STUDY OF HEAVY COMMERCIAL VEHICLE CRASH RECONSTRUCTION
WITH COMPARATIVE ANALYSIS OF PASSENGER VEHICLES
A Thesis
Presented to the faculty of the Department of Mechanical Engineering
California State University, Sacramento
Submitted in partial satisfaction of the requirements for the degree of
MASTER OF SCIENCE
in
Mechanical Engineering
by
Dhanashri Patel
SPRING 2020
© 2020
Dhanashri Patel
ALL RIGHTS RESERVED
ii
STUDY OF HEAVY COMMERCIAL VEHICLE CRASH RECONSTRUCTION
WITH COMPARATIVE ANALYSIS OF PASSENGER VEHICLES
A Thesis
by
Dhanashri Patel
Approved by:
______, Committee Chair Jose Granda
______, Second Reader Troy D. Topping
______Date
iii
Student: Dhanashri Patel
I certify that this student has met the requirements for format contained in the University format manual, and that this thesis is suitable for electronic submission to the Library, and credit is to be awarded for the thesis.
______, Graduate Coordinator ______Troy D. Topping Date
Department of Mechanical Engineering
iv
Abstract
of
STUDY OF HEAVY COMMERCIAL VEHICLE CRASH RECONSTRUCTION
WITH COMPARATIVE ANALYSIS OF PASSENGER VEHICLES
by
Dhanashri Patel
Commercial motor vehicles crash investigation and reconstruction technology is a great interest to vehicle design engineers. It is of interest to identify the cause behind heavy vehicle crashes for safer designs. Regulatory agencies like the Commercial Vehicle Safety
Alliance (CVSA) is concerned with the public safety. To this aim, this thesis seeks to research this area using state of the art technology and analysis methods based on experimental results and applicable to commercial vehicle crashes for reconstruction purposes.
Northwestern University Public Safety department has been a leader in developing the theory in vehicle crashes for passenger cars and the field of heavy vehicles. Heavy vehicles field is still in need of further development. Also, The National Highway Transportation
Safety Administration, NHTSA, a government agency has been conducting tests of vehicles of several manufacturers and models to determine crash factors and determine heavy commercial vehicle behavior under collisions. CVSA (Commercial Vehicle Safety
Alliance) is another entity, which is a non-profit organization. It is establishing safety
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standards for motor vehicles, drivers by providing necessary education and training programs. For this reason, already existing methods and data allows a base structure for developing new methods.
Important factors for reconstruction of heavy commercial vehicles have been identified and studied. With the use of already available data, a detailed approach of analysis for important crash factors has been developed.
______, Committee Chair Jose Granda
______Date
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ACKNOWLEDGMENTS
I would like to thank Professor Jose Granda for his interest in the topic I selected for this thesis. I would also like to express my sincere gratitude to him for his continuous support during my time at California State University, Sacramento.
Next, I would like to thank Professor Troy Topping for his guidance.
Finally, I would like to thank my husband and parents, whose support and understanding have made me what I am today.
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TABLE OF CONTENTS
Page
Acknowledgments...... vii
List of Tables ...... xi
List of Figures ...... xii
Chapter
1. INTRODUCTION ...... 1
1.1 Purpose ...... 1
1.2 Problem Statement ...... 2
2. HEAVY-DUTY VEHICLE LAYOUT AND CLASSIFICATION ...... 3
2.1 Heavy-Duty Vehicle layouts ...... 3
2.2 Vehicle Classification ...... 6
3. BRAKE SYSTEM ...... 10
3.1 Hydraulic Barking System ...... 10
3.2 Air Braking System...... 10
3.3 Reason to Use Air Braking System in Heavy-Duty Vehicles...... 11
3.4 Brake System in Heavy-Duty Vehicles ...... 11
4. POST-CRASH INSPECTION ...... 17
4.1 Commercial Vehicle Safety Alliance ...... 18
4.2 Brake System ...... 20
4.2.1 Brake Adjustment Limits ...... 21
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4.3 Measure Applied Push-Rod Stroke ...... 24
4.4 Axle Weight Measurement ...... 26
4.4.1. Dimensional Measurement ...... 28
5. CENTER OF MASS LOCATION ...... 30
5.1 Center of Mass Lateral Location...... 34
5.2 Center of Mass Vertical Location ...... 35
6. BRAKING DISTANCE AND SPEED ...... 37
6.1 Single Adjusted Drag Factor Method ...... 38
6.2 Resultant Drag Factor Method ...... 39
7. ENERGY ANALYSIS FOR COMMERCIAL VEHICLES ...... 45
7.1 Law of Conservation of Energy ...... 45
7.2 Conservation of Energy in Vehicles ...... 45
8. GENERAL ANALYSIS FOR CONSERVATION OF MOMENTUM ...... 48
8.1 Momentum ...... 48
8.2 PDOF and ∆ ...... 52
9. MOMENTUM ANALYSIS FOR COMMERCIAL VEHICLE ...... 58
9.1 Velocity/ Speed Factor for Commercial Vehicles ...... 58
9.2 PDOF Factor for Commercial Vehicles ...... 59
9.3 Case Study ...... 61
9.3.1 Rear End Collision Case 1 ...... 61
9.3.2 Rear End Collision Case 2 ...... 67
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9.3.3 Rear End Collision Case 3 ...... 71
9.3.4 Side Collision Case 4 ...... 76
9.3.5 Side Collision Case 5 ...... 81
9.3.6 Side Collision Case 6 ...... 86
10. CONCLUSIONS...... 92
References ...... 93
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LIST OF TABLES
Tables Page
1. Types of Articulated Vehicles ...... 5
2. Vehicle Classification System as Per Weight Rating Value ...... 7
3. Vehicle Classification System as Per Number of Axles ...... 9
4. Brake Adjustment Limits for Clamp Type Brake Chamber ...... 23
5. ECM Report from Freightliner Truck Tractor ...... 62
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LIST OF FIGURES
Figures Page
1. Heavy Duty Truck Tractor with Semitrailer ...... 2
2. Single Unit Truck ...... 3
3. Conventional Truck Tractor ...... 3
4. Cab-Over- Engine Truck Tractor ...... 4
5. Single Axle Dolly Converter ...... 6
6. Typical Truck Tractor Semitrailer Braking System...... 12
7. Typical Truck Tractor Air Supply System...... 13
8. Wheel and Sensor Block Mounting at Monitored Wheel ...... 15
9. Position of Wheel Sensor and Tooth Wheel ...... 16
10. Free-Stroke and Reserve Stroke Length for Properly Adjusted and Out of
Adjustment Brakes ...... 21
11. Initial Measurement with Service Brake Released ...... 24
12. Measurement Taken with Full-Service Brake Application ...... 25
13. Brake Chamber, Pushrod and Slack Adjuster System ...... 26
14. Truck Tractor with Semitrailer ...... 27
15. Variables Needed to Calculate the Vehicle’s Center of Mass Height ...... 30
16. Wheelbase Measurement and Three Weights for Five Axle Truck Tractor with
Semitrailer ...... 32
17. Track Width of Dual Tire Axle ...... 35
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18. The Values Used to Estimate the Height of Truck Tractor’s Center of Mass ...... 36
19. Resultant Drag Factor for a Two Axle Single Unit Truck ...... 40
20. Variables for Semitrailer ...... 41
21. Variables for Truck Tractor ...... 42
22. Vector Diagram ...... 49
23. Change in velocity Vector for 1 ...... 52
24. X and Y component of ∆ 1 ...... 53
25. Vector Diagram for 1 and 1 ...... 54
26. Change in velocity Vector for 2 ...... 55
27. X and Y component of 2 ...... 55
28. Vector Diagram for to 2 and 2 ...... 56
29. PDOF Angle Based on Calculated Δθ ...... 57
30. PDOF Passes Between the Center of Mass and the Fifth Wheel of the Tractor ... 59
31. PDOF Passes Between the Center of Mass and the Fifth Wheel of the Tractor
Semi-trailer ...... 60
32. PDOF Passes Through Center of Mass on the Tractor with No Semitrailer ...... 60
33. PDOF Passes Through Center of Mass on the Tractor with Semitrailer ...... 61
34. Collision System ...... 63
35. In-Line Collinear Crash ...... 63
36. Pre-impact system for case 1 ...... 65
37. System right after collision at time 0.15 sec. for case 1 ...... 65
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38. System at tme 0.25 sec. case 1 ...... 66
39. Pre-impact system for case 2 ...... 67
40. System right after collision at time 0.5 sec. for case 2 ...... 68
41. System at 1.55 sec. for case 2 ...... 68
42. Momentum diagram for case 2 ...... 69
43. Pre-impact system for case 3 ...... 72
44. System right after collision at time 0.55 sec. for case 3 ...... 73
45. System at time 1.55 sec. for case 3 ...... 73
46. Momentum diagram for case 3 ...... 74
47. Pre-impact system for case 4 ...... 77
48. System right after collision at time 0.75 sec. for case 4 ...... 78
49. System at time 1.55 sec. for case 4 ...... 78
50. Momentum diagram for case 4 ...... 79
51. Pre-impact system for case 5 ...... 82
52. System right after collision at time 0.70 sec. for case 5 ...... 83
53. System at time 1.55 sec. case 5 ...... 83
54. Momentum diagram for case 5 ...... 84
55. Pre-impact system for case 6 ...... 87
56. System right after collision at time 0.55 sec for case 6 ...... 88
57. System at time 1.55 sec. for case 6 ...... 88
58. Momemtum diagram for case 6 ...... 89
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1
1. INTRODUCTION
According to the National Highway Traffic Safety Administration report, deaths from a heavy-duty truck (commercial vehicle) crashes raised in year 2017 than in previous years.
Deaths due to heavy-duty trucks reported 9% more than the prior year. To fulfill the ongoing demand of the consumer economy in a small period, the use of heavy commercial trucks is increasing.
1.1 Purpose
The purpose of this thesis is to help readers understand the basic dynamics of heavy-duty trucks and the basic method to accurately reconstruct the heavy-duty truck crash. The technique of reconstruction and investigation of heavy-duty trucks crash is a little bit similar to the crash involving automobiles like cars. For the heavy-duty truck crash investigation and reconstruction, scenic data documentation is required. As compared to automobiles, heavy-duty trucks collision site consists of more scratches, tire marks, and liquid debris.
This type of data is collected in the form of dimensional measurements and photographs for evidence. After the crash, this kind of inspection plays an important role in collision reconstruction. Investigation and reconstruction of heavy truck is different in the case of weight, braking system operation, electronic monitoring of the vehicle than that of automobiles.
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Figure 1. Heavy Duty Truck Tractor with Semitrailer
1.2 Problem Statement
It is of interest to identify the cause behind heavy vehicle crashes for safer designs.
Regulatory agencies like the Commercial Vehicle Safety Alliance (CVSA) is concerned with the public safety. To this aim, this proposal seeks to research this area using state of the art technology and analysis methods based on experimental results and applicable to commercial vehicle crashes for reconstruction purposes.
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2. HEAVY-DUTY VEHICLE LAYOUT AND CLASSIFICATION
2.1 Heavy-Duty Vehicle layouts
Heavy-duty vehicles are trucks; self-propelled motor vehicle used for commercial purposes. The single unit truck shown in Fig. 2 below is one of the examples of heavy-duty trucks.
Figure 2. Single Unit Truck
A truck tractor is similar as truck including accessories to draw other vehicles. This accessory includes engine, steer axle etc. There are two types of truck tractors available; conventional chassis and cab over engine. Fig. 3 shows conventional chassis trucks in which the engine is located ahead of the cab location.
Figure 3. Conventional Truck Tractor
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In cab-over-engine, type truck tractor, the engine is located directly under the cab location.
The cab over engine truck tractor is shown in Fig. 4
Figure 4.Cab-Over- Engine Truck Tractor
Articulated vehicle is a vehicle that consists of a truck tractor or motor vehicle unit which is connected by pivoting hitch to a trailer. Typically, articulated vehicles are five-axle truck tractor trailers. There are multiple articulated trucks available which consists of more than one trailer. Types of multiple articulated vehicle depends upon the number of trailers attached and length of trailers are shown in Table 1.
5
Table 1.Types of Articulated Vehicle
1) Turnpike Double: Truck
configuration consists of two 40-
53 ft. long trailers
2) Rocky Mountain Doubles:
It consists of front semitrailer 40-53
ft. long and rear semitrailer 26-29 ft.
in length.
3) Triple Trailer: It consists
of three semitrailers of length 27-28
ft.
4) Western Doubles/ Double Bottom:
It consists of two semitrailers of
length 26-29 ft. Each semitrailer is
called as Pup.
6
In heavy-duty trucks, dolly converter and fifth wheel can be considered as a coupling device. The fifth wheel is used to attach the front of the semitrailer to the tractor. Dolly converter is called as an auxiliary front axle which is composed of one or two axles and fifth wheel. Single axle dolly converter allows semitrailer to act as a full trailer which is shown in Fig. 5.
Figure 5. Single Axle Dolly Converter
2.2 Vehicle Classification
Heavy-duty trucks are commonly classified as per their Gross Vehicle Weight Rating value
(GVWR) and number of axles. GVWR is the actual loaded weight of single motor vehicles including cargo. Gross combination weight rating values (GCWR) is used for the articulated vehicle.
The Table 2 [6],[7] given below illustrates the vehicle classification as per their weight rating value. These categories are used by the Federal Highway Administration (FHWA), the U.S. Census Bureau, and the U.S. Environmental Protection Agency (EPA).
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Table 2. Vehicle Classification System as Per Weight Rating Value
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Federal Highway Administration classifies vehicles in 13 different categories as per the number of axles. This classification is generally used in the pavement design community.
It is important to note that vehicle classification varies from state to state because vehicle characteristics often change in sizes and weight as per state. The below Table 3 shows vehicle classification specified by FHWA. In this way, classification vehicles are often used in the collision reconstruction and vehicle design community.
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Table 3. Vehicle Classification System as Per Number of Axles
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3. BRAKE SYSTEM
There are two types of braking systems used in vehicles. They are explained below.
3.1 Hydraulic Barking System
The hydraulic system utilizes fluid to apply pressure and gerate forces that operate the brakes. This system is used in other applications such as to move the position of a forklift arm. In automobiles, brake fluid is stored in a reservoir tank. During the braking action, this fluid is pushed into the tubing to provide the necessary pressure to the wheel cylinder located at each wheel in order to activate the brake pads pressing the brake disks or drums.
The hydraulic braking system works well with lightweight vehicles or automobiles due to their lightweight and small construction. This type of brake system occupies a minimal amount of space. Hydraulic brakes consider as standard brake system for class 5 and class
6 vehicles.
3.2 Air Braking System
In this type of braking system, air used as a medium to generate pressure instead of a fluid
When the driver pushes the brake pedal, pressure is let in the lines that go from the air compressor, governor and reservoir tank. This pressure activates the push road and stack adjustos to turn a mechanical cam system that produces forces against the brake shoes and these produce friction forces with the drums. This type of braking system is generally used
11 for class 7 and greater vehicles. The Air braking system in heavier trucks is explained in detail in the following section of this chapter.
3.3 Reason to Use Air Braking System in Heavy-Duty Vehicles
A significant reason why the air braking system chose over the hydraulic braking system in a heavier truck is their strong stopping power during their work mode and failure mode.
In the hydraulic braking system, the leak in the brake line lowers the fluid pressure. As liquid is not compressible, this lowers the pressure induce low forces on the brake pads to slow down the motion of the vehicle.
In the case of an air braking system, if there is a leak in the air brake lines then the air pressure decreases which activates the parking or emergency brakes at wheels to bring the vehicle to complete stop.
3.4 Brake System in Heavy-Duty Vehicles
Braking is one of the most important crash factors during heavy truck crash reconstruction.
It is necessary to have detailed knowledge of the braking system in heavy-duty vehicles.
For heavy-duty vehicles (class 8 trucks and tractors), the air brake system is used.
The air brake system uses air to transmit pressure on the service brakes. Compression pressure (85-150 psi) provides enough power to brake heavy trucks. The typical air brake system for heavy truck tractor and trailer is shown in Fig. 6 [8].
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Figure 6. Typical Truck Tractor Semitrailer Braking System
In the air brake system, when the driver applies force to the brake pedal, air pressure from the pressurized tank is passed through valves and lines to the brake chamber located at wheel brake. The brake system consists of sub-systems given below.
1) Air supply: Air supply system in Fig. 7 [8] consists of an air compressor which
provides pressurized air. Governor is connected to an air compressor to control the
range of pressure as per the requirement of the system. This air goes to the air dryer
13
to remove moisture from the air. This compressed conditioned air goes to the
reservoir tank to store it.
Figure 7. Typical Truck Tractor Air Supply System
2) Service brake sub-system: Air brake system of heavy commercial vehicles requires
a “split” service brake system. Split system is designed in such a way that if one
part of the system fails, the other part provides necessary vehicle braking. In a truck
tractor, the split system is divided into two parts as the primary and secondary
service brake system. Primary service brake system which controls the service
brake for the rear axle. Secondary service brake system which controls the service
brake for the front axle.
3) Emergency brake sub-system: The parking brake can also be called as an
emergency brake. It is used for parking vehicles or complete stop of vehicle in
emergency (brake system failure) situation. It consists of spring brake actuators on
rear axles to achieve the emergency brake function.
4) Trailer supply and control subsystem: For applications where there is a cab and a
tractor, , the brake system operates from the tractor. The trailer supply valve is a
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dash-mounted driver-controlled pull-push valve. The trailer control system consists
of a hand control valve which is a hand-actuated valve that provides pressure to the
brake system of the trailer.
5) Foundation brake sub-system: Foundation brakes is an actual braking mechanism
which is a set of actuators, mechanical brake including frictional material. Due to
the friction between the drum and foundation brake lining converts kinetic energy
into heat energy which can be dissipated after applying brakes. This sub-system
also converts pressurized air into the mechanical force required to apply the brakes.
6) Anti-lock braking sub-system (ABS): The anti-lock braking system is an additional
accessory with the air brake system. It plays an important role in heavy commercial
vehicles with more than 10,000 lbs. gross value weight rating. It provides additional
control to the braking system during a 100% slip of wheels by monitoring the
rotational speed of the wheel and modulating air pressure in the brake chamber.
The anti-lock braking system is helpful during emergency stop situations. It also
supports the vehicle to stabilize during skidding.
ABS is a set of wheel speed sensors, electronic control unit (ECU), modulator
valves, electrical switches and wires, etc.
ECU is an electronic control unit that makes decisions after receiving information
from the wheel speed sensors. It will then send a signal to the modulator valve to
control the vehicle. As per commands received from ECU, modulator allows brakes
pressure to act proportionally to the need of braking by decreasing the pressure
15
required or holds braking pressure at the current level or increase the braking
pressure. The basic function of the modulator valve is to regulate air pressure.
Wheel speed sensors are mounted in the wheel brake. A tooth wheel is located on
the wheel hub to activate the sensor to recognize the speed of the wheel. Tooth
wheel and sensor block mounting at the wheel are is shown in Fig. 8 [10] below.
Figure 8. Wheel and Sensor Block Mounting at Monitored Wheel
The sensor is a set of coils and magnets. As the wheel rotates, the teeth pass through a magnetic field of the sensor. As per the passing of the tooth and gap, the magnetic field builds and collapses. The pulses created by the magnet have a frequency in the proportion to the wheel speed. The sensor is located close enough to the toothed wheel so that is can sense the presence of the tooth or gap shown in Fig. 9 [10].
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Figure 9. Position of Wheel Sensor and Tooth Wheel
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4. POST-CRASH INSPECTION
Post-crash inspection of heavy trucks is necessary to perform successful determination of:
causes behind collision which involves the mechanical condition of the
truck and their elimination.
Inspection of dimensional and inertial parameters of the truck.
Speed of vehicle from its skid marks.
Heavy vehicle instability due to several reasons.
In heavy truck collision, inspection and reconstruction plays an important role. Inspection at the site of collision helps to gather information for reconstruction. Investigators perform the post-collision inspection. The evidence from pos collision data in the form of measurements and positions is used to match that evidence to the mathematical analysis done with the help of analytical equations for the reconstruction of heavy truck collisions.
The purpose of post-collision inspection falls under two categories:
1) For compliance with regulations: Inspection in this category is performed by a
commercial vehicle inspector. These inspectors make decisions as per the severity
of collisions. These decisions include at scene inspection, relocation of heavy truck
for later time.
2) Reconstruction of traffic collision purpose: During the inspection, the inspector
gathers data regarding brake pushrod travel distance which can be used to calculate
the pressure at the time of the accident and the working conditions of the brakes.
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4.1 Commercial Vehicle Safety Alliance
CVSA is a non-profit organization. It is an association of state, provincial, and federal officials responsible for administration and enforcement of motor carrier safety laws in the
US, Canada, and Mexico. It promotes commercial collision and incidents free environment.
This organization establishes safety standards for motor carriers, drivers, and inspectors by providing necessary education training programs.
CVSA has designed certain inspection criteria for heavy trucks to determine its condition when it is not in working condition. These criteria are called North American Standard
Out-of-Service Criteria.
There are six levels through which CVSA inspection goes:
Level 1)- North American Standard Inspection: This type of inspection involves
proper examination of driver and vehicle. Inspecting physical test components
of a vehicle.
Level 2)- Walk Around Vehicle/ Driver Inspection: In this type of inspection,
examination of a driver or vehicle happens without physically touching the
vehicle.
Level 3)- Driver Credential/ Administrative Inspection: This type of inspection
includes the proper examination of a driver by checking driver’s license,
his working hours, driver’s previous records.
Level 4)- Special Inspection: In this type of inspection, a particular item goes
through examination in the support of the study.
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Level 5)- Vehicle Only inspection: This type of inspection includes a detailed
examination of each item that comes under North American Standard
Inspection, in the absence of the driver.
Level 6)- North American Standard Inspection for Transuranic Waste and
Radioactive Material: This type of inspection includes examination of
radioactive shipments that include inspection procedures.
CVSA inspection covers only critical items listed below:
Brake system adjustment
Locking device
Truck body and frame
Wipers and windshield
Headlight, turn signal
Steering parts
Wheels and rims
Tires with wear and load limits
Exhaust system
Suspension system
Hazardous material
After all types of proper inspection, the inspector will make a report for collision reconstruction to determine and eliminate the factors behind collision considering violations.
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As per North American Standard Out of Service Criteria, when the vehicle is out of service due to its mechanical condition or loading. If a collision happens under these circustances the collision contributing factor is considered as a mechanical condition not as a violation.
As the braking system plays an important role during the collision, we will discuss about the brake system out of service criteria in the following section.
4.2 Brake System
When defective brakes have a level of greater than or equal to 20% of the normal service brake on the vehicle, the truck is considered as out of service as per North American
Standard out of service criteria. Brakes can be considered as defective if they meet one of the following criteria.
1. Absence of effective braking after applying service brakes
2. Missing or broken mechanical components
3. Loose brake components
4. Audible air leak at brake chamber
5. Brake adjustment limit
6. Brake pads
In the following subsection, we will study the brake adjustment limit in detail.
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4.2.1 Brake Adjustment Limits
During brake application, trucks use only a fraction of their braking capacity and a major amount is used during heavy braking.
This small amount of braking called as reserve braking ability. This ability becomes low or zero when brakes become “out of adjustment”. Strokes of brake chamber has divided into three stages.
1. Free- Stroke: In this portion, brake shoes move from their released position and
comes into contact with the brake drum. For free-stroke, stroke length gets longer
when brakes are out of adjustment.
2. Normal Braking: In this portion of the stroke, the length of stroke remains the same
when brakes are out of adjustment.
3. Reserve- Stroke: This portion of the stroke is used during heavy braking. For
reserve stroke, stroke length becomes small when brakes become out of adjustment.
Below Fig. 10 shows the changing length of strokes for properly adjusted and out
of adjustment brake system.
Figure 10. Free-Stroke and Reserve Stroke Length for Properly Adjusted and Out of Adjustment Brakes
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During the inspection, it is necessary to check brake adjustments when brakes are cold.
Stroke measurements will be longer when brakes are hot due to heat expansion of the brake drum. The first step of inspection is to identify the chamber and its size. Brake chamber size can be measured by size markings on the chamber and with the help of specially designed calipers to achieve accurate size. As shown in the Table 4 [10] below, commercial vehicle brake range in size from 6-36 with 30 the most common size in use. Steer axle brakes are generally smaller, ranging in sizes from 12-20. Larger size chambers are typically used on rear axles that carry heavy loads.
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Table 4. Brake Adjustment Limits for Clamp Type Brake Chamber
Once you have determined the accurate chamber size, it is important to check whether it is standard or long chamber.
Long-stroke chamber can be identified easily by visual inspection in the following three ways:
1. Long-stroke brake chambers have square airline ports and standard brake chambers
have round airline ports.
2. Marking on the brake chamber body indicates the size of the chamber stroke.
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3. Trapezoidal shaped tag places under the clamp bolt shows brake chambers
maximum stroke dimension.
In this way, accurate size and stroke length of the chamber helps to make sure the brakes comply.
4.3 Measure Applied Push-Rod Stroke
To measure applied push-rod stroke requires getting under the truck. There are two methods available to make measurements. Both methods are shown in Fig. 11 [1] below.
Figure 11. Initial Measurement with Service Brake Released
1. Method 1: Mark the pushrod at the brake chamber or another fixed reference point.
Using a marker pen or other similar instruments. Make sure marks are precise.
2. Method 2: Measure the released position of the push rod and write down the
distance from a point on the pushrod to a fixed point near the brake chamber.
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The resultant value is the first measurement after choosing any method. Now either raise or lower the vehicle’s air pressure by running engine or pushing the brake pedal until you get 90-100 psi indicated in both primary and secondary air tanks. Once you have got an accurate pressure, apply and hold a full-service brake pedal.
Determine the applied pushrod stroke according to the method you have selected to obtain the resultant value. With the use of the above two methods, your goal is to measure the distance from the previously selected point on the pushrod to the previously selected fixed point near the brake chamber to get applied pushrod stroke measurement. Fig. 12 [1]
Shows the measurement taken with a full-service brake application.
Figure 12. Measurement Taken with Full-Service Brake Application
To get a second measurement, we need to measure the applied position of the pushrod. The result of the subtraction of measurement number one from the second measurement is the resultant value of the applied push rod stroke.
Now based on your previously determined applied pushrod stroke, brake chamber size, and type, compare the measurements against the correct adjustment limit for a specific brake
26 chamber. If the applied pushrod stroke is longer than the adjustment limit, then the brake is considered as out of adjustment.
Fig. 13 [15] shows the brake chamber, push-rod, and slack adjuster system.
Figure 13. Brake Chamber, Pushrod and Slack Adjuster System
4.4 Axle Weight Measurement
The stability and stooping distance of the truck is dependent on several factors, one main one is the load (% of Weight) on each tire. This is important because the friction forces between the truck tires and the road depend on the normal force and the coefficient of friction (drag factor is the term used in accident reconstruction) For reconstruction purposes, weight plays a very important role. From a practical matter, during an inpection, individual scales can be put under the tires to measure the loading forces applied at each tire and thus determine the the loads by axle.
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At large platform weight stations, the truck is placed at different points to capture the axle weight. Small platform weight stations have different sections that will weigh the steer, drive, and semitrailer tires at one time.
On a large platform scale, unknown axle weight can be calculated with the known gross vehicle weight in combination with the help of weight on the other axles. To understand this, consider the truck shown in Fig. 14 [12] below.
Figure 14. Truck Tractor with Semitrailer
Consider the truck’s total weight is 33,320 lbs. The driver then pulls just the steer axle off the pad and total weight on the drive and semitrailer tandems of 22,520 lbs. Finally, the tractor is moved off the scales with only the semitrailer wheels on the scale, which gives a weight of 10,540 lbs. Therefore, front axle weight is the result of the subtraction of weight of the drive combined semitrailer tandem from the total gross weight of the truck, which gives 33,320-22,520= 10,800 lbs. Weight on the drive tandem is a result of subtraction of
28 weight of semitrailer tandem alone from the combined weight of the drive and semitrailer tandem, which gives 22,520- 10,540= 11,980 [1].
Using a previously weighed truck, the truck tractor was disconnected from the semitrailer, and axles were weighed separately. The whole assembly of the tractor was first placed on the scale and weighed 18,740 lbs. The steer axle was then pulled off and drive tandem weighed 7900 lbs. The steer axle weight would be the total of tractor weight minus the drive tandem weight (18,740- 7900= 10,840). The weight of the semitrailer would be the total weight of the combination vehicle minus the tractor only weight (33,320- 18,740=
14,580). The weight of the semitrailer resting on the fifth wheel would be the weight of the semitrailer only. It would be the initial weight of the combination vehicle minus the semitrailer tandem weights (14,580-10,540= 4040) lbs. [1]
4.4.1. Dimensional Measurement
Axle spacing and the dimensions of the truck are necessary to measure during the post- crash inspection. These would help us determine the center of mass which in turns controls the weight distribution. Truck tractor and semitrailer specifications are based upon measurement of the truck tractor wheelbase or length of the semitrailer. Because truck tractors and semitrailers are equipped with sliding axles, the location of the axle at the time of the crash may be different from when the tractor or semitrailer came when it is inspected.
In some cases of heavy commercial truck collision, reconstruction happens several years later. In such cases, sometimes, it discovers that weight is wheelbase dependent.
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Therefore, dimensional measurement becomes important after crash inspection.
Heavy truck post-crash inspection form gives general information regarding truck-like tractor’s location, company name, year, gross vehicle weight ratio, model number as well as details regarding its engine. This form also gives details about the semitrailer like its type, year, load, GVWR, model, etc. Investigator also involves small details like dimensions of the truck. This form also gives information of collision inspector or truck investigator with date
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5. CENTER OF MASS LOCATION
The Center of mass is a unique point where all of the system’s mass is to be concentrated.
Center of mass of an object can be located with three measurements:
1. Longitudinal dimension (x)
2. Lateral dimension (y)
3. Vertical dimension (z)
Calculation of center of mass location for straight truck, truck tractor, semitrailer is similar to that of passenger cars in the sense that the equilibrium of moments of the external forces about the center of mass has to be zero.
Procedure for calculating the center of mass location of a vehicle is given below:
Axle weight is necessary to know the center of mass location. We can get information about the axle weight from the manufacturer catalog or using the method of axle weight measurement described in chapter 4 of this thesis. For this information, we can derive a longitudinal center of mass location.
Figure 15. Variables Needed to Calculate the Vehicle’s Center of Mass Height
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In the above Fig. 15 [4],
= Static weight on the front axle.
= Weight on rear axle
Ɩ= Wheelbase of vehicle which is a distance from front to rear axle. r= Radius of wheels with tires
= Weight of one axle in case of other axle is lifted to a height (h) h= Height in which axle is lifted.
= Total weight of vehicle= +
The longitudinal position of the center of mass is a decimal fraction of the wheelbase behind the front axle, is = . Ɩ
Where, static weight on front axle: =
The static axle weight of the vehicle front and rear axle are obtained with the use of the axle weight measurement method described in section 3 of this thesis.
Similarly, For heavy-duty 5-axle truck tractor semitrailer.
Wheelbase measurement and three weights are shown below Fig. 16 [13]
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Figure 16. Wheelbase Measurement and Three Weights for Five Axle Truck Tractor with Semitrailer
For Heavy vehicle truck semitrailer,
Ɩ= Wheelbase length
= Weight of steer axle
= Weight of drive tandem
= Weight of semitrailer tandem.
Let’s calculate the longitudinal center of mass location assuming that data regarding axle weight is already obtained.
Below is the equation of the distance of the center of mass behind the front (steer axle) of a truck tractor.
= ( ). Ɩ
Where, W= Total weight of vehicle
In this way, we can calculate the center of mass location of straight empty truck.
There are two ways to calculate the center of mass location for a loaded truck.
a) With the vehicle loaded weight both axles of the vehicle separately.
b) Calculate the center of mass location by combining the load and empty truck.
33
If the load is uniformly distributed within the rectangular truck in the forms like liquid, same size, and weight boxes grains then the load’s center of mass is almost equal to its center of volume longitudinally, laterally, and vertically.
To calculate the combined center of mass for the empty truck and center of mass of its load, we need the following data:
a) A distance of the center of mass location of vehicle behind the front axle,
b) Weight of empty vehicle,
c) Distance of center of mass location of the load behind the front axle,
d) Weight of the load,
Using all of the above available data, we can calculate the equation for the longitudinal location of the combined center of mass of the vehicle in addition to load behind the vehicle’s front axle ( ).
( . ) ( . ) =
Where, = Combined truck and load center of mass position behind the front axle.
If in the truck, more than one load is involved or parts of load are treated separately, the product of the weight of each load times the distance its center of mass is behind the front axle is added to the front axle. This sum is divided by the combined weight of vehicles and loads.
For liquid and granular items, the center of mass is a product of the volume of the load and density of that product. The volume of the load calculations can be done from the dimensions of the container or truck interior.
34
In the case of rectangular trucks, the volume of load can be calculated as a product of length, width, and height. Length and width are usually depending on the dimensions of the truck. But height can be unknown. To find out height, we may need to depend on the informants. The density of the product is considered as already known to the person doing calculations.
If a vehicle is having more than two axle weights, then each axle times its relative distance the reaction point is behind the front axle. For five-axle truck semitrailers consisting three- axle weights, it will be +
This sum is divided by the total weight of a combination vehicle. The result of this division is a distance of the center of mass from the front axle is as follows [1]
=
Where, = Distance of center of mass from the front axle.
= Distance center of drive tandem is behind the front axle.
= Distance center of semitrailer tandem is behind the front axle.
= Weight of drive tandems.
= Weight of semitrailer tandems
= Total weight of combination vehicle
5.1 Center of Mass Lateral Location
To calculate the lateral location of the center of mass of the vehicle, we use front and rear weights and wheelbase length similarly. We can calculate the lateral location of the
35 vehicle’s center of mass using left and right weights and track width. Left and right sides of the vehicle can be weighed separately and track width can be measured from the geometric center of dual tires. The track width of the dual tire axle is shown in Fig. 17 below.
Figure 17. Track Width of Dual Tire Axle
If load on left and right wheels are similar with a negligible difference then assume that lateral location of the center of mass is on vehicle’s longitudinal centerline.
5.2 Center of Mass Vertical Location
Data for the height of the center of mass above ground is obtained by weighing in the front or rear axles while another axle is lifted several feet above the ground shown in Fig. 18.
The amount of hoist ‘h’, is the distance the axle is lifted not its height above the ground.
36
‘r’ is the radius of the wheel with tire. The distance of the tire from the ground is the same as the axle lift and can often be more easily measured with the use of all the available data, calculations of the center of mass is done.
The equation for the height of the center of mass with rear end lifted given as follows [1],
.√ . = + .
Where, Ɩ= Wheelbase
= Height of the center of mass
h= Height of the rear wheels were hoisted
r= Radius of tire and wheel together
= Level weight on front axle
= Weight on front axle after hoisting
The equation for the height of the center of mass with front lifted can be,
.√ ∗ ( ) = + .
The height of the center of mass expressed as a decimal fraction of the wheelbase is,
=
Figure 18. The Values Used to Estimate the Height of Truck Tractor’s Center of Mass
37
6. BRAKING DISTANCE AND SPEED
The purpose of this topic is to explain methods used during the reconstruction of braking distance from the initial speed of heavy commercial trucks. It also explains the method to determine the speed of the vehicle from its skid marks. Two methods to estimate braking distance and speed from skid marks are introduced in this section and the brake force model method is explained in the following section.
1) Single drag factor
2) Resultant drag factor
The sliding coefficient of friction is briefly explained below for heavy trucks.
Heavy truck tires develop less coefficient of friction than automobile tires as a truck tire is made of a harder rubber compound and has higher inflation pressure around 105 psi. When all wheels of the truck are locked and sliding, the drag factor for heavy trucks is equivalent to 85% of the sliding coefficient of friction of passenger cars as per the Heuser theory [1].
In reality, to maintain steering action brake chamber on steer axle cannot lock steer tires.
So, the drag factor for heavy trucks without brake deficiency is less than 85%.
As per Radlinski, the coefficient of friction for a heavy truck tire is 75% of the passenger car’s coefficient of friction when the vehicle is stopped but under control while in panic mode. But to keep the vehicle under control in a panic stop, all wheels cannot be locked and sliding. This reality leads the conclusion that 75% is not a perfect number for the coefficient of friction for truck tires in comparison with car tires. After several skid tests, the range 75-85% for heavy truck tire is finalized to be used as percent of passenger car tires.
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The “Accident Reconstruction Journal” [14] has published information related to several skid testing of heavy vehicles in comparison with passenger cars. As per several skid tests, in wet condition heavy vehicle tire has a coefficient of friction of 70-80% to that of passenger car tires under the same situation. In winter with ice and snow condition, truck had 0.23 and 0.15 average braking coefficient of friction and traction coefficient. Some skid tests were conducted to determine the effects of the ABS system on braking capacity.
Skid test results supported the fact that ABS for tractor-semitrailer is beneficial in dry road conditions stopping.
6.1 Single Adjusted Drag Factor Method
The single adjusted drag factor method expresses all the parameters that affect the deceleration of heavy trucks into a single value. Those parameters include heavy vehicle tire composition, mechanical condition of brake components, well maintained, and adjusted heavy truck braking system. Range 75-85 % is used to determine the heavy truck tire coefficient of friction in comparison to that of a passenger car tire. In this method, a single adjusted value will be applied to the total braking distance of the heavy truck in case of no brake deficiency.
In case of any brake deficiency, the brake force model will be used. Brake deficiency is recognized in the following conditions:
Deficiency of mechanical components that causes a reduction in effective braking
action.
Absence of proper braking action when a service brake is applied.
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Air leaking at the brake chamber.
Pressurized air lines when the service brake is applied.
Corrosion in camshaft bearings.
Driver’s input is necessary while determining the difference between continuous and repeated brake pedal application in the case of skid marks inspection. When the examination of the brake system does not reveal any mechanical defects, then the longest skid mark is used to compute the vehicle speed. The longest skid mark must be consistent with the deceleration distance of the truck. Individual, skid mark lengths indicate that not all wheels locked or locked at the same time and each braked wheel contributed near its maximum braking capacity when the air line pressure was sufficiently high to lock the first wheel.
This method considers brake force as a part of maintenance, vertical weight component, and composition of tires.
6.2 Resultant Drag Factor Method
This method is used for configuration vehicles. In configuration vehicles, braking and accelerating force creates a weight shift from one axle to another. This weight shifting causes different drag factors on each axle as the drag factor is based on the center of gravity.
The resultant drag factor equation for a particular vehicle configuration is given below.
Resultant drag factor for two-axle single unit truck shown in Fig. 19 [1]
=
40
Figure 19. Resultant Drag Factor for a Two Axle Single Unit Truck
Where, = Estimated drag factor for front axle
= Estimated drag factor for rear axle.
= Horizontal distance of the center of mass to the front axle. It is a decimal fraction of the wheelbase ∕ .
= Height of center of mass as a decimal fraction of the wheelbase ( ∕ ).
Equations for front axle ( ) and rear axle ( ) during the braking event can be estimated:
= +
= ∫ −
Where,
= Total weight of the vehicle
= Horizontal distance of the center of mass to the front axle as a decimal fraction of the wheelbase ( ∕ )
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= Horizontal center of mass to the rear axle as a decimal fraction of the
wheelbase ( ∕ ).
Resultant drag factor for truck tractor with semitrailer:
Fig. 20 and 21 shows the semitrailer part and tractor part of the truck tractor
semitrailer respectively.
Below Fig. 20 shows variables for semitrailer which are involved in estimating
resultant drag factor for truck tractor with a semitrailer.
Figure 20. Variables for Semitrailer
Below Fig. 21 shows variables for the tractor which are involved in estimating
resultant drag factor for truck tractor with a semitrailer.
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Figure 21. Variables for Truck Tractor
( ) =
In this equation, the following abbreviations are used [1]:
= [1 +
= 1 +
D = [1 − ]
E=
J= + +
K= 1 − + −
Where,
= Weight of tractor
= Weight of the semitrailer and load
= Height of center of mass as a decimal fraction of wheelbase ( ∕ ).
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= Height of combined center of mass as a decimal fraction of wheelbase- the wheelbase of the trailer is the distance from the center of rear axle to the center of
the kingpin ( ∕ ).
= Horizontal distance of the center of mass to front axle same as a decimal fraction of the wheelbase
= Horizontal distance from the center of the kingpin to the combined center of mass same as
= Horizontal distance of the fifth wheel center ahead of the rear axle same as a decimal fraction of the tractor wheelbase
= Height of center of the fifth wheel same as a decimal fraction of the tractor wheelbase
= Height of the kingpin as a decimal fraction of the semitrailer’s wheelbase
= Estimated drag factor for the tractor front axle
= Estimated drag factor for the semi-trailer
Loads can be calculated with the use of following equations,
Vertical load on the front axle, = + K
Load on tractors rear axle, = ( − ) + + −
Semitrailer axle, = −
Vertical load on fifth wheel, = +
The horizontal force, , between tractor and semitrailer at the fifth wheel,
= −
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In this way, the equation for the resultant drag factor for Truck and Full trailer can also be developed.
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7. ENERGY ANALYSIS FOR COMMERCIAL VEHICLES
Energy analysis for commercial vehicles will discuss the process of calculating the total amount of energy used during the crash. Energy analysis is based on the law of physics known as the Principle of Conservation of Energy.
7.1 Law of Conservation of Energy
The Law of conservation of energy states that the total energy of an isolated system remains constant. It means energy can never be created or destroyed; it is just transferred from one form into another form.
7.2 Conservation of Energy in Vehicles
The law of conservation of energy can be used to develop an energy balance equation.
A vehicle crash is considered as an example of an inelastic collision. In this type of collision, no kinetic energy is conserved due to the action of internal friction, heat, and sound. Rather, energy is discipated. Also, the vehicle is neither stretched nor compressed so there is zero elastic energy.
In case of vehicle crash whether in passenger cars or heavy commercial vehicles , vehicles involved come to rest and at that point, their final velocity becomes zero. Therefore, their final kinetic energy becomes zero.
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The Law of Conservation of Energy in case of passenger vehicles works the same as that for commercial vehicles with a minute difference. In the case of commercial vehicles, it is difficult to consider all different ways in which energy is transformed or dissipated. For commercial vehicles like big trucks with more than one trailer or railway truck with several boxes, energy does not seem to be balanced because some amount of initial energy dissipates in several different ways like vibration or thermal energy generated from friction in between two couplings.
The law of conservation of energy can be used to create an energy balance equation. To develop an energy balance equation, consider an isolated system where energy remains the same but may change form. Let’s apply the energy balance equation for passenger vehicles and roads. Assume the case where the vehicle was traveling with some kinetic energy on the flat road surface. After some time, the driver applied the brakes, and the vehicle slowed down to a new velocity. In this case, some initial kinetic energy was transformed into thermal energy due to the friction between vehicle tires and road. After applying brakes, the vehicle had some energy left because the vehicle still had some motion. Energy balance equation for this case can be written as,
= +
Where,
= total initial kinetic energy
=energy transformed into thermal energy
= remaining energy after skidding (kinetic energy)
In this way, the general energy balance equation can be written as,
47
= + + +……….+
Where,
= total initial kinetic energy
=first amount of energy transformed
= second amount of energy transformed
= third amount of energy transformed
= (last) amount of energy transformed
Energy is work done by vehicle while skidding,
=
Where,
= weight of vehicle
= drag factor
= displacement due to motion after skidding
While a vehicle is traveling with some velocity,
=
= weight of vehicle
= gravitational acceleration
= velocity at that point
During skidding if a vehicle stops after traveling some distance at the place which is up in height then some amount of energy is transformed into gravitational potential energy due to change in height, energy becomes,
= = ℎ
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8. GENERAL ANALYSIS FOR CONSERVATION OF MOMENTUM
8.1 Momentum
Momentum is a product of mass and velocity. It is a vector quantity as it has direction and magnitude.
=
(lb. s / N.s) = kg . (ft/s or m/s)
But, =
Hence, = ( ).
During a collision between two bodies, there is a change of momentum that is transferred with each other. Transfer of momentum is called impulse which is a product of force and time. The direction of the force (PDOF) plays a crucial role in complete momentum analysis. In the absence of external forces the Law of Conservation of Momentum must be applied to traffic collision reconstruction. As per that law, the total momentum before the collision must be the same, equal to the total momentum after the collision.
It is expressed as [2] [3],
+ = +
As, =
Where, and = Velocity before collision of two vehicles consecutively
and = Velocity after the collision of two vehicles consecutively
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= Momentum
Equation becomes, + = +
If we consider, both vehicles collided collinearly with head to head touch and vehicle still stay together after impact then,
Velocity of both vehicles after impact becomes the same, = =
Now equation becomes, + = + = ( + )
Therefore, =
As momentum is a vector quantity. It depends on the magnitude and direction to express it.
Figure 22. Vector Diagram
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Fig. 22 gives some considerations to derive the momentum equation.
1) Vehicle 1 is on X-axis making angle as zero degrees or 180 degrees. All angles
are measured counterclockwise from the positive X-axis set as the zero reference.
2) Here, the origin is considered to be a first point of collision for both vehicles.
3) Velocity of vehicle 2 at the first point of collision is solved.
4) As, = × but from the next equation, momentum ‘ ’ is considered as a product
of weight and velocity. This is possible because if we divide both sides of the
equation expressing the Principle of Conservation of Momentum, by gravity, we
can use the weights instead of the masses. This change does not affect in any way
the application of the Principle of Conservation of Momentum.
Momentum for both vehicles at their first point of collision will be written as follows.
At first contact, momentum ‘ ’ in X and Y direction for vehicle 1 is expressed as,
For X- direction, = = ……. (1)
For Y- direction, = = sin ……. (2)
At first contact, momentum ‘ ’ in X and Y direction for vehicle 2 is expressed as,
For X- direction, = = cos ……. (3)
For Y- direction, = = sin ………. (4)
Momentum of vehicle 1 and 2 after impact will be as follows,
After impact momentum in X and Y direction for both vehicles is expressed as,
For vehicle 1,
In X-direction, = = ……….(5)
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In Y-direction, = = ……….(6)
For vehicle 2,
In X-direction, = = ……….(7)
In Y-direction, = = ……….(8)
Total first contact momentum in X- direction is the sum of equation (1) and (3),
+ = + ………. (9)
Total after impact momentum in X-direction is the addition of equation (5) and (7),
+ = + ……. (10)
Equation (9) and (10) are equal,
+ = +
Therefore,
= ( + - ) / ( ) ……..(11)
Equation (11) gives the velocity of vehicle 1 at first contact,
Total momentum at first contact in Y-direction is a sum of equation (2) and (4),
+ = sin + sin ………. (12)
Total momentum after impact in Y-direction is a sum of equation (6) and (8),
+ = + ………. (13)
Equation (12) and (13) are equal and we can solve these equations to get ,
= ( + − sin )/( sin ) ………..(14)
As a first contact (origin) vehicle 1 makes an angle 0 degree or 180 degrees. Whereas, values of 0 and 180 are zero. Hence equation (14) becomes,
= ( + )/( sin ) ……….(15)
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8.2 PDOF and ∆
Momentum changes for vehicles involved in a collision along the principal direction of force (PDOF). Also, velocity change Delta-V takes place during momentum change. This velocity change called ∆ (delta).
From equation (11) and (14) we obtained the magnitude of first contact velocity with known direction.
Now, we will derive equations for change in velocity in a collision for each vehicle.
Figure 23. Change in velocity Vector for
In the above Fig 23, we have considered a velocity vector for before collision will always be on X-axis making an angle 0 or 180 degrees. Velocity after impact for vehicle
1 is is in fourth quadrant and change in velocity is shown by ∆ .
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Figure 24. X and Y component of ∆
Fig 24 above gives X-component and Y-component of ∆ from trigonometry.
= -
= -
Right angle triangle is made from ∆ , ∆ and ∆ . Now we can apply Pythagoras theorem,
= +
Hence, = ( cos − cos ) ( sin − sin )
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Figure 25. Vector Diagram for and
From Fig. 25 above we got,
As, =
Therefore, = =
In the above equation, is the value of angle but for proper angle of , in case of negative values of and we can use equation,
= + 180
Similarly, we can calculate and for vehicle 2 [3].
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Figure 26. Change in velocity Vector for
Figure 27. X and Y component of
56
Figure 28. Vector Diagram for to and
From above Fig. 26, 27, 28 [3],
= ( cos − cos ) ( sin − sin )
=
For both positive and , value of is equal to .
Also, and are in the opposite direction with an angle difference of 180 degrees.
PDOF angle can be determined as follows shown in Fig. 24 [3] below.
Note that positive and negative directions for both X ( and ) and Y
( ) components. Use absolute value that is a positive value for .
57
Figure 29. PDOF Angle Based on Calculated Δθ
58
9. MOMENTUM ANALYSIS FOR COMMERCIAL VEHICLE
9.1 Velocity/ Speed Factor for Commercial Vehicles
Conservation of momentum is used during the reconstruction of crashes to derive the velocity of vehicles involved. As explained above, this method considers before and after impact angle, weight, post-impact velocity for each vehicle. Conservation of momentum is a very reliable method when a collision happens in between vehicles which have similar weights or small weight difference.
This method becomes sensitive when it is applied to the crash between a commercial truck and a passenger car because of small changes in the pre-impact and post-impact angles.
Also, the car is lightweight in comparison with a commercial truck, it becomes sensitive to determine the velocity of the car. Due to the very light weight of the car compared to the truck, the car has a large change of momentum after the collision. In this type of crash, a small difference in the post-impact angle can have a large effect on the derived velocity of the car and a small effect on the calculated velocity of the truck. Therefore, we should be extra careful using the conservation of momentum methods involving car and truck crashes.
This method is reliable when a crash or collision happens between two trucks with small weight differences. Velocity calculations become less complex in such a case.
We will study a case involving the crash of two trucks in the upcoming section of this chapter. The challenge there is to keep track of the tracktors and the main trailers, four rigid bodies minimum.
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9.2 PDOF Factor for Commercial Vehicles
Collision behavior is different for commercial vehicles like tractor with trailers attached to it based on the Principal Direction of Force than that for single vehicles. Tractor with trailer tends to rotate about its fifth wheel pivot point rather than rotating about tractor’s center of mass. When the principal force of direction passes between the tractor’s center of mass and fifth wheel, tractor behaves like a single unit vehicle shown below in Fig. 30 [1]
Figure 30. PDOF Passes Between the Center of Mass and the Fifth Wheel of the Tractor
When PDOF passes between the tractor’s center of mass and fifth wheel for a tractor with a semitrailer attached to it, the tractor will have a rotation in the opposite direction shown in Fig. 31 below [1].
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Figure 31. PDOF Passes Between the Center of Mass and the Fifth Wheel of the Tractor Semi-trailer
When PDOF is passing through the center of mass of tractor, there will be no rotation or very small rotation of the tractor shown in Fig. 32 below [1].
Figure 32. PDOF Passes Through Center of Mass on the Tractor with No Semitrailer
Same as above PDOF is passing for tractor with semitrailer then such a centered PDOF can produce strong rotation shown in Fig. 33 [1] below. As the force on the tractor shifts towards the hitch pivot, the tractor tends to become more stable without rotation.
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Figure 33. PDOF Passes Through Center of Mass on the Tractor with Semitrailer
The behavior of the vehicle with a full trailer in collision depends on the articulation angle
(angle in between tractor and full trailer), motion of trailers, and load on the trailer hitch.
In the case of a tractor with a semitrailer, where semitrailer with half of its weight on the fifth wheel is the principal factor for the behavior of towing vehicle in case of collisions.
9.3 Case Study
9.3.1 Rear End Collision Case 1
In this example, Freightliner truck tractor semitrailer (Vehicle 2) and Western Star truck tractor flatbed semitrailer (Vehicle 1) traveling collinearly in the same direction with a combined weight of 56800 lbs. and 60150 lbs. consecutively. Western Star truck flatbed tractor semitrailer crashed into the rear of Freightliner truck semitrailer on an interstate highway. Both vehicles stayed engaged the entire distance from first contact to final rest.
Let’s do momentum analysis for this type of collision. For momentum analysis we need to
62 apply conservation of momentum for this case. In this case we have a pre-impact velocity of Freightliner truck semitrailer as there was an electronic control module that was installed on it. It was downloaded during crash inspection of the crash. As per the report shown in the Table 5 [1].
Table 5. ECM Report from Freightliner Truck Tractor
pre-impact velocity of Freightliner truck would be 22.5 mi/h which was constant before impact. The report shows within a fraction of seconds there was a sudden increment in the velocity of Freightliner truck tractor to 46.5 mi/h which was the velocity after impact.
The post-impact velocity of the Western Star truck is similar to the post-impact velocity of the Freightliner truck that is 46.5 mi/h.
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Collision System shown below in Fig. 34
Figure 34. Collision System
In-Line Collinear Crash in Fig. 35 [16]
Figure 35. In-Line Collinear Crash
As, = +
So, = +
And therefore, = +
According to the law of conservation of momentum, =
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Therefore, + = +
In this case, = = = = 0
After putting values of , , , into above equation becomes,
+ = + ……………………….(1)
In this case, both vehicles have the same post-impact velocity as both vehicles were engaged whole distance after collision till the final rest.
Therefore, = =
So above equation (1) becomes,
+ = +
+ + = ( + )
We have values of, 1) = 60150 lbs.
2) = 56800 lbs.
3) = 22.5 mi/h= 33ft/s (velocity of vehicle 2)
4) = 46.5 mi/h= 68.2ft/s
( ) Putting values in the equation =
We got, the velocity of vehicle 1 before impact, = 101.4 ft/s (69 mi /h)
The impact velocity of vehicle 1 has been cross verified with the computer simulation model.
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Pre-impact system
Figure 36. Pre-impact system for case 1
System right after collision at time 0.15 sec.
Figure 37. System right after collision at time 0.15 sec. for case 1
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System at 0.25 sec.
Figure 38. System at tme 0.25 sec. case 1
Let’s compare impact speed with computer simulations and momentum equations with different case scenarios. For all of the following cases, both vehicles are the same as that of in case 1. For vehicle 1(combination weight= 60150 lbs.), weight of semitrailer
[( ) =52150 lbs.] and cab [( ) =8000 lbs.]. For vehicle 2 (combination weight=
56800 lbs.), weight of semitrailer [( ) =49800 lbs.] and cab [( ) =7000 lbs.].
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9.3.2 Rear End Collision Case 2
Rear end collision with one stopped vehicle where vehicle 1 moving with velocity,
29.333 ft/sec, or 20 mi/hr. has rear-ended stationary vehicle 2. Therefore, = 29.333 ft/sec and =0.
Computer Simulations:
Pre-impact system
Figure 39. Pre-impact system for case 2
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System right after collision at 0.5 sec
Figure 40. System right after collision at time 0.5 sec. for case 2
System at time 1.55 sec.
Figure 41. System at 1.55 sec. for case 2
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Mathematical method:
After running simulations for the crash of two heavy commercial vehicles, we have got data as follows.
( ) = Post impact velocity of semitrailer for vehicle 1 =13.734 ft/sec.
( ) = Post impact velocity of cab for vehicle 1 =13.65 ft/sec
( ) = Post impact angle of rotation of semitrailer for vehicle 1 =3.24 degrees
( ) = Post impact angle of rotation of cab for vehicle 1 = -1.93 degrees
( ) = Post impact velocity of semitrailer for vehicle 2 = 16.555 ft/sec.
( ) = Post impact velocity of cab for vehicle 2 = 16.589 ft/sec.
( ) = Post impact angle of rotation of semitrailer for vehicle 2 = -1.941 degrees
( ) = Post impact angle of rotation of cab for vehicle 2 = 0.649 degrees
As per conservation of momentum shown in Fig. 42 [16],
Figure 42. Momentum diagram for case 2
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Momentum before collision = Momentum after collision
=
= + and = +
Therefore, + = +
As, = 0 therefore = 0
+ 0 = +
Therefore, = +
X- component:
cos = cos + cos
As, = 0 therefore cos = 1
= cos + cos ………. (1)
Y- component:
sin = sin + sin
As, = 0 therefore sin = 0
Hence equation becomes, - sin = sin ……… (2)
From equation (1),
= ( cos + cos ) / …………….(3)
As, = [( ) + ( ) ], = [( ) + ( ) ]/ 2 , = [( ) + ( ) ]
Expansion of term, cos as follows,
cos = [( ) + ( ) ]. [ [( ) + ( ) ]/2] . cos [( ) + ( ) ]
= [( ) + ( ) ]. [ [( ) + ( ) ]/2] .
[ cos( ) . cos( ) − sin( ) . sin( ) ]
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Similarly,
cos = [( ) + ( ) ]. [ [( ) + ( ) ]/2]. cos [( ) + ( ) ]
cos = [( ) + ( ) ]. [ [( ) + ( ) ]/2] .
[ cos( ) . cos( ) − sin( ) . sin( ) ]
After putting values in the expansions, we have got cos = 821806.593
cos = 939293.648
Also, = 60150
After putting values in equation (3), we have got = 29.27 ft/sec or 19.95 mi/hr.
The value of impact velocity of vehicle 1 with the momentum equation has been cross verified with computer simulations.
9.3.3 Rear End Collision Case 3
Rear end collision with both vehicles moving where vehicle 1 moving with velocity, 29.333 ft/sec, or 20 mi/hr. has rear-ended vehicle 2 moving with velocity 7.333 ft/sec or 5 mi/hr.
Therefore, = 29.333 ft/sec and =7.333 ft/sec.
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Computer Simulations:
Pre-impact system
Figure 43. Pre-impact system for case 3
73
System right after collision at time 0.55 sec.
Figure 44. System right after collision at time 0.55 sec. for case 3
System at time 1.55 sec.
Figure 45. System at time 1.55 sec. for case 3
74
Mathematical method:
After running simulations for the crash of two heavy commercial vehicles, we have got data as follows.
( ) = Post impact velocity of semitrailer for vehicle 1 =19.378 ft/sec.
( ) = Post impact velocity of cab for vehicle 1 =19.238 ft/sec
( ) = Post impact angle of rotation of semitrailer for vehicle 1 =9.070 degrees
( ) = Post impact angle of rotation of cab for vehicle 1 = -6.112 degrees
( ) = Post impact velocity of semitrailer for vehicle 2 = 18.341 ft/sec.
( ) = Post impact velocity of cab for vehicle 2 = 18.962 ft/sec.
( ) = Post impact angle of rotation of semitrailer for vehicle 2 = -6.179 degrees
( ) = Post impact angle of rotation of cab for vehicle 2 = 3.385 degrees
As per conservation of momentum as shown in Fig. 46 [16],
Figure 46. Momentum diagram for case 3
75
Momentum before collision = Momentum after collision
=
= + and = +
Therefore, + = +
Hence, + = +
X- component:
cos + cos = cos + cos
As, and = 0 therefore cos and cos = 1
+ = cos + cos ………. (1)
Y- component:
sin + sin = sin + sin
As, and = 0 therefore, sin and sin = 0
Hence equation becomes, - sin = sin ……… (2)
From equation (1),
=[ ( cos + cos ) – ( ) ] / …………….(3)
As, = [( ) + ( ) ] , = [( ) + ( ) ] / 2 , = [( ) + ( ) ]
Expansion of term, cos as follows,
cos = [( ) + ( ) ]. [ [( ) + ( ) ]/2] . cos [( ) + ( ) ]
= [( ) + ( ) ]. [ [( ) + ( ) ]/2] .
[ cos( ) . cos( ) − sin( ) . sin( ) ]
Similarly,
cos = [( ) + ( ) ]. [ [( ) + ( ) ] /2] . cos [( ) + ( ) ]
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cos = [( ) + ( ) ]. [ [( ) + ( ) ] /2] .
[ cos( ) . cos( ) − sin( ) . sin( ) ]
After putting values in the expansions, we have got cos = 1157892.007
cos = 1057286.39
Also, = 60150 and = 416514.4
After putting values in equation (3), we have got = 29.9 ft/sec or 20.38 mi/hr.
The value of impact velocity of vehicle 1 with the momentum equation has been cross verified with computer simulations.
9.3.4 Side Collision Case 4
Side collision with one vehicle is stationary where vehicle 1 moving with velocity, 29.333 ft/sec, or 20 mi/hr. has collided stationary vehicle 2 making 90 degree angle with vehicle
1. Therefore, = 29.333 ft/sec , =0 ft/sec and = 90 degrees.
77
Computer Simulations:
Pre-impact system
Figure 47. Pre-impact system for case 4
78
System right after at time 0.75 sec. collision
Figure 48. System right after collision at time 0.75 sec. for case 4
System at time 1.55 sec.
Figure 49. System at time 1.55 sec. for case 4
79
Mathematical method:
After running simulations for the crash of two heavy commercial vehicles, we have got data as follows.
( ) = Post impact velocity of semitrailer for vehicle 1 =15.423 ft/sec.
( ) = Post impact velocity of cab for vehicle 1 =15.735 ft/sec
( ) = Post impact angle of rotation of semitrailer for vehicle 1 =11.11 degrees
( ) = Post impact angle of rotation of cab for vehicle 1 = -13.17 degrees
( ) = Post impact velocity of semitrailer for vehicle 2 = 14.417 ft/sec.
( ) = Post impact velocity of cab for vehicle 2 = 17.208 ft/sec.
( ) = Post impact angle of rotation of semitrailer for vehicle 2 = (75.91-90)= -14.09 degrees
( ) = Post impact angle of rotation of cab for vehicle 2 = (102.09- 90)= 12.09 degrees
As per conservation of momentum as shown in Fig. 50 [16],
Figure 50. Momentum diagram for case 4
Momentum before collision = Momentum after collision
=
80
= + and = +
Therefore, + = +
Hence, + = +
Lets consider single equation approach,
sin + sin = sin + sin
sin( − ) + sin( − ) = sin( − ) + sin( − )
…………………………………….(1)
As, = [( ) + ( ) ] , = [( ) + ( ) ] / 2 , = [( ) + ( ) ]
And = [( ) + ( ) ] , = [( ) + ( ) ] / 2 , = [( ) + ( ) ]
Enter known values in equation (1),
(60150) . . sin(0 − ) + (56800) . . sin(90 − ) = (60150) . [[( ) + ( ) ]/2 ]
.sin [ [( ) + ( ) ] - ] + (56800) . [ [( ) + ( ) ] / 2 ] . sin [ [( ) + ( ) ] - ]
As, = 15.57 and = 15.81 above equation will become,
(60150) . . sin(0 − ) + (56800) . . sin(90 − ) = (60150) . (15.57) . sin [ [( )
+ ( ) ] - ] + (56800) . (15.81) . sin [ [( ) + ( ) ] - ] ………………….(2)
Expansion of sin [ [( ) + ( ) ] - ]= sin( ) . ( ) . ( ) + ( ) .
( ) . ( ) - ( ) . ( ) . ( ) + sin( ) . ( ) . ( )
………………(3)
Similarly,
81
sin [ [( ) + ( ) ] - ]= sin( ) . ( ) . ( ) + ( ) . ( ) .
( ) - ( ) . ( ) . ( ) + sin( ) . ( ) . ( ) ………….(4)
Enter = 90 degree into equation (2) to drop terms of as sin(90 − 90)= sin(0)=0
(60150) . . sin(0 − 90) = (60150) . (15.57) . sin [ [( ) + ( ) ] - 90 ] + (56800) .
(15.81) . sin [ [( ) + ( ) ] - 90 ] ……………………………………………….(5)
As per expansion in equation (3) and (4),
sin [ [( ) + ( ) ] - 90 ] = - ( ) . ( ) + sin( ) . ( )
and sin [ [( ) + ( ) ] - 90 ] = - ( ) . ( ) + sin( ) . ( )
Therefore equation (5) will become,
(60150) . . (-1)= (60150) . (15.57) . [− ( ) . ( ) + sin( ) . ( ) ] +
(56800) . (15.81) . [− ( ) . ( ) + sin( ) . ( ) ]
Therefore,
= [(60150) . (15.57) . [− ( ) . ( ) + sin( ) . ( ) ] + (56800) .
(15.81) . [− ( ) . ( ) + sin( ) . ( ) ] ] / (-60150) ……………...(6)
After putting values in equation (6), we have got = 30.39 ft/sec.
The value of impact velocity of vehicle 1 with the momentum equation has been cross verified with computer simulations and the results is nearly accurate.
9.3.5 Side Collision Case 5
Side collision with both vehicles are moving where vehicle 1 moving with velocity, 29.333 ft/sec, or 20 mi/hr. has collided vehicle 2 moving with velocity 14.66 ft/sec or 10 mi/hr and
82
making 90 degree angle with vehicle 1. Therefore, = 29.333 ft/sec , =14.66 ft/sec and
= 90 degrees.
Computer Simulations
Pre-impact system
Figure 51. Pre-impact system for case 5
83
System right after collision at time 0.70 sec.
Figure 52. System right after collision at time 0.70 sec. for case 5
System at time 1.55 sec.
Figure 53. System at time 1.55 sec. case 5
84
Mathematical method:
After running simulations for the crash of two heavy commercial vehicles, we have got data as follows.
( ) = Post impact velocity of semitrailer for vehicle 1 =25.855 ft/sec.
( ) = Post impact velocity of cab for vehicle 1 =25.509 ft/sec
( ) = Post impact angle of rotation of semitrailer for vehicle 1 =18.698 degrees
( ) = Post impact angle of rotation of cab for vehicle 1 = 2.159 degrees
( ) = Post impact velocity of semitrailer for vehicle 2 = 12.758 ft/sec.
( ) = Post impact velocity of cab for vehicle 2 = 20.448 ft/sec.
( ) = Post impact angle of rotation of semitrailer for vehicle 2 = (46.375-90)= -43. 62 degrees
( ) = Post impact angle of rotation of cab for vehicle 2 = (72.22- 90)= -17.77 degrees
As per conservation of momentum shown in Fig. 54 [16],
Figure 54. Momentum diagram for case 5
85
Momentum before collision = Momentum after collision
=
= + and = +
Therefore, + = +
Hence, + = +
Lets consider single equation approach,
sin + sin = sin + sin
sin( − ) + sin( − ) = sin( − ) + sin( − )
…………………………………….(1)
As, = [( ) + ( ) ] , = [( ) + ( ) ] / 2 , = [( ) + ( ) ]
And = [( ) + ( ) ] , = [( ) + ( ) ] / 2 , = [( ) + ( ) ]
Enter known values in equation (1),
(60150) . . sin(0 − ) + (56800) . . sin(90 − ) = (60150) . [[( ) + ( ) ]/2 ]
.sin [ [( ) + ( ) ] - ] + (56800) . [ [( ) + ( ) ] / 2 ] . sin [ [( ) + ( ) ] - ]
As, = 25.68 and = 16.603 above equation will become,
(60150) . . sin(0 − ) + (56800) . . sin(90 − ) = (60150) . (25.68) . sin [ [( )
+ ( ) ] - ] + (56800) . (16.603) . sin [ [( ) + ( ) ] - ] ………………….(2)
Expansion of sin [ [( ) + ( ) ] - ]= sin( ) . ( ) . ( ) + ( ) .
( ) . ( ) - ( ) . ( ) . ( ) + sin( ) . ( ) . ( )
………………(3)
Similarly,
86
sin [ [( ) + ( ) ] - ]= sin( ) . ( ) . ( ) + ( ) . ( ) .
( ) - ( ) . ( ) . ( ) + sin( ) . ( ) . ( ) ………….(4)
Enter = 90 degree into equation (2) to drop terms of as sin(90 − 90)= sin(0)=0
(60150) . . sin(0 − 90) = (60150) . (25.68) . sin [ [( ) + ( ) ] - 90 ] + (56800) .
(16.603) . sin [ [( ) + ( ) ] - 90 ] ……………………………………………….(5)
As per expansion in equation (3) and (4),
sin [ [( ) + ( ) ] - 90 ] = - ( ) . ( ) + sin( ) . ( )
and sin [ [( ) + ( ) ] - 90 ] = - ( ) . ( ) + sin( ) . ( )
Therefore equation (5) will become,
(60150) . . (-1)= (60150) . (25.68) . [− ( ) . ( ) + sin( ) . ( ) ] +
(56800) . (16.603) . [− ( ) . ( ) + sin( ) . ( ) ]
Therefore,
= [(60150) . (25.68) . [− ( ) . ( ) + sin( ) . ( ) ] + (56800) .
(16.603) . [− ( ) . ( ) + sin( ) . ( ) ] ] / (-60150) ……………...(6)
After putting values in equation (6), we have got = 31.50 ft/sec.
The value of impact velocity of vehicle 1 with the momentum equation has been cross verified with computer simulations and the results is nearly accurate.
9.3.6 Side Collision Case 6
Side collision with both vehicles are moving where vehicle 1 moving with velocity, 29.333 ft/sec, or 20 mi/hr. has collided vehicle 2 moving with velocity 29.333 ft/sec or 20 mi/hr
87
and making 90 degree angle with vehicle 1. Therefore, = 29.333 ft/sec , =29.333 ft/sec and = 90 degrees.
Computer Simulations:
Pre-impact system
Figure 55. Pre-impact system for case 6
88
System right after collision at time 0.55 sec.
Figure 56. System right after collision at time 0.55 sec for case 6
System at time 1.55 sec.
Figure 57. System at time 1.55 sec. for case 6
89
Mathematical method:
After running simulations for the crash of two heavy commercial vehicles using Working
Model 2D, we have got data as follows.
( ) = Post impact velocity of semitrailer for vehicle 1 =25.469 ft/sec.
( ) = Post impact velocity of cab for vehicle 1 =25.545 ft/sec
( ) = Post impact angle of rotation of semitrailer for vehicle 1 =40.89 degrees
( ) = Post impact angle of rotation of cab for vehicle 1 = 17.427 degrees
( ) = Post impact velocity of semitrailer for vehicle 2 = 25.565 ft/sec.
( ) = Post impact velocity of cab for vehicle 2 = 27.075 ft/sec.
( ) = Post impact angle of rotation of semitrailer for vehicle 2 = (56.32-90)= -33.68 degrees
( ) = Post impact angle of rotation of cab for vehicle 2 = (82.77- 90)= -7.23 degrees
As per conservation of momentum shown in Fig. 58 [16],
Figure 58. Momemtum diagram for case 6
90
Momentum before collision = Momentum after collision
=
= + and = +
Therefore, + = +
Hence, + = +
Lets consider single equation approach,
sin + sin = sin + sin
sin( − ) + sin( − ) = sin( − ) + sin( − )
…………………………………….(1)
As, = [( ) + ( ) ] , = [( ) + ( ) ] / 2 , = [( ) + ( ) ]
And = [( ) + ( ) ] , = [( ) + ( ) ] / 2 , = [( ) + ( ) ]
Enter known values in equation (1),
(60150) . . sin(0 − ) + (56800) . . sin(90 − ) = (60150) . [[( ) + ( ) ]/2 ]
.sin [ [( ) + ( ) ] – ] + (56800) . [ [( ) + ( ) ] / 2 ] . sin [ [( ) + ( ) ] –
]
As, = 25.507 and = 26.32 above equation will become,
(60150) . . sin(0 − ) + (56800) . . sin(90 − ) = (60150) . (25.507) . sin [ [( )
+ ( ) ] – ] + (56800) . (26.32) . sin [ [( ) + ( ) ] – ] ………………….(2)
Expansion of sin [ [( ) + ( ) ] – ]= sin( ) . ( ) . ( ) + ( ) .
( ) . ( ) – ( ) . ( ) . ( ) + sin( ) . ( ) . ( )
………………(3)
91
Similarly,
sin [ [( ) + ( ) ] – ]= sin( ) . ( ) . ( ) + ( ) . ( ) .
( ) – ( ) . ( ) . ( ) + sin( ) . ( ) . ( ) ………….(4)
Enter = 90 degree into equation (2) to drop terms of as sin(90 − 90)= sin(0)=0
(60150) . . sin(0 − 90) = (60150) . (25.507) . sin [ [( ) + ( ) ] – 90 ] + (56800) .
(26.32) . sin [ [( ) + ( ) ] – 90 ] ……………………………………………….(5)
As per expansion in equation (3) and (4),
sin [ [( ) + ( ) ] – 90 ] = - ( ) . ( ) + sin( ) . ( )
and sin [ [( ) + ( ) ] – 90 ] = - ( ) . ( ) + sin( ) . ( )
Therefore equation (5) will become,
(60150) . . (-1)= (60150) . (25.507) . [− ( ) . ( ) + sin( ) . ( ) ]
+ (56800) . (26.32) . [− ( ) . ( ) + sin( ) . ( ) ]
Therefore,
= [(60150) . (25.507) . [− ( ) . ( ) + sin( ) . ( ) ] + (56800) .
(26.32) . [− ( ) . ( ) + sin( ) . ( ) ] ] / (-60150) ………………(6)
After putting values in equation (6), we have got = 32.60 ft/sec.
The value of impact velocity of vehicle 1 with the momentum equation has been cross verified with computer simulations and the results is closely accurate.
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10. CONCLUSIONS
• Momentum analysis for heavy vehicles has been studied in detail. Experiments
considered collision of two vehicles with a tractor-trailer combination. This
presents a challenge to the momentum analysis since we need to keep track of the
vehicle dynamis of four rigid bodies, the two tracktors and the two trailers. ..
• Several cases using a mathematical approach for heavy commercial vehicles crash
has been compared to a computer simulation approach.
• Computer simulations were used under different scenarios of heavy truck
collisions. The main question is what were the conditions under which the collision
occurred . This means that we must reconstruct the accident in a backwards process
and compare it to a computer simulation that it is a forward process. Impact speeds
(speeds as the vehicles were going pre crash) have been cross verified with the use
of the momentum equations.
• These computer simulation models track the center of mass of semitrailer and cab
for both vehicles.
• The simulation shows the Delta-V for the velocity of the cab and semitrailer of the
vehicles.
93
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