BRIDGE RAIL DESIGN PROCEDURE EMAD BADIEE NASIM UDDIN, CHAIR IAN EDWARD HOSCH LEE MORADI

Transcription

1 BRIDGE RAIL DESIGN PROCEDURE by EMAD BADIEE NASIM UDDIN, CHAIR IAN EDWARD HOSCH LEE MORADI A THESIS Submitted to the graduate faculty of The University of Alabama at Birmingham, in partial fulfillment of the requirements for the degree of Master of Science BIRMINGHAM, ALABAMA 2014

2 Copyright by Emad Badiee 2014

3 BRIDGE RAIL DESIGN PROCEDURE EMAD BADIEE CIVIL ENGINEERING ABSTRACT The AASHTO Bridge Specifications recommend a yield line theory analysis to determine the structural capacity of concrete bridge railing based on static strength of concrete. However, this analysis technique has been shown to significantly underestimate the capacity of concrete bridge rails to withstand high speed truck impacts. Traditionally this shortcoming has been mitigated by artificial reductions in bridge rail design loads implemented into the design specifications. Fear of litigation associated with failure of a bridge rail to contain and redirect an errant vehicle has made continuing this policy unacceptable for most state highway agencies. On the other hand, existing barrier design guidelines contained in the Bridge Specifications are based upon National Cooperative Highway Research Program (NCHRP) Report 350. This document has been superseded by the Manual for Assessing Safety Hardware (MASH). The updated performance guidelines incorporate heavier vehicles, higher impact angles, and in one case, higher impact speeds. Full-scale crash testing has shown that the new testing criteria will require stronger and taller barriers. One study has attempted to generate new height and design load requirements for inclusion in the updated Bridge Design Specifications. The load recommended for implementation proved to be extremely high and was not well received by AASHTO s T7 committee on Guardrails and Bridge Rails. Thus, there is a national need for a more thorough evaluation of bridge rail design loads and minimum barrier heights required to meet the MASH guidelines. This thesis presents an improved method iii

4 based on modified Yield Line Theory and dynamic strength of concrete to estimate the design impact loads to realistic levels without adjustment of the underlying analysis technique. The net effect of applying the new method to the design would be large decreases in the size and cost of bridge railing necessary to withstand the elevated loads. The objective of the research proposed herein includes: (1) Developing improved methods for estimating the structural capacity of bridge rails and cantilevered deck systems based on dynamic strength of concrete, and including the contribution of deflection of deck overhang, moment of inertia of the barrier and deck overhang sections, and mass of the vehicle and barrier and (2) Identifying appropriate design loads for use in the new methods that are representative of MASH recommended crash test conditions TL-2 through TL-5. Keywords: NCHRP, MASH, AASHTO, Bridge railing, Yield Line Theory, Moment of Inertia. iv

5 ACKNOWLEDGMENTS I am greatly indebted to my advisor and committee chair, Dr. Nasim Uddin, for all of his time, support, advice, and relentless patience in the development of this project. It has certainly been a rewarding experience, and I am grateful for the opportunity provided to me. As well I wish to thank Dr. Dean L Sicking and Dr. Lee Moradi for their cooperation, expertise, and insight. I also wish to thank Dr. Ian Hosch for all of his time and guidance. Last but not least, I express my gratitude to my family, especially my parents and brothers, for all of their love and support. Finally, I thank UAB for this opportunity, which allowed me to grow personally and professionally. v

6 TABLE OF CONTENTS Page ABSTRACT... iii ACKNOWLEDGMENTS.. v 1 INTRODUCTION Background Problem Statement Objective LITERATURE REVIEW Guardrail Design Barrier Strength WORK METHOD Objective Test Levels Modified Yield Line Method Unit Mass Velocity Distributed Impact Force Dynamic Increase Factor Moment of Inertia Barrier Segment I Segment II Segment III Barrier Section Moment of Inertia Deck Overhang...34 vi

7 TABLE OF CONTENTS (Cont.) Page Displacement Barrier Deck Overhang Superposition Strain Energy Absorption due to Total Horizontal Displacement Moment Capacity of the Barrier Vertical Moment Capacity, Mw Segment I Segment II Segment III Horizontal moment capacity, Mc Segment I Segment II and III Top Beam Moment Capacity Internal Virtual Work along Yield Line, Eyield External Virtual work by Applied Load Critical length of Yield-Line failure pattern, Lc Nominal Railing Resistance to Transverse Loads, Rw ENERGY METHOD Objective Moving Vehicle Energy, SI Barrier Strain Energy vii

8 TABLE OF CONTENTS (Cont.) Page Barrier Strain Energy Capacity in Elastic Region, E Absorbed Energy by the Barrier, Δ Absorbed Energy by Barrier in Plastic Region, Δ Absorbed Energy by Vehicle Deformation, Δ Deck Overhang Strain Energy Deck Overhang Strain Energy Capacity in Elastic Region, E Absorbed Energy by Deck Overhang, Δ LS-DYNA SIMULATION Implementation of LS-DYNA NCAC Model NCAC Single Unit Truck NCAC Rigid Barrier Objective of NCAC Model Proposed Model Deck Overhang NEW JERSEY Concrete Barrier RESULTS AND DISCUSSION Work Method Results Compression Energy Method Moving Vehicle Energy, IS Barrier Strain Energy Barrier Strain Energy Capacity, E Absorbed Energy by Barrier in Elastic Region, Δ Absorbed Energy by Barrier in Plastic Region, Δ Absorbed Energy by Vehicle, Δ viii

9 TABLE OF CONTENTS (Cont.) Page Deck Overhang Strain Energy Deck Overhang Strain Energy Capacity, E Absorbed Energy by Deck Overhang in Elastic Region, Δ Results from Proposed Model SUMMARY AND CONCLSION Work Method Work Method Conclusions Energy Method Energy Method Conclusions LS-DYNA Model LS-DYNA Model Conclusions Recommendations for future studies LIST OF REFERENCES APPENDIX A ix

10 LIST OF TABLES Page Table 1: Test Levels Configurations. (AASHTO, A )... 2 Table 2: AASHTO Specifications Test levels Details. (AASHTO, A ) Table 3: AASHTO Specifications Test Levels Details Table 4: Work Method Results Comparison Table 5: Work Method Results Comparison x

11 LIST OF FIGURES Page Figure 1: Test Level 2 Vehicle Impact. (After Sheikh, 2007)... 3 Figure 2: Test Level 3 Vehicle Impact. (After Bligh, 2010)... 4 Figure 3: Test Level 4 Vehicle Impact. (After Sheikh, 2011)... 5 Figure 4: Concrete NEW JERSEY Type Barrier. (After Barker, 2013)... 6 Figure 5: Front View of Yield Lines Failure Pattern. (After Hirsh, 1978)... 7 Figure 6: Yield Line Pattern due to Collision Force. (After Hirsch, 1978.) Figure 7: External Virtual Work Done by Vehicle Collision. (After Calloway, 1993) Figure 8: Top View of Plastic Hinge for Top Beam. (After Calloway, 1993) Figure 9: Moment Capacities of Barrier Wall. (After Calloway, 1993) Figure 10: Top View prior to the Impact Figure 11: Impact Moment Top View Figure 12: Strain Rate According to Real Loads.(After Pajak, 2011) Figure 13: Strain Rate Effect on Compressive Strength of Concrete.(After Pajak, 2011.) Figure 14: Strain Rate Effect on Tensile Strength of Concrete.(After Pajak, 2011.) Figure 15: 3D View of Impact Force on Barrier Figure 16: Barrier Segments Figure 17: Barrier Top Segment Vertical Rebar in XY Plane Figure 18: Plan View of Barrier Top Segment in ZX Plane Figure 19: Vertical Rebar of Barrier Second Segment in XY Plane Figure 20: Plan View of Barrier Middle Segment in ZX Plane xi

12 LIST OF FIGURES (Cont.) Page Figure 21: Barrier Bottom Segment with Vertical Rebar in XY plane Figure 22: Plan View of Barrier Bottom Segment in ZX Plane Figure 23: SUPER POSITION Method for Impact Force Figure 24: 3D View of Barrier and Deck Figure 25: Deck Overhang Section in ZY Plane Figure 26: Barrier Elastic Deflection due to Vehicle Collision Force Figure 27: Deck Overhang Vertical Deflection Figure 28: Deck Overhang Deflection Configuration Figure 29: Superposition Method for Barrier in Horizontal Deflection Figure 30: Barrier Segment I in XY Plane. (After Barker, 2013) Figure 31: Barrier Segment II in XY Plane. (After Barker, 2013) Figure 32: Barrier Segment III in XY Plane. (After Barker, 2013) Figure 33: Front View of Yield Line Failure Pattern. (After Calloway, 1993) Figure 34: Top View of Yield Line Failure Pattern. (After Calloway, 1993) Figure 35: External Virtual Work by the Impact Load. (After Calloway, 1993) Figure 36: Yield Line Pattern Front View. (After Barker, 2013) Figure 37: Transferred Collision Force Between Barrier and Deck. (After Barker, 2013) Figure 38: Moving Vehicle Kinetic Energy Figure 39: Stress-Strain Curve of Concrete Figure 40: External Virtual Work by the Impact Load. (After Calloway, 1993) xii

13 LIST OF FIGURES (Cont.) Page Figure 41: Initial Vehicle Deformation Figure 42: Final Vehicle Deformation Figure 43: Vehicle Deformation Side View Figure 44: Vehicle Deformation of TL-4. (After Sheikh, 2011) Figure 45: Single Unit Truck Top View Figure 46: Single Unit Truck Side View Figure 47: Single Unit Truck Front View Figure 48: NCAC Barrier Top View Figure 49: NCAC Barrier Front View Figure 50: NCAC Barrier 3D View Figure 51: NCAC Model Top View Figure 52: NCAC Model 3D View Figure 53: NCAC Model Side View Figure 54: Stress-Strain Curve of 60 ksi Steel Figure 55: Section Details. (After Barker, 2013) Figure 56: Model Section Figure 57: Model 3D Section Figure 58: Section Element Formation Figure 59: Model 3D view Figure 60: Model Top View Figure 61: Model Side View xiii

14 LIST OF FIGURES (Cont.) Page Figure 62: Equivalent Distributed Force Figure 63: Change in Kinetic Energy of the Vehicle Based on TL Figure 64: 1 Foot Strip of Deck Overhang Section in ZY Plane Figure 65: Maximum Displacement of Barrier Figure 66: Barrier Maximum Displacement with Scale of Figure 67: Maximum Displacement of Deck Overhang Figure 68: Deck Overhang Maximum Displacement with Scale of Figure 69: Maximum Effective Stress in Barrier Figure 70: 3D View of Barrier Maximum Effective Stress Figure 71: Maximum Effective Stress in Deck Overhang Figure 72: Bottom View of Deck Overhang, Maximum Effective Stress Figure 73: Axial Force Resultant Distribution Figure 74: Rebar Front View xiv

15 1 INTRODUCTION 1.1 Background Recently, the application of using median and parapet to divide the highways and roadways has become an important subject in highway designs. The main object of median is to provide a recovery area for errant vehicles to reach steady-state without interrupting traffic flow. However, many structures have been installed in the median, for instance, bridge supports and piers. Although these types of structures are often located a short distance from the roadway, they can cause serious accidents in the median as well as along the roadway. Barriers, a reasonable distance from these structures are installed in order to minimize hazards and avoid serious accidents. As an example, steel guardrail envelopes are used to protect vehicles from impacting the bridge structure, mostly on piers faces, and upstream. In order to protect the structures from vehicle collision, the design of all barriers and guardrails must be such that the vehicle would not be able to penetrate and pass through the barrier and reach the structure surface. In the case of semirigid barriers, full scale crash testing, based on the AASHTO manual, determines the placement of the barrier (Reid, 2008). There are six test levels defined in AASHTO Standard Bridge Specifications that are vary in type of vehicle, impact angle, and vehicle velocity. Each test level presents a certain vehicle crashing the barrier with certain impact angles and velocities.[a ]. 1

16 Table 1: Test Levels Configurations. (AASHTO, A ) In the table above, W represents the vehicle mass in kips; B is the location of the center of gravity from the front edge of vehicle; θ represents the crash angle in degrees; and TL-1 through 5 represent the test levels. Test level 1 through 3 does not apply to tractor-trailers and trailers. Figure 1 and Figure 2 represent test level 2 and 3 procedures respectively for 4.5 kips pickup trucks crashing the barrier with a velocity of 45 mph and with 25 degrees as the angle of crash. Figure 3 presents AASHTO test level 4. 2

17 Figure 1: Test Level 2 Vehicle Impact. (After Sheikh, 2007) 3

18 Figure 2: Test Level 3 Vehicle Impact. (After Bligh, 2010) 4

19 Figure 3: Test Level 4 Vehicle Impact. (After Sheikh, 2011) Overtime, various types of barriers and parapets have been produced with specified missions and purposes. The main objective of concrete barriers in vehicle 5

20 collision is to redirect and to control the vehicle such that said vehicle would not be able to redirect the flow of traffic. Therefore, the barrier must meet the design criteria necessary to absorb the impact energy during vehicle collision as well as to redirect the vehicle in a controlled manner. For this purpose the barrier must satisfy both geometric as well as strength design. Geometric design is based on the redirection of the vehicle whether it is in a controlled manner or not, whereas strength design depends on the vehicle type and its velocity and its computation is based on traffic flow and test level in formation. NEW JERSEY type concrete barriers are a type of concrete barrier that meets the criteria of TL 4. Figure 4 represents the appropriate dimensions of NEW JERSEY barrier based on TL 4. Figure 4: Concrete NEW JERSEY Type Barrier. (After Barker, 2013) The strength design of the barrier is based on the yield line equation and the limit states. It has been assumed that the vehicle collision produces distributed impact force F t, 6

21 along the length L t, which causes the yield lines failure pattern in the barrier (Barker, 2013). Figure 5: Front View of Yield Lines Failure Pattern. (After Hirsh, 1978) 1.2 Problem Statement The yield line method indicates that the external virtual work done by the applied loads is equal to the internal virtual work done by the resisting moments along the yield lines. The calculated impact force F t, based on the yield line equation, is either equal to or greater than the real impact force of the truck. Also, in case of barriers installed on a bridge deck overhang, since the deck overhang strain energy absorption is neglected, the yield line design does not reflect the actual condition of the impact. In reality large amounts of the vehicle impact energy will be absorbed by the deck overhang vertical deflection. The yield line design calculates the required capacity of the barrier at a higher value than what it should be; therefore, the yield line design can be considered as more conservative. 7

22 1.3 Objective The objective of this research is to first propose a modified yield line method based on work theory by taking into account the effect of deck overhang vertical deflection in yield line method, and to calculate the revised critical length of the barrier in order to calculate the capacity of said barrier. The second objective of this research is to propose an energy based method to determine the contribution of each component in the system in terms of their energy absorption and to compare that to the energy absorption capacity of each component by using the conservation of energy equation for impact due to transferred energy and all losses. The third objective of this research is to develop a LS-DYNA model in order to validate proposed methods, also the model developed can be used as a simulation tool for other aspects of test levels and the interaction of barrier and deck overhang systems in future research. In order to do that, a modified version of yield line analysis with contribution from deck overhang and barrier deflections in the elastic region of section is presented in chapter 3. Since TL-4 is a type of dynamic loading, this procedure is followed by the energy method presented in chapter 4, a method based on conservation of energy. Chapter 5 indicates the simulation of TL-4 in LS-DYNA code with equivalent force applied to the model. In conclusion, the results of all 3 methods are summarized and compared in chapter 7. 8

23 This is the first time that a research has been done on NEW JERSEY concrete barriers capacity, that research proved that the current design is more conservative than what it needs to be. Also for the first time, LS-DYNA model of AASHTO Standard NEW JERSEY barrier has been developed in order to validate the analytical methods to be applied to future research on barriers and cantilever overhang systems as well as different aspects of test levels. 9

24 2 LITERATURE REVIEW 2.1 Guardrail Design Over the years, various types of guardrails with different purposes have been developed. The main purpose of concrete barriers is to absorb collision impact energy and to redirect the vehicle in a controlled manner. Guardrail design consists of geometric design as well as strength design. Geometric design refers to its aesthetic (a branch of philosophy) and also governs the redirecting of the vehicle after the collision whether it is in a controlled manner or not. The strength design of the barriers depends on the size, geometry and velocity of the vehicle and the traffic volume of either the bridge or roadway. For a given condition of the roadway, the barrier strength and performance can be selected from the AASHTO Standard Bridge Specifications [A13.7.2]. Hirsh (1978) analyzed the lateral load capacity for the barriers with uniform thickness. He expressed the strength of the barrier to lateral load based of the formation of yield line analysis and limit states. Yield line implies an assumed failure pattern of the barrier caused by the vehicle impact force F t, over the length of distributed impact force L t. For an assumed yield line failure pattern, the external virtual work done by the applied load - which is the vehicles impact force - must be equate to the internal virtual work done by the resisting moments of the barrier along the yield line. The calculated 10

25 applied load based on the yield line method is either equal to or greater than the actual impact force. Therefore, it is important to minimize the impact load (Hirsch, 1978). Figure 6: Yield Line Pattern due to Collision Force. (After Hirsch, 1978.) Calloway (1993) presented the equation for external virtual work done by the applied load with respect to vehicle horizontal collision force F t, deformation of the barrier in horizontal direction δ, critical length of the barrier L c and the length of distributed collision force L t. Callaway implied that the shaded area of Figure 7 presents the integral of total horizontal deformation through the distributed length of vehicle collision. Figure 7: External Virtual Work Done by Vehicle Collision. (After Calloway, 1993) 11

26 The internal virtual work along the yield lines is the summation of all rotations and moments caused by the barrier displacement. It has been assumed that the barrier acts as a rigid wall so that all the rotations occur in yield line paths. The rotation of the top part of the barrier can be expressed as, Θ = tan Θ= 2δ L c (2-1) The barrier can be divided into 2 segment, top beam and the uniform thickness wall below that. The top beam of the barrier develops plastic moment of M b Figure 8: Top View of Plastic Hinge for Top Beam. (After Calloway, 1993) The horizontal reinforcement of the wall develops moment resistance in vertical direction M w, and the vertical reinforcement of the wall develops moment resistance in horizontal direction M c. These two moment capacities develop inclined moment capacity M α, along the yield line. 12

27 Figure 9: Moment Capacities of Barrier Wall. (After Calloway, 1993) Therefore, the nominal railing resistance is achieved by adding the virtual works provided by moments. The nominal railing resistance to transverse impact collision force can be expressed by equating the external virtual work by applied loads to internal virtual work along the yield lines. In order to minimize the calculated vehicle collision, the yield line equation must be written based on vehicle collision force and be differentiated with respect to critical length. The next step is to set the result equal to zero to find the critical length of the barrier. The vehicle collision force can be achieved by substituting the critical length in the yield line equation. The achieved value of the vehicle collision force is denoted as nominal railing resistance of the barrier to transverse loads R w (Calloway, 1993). 13

28 2.2 Barrier Strength Calloway (1993) investigated barriers strength with non-constant thickness based on yield line approach. The equation for railing resistance was developed by integrating the various moments and rotations along the thickness. Barker (2013) compared Calloway s and Hirsch s approaches in terms of critical length and nominal railing resistance to transverse loads, in order to obtain barrier wall moment capacities. The recommended procedure based on the comparison was to use Hirsch s equations with the average value for vertical and horizontal moment capacities. In case of using the average values in Hirsch s equation, Calloway concluded that the calculated nominal railing resistance to transverse loads is 4% less than the actual nominal railing resistance, which leads to a conservative design (Barker, 2013). 14

29 3 WORK METHOD 3.1 Objective Yield Line approach is an energy method used in AASHTO Standard Bridge Specifications (American Association of State Highway and Transportation Officials), in order to design and to calculate the moment capacity of the barrier. The solution of this approach can be obtained by equating the external virtual work done by the applied loads to the internal virtual work done by resisting moments along the yield lines. W = W yield (3-1) In the equation above, W represents the external virtual work done by applied load and W yield is the internal virtual work done by resisting moments. However, in the case of barriers settled on the bridge deck overhangs, conservative design of AASHTO Standard Bridge Specifications does not reflect the strain energy absorption of the deck overhang in a vertical direction due to the vehicle impact in the Yield Line approach, so that the deck overhang energy absorption during vehicle impact has been completely ignored. Based on AASHTO design criteria, a new modification can be applied to the Yield Line approach due to the strain energy absorbed in the deck overhang during vehicle impact. The modification can be developed by adding a new term Strain Energy Absorption due to the deck overhang deflection to the Yield Line approach. Therefore, the modified equation can be expressed as W W d = W yield (3-2) 15

30 In equation above W d presents the Strain Energy Absorption due to the deck overhang deflection. 3.2 Test Levels Six different test levels are defined in AASHTO Standard Bridge Specifications based on different vehicles, angles and velocities for designing barriers. AASHTO design for NEW JERSEY concrete barriers is based on Test Level four (TL-4) which represents an 18 kips single unit truck hitting the barrier with the velocity of 50 miles per hour and with an angle of impact of 15 degrees. Table 2: Test Levels Configurations. (AASHTO, A ) 16

31 3.3 Modified Yield Line Method Unit Mass Velocity Since vehicle impact is a type of dynamic loading applied to barriers, it is possible to use conservation of momentum and energy based equation, in order to obtain strain energy absorption due to the deck overhang deflection. Momentum is the quantity of motion of a moving body, measured as a product of its mass and velocity. So that the conservation of momentum equation for TL-4 can be written as: m 1 v 1 sin Θ + m 2 v 2 = (m 1 + m 2 ) V (3-3) m 1 represents the mass of the vehicle v 1 is the initial velocity of the vehicle, Θ is the angle of the impact m 2 is the unit mass of the barrier, v 2 is the initial velocity of the barrier before impact which is zero and V is the combined mass velocity perpendicular to the barrier, where the barrier and vehicle can be considered as a unit mass for a short period of time after the impact. In order to calculate the Combined Mass Velocity, the left side of the conservation of momentum equation, which is the total momentum prior to the impact must be divided by the total mass. V= m 1v 1 sin Θ m 1+ m 1 (3-4) 17

32 Figure 10: Top View prior to the Impact Distributed Impact Force Impulse is change in momentum over time and can be expressed for both barrier and vehicles as well. Vehicle impulse is: m 1 v 1 + F t d t = m 1 V (3-5) And for the barrier is: m 2 v 2 + F t d t = m 2 V (3-6) Where F t is the vehicle impact force perpendicular to the barrier, and d t the initial impact duration is equal to Δt, the time that the barrier and vehicle can be considered as a unit mass. d t = Δt (3-7) 18

33 So that by rewriting equation (3-6), m 2 v 2 + F t Δt = m 2 V (3-8) In order to calculate the impact force, F t, barrier energy before impact must go to the right side of the equation, F t Δt = m 2 V m 2 v 2 (3-9) Next divide the equation by the impact time, Δt. F t = m 2V m 2 v 2 Δt (3-10) Since the barrier does not have any motion prior to impact, the term m 2 v 2 is equal to zero and the equation can be rewritten as: F t = m 2V Δt (3-11) The Combined Mass Velocity V, has already been calculated. Figure 11: Impact Moment Top View 19

34 3.3.3 Dynamic Increase Factor Pajak (2011) proposed that concrete in compression, for all range of strain rates must be investigated in two domains of strain rate. The first domain is the strain rate that the answer is changing which is called the transition strain rate. In this region, the dynamic increase factor, DIF, can achieve up to 1.8 DIF is defined as the ratio of dynamic strength to quasi static strength. The second domain is denoted as pronounced strength where DIF in this domain is equal to 3.5. There is a shift in DIF for higher strain rates ( 10 1/s). The sensitivity of concrete in tension is significantly different than in compression. The DIF factor in tension can reach 13. Generally concrete behavior under different tensile strain rates is more uniform than in compression. In overall view, the author implied that the behavior of concrete in tension and compression in smaller strain rates (up to 10 1 ) can be considered the same and that the significant difference starts at higher strain rates. The author also proposed that the size and geometry of concrete specimens probably do not have any effect on strain rate. The values of DIF factors are presented based on normal strength concrete. In order to reflect the dynamic loading in the analysis for concrete, modulus of elasticity and compressive strength of the concrete must be multiplied by the DIF factor. The first step is to determine the strain rate based on the loading type. The author proposed a chart to determine the strain rate in dynamic loading based on loading type. As shown in Figure 12 the chart represents that the strain rate, according to real loads, is approximately 10 4 to 10 3 for vehicle impact (Pajak, 2011). 20

35 Figure 12: Strain Rate According to Real Loads.(After Pajak, 2011) Strain rate has effect on both compressive and tensile strength of concrete. Figure 13 and Figure 14 present this effect on compressive and tensile strength of concrete respectively. Figure 13: Strain Rate Effect on Compressive Strength of Concrete.(After Pajak, 2011.) 21

36 Figure 14: Strain Rate Effect on Tensile Strength of Concrete.(After Pajak, 2011.) Change in strain rate curves for both tensile and compressive strength of concrete, is approximately equal up to 10 1 [1/s] so that the DIF is approximately 1.05 to 1.20 and in this study, it has been conservatively assumed that the DIF is equal to 1.05 (Pajak, 2013). Average DIF=1.05 E new = 1.05 E / f c new = 1.05 f c / (3-12) 22

37 3.3.4 Moment of Inertia Barrier Since the impact force, F t, is perpendicular to the barrier face, the barrier moment of inertia must be calculated about the rotation axis which is the Z axis in the XZ plane. Also the barrier thickness varies from top to bottom (Y axis), so that, in order to calculate the moment of inertia, the barrier must be divided into three segments with three individual moments of inertia and must use the weighted-mean for the total moment of inertia for the section. Figure 16 represents different segments of a barrier. Figure 15: 3D View of Impact Force on Barrier 23

38 Figure 16: Barrier Segments Segment I Figure 17 represents the top segment as a trapezoidal consists of two heights, 6 inches at the top and 8 inches at the bottom. In order to calculate the moment of inertia in ZX plane, the segment must be simplified to a rectangular segment with height of the average of the trapezoid s two heights. Also the only rebar contributing to strain energy absorption about the Z axis are vertical rebar and stirrups. Modified height = = 7 in. 24

39 Figure 17: Barrier Top Segment Vertical Rebar in XY Plane Since the strain energy will be calculated for a unit length of the barrier, the length of the section is equal to 1 foot (12 inches), so that, based on Standard NEW JERSEY type barrier; the modified segment will be a rectangle with 7 inches in height and 12 inches in width. Figure 18 represents the modified section. 25

40 Figure 18: Plan View of Barrier Top Segment in ZX Plane The distance between the bottom edge of the section to the center of gravity of the bottom rebar, based on 2 inches cover, is 2.25 inches and the distance from the bottom edge to the center of gravity of the top rebar is approximately 4.75 inches; also the area of No.4 rebar is 0.20 in 2. The total amount of steel area used in this section is (2 0.20) + ( ) = 0.8 in2 In order to calculate the center of gravity, the steel area must be transformed to concrete. Therefore, n, which is the ratio between the modulus of elasticity of steel to the modulus of elasticity of concrete, must be multiplied by the steel area. Since rebar is 60 ksi steel, n in expressed as, n= E steel ksi = = E concrete ksi The center of gravity of the section is, CG = Ad A (3-13) 26

41 A represents the area of rebar and concrete respectively and d represents the distance between the center of gravity and the bottom edge of the section of rebar. Therefore, CG = ( A d) rbr.+( A d) conc. A rbr.+a conc. (3-14) So that, ( )+( )+( ) CG = = 3.5 in. ( )+( )+(7 12) Moment of inertia of the section is equal to, I = b h (A d2 ) conc. + (A d 2 ) rbr. (3-15) The distance from the center of gravity of the concrete to the section center of gravity is zero, so that the second term of the above equation is equal to zero. I= in4 ft + [ ( ) 2 ] + [ ( ) 2 ] = In order to convert the moment of inertia to ft4 ft, it must be multiplied by So, I = in4 ft = ft4 ft Segment II 27

42 The figure below represents the mid segment of a trapezoid with heights of 8 inches at the top and 15 inches at the bottom. In order to calculate the moment of inertia in the ZX plane, the segment must be simplified to a rectangular segment with its height the average of the trapezoid s two heights. Figure 19: Vertical Rebar of Barrier Second Segment in XY Plane The only rebar contributing in strain energy absorption about the Z axis are vertical rebar and stirrups in the Y direction; also, the effect of stirrups are negligible since that effect is small. Modified height = = 7 in. As was the case with the top section, the strain energy will be calculated for a unit length of barrier. The length of the section will be equal to one foot (12 inches), so that based on AASHTO Standard NEW JERSEY type barrier, the modified segment will be a rectangle with 11.5 inches in height and 12 inches in wide. Figure 20 presents the modified section. 28

43 Figure 20: Plan View of Barrier Middle Segment in ZX Plane The distance between the bottom edge of the section to the center of gravity of the bottom rebar, based on 2 inches cover, is 2.25 inches and the distance from the bottom edge to the center of gravity of the top rebar is approximately 9.25 inches, and the area of No.4 rebar is 0.20 in 2. The total amount of steel area used in this section is, (2 0.20) + ( ) = 0.8 in2 The center of gravity of the section is, CG = Ad A (3-16) A represents the area of rebar and concrete respectively and d represents the distance between the center of gravity and the bottom edge of the section for rebar. Therefore; CG = ( A d) rbr.+( A d) conc. A rbr.+a conc. (3-17) 29

44 So that, ( )+( )+( ) CG = = 5.75 in. ( )+( )+( ) The moment of inertia of the section is, I = b h (A d2 ) conc. + (A d 2 ) rbr. (3-18) The distance from the center of gravity of concrete to the section center of gravity is zero, so, the second term of the equation above is zero. I= [ ( ) 2 ] + [ ( ) 2 ] = in4 ft In order to convert the moment of inertia to ft4 ft, it must be multiplied by I = in4 ft = ft4 ft Segment III The figure below represents the bottom segment as a rectangle with a height of 15 inches. 30

45 Figure 21: Barrier Bottom Segment with Vertical Rebar in XY plane. The only rebar contributing in strain energy absorption about the Z axis in the ZX plane are vertical rebar and stirrups in the Y direction. The effect of stirrups is negligible. It is the same as top and mid sections. The strain energy will be calculated for a unit length of the barrier. The length of the section will be equal to one foot (12 inches), and based on AASHTO Standard NEW JERSEY type barrier; the segment will be a rectangle with 15 inches in height and 12 inches in wide. Figure 22 presents the modified section. 31

46 Figure 22: Plan View of Barrier Bottom Segment in ZX Plane The distance between the bottom edge of the section to the center of gravity of the bottom rebar, based on 2 inches cover, is 2.25 inches and the distance from the bottom edge to the center of gravity of the top rebar is approximately inches, and the area of No.4 rebar is 0.20 in 2. The total amount of steel area used in this section is, (2 0.20) + ( ) = 0.8 in2 The center of gravity of the section is, CG = Ad A (3-19) A represents the area of rebar and concrete respectively and d represents the distance between the center of gravity and the bottom edge of the section of rebar. Therefore, 32

47 CG = ( A d) rbr.+( A d) conc. A rbr.+a conc. (3-19) So that, ( )+( )+( ) CG = = 7.5 in. ( )+( )+(15 12) Moment of inertia of the section is, I = b h (A d2 ) conc. + (A d 2 ) rbr. (3-20) The distance from center of gravity of the concrete to the section center of gravity is zero, so that, the second term of the equation above is equal to zero. I= in4 ft + [ ( ) 2 ] + [ ( ) 2 ] = In order to convert the moment of inertia to ft4 ft, it must be multiplied by I = 3375 in4 ft = ft4 ft Barrier Section Moment of Inertia In order to calculate section moment of inertia, a weighted-mean method of three segment s moment of inertia is used. The top, middle and bottom segments height of the barrier are 21, 10 and 3 inches respectively. So that, 33

48 I 2 = = ft4 ft Deck Overhang Since the initial location of impact force is on the top and perpendicular to the barrier s face, in order to calculate the strain energy absorption of the deck overhang, Impact force must be translated to the deck overhang. The SUPER POSITION method indicated that by translating the impact force from the top of the barrier to midpoint of the deck overhang thickness, a clockwise moment will be produced and is applied to the midpoint of the deck overhang thickness. Its magnitude is barrier height H, plus half of deck overhang thickness t, times impact force F t. Figure 23 represents the SUPER POSITION method procedure. Figure 23: SUPER POSITION Method for Impact Force The deck overhang moment of inertia I 1, must be calculated about the rotational axis of the deck overhang which is the Z axis in ZY plane. The calculations for its moment of inertia, are delineated on subsequent pages. 34

49 Figure 24: 3D View of Barrier and Deck Since the strain energy is calculated for a unit length of the deck overhang, the length of the section is equal to one foot (12 inches), so that, based on AASHTO Standard bridge deck; the deck overhang segment will be a rectangle with 9 inches in height and 12 inches in wide. Figure 25 presents the deck overhang section. Figure 25: Deck Overhang Section in ZY Plane 35

50 The distance between the bottom edge of the section to the center of gravity of the bottom rebar, based on 2 inches cover, is inches and the distance from the bottom edge to the center of gravity of the No.5 top rebar is = inches; also, the distance from the bottom edge to the center of gravity of the No.3 top rebar is = inches, based on 2.5 inches cover from the top. The total amount of steel area used in this section is, (4 0.31) + (2 0.11) = 1.46 in 2 2 Also the center of gravity of the section is, CG = Ad A (3-21) A represents the area of rebar and concrete respectively and d represents the distance between the center of gravity and the bottom edge of the section for rebar. Therefore, CG = ( A d) rbr.+( A d) conc. A rbr.+a conc. (3-22) So that, ( )+( )+( )+( ) CG = ( )+( )+( )+(9 12) = 5.85 in. The moment of inertia of the section is I = b h (A d2 ) conc. + (A d 2 ) rbr. (3-23) 36

51 The distance from the center of gravity of the concrete and the section center of gravity is zero. Therefore, the second term in the equation above is equal to zero. I= [9 12 ( ) 2 ] +[ ( ) 2 ] + [ ( ) 2 ] + [ ( ) 2 ] I = in4 ft In order to convert the moment of inertia to ft4 ft, it must be multiplied by So, I 1 = in4 ft = ft4 ft Barrier Displacement In order to calculate the horizontal displacement of the barrier caused by barrier deflection Δ 2, the barrier is assumed to act as a cantilever beam with vehicle impact force on its span. The location of impact load depends on the location of the center of gravity of the vehicle. Δ 2 = F ta 2 6EI barrier (3H a) (3-24) In the equation above, a represents the effective length of the barrier where force is applied, E is the modulus of elasticity and H is the barrier height. 37

52 Figure 26: Barrier Elastic Deflection due to Vehicle Collision Force Since the vehicle is Single Unit Truck and its center of gravity is 49 inches above the ground, the collision force is applied to the top of the barrier with a height of 34 inches. So a = H; therefore, Δ 2 = F th 3 3EI barrier = F th 3 3EI 2 (3-25) Deck Overhang It has been assumed that the deck deflects vertically from its edge to the outer edge of the nearest girder; however, the girders are considered to be rigid, so that the deflection in vertical axis is negligible. The figure below represents the horizontal deflection of the barrier due to vertical deflection within the deck overhang. 38

53 Figure 27: Deck Overhang Vertical Deflection Figure above I 1 represents the deck overhang moment of inertia. I 2 is the barrier moment of inertia, l is the effective length of deck overhang, and M the produced moment which causes the vertical deflection in the deck overhang. In order to calculate Δ 1 - the horizontal displacement of the barrier caused by deck overhang deflection the deck overhang is assumed to be cantilever a beam with a moment produced by the Vehicle impact force on the free end. The magnitude of this moment is calculated as M=F t a, and since the vehicle is Single Unit Truck, a is the barrier height (H plus half of the deck overhang thickness t). Therefore, in order to calculate the horizontal displacement of the barrier, the deflection angle of the deck overhang must be calculated. Θ= Ml = F t t(h + 2 ) l EI EI 1 (3-26) I 1 represents the moment of inertia of the deck overhang and l stands for the overhang length. 39

54 Figure 28: Deck Overhang Deflection Configuration Based on geometry, deck overhang deflection Angle θ, is equal to the barrier displacement angle in the horizontal direction, so that, Sinθ = Δ 2 H+ t 2 (3-27) Therefore Δ 2 is: Δ 2 = (H + t )Sinθ (3-28) 2 And by substituting θ, Δ 2 = (H + t t(h + t 2 )Sin(F 2 ) l ) (3-29) EI 1 Since deck overhang deflection angle θ is a small angle, Sinθ = θ, Δ 2 = H + t 2 Sin F t(h + t 2 ) l EI 1 = H+ t 2 2 F t l (3-30) EI 1 40

55 Superposition The Total displacement in the horizontal axis was calculated by using the super position method which is expressed as The total displacement in horizontal direction perpendicular to the barrier face (Δ), is equal to horizontal displacement of the barrier perpendicular to its face before cracking (Δ 2 ), plus horizontal displacement of the barrier perpendicular to its face due to vertical displacement of the deck overhang ( Δ 1 ). Figure 29 represents the super position procedure. Figure 29: Superposition Method for Barrier in Horizontal Deflection Therefore, the total displacement of the barrier in horizontal axis Δ, is, Δ =Δ 2 + Δ 1 = F th 3 3EI 2 + H+ t 2 2 F t l (3-31) EI Strain Energy Absorption due to Total Horizontal Displacement The only unknown variable in the horizontal displacement equation is collision force F t, therefore the force-displacement curve is linear and the amount of strain energy absorbed due to this displacement is the area under the curve and can be expressed as, W d = 1 F 2 tδ (3-32) 41

56 By substituting Δ, W d = F t 2 H 3 + F t 2 l H+ t 2 2 (3-33) 6EI 2 2EI Moment Capacity of the Barrier Vertical Moment Capacity, M w The horizontal reinforcement of the barrier provides moment capacity in the vertical axis that is represented by M w. Based on the vehicle impact strain rate, the compressive strength of the concrete must be multiplied by the DIF factor. / f c new = = 4.2 ksi Since the barrier thickness varies from top to bottom, in order to calculate M w, the barrier section must be divided into 3 segments Segment I The positive and negative moment capacities of the top segment are approximately equal and can be calculated as, A s =2- No. 3 / s =2(0.11) = 0.22 in. 2 d avg = = 3.56 in. a= A s f y 0.85 DIF f c / b = = in. ΦM n1 = ΦA s f y (d a 2 ) = k.ft 42

57 Figure 30: Barrier Segment I in XY Plane. (After Barker, 2013) Segment II For this segment, the moment capacity is more complicated. Positive moment is, A s =1- No. 3s= 0.11in. 2 d pos = = 6.75 in. a= = in. ΦM npos = ( )= k.ft 2 Figure 31: Barrier Segment II in XY Plane. (After Barker, 2013) 43

58 And for negative moment, d neg = = 6 in. ΦM nneg = ( )= k.ft 2 So that, the average value of positive and negative moment capacities is, ΦM n2 = ΦM npos+φm nneg 2 = k.ft Segment III For this segment the positive and negative moment capacity are equal and A s =1- No. 3s= 0.11in. 2 d = = in. a= = in. ΦM n3 = ( ) 2 = k.ft Figure 32: Barrier Segment III in XY Plane. (After Barker, 2013) The total moment capacity of the barrier about the vertical axis is the sum of moments in all three segments. 44

59 M w = ΦM n1 + ΦM n2 + ΦM n3 = k.ft Horizontal moment capacity, M c The vertical reinforcement of the barrier provides moment capacity in a horizontal direction which is represented by M c.since the barrier thickness varies from top to bottom, in order to calculate M w, the barrier must be divided into 3 segments (the same segments as in previous figures). The yield lines that cross the vertical reinforcement produce tension; therefore, only negative moment capacity needs to be calculated. Figure 33: Front View of Yield Line Failure Pattern. (After Calloway, 1993) Segment I A s = 0.39 in. 2 d = = 4.75 in. a= A s f y / = 0.85 DIF f c b in. M C1 = ΦA s f y (d a ) = ( ) = k.ft

60 Segment II and III A s = 0.39 in. 2 d = = 8.75in. a= A s f y 0.85 DIF f c / b = = in. M C I+II = ΦA s f y (d a ) = ( ) = k.ft 2 2 The weighted-mean method was used in order to calculate the total moment capacity in the horizontal direction of the barrier, M c = M c1 (21)+M c2( ) 34 = k.ft/ft Also, in order to develop the horizontal moment capacity, M c, through the whole length, moment capacity along the Z-axis, must be multiplied by barrier length. So that, M c =(M c /ft)* length (ft) (3-34) Top Beam Moment Capacity Since NEW JERSEY type barrier does not have a top beam, the moment capacity along the top beam,m b, is equal to zero Internal Virtual Work along Yield Line, E yield The internal virtual work along the yield lines is the sum of rotations and moment capacities through the path within which they act. At the top segment of the wall, the rotation θ is 46

61 θ tan θ = 2δ L c (3-35) Figure 34: Top View of Yield Line Failure Pattern. (After Calloway, 1993) Assuming the negative and positive plastic moment capacities are equal, the internal virtual work done by the top beam would be, U b =4M b Θ = 8M bδ L c (3-36) As mentined before, NEW JERSEY type barrier does not have a top beam, therefore U b = 0. The internal virtual work done by horizontal rebar can be expressed as, U w =4M w Θ = 8M wδ L c (3-37) The projection of displacement on the vertical path about the inclined yield line is δ H ; therefore, the internal virtual work done by the vertical rebar U c can be expressed as U c = M cl c δ H (3-38) And the total internal virtual work along yield lines is W yield = 8M bδ L c + 8M wδ L c + M cl c δ H (3-39) 47

62 3.3.9 External Virtual work by Applied Load Figure 35 presents the deformations of the barrier top beam with respect to its original position. The shaded area represents the integral of this deformation through which the distributed vehicle collision force w t =F t /L t, acts. Based on that, for the total displacement δ, the displacement x along the length is, X= L c L t L c δ (3-40) Figure 35: External Virtual Work by the Impact Load. (After Calloway, 1993) L t is the length of distributed impact force F t, and L c is the barrier critical length. The shaded area can be presented as, Area= 1 2 (δ + X)L t = δ L t L c (L c L t 2 ) (3-41) The external virtual work done by distributed impact force w t, is expressed as, W = w t (area)= F t Δ L c (L c L t 2 ) (3-42) 48

63 Critical length of Yield-Line failure pattern, L c As already discussed at the beginning of the chapter, in modified yield line, deck overhang and barrier act as individual cantilever beams with fixed ends bonded together with a 90 degree angle at the deck overhang. In order to obtain more accurate results, barrier and deck overhang lengths must be minimized to act more like cantilever beams. Therefore, modified yield line approach becomes, W - W d = W yield (3-43) And by setting the above equation to zero, modified yield line approach can be expressed as, W - W d - W yield = 0 (3-44) By substituting all the terms, F t β 2 8 M wβ L c 8 M bβ L c F tl t β L c M c β 2 L c H = 0 (3-45) Where, β = F tb 2 H 2 E I 2 F tb 3 6 E I 2 + F t b H L 2 E I 1 + F t b l t 2 E I 1 (3-46) In equations above, F t is the transverse impact force represented in kips; m 1 is the vehicle mass presented in kips; m 2 is the total mass of barrier presented in kips; 49

64 b is the vehicle center of gravity height represented in feet; L t is the length of the distributed vehicle collision force presented in feet; L c is the critical length of the barrier represented in feet; l is the effective deck overhang length represented in feet; H is the barrier height represented in feet; Δ is the total displacement of the barrier in the horizontal direction caused by vehicle impact and represented in feet; t is the deck overhang thickness represented in feet; E is the modulus of elasticity of the concrete represented in kips per square feet (ksf); I 1 is the moment of inertia of the deck overhang section about rotation axis represented in ft 4 ; I 2 is the moment of inertia of the barrier section about rotation axis represented in ft 4 ; V is the combined mass velocity after the impact represented in mile per hour (mph); M w is the moment capacity of the barrier along vertical axis provided by reinforcement in the horizontal direction presented in kips-ft; M c is the moment capacity of the barrier along horizontal axis provided by reinforcement in the vertical direction represented in kips-ft; M b is the barrier top beam moment capacity represented in kips-ft; θ is the angle of impact represented in radians; Δt is the impact duration which the vehicle and barrier are considered as a unit mass with the same velocity presented in second. In order to find the critical length of the yield line pattern L c, the modified yield line equation must be solved for the impact force, F t. So that, 50

65 F t = 2(M cl c 2 +8HMw +8HM b ) H (L t L c ) (3-47) The only unknown variable in this equation needed to determine the inclination of yield line, α, is the critical length of the barrier. The value of the critical length of barrier L c, that minimizes the F t is determined by differentiating this equation, with respect to the critical length of the barrier, and equating the result to zero. df t dl c = 0 (3-48) And by substituting the terms, df t = 2 M 2 cl c + 16 H Mw + 16 H M b dl c H ( L t L c ) 2 4 L c M c H ( L t L c ) = 0 (3-49) This differentiation results in two quadratic results that must be solved for critical length of barrier L c, L c1 = L tm c + M c (M c L 2 t +8HM w +8HM b ) (3-50) M c And L c2 = L tm c M c (M c L 2 t +8HM w +8HM b ) (3-51) M c and by substituting variables, the second equation for critical length always has a negative value. Therefore; L c = Max {L c1, L c2 } (3-52) 51

66 Figure 36: Yield Line Pattern Front View. (After Barker, 2013) Nominal Railing Resistance to Transverse Loads, R w Minimum value of F t is obtained by substituting the calculated value of L c in the collision force equation. This value is denoted as nominal railing resistance to transverse loads, R w, Min F t = R w (3-53) According to AASHTO Bridge Specifications, nominal railing resistance to transverse load R w, must be greater than allowable collision force F t, for the specified test level. The following equation describes their relationship: R w > F t (TL) (3-54) 52

67 Table 2: AASHTO Specifications Test levels Details. (AASHTO, A ) The nominal railing resistance R w, has been transferred through a cold joint by shear friction. Figure 37 represents the free-body diagram of this procedure. Assuming the nominal railing resistance R w, sheared out with the slope of 1 : 1 from the critical length of the barrier L c, the shear which is produced by the vehicle impact force V CT, or Tensile force T, at the bottom of the barrier is calculated as, V CT = T = R w L c +2H (3-55) 53

68 Figure 37: Transferred Collision Force Between Barrier and Deck. (After Barker, 2013) And based on AASHTO Bridge Specifications, the nominal shear resistance of the interface plane V n, is the minimum of, V n = min { c A cv + µ A vf F y + P c, K 1 DIF f c / A cv, K 2 A cv } (3-56) where: A cv is the shear contact area which is: = 180 in.2 ft 54

69 A vf is the dowel area across shear plane which is 0.39 in.2 ft ; c is the cohesion factor which is based on [A ] ksi; f c / is the compressive strength of the weaker concrete which is 4 kips per square inches (ksi); F y is the yield strength of reinforcement which is 60 ksi; P c is the permanent compressive force which is equal to 0.320kips/ft; µ is the friction factor based on [A ] which is 0.6; K 1 is the fraction of concrete strength available to resist interface shear which as specified in [A ] is 0.2; K 2 is the limit interface shear resistance factor based on [A ] which is 0.8 ksi Factors c, µ, K 1 and K 2 are for normal concrete placed against hardened concrete, clean and without any laitance but not roughened. So that for one foot of design of barrier, K 1 DIF f c / A cv = 2(1.05)(4)(180)=144 kips/ft K 2 A cv = 0.8(180)= 144 kips/ft ca cv + µ A vf f y + P c =0.075(180) [0.39(60) ]= =27.73 kips/ft Therefore: V n = min {27.73, 144, 144} = kips/ft The nominal shear resistance must be greater than the shear produced by the truck collision V CT. (Barker, 2013) 55

70 V n > V CT (TL) (3-57) Results are presented in chapter six. 56

71 4 ENERGY METHOD 4.1 Objective The first objective of this chapter is to propose a method that involves the conservation of energy from the impact. Based on that, the total energy interacting in the system before the impact - Impact Severity - must be equal to the sum of total energy absorbed by each component due to all deformations and displacements in the system and all losses. The second objective of this chapter is to calculate the energy absorption capacity of barrier and cantilever overhang in elastic regions in order to compare them with the amount of energy absorbed by each component in order to investigate whether the barrier or overhang fails under TL-4 conditions or not, and if they do not fail, whether they exceed their elastic limit and reach the plastic limit or not. In this method it has been proposed that the moving vehicle energy prior to the impact perpendicular to the barrier - Impact Severity - is less than the sum of the following: - strain energy absorbed by the barrier in elastic region of concrete Δ 1 - strain energy absorbed by the barrier in the plastic region of concrete Δ 2 - energy absorption by the vehicle due to its deformation Δ 3 - strain energy absorbed by deck overhang in elastic region of concrete Δ 4. Therefore: IS < (Δ 1 or Δ 2 )+Δ 3 +Δ 4 (4-1) 57

72 4.2 Moving Vehicle Energy, SI Since the vehicle has translational displacement due to constant velocity, the total energy of the moving vehicle prior to the impact can be expressed as its overall kinetic energy. This kinetic energy has 2 components to the barrier, perpendicular and parallel. The only component that interacts with the barrier and deck overhang system is the perpendicular component of the vehicle s kinetic energy (IS). Based on TL-4, with 15 degree angle of crash, kinetic energy can be expressed as, IS = w 1 v 2 sin 2 θ 2g c (4-2) In equation above, w is vehicle mass in pound, v is the initial velocity, θ is the impact angle, and g c is the gravity acceleration. 58

73 Figure 38: Moving Vehicle Kinetic Energy 4.3 Barrier Strain Energy Barrier Strain Energy Capacity in Elastic Region, E 1 Figure 39 presents the stress-strain curve of concrete. As is shown, concrete in tension starts with linear behavior followed by nonlinear behavior after cracks occur. The red dot represents the change in phase of concrete. The strain energy capacity of reinforced concrete in linear zone,e 1, based on section properties can be expressed as, l l = M c 2 dx 0 2E I 2 0 2E I 2 E 1 = M B 2 dx (4-3) l is the barrier height, M B represents the moment capacity of the barrier from top view which is the same as M c, E is the modulus of elasticity of the concrete, and I 2 is the barrier moment of inertia. If the total applied energy on the section exceeds strain energy capacity, concrete starts to crack, which is the introduction to nonlinear behavior. 59

74 Figure 39: Stress-Strain Curve of Concrete Absorbed Energy by the Barrier, Δ 1 Conservation of momentum for the impact can be expressed as, m 1 v 1 sinθ + m 2 v 1 = (m 1 + m 2 ) V (4-4) In the equation above, m 1 represents the mass of the vehicle, v 1 is the vehicle initial velocity, m 2 is the mass of barrier, and V represents the combined mass velocity where the barrier and vehicle are considered as a unit mass. In the conservation of momentum equation the barrier weight is 0.32 kips/ft and the barrier length assumed to be the same as vehicle length is 26.6 feet, Therefore, the barrier mass calculates as, m 2 = = 8.83 kips = 8837 lb By calculating V from the conservation of momentum equation, the kinetic energy absorbed by the barrier just after the impact can be expressed as, Δ 1 = E barrier, after impact = w 2V 2 (4-5) 2g c If this amount energy exceed the capacity of the section in elastic limit, the cracks start to occur which is the introduction to the plastic region of the section. 4.4 Absorbed Energy by Barrier in Plastic Region, Δ 2 As discussed before, concrete in tension has linear behavior with elastic deflection followed by occurring cracks which is the introduction to nonlinear behavior. Δ 2 60

75 represents the strain energy absorbed by the barrier in the nonlinear zone. Since AASHTO Standard Bridge Specifications designs NEW JERSEY concrete barrier based on a virtual failure pattern, which represents the cracks and failure modes in concrete, the same approach can be used in this step as well. Based on Yield Line approach, barrier strain energy in nonlinear zone can be expressed as internal virtual work along yield lines, Δ 2 = 8M bδ L c + 8M wδ + M cl c δ L c H (4-6) In the equation above, M b represents the top beam moment capacity, M w represents the vertical moment capacity of the wall provided by horizontal rebar, M c is the horizontal moment capacity of the wall provided by vertical rebar, δ is the horizontal displacement of the barrier due to vehicle collision force, L c is the critical length of the barrier, and H is the barrier height. Figure 40: External Virtual Work by the Impact Load. (After Calloway, 1993) Since NEW JERSEY concrete barrier does not have a top beam, top beam moment capacity is equal to zero. Also, based on AASHTO yield line design, the critical 61

76 length of the barrier L c, is equal to 7.17 feet and M w and M c are kips-ft and kips-ft/ft (Barker, 2013). The only unknown variable in the above equation is horizontal displacement of the barrier due to vehicle collision force and in order to calculate that, LS-DYNA code, based on Finite Element Method, was used by setting equivalent distributed force to the barrier. The perpendicular component of vehicle impact force to the barrier is 54 kips which is distributed to a length of 3.5 feet. The position of this force is the same position as the center of gravity of the single unit truck which is 49 inches from the ground. Since the NEW JERSEY barrier height is 34 inches, the force will apply to a defined area on top of the barrier. AASHTO Standard Bridge Specifications recommend that the length of this distributed load is 3.5 feet and based on the truck geometry the height of this area is 0.41 foot. Initial assumption is that the barrier will not exceed the elastic limit and the total energy absorbed by the barrier will be absorbed in the elastic region of section. By having the elastic capacity and calculating the energy transferred to the barrier, it will be possible to discuss whether Δ 2 must be calculated or not. 4.5 Absorbed Energy by Vehicle Deformation, Δ 3 There are two method proposed in this section in order to calculate absorbed energy by vehicle deformation after the impact. The first method is using the National Crash Analysis Center (NCAC) model provided in LS-DYNA code for TL-4 and the second method has been proposed based on the conservation of energy equation for the system before and after the impact. Figure 41 to Figure 43 represent the first method which is vehicle deformation in LS-DYNA code based on National Crash Analysis 62

77 Center (NCAC) model for TL-4 which will be discussed in details in chapter 5 (NCAC, 2008). Figure 41: Initial Vehicle Deformation 63

78 Figure 42: Final Vehicle Deformation 64

79 Figure 43: Vehicle Deformation Side View The Second method proposes that vehicle deformation absorbs a large amount of energy which means that this amount of energy will not be absorbed by the barrier or deck overhang but will be absorbed directly by the deformation of the vehicle. Figure 44 presents a sample of vehicle deformation due to impact. This procedure is described as follows: 65

80 Figure 44: Vehicle Deformation of TL-4. (After Sheikh, 2011) By rewriting the kinetic energy equation for the vehicle perpendicular to the barrier after the impact and taking into account the combined mass velocity, E vehicle, after impact = w 1V 2 (4-7) 2g c Since the velocity of the vehicle perpendicular to the barrier after impact becomes zero, this energy must be absorbed by the vehicle due to its deformation. Therefore, Δ 3 = E vehicle, after impact = w 1V 2 (4-8) 2g c 4.6 Deck Overhang Strain Energy Deck Overhang Strain Energy Capacity in Elastic Region, E 4 The vertical displacement of the deck overhang presents strain energy absorption. This energy cannot exceed more than deck overhang section capacity; otherwise the 66

81 deformation passes the elastic limit and starts to crack, which is the introduction to the plastic limit. Therefore, the maximum moment capacity of the deck overhang must be used in order to calculate the maximum allowable energy absorption by the deck overhang. So that Δ 4 can be expressed as, l 0 2E I 1 E 4 = M D 2 dx (4-9) l is the length of deck overhang, M D represents the moment capacity of the deck overhang from side view, E is the modulus of elasticity of the concrete, and I 1 is the deck overhang moment of inertia. As discussed before, AASHTO Standard Bridge Specifications indicates that, based on TL-4, the impact load will be distributed to a length of 3.5 feet. Therefore, the moment capacity and moment of inertia of the deck overhang must be calculated for 3.5 feet Absorbed Energy by Deck Overhang, Δ 4 Based on conservation of energy for the barrier, deck overhang, and vehicle system due to impact, IS = (Δ 1 or Δ 2 ) +Δ 3 +Δ 4 (4-10) As already discussed, it has been assumed that the barrier does not exceed the elastic limit and does not crack. Therefore: Δ 2 = 0 (4-11) 67

82 Therefore, the conservation of energy can be rewritten as, IS = Δ 1 +Δ 3 +Δ 4 (4-12) so that, absorbed energy by deck overhang Δ 4, can be expressed as, Δ 4 = IS ( Δ 1 +Δ 3 ) (4-13) Results are presented in chapter six. 68

83 5 LS-DYNA SIMULATION The objective of this chapter is to develop a simulation tool by LS-DYNA code for TL-4 and compare the results obtained from the simulation and analytical methods (Work and energy methods) in order to verify the results of the analytical methods. After testing this simulation tool experimentally, it can be used in future research in order to validate different case scenarios and different test level aspects such as vehicle speed, type and angle of impact. 5.1 Implementation of LS-DYNA In order to simulate TL-4, it is important to use accurate data. Finite element simulation of vehicle impact was modeled by LS DYNA code. This model represents the NEW JERSEY barrier settled on the edge of an AASHTO standard bridge deck. The simulation consists of two parts which are explained in details. 5.2 NCAC Model The first model was obtained from the National Crash Analysis Center (NCAC) website that represents TL-4. NCAC defined the barrier as JERSEY type barrier with rigid material. The distance from the vehicle front to the impact point is defined as 6 feet. The angle of impact is degree and also the length of the barrier is defined as 118 feet. The NCAC model is based on SI units and has 302 time steps of 0.9 seconds each. 69

84 5.2.1 NCAC Single Unit Truck The truck presented in NCAC model is a 18 kips Ford single unit truck 27.6 feet long, feet in height and 8 feet in width. The materials used in the truck are *MAT- 001ELASTIC, *MAT-009-NULL, *MAT-020-RIGID, *MAT-024-PIECEWISE- LINEAR-PLASTICITY, *MAT-S01-DAMPER-VISCOUSE, and *MAT-SPRING- NONLINEAR-ELASTIC. The element types used in the truck model are beam, discrete, mass, seatbelt-accelerometer, shell, and solid elements, also the total number of elements are consisting of nodes. The vehicle has a translational velocity of mm/sec (51.4 miles per hour) as well as rotational velocity with the same magnitude defined for wheels. The gravity force on the truck is defined based on gravity acceleration, 9806 mm/s 2. The truck model consists of 151 individual parts with individual sections. The air bag definition was based on SIMPLE-AIRBAG-MODEL and 6 different types of contacts were defined in the truck. Figure 45: Single Unit Truck Top View 70

85 Figure 46: Single Unit Truck Side View 71

86 Figure 47: Single Unit Truck Front View NCAC Rigid Barrier The barrier presented in NCAC model is a JERSEY type barrier with dimensions of 118 feet in length, 2.66 feet in height, 2 feet at the bottom, and 0.5 feet at the top widths. The material used in barrier is *MAT-020-RIGID and the element type is Belytschko-Tsay shell element; also the total number of elements are consisting of nodes. The NCAC barrier is settled on a rigid plane. 72

87 Figure 48: NCAC Barrier Top View Figure 49: NCAC Barrier Front View 73

88 Figure 50: NCAC Barrier 3D View Objective of NCAC Model Since the only moving object in the NCAC model is the truck, and the barrier is defined as a rigid body, the outputs of kinetic energy represent the kinetic energy of the vehicle in every time step. Based on conservation of energy, the total amount of energy absorbed by the vehicle deformation can be tracked by plotting the kinetic energy versus time for the impact. Figure 51 to Figure 53 represent the NCAC TL-4 model. 74

89 Figure 51: NCAC Model Top View 75

90 Figure 52: NCAC Model 3D View 76

91 Figure 53: NCAC Model Side View 5.3 Proposed Model The second model is created by LS-PrePost 4.2 software that consists of AASHTO standard NEW JERSEY barrier installed on the standard bridge deck. Instead of using the truck in this model, a simplified approach was used by using an equivalent static force of single unit truck hitting the barrier based on TL-4. AASHTO indicates that this equivalent force is 54 kips distributed in a 3.5 feet length perpendicular to the barrier. It has been assumed that the force applies to an area with 0.41 feet height and 3.5 feet in length with the duration of seconds. The units used in the model are SI units. 77

92 The element type used for modeling the concrete was fully integrated quadratic 8 node solid element with nodal rotations with a length of approximately 25.4 mm (1 inch), and the material model used for concrete was *MAT 159-CSCM-CONCRETE. This material can reflect the cracks in concrete caused by plastic deformation. This material was defined with the properties of normal strength concrete. MAT159 input keyword for normal strength concrete was defined as, The element type used for modeling the rebar was Hughes-Liu with cross section integration beam elements, approximately 25.4 mm (1 inch) in length. The material model used for rebar based on 60 ksi steel was *MAT-024-PEACEWISE-LINEAR- PLASTICITY. The section dimensions for rebar No.3, No.4, and No.5 were defined individually. Figure 54 represents the stress-strain curve for 60 ksia36 steel (Getter, 2013). The horizontal axis represents the strain and the vertical axis represents the stress in Newton/mm 2. MAT024 input keyword was defined as, And input keyword for a stress-strain curve of 60 ksi steel was defined as, 78

93 Figure 54: Stress-Strain Curve of 60 ksi Steel Since the equivalent force is quasi static, the Strain rate parameters, C and P, are set to zero. The spacing of rebar varies in the section, therefore, in order to define the contact between concrete and rebar and forcing them to act along each other, the elements length was defined as 25.4 mm (1 inch) for both concrete and rebar. Then, all the nodes of 79

94 concrete and rebar with the distance of 12.7 mm were merged. This procedure forced concrete and rebar to interact together Deck Overhang Deck overhang was modeled based on AASHTO standard bridge deck. Figure 55 represents the section in detail. The thickness of overhang is 9 inches and the length measured from the center of the outer column to the edge is 39 inches. Reinforcements of the deck consist of alternating No.3 and No.5 at 7.5 inches at the top and No.5 at 9 inches at the bottom, parallel to the section. Also longitudinal reinforcements were added No. 4 at 8 inches on top and No. 4 at 18 inches at the bottom NEW JERSEY Concrete Barrier The barrier Section is 34 inches in height, with 6 inch widths on top and 15 inch width at the bottom. Rebar consists of a total of seven No.3 longitudinal and three No.4 stirrups, 6 inches apart. Figure 55 represents the section. 80

95 Figure 55: Section Details. (After Barker, 2013) 81

96 Figure 56: Model Section 82

97 Figure 57: Model 3D Section 83

98 Figure 58: Section Element Formation The length of the barrier and deck overhang was defined as 40 feet and the equivalent impact force were defined at the middle of the barrier, feet from anchored. Since the barrier and deck will not be built in same time, the only connection between barrier and deck will be stirrups. Therefore, the barrier and deck overhang were modeled individually and only connected by two 2 stirrup legs. The effect of gravity in this model was defined based on 9806 mm/s 2 ( lbm ft lbt sec2) gravity acceleration. Figure 59 to Figure 62 represent the model of the barrier and deck overhang system. 84

99 Figure 59: Model 3D view Figure 60: Model Top View 85

100 Figure 61: Model Side View 86

101 Figure 62: Equivalent Distributed Force Results are presented in chapter six. 87

102 6 RESULTS AND DISCUSSION 6.1 Work Method Modified version of yield line analysis was presented in the first chapter based on strain energy absorption of the barrier due to its horizontal displacement caused by vertical displacement of deck overhang and strain energy absorption of the barrier in horizontal direction before cracking. The calculations of work method are based on TL-4, initial impact, and for a onefoot strip of the barrier, Crash angle θ= 15 degree = 15 Π 180 = Rad Sinθ in radian is and weight of the barrier is 0.32 kips/ft. Therefore the combined mass velocity for initial impact is, V= m 1v 1 sin Θ = = mph m 1+ m ( ) Based on 0.1 second impact duration, the impact force F t, can be written as, F t = m 2V m 2 v 2 Δt = ( ) (0.32 0) 0.1 = kips The modulus of elasticity of normal concrete E, with 4000 pound per square inch compression strength f c /, can be calculated as, E = / f c 1000 = = ksf 88

103 Based on a 1.05 dynamic increase factor; 34 inches or 2.83 feet (the TL-4 standard height of NEW JERSEY barrier) H; ft 4 moment of inertia of the barrier I 2 ; ft4 ft moment of inertia of the deck overhang I 1 ; 9 inches (0.75 foot) standard deck thickness t; and 2.66 feet deck overhang length from the outer edge of the girder l; the total displacement is, Δ=Δ 2 + Δ 1 = F th 3 3EI 2 + H+ t 2 2 F t l = EI = = ft By substituting 13.84k.ft vertical moment capacity of the barrier M w ; k.ft horizontal moment capacity of the barrier M c ; zero as top beam moment capacity M b ; and 3.5 feet distributed length of impact force l, critical length is, L c1 = L tm c + M c (M c L t 2 + 8HM w + 8HM b ) M c = ( ) ( ) = 9.74f And, L c2 = L tm c M c (M c L t 2 + 8HM w + 8HM b ) M c = ( ) ( ) = -2.74ft And, L c = Max {L c1, L c2 }= Max {9.74, 2.74}= 9.74 ft 89

104 Minimum value of F t is obtained by substituting the maximum value of L c in the impact force equation, which is expressed as nominal railing resistance to transverse loads, R w, MinF t = R w (6-1) Therefore, F t = R w = 2(M cl c 2 +8HMw +8HM b ) H (L t L c ) = 2( ) 2.83 ( ) = 161 kips The AASHTO Standard Bridge Specifications test-levels table presents 54 kips as the required resistance to vehicle impact force for TL-4, while modified yield line method calculates the capacity of the barrier at 161 kips. Table 3: AASHTO Specifications Test Levels Details Transferred shear between barrier and deck overhang, V CT, is equal to tensile force, T, and is equal to, 90

105 V CT = T = R w = 161 = kips/ft L c +2H Also based on AASHTO Standard Bridge Specifications the nominal shear resistance of the interface is calculated as, V n = min { c A cv + µ A vf F y + P c, K 1 DIF f c / A cv, K 2 A cv } (6-2) And by substituting the variables, c A cv + µ A vf F y + P c = [ ( )] = kips/ft K 1 DIF f / c A cv = = kips/ft K 2 A cv = = 144 kips/ft Therefore: V n = min {27.732,151.2, 144} = kips/ft Yield line method calculates the transferred shear between barrier and deck, V CT, as 4.61 kips/ft, and the AASHTO Standard Bridge Specifications defines the nominal shear resistance, V n, as kips/ft, while the modified yield line method calculates the transferred shear between the barrier and deck, V CT, as kips/ft. This value is greater than 4.61 kips/ft but still within the limit of AASHTO Standard Bridge Specifications Results Compression Table 4 presents the results regarding the yield lines and modified yield lines methods based on TL-4. 91

106 Table 4: Work Method Results Comparison As has been shown in the above table, the critical length calculated by the yield line method is 7.17 feet while the modified yield line proved that the critical length is 9.74 feet. Also the calculated railing resistance based on yield line method is 59.1 kips and in modified yield line it has been proven that the actual railing resistance to impact force is 161 kips. AASHTO requirements for railing resistance is 54 kips for railing resistance. Therefore, the actual capacity is almost 3 times greater than AASHTO requirements which leads to a very conservative design. Yield line method calculates the shear transferred between barrier and deck overhang at 4.61 kips/ft and in modified yield line method it has been shown that the 92

107 actual shear transferred between barrier and deck overhang is kips/ft. AASHTO requirements for shear transferred is kips/ft. Therefore, the actual shear transfer is greater than the one calculation based on yield line method, but it is still within AASHTO limits. 6.2 Energy Method The calculations of energy method are presented based on TL-4, initial, and secondary impacts, also the design is based on impact length 3.5 feet Moving Vehicle Energy, IS Total energy before impact can be expressed as moving vehicle kinetic energy - Impact Severity - and is calculated as, IS = w v2 sin 2 θ = g c v = 50 mph = 72.9 ft/sec = ft-lbf Barrier Strain Energy Barrier Strain Energy Capacity, E 1 Barrier elastic strain energy absorption due to its capacity is, l 0 2E I 2 E 1 = M B 2 dx (6-3) M B is the moment capacity of the barrier provided by vertical bars; therefore, M B is the same as M c and, for 3.5 feet strip of barrier, can be written as. M B 3.5 = M c 3.5 = = k-ft 93

108 Since the barrier thickness varies from top to bottom, therefore, in order to calculate the strain energy, barrier must be divided into 3 segments. And by substituting the variables in strain energy equation, E 1 = = M c 2 E I 2 1 = 0.25 M 2 c dx 0 2E I M 2 c E I M 2 c dx E I Mc 2 E I M 2 c dx E I Mc 2 E I Mc 2 E I = ft-lbf Absorbed Energy by Barrier in Elastic Region, Δ 1 By rewriting the conservation of momentum for the secondary impact and calculating the combined mass velocity V, m 1 v 1 sinθ + m 2 v 1 = (m 1 + m 2 ) V (6-4) Therefore: V= m 1v 1 sinθ m 1 +m 2 = Sin(Rad 15) ( )+( ) = ft/sec so that, the kinetic energy absorbed by the barrier just after the impact can be calculatee as, Δ 1 = E Barrier, after impact = w 2V = = ft-lbf 2g c

109 And since Δ 1 is less than and about 40 percent of overall capacity of the barrier (E 1 ), the barrier will not reach the allowable capacity and will not exceed the elastic limit. Therefore, there will not be any crack or failure in the barrier caused by impact load Absorbed Energy by Barrier in Plastic Region, Δ 2 Since AASHTO Standard Bridge Specifications designs the NEW JERSEY concrete barrier based on yield line pattern which represents the cracks and failure modes in concrete. The same approach can be used in this step as well. Δ 2 = W yield line = 8M bδ L c + 8M wδ + M cl c δ L c H (6-5) However, since the barrier does not exceed the elastic limit of the section, Δ 2 is equal to zero. If the overall capacity (neglecting the effect of plastic limit) is less than TL-4 energy, then by obtaining the horizontal displacement of the barrier from LS DYNA simulation and substituting the variables in equation above, Δ 3 can be expressed as, Δ 2 = 8M bδ L c + 8M wδ + M cl c δ L c H = 8 0 δ δ δ = δ δ + =15.41 δ δ = 45 δft-lbf Absorbed Energy by Vehicle, Δ 3 As it has been explained in chapter five, there are two methods represented in this research in order to calculate the amount of energy absorbed by the vehicle deformation after impact. First, the conservation of energy equation and is used to calculate the energy 95

110 absorbed by vehicle deformation and the second method is designed to obtain data from the NCAC model. The first case scenario implies that since the only motion of a vehicle after the impact is parallel to the barrier and does not have motion perpendicular to said barrier anymore, perpendicular kinetic energy of the vehicle just after the impact must be absorbed by its deformation. Therefore, the perpendicular kinetic energy of the vehicle after the impact must be equal to the absorbed energy by the vehicle due to its deformation. This deformation energy can be calculated as, Δ 3 1 =E Vehicle, after impact = w 1V = = ft-lbf 2g c Δ 3 1 = 45% of Vehicle Kinetic Energy Second case scenario is to calculate the absorbed energy by the vehicle deformation and is based on the NCAC model. Appendix A represents the output data obtained from the kinetic energy of the vehicle for different time steps from the NCAC model. Based on the simulation, the impact starts from the 17 th time step. Therefore, by dividing the kinetic energy of this time step to the initial steady state kinetic energy before impact, the fraction of remaining kinetic energy in the vehicle, with respect to total energy before impact, is obtained. The same procedure must be done for all the time steps after the impact. The amount of absorbed energy by the vehicle deformation can be obtained by subtracting this fraction from one, and the total percentage of absorbed energy due to vehicle deformation in each time step can be obtained by multiplying each result by 100. Figure 63 presents the change in kinetic energy of the vehicle during the simulation. 96

111 The percentage of kinetic energy absorbed by the vehicle deformation after impact is the final fraction of absorbed energy by said vehicle, Δ 3 2 = 56 % of Vehicle Kinetic Energy Figure 63: Change in Kinetic Energy of the Vehicle Based on TL-4 Total absorbed energy by the vehicle deformation is expressed as the minimum of absorbed energy based on kinetic energy and absorbed energy based on the NCAC model. Δ 4 = min { Δ 3 1, Δ 3 2 }= min { 45%, 56% }= 45% = ft-lbf Deck Overhang Strain Energy Deck Overhang Strain Energy Capacity, E 4 Deck overhang strain energy absorption due to vertical displacement based on its capacity is, 97

112 l 0 2E I 1 E 4 = M D 2 dx (6-6) Figure 64: 1 Foot Strip of Deck Overhang Section in ZY Plane And the moment capacity of the deck overhang for 1 foot strip calculates as, As = = 0.84 in 2 d = a= A s f y 0.85 f c / b = = in = in M n =A s f y (d a ) = (6.255 ) = k-ft/ft 2 2 Therefore, the factored capacity for the 3.5 foot strip is, ΦM n = = kips-ft = ft-lbf And by substituting all the variables in the strain energy equation, 98

113 l 0 2E I 1 E 4 = M D 2 dx = M D 2 l 2E I 1 = = ft-lbf Absorbed Energy by Deck Overhang in Elastic Region, Δ 4 Based on conservation of energy being low for the barrier, deck overhang, and vehicle system due to impact, IS = ( Δ 1 or Δ 2 ) + Δ 3 + Δ 4 (6-7) And since the barrier does not exceed the elastic limit of the section, Δ 2 = 0 (6-8) So that, the conservation of energy can be rewritten as, IS = Δ 1 + Δ 3 + Δ 4 (6-9) Therefore the absorbed energy by deck overhang Δ 4, can be expressed as, Δ 4 = IS ( Δ 1 +Δ 3 ) = ( ) = ft-lbf Since Δ 4 is less than and almost 45 percent of total capacity of deck overhang (E 4 ), the deck overhang will not reach the allowable capacity and will not exceed the elastic limit. Therefore, the deck overhang will not fail or crack under impact load. 6.3 Results from Proposed Model In Work and Energy methods chapters, it has been discussed that NEW JERSEY concrete barrier designed based on TL-4 is conservative and over designed. The same 99

114 results were obtained by applying the equivalent force of TL-4 to the barrier. Neither barrier nor deck overhang exceeded the elastic limit, due to the strain energy absorbed by the barrier and deck overhang within the limits of the elastic region. As already expected, the maximum horizontal displacement occurs in the barrier at midpoint on top. Figure 65 presents the displacement verses time for this point. Figure 65: Maximum Displacement of Barrier Figure 66: Barrier Maximum Displacement with Scale of

115 Displacement and time are in millimeters and seconds respectively. As is shown the maximum displacement in the horizontal direction is almost linear up to 0.11 seconds and reaches 12 millimeters. Also, from 0.11 seconds the change in displacement is less than 1 millimeter. Maximum displacement in vertical direction occurs in the deck overhang at midspan under the load. Figure 67 presents the displacement verses time for this point. Figure 67: Maximum Displacement of Deck Overhang 101

116 Figure 68: Deck Overhang Maximum Displacement with Scale of 50 As it is shown, the maximum displacement in vertical direction is almost linear up to 0.11 seconds and reaches 1.62 millimeters. Also from 0.11 seconds the change in displacement is less than 0.5 millimeters. Stress distribution in barrier has an almost linear relationship with time up to 0.11 seconds and the position of maximum stress produced in the barrier is on the tension side under the load at 0.11 second with magnitude of 28.7 MPa. Figure 69 presents barrier maximum effective stress versus time. Stress and time are in MPa and seconds respectively. As it is shown, the maximum stress is almost linear up to 0.11 seconds and reaches 28.7 MPa. Also from 0.11 seconds the change in stress is less than two MPa. Since the system was modeled based on criteria of normal strength concrete with 30 MPa maximum stress, the concrete in barrier will not crack and or fail under the applied load. 102

117 Figure 69: Maximum Effective Stress in Barrier 103

118 Figure 70: 3D View of Barrier Maximum Effective Stress Stress distribution in the deck overhang has a linear relationship with time up to 0.11 seconds and the position of maximum stress produced in the deck overhang is at midpoint in the same position as maximum deflection at 0.11 seconds with a magnitude of almost 19.5 MPa. Figure 71 presents deck overhang maximum effective stress versus time. Stress and time are in MPa and seconds respectively. As it is shown the maximum stress is almost linear up to 0.11 second and reaches 19.5 MPa. Also from 0.11 seconds the change in stress is less than one MPa. Since the system was modeled based on criteria of normal strength concrete with 30 MPa maximum stress, the concrete in deck overhang will not crack and or fail under the applied load. 104

119 Figure 71: Maximum Effective Stress in Deck Overhang 105

120 Figure 72: Bottom View of Deck Overhang, Maximum Effective Stress Maximum axial force resultant in rebar has a linear relationship with time up to 0.11 seconds, and the positions of maximum force resultant in rebar are in stirrups and vertical rebar of barrier on the tension side, and rebar on top of deck overhang at midpoint of the model at 0.11 seconds. Figure 73 and Figure 74 present rebar axial force resultant. As it is shown, the maximum axial force resultant is Newton in rebar No.4 and No.5 and Newton in rebar No.3; therefore, the stress produced in these rebar will be, No. 3 => Ơ = No. 4 => Ơ = No. 5 => Ơ = F Area = = MPa F Area = = MPa F Area = = MPa 106

121 Since the rebar was modeled based on criteria of 60 Ksi steel which means the yield stress of rebar is 420 MPa, the critical rebar of the section (No.3, No.4 and No.5) will not yield and or fail under the applied load, therefore the section will not reach the yield point. Figure 73: Axial Force Resultant Distribution 107

122 Figure 74: Rebar Front View As was expected, the deck overhang and barrier system did not exceed the elastic limit of concrete; therefore, concrete will not crack under TL-4 impact load. Also none of the rebar resulted in yielding or failure. In overall view, the system displayed linear behavior in displacement and stress distribution up to 0.11 seconds, and after that, the change in displacement had a maximum one millimeter tolerance and the change in stress distribution has a maximum two MPa tolerance. The effective impact time is 0.11 seconds; therefore, the estimation of effective impact time in the work method chapter which was 0.1 is verified. 108