Determining Lateral Deflections of Plastic Water-Filled Barriers

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This paper presents two methods for determining the lateral deflection and the basic movement of a plastic waterfilled …

This paper presents two methods for determining the lateral deflection and the basic movement of a plastic waterfilled
barrier (PWB) system after an impact. In the first method theoretical equations are developed based on the
conservation of energy while MADYMO simulations are used in the second method of estimating lateral deflections. Crash
tests on various PWB systems carried out by different institutions including Monash University are used to validate the
results from both the theoretical calculations and the MADYMO models.

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  • 1. Determining Lateral Deflections of Plastic Water-Filled Barriers Jiang, T., Grzebieta, R.H. and Zhao, X.L. Civil Engineering, Faculty of Engineering, Monash University, Wellington Road, Melbourne 3800, AustraliaAbstract - This paper presents two methods for determining the lateral deflection and the basic movement of a plastic water-filled barrier (PWB) system after an impact. In the first method theoretical equations are developed based on theconservation of energy while MADYMO simulations are used in the second method of estimating lateral deflections. Crashtests on various PWB systems carried out by different institutions including Monash University are used to validate theresults from both the theoretical calculations and the MADYMO models.NOTATIOND Lateral deflection (m) of a road safety barrierDi Lateral deflection (m) of the ith segment of a PWB during impactDLong. Displacement (m) of the segments that moved longitudinally during impactdY Deflection (m) of a steel beam when yield occursE Young’s modulus (N/m2)EBEAM Energy (J) dissipated by the distortion of a steel beamEBR Energy (J) dissipated by the movement of PWB segmentsEBR-Lat. Work (J) done by the friction force of the segments that moved laterally during impactEBR-Long. Work (J) done by the friction force of the segments that moved longitudinally during impactEL Vehicle kinetic energy resolved perpendicular to barrier, or lateral kinetic energy, or impact severity (J)EVC Energy (J) dissipated by vehicle crushEVR Work (J) of vehicle’s rolling friction resistance forceFBA Average force applied to the barrier (N)FVA Average lateral force applied to the vehicle (N)I Second moment of area of the cross-section of a steel beam (m4)l1 Length of the cantilever beam (m)lB Length (m) of a PWB segmentM(α) Relationship between the moment M (Nm) and the rotation angle α (rad) of the joints between PWB segmentsmB Mass (kg) of an individual PWB segment filled with watermV Vehicle mass (kg).MY Yield moment (Nm)NT Overall number of segments installed in a PWB systemNBEAM Number of the steal beams used to strengthen a PWBn Number of segments that moved laterally during impactnLong. Number of segments that moved longitudinally during impactV Impact speed (m/s)z Distance (m) of the point furthest from the elastic neutral axisα Rotation angle (rad) of PWB segments and/or beamαY Rotation angle (rad) when yield occursθ Impact angle (degrees)μB Friction coefficient between a PWB and ground.μV Vehicle rolling friction coefficientσY Yield stress (N/m2) 1Jiang, T., Grzebieta, R.H. and Zhao, X.L, Determining Lateral Deflections of Plastic Water-Filled Barriers,Paper submitted to International Journal of Crashworthiness.
  • 2. A BRIEF INTRODUCTION TO PLASTIC WATER-FILLED BARRIER SYSTEMSTemporary road safety barrier systems are designed to prevent an errant vehicle access intoconstruction or maintenance work zones and can be erected and dismantled quickly. Conventionally,they are made of precast concrete. Since the late 1980’s, a new type of temporary safety barrier, i.e. aplastic water-filled barrier, has been developed and used worldwide.Normally, a PWB system consists of a certain number of identical segments/modules that are hollowinside with internal stiffening provided by web framing. Segments are connected together at their endsthrough specially designed joints to form a complete system and, once installed, are filled with waterto provide ballast against movement (Grzebieta et al. 2001). Often steel cables, tubes, or W-beams arealso used to connect each segment and provide extra strength for the system. Sometimes there are alsopedestals and mounting straps for each section, so that the barrier’s centre of gravity can be adjustedto provide high-speed protection. Usually PWBs are made from low-density polyethylene with a wallthickness ranging from 6 mm to 9 mm. Typically each segment is about 2000mm long, 600mm wideat the base and 800mm to 1000mm high with a weight of 50 kg to 75 kg when empty and 300 kg to1000 kg when filled with water. The major advantage of PWBs is that each segment is only 50 kg to 75 kg when empty and two people can install an entire system (Zou et al. 2000;Grzebieta et al. 2001).The first PWB system was developed in the late 1980s in the USA. With the release of NCHRPReport 350 in 1993, each PWB system needed to be assessed through full-scale crash tests, and all UScertified PWB systems are listed on the website of the Federal Highway Administration (FHWA)(FHWA 2004). In Australia, however, when the first PWBs were introduced in the early 1990s, theywere not required to satisfy any crash test requirements until the release of AS/NZS 3845 inDecember 1999.When selecting and installing PWBs for temporary construction or road works, it is very important fordesigners to know the barrier’s working width and how it moves when an errant vehicle crashes into itat different speeds and angles. However, most of the PWB vendors can only provide the lateraldeflection under two certification crash test conditions, i.e. a small car (820C) crashing into the barrierat an angle of 20 degrees and at the speed of the PWB’s test level; and a pickup truck (2000P)impacting at an angle of 25 degrees and at the same speed as the 820C car test. Such information isimportant but limited for barrier designers when they face real world traffic and site conditions. It isimpractical to carry out crash tests on a PWB system under all impact conditions. Therefore, methodsneed to be developed that can be used to determine lateral deflections of PWBs for different impactconditions.FULL-SCALE CRASH TESTS ON PWBSTo develop theoretical equations and computer models for determining lateral deflections generatedby such a complex impact process as a car crashing into a PWB, it is necessary to have data from alarge number of crash tests. Hence, efforts have been made to search for full-scale crash test data onPWBs in the literature and online resources. A total of twenty-three crash tests were found. All ofthese crash tests were conducted according to either NCHRP Report 350 or AS/NZS 3845.Crash test results on various PWB systems are summarised in Table 1. The impact points for thesecrash tests were basically located near the centre of the installation which was between 40 m and 70 mlong. For Test Level 2 and Test Level 3 PWB systems, specially-designed devices using either steelcables (the Triton barrier and the 426 Barrier) or steel tubes/beams (the Guardian 350 barrier, theYodock barrier and the Guardliner™ Barrier) or using both steel tubes and cables (the SB-1-TLbarrier), are needed to strengthen the PWBs. 2Jiang, T., Grzebieta, R.H. and Zhao, X.L, Determining Lateral Deflections of Plastic Water-Filled Barriers,Paper submitted to International Journal of Crashworthiness.
  • 3. Table 1 - Summary of full-scale crash tests on PWB systems Manufacturer Vehicle Impact Impact Test result TestNo and mass speed angle Lateral Number of Image Reference level Product (kg) (km/h) (deg) deflection moved laterally Vehicle response (m) segments Energy Absorption Came to rest Systems, Inc.: 807.3 72.0 20 1.0 Not available against the barrier Triton TL-2 Came to rest 1970.5 72.3 25 3.9 Not available against the barrier Energy Dimension (mm): Not redirected, Absorption 1981*533*813(TL2) 875 97.04 21 2.3 9 captured by the Syatems 1981*533*991(TL3) 1 barrier (2004); Weight (kg): TL-3 Not redirected, Empty: 64; Full: 612 2005 97.56 25 5.8 11 captured by the FHWA (TL2), 620 (TL3) barrier (2004) Connection: pin joints, steel cables Contained and on top between --- 2004 95.74 7 1.4 7 redirected parallel joints to the barrier Safety Barrier Systems: 820 70.6 20.3 0.6 Not available Not available Guardian 350 TL-2 Highway Kit Dimension (mm): 2000 71.5 25.8 1.98 Not available Not available 1830*610*1070 Weight (kg): Empty: 61; Full: 880 Contained and FHWA 2 (including tubes) 820 100 20 1.1 Not available redirected (2004) Connection: steel bars welded to the TL-3 inside of the pipes on one end and slotted on the other 2000 100 25 3.4 Not available Contained for a bolted connection 3Jiang, T., Grzebieta, R.H. and Zhao, X.L, Determining Lateral Deflections of Plastic Water-Filled Barriers,Paper submitted to International Journal of Crashworthiness.
  • 4. Table 1 - Summary of full-scale crash tests on PWB systems (continued) Manufacturer Vehicle Impact Impact Test result TestNo and mass speed angle Lateral Number of Image Reference level Product (kg) (km/h) (deg) deflection moved laterally Vehicle response (m) segments Yodock Wall Company, Inc.: Model 2001M (TL2) TL-2 2042 68.5 24.0 3.68 Not available Contained Yodock Model 2001 (TL3) Wall Dimension (mm): Company 1830*457*813(TL2) (2003 - 3 1830*610*1170(TL3) 896 97.7 19.8 1.23 Not available Contained 2004); Weight (kg): Full: 460 (TL2), 786 TL-3 FHWA (TL3) (2004) Connection: 2041 98.4 24.8 4.28 Not available Contained polyethylene couplers; steel tubes Rhino Safety Barrier LLC: Rhino Rhino Barrier 917 72.6 20 1.75 9 Redirected Safety Barrier Dimension (mm): LLC 4 TL-2 2000*690*890 (2001); Weight (kg): Contained and Empty: 54; Full: 476 2000 69.2 25 4.0 Not available Connection: steel-reinforced polyethylene pins and steel FHWA redirected Connection: bridging strips span the joint between barrier segments (2004) see images Creative Building Products: 426 Barrier 820 100.7 20 0.9 Not available Not reported Dimension (mm): 1830*610*1070 FHWA 5 TL-2 Weight (kg): (2004) Empty: 77; Full: 782 Connection: three Captured by the 2000 73 25 3.14 Not available steel cables threaded barrier through three holes 4Jiang, T., Grzebieta, R.H. and Zhao, X.L, Determining Lateral Deflections of Plastic Water-Filled Barriers,Paper submitted to International Journal of Crashworthiness.
  • 5. Table 1 - Summary of full-scale crash tests on PWB systems (continued) Manufacture Vehicle Impact Impact Test result TestNo and mass speed angle Lateral Number of Image Reference level Product (kg) (km/h) (deg) deflection moved laterally Vehicle response (m) segments Safety Barriers, Inc.: Model SB-1-TL Dimension (mm): 2130*610*1070 Weight (kg): Empty: 74.5; Captured by the FHWA 6 Full: 835.5 TL-3 2054 99.4 25.4 4.78 Not available barrier (2004) Connection: two aluminium pipes cast into each side of each segment; four steel cables are threaded through these pipes to connect all segments Barron & Rawson Pty Ltd: GuardlinerTM Barrier Dimension (mm): 891 70 20.5 0.78 12 Contained 2000*600*925 Zou and Weight (kg): Grzebieta 7 TL-2 Empty: 50; Full: 655 (including (2004) W-beam and brackets) 2170 70 25 2.64 16 Redirected Connection: Connection: four short steel stripes at each joint; W- see images beam supported by steel brackets which are secured to every segment join by two bolts 5Jiang, T., Grzebieta, R.H. and Zhao, X.L, Determining Lateral Deflections of Plastic Water-Filled Barriers,Paper submitted to International Journal of Crashworthiness.
  • 6. Table 1 - Summary of full-scale crash tests on PWB systems (continued) Test result Manufacturer Vehicle Impact Impact Test Number ofNo and mass speed angle Lateral segments Image Reference level deflection Vehicle response Product (kg) (km/h) (deg) (m) moved laterally Barron & Rawson Pty Ltd: Contained and Roadliner 2000™ S redirected with 920 49 20 0.54 9 moderate pitching Dimension (mm): and rolling 2000*600*925 Grzebieta 8 Weight (kg): TL-0 and Zou Empty: 50; (1998-2000) Full: 630 Connection: four Contained and 1570 48 25 1.2 10 short steel stripes redirected and two bolts at each joint (see images) Barrier Systems Pty Ltd: 910 50 20 0.49 8 Redirected Dimension (mm): 2000*600*940 Grzebieta 9 Weight (kg): TL-0 and Zou Empty: 50; (1998-2000) Full: 950 Contained and came Connection: see 1560 49 25 1.51 9 to rest parallel images against the barrier 6Jiang, T., Grzebieta, R.H. and Zhao, X.L, Determining Lateral Deflections of Plastic Water-Filled Barriers,Paper submitted to International Journal of Crashworthiness.
  • 7. THEORETICAL EQUATIONS FOR DETERMINING LATERAL DEFLECTIONSOF PWBSTo develop theoretical equations for calculating the lateral deflection and the basic movement of aPWB, it has to be assumed that a) the PWB installation is long enough and the vehicle hits the middleof the installation; b) the PWB redirects the vehicle and the vehicle does not roll over or ride over thebarrier; c) energy dissipated as a result of tearing of the PWB barrier is neglected and d) the lateraldeflection of a PWB caused by a vehicle impacting it at a speed of V and an angle of θ is taken to beequivalent to that generated by the vehicle crashing with the PWB laterally at a resolved speed ofV·sin θ.According to the crash test results, the deformation of a PWB after an impact can be approximated insimplistic form as shown in Figure 1. Based on the theory of energy conservation, the vehicle’s lateralkinetic energy (EL) is approximately equal to the work done by the movement of the PWB segments,the crush of the impacting vehicle, the rolling friction resistance of the vehicle and the energydissipated by the distortion of the beam (if fitted). Thus, 1 EL = m V (V sin θ ) 2 = E BR + E VC + E VR + E BEAM [1] 2where EL is the vehicle’s lateral kinetic energy (J), mV is the mass of the vehicle (kg), V is the impactspeed (m/s), θ is the impact angle (degrees), EBR is the energy dissipated by the movement of thePWB segments (J), EVC is the energy dissipated by the vehicle crush (J), EVR is the work of thevehicle’s rolling friction resistance force (J) and EBEAM is the energy (J) dissipated by the distortion ofthe beam. If EBR, EVC, EVR and EBEAM can all be determined as a function of the lateral deflection (D),the lateral deflection D can then be estimated using Equation 1. DLong. DLong. l α D .…. …… Central 1st 2nd …… (n − 1) segment 2 After impact lB …... …… C A Vsinθ R Before impact Figure 1 - Assumed deformation of a PWBDetermining EVC and EVRIn the case of a car crashing into a PWB, the impact load is low because a PWB is a flexible system.Hence, the energy dissipated by the crush of vehicle EVC can be neglected. This is evident from thecrash test result where a 1600 kg car crashed into a PWB system at a speed of 50 km/h and an angleof 25 degrees (Grzebieta and Zou 1998-2000). Figure 2 shows that only minor crush was observedafter the impact. 7Jiang, T., Grzebieta, R.H. and Zhao, X.L, Determining Lateral Deflections of Plastic Water-Filled Barriers,Paper submitted to International Journal of Crashworthiness.
  • 8. Figure 2 - Vehicle frontal crush after impactThe work done by the vehicle’s rolling friction resistance force EVR as it is being redirected can bedetermined as E VR = m V gμ V D [2]where g is 9.81 (m/s2), μV is the rolling friction coefficient of the vehicle and D is the lateral deflection(m). Generally, the rolling friction coefficient μV is less than 0.1 (Bauer et al. 2000). Thus, EVR issmall compared to the impact energy EL and can be ignored as well. Hence, Equation 1 can be furtherexpressed as 1 mV (V sin θ ) 2 = E BR + E BEAM [3] 2Determining EBRFigure 1 shows that whilst some segments were pushed and moved laterally, the other segments weredragged and displaced longitudinally towards the impact point. Hence, the energy dissipated by themovement of segments of a PWB (EBR) can be approximately considered as comprising of two parts;the work done by the friction force of the segments that moved laterally and the work done by thefriction force of the segments that moved longitudinally. Thus E BR = E BR − Lat. + E BR − Long . [4]where EBR-Lat. is the work (J) done by the friction force of the segments that moved laterally andEBR-Long. is the work (J) done by the friction force of the segments that moved longitudinally. However,to determine EBR-lat. and EBR-Long., the number of segments that moved laterally during impact (n) needsto be estimated first.Estimating the number of PWB segments that moved laterally during impactThe process of a car crashing into a PWB is very complicated. Many time dependant factors areinvolved in the whole dynamic process, such as the impacting force and angle, the interaction amongsegments, the movement of segments both laterally and longitudinally, the friction between the 8Jiang, T., Grzebieta, R.H. and Zhao, X.L, Determining Lateral Deflections of Plastic Water-Filled Barriers,Paper submitted to International Journal of Crashworthiness.
  • 9. vehicle and the PWB, the deformation of the vehicle and barrier, and even the movement of waterinside the barrier (Jiang et al. 2002). It is a challenge to develop a theoretical method for determiningthe number of segments that moved laterally during impact. Nevertheless, a simplified method wasdeveloped and reported as follows.If the initial lateral impact speed of the vehicle is assumed as (Vsinθ) and its final lateral speed is zeroat the maximum lateral displacement D, the average lateral force applied to the vehicle can becalculated as (V sin θ ) 2 FVA = m V [5] 2Dwhere FVA is the average lateral force applied to the vehicle (N). The average force applied to thePWB (FBA) is equal but opposite to FVA. The load required to move the barriers can be directlyequated to the sliding resistance of the plastic barriers themselves. Thus (V sin θ ) 2 FBA = FVA = mV = n ⋅ mB gμ B [6] 2Dwhere n is the number of segments that moved laterally during impact, mB is the mass (kg) of anindividual PWB segment filled with water and µB is the friction coefficient between the PWB and theground. Hence, the number of segments that moved during impact can be estimated using Equation 6as mV (V sin θ ) 2 n= [7] 2 D ⋅ mB gμ BTo validate Equation 7, the numbers of segments that moved laterally during impact calculated usingEquation 7 were compared to those observed in the crash tests. Table 1 shows ten crash tests wherethe number of segments that moved laterally during impact is available. The friction coefficient μBwas assumed being the result of a laboratory test performed by Grzebieta and Zou (1998-2000) wherethey dragged a New Jersey type PWB over a concrete floor. As PWBs are basically made from asimilar polyethylene material, the friction coefficient of PWBs placed on dry bitumen surfaces wasalso assumed to be 0.4.Hence, for the 820C Test Level 3 test of the Triton barrier (No. 1 in Table 1), the estimated number ofsegments that moved laterally during impact was calculated using Equation 7 as 97.04 875 × ( × sin 21°) 2 mV (V sin θ ) 2 3 .6 n= = ≈7 2 D ⋅ m B gμ B 2 × 2.3 × 620 × 9.81 × 0.4Similarly, the estimated number of segments that moved laterally during impact for the other ninetests are summarised in Table 2. Results in Table 2 show that Equation 7 basically underestimates nby an average number of two segments. Hence, the equation for determining n is empirically adjustedby adding 2 to Equation 7 as mV (V sin θ ) 2 n= +2 [8] 2 D ⋅ mB gμ B 9Jiang, T., Grzebieta, R.H. and Zhao, X.L, Determining Lateral Deflections of Plastic Water-Filled Barriers,Paper submitted to International Journal of Crashworthiness.
  • 10. Table 2 - Comparison of n from the calculations and the crash tests PWB Lateral Number of segments that Vehicle Impact Impact No. in segment deflection moved laterally, n Difference No mass speed angle Table 1 mass from test Crash test Calculated result, (=nCalculated - nTest) (kg) (km/h) (deg) (kg) D (m) result, nTest nCalculated 1 1 875 97.04 21 620 2.3 9 7 -2 2 (TL-3) 2005 97.56 25 620 5.8 11 9 -2 3 1 2004 95.74 7 620 1.4 7 3 -4 4 4 917 72.6 20 470 1.75 9 7 -2 5 7 891 70 20.5 655 0.78 12 12 0 6 (TL-2) 2170 70 25 655 2.64 16 12 -4 7 8 920 49 20 630 0.54 9 7 -2 8 (TL-0) 1570 48 25 630 1.2 10 8 -2 9 9 910 50 20 950 0.49 8 6 -2 10 (TL-0) 1560 49 25 950 1.51 9 5 -4Determining EBR-Lat.Figure 1 indicates that the energy dissipated by the segments that moved laterally during impact canbe approximately divided into the work done by the friction force of every segment that movedlaterally (without rotation), which is symbolised as EBR-Lat.-NR, and the work required to rotate thesesegments to an angle α, which is symbolised as EBR-Lat.-R. Thus E BR -Lat. = E BR -Lat.- NR + E BR -Lat.-R [9](a) Determining EBR-Lat.-NRAccording to the crash test results, the deformation of the barriers on either the left hand or the righthand side of the central segment can be considered as approximately symmetrical (see Figure 1). Thus,EBR-Lat.-NR can be determined as n −1 n 2 E BR -Lat.- NR = ∑ m B gμ B Di = mB gμ B D + 2 ( ∑ m B gμ B Di ) [10] i =1 i =1where Di is the lateral displacement (m) of the ith segment. Figure 1 shows that the lateraldisplacement D1 of the 1st segment that has moved can be approximated as 1 D1 = l B sin α [11] 2where lB is the length (m) of a PWB segment. Similarly, the lateral displacement D2 of the 2nd segmentis 3 D2 = l B sin α [12] 2and so on, where finally the lateral displacement of the barrier next to the central segment D n −1 is 2 10Jiang, T., Grzebieta, R.H. and Zhao, X.L, Determining Lateral Deflections of Plastic Water-Filled Barriers,Paper submitted to International Journal of Crashworthiness.
  • 11. (n − 2) D n −1 = l B sin α [13] 2 2The angle α can be calculated from Figure 1 using trigonometry as D D [14] sin α = = l ⎡ ( n − 1) ⎤ 2 D2 + ⎢ lB ⎥ ⎣ 2 ⎦Substituting Equations 11 to 14 into Equation 10 results in n −1 2 E BR -Lat.- NR = mB gμ B D + 2(∑ mB gμ B Di ) i =1 1 3 (n − 2) = mB gμ B D + 2mB gμ B ( l B sin α + l B sin α + ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ + l B sin α ) 2 2 2 ⎛ 1 (n − 2) ⎞ (n − 1) ⎜ + ⎟⋅ = m B gμ B D + 2 m B gμ B l B ⎝2 2 ⎠ 2 ⋅ sin α 2 mB gμ B l B (n − 1) 2 D = m B gμ B D + ⋅ 4 ⎡ (n − 1) ⎤ 2 D2 + ⎢ lB ⎥ ⎣ 2 ⎦ mB gμ B l B (n − 1) 2 D [15] = m B gμ B D + 2 ⎡ (n − 1) ⎤ 4 ⋅ D2 + ⎢ lB ⎥ ⎣ 2 ⎦where n is as given in Equation 8.(b) Determining EBR-Lat.-RThe work required to rotate one PWB segment to an angle α can be determined from Figure 3 as α lB E RBR − Lat.-R -1 = ∫ [ M (α ) + mB gμ B ⋅ ]dα 0 2 [16] α mB gμ B l B = ∫ M (α )dα + ⋅α 0 2where EBR-Lat.-R-1 is the work (J) required to rotate one PWB segment to an angle α, and M(α)represents the relationship between the moment M (Nm) and the rotation angle α (rad) of the joint. 11Jiang, T., Grzebieta, R.H. and Zhao, X.L, Determining Lateral Deflections of Plastic Water-Filled Barriers,Paper submitted to International Journal of Crashworthiness.
  • 12. F lB α mBgμB M(α) Figure 3 - Rotation of a PWB segmentIt was observed from crash test videos that the segments that moved laterally during impact rotatedsimilarly to an angle α (Grzebieta and Zou 1998-2000; Zou and Grzebieta 2004). Thus, EBR-R can beestimated as α nmB gμ B l Bα E BR -Lat.-R = nE BR -Lat.-R -1 = n ∫ M (α )dα + [17] 0 2Substituting Equations 15 and 17 into Equation 9, EBR-Lat. can be calculated as E BR -Lat. = E BR -Lat.- NR + E BR -Lat.-R [18] mB gμ B l B (n − 1) 2 D α nmB gμ B l Bα = m B gμ B D + + n ∫ M (α )dα + ⎡ (n − 1) ⎤ 2 0 2 4 ⋅ D2 + ⎢ lB ⎥ ⎣ 2 ⎦Using trigonometry, the value of α (rad) approximately equals to sinα with a difference of less than1% when α < 14°(Rosenbach et al. 1961; Bauer et al. 2000). Hence, the rotation angle α can beapproximated using Figure 1 and Equation 14 as D [19] α ≈ sin α = 2 ⎡ (n − 1) ⎤ D2 + ⎢ lB ⎥ ⎣ 2 ⎦Substituting Equation 19 into Equation 18 gives mB gμ B l B (n − 1) 2 D α E BR -Lat. = mB gμ B D + + n ∫ M (α )dα + 2 0 ⎡ (n − 1) ⎤ 4 ⋅ D2 + ⎢ lB ⎥ ⎣ 2 ⎦ [20] nmB gμ B l B D 2 ⎡ (n − 1) ⎤ 2 ⋅ D2 + ⎢ lB ⎥ ⎣ 2 ⎦Determining EBR-Long.The number of segments that moved longitudinally during impact can be estimated as n Long. = N T − n [21] 12Jiang, T., Grzebieta, R.H. and Zhao, X.L, Determining Lateral Deflections of Plastic Water-Filled Barriers,Paper submitted to International Journal of Crashworthiness.
  • 13. where nLong. is the number of segments that moved longitudinally during impact, n is the number ofsegments that moved laterally during impact and NT is the total number of segments installed in thePWB system. The displacement of the segments that moved longitudinally during impact, which isexpressed as DLong., can be approximately determined from Figure 1 as 2 (n − 1) ⎡ (n − 1) ⎤ (n − 1) DLong. = l − lB = D 2 + ⎢ lB ⎥ − lB [22] 2 ⎣ 2 ⎦ 2Thus, EBR-Long. can be calculated as E BR -Long. = nLong. mB gμ B DLong. [23] ⎡ ⎡ (n − 1) ⎤ 2 (n − 1) ⎤ = ( N T − n ) m B gμ B ⎢ D 2 + ⎢ lB ⎥ − lB ⎥ ⎢ ⎣ 2 ⎦ 2 ⎥ ⎣ ⎦Substituting Equations 20 and 23 into Equation 4, the energy dissipated by the movement of thesegments of a PWB (EBR) can be calculated such that E BR = E BR − Lat. + E BR − Long. = m B gμ B l B (n − 1) 2 D α nm B gμ B l B D m B gμ B D + + n ∫ M (α )dα + + 2 0 2 ⎡ (n − 1) ⎤ ⎡ (n − 1) ⎤ 4 ⋅ D2 + ⎢ lB ⎥ 2 ⋅ D2 + ⎢ lB ⎥ ⎣ 2 ⎦ ⎣ 2 ⎦ ⎡ ⎡ (n − 1) ⎤ 2 (n − 1) ⎤ ( N T − n ) m B gμ B ⎢ D 2 + ⎢ lB ⎥ − lB ⎥ ⎢ ⎣ 2 ⎦ 2 ⎥ ⎣ ⎦ [24]where n is as given in Equation 8.Determining EBEAMWith regard to PWBs where steel beams are used to strengthen the barrier, as illustrated in Figure 4,the energy dissipated by the distortion of the steel beam (EBEAM) during impact needs to be determined.To do so, it is assumed that the beam distorts in a manner similar to a beam with fully fixed ends asshown in Figure 5. That the distortion of the steel beam is similar to a beam restrained by fully fixedends is clearly visible in the overhead photograph (Figure 6) of the PWB crash test performed byGrzebieta and Zou (2003). Segments of a PWB Steel beam D nlB Figure 4 - Assumed deformation of a PWB with steel beams 13Jiang, T., Grzebieta, R.H. and Zhao, X.L, Determining Lateral Deflections of Plastic Water-Filled Barriers,Paper submitted to International Journal of Crashworthiness.
  • 14. To analyse the energy dissipated by the distortion of the steel beam as shown in Figure 5, it isconvenient to adopt a rigid-plastic material model. Hence the steel beam bending characteristics andthus the relationship between the bending moment M(α) and the rotation angle α can generally besimplified as bi-linear as shown in Figure 7 (Megson 1996), where MY is the yield moment (Nm) andαY is the rotation angle (rad) when yield occurs. Steel beam α D (n − 1) lB 2 Figure 5 - Distortion of a steel beam Figure 6 - Overhead view of a PWD crash test (Grzebieta and Zou 2003) M (Nm) MY α (rad) αY α Figure 7 - Bending moment versus rotation angle for a steel beam Figure 8 - A cantilever steel beam 14Jiang, T., Grzebieta, R.H. and Zhao, X.L, Determining Lateral Deflections of Plastic Water-Filled Barriers,Paper submitted to International Journal of Crashworthiness.
  • 15. Figure 8 shows a cantilever steel beam where the deflection is D. The energy required to push thebeam to a deflection D can be estimated using the curve shown in Figure 7 as α 1 E BEAM-1 = ∫ M (α )dα = M Y ⋅ α − M Y ⋅α Y [25] 0 2where the rotation angle α can be approximated using Equation 19 and Figure 8 as D α ≈ sin α = [26] D + l12 2where l1 is the length of the cantilever beam.With regard to the beam shown in Figure 5, the energy dissipated by the distortion of the steel beamcan be considered as double of the energy dissipated by the beam shown in Figure 8. Hence E BEAM = 2 E BEAM-1 = 2M Y ⋅ α − M Y ⋅ α Y [27]MY can be calculated as σYI MY = [28] zwhere σY is the yield stress of the steel (N/m2), I is the second moment of area of the cross-section ofthe steel beam (m4) and z is the distance of the point furthest from the elastic neutral axis (m)(Megsson, 1996). Once MY is determined, the deflection of the beam when yield occurs can also becalculated as FY l13 ( FY ⋅ l1 )l12 M Y l12 dY = = = [29] 3EI 3EI 3EIwhere dY is the deflection (m) of the beam when yield occurs and E is the Young’s modulus (N/m2)(Megsson, 1996). Generally, dY is small compared to l1. Hence, αY can be approximated usingtrigonometry and Equation 29 as d Y M Y l1 αY ≈ = [30] l1 3EIData from a laboratory test performed by Cichowski et al. (1961) was used to validate Equation 27.Figure 9 shows the test set-up and the test result, where a load was placed in the centre of a 3810 mmlong W-beam rail and the load-deflection curve was recorded.The yield strength for this steel (σY) was 350 MPa and the furthest point from the neutral axis (z) was40 mm (Cichowski et al. 1961). The second moment of area of the cross-section of the W-beam wasdetermined as 970000 mm4 or 970.0E-09 m4 (Deleys and McHenry 1967; Reid et al. 1997). Thus, MYfor the W-beam steel was calculated using Equations 28 as σYI 350 × 10 6 × 970 × 10 −9 MY = = = 8487.5 (Nm) z 40 × 10 −3 15Jiang, T., Grzebieta, R.H. and Zhao, X.L, Determining Lateral Deflections of Plastic Water-Filled Barriers,Paper submitted to International Journal of Crashworthiness.
  • 16. 5000 W-beam length: 150" (3810 mm) 4000 Load (lbs) . 3000 2000 1000 0 0 1 2 3 4 5 6 7 8 9 10 11 12 Deflection (inches) Figure 9 - W-beam laboratory test set-up and test result (Cichowski et al. 1961) 3810 2For the case shown in Figure 9, l1 was = 1905 (mm) . E can be taken as 210.0E+09 N/m . Thus, 2the rotation angle when yield occurs αY was calculated using Equation 30 as M Y l1 8487.5 × 1.905 αY ≈ = = 0.0265 (rad) 3EI 3 × 210 × 10 9 × 970 × 10 −9If the deflection was assumed as 152.4 mm (6 in.), α can be estimated using Equation 26 as D 0.1524 α≈ = = 0.08 (rad) D +l2 1 2 0.1524 2 + 1.905 2Thus, when the deflection was 152.4 mm (6 in.), the energy dissipated by the W-beam was calculatedby substituting αY and α into Equation 27 as E BEAM = 2M Y ⋅ α − M Y ⋅ α Y = 2 × 8487.5 × 0.08 − 8487.5 × 0.0265 = 1133 (Nm)The energy dissipated by the W-beam when the deflection was 152.4 mm (6 in.) was also determinedusing the test data as about 1300 Nm, which was the shadow area shown in Figure 9. The energydissipated by the W-beam calculated using Equation 27 compares reasonably well with the test result. (n − 1)Hence, with regard to the beam distorted as shown in Figure 5 where l1 = l B , the energy 2dissipated by the beam can be determined by substituting Equations 26 and 30 into Equation 27 as E BEAM = 2 M Y ⋅ α − M Y ⋅ α Y [31] D M Y l1 2M Y D (n − 1) M Y l B 2 = 2M Y − MY = − D 2 + l12 3EI ⎡ (n − 1) ⎤ 2 6 EI D2 + ⎢ lB ⎥ ⎣ 2 ⎦ 16Jiang, T., Grzebieta, R.H. and Zhao, X.L, Determining Lateral Deflections of Plastic Water-Filled Barriers,Paper submitted to International Journal of Crashworthiness.
  • 17. For PWBs where more than one steel beam is used to strengthen the PWB system, such as theGuardian 350 Highway Kit (No. 2 in Table 1) and the Yodock barrier (No. 3 in Table 1), the energydissipated by the distortion of the beams during impact can be estimated using Equation 31 as ⎡ ⎤ ⎢ ⎥ ⎢ 2M Y D ( n − 1) M Y l B ⎥ 2 [32] E BEAM = N BEAM ⎢ − ⎥ 2 6 EI ⎢ D 2 + ⎡ ( n − 1) l ⎤ ⎥ ⎢ ⎢ 2 B⎥ ⎣ ⎦ ⎥ ⎣ ⎦where NBEAM is the number of the steal beams used to strengthen the PWB barrier.Determining DSubstituting Equations 24 and 32 into Equation 3, the following equation is obtained. 1 mV (V sin θ ) 2 = E BR + E BEAM = 2 mB gμ B l B (n − 1) 2 D α nmB gμ B l B D mB gμ B D + + n ∫ M (α )dα + + 2 0 2 ⎡ (n − 1) ⎤ ⎡ (n − 1) ⎤ 4⋅ D + ⎢ 2 lB ⎥ 2⋅ D + ⎢2 lB ⎥ ⎣ 2 ⎦ ⎣ 2 ⎦ ⎡ ⎡ (n − 1) ⎤ 2 (n − 1) ⎤ ( N T − n ) m B gμ B ⎢ D 2 + ⎢ lB ⎥ − lB ⎥ + ⎢ ⎣ 2 ⎦ 2 ⎥ ⎣ ⎦ ⎡ ⎤ ⎢ ⎥ ⎢ 2M Y D (n − 1) M Y l B ⎥ 2 N BEAM ⎢ − ⎥ ⎢ D2 + ⎡ (n − 1) ⎤ 2 6 EI ⎥ [33] ⎢ ⎢ 2 lB ⎥ ⎥ ⎣ ⎣ ⎦ ⎦where n and MY is as given in Equation 8 and Equation 28, respectively. Although Equation 33 is in acomplicated form, D, which is the only unknown factor, can be calculated using the numericalsolution for non-linear equations in a single unknown (Bronshtein et al. 2004).Comparison with crash test resultsTo validate the equations for calculating the lateral deflection and the number of segments that movedlaterally during impact, the calculated results were compared to the results from full-scale PWB crashtests listed in Table 1. To do so, M(α) and MY in Equation 33 had to be determined first. M(α) and MYwere either determined using data of quasi-static laboratory tests or calculated using Equation 28.Details regarding the derivation of M(α) and MY are presented in Appendix A.Using the results of M(α) and MY determined, D and n can be calculated using Equations 33 and 8,respectively. The results are summarised in Table 3. Details of how these results were calculated arereported in Appendix B. 17Jiang, T., Grzebieta, R.H. and Zhao, X.L, Determining Lateral Deflections of Plastic Water-Filled Barriers,Paper submitted to International Journal of Crashworthiness.
  • 18. Table 3 – Comparison of the results from the crash tests and the calculations PWB segment data MY Vehicle Impact Impact Test result Calculated result Difference * Manufacturer Test Length (lB) × Full Number of Number of Number ofNo and and mass speed angle Lateral segments Lateral segments Lateral segments Product level Width (at base) × weight deflection that moved deflection that moved deflection that moved Height (mm) mB (kg) M(α) mV (kg) V (km/h) θ (deg) D (m) D (m) (%) laterally, n laterally laterally, n Energy Not 807.3 72.0 20 1.0 1.033 10 +3.3 --- Absorption available TL-2 1981×533×813 612 Systems, Inc: Not 1970.5 72.3 25 3.9 3.03 12 -22.3 --- MY = 0 available Triton 1 875 97.04 21 2.3 9 1.917 11 -16.6 +2 TL-3 1981×533×991 620 M(α) = 0 2005 97.56 25 5.8 11 4.945 13 -14.7 +2 --- 1981×533×991 620 2004 95.74 7 1.4 7 0.556 10 -60.3 +3 Safety Not 820 70.6 20.3 0.6 0.708 10 +17.9 --- Barrier MY = available TL-2 1830×610×1070 880 Systems: 2202.2 Not 2000 71.5 25.8 1.98 2.16 12 +9.1 --- (Nm) available 2 Guardian 350 Not Highway Kit 820 100 20 1.1 1.221 11 +11.0 --- available TL-3 1830×610×1070 880 M(α) = 0 Not 2000 100 25 3.4 3.546 14 +4.3 --- available Yodock Wall Not Company, TL-2 1830×457×813 460 MY = 2042 68.5 24.0 3.68 3.267 13 -11.2 --- available Inc: 6046.8 (Nm) Not 3 896 97.7 19.8 1.2 1.430 11 +19.2 --- available Model 2001M (TL2) TL-3 1830×610×1170 786 Not Model 2001 M(α) = 0 2041 98.4 24.8 4.02 available 4.215 13 +4.9 --- (TL3) 18Jiang, T., Grzebieta, R.H. and Zhao, X.L, Determining Lateral Deflections of Plastic Water-Filled Barriers,Paper submitted to International Journal of Crashworthiness.
  • 19. Table 3 – Comparison of the results from the crash tests and the calculations (continued) PWB segment data MY Vehicle Impact Impact Test result Calculated result Difference * Manufacturer Test Length (lB) × Full Number of Number of Number ofNo and and mass speed angle Lateral segments Lateral segments Lateral segments Product level Width (at base) × weight deflection that moved deflection that moved deflection that moved Height (mm) mB (kg) M(α) mV (kg) V (km/h) θ (deg) D (m) D (m) (%) laterally, n laterally laterally, n Rhino Safety MY = 0 917 72.6 20 1.75 9 1.377 11 -21.3 +2 Barrier 4 TL-2 2000*690*890 476 LLC: M(α)= Not Rhino Barrier 2000 69.2 25 4.0 3.082 14 -22.9 --- 12835 α available Creative Not MY = 0 820 100.7 20 0.9 1.431 11 +59.0 --- Building available 5 TL-2 1830*610*1070 782 Products: Not M(α)=0 2000 73 25 3.14 2.452 12 -21.9 --- 426 Barrier available Safety MY = 0 Barriers, Inc: Not 6 Model SB-1- TL-3 2130*610*1070 835.5 2054 99.4 25.4 4.78 4.91 11 +2.7 --- available TL M(α)=0 Barron & MY = Rawson Pty 8487.5 891 70 20.5 0.78 12 0.925 11 +18.6 -1 Ltd: (Nm) 7 GuardlinerTM TL-2 2000*600*925 655 Barrier M(α)= 2170 70 25 2.64 16 2.503 14 -5.2 -2 31880 α Barron & MY = 0 920 49 20 0.54 9 0.541 10 +0.2 +1 Rawson Pty 8 Ltd: TL-0 2000*600*925 630 M(α)= Roadliner 2000™ S 31880 α 1570 48 25 1.2 10 1.133 11 -5.6 +1 Barrier MY = 0 910 50 20 0.49 8 0.443 8 -9.7 0 9 System Pty TL-0 2000*600*940 950 Ltd: M(α) = 0 1560 49 25 1.51 9 0.987 9 -34.6 0Note: *: ‘-’ underestimate; ‘+’ overestimate. 19Jiang, T., Grzebieta, R.H. and Zhao, X.L, Determining Lateral Deflections of Plastic Water-Filled Barriers,Paper submitted to International Journal of Crashworthiness.
  • 20. DiscussionComparison between the calculated and the test results in regards to the number of segments thatmoved during impact indicates that the differences are on average within two segments. Thecalculated lateral deflections also compare reasonably well with the crash test results, except for threetests; the 95.74 km/h at 7 degrees test for the Triton barrier, the 820C TL-2 test for the 426 Barrierand the 1600C TL-0 test for the Barrier System barrier where the differences were -60.3%, +59.0%and -34.6%, respectively.For the 95.74 km/h and 7 degrees impact test on the Triton barrier, the impact severity, defined as1 mV (V sin θ ) 2 , was 10.5 kJ. The test result for the lateral deflection was 1.4 m and the calculated2lateral deflection was 0.556 m. The manufacturer has provided two curves showing the impactseverity versus lateral deflection in Figure 10, which were obtained from crash test data (EnergyAbsorption Systems Inc. 2004; FHWA 2004). Curve A was derived from crash tests where the impactpoint was located between 10 to 20 metres from the upstream end whereas Curve B was obtained forcrash tests where the impact point was at least 20 m away from the upstream end. From Curve B, thelateral deflection is about 0.6 m when the impact severity is 10 kJ, to which the deflection (0.566 m)calculated using the methodology outlined in this paper correlated well.The test result for the lateral deflection of a 820C car crashing into the 426 Barrier at 100.7 km/h andan angle of 20 degrees was 0.9 m. The calculated deflection was 1.43 m. A check of the test reportfound that “the barrier segment that was struck first shattered, but the internal cables successfullycontained the small car and minimized the deflection” (Power, pers. comm., October 26, 2004). Inaddition, when comparing the above test to the 820C Test Level 3 test for the Triton barrier, where a875 kg car crashes into the barrier at a speed of 97.04 km/h and an angle of 21 degrees, these two testscan be considered similar in terms of impact severity and barrier joint stiffness. However, the lateraldeflection recorded in the Triton barrier test was 2.3 m as opposed to 0.9 m for the 426 Barrier. Henceit would seem that the 426 Barrier test was possibly an anomaly because of the tearing of the struckbarrier. The theory in this paper assumes the barriers do not tear or shatter on impact.With regard to the 1600C TL-0 test for the Barrier System barrier, the calculated lateral deflectionunderestimated the test result by about 35%. This is because the theory in this study assumes the car isredirected, whereas in the crash test the car was contained and came to rest parallel against the barrier(Grzebieta and Zou 1998-2000). 6 Curve B Lateral deflection (m) 5 4 Curve A 3 Obtained from crash tests where impact points are 10 to 20 m 2 from the upstream end Obtained from crash tests where 1 impact points are at least 20 m 0.6 m away from the upstream end 0 0 10 20 30 40 50 60 70 80 90 100 Impact severity (kJ) 20Jiang, T., Grzebieta, R.H. and Zhao, X.L, Determining Lateral Deflections of Plastic Water-Filled Barriers,Paper submitted to International Journal of Crashworthiness.
  • 21. Figure 10 - Impact severity versus lateral deflection for the Triton barrierThat the calculated lateral deflection underestimated the test result when the car was captured ratherthan directed by the barrier is also evident from the tests for the Triton barrier, the Rhino barrier andthe 426 Barrier. The comparison of the results in Table 3 for these tests indicates that the calculatedlateral deflection underestimated the test result by on average 20%. Hence, for PWBs where the jointstiffness is small and no steel beams are used to strengthen the barrier, such as the Triton barrier, theRhino barrier, the 426 Barrier and the Barrier System barrier, the possible maximum lateral deflection,which is denoted as DM, can be determined empirically by multiplying a factor of 1.2 to the deflectioncalculated using Equation 33. DM = 1.2 D [34]Figure 11 shows the plots of the impact severity versus lateral deflection for the Triton barrier, whichwere obtained using crash test data (Curve B), using Equation 33 (Curve C) and using Equation 34(Curve D), respectively. Details of how the lateral deflections for different impact severities werecalculated are given in Appendix C. Also shown in Figure 11 is Curve B which represents the 80%value of Curve B or the -20% data band of Curve B. As can be observed in Figure 11, the results fromEquation 33 (Curve C) is within the data band between Curve B and Curve B, which indicates areasonably well comparison, whereas the results from Equation 34 (Curve D) closely agrees with thecurve representing the test result for Curve B.In conclusion, comparisons between the calculated values and the test results indicate that theequations developed in this paper can be used to determine the PWB lateral deflection and the numberof segments that move with reasonable accuracy, so long as the PWB installation is long enough, theimpact point is close to the middle of the installation, the barrier does not overly tear apart, and the cardoes not roll over or ride over the barrier. With regard to PWBs where the joint stiffness is small andno steel beams are used to strengthen the barrier, the possible maximum lateral deflection can bedetermined by multiplying to the lateral deflection by a factor of 1.2 calculated using Equation 33. 5 Curve C 4 Curve B Lateral deflection (m) 3 Obtained from crash tests where 2 impact points are at least 20 m away from the upstream end -20% of Curve B Curve D 1 Obtained using Equation 33 Obtained by multiplying 1.2 to the Curve B result from Equation 33 0 0 10 20 30 40 50 60 70 80 90 100 Impact severity (kJ) 21Jiang, T., Grzebieta, R.H. and Zhao, X.L, Determining Lateral Deflections of Plastic Water-Filled Barriers,Paper submitted to International Journal of Crashworthiness.
  • 22. Figure 11 - Comparison of the plots of impact severity versus lateral deflection from the crash test and the calculations for the Triton barrierDETERMINING THE LATERAL DEFLECTION OF A PWB SYSTEM USINGMADYMO SIMULATIONSTwo computer codes, LS-DYNA3D and MADYMO have been extensively used in road safety barrierresearch and designs since the 1990s. This section describes how MADYMO was used to simulate aPWB crash test where a 2000 Chevrolet C2500 pickup was impacted into the GuardlinerTM Barrier ata speed of 70 km/h and an angle of 25 degrees (Zou and Grzebieta 2004). Figure 12 shows the testsetup. A total of 30 segments (60 m) were installed for the test. Figure 12 - Test setup for the GuardlinerTM BarrierVehicle modelThe vehicle was modelled using a multi-body system. The basic dimensions (width, length, wheelbaseand track), mass and centre of gravity of the vehicle model were basically the same as those measuredfrom the test vehicle. Ellipsoids were used to model the bumper, front body and side body. Force-displacement functions were assigned to these ellipsoids to reflect the crush characteristics of thebumper, front body and side body, respectively. A frontal rigid barrier crash test found in theNHTSA’s crash test database (NHTSA 2003) were used to derive the force-displacement function forthe frontal and bumper ellipsoids.In the frontal rigid barrier crash test (NHTSA Test No. 2809), a 1998 model Chevrolet pickup truck,which is essentially the same as the pickup truck used in the GuardlinerTM Barrier test, collided with arigid barrier at a speed of 56.1 km/h. The vehicle mass was 2328 kg, the vehicle width was 1950 mmand the average crush depth was 690 mm (NHTSA 2003). Thus, using the results from studies ofvehicle frontal crush characteristics (Campbell 1974; Strother et al. 1986; Jiang et al. 2003), thecoefficients A and B for this vehicle can be determined as follows. 22Jiang, T., Grzebieta, R.H. and Zhao, X.L, Determining Lateral Deflections of Plastic Water-Filled Barriers,Paper submitted to International Journal of Crashworthiness.
  • 23. V − b0 (56.1 / 3.6) − 2.2 m/s b1 = = = 19.40 ( ) C 0.690 m m V b0 b1 2328 × 2.2 × 19.40 A= = = 50953 (N/m) w0 1.950 m V b12 2328 × 19.4 × 19.40 B= = = 449316 (N/m 2 ) w0 1.950Hence, the force-crush equation for this vehicle’s front body was obtain as (Jiang et al. 2004): F = W ( A + B ⋅ C ) = 1.950 × (50953 + 449316C )The force-displacement function for the bumper and the front body ellipsoids was determined byevenly distributing the above equation to each ellipsoid modelling the bumper and the front body.Barrier modelThe barrier segment model was developed based on a previous study performed by Zou et al. (2000).Each segment of the water barrier system was modelled using a multi-body system. Planes wereattached to each segment to model the barrier shape. These planes were also used for contactdefinitions. A translational joint and two revolute joints were modelled between every two adjacentsegments to account for barrier elongation, torsion and rotation, respectively. The joint stiffness forthe rotational joints was determined using the results shown in Appendix A (Figure A.3). The jointstiffness for the elongation and torsion joints were also determined using the results of quasi-staticlaboratory component tests carried out by Zou et al. (2000).As shown in Figure 13, the W-beam and the water barrier segments are connected in such a way that“two bolts securing the segment joins are passed through the feet of a steel linkage bracket fitted intothe two apertures and are fixed to and tightened by a washered nut in the aperture on the oppositeside. A deformable C section is welded to the outer face of the linkage brackets to which a 4200 mmlong W-beam is attached by a single mushroom head keyed bolt. The beams are connected byoverlapping 280mm end sections and spliced by using eight 19mm diameter - 60mm long mushroomhead bolts. The centreline height of the W-beam is fixed at 600mm.”(Zou and Grzebieta 2004). C section Linkage bracket Figure 13 - Connection between the W-beam and the segments of the GuardlinerTM Barrier 23Jiang, T., Grzebieta, R.H. and Zhao, X.L, Determining Lateral Deflections of Plastic Water-Filled Barriers,Paper submitted to International Journal of Crashworthiness.
  • 24. Figure 14 - Barrier segment and W-beam modelsAs illustrated in Figure 14, the W-beam was modelled as a 2 m long section and each sectionconsisted of four identical bodies. An ellipsoid was attached to each body to approximately model theW-beam shape. Moreover, a force-displacement function was assigned to these ellipsoids so that thecrush characteristics of the W-beam could be simulated. The force-displacement function for the W-beam ellipsoids was derived using Reid et al.’s data, where 300 mm long W-beam sections werecompressed until flattened (Reid et al. 1997). Reid et al. (1997) found that the average peak crushforce from eight such compressing tests was 31 kN. Figure 15 shows the average force versusdisplacement curve of the W-beam flattening tests (Reid et al. 1997).A translational joint and two revolute joints were modelled between adjacent bodies of the W-beamsection to replicate W-beam elongation, torsion and bending (rotation) characteristics, respectively.The translational joint and two revolute joints were also used to model the continuity of the whole W-beam installation. The stiffness of the rotational (bending) joints was determined using the resultspresented in Appendix A. The stiffness of the elongation and torsion joints was assumed as 3.0E+04N/mm and 3.0E+05 Nm/rad, respectively.At each segment join bolt location, a point-restraint was used to model the connection between the W-beam and the segments. Figure 16 shows the whole model setup. 40 30 Force (kN) 20 10 0 0 20 40 60 80 Displacement (mm) Figure 15 - Force versus displacement of W-beam flattening test 24Jiang, T., Grzebieta, R.H. and Zhao, X.L, Determining Lateral Deflections of Plastic Water-Filled Barriers,Paper submitted to International Journal of Crashworthiness.
  • 25. Figure 16 - MADYMO model setup for the GuardlinerTM Barrier impactContact definitionsTo simulate the GuardlinerTM Barrier crash test, contact between the vehicle and the ground, betweenthe barrier segments and the ground, between the vehicle and the W-beam, and between the vehicleand the barrier segments had to be defined. Key parameter values and MADYMO contact functionsbetween the ellipsoids and the planes that were used are detailed in Table 4. The friction coefficientswas determined using the data provided by Bauer et al. and ROYMECH (Bauer et al. 2000;ROYMECH 2004). Table 4 - Contact characteristics No Contacts MADYMO function Key parameter values Vehicle mass mV = 2170 kg; Impact speed: 70 km/h; 1 Vehicle and ground PLANE-ELLIPSOID Impact angle: 25 degrees Rolling friction between vehicle wheels and ground: μV = 0.02 Segment mass mB = 630 kg; Barrier segments and 2 PLANE-ELLIPSOID Friction between barrier segments ground and ground μB = 0.40 W-beam section (2m) mass: 25 kg; Friction between vehicle body and 2 Vehicle and W-beam ELLIPSIOD-ELLIPSOID W-beam: 0.15 Friction between vehicle tyres and W-beam: 0.3 Friction between vehicle body and Vehicle and barrier PLANE-ELLIPSOID barrier segments: 0.2 3 segments ELLIPSIOD-ELLIPSOID Friction between vehicle tyres and barrier segments: 0.3 25Jiang, T., Grzebieta, R.H. and Zhao, X.L, Determining Lateral Deflections of Plastic Water-Filled Barriers,Paper submitted to International Journal of Crashworthiness.
  • 26. Comparison with the crash testThe model was run on a 1.8 GHz PC requiring only 5 minutes computation for a 1.5 s impact event.Altair MotionView was used to view the model animation. Figure 17 compares the sequentialoverhead views obtained from the crash test and the simulation.Figure 17 shows that the simulation basically replicates the impact test and the response of the barrier.In addition, the lateral deflection, the number of segments that moved laterally during impact and thebarrier up-stream end movement between the crash test and the simulation were also compared.Figure 18 shows the barrier up-stream end movement in the simulation. Figure 19 shows the barrierlateral deflection. In the crash test, the lateral deflection was measured as 2.64 m, the number ofsegments that moved laterally during impact was 16 segments and the up-stream end movement wasabout 0.52 m (Zou and Grzebieta 2004). Table 5 summarises these comparisons. Time Comparison Test 0.000 s Simulation Test 0.170 s Simulation 0.370 s Test 26Jiang, T., Grzebieta, R.H. and Zhao, X.L, Determining Lateral Deflections of Plastic Water-Filled Barriers,Paper submitted to International Journal of Crashworthiness.
  • 27. Simulation Test 0.770 s Simulation Figure 17 - Sequential overhead views comparing the crash test and simulation Before test After test Figure 18 - Simulated barrier up-stream end movement 27Jiang, T., Grzebieta, R.H. and Zhao, X.L, Determining Lateral Deflections of Plastic Water-Filled Barriers,Paper submitted to International Journal of Crashworthiness.
  • 28. Figure 19 - Simulated barrier lateral deflectionTable 5 - Comparison of the barrier response between the crash test and the simulation Lateral Number of segments Up-stream end deflection moved laterally movement Crash test 2.64 m 16 segments 0.52 m MADYMO simulation 2.50 m 14 segments 0.53 mDiscussionComparison of the results presented above indicates that the MADYMO models developed in thisstudy can be used to determine the lateral deflection and the number of segments that move duringimpact for the GuardlinerTM Barrier. To further validate the MADYMO models, the impact speed andangle were changed and the models were re-run. A total of 17 cases were simulated where the impactspeed ranged from 28 km/h to 100 km/h, the impact angle ranged from 15 degrees to 45 degrees. The 1impact severity, which is defined as m V (V sin θ ) 2 , ranged from 10 kJ to about 150 kJ. For each 2case, the lateral deflection and the number of segments that moved laterally during impact were alsocalculated using Equations 33 and 8, respectively. Details of how the lateral deflections for differentimpact severities were calculated are given in Appendix D.Results obtained from the MADYMO simulations and the calculations are compared in Table 6. Datafrom the two crash tests for this barrier are also listed in Table 6. Comparison of the results from theMADYMO analyses and the theoretical calculations are also illustrated as a plot of the impactseverity versus lateral deflection in Figure 20. Table 6 - Comparison of the results from the simulations and the calculations Impact parameters Impact Theoretical results MADYMO results No Impact Impact severity a Number of Lateral Number of Lateral speed angle, θ (kJ) segments that deflection segments that deflection (m) (km/h) (degrees) moved laterally (m) moved laterally 1 25.86 25 10.0 0.523 10 0.45 8 2 36.57 25 20.0 0.902 11 0.75 11 3b 70.0 20.5 20.7 0.80 13 --- --- 4 73.14 15 30.0 1.241 12 1.39 12 5 51.72 25 40.0 1.556 12 1.49 12 6 71.45 20 50.0 1.854 13 2.00 14 28Jiang, T., Grzebieta, R.H. and Zhao, X.L, Determining Lateral Deflections of Plastic Water-Filled Barriers,Paper submitted to International Journal of Crashworthiness.
  • 29. 7 48.88 30 50.0 1.854 13 1.54 13 8 63.35 25 60.0 2.140 13 2.08 14 9 84.54 20 70.0 2.415 14 2.66 14 10 44.99 40 70.0 2.415 14 2.64 13 11 b 70.0 25 73.3 2.503 14 2.50 14 12 86.50 20 73.3 2.503 14 2.80 15 13 59.17 30 73.3 2.503 14 2.30 13 14 73.14 25 80.0 2.682 14 2.63 14 15 57.16 35 90.0 2.943 14 2.70 14 16 46.37 45 90.0 2.943 14 3.00 13 17 81.78 25 100.0 3.198 14 3.31 15 18 91.43 25 125.0 3.816 15 3.96 16 19 100.0 25 149.5 4.401 15 4.70 16 1Note: a: Impact severity (kJ) is calculated as mV (V sin θ ) 2 . mV is 2170 kg for all cases except 2000 for No. 3 where mV is 891 kg. b: No. 3 and No. 11 are the same impact parameters as in the case of the two crash tests. 5.0 4.0 Lateral deflection (m) 3.0 2.0 Test data Theoretical curve (Eq. 33) 1.0 MADYMO data (θ = 25°) MADYMO data (15°< θ <25°) MADYMO data (25°< θ <45°) 0.0 0.0 30.0 60.0 90.0 120.0 150.0 Impact severity (kJ) Figure 20 - Impact severity versus lateral deflection for the GuardlinerTM barrierAs can be observed in both Table 6 and Figure 20, the results from the MADYMO simulationscorrelated well to the test data and the theoretical data. It can be concluded that the MADYMOmodels developed here can be used to determine the lateral deflection and the number of segmentsthat move for the GuardlinerTM Barrier under different impact conditions with reasonable accuracy.The modelling techniques developed can also be used to model other PWB systems listed examples ofwhich are presented in Table 1. 29Jiang, T., Grzebieta, R.H. and Zhao, X.L, Determining Lateral Deflections of Plastic Water-Filled Barriers,Paper submitted to International Journal of Crashworthiness.
  • 30. CONCLUSIONSThis study investigated two methods that can be used to calculate the lateral deflection and thenumber of segments that move in a PWB after a vehicle has struck the barrier. In the first methodtheoretical equations were developed based on the conservation of energy whereas the second methodused MADYMO simulations to calculate how the barrier deformed.The theoretical equations and the MADYMO simulation models developed in this study werevalidated against full-scale crash tests. A total of 23 full-scale crash tests were used to validate thetheoretical equations. Comparison of the results indicate that both the theoretical method andMADYMO modelling simulation can be used to determine the PWB’s lateral deflection and thenumber of segments that move during impact with reasonable accuracy.The significance of this study is that cost-effective methods developed in this research can be used sothat lateral deflections of almost all certified PWB systems under different impact conditions can bedetermined, as confirmed in Figure 20. Had full-scale crash tests been used to obtain the curve asshown in Figure 20, the cost would have been considerable.The findings of this study would be useful for both PWB providers and users. PWB designers cannow provide more information about their PWB products such as the possible working width and thenumber of segments that would most likely move for different impact severities. Highway engineerscan also use the findings of this study to properly select and install a PWB system for different siteand traffic conditions.ACKNOWLEDGEMENTSThe authors would like to thank the Australian Research Council for providing funds to investigateroadside barrier crashes. 30Jiang, T., Grzebieta, R.H. and Zhao, X.L, Determining Lateral Deflections of Plastic Water-Filled Barriers,Paper submitted to International Journal of Crashworthiness.
  • 31. REFERENCESASTM (2002). A53/A53M-01: Standard Specification for Pipe, Steel, Black and Hot-Dipped, Zinc-coated, Welded andSeamless. Annual Book of ASTM Standards 2002: Section One - Iron and Steel Products Volume 01.01. WestConshohocken, PA, USA, American Society for Testing and Materials.Bauer, H., K. H. Dietsche, J. Crepin and F. Dinkler (2000). Automotive Handbook (5th ed.). Stuttgart, Robert Bosch GmbH.Bronshtein, I. N., K. A. Semendyayev, G. Musiol and H. Muehlig (2004). Handbook of Mathematics. New York, Springer.Campbell, K. L. (1974). "Energy Basis for Collision Severity." SAE Technical Paper 740565.Cichowski, W. G., P. C. Skeels and W. R. Hawkins (1961). Appraisal of Guardrail Installations by Car Impact andLaboratory Tests. In: Highway Research Board Proceedings Vol. 40, Highway Research Board, Washington, D.C.Deleys, N. J. and R. R. McHenry (1967). NCHRP Report 36: Highway Guardrails-A Review of Current Practice.Washington, D.C., Highway Research Board, National Research Council.Energy Absorption Systems Inc. (2004). Workzone Products. Energy Absorption Syatems, Inc. [online] Available:http://www.energyabsorption.com/products/workzone/workzone_safety.htm.FHWA (2004). Longitudinal Barriers and Miscellaneous Items. Federal Highway Administration(http://safety.fhwa.dot.gov/fourthlevel/hardware/listing.cfm?code=long). [online] Available:http://safety.fhwa.dot.gov/fourthlevel/hardware/listing.cfm?code=long.Grzebieta, R. H., J. Cameron, A. Carey and R. Zou (2001). "Water-Filled Plastic Safety Barrier Systems." Road & TransportResearch Vol 10(No 3): pp. 66-83.Grzebieta, R. H. and R. Zou (1998-2000). Water-filled Barrier Crash Test Reports. Melbourne, Australia, Department ofCivil Engineering, Monash University.Grzebieta, R. H. and R. Zou (2003). NCHRP Report 350 Test 2-10 of a 920 mm High Barron & Rawson Barrier Model2000S, Reinforced with W-Beam. Melbourne, Australia, Department of Civil Engineering, Monash University.Jiang, T., R. H. Grzebieta, G. Rechnitzer, S. Richardson and X. L. Zhao (2003). Review of Car Frontal Stiffness Equationsfor Estimating Vehicle Impact Velocities. 18th International Technical Conference on the Enhanced Safety of Vehicles,Nagoya, Japan.Jiang, T., R. H. Grzebieta and X. L. Zhao (2004). "Predicting impact loads of a car crashing into a concrete roadside safetybarrier." International Journal of Crashworthiness Vol 9(No. 1): 45-63.Jiang, T., R. H. Grzebieta, X. L. Zhao, R. Zou, G. Rundle and C. Powell (2002). Methods to Predict Dynamic Performanceof Water-Filled Plastic Barriers. International Crashworthiness Conference - ICrash2002, Melbourne, Australia.Megson, T. H. G. (1996). Structural and Stress Analysis. London, Arnold.NHTSA (2003). Vehicle Crash Test Database. National Highway Traffic Safety Administration. [online] Available:http://www-nrd.nhtsa.dot.gov/database.Powers, D. (26 October, 2004). Test Results for the 426 Water-Filled Plastic Barrier (FHWA List Code B111).Reid, J. D., D. L. Sicking, R. K. Faller and B. G. Pfeifer (1997). Development of a New Guardrail System. TransportationResearch Record No. 1599. Washington, D.C., Transportation Research Board, National Academy Press.Rhino Safety Barrier LLC (2001). Testing Videos and Technical Documentation. Rhino Safety Barrier LLC. [online]Available: http://www.rhinobarriers.com/testing.cfm.Rosenbach, J. B., E. A. Whitman and D. Moskovitz (1961). Essentials of Trigonometry, with tables. Boston, GINN ANDCOMPANY.ROYMECH (2004). Friction Factors. Roy Beardmore. [online] Available:http://www.roymech.co.uk/Useful_Tables/Tribology/co_of_frict.htm. 31Jiang, T., Grzebieta, R.H. and Zhao, X.L, Determining Lateral Deflections of Plastic Water-Filled Barriers,Paper submitted to International Journal of Crashworthiness.
  • 32. Standards Australia (1992). AS 1594 -1992 Hot-rolled steel flat rpoducts. Sydney, Standards Australia.Standards Australia and Standards New Zealand (1999). AS/NZS 3845:1999 - Road safety barrier systems.Sydney/Wellington, Standards Australia/Standards New Zealand.Strother, C. E., R. L. Woolley, M. B. James and C. Y. Warner (1986). "Crush Energy in Accident Reconstruction." SAETechnical Paper 860371.Yodock Wall Company (2003 - 2004). Highway Safety & Traffic Control. Yodock Wall Company. [online] Available:http://www.waterbarrier.com/applications/highHome.asp.Zou, R. and R. H. Grzebieta (2004). NCHRP REPORT 350 Crash Test Results for the GuardlinerTM Barrier System(Prepared for: WTG Aussindo Inc., 5 Carol Ave, Springwood, Qld 4127, Australia and Barron & Rawson Pty Ltd, 35-37Marigold Street Revesby, NSW 2212, Australia). Melbourne, Australia, Department of Civil Engineering, MonashUniversity.Zou, R., R. H. Grzebieta, G. Rundle and C. Powell (2000). Development of a Temporary Water-filled Plastic Barrier System.International Crashworthiness Conference, London, UK. 32Jiang, T., Grzebieta, R.H. and Zhao, X.L, Determining Lateral Deflections of Plastic Water-Filled Barriers,Paper submitted to International Journal of Crashworthiness.
  • 33. APPENDIX A: DETERMINING M(α) AND MYM(α) and MY for every PWB listed in Table 1 were determined and reported as follows.Triton barrier, 426 Barrier, SB-1-TL barrier and Barrier System barrierWith regard to the Triton barrier (test No. 1 in Table 1), the 426 Barrier (test No. 5 in Table 1), theSB-1-TL barrier (test No. 6 in Table 1) and the Barrier system barrier (test No. 9 in Table 1), imagesof these barriers given in Table 1 show that the joints of these PWBs consist of pins (the Triton barrierand the Barrier system barrier) or use steel cables (the 426 Barrier and the SB-1-TL barrier) toconnect segments. M(α) for these PWBs are small and thus can be assumed as negligible.As no steel beams are used to strengthen these PWBs, EBEAM or MY for these barriers is zero.Roadliner 2000™ S barrierWith regard to the Roadliner 2000™ S barrier (test No. 8 in Table 1), bolts and short steel latches areused to connect segments (see Table 1). Data of two quasi-static laboratory tests performed byGrzebieta and Zou (Grzebieta and Zou 1998-2000; Zou et al. 2000; Grzebieta et al. 2001) were usedto determine M(α). Figure A.1 shows the two joints that were tested. The barriers were empty whentested. Joint 1 and Joint 2 are quite similar except that Joint 2 has two additional short steel latches atthe bottom. Figure A.2 shows the test data from both Joint 1 and Joint 2 tests and the associated plots. Joint 1 Joint 2 Figure A.1 - Test set-up for determining the joint stiffness of a PWB 33Jiang, T., Grzebieta, R.H. and Zhao, X.L, Determining Lateral Deflections of Plastic Water-Filled Barriers,Paper submitted to International Journal of Crashworthiness.
  • 34. 7000 6000 Joint 2 Moments, M (Nm) 5000 4000 Joint 1 3000 2000 1000 0 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 Rotational angle, α (rad) Figure A.2 - Joint stiffness test data for Joint 1 and Joint 2Although the PWB used in the joint rotational stiffness tests was a modified design as shown in testNo. 8 of Table 1 (Roadliner 2000™ S), the joint design of the Roadliner 2000™ S barrier stillremained similar to Joint 2 except that two more short steel latches are added to the top of the barrier(see test No. 8 in Table 1). Hence, the joint rotational stiffness of the Roadliner 2000™ S barrier wasdetermined as M Roadliner 2000 ™ S (α i ) = M Joint 2 (α i ) + [ M Joint 2 (α i ) − M Joint 1 (α i )] [A.1]where MRoadliner 2000™ S (αi) is the moment required to rotate the joint of the Roadliner 2000™ S barrierto an angle αi (Nm), MJoint 2 (αi) is the moment required to rotate Joint 2 to an angle αi (Nm) and MJoint 1(αi) is the moment required to rotate Joint 1 to an angle αi (Nm). As shown in Figure A.3, M(α) of theRoadliner 2000™ S barrier was determined as M Roadliner 2000 ™ S (α ) = 31880 α (when α ≤ 0.2 rad or 12 degrees) [A.2]No steel beams are used to strengthen the Roadliner 2000™ S barrier. Thus, EBEAM or MY for thisbarrier is also zero. 10000 M = 6962α + 5052 9000 R 2 = 0.9988 Roadline r 2000™ S 8000 Moments, M (Nm) 7000 M =31880α M = 12835α 6000 R 2 =0.9983 R 2 = 0.9988 Joint 2 5000 4000 Rihno barrier 3000 2000 Joint 1 1000 M = 4548 α R ² = 0.9996 0 0.0 0.2 0.4 0.6 0.8 Rotational angle, α (rad) Figure A.3 - Results of joint stiffness 34Jiang, T., Grzebieta, R.H. and Zhao, X.L, Determining Lateral Deflections of Plastic Water-Filled Barriers,Paper submitted to International Journal of Crashworthiness.
  • 35. Rhino barrierWith regard to the Rhino barrier, the joint of the Rihno barrier, as shown in test No. 4 in Table 1,consists of a steel-reinforced polyethylene pin and two short galvanized steel “bridging strips”. Thejoint stiffness is mainly determined by the stiffness of these two short steel strips. Hence, M(α) of theRihno barrier can also be approximated using the data of Joint 1 and Joint 2 tests as M Rihno (α i ) = M Joint 2 (α i ) − M Joint 1 (α i ) [A.3]where MRihno (αi) is the moment required to rotate the joint of the Rihno barrier to an angle αi (Nm).Figure A.3 shows that M(α) of the Rihno barrier can be determined as M Rihno (α ) = 12835 α (when α ≤ 0.2 rad or 12 degrees) [A.4]No steel beams are used to strengthen the Rihno barrier. Thus, EBEAM or MY for this barrier is alsozero.Guardian 350 Highway Kit barrierWith regard to the Guardian 350 Highway Kit barrier (test No. 2 in Table 1), four schedule 40 ASTMA120 steel tubes are used to strengthen the Guardian 350 Highway Kit barrier (FHWA 2004). Theoutside diameter of the tube is 60.3 mm and the thickness of the tube wall is 3.91 mm (ASTM 2002).Hence, the second moment of area for this tube is (Megson 1996): π [60.3 4 − (60.3 − 2 × 3.91) 4 ] I= = 276647 (mm 4 ) = 276647 × 10 −12 (m 4 ) 64σY for the tube is 240 MPa (ASTM 2002). Thus, MY for this tube can be calculated using Equation 28as σ I 240 × 10 6 × 276647 × 10 −12 MY = Y = = 2202.2 (Nm) z (60.3 / 2) × 10 −3The image of the barrier shown in Table 1 indicates that M(α) of this barrier can be considered asnegligible.Yodock barrierEach segment of the Yodock barrier has two 1830 mm long square steel tubes, which are spliced tothe square tubes of the adjacent segment with 280 mm long 63.5 × 63.5 × 6.35 mm square steel tubesusing two bolts (Yodock Wall Company 2003 - 2004; FHWA 2004). The second moment of area forthis square tube is (Megson 1996): [63.5 4 − (63.5 − 2 × 6.35) 4 ] I= = 799945 (mm 4 ) = 799945 × 10 −12 (m 4 ) 12σY for this steel tube is also assumed as 240 MPa (ASTM 2002) and its respective MY is σYI 240 × 10 6 × 799945 × 10 −12 MY = = = 6046.8 (Nm) z (63.5 / 2) × 10 −3The segments of the Yodock barrier are connected at the ends using polyethylene couplers (YodockWall Company 2003 - 2004; FHWA 2004). Thus, M(α) of the barrier can also be considered asnegligible. 35Jiang, T., Grzebieta, R.H. and Zhao, X.L, Determining Lateral Deflections of Plastic Water-Filled Barriers,Paper submitted to International Journal of Crashworthiness.
  • 36. GuardlinerTM barrierAs shown in test No. 7 in Table 1, the GuardlinerTM barrier is assembled by securing steel W-beamrails to the Roadliner 2000™ S barrier (test No. 8 in Table 1). As mentioned above, the secondmoment of area of the cross-section of the W-beam was 970000 mm4 or 970.0E-09 m4 (Deleys andMcHenry 1967; Reid et al. 1997). The W-beam was formed from steel grade HA350 or equivalent inaccordance with AS 1594 (Standards Australia and Standards New Zealand 1999). The yield strengthfor this steel (σY) is 350 MPa (Standards Australia 1992). Also according to AS/NZS 3845 (StandardsAustralia and Standards New Zealand 1999), the furthest point from the neutral axis (z) is 40 mm.Thus, MY for this W-beam was calculated using Equations 28 as σYI 350 × 10 6 × 970 × 10 −9 MY = = = 8487.5 (Nm) z 40 × 10 −3M(α) of the GuardlinerTM barrier can also be determined using Equation A.2. 36Jiang, T., Grzebieta, R.H. and Zhao, X.L, Determining Lateral Deflections of Plastic Water-Filled Barriers,Paper submitted to International Journal of Crashworthiness.
  • 37. APPENDIX B: DERIVATION OF RESULTS IN TABLE 3With regard to the first test of the Triton barrier (test No. 1 in Table 1), the vehicle mass mV was 807.3kg, the impact speed V was 72.0 km/h, the impact angle θ was 20 degrees, 30 segments were installedin the crash test, the length of one segment of this barrier lB was 1.981 m and the weight of onesegment filled with water mB was 612 kg (Energy Absorption Systems Inc. 2004; FHWA 2004). μBwas also assumed as 0.4. As determined in the above section, M(α) and MY for this barrier can beignored. Thus, substituting these values into Equation 33 gives 1 72.0 × 807.3 × ( sin 20°) 2 = 2 3.6 (612 × 9.81 × 0.4 × 1.981)(n − 1) 2 D (612 × 9.81 × 0.4 × 1.981)nD (612 × 9.81 × 0.4) D + + + 2 2 ⎡ (n − 1) ⎤ ⎡ (n − 1) ⎤ 4⋅ D + ⎢2 × 1.981⎥ 2⋅ D + ⎢2 × 1.981⎥ ⎣ 2 ⎦ ⎣ 2 ⎦ ⎡ ⎡ (n − 1) ⎤ 2 (n − 1) ⎤ (612 × 9.81 × 0.4)(30 − n) ⎢ D 2 + ⎢ × 1.981⎥ − × 1.981⎥ ⎢ ⎣ 2 ⎦ 2 ⎥ ⎣ ⎦or 4757.35 (n − 1) 2 D 4757.35nD 18887.23 = 2401.49 D + + + 4 ⋅ D + [0.9905 (n − 1)] 2 ⋅ D + [0.9905 (n − 1)] 2 2 2 2 [B.1] 2401.49 (30 − n) ⎡ D 2 + [0.9905 (n − 1)] − 0.9905 ( n − 1)⎤ 2 ⎢ ⎣ ⎥ ⎦where n is calculated using Equation 8 as 72.0 807.3 × ( sin 20°) 2 mV (V sin θ ) 2 3.6 7.86 n= +2= +2= +2 [B.2] 2 D ⋅ m B gμ B 2 × (612 × 9.81 × 0.4) D DThe solution of Equation B.1 can be determined using the Iteration Method for solving non-linearequations in a single unknown (Bronshtein et al. 2004). The basic step is to assign a value for D first;then calculate n using Equation B.2; substitute the assigned value for D and the calculated n intoEquation B.1 and compare the right hand side value with the left hand side value of Equation B.1;repeat the above process and D can be determined until the difference is less than 0.0001. Thisprocess can be easily done using a worksheet. D and n determined for this test was 1.033 m and 9.6(rounded up to 10), respectively. The lateral deflection from the crash test was 1.0 m (EnergyAbsorption Systems Inc. 2004; FHWA 2004). The difference between the calculated lateral deflection 1.033 − 1.0and the test result was determined as × 100% = +3.3% . The difference between the 1. 0calculated number of segments that moved laterally during impact and the test result could not bedetermined because this data was not available.Similarly, the calculated lateral deflection, the calculated number of segments that moved laterallyduring impact and the difference between the calculated and the test result for the Triton barrier (testNo. 1 in Table 1), the 426 Barrier (test No. 5 in Table 1), the SB-1-TL barrier (test No. 6 in Table 1) 37Jiang, T., Grzebieta, R.H. and Zhao, X.L, Determining Lateral Deflections of Plastic Water-Filled Barriers,Paper submitted to International Journal of Crashworthiness.
  • 38. and the Barrier system barrier (test No. 9 in Table 1) were obtained using the same procedure asdescribed above.With regard to the 2000P Test Level 2 test for the GuardlinerTM barrier (test No. 7 in Table 1), thevehicle mass mV was 2170 kg, the impact speed V was 70.0 km/h, the impact angle θ was 25 degrees,30 segments were installed in the crash test, the length of one segment of this barrier lB was 2.0 mand the weight of one segment filled with water mB was 655 kg (Zou and Grzebieta 2004). μB wasdetermined as 0.4. NBEAM for this barrier was 1. MY of the W-beam rail was determined in the abovesection as 8487.5 Nm. The second moment of area of the W-beam was 970.0E-09 m4. E can be takenas 210.0E+09 N/m2. M(α) of the GuardlinerTM barrier was as given in Equation A.2. Substitutingthese values and Equations 19 and A.2 into Equation 33 results in 1 70.0 × 2170 × ( sin 25°) 2 = 2 3 .6 (655 × 9.81 × 0.4 × 2.0)(n − 1) 2 D α (655 × 9.81 × 0.4) D + + n ∫ 31880α dα + 2 0 ⎡ (n − 1) ⎤ 4 ⋅ D2 + ⎢ × 2 .0 ⎥ ⎣ 2 ⎦ (655 × 9.81 × 0.4 × 2.0) nD + 2 ⎡ (n − 1) ⎤ 2 ⋅ D2 + ⎢ × 2 .0 ⎥ ⎣ 2 ⎦ ⎡ ⎡ (n − 1) ⎤ 2 (n − 1) ⎤ (30 − n)(655 × 9.81 × 0.4) ⎢ D 2 + ⎢ × 2.0⎥ − × 2 .0 ⎥ + ⎢ ⎣ 2 ⎦ 2 ⎥ ⎣ ⎦ ⎡ ⎤ ⎢ ⎥ ⎢ (2 × 8487.5) D (n − 1)(8487.5 2 × 2.0) ⎥ 1× ⎢ − 2 6 × 210 × 10 9 × 970 × 10 −9 ⎥ ⎢ D 2 + ⎡ (n − 1) × 2.0⎤ ⎥ ⎢ ⎢ 2 ⎥ ⎥ ⎣ ⎣ ⎦ ⎦Through some calculations of the above equation, the following equation is obtained. 38Jiang, T., Grzebieta, R.H. and Zhao, X.L, Determining Lateral Deflections of Plastic Water-Filled Barriers,Paper submitted to International Journal of Crashworthiness.
  • 39. 5140.44 (n − 1) 2 D 31880 2 5140.44 nD 73268.5 = 2570.22 D + +n( α )+ + 4 ⋅ D + (n − 1) 2 2 2 2 ⋅ D 2 + (n − 1) 2 2570.22 (30 − n) [D 2 ] ⎡ 16975 D + (n − 1) 2 − (n − 1) + ⎢ ⎢ D 2 + (n − 1) 2 ⎤ − 117.88(n − 1)⎥ ⎥ ⎣ ⎦ 1285.11 (n − 1) 2 D 15940nD 2 2570.22 nD = 2570.22 D + + + + D 2 + (n − 1) 2 D + (n − 1) 2 2 D 2 + (n − 1) 2 2570.22 (30 − n) [D 2 ] ⎡ 16975 D + (n − 1) 2 − (n − 1) + ⎢ ⎢ D 2 + (n − 1) 2 ⎤ − 117.88(n − 1)⎥ ⎥ ⎣ ⎦where n is calculated using Equation 8 as 70 2170 × ( sin 25°) 2 mV (V sin θ ) 2 3.6 28.51 n= +2= +2= +2 2 D ⋅ mB gμ B 2 × (655 × 9.81 × 0.4) D DUsing the same method that was used to solve Equation B.1, the solution of the above equation wascalculated as D = 2.503 (m) and n = 13.4 ≈14.The lateral deflection and the number of segments that moved laterally from the crash test was 2.64 mand 16, respectively (Zou and Grzebieta 2004). The difference between the calculated lateral 2.503 − 2.64deflection and the test result was determined as × 100% = −5.2% . The difference 2.64between the calculated number of segments that moved laterally during impact and the test result wasdetermined as: n (from the calculation) - n (from the crash test) = 14 – 16 = -2.Similarly, the calculated lateral deflection, the calculated number of segments that moved laterallyduring impact and the difference between the calculated and the test result for the Guardian 350Highway Kit barrier (test No. 2 in Table 1), the Yodock barrier (No. 3 in Table 1), the Rhino barrier(No. 4 in Table 1) and the Roadliner 2000™ S barrier (test No. 8 in Table 1) were also obtained in asimilar way.The worksheet that was used to calculate D and n for all of the crash tests listed in Table 1 is given inTable B.1. 39Jiang, T., Grzebieta, R.H. and Zhao, X.L, Determining Lateral Deflections of Plastic Water-Filled Barriers,Paper submitted to International Journal of Crashworthiness.
  • 40. Table B.1 - Worksheet used to calculate D and n for all of the crash tests listed in Table 1 A B C D E F G H I J K L M N O P Q R S T U 1 mV V θ mB μB lB N N BEAM M (α )= M Y D n D Long. ELa E BR-Lat.-NR E BR-Lat.-R E BR-Long. E BEAM ∑E b Difference c 2 (kg) (km/h) (deg) (kg) (m) kJα (Nm) (m) (m) (J) (J) (J) (J) (J) (J) 3 kJ Eq. 22 Eq. 15 Eq. 17 Eq. 23 Eq. 32 4 Triton barrier (Nm/rad) 5 TL-2 807 72 20.0 612 0.4 1.981 30 0 0 0 1.0331 9.613 0.0623 18887.231 13087.591 2748.95 3051.47 0 18888.01 4.123E-05 6 TL-2 1971 72.3 25.0 612 0.4 1.981 30 0 0 0 3.026 11.77 0.421 70976.494 44902.946 7640.228 18433.3 0 70976.479 -2.109E-07 7 TL-3 875 97.04 21.0 620 0.4 1.981 30 0 0 0 1.9165 10.76 0.1882 40825.634 26972.207 5042.328 8811.94 0 40826.476 2.061E-05 8 TL-3 2005 97.56 25.0 620 0.4 1.981 30 0 0 0 4.945 12.93 0.9929 131498.1 78232.235 12028.3 41235.3 0 131495.86 -1.7E-05 9 2004 95.74 7.0 620 0.4 1.981 50 0 0 0 0.5563 9.777 0.0178 10525.397 7280.6925 1504.535 1739.97 0 10525.2 -1.875E-05 10 Guardian 350 Highway Kit 11 TL-2 820 70.6 20.3 880 0.4 1.830 33 4 0 2202.2 0.7076 9.768 0.0311 18979.558 13113.495 2711.589 2498.66 655.014 18978.761 -4.204E-05 12 TL-2 2000 71.5 25.8 880 0.4 1.830 33 4 0 2202.2 2.1595 12.02 0.2286 74721.728 47635.336 7953.33 16564.4 2566.99 74720.023 -2.283E-05 13 TL-3 820 100 20.0 880 0.4 1.830 33 4 0 2202.2 1.2214 10.77 0.083 37006.858 24640.318 4606.385 6371.4 1388.45 37006.552 -8.255E-06 14 TL-3 2000 100 25.0 880 0.4 1.830 33 4 0 2202.2 3.5464 13.25 0.5475 137813.42 83782.702 12628.85 37332.2 4065.11 137808.82 -3.337E-05 15 Yodock barrier 16 TL-2 2042 68.5 24.0 460 0.4 1.830 25 2 0 6046.8 3.267 12.37 0.5009 61154.507 37882.9 6121.007 11418.9 5736.71 61159.465 8.107E-05 17 TL-3 896 97.7 19.8 786 0.4 1.830 25 2 0 6046.8 1.4304 10.58 0.1159 37860.798 25272.188 4808.58 5154.83 2622.46 37858.061 -7.227E-05 18 TL-3 2041 98.4 24.8 786 0.4 1.830 25 2 0 6046.8 4.2154 12.32 0.8251 134141.42 81143.365 13105.94 32275.2 7616.65 134141.19 -1.746E-06 19 Rhino barrier d 20 TL-2 917 72.6 20.0 476 0.4 2.000 34 0 0 0 1.3765 10.48 0.0994 21812.776 14636.557 2812.684 4364.79 0 21814.031 5.755E-05 21 TL-2 2000 69.2 25.0 476 0.4 2.000 34 0 12835 0 3.082 13.46 0.3754 65993.884 40582.981 11015.39 14399.2 0 65997.625 5.669E-05 22 426 barrier 23 TL-3 820 100.7 20.0 782 0.4 1.830 32 0 0 0 1.431 10.55 0.1164 37526.767 25074.33 4787.293 7665.88 0 37527.499 1.95E-05 24 TL-2 2000 73 25.0 782 0.4 1.830 32 0 0 0 2.4515 11.76 0.3005 73440.77 46805.398 7978.062 18663.9 0 73447.395 9.02E-05 25 SB-1-TL barrier 26 TL-3 2054 99.4 25.4 836 0.4 2.130 22 0 0 0 4.91 10.95 1.0824 144052.71 88750.657 16073.49 39216 0 144040.16 -8.712E-05 27 TM. Guardlin er barrier 28 TL-2 891 70 20.5 655 0.4 2.000 30 1 31880 8487.5 0.9253 10.69 0.0441 20658.052 13844.167 4152.204 2188.89 472.367 20657.625 -2.065E-05 29 TL-2 2170 70 25.0 655 0.4 2.000 30 1 31880 8487.5 2.5031 13.39 0.2503 73268.504 45495.179 15185.77 10688.5 1901.46 73270.921 3.299E-05 30 TM. Roadliner 2000 S barrier 31 TL-0 920 49 20.0 630 0.4 2.000 30 0 31880 0 0.5413 9.45 0.0173 9968.9254 6980.1355 2109.093 879.924 0 9969.1529 2.282E-05 32 TL-0 1570 48 25.0 630 0.4 2.000 30 0 31880 0 1.1326 10.9 0.0646 24925.486 16572.829 5306.838 3048.11 0 24927.776 9.187E-05 33 Barrier System barrier 34 TL-0 910 50 20.0 950 0.4 2.000 24 0 0 0 0.4425 8.224 0.0135 10267.146 7596.7544 1874.375 796.239 0 10267.369 2.167E-05 35 TL-0 1560 49 25.0 950 0.4 2.000 24 0 0 0 0.9874 9.012 0.0606 25809.421 18315.235 4109.164 3386.76 0 25811.159 6.734E-05 36 Note: a. E L = ½ m V (V sinθ /3.6)2 b. ∑E = E BR-Lat.-NR + E BR-Lat.-R + E BR-Long. + E BEAM 37 c. Difference = ( ∑E - E L ) / E L d. In this test, no short steel “bridging strips” were used. Thus, M (α ) was ignored. 40Jiang, T., Grzebieta, R.H. and Zhao, X.L, Determining Lateral Deflections of Plastic Water-Filled Barriers,Paper submitted to International Journal of Crashworthiness.
  • 41. APPENDIX C: WORKSHEET USED TO CALCULATE D AND n FOR THE TRITON BARRIER A B C D E F G H I J K L M N O P Q R S T U 1 mV V θ mB μB lB N N BEAM M (α )= M Y D n D Long. EL a E BR-Lat.-NR E BR-Lat.-R E BR-Long. E BEAM ∑E b Difference c 2 (kg) (km/h) (deg) (kg) (m) kJα (Nm) (m) (m) (J) (J) (J) (J) (J) (J) 3 kJ Eq. 22 Eq. 15 Eq. 17 Eq. 23 Eq. 32 4 (Nm/rad) 5 Triton barrier 6 7 2004 95.74 7.0 620 0.4 1.981 50 0 0 0 0.5563 9.777 0.0178 10525.397 7280.6925 1504.535 1739.97 0 10525.2 -1.875E-05 8 807 72 20.0 612 0.4 1.981 30 0 0 0 1.0331 9.613 0.0623 18887.231 13087.591 2748.95 3051.47 0 18888.01 4.123E-05 9 2000 47 25.0 620 0.4 1.981 30 0 0 0 1.511 10.28 0.1233 30442.984 20509.754 4018.239 5917.34 0 30445.328 7.7E-05 10 875 97.04 21.0 620 0.4 1.981 30 0 0 0 1.9165 10.76 0.1882 40825.634 26972.207 5042.328 8811.94 0 40826.476 2.061E-05 11 2000 60 25.0 620 0.4 1.981 30 0 0 0 2.2433 11.09 0.2487 49612.83 32324.497 5852.936 11439.6 0 49617.052 8.51E-05 12 2000 65 25.0 620 0.4 1.981 30 0 0 0 2.552 11.38 0.312 58226.169 37476.978 6606.416 14137 0 58220.441 -9.837E-05 13 1971 72.3 25.0 612 0.4 1.981 30 0 0 0 3.026 11.77 0.421 70976.494 44902.946 7640.228 18433.3 0 70976.479 -2.109E-07 14 2000 76 25.0 620 0.4 1.981 30 0 0 0 3.2855 11.96 0.4863 79601.03 49912.142 8348.567 21346.9 0 79607.609 8.265E-05 15 2000 80 25.0 620 0.4 1.981 30 0 0 0 3.57 12.16 0.5624 88200.587 54781.939 9005.58 24417.5 0 88204.992 4.994E-05 16 2000 85.1 25.0 620 0.4 1.981 30 0 0 0 3.947 12.39 0.6703 99804.615 61239.088 9859.742 28712.7 0 99811.571 6.97E-05 17 18 19 Note: a. E L = ½ m V (V sinθ /3.6)2 b. ∑E = E BR-Lat.-NR + E BR-Lat.-R + E BR-Long. + E BEAM 20 c. Difference = ( ∑E - E L ) / E L 41Jiang, T., Grzebieta, R.H. and Zhao, X.L, Determining Lateral Deflections of Plastic Water-Filled Barriers,Paper submitted to International Journal of Crashworthiness.
  • 42. APPENDIX D: WORKSHEET USED TO CALCULATE D AND n FOR THE GUARDLINERTM BARRIER A B C D E F G H I J K L M N O P Q R S T U 1 mV V θ mB μB lB N N BEAM M (α )= M Y D n D Long. EL a E BR-Lat.-NR E BR-Lat.-R E BR-Long. E BEAM ∑E b Difference c 2 (kg) (km/h) (deg) (kg) (m) kJα (Nm) (m) (m) (J) (J) (J) (J) (J) (J) 3 kJ Eq. 22 Eq. 15 Eq. 17 Eq. 23 Eq. 32 4 (Nm/rad) 5 Guardlin erTM. barrier 6 7 2170 25.86 25.0 655 0.4 2.000 30 1 31880 8487.5 0.5233 9.435 0.0162 9999.5 7006.3587 2078.226 857.23 56.8574 9998.6725 -8.275E-05 8 2170 36.57 25.0 655 0.4 2.000 30 1 31880 8487.5 0.9019 10.63 0.0422 19997.291 13427.12 4021.58 2099.12 448.607 19996.423 -4.341E-05 9 2170 73.14 15.0 655 0.4 2.000 30 1 31880 8487.5 1.2412 11.4 0.0738 30000.403 19668.588 6023.033 3526.16 784.403 30002.188 5.949E-05 10 2170 51.72 25.0 655 0.4 2.000 30 1 31880 8487.5 1.556 12 0.1095 39998 25781.105 8071.629 5065.22 1080.38 39998.327 8.185E-06 11 2170 71.45 20.0 655 0.4 2.000 30 1 31880 8487.5 1.854 12.49 0.1486 49995.689 31796.107 10164.93 6686.68 1348.96 49996.679 1.98E-05 12 2170 48.88 30.0 655 0.4 2.000 30 1 31880 8487.5 1.8543 12.49 0.1486 50006.577 31802.557 10167.05 6688.27 1349.18 50007.06 9.666E-06 13 2170 63.35 25.0 655 0.4 2.000 30 1 31880 8487.5 2.1398 12.91 0.1907 60008.736 37737.924 12301.8 8374.96 1597.34 60012.025 5.48E-05 14 2170 84.54 20.0 655 0.4 2.000 30 1 31880 8487.5 2.4148 13.28 0.2352 69992.673 43589.926 14468.6 10110.5 1828.78 69997.79 7.311E-05 15 2170 44.99 40.0 655 0.4 2.000 30 1 31880 8487.5 2.4153 13.28 0.2353 70015.099 43602.7 14472.21 10112.8 1828.96 70016.689 2.271E-05 16 2170 86.5 20.0 655 0.4 2.000 30 1 31880 8487.5 2.5031 13.39 0.2503 73275.756 45498.872 15184.99 10686.8 1901.04 73271.728 -5.496E-05 17 2170 59.17 30.0 655 0.4 2.000 30 1 31880 8487.5 2.5031 13.39 0.2503 73277.032 45499.522 15184.85 10686.5 1900.96 73271.87 -7.044E-05 18 2170 73.14 25.0 655 0.4 2.000 30 1 31880 8487.5 2.682 13.6 0.2822 79989.163 49383.245 16670.84 11892.2 2047.28 79993.546 5.479E-05 19 2170 57.16 35.0 655 0.4 2.000 30 1 31880 8487.5 2.9425 13.9 0.3314 89989.485 55117.392 18903.97 13713.1 2254.83 89989.305 -1.997E-06 20 2170 46.37 45.0 655 0.4 2.000 30 1 31880 8487.5 2.943 13.9 0.3315 90005.474 55126.715 18908.71 13717.5 2255.4 90008.376 3.224E-05 21 2170 81.78 25.0 655 0.4 2.000 30 1 31880 8487.5 3.198 14.17 0.3828 100003.56 60802.294 21173.73 15578.8 2454.45 100009.27 5.713E-05 22 2170 91.43 25.0 655 0.4 2.000 30 1 31880 8487.5 3.8155 14.75 0.5197 124996.74 74752.907 26951.03 20375.7 2919.68 124999.3 2.046E-05 23 2170 100 25.0 655 0.4 2.000 30 1 31880 8487.5 4.401 15.22 0.6655 149527.56 88135.394 32774.09 25283.1 3342.93 149535.48 5.296E-05 24 25 Note: a. E L = ½ m V (V sinθ /3.6)2 b. ∑E = E BR-Lat.-NR + E BR-Lat.-R + E BR-Long. + E BEAM 26 c. Difference = ( ∑E - E L ) / E L 42Jiang, T., Grzebieta, R.H. and Zhao, X.L, Determining Lateral Deflections of Plastic Water-Filled Barriers,Paper submitted to International Journal of Crashworthiness.