4 Asset Management

To this point you have learned how to study traffic streams and design the roads that they use without very much concern for the real infrastructure that the roads are built from. In this chapter you will learn:

concepts related to pavement quality and performance
methods to design flexible pavements
procedures for life-cycle cost analysis of transportation facilities
how to describe the threat that climate change poses to transportation assets.

4.1 Concepts

Pavements are concrete structures with underlying layers of concrete and / or geotechnical materials that are designed to efficiently bear the loads of vehicles passing over them. There are two overarching classifications of pavements:

Flexible pavements have a surface made of asphalt cement concrete placed over gravel/aggregate base and sub-base layers. The load from vehicle tires causes the pavement to deform (mildly), which distributes the load through the underlying layers.
Rigid pavements have a surface made of reinforced portland cement concrete slabs. The load from vehicle tires is distributed to the underlying ground via a springing motion.

These two load mechanisms are illustrated in Figure 4.1). Flexible pavements tend to be less expensive to build and maintain, while rigid pavements are stronger and more durable. Flexible pavements are also quieter when tires drive on them at high speeds.

Figure 4.1: Flexible and rigid pavement load mechanisms.

All pavements deteriorate over time through use and weathering. The quality of a pavement can be described by various indices including the present serviceability index (PSI), which grades pavements on a scale of 0 (impassable) to 5 (perfectly smooth). PSI is a somewhat subjective measure and has been replaced in most contexts by the international roughness index (IRI) or the pavement condition index (PCI), which are more scientifically described.

Over time, pavements degrade as a result of traffic loads and environmental effects.

Figure 4.2: Present serviceability index (PSI) over time.

4.1.1 Types of Pavement Failure

ASTM 6433 describes how to assess pavements and measure the PCI for a stretch of roadway. Engineers who are evaluating the pavement quality must be able to identify multiple different types of pavement failure. Failure mechanisms for flexible pavements include (among others) the following methods illustrated in Figure 4.3):

Alligator cracking
Block cracking
Potholes
Rutting
Bleeding
Swelling and Sags: more useful for airports

Failure mechanisms for rigid pavements pavements include (among others) the following methods shown in Figure 4.4:

Blowup / Buckling
Corner breaking
Linear / transverse / diagonal cracking
Pop outs / Punch outs
Pumping
Spalling

These failures arise from a combination of repeated loadings by vehicles and weather patterns. Bleeding, for example, occurs when heavy vehicles repeatedly drive on hot pavement that has been under-designed for the load, or that has fatigued from overuse.

4.1.2 Estimating Loads

The load placed by traffic on roads is an important design consideration. Vehicles that regularly must support heavy construction or freight vehicles may require thicker pavements than roads that only bear passenger vehicles.

In the 1950’s, the American Association of Highway Officials (AASHO, later to become AASHTO) conduced a series of tests on a loop of roadway that would eventually become part of I-80 near Chicago, Illinois. The tests involved surplus Army trucks and drivers driving in a constant loop until the pavement wore down completely. Additionally, they had road users grade each section of pavement on a scale of 0 to 5, and used the information to develop the PSI index discussed earlier.

One outcome of this research was the development of a measure called the equivalent single axle load (ESAL). This measure describes the load of a vehicle on the pavement based on the sum of its axle weights in relation to an indexed 18,000 lb loaded truck axle, \[ ESAL = \sum_{i = 1}^n \left(\frac{w_i}{18000}\right)^4 \tag{4.1}\] where $w_i$ is the weight loaded at each single axle.

Example

A tractor in a “double-bottom” configuration is pulling two trailers placing 12 kips over the front axle and 17 kips over four single axles as shown in the image below (80k gross vehicle weight). How many ESAL’s does this truck load?

Using Equation 4.1,

\[\begin{align*} ESAL &= \sum_{i = 1}^n \left(\frac{w_i}{18000}\right)^4 \\ &= \left(\frac{12000}{18000}\right)^4 + 4 \times\left(\frac{17000}{18000}\right)^4\\ &= 3.38 \end{align*}\]

When two axles are close together (approximately 4’), they function as a tandem pair. In this case, the equation for a tandem axle becomes

\[ ESAL_{\mathrm{tandem}} = \left(\frac{w_i}{33200}\right)^4 \tag{4.2}\]

Example

A tractor in a typical semi configuration is pulling a trailer placing 12 kips over the front axle, 36 kips over the drive axles, and 32 kips over the trailing axles as shown in the figure below (80k gross vehicle weight). How many ESAL’s does this truck load?

Using Equation 4.1 and Equation 4.2, \[\begin{align*} ESAL &= \left(\frac{12000}{18000}\right)^4 + \left(\frac{36000}{33200}\right)^4 + \left(\frac{32000}{33200}\right)^4\\ &= 2.44 \end{align*}\]

The total number of ESAL’s loaded onto a pavement in its lifetime is a function of the truck weight and load configurations, the number of trucks, and the amount of growth in traffic over time. The total lifetime ESAL’s can be had with an exponential growth rate equation, \[ ESAL_{\mathrm{lifetime}} = ESAL_{\mathrm{current}} \left[\frac{(1 + i)^n - 1}{i}\right] \tag{4.3}\] where $i$ is the growth rate and $n$ is the number of periods.

Tip

Note that this is the standard $F = A(i, n)$ equation (future value of a repeating value for $n$ periods).

Example

Each workday (250 in a year), 30 semi trailers full of goods arrive at a facility and leave empty. What are the lifetime ESAL’s assuming a 10 year life and a 2% growth rate on the driveway of this facility (that is, the empty and full trucks both travel on the same pavement)? An empty semi exerts loads of 12 kips on the front axle, 18 kips on the drive axles, and 10 kips on the trailing tandem axles.

We already calculated that the full semi is 2.44 ESAL’s. The empty semi is \[\begin{align*} ESAL &= \left(\frac{12000}{18000}\right)^4 + \left(\frac{18}{33200}\right)^4 + \left(\frac{10}{33200}\right)^4\\ &= 0.29\mathrm{\ ESALs} \end{align*}\]

If 30 full and 30 empty semis travel on the road each of 250 days in a year, that makes for 20,514 annual ESALs in the base year. Given the 2% growth rate and Equation 4.3), the 10-year lifetime ESALs are

\[\begin{align*} ESAL_{\mathrm{lifetime}} &= ESAL_{\mathrm{current}} \left[\frac{(1 + i)^n - 1}{i}\right]\\ &= 20,514 \times \left[\frac{(1 + 0.02)^{10} - 1}{0.02}\right]\\ &= 224,626 \end{align*}\]

Think

For large, federally funded roads in the United States, current practice has progressed / advanced beyond the ESAL method. AASHTO publishes both charts and software that allows engineers to forecast pavement traffic loading as a result of vehicle counts and an understanding of pavement mechanics. The ESAL method is still used in cities and counties — and internationally. It also provides a good introduction to more advanced methods you might encounter later, specifically in CE 563

4.2 Flexible Pavement Design

Flexible pavements almost always consist of three layers:

Hot mix asphalt concrete
Base course: Coarse aggregate, sometimes treated
Sub-base course: finer gravel

Beneath this is the subgrade, which is made of local soil or sometimes soil that has been brought in as a replacement for poor quality local soil.

The total strength of the pavement can be described with a structural number, \[ SN = a_1 d_1 + a_2 d_2 + a_3 d_3 \tag{4.4}\] where $a_1, a_2, a_3$ are structural coefficients for each layer and $d_1, d_2, d_3$ are layer thicknesses. Thicker pavements made of stronger materials will support more loading and/or be more resilient. The challenge for designers is to identify the required structural number based on local conditions and traffic demand (in terms of ESALs). The equation relating all of these factors is rather unwieldy, \[ \log_{10}(W_{18}) = Z_R S_0 + 9.36 \log_{10}(SN + 1) - 0.2 + \frac{\log_{10}(\Delta PSI) / 2.7}{0.4 + (1094 / (SN + 1)^{5.19})} + 2.32 \log_{10}(M_r) - 8.07 \tag{4.5}\] So instead we use the nomograph shown in Figure 4.5. The parameters of this equation are:

$W_{18}$: the expected ESALs in millions
$Z_R$: pavement reliability expectation. For major roads, this will be over 90%.
$S_0$: standard deviation of all factors in the design: if we are more certain about $W_18$ and the other parameters, this can be smaller. A value around 0.4 is typical.
$M_r$: the road bed resilient modulus in ksi.
$\Delta PSI$: What is the allowable deterioration in pavement serviceability over the life of the pavement?

Figure 4.5: AASHTO flexible pavement design nomograph.

The steps to determine a structural number for a pair of given pavements are illustrated on the nomogoraph.

Example

You are designing a road in an area with a roadbed resilient modulus of 5 ksi. The lifetime demand estimate is 5 million ESALs, with a standard deviation of 0.35. The road requires 95% reliability and an allowable serviceability loss of 2 PSI. What is the required structural number for this pavement?

This example is illustrated on Figure 4.5 with the fainter line.

Find the 95% reliability on the $R$ line at the left.
Trace with a straightedge through the point identified in step 1 and the estimated standard deviation of 0.35 on the $S_0$ line.
Come to the first turn line $T_L$.
Find the required demand $W_{18} = 5.0$ million ESALs.
Draw a line with a straightedge from your point on the first $T_{L}$ through the point identified in step 4 to the second $T_L$.
Identify the resilient modulus of 5 ksi on the $M_r$ line.
Draw a line from where you stopped on the second turn line through the point identified in step 6 to the vertical axis of the chart.
Draw a horizontal line from the vertical axis to the curve representing the design serviceability loss, in this case 2.0.
Draw a vertical line dropping from the point in step 8 down to the horizontal axis. This is the design structural number.

The design structural number for this pavement is $SN = 5$.

Once the structural number for the entire pavement has been determined using Figure 4.5, we need to determine the thickness of each pavement layer from Equation 4.4. Each layer sits on top of what’s beaneath it, which means we can continue to use the nomograph to determine the $SN$ of each layer. This depends on the materials that are selected for the base and sub-base courses.

Base course materials might be specified in any number of ways, including California Bearing Ratio (CBR), Texas Triaxial, or resilient modulus. Figure Figure 4.6 (a) shows these values for base layers and Figure 4.6 (a) for sub-base layers.

Example

The pavement described in the previous example is using a base layer with a resilient modulus of 25 ksi and a sub-base layer with a resilient modulus of 11 ksi. Determine the structural numbers for each layer.

The total structural number was already determined to be 5. The demand, reliability, and certainty all remain unchanged from the previous example, so we can start with step 5. Because the top layer sits on top of the base layer, we draw a line through the $M_r$ for the base layer, which is 25 (illustrated in blue). The structural number for the top layer by itself is approximately $SN_1 = 2.8$.

The structural number for the base layer is determined by the resilient modulus of the sub-base layer, which in this case is 11 ksi. Tracing with the red line in the figure above, we determine that the structural number for layer 2 is $SN_2 = 3.8$. Note that this is the effective structural number for layer 1 and layer 2. To summarize:

Total $SN = 5$: top + base + subbase
Base $SN_2 = 3.8$: top + base
Top $SN_1 = 2.8$: top only

Now that we know the required structural number at each layer, we can determine the thickness of each layer. The structural coefficients $a_2, a_3$ can be determined from Figures Figure 4.6 (a) and Figure 4.6 (b). The top layer coefficient is determined based on the elastic modulus of the asphalt used, shown in Figure 4.7.

Figure 4.7: Structural coefficient for top layer of flexible pavement.

Example

The HMA used in this example has an elastic modulus of 375,000 psi. Determine the structural coefficient for each layer of pavement in this example.

Using Figure 4.6 and Figure 4.7, we can look up the following values:

$a_1$ = 0.41
$a_2$ = 0.12
$a_3$ = 0.08

The thickness of each layer is computed from the structural number, $d_1 = SN_1 / a_1$. There are two additional considerations, however. First, regulations specify a minimum recommended thickness regardless of the structural number or the strength of the material; these are shown in Table 4.1. Second, the accuracy of road contractors is fairly limited: always round a design thickness up to the next half inch for the hot mix asphalt and the next full inch for base and sub-base layers.

Table 4.1: Minimum thickesses for flexible pavement design.

Traffic (ESAL)	HMA (inches)	Aggregate Base
<50k	1.0	4
50k - 150k	2.0	4
150k - 500k	2.5	4
500k - 2 million	3.0	6
2 million to 7 million	3.5	6
Over 7 million	4.0	6

Example

Determine the thickness of each layer in the pavement design.

We work down through the pavement. The required depth for the top layer is \[\begin{align*} SN_1 &= a_1 d_1\\ 2.8 &= 0.41 * d_1\\ d_1 &= 6.8\mathrm{\ inches} \end{align*}\]
This is thicker than the 3.5 inch minimum for a road with 5 million ESALs, but we need to round up to 7 inches.

The required depth for the base layer is \[\begin{align*} SN_2 &= a_1 d_1 + a_2 d_2\\ 3.8 &= 0.41 * (7) + 0.12 * d_2\\ d_2 &= 7.75 \mathrm{\ inches}. \end{align*}\] Again, this is thicker than the minimum of 6, but we need to round to 8 inches.

The required depth for the sub-base layer is \[\begin{align*} SN &= a_1 d_1 + a_2 d_2 + a_3 d_3\\ 5 &= 0.41 * (7) + 0.12 * (8) + 0.08 * d_3\\ d_3 &= 14.6 \mathrm{\ inches} \end{align*}\] Or 15 inches. We should check the total structural number to ensure that our design meets the requirements: \[\begin{align*} SN &= 0.41 * (7) + 0.12 * (8) + 0.08 * (15)\\ &= 5.03 \end{align*}\]

Tip

The structural number formula in Equation 4.4) is more precisely \[ SN = \sum_{i = 1}^3 a_i d_i m_i \tag{4.6}\] where the additional $m_i$ term adjusts for the moisture content of the base and sub-base layers. For the top layer, $m_1 = 1$ always because asphalt concrete does not absorb moisture to any appreciable degree. If you must design a pavement that is likely to be submerged or have poor drainage, then you may need to consider other values for $m_2, m_3$ in the design.

The structural number equation on the FE reference sheet doesn’t include the $m_i$ terms, but some texts will include them.

4.3 Life Cycle Cost Analysis

Pavements do not last forever. Within a few years of installation (maybe less), the pavement will start to develop cracks. These cracks allow moisture to penetrate the surface of the pavement, and can start to erode the base layer. The presence of moisture in the pavement will also cause additional damage during freeze-thaw cycles. If left un-maintained, the road will begin to deteriorate more rapidly. Figure 4.8) shows the non-linear nature of this deterioration curve in empirical data, with the pavement condition index (PCI) for surfaces less than two years old being in the high 90’s and surfaces in more than ten years old showing sometimes substantial degradation.

Figure 4.8: Observed deterioration in pavement quality on Interstates in Virginia, from Pantuso et al. (2021). a) observed deterioration, b) modeled deterioration.

Various models can be used to predict pavement quality after a number of years. One fairly simple “sigmoid” model (Pantuso et al., 2021) uses a set of two estimated regression coefficients $d, e$, \[ PCI_{\mathrm{predicted}} = 100 - d * \exp(-e/age) \tag{4.7}\] The shape of this curve for three different sets of variables are given in Figure 4.9.

Figure 4.9: Statistical pavement deterioration curves. Graphics created from models in Pantuso et al. (2021).

Example

You are forecasting the deterioration of a new road. An analysis of similar roads in your region generated degradation coefficients of $d = 52.01$ and $e = 8.214$. What is the expected PCI in year 7, assuming the pavement starts at a PCI of 100?

Using Equation 4.7, we can calculate the PCI as \[\begin{align*} PCI_{\mathrm{predicted}} &= 100 - d * \exp(-e/age)\\ &= 100 - 52.01 \exp(-8.214 / 7) \\ &= 83.9 \end{align*}\]

Multiple strategies can be used to rehabilitate pavements and / or prolong their useful life. For block cracking, sealing the cracks with tar, a sealcoat, or a chip seal keeps moisture out and is relatively inexpensive. For more substantive cracking, an overlay of additional hot mix asphalt can protect the pavement. When the damage to the pavement surface is substantial but the underlying layers are intact, the road may potentially be a candidate for in-place recycling, where a train of machines (see Figure 4.10) rips up and mills the old pavement, re-melts the asphalt, and re-lays a new layer of pavement. Eventually, however, a full-depth reconstruction of the entire road will be necessary.

Figure 4.10: Diagram of pavement recycling equipment. Image from Texas DOT

Costs for each of these options are highly localized and variable. But the cost of a full-depth construction may be several orders of magnitude higher than a chip seal.

In a life cycle cost analysis, an engineer considers not just the cost of an initial road design, but the cost of future repairs, salvage, and total service life. In order to compare different alternative life cycles on an equivalent plane, it is necessary to convert future anticipated expenditures into present values: \[ F = P(1 + i)^n \tag{4.8}\] Where a present value $P$ is equivalent to a future value $F$ given a discount rate $i$ and a number of time periods $n$.

Example

This example is taken from the FHWA Life Cycle Cost Analysis manual.

Two alternatives for building a road exist:

An initial design that costs $26M requires a $15M rehabilitation after 20 years. After 35 years, the amount of life left in the pavement will be worth approximately $3.75M to the agency and $7.5 to users. The construction and rehabilitation activities are substantial, and will result in $11M of user costs for the initial construction and $30M at the 20-year rebuild.
An initial design that costs $20M requires a $6M rehabilitation in year 12, 20, and 28. After 35 years, the amount of life left in the pavement will be worth approximately $0.75M to the agency and $3.5M to users. The construction and rehabilitation activities are less substantial, and will result in $8M of user costs for the initial construction, and $10M, $16M, and $28M for each rehabilitation.

All prices are given in constant (nominal) dollars. Assuming a discount rate of 4%, which alternative is less costly?

The results are calculated in the table below. Based on this analysis, Alternative A has lower present life cycle costs. It should be noted, however, that the agency costs for Alternative B are lower. Theoretically this shouldn’t matter, but in practice an agency may have its own budgeting priorities, and user costs are somewhat hard to measure.

Present LCCA of Two Pavement Alternatives
	Alternative A			Alternative B
Year	Agency Costs	User Costs	Present Costs	Agency Costs	User Costs	Present Costs
0	26.00	11.0	37.000000	20.00	8.0	28.000000
12	0.00	0.0	0.000000	6.00	10.0	9.993553
20	15.00	30.0	20.537413	6.00	16.0	10.040513
28	0.00	0.0	0.000000	6.00	28.0	11.338234
35	-3.75	-7.5	-2.850924	-0.75	-3.5	-1.077016
Total	37.25	33.5	54.686489	37.25	58.5	58.295284

4.3.1 Climate Change Adaptation

Climate affects pavements in a number of ways:

At high temperatures, asphalt binders become softer. This leads to rutting and bleeding.
Sunlight and UV radiation deteriorates asphalt over time.
At low temperatures, asphalt binders become brittle, leading to cracking.
Precipitation can introduce additional moisture to the base and sub-base layers, threatening the stability of the pavement structure.
Freeze-thaw cycles are a major factor in cracking, buckling, and deterioration of underlying pavement layers.
Winter road maintenance activities such as plowing and using chains or studded tires can degrade the pavement surface.
Salting the pavement also leads to oxidation of reinforcing bars and other chemical deterioration.

As a result of these effects, pavement standards in different regions can vary substantially. Figure 4.11 shows the nine different climate zones established by CalTrans.¹

Figure 4.11: California climate regions for pavement design.

Climate change therefore poses a particular challenge to pavement engineers. Hotter, drier summers may mean more rutting and bleeding, and faster pavement degradation. In many places the ground freezes solid for the entirety of winter; warming temperatures may mean that these areas will experience multiple freeze-thaw cycles in their base structure.

The impacts of climate change on transportation assets are not limited to pavements. In its Asset Management, Extreme Weather, and Proxy Indicators Pilot Project, the Arizona Department of Transportation includes a number of potential first-order and second order impacts of climate change. These effects are illustrated in Figure 4.12. Some of these effects are acute and catastrophic (such as a wildfire or a flood), while others are longer-term and more marginally damaging.

Figure 4.12: Impacts of climate change on roadway infrastructure. From Arizona Department of Transportation.

Think

Most pavement deterioration models in the recent literature are strictly empirical: they use past observed deterioration to forecast future deterioration. What might be the limits of this for climate change adaptation? What might pavement researchers consider?

4.4 Freight

As introduced at the outset of these notes, transportation is the movement of people and goods. To this point many of our problems have dealt with neither of these things, but rather we have looked at how vehicles move on the roadway. In the next unit we will look at how people choose which transportation modes to use, and how to design transportation systems to maximize people throughput. But for now we will talk about goods.

Freight is the commercial transport of goods from place to place. It can be difficult to model and understand freight and freight movements, and that makes it difficult for engineers to plan for and design transportation systems to accommodate freight.

Freight uses modes that we haven’t addressed to this point, and that might be hidden from typical analysis. The Mississippi River carries more freight by weight on barges than any interstate highway. Of course, trucks usually have to get the cargo from the shipper to the barge and off again. Truck traffic that supports port and rail operations is called drayage. Some of the most common freight modes are pictured in Figure 4.13). Intermodal freight shipments became considerably easier with the invention of the shipping container, a standard metal box that can be easily moved between ships, rail cars, and truck beds with cranes and forklifts. A plot of the some of the busiest shipping ports by volume is given in Figure 4.14; most of the world’s busiest ports are in Asia, and the growth of international shipping from Asia to the rest of the world has been impressive. It is not too great a claim to say that freight containerization has been the most important technological innovation for the global economy in the last 100 years.

Contracts between freight operators and shippers are usually privately arranged and not always rational. The cost to ship a truckload of hogs from a North Carolina farm to a Wisconsin meat packer is not printed in a neat schedule of prices. So even if the average cost of shipping by rail is cheaper than truck, a shipper might have a relationship with a truck-based freight operator that leads them to pick the truck at a higher price, or a lower price that an external analyst cannot observe.

The commodities shipped as freight are all different from each other, and require different considerations for time of delivery, method of shipping, etc. Coal is cheap and heavy; electronics and pharmaceuticals are light but expensive and fragile. Produce and many food products must be refrigerated from the farm to the store, and operational inefficiencies on rail networks mean that produce in the United States rarely moves by rail.

Many freight operators are large enough to work as de facto monopolies. For passenger travel, you can sample 1% of households and get a good picture on how people get to work and back; if you sample 1% of freight companies and don’t get Union Pacific you miss a lot of goods movement. Also UPS and FedEx are different enough from each other that understanding one won’t help you understand the other.

Figure 4.14: Container port volumes rankings. Data from Lloyd’s list via Wikipedia

Homework

HW 1: ESAL calculations of an SUV vs. Semi

An SUV weighs 5,000 lbs. How many of these SUVs would it take to have the same total ESALs as the fully-loaded truck whose tractor has a single axle that carries 12,000 lbs, a set of tandem axles that carry 36,000 lbs, and a set of tandem axles that carry 32,000 lbs?

HW 2: ESAL Calculations of a Truck With Two Trailers

A logistics firm uses 25,000-lb tractors that have a single front steering axle (load 10,000 lbs) and a tandem drive axle. It pulls two trailers behind each tractor, each having a tare weight of 10,000 lbs. The front trailer carries a 30,000-lb load and attaches to the tractor with no front axle, but a single rear axle. The rear trailer has two single axles and a 60,000-lb load. The weight can be assumed be distributed equally in each trailer.

If the firm sends out 80 trucks per day on 300 days per year, what is the annual ESAL for this firm?

HW 3: ESAL Calculations With WIM Data

WIM Data
kips/axle	Axle Type	Design lane freq/day N (i)	Growth rate r
1.5	Single	19699	0.018
1.5	Single	2500	0.062
3.5	Single	2500	0.062
3.5	Single	2344	0.022
10.0	Single	2550	0.045
12.0	Tandem	1344	0.022
20.0	Tandem	2550	0.045
34.0	Tandem	2550	0.045

The table above shows the count of axles by weight and type on an average day in the design lane of US-6, along with the observed growth rate in those axles. Using the fourth power formula and a 320-day year, calculate the total 25-year ESAL value for this design lane. You should use a spreadsheet, but you should show sample calculations of ESALs and growth rate factor $G$ for the first row of data.

HW 4: Subbase Thickness

The subgrade for a flexible pavement with a predicted load of 20 million ESAL has been prepared to a CBR of 5. The structural coefficient for the 2-inch hot mix asphalt is $a_1=0.4$. Below this there should be 6 inches of asphalt base with a structural coefficient of $a_2= 0.3$. The structural coefficient of the subbase is $a_3= 0.1$.
The resiliability target is set to $R = 90$ percent, standard deviation is 0.4, and 2e+07 ESAL are expected. The change in PSI is 2.5. What is the required depth of aggregate sub-base? Show your answers on an included nomograph. Express your answer in inches. Note that for sub-grade,$M_R = 1500 * CBR$.

HW 5: Layer Thicknesses

You are rehabilitating a road that expects 10 million ESAL over the next 15 years.
The state requires 90 percent reliability with a standard deviation of 0.4 and a change in the PSI no greater than 1.5. The subgrade has a CBR of 6. determine the thickness of hot mix asphalt concrete ($a_1 = 0.4$) that will be needed for a the top layer (with no base course), if the subbase is 12 inches of crushed packed stone with a structural layer coefficient of 0.1.

HW 6: A Pavement Life That Falls Short

Give three reasons why a pavement with a valid 20-year design might “fail” before 20 years.

HW 7: Flexible Pavement Design

Data for Layer Design and Cost Analysis
Material	Layer	a(i)	M(R)	Unit weight (lb/cu ft)	Unit Cost / ton (Delivered)
Hot mix asphalt concrete	Top / wearing	0.40	NA	140.1	89.62
Untreated aggregate	Base	0.10	21000	131.4	18.67
Cement-treated aggregate	Base	0.16	200000	133.2	25.88
Untreated aggregate	Subbase	0.07	11000	129.9	14.25
^a The subgrade soil has a CBR of 15.
^b Excavation costs are $4.87 per cubic yard of soil displaced.
^c The subgrade soil has an M(R) value of 7750 psi.

You are designing the pavement for a 12-foot lane on a highway that is 1-mile long, expecting 39 million ESALs. The cost of your design is a function of the amount of materials used, with the material costs given in the table above. Your design alternatives should include:

A full-depth Hot Mix Asphalt Concrete pavement that uses a design Structural Number of 4.5.
A layer design that follows the guidelines for layer thickness given in Section 4.2. Use $SN_1 = 2.8, SN_2 = 3.8, SN_3 = 4.5$ as your design structural numbers.
Design a pavement depth $d_1, d_2, d_3$ that meets the total structural number of 4.5, seeking to minimize the total cost and maintaining minimum depths. You may optionally use the cement-treated aggregate base that is stronger but more expensive.

Assume that you will excavate the full depth of the pavement. Which of your layer designs is the most economical? Show your analyses and state your conclusions clearly. Note: This problem is best solved with a spreadsheet containing calculation notes, or with well-documented computer code.

HW 8: Life Cycle Cost Analysis

You are conducting a LCCA analysis for a new rural highway. Using observations from other similar roads in the same climate region, you have estimated the following pavement deterioration function: \[PCI_n = PCI_0 - 90\exp(-8/n)\] where $n$ is the number of years since the last treatment and $PCI_0$ is the initial pavement condition index after the last treatment. You have the following treatment options:

A full-depth reconstruction of the road returns the PCI to 100, and costs $400M
An in-place surface recycling adds 15 PCI to the previous year’s PCI and costs $100M
A HMA overlay adds 7 PCI to the previous year and costs $30M
A seal coat adds 2 PCI and costs $5M.

In year 20, any PCI over 70 is considered recovered cost. That is, if $PCI_{20} > 73.5$, add a positive cash flow of $3.5 M in year 20. But for any year where the PCI is below 70, the difference is considered an expenditure (if the PCI is 68 in a year, add a 2 M deficit). For each plan, plot the expected PCI in each year and calculate the NPV. Use a discount rate of 4%. Consider the following treatment plans:

Full-depth reconstruction in year 0. No additional treatments. Hint: your NPV should be $520-540 M.
Full-depth reconstruction in year 0, HMA overlay in year 6, in-place recycling in year 11, and a seal coat in year 16.
Design your own pavement treatment plan, trying to minimize the Net Present Value (NPV).

HW 9: Climate Change Adaptation

Re-calculate the $NPV$ and $PCI$ values from your plans above, assuming that the value changes from 8 to 6.5. What impact would this have on the sustainability of your pavement plans? Do your plans still have the same ordering of NPV?

HW 10: Choosing The Right Freight Mode

Shipping Prices
Item	Value [USD]	Weight [lbs]	Rate [USD/ton]
Tulip bulbs	0.6	8.8e-02	2400
LED TV	1500.0	5.0e+01	4000
Coal [ton]	20.0	2.0e+03	10

?tbl-hw-freight-values shows the weight, value, and typical shipping rates for several goods.

Compute the value of a ton of each good.
The shipping rates for an unspecified freight mode $X$ is $1000 per ton, regardless of the cargo. Why might a shipper choose to remain with their existing mode instead of switching to mode $X$?

HW 11: Inventory and Transportation Costs

A piano from a manufacturer in rural Indiana costs $5,000.00, weighs 800 pounds with packing, and is packaged in a custom container 72 inches wide, 54 inches tall, and 27 inches deep. The pianos should not be stacked or rotated vertically.

The trucks available to ship the pianos have a back box 8’ wide, 8’ tall, and 14’ long. The trucks have an empty weight of 10,000 pounds, and a bridge near the factory has a weight limit of 30,000 pounds per truck.

How many pianos fit into the truck?
Will the trailer weigh out before it cubes out?

HW 12: Freight Analysis Framework

The Freight Analysis Framework (FAF) is a statistical model output created for the Federal Highway Administration (FHWA) by staff at Oak Ridge National Laboratory. The model is estimated from data collected in the Commodity Flow Survey and the 5-year economic census. The most recent update was from data collected in 2017, and was released as FAF5. This product includes a forecast of future freight flows.

Go to the data tabulation tool for the FAF at the Oak Ridge National Laboratory website, and view the “Data Tabulation Tool” Select freight flows originating in Utah (Origin 491 - Salt Lake City and 499 - Rest of UT), and answer the following questions:

How many total kilotons of freight did Utah produce in the 2022 FAF forecast?

Now, go find the state-level summary statistics tables, and answer the following questions: b. What percent of that freight (by weight) remained in Utah? c. What state received the largest amount of Utah’s domestic exports (by weight)? d. Of the freight leaving Utah and going to other US States (not exported internationally), what is the mode share by weight? That is, what percent of freight leaving Utah goes by truck versus pipeline versus air? e. Of the freight leaving Utah and going to other US States (not exported internationally), what is the mode share by value? (Find the sheet in the excel workbook labeled “Value”) f. What are the top three commodities shipped to Utah from other states by weight? g. What are the top three commodities shipped to Utah from other states by value?

References

Pantuso, A., Flintsch, G. W., Katicha, S. W., & Loprencipe, G. (2021). Development of network-level pavement deterioration curves using the linear empirical bayes approach. International Journal of Pavement Engineering, 22(6), 780–793. https://doi.org/10.1080/10298436.2019.1646912

To my knowledge, Utah uses a single climate region, but I might be wrong.↩︎