node GRight The value of m is determined by applying AASHTO LRFD Equation 5653 where the values of A1 and A2 are determined using Figure 322 Since the calculated value of m is less than 2 use the calculated value of 113 400 in Cap Beam Node G 360 in Column 600 in Column 252 in 292 in 252 in x 360 in Bearing Face A1 Figure 322 Confinement Modification Factor m at Node G FHWANHI130126 Design Example 3 InvertedTee Moment Frame StrutandTie Modeling STM for Concrete Structures Straddle Bent Cap 334 Applying the Confinement Modification Factor Experimental results by Bayrak et al and evaluation of existing deep beam tests suggest that the benefits of the triaxial confinement effect are applicable to all faces of a node due to the confining effect of the concrete surrounding the node In this design example it is deemed appropriate to apply the confinement modification factor to all of the faces of Node GRight because the adjoining Node GLeft provides restraint to the back face of the node the column provides restraint to the bearing face of the node Please note that not taking advantage of the triaxial confinement effect would result in a more conservative design by reducing the allowable concrete stresses at the node faces It is up to the designer to determine if using the confinement modification factor is appropriate Please see the references for additional information Each face of the subdivided node is now checked using the nodal strength checks of AASHTO LRFD Article 5825 Begin by applying AASHTO LRFD Equations 58231 and 582511 The concrete efficiency factors v may be found in AASHTO LRFD Table 58253a1 For the bearing face For the back face For the struttonode interface FHWANHI130126 Design Example 3 InvertedTee Moment Frame StrutandTie Modeling STM for Concrete Structures Straddle Bent Cap 335 Node GLeft CCC The pressures acting on the back face of the left portion of Node G are the same as the right portion due to equilibrium The pressure acting on the bearing face of Node GLeft is the same as the right portion recall that Strut GG was divided by maintaining equal pressures in each portion of the subdivided strut Therefore only the struttonode interface of Node GLeft need be checked For the struttonode interface Hence Node G is sufficient to resist the applied forces Node L CCCCCT For Node L the geometry and subdivision of forces is carried out in exactly the same way as for Node G Calculations show that all of the nodal faces at Node L have adequate strength to resist the applied forces FHWANHI130126 Design Example 3 InvertedTee Moment Frame StrutandTie Modeling STM for Concrete Structures Straddle Bent Cap 336 Node C CCT The nodal strength checks for Node C are performed next The diagonal Strut CH entering the node is highly stressed and large compressive forces act on a relatively small area on the back face of Node C This node is therefore a critical node to check Because struts enter the node on its left and right faces this node is subdivided into two parts see Figure 323 The total length of the top nodal face is assumed to be the same as the width of the corresponding hanger tie Tie CI The width of the top face is consequently taken as 517 ft or 620 in The height of the back face is taken as double the distance from the top of the bent cap beam to the centroid of the top chord reinforcement 92 in 620 in 92 in Top Face of Cap Beam 196 k 08 k SelfWeight 2141 2116 095 093 Per Global STM 595 in 25 in 4965 k 209 k 924 in Figure 323 Geometry of Node C Here the length of the top face for each nodal subdivision is based on the magnitude of the vertical component of each diagonal strut in relation to the net vertical force from Tie CI and the applied selfweight at Node C this approach is exactly the same as was done for Node G The length of each nodal top face is where 10105 kips is the force in Strut CH 204 kips is the selfweight applied at Node C 5174 kips is the force in Tie CI 3071 is the angle of inclination of Strut CH FHWANHI130126 Design Example 3 InvertedTee Moment Frame StrutandTie Modeling STM for Concrete Structures Straddle Bent Cap 337 Note the difference in size between the left and right portion of Node C Before revising the diagonal strut angles to reflect the subdivided node the struts adjacent to Node C are resolved to reduce the number of forces acting on the node Struts BC and CH and Struts CD and CJ are resolved into two struts acting on the left and right faces of Node C respectively The resolved geometry is given in Figure 323 The resolved strut acting on the left face of Node C is determined thus where 4639 kips is the horizontal component of the force in Strut BC 8684 kips is the horizontal component of the force in Strut CH 5158 kips is the vertical component of the force in Strut CH 0 kips is the vertical component of the force in Strut BC The angle of inclination of the resolved strut is Similarly the resolved strut force and inclination angle are determined for the right face of Node C Subsequently the resolved strut inclination angles are revised to reflect the subdivided nodal geometry The angle of inclination for the resolved strut on the left of Node C is calculated thus and the angle of inclination for the resolved strut on the right of Node C is FHWANHI130126 Design Example 3 InvertedTee Moment Frame StrutandTie Modeling STM for Concrete Structures Straddle Bent Cap 338 Finally the widths of the new struttonode interfaces are calculated Node CLeft CCT Node C has no bearing surface therefore no bearing check is necessary Longitudinal reinforcement is provided along the entire length of the bent cap beam So long as this reinforcement is detailed to develop its yield stress in compression the longitudinal reinforcing will contribute to the strength of the back face of Node C Recall that the top chord reinforcing determined in Design Step 7 was 6 No 11 reinforcing bars Using this amount of reinforcing the back face of the node may be checked using AASHTO LRFD Equations 58231 and 582511 modified by including the strength of the reinforcing steel and subtracting the area of reinforcing steel from the area of concrete Alternatively AASHTO LRFD Equation 56443 may be used For the back face FHWANHI130126 Design Example 3 InvertedTee Moment Frame StrutandTie Modeling STM for Concrete Structures Straddle Bent Cap 339 For the struttonode interface Node CRight CCT Checks of the right portion of Node C are carried out in the same manner as for the left portion The check for the back face of the node is identical to the check for the left portion of Node C so it will not be reproduced here For the struttonode interface The struttonode interface calculations indicate that the node face does not have the strength required to resist the resolved strut force However by examining the resolved strut its angle of inclination is practically zero and may essentially be neglected making the check at this node face the same as the back face check for the left portion of Node C Therefore SAY OKAY FHWANHI130126 Design Example 3 InvertedTee Moment Frame StrutandTie Modeling STM for Concrete Structures Straddle Bent Cap 340 Node I Ties Only Node I is located directly below Beam Line 1 Referring to the global strutandtie model in Figure 310 only ties intersect at this node Nodal checks are therefore unnecessary because no compressive force act on the node However the strength of the bearings at Beam Line 1 must be checked to ensure adequate strength against bearing failure These checks should be performed as part of the local strutandtie model evaluation Node K CTT Node K directly below Beam Line 3 is shown in Figure 324 The length of the bottom face of the node is conservatively chosen to the width of the bearing pad W Alternatively the designer may wish to reduce the nodal stresses by accounting for the lateral distribution of the applied beam load through the depth of the ledge by considering this distribution the length of the bottom face of the node would increase This approach is not necessary to satisfy the nodal strength check for this design example 340 in 120 in Bent Cap Beam Face 9305 k 3130 4493 k Tie KL 2367 k Tie JK 10281 k Bearing Pad Top of Beam Ledge Figure 324 Geometry of Node K In spite of the presence of a bearing pad on the ledge a bearing force does not act directly on the node hence the confinement modification factor cannot be applied at Node K In addition recall that the nodes of the global strutandtie model are assumed to be confined within the stem of the invertedtee beam and not the ledges Note that a FHWANHI130126 Design Example 3 InvertedTee Moment Frame StrutandTie Modeling STM for Concrete Structures Straddle Bent Cap 341 width of 400 in is used for the width of the struttonode interface design check below The distance lb will be restricted to the width of the bearing pad 340 in The width of the struttonode interface is first determined For the struttonode interface The Back Face of the Node Recall that AASHTO LRFD Article C58253b states that there is no research that showed bond stress of reinforcement controlling the strength of a nodal region Therefore it is not necessary to check the strength of the back face of Node K Hence Node K is sufficient to resist the applied forces Nodes A and F CTT Nodes A and F are referred to as curvedbar nodes Curvedbar nodes are not included in the AASHTO LRFD specifications and because these nodes are not highly stressed these nodes will not be checked CurvedBar Nodes A curvedbar node results when a largediameter reinforcing bar ie a No 11 or No 18 bar is bent around a corner The geometry of a curvedbar node is shown in Figure 325 on the following page This type of node is not yet included in the AASHTO LRFD specifications because it has not yet been vetted by as much experimental data as for CCC CCT and CTT nodes For additional information refer to TxDOT research report 5 525301 Williams et al 2011 FHWANHI130126 Design Example 3 InvertedTee Moment Frame StrutandTie Modeling STM for Concrete Structures Straddle Bent Cap 342 Assumed Tie Location Strut or Strut Resultant Circumferential Bond Stress Radial Compressive Stress θc θc 45 Ast fy Ast fy tan θc Figure 325 Geometry of a CurvedBar Node Nodes Cs and Fs Local STM CCT Nodes Cs and Fs of the local strutandtie model at Beam Line 1 Figure 311 are the most critical nodes of the three local strutandtie models Since the nodes are identical the local strutandtie model is symmetrical about the cap beam centerline only one needs to be checked The geometry of Node Cs is given in Figure 326 on the following page The length of the bearing face of the node is taken as the width of the bearing pad or 80 in and the height of the back face is taken as double the distance from the top surface of the ledge to the top horizontal leg of the ledge stirrup or 48 in The width of the node in and out of the page is assumed to be the length of the bearing pad or 340 in FHWANHI130126 Design Example 3 InvertedTee Moment Frame StrutandTie Modeling STM for Concrete Structures Straddle Bent Cap 343 af 105 in 6092 24851 k 1382 k 80 in 48 in 24 in Figure 326 Node Cs of Local StrutandTie Model at Beam Line 1 To simplify calculation the confinement modification factor m is conservatively taken as 10 It will be demonstrated that all of the nodal faces have sufficient strength to resist the applied loads without consideration of the effects of triaxial confinement The demand on the bearing face of the node is equal to the reaction from the trapezoidal box beam The width of the struttonode interface is first determined For the bearing face FHWANHI130126 Design Example 3 InvertedTee Moment Frame StrutandTie Modeling STM for Concrete Structures Straddle Bent Cap 344 For the struttonode interface No direct compressive force acts on the back face of the node therefore checking the back face is not necessary Hence Nodes Cs and Fs are sufficient to resist the applied forces Other Nodes Nodes D E J A and F of the global strutandtie model shown in Figure 310 may be checked using the methods presented previously All of the nodes in the global strut andtie model have sufficient strength to resist the applied forces for the studied load case Nodes B and H in the global strutandtie model are smeared nodes interior nodes with no definable geometry hence no strength checks are required Nodes Gs and Hs in the local strutandtie models are also smeared nodes and require no strength checks By inspection the struts entering these nodal regions have adequate space to spread out and are therefore deemed not to be critical Design Step 10 Proportion Crack Control Reinforcement The minimum requirements for crack control reinforcement are now checked and compared against the vertical tie reinforcement that was determined in Design Step 7 To maintain consistency in detailing No 6 stirrups will be used for the cap beam The required spacing of the crack control reinforcement is found by applying the provisions of AASHTO LRFD Article 5826 The value of bw is taken as 400 in AASHTO LRFD Equations 58261 and 58262 FHWANHI130126 Design Example 3 InvertedTee Moment Frame StrutandTie Modeling STM for Concrete Structures Straddle Bent Cap 345 Recall that the stirrup spacing specified for Tie CI at Beam Line 1 will be used for the entire length of the ledge except in the region surrounding Tie EK where bundled stirrups are used 4legged stirrups versus 2legged stirrups The minimum stirrup spacing required for strength requirements was 57 in therefore the stirrups provided in the ledge region will satisfy the crack control reinforcement requirement However the required crack control reinforcement governs in the region of Tie BH and must also be provided over the remaining length of the bent cap beam ie over the columns The required spacing of No 6 reinforcing bars provided as skin reinforcement parallel to the length of the bent cap beam is calculated next using AASHTO LRFD Equation 58262 Skin reinforcement of 2 No 6 reinforcing bars will be provided at a spacing of less than 73 in Design Step 11 Provide Necessary Anchorage for Ties The reinforcement in the top and bottom chords of the global strutandtie model must be properly anchored at each end of the bent cap beam in accordance with AASHTO LRFD Article 51082 Continuity of the reinforcement over the length of the bent cap beam will be provided by using longitudinal lap splices Proper anchorage of the horizontal ledge reinforcement in the local strutandtie models must also be ensured The bottom chord reinforcement of the cap beam must be fully developed at Nodes G and L If straight reinforcing bars are to be used the required tension development length is determined using AASHTO LRFD Equation 510821a1 FHWANHI130126 Design Example 3 InvertedTee Moment Frame StrutandTie Modeling STM for Concrete Structures Straddle Bent Cap 346 where ldb is the basic development length defined by AASHTO LRFD Equation 510821a2 The required development length will be compared against the available development length at the ends of the bent cap beam as shown in Figure 327 below In Figure 327 it is determined that the available development length is approximately 694 in Examining AASHTO LRFD Article 510821b no modification factors are required which would increase the required development length uncoated or black reinforcing bars are assumed Conservatively the calculated development length will not be reduced as allowed by AASHTO LRFD Article 510821c Thus Longitudinal Column Reinforcement Extended Nodal Zones Assume Prismatic Struts 60 in 240 in 600 in Column Available Length 694 in Critical Section Nodal Zones Figure 327 Bottom Chord Reinforcement Anchorage at Node G FHWANHI130126 Design Example 3 InvertedTee Moment Frame StrutandTie Modeling STM for Concrete Structures Straddle Bent Cap 347 The required development length is greater than the available length as illustrated in Figure 327 Therefore try providing hooked ends and checking the required development length for a hooked bar The required development length of a hooked bar is given by AASHTO LRFD Equation 510824a1 where lhb is given by AASHTO LRFD Equation 510824a2 Consequently Adequate length is available to develop a hooked reinforcing bar in tension therefore provide standard hooks at the ends of the bottom chord reinforcing Proper development of the top chord reinforcing is satisfied by the fact that the top chord reinforcing is continuous around the corners of the moment frame and by inspection it will be adequately developed at Nodes A and F Finally proper anchorage of the ledge reinforcement Tie CsFs of the local strutandtie models must be checked The top horizontal portion of the ledge reinforcement is terminated in a 90degree hook Recall that the available development length at Nodes Cs and Fs of the local strutandtie model is measured from the location where the centroid of the reinforcing enters the extended nodal zone illustrated in Figure 328 on the following page FHWANHI130126 Design Example 3 InvertedTee Moment Frame StrutandTie Modeling STM for Concrete Structures Straddle Bent Cap 348 6092 2485 k 1382 k 80 in 48 in 24 in 35 in Nodal Zone Extended Nodal Zone Assume Prismatic Strut 20 in Clear Available Length Critical Section Figure 328 Ledge Reinforcement Anchorage at Node Cs The available development length is The required development length of a No 6 reinforcing bar with a 90degree hook is Note that the reinforcement confinement factor λrc is taken as 08 per AASHTO LRFD Article 510824b Sufficient development length is available for the ledge reinforcing using 90degree hooked ends FHWANHI130126 Design Example 3 InvertedTee Moment Frame StrutandTie Modeling STM for Concrete Structures Straddle Bent Cap 349 Design Step 12 Draw Reinforcement Layout At this point the strutandtie analysis for the example moment frame invertedtee straddle bent cap beam is complete The designer is reminded that other load cases must be checked to ensure adequate strength is provided for all imposed loads Sketches of the final beam dimensions and reinforcing layouts follow The designer must also consider reinforcing details such as reinforcing lap splice locations possible reinforcing conflicts such as between the column reinforcement and bottom mat of reinforcement in the cap beam and minimum reinforcement spacing Specific Agency Policies and Practices Any reinforcement details not shown in the following figures that were not described within this design example may be adjusted based on specific state or agency policies and practices FHWANHI130126 Design Example 3 InvertedTee Moment Frame StrutandTie Modeling STM for Concrete Structures Straddle Bent Cap 350 600 ft ELEVATION Elev 8141 CJ Elev 7635 CJ Elev 5633 CJ Elev 5533 CJ Elev 7547 Elev 8041 No 11 Bars 30 in 30 in 28 Spa 70 in 1633 ft 39 Spa 55 in 1833 ft A Bars B Bars 30 in 30 in 12 Spa 60 in A Bars B Bars A Bars Double No 6 Stirrups 4 legs B Bars No 6 Ledge Stirrups C Bars Closed No 6 Column Ties 7 Spa 50 in A Bars 4 Spa 70 in 233 ft No 6 Stirrups 2 legs Typ Unless Noted Otherwise No 6 Bars 9 No 6 Bars 6 No 11 Bars Typ A B A B 55 in 70 in C Bars 35 in 554 ft C Bars 35 in 554 ft Figure 329 Elevation View of Reinforcement Layout FHWANHI130126 Design Example 3 InvertedTee Moment Frame StrutandTie Modeling STM for Concrete Structures Straddle Bent Cap 351 20 in Clear 334 ft 6 No 11 20 in Clear 600 ft 13 No 11 No 6 Bars 8 Eq Spa SECTION AA 20 in Clear 334 ft 20 in Clear 133 ft 133 ft 6 No 11 13 No 11 No 11 Bars SECTION BB 600 ft 233 ft No 6 Bars Eq Spa No 6 Bar Typ Figure 330 Sections AA and BB of Figure 329 FHWANHI130126 Design Example 3 InvertedTee Moment Frame StrutandTie Modeling STM for Concrete Structures Straddle Bent Cap 352 References 1 Williams CS Deschenes DJ and Bayrak O StrutandTie Model Design Examples for Bridges Implementation Report 5525301 Center for Transportation Research Bureau of Engineering Research University of Texas at Austin October 2011 272 pp FHWANHI130126 Design Example 4 Drilled Shaft Footing StrutandTie Modeling STM for Concrete Structures 41 Design Example 4 Drilled Shaft Footing Table of Contents Page Design Step 1 Define StrutandTie Model Input 44 Design Step 2 Determine the Locations of the B and DRegions 46 Design Step 3 Define Load Cases 46 Design Step 4 Analyze Structural Components Load Case 1 47 Design Step 41 Determine Loads 48 Design Step 42 Develop StrutandTie Model 411 Design Step 43 Proportion Ties 419 Design Step 44 Perform Nodal Strength Checks 422 Design Step 45 Proportion Shrinkage and Temperature Reinforcement 425 Design Step 46 Provide Necessary Anchorage for Ties 425 Design Step 5 Analyze Structural Components Load Case 2 429 Design Step 51 Determine Loads 429 Design Step 52 Develop StrutandTie Model 432 Design Step 53 Proportion Ties 435 Design Step 54 Perform Nodal Strength Checks 438 Design Step 55 Proportion Shrinkage and Temperature Reinforcement 438 Design Step 56 Provide Necessary Anchorage for Ties 438 Design Step 6 Draw Reinforcement Layout 441 FHWANHI130126 Design Example 4 Drilled Shaft Footing StrutandTie Modeling STM for Concrete Structures 42 Design Example 4 presents the application of the strutandtie method STM to the analysis and design of a drilled shaft footing The footing supports a single column and is supported by four drilled shafts Existing research demonstrates that the strutandtie method is appropriate for the design of footings such as these see the references at the end of this design example This design example demonstrates the development and application of threedimensional strutandtie models to effectively model the complex distribution of forces in deep footings The example features the elements of strutandtie design of concrete members listed below Design Step 1 Define StrutandTie Model Input Design Step 3 Define Load Cases Design Step 4 Analyze Structural Components Load Case 1 Design Step 41 Determine Loads Design Step 42 Develop StrutandTie Model Design Step 43 Proportion Ties Design Step 2 Determine the Locations of the B and DRegions Continued on Next Page FHWANHI130126 Design Example 4 Drilled Shaft Footing StrutandTie Modeling STM for Concrete Structures 43 Design Step 44 Perform Nodal Strength Checks Design Step 45 Proportion Shrinkage Temperature Reinforcement Design Step 46 Provide Necessary Anchorage for Ties Design Step 6 Draw Reinforcement Layout Design Step 5 Analyze Structural Components Load Case 2 Design Step 51 Determine Loads Design Step 52 Develop StrutandTie Model Design Step 53 Proportion Ties Design Step 54 Perform Nodal Strength Checks Design Step 55 Proportion Shrinkage Temperature Reinforcement Design Step 56 Provide Necessary Anchorage for Ties Continued from Previous Page Please note that this example is based on an example problem developed in an implementation project sponsored by TxDOT Report No 5525301 Williams et al 2012 Figures included in this design example are adapted from this report The example has been revised herein to provide additional explanations and to provide compliance with the STM provisions of the 8th Edition of AASHTO LRFD as appropriate and differing material strengths have been adopted for reasons that will be discussed shortly FHWANHI130126 Design Example 4 Drilled Shaft Footing StrutandTie Modeling STM for Concrete Structures 44 There is a shortage of documented research on the application of the strutandtie design method to the design of deep pile caps or drilled shaft footings As a result several approximations and assumptions must be made to design these structures The implications of these assumptions on the design are analyzed and discussed before they are used these judgements tend to err on the side of conservatism Design Step 1 Define StrutandTie Model Input Figure 41 illustrates the overall geometry of the drilled shaft footing under consideration The footing is determined to be 500 ft thick 1600 ft wide and 1600 ft long The function of the footing is to transfer loads imposed by a 750 ft by 625 ft rectangular column to four drilled shafts that are each 400 ft in diameter This means of load transfer provides an opportunity to demonstrate the AASHTO LRFD strutandtie design specifications in a threedimensional context The design compressive strength of the concrete fc is taken as 50 ksi and the yield strength of the steel reinforcing fy is taken as 750 ksi The compressive strength of the concrete represents an average of concrete compressive strengths for footings used nationally and differs from the original design example developed by Williams et al 2012 The reinforcement strength is chosen to be 750 ksi in order to take advantage of smaller required reinforcing areas The AASHTO LRFD design specifications permit the use of higherstrength reinforcing bars strengths in excess of 600 ksi Grade 75 reinforcing is used in this example to illustrate the use of the AASHTO LRFD design specifications with highstrength material that is now available The appropriate strengths should be chosen per the OwnersAgencys requirements FHWANHI130126 Design Example 4 Drilled Shaft Footing StrutandTie Modeling STM for Concrete Structures 45 C Drilled Shaft L C Column C Footing L L PLAN ELEVATION OH 075 425 Wcol 750 425 H 500 650 OH 075 DDS 400 DDS 400 L1 1600 SDS 1050 L1 1600 800 800 L1 1600 800 800 OH 075 OH 075 650 DDS 400 DDS 400 OH 075 OH 075 650 DDS 400 DDS 400 Wcol 750 Dcol 625 Figure 41 Plan and Elevation Views of Drilled Shaft Footing Adapted from Williams et al 2012 FHWANHI130126 Design Example 4 Drilled Shaft Footing StrutandTie Modeling STM for Concrete Structures 46 Design Step 2 Determine the Locations of the B and DRegions The subject drilled shaft footing is characterized by disturbances introduced by the column and drilled shafts Because of their close proximity relative to the depth of the footing itself the classical Bernoulli beam theory assumption that plane sections remain plane is inappropriate As such the entire footing is classified as a DRegion Design Step 3 Define Load Cases In a departure from previous design examples this design example presents two load cases Each consists of a system of forces imposed by the column on the drilled shaft footing which are in turn resisted by the drilled shafts In Load Case 1 the column is subjected to combination of axial force and strongaxis bending moment refer to Figure 42 When the load is transferred through the footing all of the drilled shafts will remain in compression as shown in Figure 42 The actual calculation of the support reactions is discussed in Design Step 41 Pu 2849 kips Muxx 9507 kipft z x y R1 R4 R2 R3 Figure 42 Factored Loads for Load Case 1 Adapted from Williams et al 2012 In Load Case 2 the column is again subjected to axial force and strongaxis bending moment see Figure 43 However in this case the strongaxis bending moment is approximately 16 percent less than that in Load Case 1 and the magnitude of the axial force is less than half that in Load Case 1 This load case will result in tension in two of FHWANHI130126 Design Example 4 Drilled Shaft Footing StrutandTie Modeling STM for Concrete Structures 47 the drilled shafts as shown in Figure 43 The calculation of these reactions is discussed in Design Step 51 Pu 1110 kips Muxx 7942 kipft z x y R1 R4 R2 R3 Figure 43 Factored Loads for Load Case 2 Adapted from Williams et al 2012 Many designers include the footing selfweight due to the possibility of earth settlement However for this design example the possibility of earth settlement is considered to be insignificant and the footing selfweight is not applied in either load case Each load case is presented independently ie all of the design steps will be performed for Load Case 1 before performing design for Load Case 2 Evaluating Multiple Load Cases This design example considers only the two load cases presented The designer is reminded that a complete design of the drilled shaft footing is contingent on examination of all of the critical load cases Design Step 4 Analyze Structural Components Load Case 1 The forces imposed by the column will flow through the footing to each of the four drilled shafts In order to properly model the flow of forces the axial force and moment applied by the column on the footing must be rectified into a system of equivalent forces as FHWANHI130126 Design Example 4 Drilled Shaft Footing StrutandTie Modeling STM for Concrete Structures 48 was done for the cantilever bent in Design Example 2 and the straddle bent columns in Design Example 3 This set of forces will be applied to the strutandtie model and should by definition produce the same axial load and moment as shown in Figure 42 An illustration of this procedure is given in Figure 44 z y C T C T APPLIED LOADS EQUIVALENT LOADS Figure 44 Developing an Equivalent Force System Design Step 41 Determine Loads To develop the system of equivalent forces the elastic stress distribution within the column must be determined The location of each of the forces in the equivalent force system is found relative to the column crosssection The assumed layout of the column reinforcing is given in Figure 45 Then the magnitude of each force may be found by establishing equilibrium The column crosssection and corresponding linear stress distribution are shown in Figure 46 The positions of the four loads that will comprise the equivalent force system are shown The two loads acting on the left side of the column are compressive pushing down on the footing and the two loads acting on the right of the column are tensile pulling up on the footing FHWANHI130126 Design Example 4 Drilled Shaft Footing StrutandTie Modeling STM for Concrete Structures 49 11 Equal Spaces 10 No 11 Bars x y Wcol 750 225 Clear 12 No 11 Bars 11 Equal Spaces Dcol 625 Figure 45 Preliminary Column Reinforcing Layout The locations of the compressive forces are based on the linear stress diagram The line of action for the compressive forces coincides with the location of the center of gravity of the compressive side of the stress diagram This line of action is located 172 ft from the left face of the column The loads are located transversely at the quarter points of the column depth 625 ft which is 156 ft from the top and bottom of the column section in Figure 46 FHWANHI130126 Design Example 4 Drilled Shaft Footing StrutandTie Modeling STM for Concrete Structures 410 D A Applied Load Neutral Axis C B C T 1549 psi Centroid of 6 No 11 Bars Centroid of 6 No 11 Bars Column Bars Considered to Carry Forces in Ties BI and CJ of the strutandtie model Dcol 625 156 156 313 Wcol 750 344 030 158 172 344 235 x y Figure 46 Linear Stress Distribution over Column CrossSection and Equivalent Force System Load Locations The longitudinal reinforcing steel in the column shown in Figure 45 and Figure 46 is an assumption The size and configuration of the reinforcement must be determined in final design which is beyond the scope of this design example The reinforcement on the right face of the column will resist the tension resulting from the applied bending moment The two tensile forces which will balance the two compressive forces are therefore located at the centers of gravity of the tensionface reinforcement located at 030 ft from the right face of the column The reinforcing on this face is divided in half and each half is assumed to resist each tensile force This results in each tensile force being located at the center of gravity of onehalf of the tensionface reinforcement As shown in Figure 46 each tie in the column will consist of six reinforcing bars The magnitudes of the compressive and tensile forces must now be determined so that the equivalent force system produces the same axial force moment and linear stress distribution This is accomplished by establishing equilibrium between the original and equivalent force systems In the following system of equations Fcomp is the total compressive force acting on the system or the sum of the compressive forces acting at Points A and D and Ftens is the total tensile force acting on the system or the sum of the tensile forces acting at Points B and C FHWANHI130126 Design Example 4 Drilled Shaft Footing StrutandTie Modeling STM for Concrete Structures 411 Solving yields where 750 ft is the width of the column and 172 ft and 030 ft are the distances to the centers of gravity of the compressive and tensile forces respectively Therefore the four loads that will act on the strutandtie model from the column are determined The footing must now be analyzed to determine the reaction forces the forces in each drilled shaft The reactions are assumed to act at the center of each of the drilled shafts Since all four drilled shafts are spaced equidistant from the column the axial force is assumed to be distributed equally to each drilled shaft Moment equilibrium of the footing is enforced by equating the column moment to each drilled shafts axial force times its orthogonal distance from the column see Figure 46 The drilled shaft reactions are determined thus where the value of 1050 ft is the drilled shaft spacing parallel to the applied moment Note that all drilled shaft reactions are of the same sign indicating that they act in the same direction By inspection all of the drilled shaft reactions are compressive Design Step 42 Develop StrutandTie Model The strutandtie model for Load Case 1 is shown in Figure 48 and Figure 49 Figure 48 should be worked with Figure 49 and the coordinates of each node in the strut andtie model are presented in Table 41 Development of the threedimensional strut andtie model is considered successful only if equilibrium is satisfied at every node and FHWANHI130126 Design Example 4 Drilled Shaft Footing StrutandTie Modeling STM for Concrete Structures 412 the truss reactions determined from a linear elastic analysis of the truss model are equivalent to the reactions found in Design Step 41 In order to successfully develop the threedimensional truss model the designer first must determine the lateral x y location of each applied load and support reaction Then the designer must determine the vertical z position of the planes where the upper and lower nodes of the strutand tie model lie The lateral locations of the applied loads and drilled shaft reactions were determined in Design Step 41 Table 41 Coordinates of Nodes in StrutandTie Model for Load Case 1 Node xCoordinate yCoordinate zCoordinate A 957 597 450 B 957 1145 450 C 644 1145 450 D 644 597 450 E 1325 275 045 F 1325 1325 045 G 275 1325 045 H 275 275 045 I 957 1145 045 J 644 1145 045 Note The origin is located in the bottom corner of the footing nearest to Node H The location of the bottom horizontal ties relative to the bottom of the footing are determined first These ties Ties EF FG GH and EH represent the bottom mat of reinforcing steel within the footing Their locations should therefore be based on the location of the center of gravity of the reinforcing Four inches of clear cover will be provided from the bottom of the footing to the bottom layer of reinforcing as shown in Figure 47 Assuming the same number of reinforcing bars will be used in both directions and No 11 reinforcing bars will be used the center of gravity of the bottom mat of reinforcing is located at above the bottom face of the footing FHWANHI130126 Design Example 4 Drilled Shaft Footing StrutandTie Modeling STM for Concrete Structures 413 No 11 Bar No 11 Bars z y 541 400 Clear Figure 47 Location of Bottom Mat of Reinforcing Nodes I and J are located in the same plane as Nodes E F G and H Based on the analysis for this load case Members IF and JG are struts since they must balance the horizontal compression at Nodes I and J due to Struts AI and DJ respectively The location of Nodes E F G H I and J coincides with the plane of the bottom mat of reinforcing steel in the shaft cap FHWANHI130126 Design Example 4 Drilled Shaft Footing StrutandTie Modeling STM for Concrete Structures 414 z x y A D E H C B F G I J FA 17636 kips FD 17636 kips FC 3391 kips FB 3391 kips R2 2595 kips R3 2595 kips R4 11650 kips R1 11650 kips 12953 kips 9484 kips 11829 kips 10607 kips Figure 48 Isometric View of StrutandTie Model for Load Case 1 Work This Figure with Figure 49 FHWANHI130126 Design Example 4 Drilled Shaft Footing StrutandTie Modeling STM for Concrete Structures 415 A BI CJ D E F G H L1 1600 L1 1600 x y Figure 49 Plan View of StrutandTie Model for Load Case 1 Work This Figure with Figure 48 The distance between the horizontal strut Strut AD and the top face of the footing may be determined in several ways Williams et al 2012 reports a detailed discussion on this issue based on their research Existing research on this topic recommends different locations of the top struts The options discussed in Williams et al 2012 include Option 1 Position Nodes A and D at the top surface of the footing o This is an obvious choice for the node locations however effective triaxial confinement of the nodes cannot be guaranteed and more conservative estimates of the nodal strength must be used o Additionally positioning the nodes at the top of the footing will create an artificially deep strutandtie model increasing hSTM which would result in underpredicting the bottom tie forces Option 2 Assume depth of the horizontal Strut AD is equal to h4 where h is the overall depth of the footing o This would place the center of the strut at h8 from the top face of the footing FHWANHI130126 Design Example 4 Drilled Shaft Footing StrutandTie Modeling STM for Concrete Structures 416 o This option is recommended in Park et al 2008 and in Windisch et al 2010 o This option is particularly applicable for the depth of a flexural compression zone of an elastic column at a beamcolumn joint o However this option is not highly applicable for this specific design example Option 3 Position the Nodes A and D based on the depth of the Whitney compression stress block o This would determine the strut locations based on the depth of the compression zone determined by a flexural beam theory analysis of the footing o Because the footing is a deep member with many loads and disturbances it is subject to very nonlinear strain distributions so applying a beam theory analogy would not be appropriate o This approach was used in other design examples however it is not used for this design example because the strains are not only nonlinear but they are nonlinear in three dimensions Option 4 Align Nodes A and D with the location of the top mat of reinforcing o If horizontal ties were located in the top of the footing this would be another viable location for the nodes o This method is used to develop the strutandtie model for Load Case 2 To summarize there are numerous options that the designer may consider when placing the top struts The designer may consider the above options but can also consider other options that apply to the specific design and that will result in a conservative design Consideration should also be given to the fact that moving the nodes deeper into the footing farther from the top surface decreases the effective height of the strutandtie model which will increase the demands in the bottom ties This will also increase the effect of the triaxial confinement of the nodes For the purposes of this design example 01h was chosen as the distance between the top chord compression of the space truss and the top face as shown in Figure 410 This value was selected for several reasons A value of 01h is anticipated to produce conservative results A value of 01h is within the lower and upper bound of the four options presented above A value of 01h results in a clean value of 6 inches FHWANHI130126 Design Example 4 Drilled Shaft Footing StrutandTie Modeling STM for Concrete Structures 417 z y A D 60 75 49 NOT TO SCALE Top Mat of Steel Option 4 Top of Footing Option 1 01h Chosen Position h8 Option 2 H 500 Depth of Compression Stress Block Option 3 Figure 410 Locations of Ties in Top of Footing In summary the distance from the bottom horizontal ties of the strutandtie model to the bottom of the footing is taken as 541 in and the distance from the top of the footing to Nodes A and D is assumed to be 600 in Thus the total height of the strutandtie model is Defining and Refining the Model Defining the basic geometry of the strutandtie model may be accomplished reasonably simply However establishing the struts and ties within the model can be more difficult Further refinement of the strutandtie model is based upon the following Recognizing the most probable load paths flow of forces Considering standard concrete construction details Understanding the behavior of footings Iterating by trialanderror to establish equilibrium The logic used to develop the strutandtie model for this design example is discussed in Williams et al 2012 and is presented here for the readers benefit To begin the tensile forces acting at Nodes B and C will require vertical ties that pass through the depth of the footing to Nodes I and J Although the forces in these ties are simply the column loads they are included in this model since establishing the requirement of the ties leads into how the geometry of the entire model is developed The determination of the column ties is presented for completeness Ties should almost always be oriented parallel or perpendicular to the primary axes of the structural component since inclined reinforcement is rarely used in reinforced concrete construction The forces in these vertical ties must be equilibrated by FHWANHI130126 Design Example 4 Drilled Shaft Footing StrutandTie Modeling STM for Concrete Structures 418 compressive stresses originating at Nodes A and D which leads to the placement of Struts AI and DJ In turn Struts AE AF DG and DH represent the flow of compressive forces from Nodes A and D to the supports at Nodes E F G and H Finally equilibrium is established at Node D by adding Strut AD The flow of compressive forces to each of the drilled shafts at Nodes E F G and H will induce tension in the bottom of the footing which must be equilibrated with ties Thus Ties EF FG GH and EH are established The remaining horizontal struts are added to the bottom of the strutandtie model to establish lateral equilibrium at Nodes F G I and J As with all strutandtie models recall that the angle between a tie and an adjacent strut must be greater than or equal to 25 degrees to comply with AASHTO LRFD Article C5822 The strutandtie model shown in Figure 48 satisfies this requirement ThreeDimensional Equilibrium The designer is reminded that equilibrium must be maintained in all three orthogonal directions laterally and vertically at every node in the strutandtie truss model There should be enough truss members intersecting at each node to maintain equilibrium in the x y and z directions and the designer needs to perform a stability check using a three dimensional truss model As a check keep in mind that a symmetrical footing geometry and symmetrical loading should result in a symmetrical strutandtie model Once the strutandtie model geometry has been defined the truss member forces and drilled shaft reactions are determined through a linear elastic analysis either manually or using a structural analysis software package The reactions at each of the drilled shaft locations should be equal to the reactions determined in Design Step 41 and equilibrium should be satisfied at each node If equilibrium cannot be established the strutandtie model must be revised and reanalyzed Another valid strutandtie model is presented in Figure 411 below Although it was possible to establish equilibrium at each node the overall model does not accurately represent the flow of compressive forces from Nodes A and D to each of the drilled shafts at Nodes E F G and H FHWANHI130126 Design Example 4 Drilled Shaft Footing StrutandTie Modeling STM for Concrete Structures 419 A D H B F G I J C E z x y Figure 411 Alternative StrutandTie Model for Load Case 1 Creating ThreeDimensional StrutandTie Models Using a structural analysis software package to develop threedimensional strutandtie models is recommended Truss models may be easily defined and refined until a satisfactory truss geometry is found Multiple strutandtie models may exist for a given loading and geometry Based on the unique design requirements the designer should seek to determine which model best represents the flow of forces within the component Design Step 43 Proportion Ties The forces shown in Figure 48 will be used to proportion the horizontal and vertical ties in the footing The bottom mat of reinforcement will be proportioned first To be consistent with earlier assumptions No 11 bars will be used in both directions in the reinforcing mat Ties EF and GH The forces in Ties EF and GH are equal because of the symmetry of the loading The number of reinforcing bars required is determined as in the previous design examples based on AASHTO LRFD Equations 58231 and 582411 FHWANHI130126 Design Example 4 Drilled Shaft Footing StrutandTie Modeling STM for Concrete Structures 420 Using No 11 reinforcing bars Therefore use 9 No 11 bars Ties FG and EH Because the loading on the column is potentially reversible the same reinforcement will be provided for Ties FG and EH The force in Tie FG is greater than in Tie EH Therefore Tie FG is used to proportion the reinforcement Reinforcement is designed based on AASHTO LRFD Equations 58231 and 582411 Using No 11 reinforcing bars Therefore use 12 No 11 bars for Ties FG and EH For consistency and symmetry use 12 No 11 bars for Ties EF and GH as well FHWANHI130126 Design Example 4 Drilled Shaft Footing StrutandTie Modeling STM for Concrete Structures 421 z y 45 45 Length over which bars could be spaced 478 DDS 400 Figure 412 Locating Ties FG and EH Bars over Drilled Shafts Vertical Locations of the Bottom Ties Referring to Figure 47 recall that the bottom horizontal ties were assumed to be located at the center of gravity of the bottom mat of reinforcing where the assumption was made that the number of reinforcing bars provided in both directions was equal As previously stated 12 No 11 bars will be used for Ties FG EH EF and GH for consistency and symmetry However if the computed value of 12 reinforcing bars was used in one direction and the computed value of 9 bars was used in the other direction the actual location of the center of gravity of the reinforcing mat would be The difference between the assumed and actual center of gravity would be Ties BI and CJ Required reinforcing in the ties is determined using AASHTO LRFD Equations 58231 and 582411 The forces in Ties BI and CJ are equal so the required reinforcing is determined for each FHWANHI130126 Design Example 4 Drilled Shaft Footing StrutandTie Modeling STM for Concrete Structures 422 Using No 11 reinforcing bars Therefore a minimum of 4 No 11 bars should be provided However recall that the column face reinforcement consists of 12 No 11 bars on each column face This reinforcement must be continued into the footing in order to develop its capacity either by continuing the reinforcing into the footing or using lap splices Therefore the 12 No 11 bars will be able to satisfy the requirements of Ties BI and CJ 6 No 11 bars each Design Step 44 Perform Nodal Strength Checks The nodal regions in threedimensional strutandtie models have intricate geometries that complicate the procedure of checking them for adequate strength Even the simplest threedimensional nodes will have complicated geometries and would require either excessive computational time or threedimensional CADD drafting of the node Many attempts have been made to approximate nodal geometries in threedimensional models see the references at the end of this design example but the computational effort required to accurately determine the nodal geometries severely impacts the time required to perform a strutandtie analysis Williams et al 2012 discuss the variety of assumptions made by various researchers in the literature These are briefly presented below Interested readers are referred to Williams et al 2012 and the discussed literature for additional discussion Typically the nodal stresses are limited to a prescribed value Various limits on the stresses include Assume that all of the nodal regions are sufficiently strong so long as the bearing stresses at the columns and piles are limited to 085fc Limit bearing stresses to 04fc or 06fc along with providing proper reinforcement detailing in the nodal regions Limit bearing stresses to 10fc but only if the shear spantodepth ratio ad is about equal to 10 FHWANHI130126 Design Example 4 Drilled Shaft Footing StrutandTie Modeling STM for Concrete Structures 423 Other research concludes that a nodal bearing stress limit may not be a good indicator of the footing strength Conclusions that may be drawn from a literature review are Research in this area of strutandtie design does not recommend determining the actual nodal geometries recognizing the fact that such a procedure would be computationally cumbersome and A majority of the literature recommends adopting a design method that requires a limit on bearing stresses combined with proper reinforcement detailing The procedure that will be used in this design example is as follows Position all nodes within the confines of the footing or pile cap In particular nodes directly under columns should not be positioned at the columntofooting interface Limit the concrete bearing stress on the footing or pile cap to vfc where v is the concrete efficiency factor defined in AASHTO LRFD Article 58253 Because the value of v is limited to a maximum of 085 this stress limit is in line with the recommendations from existing research Omit the confinement modification factor m for additional conservatism The introduction of these stress limits simplifies the analysis by not requiring determination of the nodal geometries Detailed calculations for struttonode interfaces are provided in the other design examples and in the accompanying training course Referring to Figure 48 the greatest bearing stresses will occur at Nodes A D E and H because of the large magnitudes of the compressive forces in Struts AE AD and DH These nodes will be checked using the procedure detailed above Check Nodes E and H CTT Due to symmetry the forces and thus the bearings areas at Nodes E and H are the same and require only one design check The bearing area of one of the 400 ft diameter drilled shafts is Because Nodes E and H are CTT nodes the corresponding concrete efficiency factor is determined using AASHTO LRFD Table 58253a1 FHWANHI130126 Design Example 4 Drilled Shaft Footing StrutandTie Modeling STM for Concrete Structures 424 The bearing force to be resisted is taken as the reaction at the drilled shafts or 11650 kips The allowable bearing force at the nodes is determined based on AASHTO LRFD Equation 582511 Modified Note the omission of the confinement modification factor m in AASHTO LRFD Equation 582511 Hence the nodal strength is adequate according to the proposed procedure Nodes A and D CCC Due to symmetry the forces and bearing areas at Nodes A and D are identical The locations of the loads are assumed to be at the centers of the shaded bearing areas shown in Figure 46 The bearing area for each node is Nodes A and D are CCC nodes and the strengths of their bearing areas may be determined using AASHTO LRFD Table 58253a1 and AASHTO LRFD Equation 582511 Modified Hence the nodal strength is adequate according to the proposed procedure Since these critical nodal strengths are sufficient to resist the applied loads all nodal strengths in the strutandtie are adequate to resist the applied loads Adebar 2004 Widianto and Bayrak 2011 and Schlaich et al 1987 FHWANHI130126 Design Example 4 Drilled Shaft Footing StrutandTie Modeling STM for Concrete Structures 425 Design Step 45 Proportion Shrinkage and Temperature Reinforcement The minimum crack control reinforcement requirements do not apply to slabs and footings per AASHTO LRFD Article 5826 However shrinkage and temperature reinforcement should still be provided to control cracking Shrinkage and temperature reinforcement may be determined using the provisions of AASHTO LRFD Article 5106 The required area of reinforcing on each face and in each direction is determined based on AASHTO LRFD Equations 51061 and 51062 limited as follows where As area of reinforcement in each direction on each face in2ft b least width of component section in h least thickness of component section in fy minimum yield strength of reinforcement 750 ksi For the drilled shaft footing the value of b is taken as 1600 ft or 19200 in and the value of h is taken as 500 ft or 6000 in Therefore the required reinforcement is determined Providing 1 No 6 bar per foot yields 044 in2ft which is acceptable No 6 bars will be provided on all faces except for the bottom face where No 11 bars will be spaced evenly between the drilled shafts The maximum spacing of the temperature and shrinkage reinforcement is 120 in for footings greater than 180 in thick per AASHTO LRFD Article 5106 Because the areas provided for a No 6 bar or No 11 bar spaced at 120 in exceeds the area of steel required the maximum spacing requirement controls Design Step 46 Provide Necessary Anchorage for Ties Each tie in the strutandtie model must be fully developed at the point where the center of gravity of the reinforcement exits the extended nodal zone in accordance with AASHTO LRFD Article 51082 The threedimensional nodes and extended nodal FHWANHI130126 Design Example 4 Drilled Shaft Footing StrutandTie Modeling STM for Concrete Structures 426 zones are difficult to define making definitive calculation of the available development length impossible Therefore the designer must decide on an approach to finding a conservative estimate of the available development length Ties EF FG GH and EH A conservative assumption will be made to estimate the available development length in relation to the dimensions of the drilled shafts The circular drilled shafts will be idealized as square shafts of the same crosssectional area and the critical section for development will be taken as the interior edge of an equivalent square shaft refer to Figure 413 The side dimension lb of the equivalent square shaft is given by Now that the location of the critical section has been defined the available length is determined by where OH is the distance that the footing overhangs the drilled shaft Therefore If straight reinforcing bars are to be used the required tension development length is computed based on AASHTO LRFD Equation 510821a1 where ldb is the basic development length defined by AASHTO LRFD Equation 510821a2 Examining AASHTO LRFD Article 510821b no modification factors are required which would increase the required development length uncoated or black reinforcing bars are assumed Conservatively the calculated development length will not be reduced as allowed by AASHTO LRFD Article 510821c Thus for No 11 reinforcing bars FHWANHI130126 Design Example 4 Drilled Shaft Footing StrutandTie Modeling STM for Concrete Structures 427 z y 900 3 Min Clear Critical Section Available Length lb 4254 DDS 400 SECTION AA A A Figure 413 Available Development Length over the Drilled Shafts FHWANHI130126 Design Example 4 Drilled Shaft Footing StrutandTie Modeling STM for Concrete Structures 428 The required development length is greater than the available length Therefore try providing hooked ends and checking the required development length for a hooked bar AASHTO LRFD Equations 510824a1 and 510824a2 are as follows where Based on AASHTO LRFD Article 5428 the concrete density modification factor Lambda λ equals 10 for normal weight concrete Consequently ldh can be computed as follows Adequate length is available to develop a hooked reinforcing bar in tension therefore provide standard hooks at the ends of the bottom mat reinforcing Ties BI and CJ Vertical Ties BI and CJ consist of the reinforcing bars extending from the column into the footing Standard practice accomplishes this by using Lshaped bars extending from the footing lapped with straight bars in the column The required development length for No 11 reinforcing bars with a 90degree hook was calculated as 2995 in The tie reinforcement must be fully developed at the point where the center of gravity of the tie reinforcement leaves the extended nodal zone Unfortunately the depths of the nodal regions and by extension the extended nodal regions cannot be determined with certainty because Nodes I and J are smeared nodes they have no bearing plates or geometric boundaries which define their limits refer to Figure 414 The available development length is therefore unknown Considering that hooked reinforcing bars have been used successfully in practice for many years it is assumed that the hooked No 11 bars will provide adequate development for Ties BI and CJ Because the loading is potentially reversible all of the vertical column reinforcement entering the footing will terminate in standard 90degree hooks FHWANHI130126 Design Example 4 Drilled Shaft Footing StrutandTie Modeling STM for Concrete Structures 429 z x Node I Geometry Cannot be Defined Available Length H 500 No 11 Bar ldh 2995 3 Clear Figure 414 Vertical Tie Unknown Available Development Length Design Step 5 Analyze Structural Components Load Case 2 Now that the design steps have been completed for Load Case 1 the same procedure will be used for Load Case 2 Since the same procedure is used any differences between the designs are noted Design Step 51 Determine Loads Refer back to Figure 43 for the loadings of Load Case 2 The axial force and bending moment are resolved into equivalent forces that will be applied to the strutandtie model which is analogous to the method used in Design Step 41 The linear stress distributions resulting from the applied loads are shown in Figure 415 The equivalent force system again consists of four vertical forces which correspond to the drilled shaft reactions However now two of the forces are compressive and two are tensile The compressive forces act at the compressive stress resultant of the linear stress diagram The compressive forces act a distance of 147 ft from the left face of the column and at the quarterpoints of the column depth from the top and bottom faces The positions of the tensile resultants are the same as in Design Step 41 each tensile resultant is located at the center of gravity of a group of 6 No 11 column bars on the tension face of the column FHWANHI130126 Design Example 4 Drilled Shaft Footing StrutandTie Modeling STM for Concrete Structures 430 D A Applied Load Neutral Axis C B C T 1106 psi Centroid of 6 No 11 Bars Centroid of 6 No 11 Bars Column Bars Considered to Carry Forces in Ties BI and CJ of the strutandtie model Dcol 625 156 156 313 Wcol 750 294 030 158 147 294 309 x y Figure 415 Linear Stress Distribution over Column CrossSection and Equivalent Force System Load Locations Equilibrium must be established once again to determine the magnitude of the compressive and tensile forces applied by the column In the following system of equations Fcomp is the total compressive force acting on the system or the sum of the compressive forces acting at Nodes A and D and Ftens is the total tensile force acting on the system or the sum of the tensile forces acting at Nodes B and C Solving yields Using Fcomp and Ftens the four loads that will act on the strutandtie model from the column are determined FHWANHI130126 Design Example 4 Drilled Shaft Footing StrutandTie Modeling STM for Concrete Structures 431 These forces are shown acting on the strutandtie model in Figure 418 The drilled shaft reactions are obtained from overall equilibrium of the drilledshaft footing under the applied loads Muxx 7942 kipft x y Pu 1110 kips z R1 R2 R4 R3 Figure 416 Loads Acting on Drilled Shaft Footing for Load Case 2 Note that the signs of R1 and R4 and R2 and R3 are opposite By the assumed sign convention the reactions R2 and R3 are tensile indicating that the drilled shafts at Nodes F and G experience uplift under this load case FHWANHI130126 Design Example 4 Drilled Shaft Footing StrutandTie Modeling STM for Concrete Structures 432 Design Step 52 Develop StrutandTie Model Development of the strutandtie model is based on the same methodology that was used in Design Step 42 The strutandtie model for Load Case 2 is shown in Figure 4 18 and the coordinates of each node are presented in Table 42 Table 42 Coordinates of Nodes in StrutandTie Model for Load Case 2 Node xCoordinate yCoordinate zCoordinate A 957 572 460 B 957 1145 460 C 644 1145 460 D 644 572 460 E 1325 275 045 F 1325 1325 045 G 275 1325 045 H 275 275 045 I 957 1145 045 J 644 1145 045 K 1325 275 460 L 1325 1325 460 M 275 1325 460 N 275 275 460 Note The origin is located in the bottom corner of the footing nearest to Node H Prior to placement of the individual struts and ties the vertical positions of the top and bottom nodes of the strutandtie model must be determined The bottom ties of the model Ties EF FG GH and EH coincide with the center of gravity of the bottom mat of reinforcement The distance from the bottom of the footing to these ties will be the same as for Load Case 1 or 541 in In addition a set of horizontal ties is required near the top of the footing to resist the tension created by the overturning moment The tension reactions in two of the drilled shafts indicate the need for these ties which will be located at the center of gravity of the top mat of reinforcement FHWANHI130126 Design Example 4 Drilled Shaft Footing StrutandTie Modeling STM for Concrete Structures 433 z y 475 400 Clear No 6 Bars No 6 Bars Figure 417 Locating Top Mat of Footing Reinforcement The top mat of reinforcing will consist of two orthogonal layers of No 6 reinforcing bars An equal number of reinforcing bars will be provided in each layer and a clear cover of 400 in will be used measured from the top face of the footing Examining Figure 417 the center of the gravity of the top mat of reinforcing will be located at The height of the strutandtie model hSTM is therefore In order to develop the rest of the strutandtie model the individual struts and ties should follow the most intuitive load path establishing equilibrium at each node The tensile forces at Nodes B and C require vertical Ties BI and CJ to transfer the tension through the footing depth Similarly two additional ties Ties FL and GM are required to resist the tensile drilled shaft reactions at Nodes F and G These ties are required to anchor the footing to the drilled shafts Note that Ties BI and FL together form a noncontact lap splice which would tie the reaction at Node F to the applied load at Node B Thus compressive stress will develop between these two nodes requiring a strut to transfer the stress between the nodes This is idealized by Strut IL The forces in Ties CJ and GM similarly require Strut JM FHWANHI130126 Design Example 4 Drilled Shaft Footing StrutandTie Modeling STM for Concrete Structures 434 z x y A D E H C B F G J FA 10268 kips FD 10268 kips FC 4718 kips FB 4718 kips R2 1007 kips R3 1007 kips R4 6557 kips R1 6557 kips 543 kips 10610 kips 9690 kips 902 kips 9958 kips 6337 kips 697 kips 4699 kips 4718 kips 4699 kips 437 kips 6337 kips I 697 kips 437 kips 6362 kips 9958 kips 5838 kips 1007 kips 10770 kips 10770 kips 1420 kips 1420 kips 4718 kips K L M N 1007 kips Figure 418 Isometric View for StrutandTie Model for Load Case 2 Vertical equilibrium will be satisfied at Nodes I and J by drawings Struts AI and DJ and Struts AE and DH satisfy equilibrium at Nodes A and D Because of the compressive FHWANHI130126 Design Example 4 Drilled Shaft Footing StrutandTie Modeling STM for Concrete Structures 435 stresses from these struts flowing to the drilled shafts tension develops in the bottom of the footing which will be carried by Ties EF FG GH and EH Likewise Struts IL and JM connecting the vertical ties create tension in the top of the footing requiring Ties KL LM MN and KN Once again the strutandtie model is checked to ensure that the angles between struts and ties are always greater than or equal to 25 degrees The strutandtie model is analyzed in the same manner as was done for Load Case 1 Recall that a linear elastic analysis of the truss model should yield the same reactions at the drilled shafts as those found in Design Step 51 Vertical Locations of the Bottom Ties The reader is encouraged to examine Figure 48 and Figure 418 together Note the differences and similarities in the strutandtie models for each load case These models are the result of several iterations to the threedimensional truss geometries that ultimately result in models that reflect the flow of forces in the footing Taking time to visualize and sketch the possible flow of forces in the footing from a given loading may help reduce the time needed to modify the truss geometry Analysis of this strutandtie model is now complete Design Step 53 Proportion Ties The calculated forces in the strutandtie models of Load Case 1 and Load Case 2 should be compared to determine the controlling design forces for the struts and ties The bottom tie forces Ties EF FG GH and EH in Load Case 1 control so design of those ties will not be reexamined The vertical tie forces Ties BI and CJ for Load Case 2 control so these ties will be redesigned The remaining ties Ties FL GM KL LM MN and KN are unique to Load Case 2 and must be designed Ties KL LM MN and KN The force in Tie LM controls the design of the top horizontal ties The amount of reinforcing required is found using AASHTO LRFD Equations 58231 and 582411 Using No 6 reinforcing bars FHWANHI130126 Design Example 4 Drilled Shaft Footing StrutandTie Modeling STM for Concrete Structures 436 Therefore a minimum of 3 No 6 bars are required to carry the tie force Recall that the temperature and shrinkage steel defined in Design Step 45 was No 6 bars in each direction at about 120 in spacing At a maximum spacing of 120 in about four No 6 bars are located above each drilled shaft The number of bars available to carry the tie force is thus greater than the number required thus the temperature and shrinkage steel is adequate to resist the tie forces Using even spaces the actual spacing of the No 6 bars is approximately 110 in s 11 Bars Considered to Carry the Tie Force x y Figure 419 Reinforcing Carrying Forces in Ties KL LM MN and KN Ties BI and CJ The forces in these ties are larger for Load Case 2 than for Load Case 1 Therefore their design is reevaluated Considering the reinforcement from Load Case 1 6 No 11 bars are provided which extend from the column into the footing for each tie The tie strength must be checked against the new demand using AASHTO LRFD Equations 58231 and 582411 Using 6 No 11 reinforcing bars FHWANHI130126 Design Example 4 Drilled Shaft Footing StrutandTie Modeling STM for Concrete Structures 437 Therefore the 12 No 11 reinforcing bars in the column 6 for each tie are adequate to resist the tie forces Ties FL and GM Finally the reinforcement for Ties FL and GM is defined These ties represent the reinforcing bars which anchor the drilled shafts into the footing The assumed layout of the reinforcement is typical of standard drilled shaft construction and is shown in Figure 420 No 9 reinforcing bars are a common size reinforcing bar specified in drilled shaft design The reinforcement in the drilled shafts at Nodes F and G will be extended into the footing in order to satisfy the reinforcement requirements for Ties FL and GM based on AASHTO LRFD Equations 58231 and 582411 Using No 9 reinforcing bars DDS 400 400 Clear No 3 Spiral 20 No 9 Bars Figure 420 Assumed Drilled Shaft Reinforcing Layout A minimum of 2 No 9 drilled shaft bars must be extended into the footing However all of the drilled shaft reinforcement will be extended into the footing consistent with typical construction practice However this reinforcement must be adequately anchored in order to contribute to the strength of Ties FL and GM This requirement will be checked in Design Step 56 FHWANHI130126 Design Example 4 Drilled Shaft Footing StrutandTie Modeling STM for Concrete Structures 438 Design Step 54 Perform Nodal Strength Checks Recall the discussion in Design Step 44 regarding the complicated geometries of nodal regions in threedimensional strutandtie models The simplified nodal strength procedure that was posited in that discussion will be applied in this design step Comparing the truss member forces in the strutandtie models for Load Case 1 and Load Case 2 the compressive forces bearing on the footing are greater for Load Case 1 than for Load Case 2 Therefore the compressive bearing stress checks for Load Case 2 will not control design of the nodal regions so no further strength checks are required Design Step 55 Proportion Shrinkage and Temperature Reinforcement The required temperature and shrinkage reinforcement was determined in Design Step 45 and does not need to be revisited Design Step 56 Provide Necessary Anchorage for Ties Proper anchorage of the bottom mat reinforcement Ties EF FG GH and EH and vertical Ties BI and CJ was discussed in Design Step 46 These ties will be sufficiently anchored with the use of 90degree hooks Anchorage of the ties unique to Load Case 2 Ties KL LM MN KN FL and GM is discussed in this design step Ties KL LM MN and KN The horizontal top mat reinforcement must be properly anchored at Nodes K L M and N These four nodes are smeared nodes with no bearing plates or boundaries that define their geometries Thinking threedimensionally the diagonal struts that enter these four nodes Struts AK DN IL and JM will create large extended nodal zones so to maintain conservatism the critical section for development of the reinforcement is assumed to be the same as the bottom mat of reinforcement This location is the plane above the edge of the equivalent square drilled shaft that was determined in Design Step 46 The available development length is therefore the same as for the bottom reinforcement or 5123 in Note that this requires 300 in of clear cover at each end of the reinforcing bars The required development length of a straight No 6 bar is now checked Checking AASHTO LRFD Articles 510821b and 510821c there are modification factors applicable to this reinforcement These are lambda sub r l λrl and lambda sub r c λrc λrl horizontal reinforcement placed such that more than 120 in of concrete is cast below it FHWANHI130126 Design Example 4 Drilled Shaft Footing StrutandTie Modeling STM for Concrete Structures 439 λrc reinforcement confinement factor determined according to the provisions of AASHTO LRFD Article 510821c below The reinforcement confinement factor is limited as follows based on AASHTO LRFD Equations 510821c1 510821c2 and 510821c3 where where cb the smaller of the distance from the center of the bar being developed to nearest concrete surface and onehalf of the centertocenter spacing of the bars being developed in ktr transverse reinforcement index Atr total crosssectional area of transverse reinforcement in spacing s which crosses the potential plane of splitting of the reinforcing being developed in2 s maximum centertocenter spacing of transverse reinforcement within ld in n number of bars developed along the plane of splitting Illustrations of these variables are given in AASHTO LRFD Figure C510821c1 The value of cb is found thus The spacing s of the transverse reinforcement is equal to 120 in recall there are No 6 bars spaced at about 120 in in both directions therefore the area Atr 044 in2 The number of bars developed along the plane of splitting is taken as 1 Therefore FHWANHI130126 Design Example 4 Drilled Shaft Footing StrutandTie Modeling STM for Concrete Structures 440 The other development length modification factors are taken as 10 Therefore the required development length of the No 6 bars is computed as follows based on AASHTO LRFD Equations 510821a1 and 510821a2 Therefore proper anchorage is provided for the straight No 6 bars with 300 in of clear cover at each end Ties FL and GM Ties Fl and GM must be properly anchored at Nodes L and M In Design Step 53 it was determined that a minimum of 2 No 9 bars must extend from the drilled shafts into the footing to satisfy the reinforcement requirements for these ties Considering typical drilled shaft construction all of the drilled shaft reinforcing bars will extend into the footing The required development length for a No 9 reinforcing bar in tension is computed based on AASHTO LRFD Equations 510821a1 and 510821a2 The reinforcement location factor λrl coating factor λcf and density modification factor λ will be taken as 10 Conservatively the reinforcement confinement factor λrc will also be taken as 10 The development length required will be reduced by the excess reinforcement modification factor λer determined using AASHTO LRFD Equation 510821c4 The required tension reinforcement was determined in Design Step 53 Twenty No 9 reinforcing bars are provided Therefore Hence FHWANHI130126 Design Example 4 Drilled Shaft Footing StrutandTie Modeling STM for Concrete Structures 441 Similar to Nodes I and J Nodes L and M are smeared nodes whose geometries cannot be determined Visualizing the threedimensional geometries of the nodal regions the straight No 9 bars should be adequate to anchor the ties if the ends of the bars are extended as close as practical to the top of the footing The ends of the bars should maintain the 400 inch minimum clear cover at the top face of the footing The No 9 bars will be extended in all four drilled shafts considering constructability concerns and the potential reversibility of the applied loads The geometry of the drilled shaft reinforcing is shown in Figure 421 DDS 400 All No 9 bars extended into footing to within 400 in of the top face of footing Figure 421 Longitudinal Drilled Shaft Reinforcing Design Step 6 Draw Reinforcement Layout At this point the strutandtie analysis of the drilled shaft footing is complete Sketches of the final reinforcement layouts are provided in Figure 422 through Figure 428 FHWANHI130126 Design Example 4 Drilled Shaft Footing StrutandTie Modeling STM for Concrete Structures 442 1600 90Degree Hooks 1600 x y Figure 422 Reinforcement Details 1 Anchorage of Ties FHWANHI130126 Design Example 4 Drilled Shaft Footing StrutandTie Modeling STM for Concrete Structures 443 z y 750 40 Clear 167 167 033 033 167 167 033 033 075 11 Eq Spa 400 No 11 Bars 11 Eq Spa 400 No 11 Bars 075 1600 No 11 Bars No 11 Bar A A 500 No 9 Bars 3 of 20 Bars Shown Typ 7 Eq Spa 650 No 11 Bars Figure 423 Reinforcement Details 2 Elevation View of Primary Reinforcement FHWANHI130126 Design Example 4 Drilled Shaft Footing StrutandTie Modeling STM for Concrete Structures 444 z y 40 Clear 15 Eq Spa 1500 No 6 Bars A A 050 050 No 6 Bars No 6 Bars No 6 Bar 30 Clear Location of No 11 Bars of Bottom Mat 500 5 Eq Spa Figure 424 Reinforcement Details 3 Temperature and Shrinkage Reinforcement z x 500 40 Clear 075 11 Eq Spa 400 No 11 Bars 7 Eq Spa 650 No 11 Bars 11 Eq Spa 400 No 11 Bars 075 No 11 Bar Figure 425 Reinforcement Details 4 Section AA of Figure 424 FHWANHI130126 Design Example 4 Drilled Shaft Footing StrutandTie Modeling STM for Concrete Structures 445 40 Clear 15 Eq Spa 1500 No 6 Bars 050 050 No 6 Bars No 6 Bars No 6 Bar 30 Clear Location of No 11 Bar of Bottom Mat No 6 Bars z x 5 Eq Spa Figure 426 Reinforcement Details 5 Temperature and Shrinkage Reinforcement of Section AA in Figure 424 1600 050 050 11 ES 400 No 11 Bars 11 ES 400 No 11 Bars 1600 30 End Cover 15 Eq Spa 1500 No 6 Bars Side Face Reinforcement 7 Eq Spa 650 No 11 Bars 7 Eq Spa 650 No 11 Bars 11 ES 400 No 11 Bars 11 ES 400 No 11 Bars 15 Eq Spa 1500 No 6 Bars Side Face Reinforcement 075 050 075 075 075 050 y x Figure 427 Reinforcement Details 6 Bottom Mat Reinforcement FHWANHI130126 Design Example 4 Drilled Shaft Footing StrutandTie Modeling STM for Concrete Structures 446 1600 1600 30 End Cover 17 Eq Spa 1526 No 6 Bars 15 Eq Spa 1500 No 6 Bars Side Face Reinforcement 15 Eq Spa 1500 No 6 Bars Side Face Reinforcement 17 Eq Spa 1526 No 6 Bars 050 050 050 050 40 Side Cover y x Figure 428 Reinforcement Details 7 Top Mat Reinforcement FHWANHI130126 Design Example 4 Drilled Shaft Footing StrutandTie Modeling STM for Concrete Structures 447 References 1 Adebar Perry Discussion of An evaluation of pile cap design methods in accordance with the Canadian design standard Canadian Journal of Civil Engineering 316 2004 1123126 2 Adebar Perry Daniel Kuchma and Michael P Collins StrutandTie Models for the Design of Pile Caps An Experimental Study ACI Structural Journal 871 1990 8192 3 Adebar Perry and Luke Zongyu Zhou Design of Deep Pile Caps by StrutandTie Models ACI Structural Journal 934 1996 43748 4 Cavers William and Gordon A Fenton An evaluation of pile cap design methods in accordance with the Canadian design standard Canadian Journal of Civil Engineering 311 2004 10919 5 Park JungWoong Daniel Kuchma and Rafael Souza Strength predictions of pile caps by a strutandtie model approach Canadian Journal of Civil Engineering 3512 2008 1399413 6 Paulay T and Priestley M J N Seismic Design of Reinforced Concrete and Masonry Buildings New York John Wiley and Sons 1992 768 pp 7 Schlaich Jörg Kurt Schäfer and Mattias Jennewein Toward a Consistent Design of Structural Concrete PCI Journal 323 1987 75150 8 Souza Rafael Daniel Kuchma JungWoong Park and Túlio Bittencourt Adaptable StrutandTie Model for Design and Verification of FourPile Caps ACI Structural Journal 1062 2009 14250 9 Widianto and Oguzhan Bayrak Example 11 Deep Pile Cap with Tension Piles SP273 Further Examples for the Design of Structural Concrete with StrutandTie Models Ed KarlHeinz Reineck and Lawrence C Novak Farmington Hills Michigan American Concrete Institute 2011 288 pp 10 Williams CS Deschenes DJ and Bayrak O StrutandTie Model Design Examples for Bridges Implementation Report 5525301 Center for Transportation Research Bureau of Engineering Research University of Texas at Austin 2012 272 pp 11 Windisch Andor Rafael Souza Daniel Kuchma JungWoong Park and Túlio Bittencourt Discussion of Adaptable StrutandTie Model for Design and Verification of FourPile Caps ACI Structural Journal 1071 2010 11920 FHWANHI130126 Design Example 4 Drilled Shaft Footing StrutandTie Modeling STM for Concrete Structures 448 This page is intentionally left blank