Selection and Placement Guidelines for Test Level 2 Through Test Level 5 Median Barriers (2022)

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41   C H A P T E R 5 5.1 Median Barrier Guidelines The main reason for using a median barrier is to minimize the chance of a vehicle fully cross- ing the median and striking or being struck by a vehicle in the opposing lanes of traffic. Like- wise, median barriers are only considered for medians where roadside barriers are not needed for clear-zone reasons (e.g., shielding is not needed for either fixed objects or terrain features). When there are fixed objects or terrain features within the median, the single-faced barrier shielding guidelines should be considered. A median barrier should only be installed if it reduces the expected number of fatal and serious injury (KA) crashes on the segment. The 1967 Yellow Book explicitly states that guard- rail and median barriers “should only be used where the result of striking the object or leaving the roadway would be more severe than the consequences of striking the rail.” (AASHO 1967, 29) In other words, the number of KA median barrier crashes and KA CMCs in a median with a median barrier installed must be less than the number of KA CMCs on the same median segment where no median barrier is installed. Applying the condition that a median barrier should only be installed if it reduces the number of KA crashes on the segment results in the following inequality: ≥OUTCOME OUTCOMECMC BAR+CMC The right side of this relationship accounts for those vehicles that interact with the barrier and are contained or redirected as well as those that penetrate, rollover or vault over the median barrier and continue across the median, enter the opposing lanes, and strike or are struck by a vehicle in the opposing lanes. The left side of the inequality represents encroachments that fully cross the unshielded median and are involved in a collision with a vehicle in the opposing lanes. This inequality can be further simplified as follows: > ∴1 OUTCOME OUTCOME Install a median barrierBAR+CMC CMC where OUTCOMECMC = The number of KA CMCs when a barrier is not installed. OUTCOMEBAR+CMC = The number of KA crashes with a longitudinal barrier plus those KA crashes that breach the barrier and continue across the median to be involved in a CMC. Guidelines

42 Selection and Placement Guidelines for Test Level 2 Through Test Level 5 Median Barriers Applying the NCHRP Project 15-65 methodology explained earlier, the frequency of KA crashes for an unshielded median (OUTCOMECMC) and the shielded median (OUTCOMEBAR+CMC) can be estimated as follows: ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) [ ] [ ] [ ] [ ] [ ] [ ] ( ) ( ) = − δ = −    = + − δ =     • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • OUTCOME BEF EAF L 5280 PSL 65 P P 1 THR BEF EAF L 5280 PSL 65 L P MW P 1 THR L OUTCOME BEF EAF L 5280 PSL 65 P P P P 1 THR THR BEF EAF L 5280 PSL 65 L P MW 2 P L CMC S S S S 3 3 c CMC SEV EOL CMC S S S S 3 3 CMC Y SEV CMC EOL S BAR S S S S 3 3 c BAR SEV c CMC SEV EOL CMC BAR S S S S 3 3 BAR Y SEV BAR S CMC BAR CMC ( ) ( )( ) ( )[ ] [ ] ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) = > = + >           + −                −      > + − − • • • • • • • • • • • • • • • RR 1 OUTCOME OUTCOME OUTCOME OUTCOME THR OUTCOME 1 L P MW 2 P L L P MW P 1 THR L THR L P MW P 1 THR L Recognizing that the median barrier is continuous along the whole segment, therefore, L = L = L . 1 P MW 2 P P MW P 1 THR P MW P 1 THR CMC+BAR CMC BAR+CMC CMC BAR CMC BAR CMC BAR Y SEV BAR S CMC Y SEV CMC EOL S BAR CMC Y SEV CMC EOL S S CMC BAR Y SEV BAR Y SEV CMC EOL Y SEV CMC EOL where MW = The median width in feet. The values for OUTCOMEBAR+CMC vary by median barrier material type (e.g., cable, metal beam, or concrete) as discussed above. Conversely, THRBAR is a function of the test level of the median barrier considered and the PT in the traffic mix. The recommendations shown in Figure 23 and Table 12 were derived based on the inequality shown above with consideration of median widths varying between 2 ft to 100 ft, barrier placement, barrier material, and median barrier TL. These guidelines apply to median barriers placed in a traversable median that is free of fixed objects. Single-faced barrier guidelines for obstructed roadsides and medians are described in the next section. Median barriers may be placed anywhere within the median where analysis, crash testing, or in-service performance evaluation has shown the barrier will likely contain and redirect errant vehicles. Bligh et al. studied MASH concrete and W-beam barriers in NCHRP Project 22-22(02) and found that MASH concrete barriers can be considered effective at any offset from the traveled way across the slope and ditch configurations, whereas MASH W-beam barriers have limited locations where the effectiveness is maintained. (Bligh 2020a) MASH W-beam barriers may be placed before the shoulder/slope breakpoint or generally within 4 ft of the center of a

Guidelines 43   Figure 23. Guidelines for median barrier need determination and material selection. MASH Test Level Traffic Conditions 2 or higher 0 PT and posted speed ≤ 45 mph 3 or higher 0 < PT ≤ 10 4 or higher 10 < PT ≤ 15 5 or higher >15 PT or a designated truck or hazardous material route Table 12. Guidelines for selection of longitudinal barrier test level. ditch. Cable barriers may be placed at any location outlined by Marzougui et al. in NCHRP Report 711. (Marzougui 2012a) As additional research is developed, each barrier may be placed at locations determined to satisfy the criteria demonstrated in MASH crash tests. Generally, the need for a median barrier is determined first, then the TL of the barrier is determined. One could, however, determine which TL is appropriate for situations warranting a barrier and then determine if the barrier is warranted. To determine the need for a median barrier, plot the point corresponding to the design year bi-directional traffic volume and median width in Figure 23. The area where this point plots indicates whether a barrier is needed and the barrier material most appropriate for the site and traffic conditions. For example, a cable median barrier would be most appropriate for a 50-ft-wide traversable median with an AADT of

44 Selection and Placement Guidelines for Test Level 2 Through Test Level 5 Median Barriers 40,000 vehicles/day, as shown in Figure 23. On the other hand, a 50-ft-wide traversable median on a controlled-access highway with bi-directional traffic volume in the design year of only 10,000 vehicles/day is better left with no median barrier. A rigid concrete, metal beam, or cable median barrier would reduce the median-related crash risk on a controlled-access highway with a 20-ft traversable median and 90,000 vehicles/day as shown in Figure 23. The particular choice between concrete, metal beam, or cable in this situation would be made based on available deflection area, cost, and other factors, but in this area, a median barrier of any material would reduce the risk compared to not having a median barrier. Once the need for a median barrier has been established using Figure 23, the appropriate test level can be determined, as will be discussed shortly. Two small areas located on the left side of Figure 23 warrant special attention. In these areas, either cable or metal-beam median barriers could reduce the risk of a fatal or serious injury crash compared with not having a median barrier, but median barriers in these areas may allow dynamic deflection of the barrier into the opposing lanes. If the design objective is to accom- modate all the barrier deflection within the median, cable median barriers should not be used in medians narrower than 16 ft and metal-beam barriers should not be used in medians narrower than 10 ft because they could deflect into the opposing lanes in a crash. If deflection into the opposing lanes is an acceptable design objective, cable or metal-beam barriers can be used in these areas of Figure 23. The need for a median barrier in an unobstructed traversable median was determined above using Figure 23. Table 12 is used to select the appropriate median barrier TL as a function of the PT in the traffic mix in the design year. For the example discussed above, a TL3 cable median barrier would be appropriate for a 50-ft-wide median with an AADT of 40,000 vehicles/day and a PT less than 10. For a PT of 12, a TL4 cable barrier would be appropriate. If a TL4 cable median barrier is not available, a TL3 cable barrier should be used, as neither metal-beam nor concrete barriers would provide a lower risk than leaving the median unprotected for these parameters. If a traffic volume and traversable median width for a particular roadway were to plot within the section shown as concrete/metal beam/cable in Figure 23 (e.g., 15-ft-wide median with an AADT of 90,000 veh/day), a cable, metal-beam, or concrete barrier would reduce the risk of a fatal or serious injury crash when compared to an unprotected median. If the PT was 18 at this particular site, a TL5 concrete barrier would be a good choice based on Figure 23 and Table 12. 5.2 Roadside Barrier Guidelines A longitudinal barrier should only be installed if it reduces the number of KA crashes on the segment compared to the unshielded road segment. In other words, the number of KA longitudinal barrier crashes on a segment must be less than the number of KA fixed object and/or terrain crashes that would have occurred without a shielding barrier. In terms of Equa- tions 1 and 2, the longitudinal barrier should be installed only when the OUTCOME of the barrier and the terrain crashes are less than the OUTCOME of the unshielded terrain (i.e., OUTCOMEGR+TER<OUTCOMETER). If there are fixed objects present, the OUTCOME of the barrier, terrain, and fixed object crashes should be less than the OUTCOME of the unshielded terrain and fixed object crashes (i.e., OUTCOMEGR+TER+FO<OUTCOMETER+FO). Both of these relationships can be rearranged algebraically, such that a longitudinal barrier is installed when either of these inequalities holds: < <OUTCOME OUTCOME 1 OR OUTCOME OUTCOME 1GR+TER TER GR+TER+FO TER+FO

Guidelines 45   Recalling Equation 2, the OUTCOME for terrain and the OUTCOME for longitudinal barrier shielding terrain with and without fixed objects are as follows: ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ) ( ) [ ] [ ] [ ] [ ] [ ] [ ] ( ) ( ) ( ) ( ) ( ) ( ) ( ) = − δ δ = = = −    = + − δ δ = = = =       + − δ          • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • OUTCOME BEF EAF L 5280 PSL 65 P P 1 THR Letting 1, and P L P W L yields: BEF EAF L 5280 PSL 65 L P W P 1 THR L OUTCOME BEF EAF L 5280 PSL 65 P P P P 1 THR THR Letting 1, P L P W L and P L P W L yields: BEF EAF L 5280 PSL 65 L P W P L L P W P 1 THR THR L TER S S S S 3 3 c TER SEV TER TER TER TER c FO TER Y TER S S S S S 3 3 TER Y TER SEV TER TER S TER+BAR S S S S 3 3 c BAR SEV c TER SEV TER TER TER BAR TER c TER TER Y TER S c BAR BAR Y BAR S S S S S 3 3 BAR Y BAR SEV BAR S TER Y TER SEV TER TER TER BAR S BAR ( ) ( ) )[ ] [ ] ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) = > = + >       + −      −      = > + − − • • • • • • • • • • • • • • • • RR 1 OUTCOME OUTCOME OUTCOME OUTCOME THR OUTCOME 1 L P W P L L P W P 1 THR THR L L P W P 1 THR L Recognizing that the barrier is continuous along the whole segment, therefore, L = L = L : RR 1 P W P P W P 1 THR THR P W P 1 THR 3 TER+BAR TER TER+BAR TER BAR TER BAR TER BAR Y BAR SEV BAR S TER Y TER SEV TER TER BAR S TER Y TER SEV CMC TER S S TER BAR TER+BAR TER Y SEV BAR Y TER SEV TER TER BAR Y TER SEV TER TER BAR Similarly, shielding with a longitudinal barrier should be considered for median and roadside slopes where there are both fixed objects and foreslopes present when the inequality holds true, as follows (i.e., Equation 4): ( ) ( ) ( ) ( ( [ ] [ ] [ ] [ ] [ ] ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) = − δ δ = = = + − = + −          • • • • • • • • • • • • • • • • • OUTCOME BEF EAF L 5280 PSL 65 P P 1 THR Letting 1, THR 0, and P L P W L L P L P W P W L yields: BEF EAF L 5280 PSL 65 L P W L L P L P W P W L P FO S S S S 3 3 c FO SEV FO FO FO FO FO c FO FO y F FO S TMax x TMax y F FO Y B FO S S S S S 3 3 FO y F FO S TMax x TMax y F FO Y F FO S SEV FO

46 Selection and Placement Guidelines for Test Level 2 Through Test Level 5 Median Barriers ) ( ( ( ) ( )( ) ) (( ) (( ) [ ] [ ] [ ] [ ] ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) = > = + + + >       + −      + + −                  −      + + −                  = > + − δ + + −       − + + − • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • RR 1 OUTCOME OUTCOME OUTCOME OUTCOME THR OUTCOME THR THRU OUTCOME OUTCOME THR L P W P L L P W P 1 THR L THR L P W L L P L P W P W L P THR THR L P W P 1 THR L L P W L L P L P W P W L P THR RR 1 L P W P L P W P 1 THR THR L P W L P L P W P W THR THR L P W P 1 THR L P W L P L P W P W P THR 4 FO+TER+BAR/FO+TER FO+TER+BAR FO+TER BAR TER BAR FO BAR TER TER FO TER BAR Y BAR SEV BAR S TER Y TER SEV BAR TER S BAR FO y F FO S TMAX x TMAX y F FO Y B FO S SEV FO TER BAR TER Y TER SEV TER TER S FO y F FO S TMAX x TMAX y F FO Y B FO S SEV FO TER FO+TER+BAR/FO+TER BAR Y BAR SEV BAR TER Y TER SEV BAR TER TER BAR FO y F FO TMAX x TMAX y F FO Y B FO TER BAR TER Y TER SEV TER TER FO y F FO TMAX x TMAX y F FO Y B FO SEV FO TER The solution of these relationships uses variables derived and discussed in the appendices of this report. Additionally, an understanding of the appropriate probable encroachment angle is needed (i.e., θ). Encroachment angles were examined in NCHRP Project 17-43, “Long-term Roadside Crash Data Collection Program.” (Gabler, forthcoming) The most current NCHRP Project 17-43 beta data set (i.e., NCHRP1743_Beta_20190624.xlsx) was used to determine the following encroachment angle statistics: • 85th percentile encroachment angle: 22 degrees • 50th percentile encroachment angle: 11 degrees • 15th percentile encroachment angle: 5 degrees The 15th percentile represents the shallowest angles in the data set, whereas the 85th percen- tile represents the steepest angles. Using the value of the 15th percentile at the leading end of the guardrail (i.e., θ15) and the 85th percentile at the trailing end (i.e., θ85) maximizes the length of the longitudinal barrier. 5.2.1 Shielding Terrain Free of Fixed Objects Glennon and Tamburri observed in 1967 and Zegeer et al. observed in 1987 that when slopes are free of all other features, the addition of a W-beam guardrail does not reduce the risk of observing a KA crash. (Glennon 1967; Zegeer 1987) If typical values are substituted into Equa- tion 3, the relative risk of the guardrail to the unprotected slope is always greater than 1, usually much greater. This indicates that for foreslopes between 12:1 and 2:1 and offsets to the bottom of the slope up to 100 ft wide, shielding with a guardrail is likely to do more harm than good if the slope is free of fixed objects and there are no fixed objects at the bottom of the slope. This analytically confirms Glennon and Tamburri’s as well as Zegeer’s conclusions that shielding slopes that are otherwise free of fixed objects with longitudinal barriers does not reduce the

Guidelines 47   risk of a KA crash, even for foreslopes as steep as 2:1. There is no question that a rollover is a more severe crash than a crash with a W-beam guardrail. A crash with a guardrail, however, is much more probable than a rollover on an unprotected slope due to the proximity of a guardrail to the roadway edge and the independent probability of rollover if a vehicle interacts with the sloped terrain. Longitudinal barriers should not be used to shield foreslopes flatter than 2:1 if the foreslope is smooth and otherwise free of fixed objects and there are no hazardous features at the bottom of the slope. 5.2.2 Shielding Terrain with Fixed Objects Many roadsides and medians do not have slopes free of features like trees, poles, or bridge piers, so it is often necessary to assess the need to remove fixed objects or shield them with longitudinal barriers on sloping terrain. As a general rule, barriers should be used to shield features when the probability of a KA crash on the segment is reduced with the installation of a longitudinal barrier as compared with the probability of a KA crash without the installation. When considering shielding fixed objects, one should simultaneously consider reducing the density and/or increasing the offset to alleviate the need for a guardrail. Installing hundreds of feet of longitudinal barrier close to the road to shield a small isolated feature like an isolated pole may increase rather than decrease the risk to vehicle occupants. In other words, a longitu- dinal barrier should only be installed in situations where it will do more good than harm. These are described as risk-beneficial conditions. Roadside features that may need shielding can be categorized as follows: • Isolated narrow fixed objects like single trees, utility poles, bridge piers, and traffic signal supports (i.e., small dimensions both parallel and perpendicular to the road). • Multiple narrow fixed objects like a line of utility poles, a series of bridge piers, or a row of roadside trees (i.e., large effective dimension parallel to the road and small dimension perpendicular to the road). • Continuous parallel features like canals, rivers, and walls parallel to the roadway (i.e., very large dimension parallel to the road and modest dimension perpendicular to the road). • Continuous perpendicular features like canals, drainage features, and rivers that are more or less perpendicular to the roadway (i.e., modest dimension parallel to the road and large dimension perpendicular to the road). • General features like buildings and industrial equipment (i.e., moderate dimensions parallel and perpendicular to the road). A general procedure for determining whether a longitudinal barrier is beneficial can be deter- mined using Equation 4 based on the following assumptions: • The shielding barrier is a metal-beam barrier (i.e., PSEV BAR = 0.0084, ∂BAR = 0). • The shielding barrier has a negligible probability of penetration by passenger vehicles (i.e., THRBAR = 0). • The shielding barrier is located 4-ft from the edge of travel (i.e., WF BAR = 4 ft). • The shielding barrier is intended to intercept 95% of encroachment trajectories. • The shielding barrier includes an approach barrier on the upstream end with a terminal that extends 12.5 ft upstream of the end of the length of need. • The terminal is an NCHRP Report 350 or MASH tangent (i.e., PSEV TRM = 0.0500, ∂BTRM = 0). • The foreslope begins at the back of the shielding barrier and ends at the face of the fixed object (i.e., PSEV TER = 0.0589, ∂TER = 1). All roadside obstacles are assumed to have a crash severity of 0.0589 as discussed in • PROBABILITY OF CRASH SEVERITY (PSEVJ) (i.e., PSEV FO = 0.0589, ∂FO = 0).

48 Selection and Placement Guidelines for Test Level 2 Through Test Level 5 Median Barriers Figure 24 defines the variables used. Notice that the lateral distance to the back of the feature (WB FO) includes a term Wv sinθ15, and a term Wv cosθ15 is added to the length of the feature (Lj FO) to account for the width of the vehicle. Figure 25 provides the results for traversable and non-traversable slopes for two different relative risk ratios, 1 and 0.75. The y-axis of Figure 25 is equal to the width of the fixed object plus the Wv sinθ15, which is equal to the equivalent width of the fixed object (WE FO). A relative risk of 1.0 indicates that the shielding barrier results in essentially the same risk as does the roadside feature. A relative risk of 0.75 indicates that the shielding guardrail results in a 25% risk reduction with respect to the unshielded roadside feature. While the shielding barrier may be risk-beneficial (i.e., 1.0), it may or may not be cost-beneficial, as will be discussed shortly. Figure 24 illustrates a site with an isolated individual narrow fixed object like a utility pole being considered for shielding with a barrier. The length and width of the utility pole are less than 1 ft so, referring to the solid line in the top left portion of Figure 25, the utility pole need not be shielded if it is more than 10 ft from the edge of the lane, for a relative risk of 1. This means that a guardrail will likely do more harm than good in this situation. While shielding would be risk-beneficial at an offset less than 10 ft, the risk reduction, as shown in the bottom left por- tion of Figure 25, would be about 25%, which would likely not be cost-effective, as described in Section 5.3. Figure 26 illustrates a site with multiple narrow fixed objects like a line of utility poles being considered for shielding with a barrier. As for the isolated narrow fixed object, the vehicle width is accounted for in the lateral distance to be back of the feature (WB FO) and the length of the feature (Lj FO) to account for the width of the vehicle, as shown in Figure 26. If the spacing between the narrow fixed objects is large enough, each feature should be considered a separate individual feature. If the spacing is smaller, the multiple narrow fixed objects are considered one long object. Figure 27 shows the critical spacing that determines when multiple narrow fixed objects should be treated as individual narrow fixed objects or a composite fixed object with a length equal to the sum of all the spacing between the objects and the length of the objects. Figure 27 is based on the work of Johnson and Gabler in which they developed risk corridors that define the length in advance of a feature where shielding would need to be provided to intercept a fixed percentage of the encroachments that would interact with the fixed object. (Johnson 2015) Using Johnson and Gabler’s approach, the length of guardrail needed to intercept 95% of the trajectories that would interact with the fixed object on roadways with 55 mi/hr or higher posted speed limits can be found as: L W W W R BFO V V = + θ θ cos tan 15 15 Figure 24. Evaluating shielding for an isolated fixed object.

Guidelines 49   If a row of multiple narrow fixed objects is less than this distance, the multiple narrow fixed objects can be considered one object where the length of the fixed object is the sum of all the spacings and the length of the objects. For example, a row of five utility poles spaced 200 ft apart is considered a single 800-ft-long (i.e., 200[5 − 1] = 800 ft) feature if it is more than 12 ft from the edge of travel. In this situation, the width of the feature is less than 1 ft so it lies on the X-axis of Figure 25. The small dashed line in Figure 25 represents a feature length of at least 100 ft and indicates that such fixed objects should be shielded even if they are 64 ft from the edge of travel, for a relative risk of 1. For a relative risk of 0.75 (i.e., 25% risk reduction), the row of utility poles should be considered for shielding if closer than 30 ft from the edge of travel on a traversable slope and 40 ft on a non-traversable slope. Features that are continuous and parallel to the road like rivers, canals, and walls can be evaluated with Figure 25 as well. In these cases, the width of the vehicle need not be accounted for because the vehicle cannot get behind the feature like it can for a fixed object 0 10 20 30 40 50 60 70 80 90 100 10 15 20 25 30 35 40 45 50 55 60 65 70 W E FO - Ef fe ct iv e W id th o f F ix ed O bj ec t ( ft) WF FO - Lateral Offset to Front of Fixed Object (ft) Traversable Slopes | Relative Risk = 1.0 0 10 20 30 40 50 60 70 80 90 100 10 15 20 25 30 35 40 45 50 55 60 65 70 W E FO - Ef fe ct iv e W id th o f F ix ed O bj ec t ( ft) WF FO - Lateral Offset to Front of Fixed Object (ft) Nonrecoverable Slopes/ Relative Risk = 1.0 0 10 20 30 40 50 60 70 80 90 100 10 15 20 25 30 35 40 45 50 55 60 65 70 W E FO - Ef fe ct iv e W id th o f F ix ed O bj ec t ( ft) WF FO - Lateral Offset to Front of Fixed Object (ft) Traversable Slopes | Relative Risk = 0.75 0 10 20 30 40 50 60 70 80 90 100 10 15 20 25 30 35 40 45 50 55 60 65 70 W E FO - Ef fe ct iv e W id th o f F ix ed O bj ec t ( ft) WF FO - Lateral Offset to Front of Fixed Object (ft) Nontraversable Slopes | Relative Risk = 0.75 LFO = 1 LFO = 10 LFO = 20 LFO = 50 LFO = 100 Figure 25. Fixed object risk-beneficial curves for relative risks of 1.0 and 0.75.

50 Selection and Placement Guidelines for Test Level 2 Through Test Level 5 Median Barriers Figure 26. Evaluating shielding for multiple fixed objects. 0 100 200 300 400 500 600 700 0 10 20 30 40 50 SF O - S pa ci ng b et w ee n M ul ti pl e Fi xe d O bj ec ts ( ft ) WB FO - Offset from Edge of Lane to Back of the Fixed Object (ft) Treat Multiple Fixed Objects as One Continuous Object Not Risk Beneficial to Shield Figure 27. Critical spacing of multiple narrow fixed objects.

Guidelines 51   like a bridge pier. For example, the lower-left portion of Figure 25 indicates that a 10-ft-wide canal (WB FO – WF FO =10 ft) parallel to a roadway for 100 ft along its length with a traversable slope should be considered for shielding if it is closer than 32 ft to the edge of the lane, for a rela- tive risk of 0.75. Similarly, a 20-ft-wide (i.e., parallel to the road) open-channel drainage ditch that crosses a 50-ft-wide traversable (i.e., WB FO – WF FO = 50 ft) median should be considered for shielding at the 0.75 relative risk level. The selection of relative risk levels is a policy decision for highway agencies. A relative risk of unity indicates that the shielding barrier does no more harm than the unshielded feature, but it also does not reduce the risk of a KA crash. A relative risk of 0.75 means that 25% fewer KA crashes are expected for the shielded location. It was found that relative risks of 0.50 were seldom possible except for very long and very wide features. Charts like Figure 25 could be produced for any relative risk between approximately 0.7 and 1.0. The relative risk can also be used directly in a benefit–cost approach recognizing that the reduction in KA crashes is equal to: ( )− = − •OUTCOME OUTCOME 1 RR OUTCOMENULL ALT NULL The approach outlined above, therefore, can be used directly where a highway agency chooses an explicit relative risk goal (e.g., 0.75 or less) or determines the need for shielding based on when the barrier does no more harm than the unshielded object (i.e., relative risk = 1) and then determines whether the shielding barrier is cost-beneficial. In either case, the decision to shield or not to shield a roadside feature is based on the quantified risk of observing a KA crash. 5.3 Cost–Benefit Guidelines Both benefit–cost and cost-effectiveness analyses are discussed here. While benefit–cost has a long history in roadside design, cost-effectiveness analysis is suggested for the implementa- tion of these findings for the reasons discussed below. When these results are implemented, it is suggested that an abridged version of one or both of these two subsections be considered as an appendix to the AASHTO RDG. 5.3.1 Benefit–Cost Analysis A common technique for maximizing value used in many technical fields is benefit–cost analysis. (Newnan 1977) In the context of roadside safety, the benefit is usually considered to be the reduction in societal costs associated with roadside crashes and the costs are the construc- tion, maintenance, and repair costs expended by the highway agency to achieve that benefit. Since benefits are defined as the reduction in the societal cost of crashes, estimating the number and severity of crashes is at the heart of the benefit–cost method in roadside safety. To compare design alternatives, an average annual crash cost is calculated by estimating the number and severity of crashes for the considered alternative and the existing condition (i.e., the null alternative) and then converting the estimate to social costs using the willingness-to- pay concept. These crash costs are then annualized over the project life at some predefined rate of return. Any direct highway agency costs (i.e., initial installation, annual maintenance, and periodic repairs) are likewise annualized and the benefit–cost ratio (BCR) is calculated. The BCR is calculated as follows: = +     • • • BCR OR C VSL DC AP MC ALT/NULL KA ALT i,n ALT

52 Selection and Placement Guidelines for Test Level 2 Through Test Level 5 Median Barriers where BCR = The benefit–cost ratio of the barrier alternative with respect to the null alternative. ORALT/NULL = The outcome reduction is the estimated difference in the annual frequency of fatal and serious injury crashes for the shielded median (ALT alternative) and the unshielded alternative (NULL alternative). See Table 13 for cable median barrier, Table 14 for metal-beam median barrier, and Table 15 for concrete median barrier. Figure 25 for fixed objects. ORALT/NULL = OUTCOMENULL - OUTCOMEALT ORALT/NULL = (1-RRALT/NULL) OUTCOMENULL VSL = The value of statistical life in dollars based on the US DOT recommendation or the agency value for a fatal crash. (Monje 2016) CKA = A unitless coefficient that transforms the VSL to the average cost of a KA crash. APi,n = The capital recovery factor as a function of the interest rate, i, and service life, n, where ( ) ( ) = + + −     AP P 1 1 1 i,n i i i n n . (Newman 1977) DCALT = The direct cost of constructing and maintaining the barrier alternative over the service life of the alternative. The direct cost of the null alternative (i.e., the unshielded) is presumed to be zero. MCALT = The annual maintenance cost of the longitudinal barrier. RR = The relative risk of the considered alternative with respect to the null alternative where RRALT/NULL = OUTCOMENULL / OUTCOMEALT A BCR equal to 1 means that the investment is just equal to the benefit obtained. A value of 1 is the minimum BCR where the alternative should be considered. Most highway agencies expect BCR values between 2 and 4 to maximize the benefit of scarce agency resources. Calculating the expected frequency of KA outcomes for a shielded and unshielded median was discussed earlier. The differences in the frequency of these outcomes for shielded and unshielded alternatives (ORALT/NULL) are shown in Table 13 through Table 15 for cable, metal-beam, and concrete median barriers, respectively, and in Figure 25 for fixed objects. Linear interpolation between cells is acceptable for non-tabulated values of median width and traffic volume. For example, a median barrier in a 70-ft traversable median on a four-lane highway with 45,000 veh/day results in a higher frequency of KA crashes than the unshielded median, so a median barrier would not be risk-beneficial, as indicated in Table 13 by a blank cell. On the other Bi-Direction AADT (veh/day) Traversable Median Width (ft) 25 30 35 40 45 50 60 70 80 90 100 25,000 0.0003 30,000 0.0009 0.0005 35,000 0.0015 0.0010 0.0003 40,000 0.0022 0.0017 0.0009 0.0004 45,000 0.0030 0.0023 0.0015 0.0009 0.0002 50,000 0.0039 0.0032 0.0022 0.0016 0.0008 0.0003 55,000 0.0048 0.0040 0.0030 0.0023 0.0014 0.0008 60,000 0.0060 0.0051 0.0039 0.0031 0.0021 0.0014 0.0003 65,000 0.0072 0.0062 0.0048 0.0039 0.0028 0.0021 0.0009 0.0001 70,000 0.0086 0.0074 0.0060 0.0050 0.0037 0.0028 0.0014 0.0006 0.0000 75,000 0.0099 0.0087 0.0071 0.0059 0.0046 0.0036 0.0021 0.0011 0.0005 0.0002 0.0001 80,000 0.0114 0.0100 0.0082 0.0070 0.0055 0.0045 0.0027 0.0016 0.0009 0.0006 0.0004 85,000 0.0121 0.0107 0.0089 0.0076 0.0060 0.0049 0.0031 0.0019 0.0011 0.0008 0.0006 90,000 0.0121 0.0107 0.0089 0.0076 0.0060 0.0049 0.0031 0.0019 0.0011 0.0008 0.0006 ≥95,000 0.0165 0.0147 0.0124 0.0108 0.0089 0.0075 0.0051 0.0035 0.0025 0.0019 0.0016 Table 13. KA outcome reduction—cable median barriers (KA CMC/mi/yr).

Guidelines 53   hand, if a cable median barrier were installed on a four-lane divided highway with 45,000 veh/day and a 45-ft-wide median, Table 13 indicates that 0.0002 fewer KA CMCs can be expected annually. In this case, the cable median barrier is risk-beneficial although it may or may not be cost-beneficial. In other words, the cable median barrier reduces the risk of a CMC, but it remains to be determined whether providing the median barrier will be a good return on the funds invested. The value of statistical life (VSL) is roughly equivalent to the fatal crash cost. The VSL is periodically defined by the U.S. DOT for use in policy analyses. (Monje 2016) The 2020 VSL is estimated to be $12.3 million based on the published 2016 update procedure. (Monje 2016) Many highway agencies establish their own local values for either VSL or the fatal crash cost, and these should be used as appropriate. CKA is a coefficient that transforms the VSL into the average cost of a KA crash. CKA is a func- tion of the particular type of crash scenario so there is a specific value for CMCs as opposed to other types of crashes. The Highway Safety Information System data for the State of Washington included 8,638 crossover-centerline crashes that occurred on highways with posted speed limits of 55 mi/hr or greater. Of the 8,638 cross-over-centerline crashes, 431 were fatal and 1,094 were serious injury crashes. Miller determined that, on average, the fatal crash cost (K) is 2,600,000/180,000 =14 times larger than the serious injury crash cost (A). (Blincoe 2002; Miller 1989) The weighted average KA crash cost coefficient of crossover crashes based on the Washington State data is, therefore: ( ) ( ) = + + =C 431 1,094 /14 431 1,094 0.33KA Although other methods are available, the annualized cost method has been used because the societal benefits are an annually recurring benefit, as are direct maintenance costs. Other techniques like present-worth or future-worth could be used with the same result. Usually, Bi-Direction AADT (veh/day) Traversable Median Width (ft) 10 15 20 25 30 35 40 45 50 55 60 50,000 0.0009 55,000 0.0033 0.0023 0.0022 0.0008 60,000 0.0062 0.0049 0.0047 0.0031 0.0015 65,000 0.0092 0.0077 0.0073 0.0055 0.0038 0.0011 70,000 0.0128 0.0110 0.0104 0.0083 0.0063 0.0034 0.0016 75,000 0.0164 0.0144 0.0135 0.0113 0.0090 0.0057 0.0037 0.0011 80,000 0.0183 0.0162 0.0152 0.0128 0.0104 0.0070 0.0049 0.0021 0.0003 85,000 0.0183 0.0162 0.0152 0.0128 0.0104 0.0070 0.0049 0.0021 0.0003 ≥90,000 0.0292 0.0263 0.0247 0.0216 0.0184 0.0141 0.0113 0.0078 0.0054 0.0033 0.0015 Table 14. KA outcome reduction—metal-beam median barriers (KA CMC/mi/yr). Bi-Direction AADT (veh/day) Traversable Median Width (ft) 10 15 20 25 30 35 80,000 0.0064 85,000 0.0064 90,000 0.0064 ≥95,000 0.0270 0.0162 0.0135 0.0127 0.0094 0.0061 Table 15. KA outcome reduction—concrete median barriers (KA CMC/mi/yr).

54 Selection and Placement Guidelines for Test Level 2 Through Test Level 5 Median Barriers highway agencies will determine appropriate values for the rate of return (i) and design life (n) to be used in economic analyses. Generally, the design life should be on the order of 25 to 30 years for typical roadside hardware and rates of return between 2% and 4% are typical for the rate of return. The next step is to calculate the direct costs. The direct cost of the null alternative is assumed to be zero. The null alternative is the already-existing condition. The direct cost of the longitudinal barrier being considered as an alternative is the direct cost of construction added to the present worth of the annual maintenance cost. For example, assume high-tension cable median barrier has a direct installation cost of $125,000 per mile of median and an annual maintenance cost of $2,500 per mile per year. Further, assume that no major earthwork is required before installing the median barrier, the design life is 30 years, and the rate of return is 2%. The present worth of the direct cost of installing the high-tension cable median barrier is, therefore: DC AP MC PA $125,000 0.0446 $2,500 $8,075ALT i,n ALT i,n+ = + =• • • For a highway with a 45-ft-wide traversable median and bi-direction design year traffic of 45,000 veh/day, the expected annual reduction in CMCs (ORALT/NULL) is 0.0002, as shown in Table 13. Given a design year VSL of $12.3 million and the direct costs above, the BCR can be calculated as follows: = +     = +     = />• • • • • • BCR OR C VSL DC AP MC 0.0002 0.33 12,300,000 125,000 0.0446 2,500 0.1 1ALT/NULL KA ALT i,n ALT A high-tension cable median barrier is not cost-beneficial on a 45-ft-wide median with 45,000 veh/day under these economic assumptions even though the cable median barrier does reduce the risk. A high-tension cable median barrier would have a BCR greater than 1 anywhere in Table 13 with a value greater than 0.0020, as follows: +    = +    = = • • • • BCR DC AP MC C VSL OR 1.0 125,000 0.0446 2,500 0.33 12,300,000 OR 0.0020 ALT i,n ALT KA ALT/NULL ALT/NULL For the 45-ft median highway discussed here, a high-tension cable median barrier will not become cost-beneficial until the traffic volume exceeds 60,000 veh/day. If the traffic volume increases to 75,000 veh/day on this same highway, the BCR will increase to a value of just over 2. The same analysis steps can be used for metal-beam median barriers using Table 14 or concrete median barriers using Table 15. Due to the wide variety of roadside features and circumstances, risk reduction tables like those shown for median barriers in Table 13 through Table 15 are not available. Benefit-cost analysis can still be performed, however, knowing the relative risk of the considered alternative to the null alternative. For example, a less than 1-ft-diameter utility pole on a traversable slope is risk- beneficial (i.e., relative risk ≤1) if the pole is closer than 10 ft from the edge of the lane. While a shielding barrier may reduce the risk of a KA crash somewhat, it is not clear if shielding would be cost-beneficial. A pole shielded by a W-beam guardrail located 8 ft from the traveled way with a traversable slope has a relative risk of 0.75 (see Figure 25). If the AADT for this two-lane undivided highway is greater than 5,000 vehicles per day 1.1911 encroachments/mi/edge/yr can be expected. The BCR is calculated as:

Guidelines 55   ( ) = +     = +     = − +     = /> • • • • • • • • • BCR OR C VSL DC AP MC BCR OR C VSL DC AP MC BCR 1 0.75 0.33 12,300,000 125,000 0.0446 2,500 0.1 1 ALT/NULL KA ALT i,n ALT ALT NULL KA ALT i,n ALT 5.3.2 Cost-Effectiveness Analysis Cost-effectiveness analysis is very similar to benefit–cost analysis but instead of mon- etizing benefits, the outcome itself (i.e., the annual reduction in KA crashes) is used. For example, instead of monetizing the societal cost of the crash reduction (i.e., benefit) result- ing from shielding a median with a barrier, the number of fatal and serious injury crashes avoided could be used directly. The annualized cost of the median shielding improvement divided by the annual reduction in the number of KA crashes would be the incremental cost- effectiveness ratio. The incremental cost-effectiveness ratio (ICER) is defined as follows: (Newnan 1976) = − − ICER DC DC PO PO i/j i j j i where ICERi/j = The incremental cost-effectiveness ratio of alternative j with respect to alternative i. POi, POj = Performance outcome for alternatives i and j over the project life. DCi, DCj = The annualized cost of the direct (i.e., construction, maintenance, and repair) costs for alternatives i and j. In the context of comparing median shielding alternatives, the ICER is calculated as follows: = + −     • ICER DC AP MC OUTCOME OUTCOME ALT i,n ALT ALT NULL Like benefit–cost analysis, present-worth, future-worth, and annual cost analyses could all be used with similar results, but annual cost-effectiveness analysis is used here because the reduction in KA crashes is an annual value. As before, the null alternative is the unshielded existing median, so there is no direct cost associated with alternative j. The KA crash reduc- tions are tabulated in Table 13 for cable median barriers, Table 14 for metal-beam barriers, and Table 15 for concrete median barriers. Returning to the example of a four-lane divided highway with an AADT of 45,000 veh/day and a 45-ft-wide traversable median, the ICER can be calculated as follows: = +   = • ICER 125,000 0.0446 2,500 0.0002 $40 million per KA crash avoidedCABLE An ICER of $40 million to avoid one KA crash, given that the average KA CMC cost (i.e., CKA VSL = 0.33 • 12.3) is $4.1 million, would appear to be a poor use of funds. Alternatively, if the

56 Selection and Placement Guidelines for Test Level 2 Through Test Level 5 Median Barriers AADT increases to 60,000 veh/day, the number of KA crashes avoided is 0.0021 from Table 13, and the ICER is: = +   = • ICER 125,000 0.0446 2,500 0.0021 $38 million per KA crash avoidedCABLE This is just a little less than the KA crash cost, so it is likely a reasonable expenditure. Notice that these were also the conditions that resulted in a BCR = 1. The values in Table 13 through Table 15 and Figure 25, therefore, can also be used to calculate the incremental cost-effectiveness ratio. One of the advantages of the ICER method of economic analysis is that it does not require the user agency to monetize fatal and serious injury crashes. Better alternatives have lower ICER values and can be chosen on that basis alone. The ICER can be viewed as a priority rank for various projects with the higher values representing a high priority and better use of funding.