Chap 10 Sample Problem 1:
Student Notes
Brief Description of the Project/Case
Rural two-lane tangent roadway segment
What is the predicted average crash frequency of the roadway segment for a particular year?
Road Features:
  • 1.5 mile length
  • Tangent roadway section
  • AADT = 10,000 veh/day
  • 2% grade
  • 6 driveways per mile
  • 10’ lanes
  • 4’ gravel shoulder
  • Roadside hazard rating = 4
Assumptions:
  • Collision type distributions used are the default values from Table 10-4
  • The calibration factor is assumed to be 1.10

Step 1 – Identify data needs for the facility
The following data are provided:
Existing Road (study segment length of 1.5 miles):
  • Tangent roadway section
  • AADT = 10,000 veh/day
  • 2% grade
  • 6 driveways per mile
  • 10’ lanes
  • 4’ gravel shoulder
  • Roadside hazard rating = 4

Step 2 – Divide Locations into Homogeneous Segments
For this study conditions are for a single homogeneous segment.
Step 3 – Apply the appropriate SPF

Predicted Segment Crashes for Base Conditions:
Using Equation 10-6, the two-lane, two-way segment SPF for base conditions can be determined as follows:



Step 4 – Apply CMFs as needed
Segment Base Conditions:
Tangent section
Level grade
5 drives per mile
12’ lanes
6’ paved shoulder
RHR = 3
CMFs will be needed for any conditions that do not meet these segment base conditions.
Lane Width CMF:

CMFra from table 10-8 for 10’ lane and 10,000 AADT = 1.30
pra= 0.574 - see discussion below
= 1.17
pra : is the proportion of total crashes constituted by related crashes. The proportion of related crashes (i.e., single vehicle run-off-road, and multiple vehicle head-on, opposite direction sideswipe, and same direction sideswipe crashes) is estimated to be 0.574 based on the default crash distribution in Table 10-4. The value pra may be updated from local data as part of the calibration process.
Shoulder width and Type CMF:

Use values from Tables10.4,10-9, and 10-10
For 4’ shoulders and AADT of 10,000, CMFwra = 1.15 Table 10-9
For 4’ gravel shoulders, CMFtra = 1.01 Table 10-10
pra= 0.574
= 1.09
Horizontal curve –tangent section CMF =1.00 (base condition)
Superelevation – tangent section CMF = 1.00
Grade – Table 10-11 for a 2% grade CMF=1.00
Driveway Density -

= 1.01
Centerline Rumble Strips - no rumble strips, CMF = 1.00 (base condition)
Passing Lanes - no passing lanes, CMF = 1.00 (base condition)
Two-way Left-Turn Lanes - no TWLT lanes, CMF = 1.00 (base condition)
Roadside Design - RHR= 4. CMF calculated by equation 10-20


CMF = 1.07
Lighting – no lighting, CMF = 1.00 (base condition)
Automated Speed Enforcement - no automated speed enforcement, CMF = 1.00 (base condition)
CMF comb = 1.17 x 1.09 x 1.01 x 1.07 = 1.38
Step 5 - Apply Local Calibration Factor
N predicted rs = Nspfrs x Cr x (CMF1 x CMF2 x … CMFn)
= 4.008 x 1.10 x (1.38) = 6.084 crashes per year
Chap 10 Sample Problem 2:
Student Notes
Brief Description of the Project/Case
Rural two-lane curved roadway segment
What is the predicted average crash frequency of the roadway segment for a particular year?
Road Features:
  • 0.1 mile length – horizontal curve
  • Curved roadway section
  • AADT = 8,000 veh/day
  • 1% grade
  • 0 driveways per mile
  • 1200 ft horizontal curve radius
  • No spiral transition
  • 11’ lanes
  • 2’ gravel shoulder
  • Roadside hazard rating = 5
  • 0.04 superelevation rate
Assumptions:
  • Collision type distributions used are the default values from Table 10-4
  • The calibration factor is assumed to be 1.10
  • Design speed = 60 mph
  • Max superelevation rate emax = 6 percent
/ Note: the solution here and in the spreadsheet for this problem will differ from that presented in the HSM, page 10-42. In the given problem the default distribution for SVROR,HO,SSOP, and SSsame is given as 0.78. The problem is solved here and in the spreadsheet using the default HSM crash distribution for SVROR,HO,SSOP and SSsame of 0.574.
Step 1 – Identify data needs for the facility
The following data are provided:
Existing Road (study segment length of 1.5 miles):
  • Curved roadway section
  • AADT = 8,000 veh/day
  • 1% grade
  • 0 driveways per mile
  • 11’ lanes
  • 2’ gravel shoulder
  • Roadside hazard rating = 5
  • Super 0.04

Step 2 – Divide Locations into Homogeneous Segments
For this study conditions are for a single homogeneous segment.
Step 3 – Apply the appropriate SPF

Predicted Segment Crashes for Base Conditions:
Using Equation 10-6, the two-lane, two-way segment SPF for base conditions can be determined as follows:



Step 4 – Apply CMFs as needed
Segment Base Conditions:
Tangent section,
Level grade
5 drives per mile
12’ lanes
6’ paved shoulder
RHR = 3
CMFs will be needed for any conditions that do not meet these segment base conditions.
Lane Width CMF:

CMFra from table 10-8 for 11’ lane and 8,000 AADT = 1.05
pra= 0.574 - see discussion below
= 1.03
pra : is the proportion of total crashes constituted by related crashes. The proportion of related crashes (i.e., single vehicle run-off-road, and multiple vehicle head-on, opposite direction sideswipe, and same direction sideswipe crashes) is estimated to be 0.574 based on the default crash distribution in Table 10-4. The value pra may be updated from local data as part of the calibration process.
Shoulder width and Type CMF:

Use values from Tables 10.4,10-9, and 10-10
For 2’ shoulders and AADT of 8,000, CMFwra = 1.30 Table 10-9
For 2’ gravel shoulders, CMFtra = 1.01 Table 10-10
pra = 0.574
= 1.18
Horizontal curve, length, radius CMF:



CMF = 1.43
Horizontal Curves Superelevation CMF:
CMF = 1.06 + 3 x (SV – 0.02)
CMF = 1.06 + 3 x ((0.06-0.04) – 0.2)
CMF = 1.06
Grade – Table 10-11 for a 1% grade CMF=1.00
Driveway Density - number of drives = 0 ≤ 5

Centerline Rumble Strips - no rumble strips, CMF = 1.00 (base condition)
Passing Lanes - no passing lanes, CMF = 1.00 (base condition)
Two-way Left-Turn Lanes - no TWLT lanes, CMF = 1.00 (base condition)
Roadside Design - RHR= 5. CMF calculated by equation 10-20


CMF = 1.14
Lighting – no lighting, CMF = 1.00 (base condition)
Automated Speed Enforcement - no automated speed enforcement, CMF = 1.00 (base condition)
Combined CMF
CMF comb = 1.03 x 1.18 x 1.43 x 1.06 x 1.14 = 2.10
Step 5 - Apply Local Calibration Factor / Note: superelevation variance (SV) = super max – actual super - example
(0.06-0.04=0.02)
See HSM page 10-28Formula varies according to level of SV
N predicted rs = Nspfrs x Cr x (CMF1 x CMF2 x … CMFn)
= 0.214 x 1.10 x (2.10) = 0.494 crashes per year
Chap 10 Sample Problem 3:
Student Notes
Brief Description of the Project/Case
Three leg stop-controlled intersection located on a rural two-lane roadway.
What is the predicted average crash frequency of the stop-controlled for a particular year?
Road Features:
  • 3 legs
  • Minor road stop control
  • No right turn lanes on major road
  • No left-turn lanes on major road
  • 30-degree skew angle
  • AADT of major road = 8,000 veh/day
  • AADT of minor road = 1,000 veh/day
  • Intersection lighting present
Assumptions:
  • Collision type distributions used are the default values from Table 10-6
  • The calibration factor is assumed to be 1.50
  • The proportion of crashes that occur at night are not known, so the default proportion for nighttime crashes is assumed.

Step 1 – Identify data needs for the facility
The following data are provided:
Existing Intersection:
  • 3 legs
  • Minor road stop control
  • No right turn lanes on major road
  • No left-turn lanes on major road
  • 30-degree skew angle
  • AADT of major road = 8,000 veh/day
  • AADT of minor road = 1,000 veh/day
  • Intersection lighting present

Step 2 – Divide Locations into Homogeneous Intersections
For this study conditions are for a single intersection.
Step 3 – Apply the appropriate SPF

Predicted Segment Crashes for Base Conditions:
Using Equation 10-6, the intersection SPF for base conditions can be determined as follows:



Step 4 – Apply CMFs as needed
Intersection Base Conditions:
Skew angle 0 degrees
No intersection left turn lanes
No intersection right turn lanes
No lighting
CMFs will be needed for any conditions that do not meet these segment base conditions.
Intersection Skew Angle CMF:
CMF can be calculated from Equation 10-22


CMF = 1.13
Intersection Left-turn Lanes CMF: since no left-turn lanes in this problem CMF = 1.00 – base condition
Intersection Right-turn Lanes CMF: since no right-turn lanes in this problem CMF = 1.00 – base condition
LightingCMF:
CMF can be calculated using Equation 10-24
CMF = 1 – 0.38 x pni
From table 10-15 for 3-leg stop controlled intersection pni default is: 0.26
CMF = 1 -0.38 X 0.26 = 0.90
Combined CMF
CMF comb = 1.13 x 0.90 = 1.02
Step 5 - Apply Local Calibration Factor
N predicted rs = Nspfrs x Cr x (CMF1 x CMF2 x … CMFn)
= 1.867 x 1.50 x (1.02) = 2.857 crashes per year
Chap 10 Sample Problem 4:
Student Notes
Brief Description of the Project/Case
Four leg signalized-controlled intersection located on a rural two-lane roadway.
What is the predicted average crash frequency of the stop-controlled for a particular year?
Road Features:
  • 4 legs
  • Signalized intersection
  • 1 right turn laneon one approach
  • 1 left-turn lanes on each of two approaches
  • 90-degree angle
  • AADT of major road = 10,000 veh/day
  • AADT of minor road = 2,000 veh/day
  • No lighting present
Assumptions:
  • Collision type distributions used are the default values from Table 10-6
  • The calibration factor is assumed to be 1.30

Step 1 – Identify data needs for the facility
The following data are provided:
Existing Intersection:
  • 4 legs
  • Signalized intersection
  • 1 right turn lane on one approach
  • 1 left-turn lanes on each of two approaches
  • 90-degree angle
  • AADT of major road = 10,000 veh/day
  • AADT of minor road = 2,000 veh/day
  • No lighting present

Step 2 – Divide Locations into Homogeneous Intersections
For this study conditions are for a single intersection.
Step 3 – Apply the appropriate SPF

Predicted Segment Crashes for Base Conditions:
Using Equation 10-10, the intersection SPF for base conditions can be determined as follows:



Step 4 – Apply CMFs as needed
Intersection Base Conditions:
Skew angle 0 degrees
No intersection left turn lanes
No intersection right turn lanes
No lighting
CMFs will be needed for any conditions that do not meet these segment base conditions.
Intersection Skew Angle CMF:
CMF for skew angle at four-leg signalized intersections is 1.00 for all cases.
Intersection Left-turn Lanes CMF:
From Table 10-13 for a signalized intersection with a left-turn lanes on two approaches the CMF = 0.67
Intersection Right-turn Lanes CMF:
From Table 10-14 for a signalized intersection with a right-turn lane on one approach the CMF = 0.96
LighthingCMF:
Since there is no intersection lighting present CMF = 1.00 - base condition
Combined CMF
CMF comb = 0.67 x 0.96 = 0.64
Step 5 - Apply Local Calibration Factor
N predicted rs = Nspfrs x Cr x (CMF1 x CMF2 x … CMFn)
= 6.796 x 1.30 x (0.64) = 5.654 crashes per year
Chap 10 Sample Problem 5:
Brief Description of the Project/Case
The project consists of three sites: a rural two-lane tangent segment, a rural two-lane curved segment, and a 3 leg intersection with minor leg stop-control. This project combines Sample problems 1,2, and 3.
What is the expected average crash frequency of the project for a particular year incorporating both the predicted average crash frequencies from Sample problems 1,2, and 3 and the observed crash frequencies using the site-specific EB method.
The Facts:
  • 2 roadway segments, 1 tangent and 1 curved
  • 1 intersection (3ST)
  • 15 observed crashes: tangent segment 10 crashes; curved segment 2 crashes; intersection 3 crashes
Outline of Solution:
To calculate the expected average crash frequency, site specific observed crash frequencies are combined with the predicted average crash frequencies for the project using the project using the site-specific EB Method (observed crashes are assigned to specific segments or intersections).
See Spreadsheet results.
Chap 10 Sample Problem 6:
Brief Description of the Project/Case
The project consists of three sites: a rural two-lane tangent segment, a rural two-lane curved segment, and a 3 leg intersection with miner leg stop-control. This project combines Sample problems 1,2, and 3.
What is the expected average crash frequency of the project for a particular year incorporating both the predicted average crash frequencies from Sample problems 1,2, and 3 and the observed crash frequencies using the project-level EB method.
The Facts:
  • 2 roadway segments, 1 tangent and 1 curved
  • 1 intersection (3ST)
  • 15 observed crashes: no information to assign crashes to specific sites
Outline of Solution:
Observed crash frequencies for the project as a whole are combined with predicted average crash frequencies for the project as a whole using the project level EB Method.
See Spreadsheet results.