Papagiannakis Solutions for text book

Instructor’s Solution Manual For

Pavement Design and Materials 2nd Print By A.T.Papagiannakis and E.A. Masad st

1 Edition Published by Wiley and Sons Inc. ©2009

Dec. 2009

Acknowledgement: The authors wish to express their sincere thanks to all the graduate students of Washington State University, Texas A&M University and the University of Texas at San Antonio that contributed to the solutions in this manual. They are in alphabetical order: • Silvia Caro • Veronica Castelo-Blanco • Chien-Wei Huang • Emad Kassem • Enad Mahmoud • Arash Rezai • Jessica Woods and • Habtamu Zelelew

Acknowledgement: The authors wish to express their sincere thanks to all the graduate students of Washington State University, Texas A&M University and the University of Texas at San Antonio that contributed to the solutions in this manual. They are in alphabetical order: • Silvia Caro • Veronica Castelo-Blanco • Chien-Wei Huang • Emad Kassem • Enad Mahmoud • Arash Rezai • Jessica Woods and • Habtamu Zelelew

Chapter 2 2.1

Truck shown in Figure 2.9(a). Use Table 2-8 for single axle and penalize load by 10% for single tires: Steering: 63.4 kN x 1.1 = 69.7 4 kN therefore ESAL = 0.575 Use Table 2.9 for tandem axles: Drive axle: 128.2 kN therefore Trailer axle: 148.3 kN therefore Total for vehicle

ESAL = 0.557 ESAL = 1.009 ESAL = 2.14

Truck show in Figure 2.9(b). Use Table 2-8 for single axle and penalize load by 10% 10 % for single tires: Steering: 53.2 kN x 1.1 = 58.5 4 kN therefore ESAL = 0.285 Use Table 2.9 for tandem axles: Drive axle: 92.5 kN therefore Trailer axle: 52 kN therefore 2nd Trailer axle: 50.2 kN therefore Total for vehicle

ESAL = 0.261 ESAL = 0.139 ESAL = 0.012 ESAL = 0.697 rounded to 0.7

2.2

Compute the dynamic load Coefficient of Variation (CV ) using Equ. 2.2 and the constants given on table 2.3. Air-spring suspension: CV = 70 0.346 2.2 0.798 = 8.16%

Rubber-spring suspension: CV = 70

0.456

2.20.728 = 12.32%

Compute standard dynamic load deviations: Tractor: 0.0816 x 75 = 6.12 kN Trailer: 0.1232 x 70 = 8.62 kN

Assuming that the dynamic load is normally distributed, the following ranges are obtained for 90% confidence: Tractor: 75 ± 1.65 x 6.12 = [85.1, 64.9] kN Trailer: 70 ± 1.65 x 8.62 = [74.2, 55.8] kN Where 1.65 is the standard normal deviate for 90% confidence.

2.3

The maximum dynamic load range tolerated for the tandem tractor axle is 20/1.65 = 12.1 kN, or 6.06 kN for each individual axle. This reflects a CV of 6.06/75 = 8.08% Similarly, the maximum dynamic load range tolerated for the tandem trailer axle is 20/1.65 = 12.1 kN, or 6.06 kN for each individual axle. This reflects a CV of 6.06/70 = 8.66%. Solve Equ. 2.2 for speed: Tractor air-spring: 8.08 = V 0.346 2.2 0.798 Trailer rubber-spring: 8.66 = V 0.456 2.2 0.728 Solution gives speeds of 68 km/h and 32.3 km/hr. The latter governs.

2.4

WIM error computations: WIM Pass 1 Test Vehicle 1 steering drive, axle 1 drive, axle 2 trailer, axle 1 trailer, axle 2 GVW

-16.78 5.41 -9.30 -13.92 -15.44 -9.40

WIM Pass 2 11.76 -14.10 3.85 -16.50 -30.73 -10.17

WIM Pass 3 -5.66 7.26 7.97 -4.59 -35.32 -5.45

WIM Pass 4 9.15 -17.09 -20.05 -30.70 2.60 -13.05

WIM Pass 5 1.31 -30.34 7.44 -3.87 -0.31 -5.51

Test Vehicle 2 steering drive, axle 1 drive, axle 2 trailer, axle 1 trailer, axle 2 GVW

-12.04 -17.98 -21.99 -5.34 -7.41 -12.72

4.17 10.59 -17.25 -7.07 -11.68 -5.06

1.85 12.84 23.89 0.58 1.57 8.24

3.01 -22.47 -12.97 -2.60 -8.83 -9.38

-0.93 -10.43 -21.04 -2.60 -4.56 -8.18

Four of 50 WIM measurements violate the ±30% rule for weighing individual axles (shaded cells), which represent a 92% < 95% conformity and hence, this system does not satisfy Type II WIM requirements. The average error in axle load measurements is -6.73% which suggests a calibration factor adjustment factor of 1.0722. 2.5:

Consider the vehicle shown in Figure 2.10. Single axle loads and tandem axle load limits are satisfied, since the single axle load 68 < 89 kN, and the tandem axle loads of 108 < 151 kN and 135 < 151 kN. The GVW is 311 < 356 kN. Check whether this vehicle satisfies the bridge formula requirement given by Equ. 2.8, by testing first the tractor that has a wheel base of L = 4.8 m, and rides on N = 3 axles:

 0.3048 4.8 3  + 12 3 + 36  = 165 kN 3 −1  

W = 2.224 

The weight of the tractor is 176 kN, which exceeds the maximum allowable calculated above. Hence, this vehicle is not legally loaded-no further bridge formula checks are needed.

Chapter 3 3.1:

Provide a summary of the subgrade, subbase and the base properties that are needed as input to the proposed design guide from the NCHRP 1-37A design approach. Materials inputs required for critical response computations Seasonally adjusted resilient modulus (Enter κ 1 , κ 2 and κ 3 ; or R-value; or

CBR(%)). Poisson’s ratio (Default=0.35) Unit weight (Maximum dry unit weight and Specific gravity of solids) Coefficient of lateral pressure ( κ 0 Default=0.5) Gradation parameters and base erodibility (for rigid pavement design) Plasticity index, gradation parameters, effective grain size, specific gravity, saturated hydraulic conductivity, optimum moisture contents, parameters to define the soil water characteristic curve

3.2 Plot the relationships between resilient modulus and CBR given in Equations 3.21 and 3.22 for a range of CBR values from 5 to 50. Comment on the predictions of these two equations.

Only Eq. 3.21 deals with the relationship between Mr and CBR; while Eq. 3.22 is the equation about R-Value. Hence, I select Eq. 3.23, which is a relationship between Mr and R-Value, and use Table 3.9 to regress the relationship between CBR and R-Value. After obtaining R-value, Mr can be calculated from Eq. 3.23 with R-value. The figure shows that Mr predicted by Eq. 3.21 and 3.23 are close, when CBR less than 30. After that, the difference between these two equation increases with increasing CBR.

120 3

2

y = 0.0816x - 4.9171x + 89.284x - 412.35 100

2

R = 1

80 R

60 40 20 0 0

5

10

15

20 CBR

25

30

35

70 y = 1.3476x + 11.334 2

60

R = 0.3355

50 40 R

30 20 10 0 0

5

10

15

20

25

30

35

CBR

1200000 Eq. 3.21 1000000

Eq. 3.22 Linear regression Eq. 3.22 p olynomial regression

800000 r M 600000

400000

200000

0 0

10

20

30

40

50

60

CBR

3.3 Discuss the benefits of subgrade and base stabilization using lime.

The benefits of using lime to stabilize subgrade and base are: 1. The stabilized subgrade and base have a smaller volume, higher strength, better resilient properties and better workability; 2. The stabilized subgrade and base have greater capability to resist fatigue and fracture, and lower sensitivity to changes in moisture content.

3.4 Based on your background in geotechnical engineering, describe the influence of moisture content on the resilient modulus.

When the moisture content is below the optimum moisture content, the resilient modulus will increase with increasing moisture content, because the increasing suction will increase the apparent cohesion between particles. If the moisture content is higher than the optimum moisture content, the resilient will decrease with increasing moisture content, because the pore pressures decreases the effective stress. 3.5 Using the data given in Example 3.1, calculate the difference in the estimates of the resilient modulus obtain from equations 3.4 and 3.6. Plot the difference versus the ratio of τ oct to θ . What do you conclude from this plot?

Using equations 3.4 and 3.6 with the material coefficients κ 1 to κ 3 which are regressed in Example 3.1 to calculate the resilient modulus. The results are shown in the following table and figures. The results show that when the ratio of τ oct to θ is high, the prediction of resilient modulus calculated by equation 3.4 is larger than that calculated by equation 3.6; while when the ratio of τ oct to θ is low, the prediction of resilient modulus calculated by equation 3.4 is smaller than that calculated b y equation 3.6. Moreover, at same ratio of τ oct to θ , increasing θ will increasing the difference between equation 3.4 and 3.6. Hence, when the ratio of τ oct to θ is high, equation 3.4 may over estimate the resilient modulus, and when the ratio of τ oct to θ is low, equation 3.4 may under estimate the resilient modulus.

Sequence No. 1 2 3 4 5 6 7 8 9 10 11 12 13 14

θ

τoct

ratio

Mr (Eq. 3.4)

Mr (Eq. 3.6)

difference

82.8 103.5 124.2 138 172.4 206.9 275.6 344.6 413.5 379.1 413.6 517 517.1 551.6

9.75807358 19.51614716 29.27422074 16.26345597 32.47977148 48.74322745 32.47977148 65.00668342 97.4864549 32.47977148 48.74322745 97.4864549 48.74322745 65.00668342

0.11785113 0.188561808 0.23570226 0.11785113 0.188397746 0.23558834 0.11785113 0.188643887 0.235759262 0.085676 0.11785113 0.188561808 0.094262672 0.11785113

58949.48542 73440.62789 87888.06616 97499.19653 121397.3871 145292.8424 192705.9667 240146.1789 287374.6914 263810.8264 287443.1469 358103.2988 358171.5253 381698.1283

69514.9416 68964.37786 73863.80361 112821.2162 111911.6336 119851.4768 217355.2312 215649.2669 230979.8737 341280.6108 319377.0797 316847.5884 438076.8124 419609.3908

10565.45618 -4476.250029 -14024.26254 15322.01963 -9485.753457 -25441.36558 24649.26458 -24496.91194 -56394.81764 77469.78441 31933.93282 -41255.71037 79905.2871 37911.26253

100000 80000 ) 60000 4 . 3 . q 40000 E 6 . 20000 3 . q E 0 ( e 0 c n -20000 e r e f f i D -40000

0.05

0.1

0.15

-60000 -80000 Ratio

The results are shown in the Figure above.

0.2

0.25

Chapter 4 4.1 Name three common minerals and identify their geological origin. Which one is more porous, and why?

Granite: Igneous rock Limestone: Sedimentary rock Quartzite: Metamorphic rock Limestone has the highest porosity among the three minerals mentioned above; in general sedimentary rocks will have a higher porosity than igneous and metamorphic rock. The reason behind that is the formation process.

4.2 Name three common types of igneous rocks used as aggregates.

Basalts, Granite, and Rhyolite

4.3 Given the following measurements on a sample of fine aggregate, calculate the bulk dry, bulk SSD, and apparent specific gravities: Aggregate saturated surface dry weight = 459.34g Weight of flask and water = 2345.67 g Weight of flask, water, and sample = 2640.35 g Aggregate weight after being dried in oven = 454.12 g

According to the equations in the book (page 85) A Apparent specific gravity = B + A − C Bulk specific gravity =

A

 B + D − C

Bulk specific gravity, SSD = Where

D B + D − C

(4-6)

(4-7)

A = weight of oven-dry specimen in air, B = weight of pycnometer filled with water, C = weight of pycnometer with specimen and water to calibration mark, D = weight of saturated surface dry specimen.

And so, A = 454.12 g, B = 2345.67 g, C = 2640.35, and D = 459.34 g

(4-8)

Bulk dry specific gravity =

A B + D − C

Bulk SSD specific gravity =

Apparent specific gravity =

=

D B + D − C A

B + A − C

454.12 2345.67 + 459.34 − 2640.35

=

=

=

454.12 164.66

= 2.758

459.34 459.34 = = 2.790 2345.67 + 459.34 − 2640.35 164.66

454.12 454.12 = = 2.848 2345.67 + 454.12 − 2640.35 159.44

4.4 Plot typical gradation curves for two samples, one with fineness modulus of 2.2 and the other of 3.2.

Fineness modulus is defined as one hundredth of the sum of the cumulative percentages held on the standard sieves (Nos. 4, 8, 16, 30, 50, and 100). Gradation curve uses the percent passing, the percent retained is simply 100 minus the percent passing. The following table shows two gradation curves with fineness modulus of 2.20 and 3.20, Excel spreadsheet was used, the values of percent passing on the sieves #4 through #100 were changed to get the wanted fineness modulus, however this solution is not unique and different curves could have the same fineness modulus. Sieve Size No. mm 1 in 25 3/8 in 9.5 #4 4.75 #8 2.36 #16 1.18 #30 0.6 #50 0.3 #100

0.15

Fineness Modulus = 2.20 Fineness Modulus = 3.20 Percent Passing Percent Retained Percent Passing Percent Retained 100 0 100 0 100 0 100 0 95 5 85 15 90 10 74 26 80 20 56 44 65 35 40 60 40 60 20 80 10

90

5

95

Summation

220

Summation

320

Fineness Modulus

2.2

Fineness Modulus

3.2

The following figure shows the gradation curves of the two samples with the different fineness modulus:

100 ) 90 % ( 80 g n 70 i s s a 60 P t 50 n e c r 40 e P 30 l a 20 t o T 10 0

FM = 2.20 FM = 3.20

0.1

1 10 Aggregate Diameter (mm)

100

4.5 Calculate and plot the gradation of the sieve analysis data shown in Table 4.6 on a smeilog plot. Table 4.6 Sieve Size

Amount Retained, g

25 mm (1 in.)

0

9.5 mm (3/8 in.)

35.2

4.75 mm (No. 4)

299.6

2.00 mm (No. 10)

149.7

0.425 mm (No. 40)

125.8

0.075 mm (No. 200)

60.4

Pan

7.3

Cumulative Amount Retained, g

Cumulative Percent Retained

Percent Passing

Cumulative amount retained on a certain sieve is just the algebraic sum of the amounts retained on that sieve and all sieves above it.

Cumulative amount retained on 25 mm sieve = 0 g Cumulative amount retained on 9.5 mm sieve = 35.2 + 0 = 35.2 g Cumulative amount retained on 4.75 mm sieve = 0 + 35.2 + 299.6 = 334.8 g . . . Cumulative amount retained on “Pan” = 0 + 35.2 + 299.6 + 149.7 + 125.8 + 60.4 + 7.3 = 678 g Cumulative Percent retained on each sieve is 100 multiplied by the percentage of cumulative amount retained on the sieve divided by the cumulative amount retained on the pan (i.e. the total mass of the sample) Cumulative Percent retained on 25 mm = 100 * 0/678 = 0 % Cumulative Percent retained on 9.5 mm = 100 * 35.2/678 = 5.2 % Cumulative Percent retained on 4.75 mm = 100 * 334.8/678 = 49.4 % . . . Cumulative Percent retained on “Pan” = 100 * 678/678 = 100 % The percent passing is simply 100 minus the cumulative percent retained, the following table summarize all the calculations, followed by the gradation curve plot:

Sieve Size

Amount Retained, g

Cumulative Amount Retained, g

Cumulative Percent Retained

Percent Passing

0

0

0

100

9.5 mm (3/8 in.)

35.2

35.2

5.2

94.8

4.75 mm (No. 4)

299.6

334.8

49.4

50.6

2.00 mm (No. 10)

149.7

484.5

71.5

28.5

0.425 mm (No. 40)

125.8

610.3

90.0

10.0

0.075 mm (No. 200)

60.4

670.7

98.9

1.1

7.3

678

100

0

25 mm (1 in.)

Pan

100 % ( g 80 n i s s a P 60 t n e c r 40 e P l a 20 t o T

Aggregate Gradation

0 0.01

0.1 1 10 Aggregate Diameter (mm)

100

4.6 Repeat the solution to problem 4.5, using the Fuller-Thompson approach. Also plot the Fuller line using an exponent of 0.45. How dense do you think this aggregate packs?

Fuller and Thompson approach is basically plotting the percent passing versus the particle size, raised to an exponent n, a value of 0.45 will be used for n as indicate by the problem statement, the percent passing is the same from problem 4.5 solution, the difference here will be that the aggregate diameter (sieve size) will be raised to the power 0.45 (see the table below) Sieve Size

Sieve Size (mm)

25 9.5 4.75 4.75 mm (No. 4) 2 2.00 mm (No. 10) 0.425 0.425 mm (No. 40) 0.075 0.075 mm (No. 200) 0 Pan * values from Problem 4.5 Solution 25 mm (1 in.) 9.5 mm (3/8 in.)

Sieve Size^0.45 (mm)

Percent Passing*

4.256699613 2.754074109 2.016100254 1.366040257 0.680416785 0.31172926 0

100 94.80825959 50.61946903 28.53982301 9.985250737 1.076696165 0

The Fuller line follows the following equation (Book page 77): P = 100(d / D )

n

Where P is the percentage of aggregates passing the sieve size d , D is the maximum aggregate size in the blend. According to ASTM C 125, the maximum size refers to the smallest sieve through which 100 percent of the aggregate sample particles pass, and so for this problem D = 25 mm, to plot the line various values ranging from 0 to 25 are plugged in the equation for the d value (see the table below)

d (mm) 25 20 15 10 5 0

d^0.45 4.256699613 3.850024543

P (%) 100 90.44623519

3.382525401 2.818382931 2.063177068 0

79.46356822 66.2105196 48.46893733 0

) 100 % ( g 80 n i s s a P 60 t n e c 40 r e P l 20 a t o T 0

Fuller line Aggrgetae Gradation

0

1

2

3

4

5

Aggregate Diameter (mm) - raised to power n

4.7 Table 4.7 shows the grain size distribution for two aggregates and the specification limits for an asphalt concrete. Determine the blend proportion required to meet the specification and the gradations of the blend. On a smeilog gradation graph, plot the gradations of aggregate A, aggregate B, the selected blend, and the specification limits 19

12.5

9.5

4.75

2.36

0.60

0.30

0.15

0.075

mm

mm

mm

mm

mm

mm

mm

mm

mm

Spec. Limits

100

80- 100

70- 90

50 - 70

35 - 50

18- 29

13 - 23

8 - 16

4 - 10

Aggregate A

100

85

55

20

2

0

0

0

0

Aggregate B

100

100

100

85

67

45

32

19

11

A quick glance over the specification limits and aggregate A and aggregate B gradations it can be easily seen that a 50/50 blend of A and B will be a very good starting point. Sieve Size (mm)

19

12.5

9.5

4.75

2.36

0.6

0.3

0.15

0.075

Specification Limits Aggregate A

100

80-100

70-90

50-70

35-50

18-29

13-23

8-16

4-10

100 100 100

85 100 92.5

55 100 77.5

20 85 52.5

2 67 34.5

0 45 22.5

0 32 16

0 19 9.5

0 11 5.5

Aggregate B 50/50 Blend

Checking the blend gradation against the specification limits, all the sieve sizes are within the limits except for the sieve 2.36 mm, where the blend is little bit below the lower specification, in order to elevate this value a more percentage of aggregate B is needed, and so a 45/55 Blend is used as a second iteration. Sieve Size (mm)

19

12.5

9.5

4.75

2.36

0.6

0.3

0.15

0.075

Specification Limits Aggregate A

100

80-100

70-90

50-70

35-50

18-29

13-23

8-16

4-10

100 100 100

85 100 93.25

55 100 79.75

20 85 55.75

2 67 37.75

0 45 24.75

0 32 17.6

0 19 10.45

0 11 6.05

Aggregate B 45/55 Blend

The 45/55 blend is within the specification limits for all the sieve sizes. The following figure contain the curves of Aggregate A, Aggregate B, 45/55 Blend, Lower Limit, and Upper Limit.

) % ( 100 g n 80 i s s a P 60 t n e 40 c r e P 20 l a t o T 0

0.01

0.1

1

10

Aggregate Diameter (mm)

Aggregate A

Aggregate B

Lower Limit

Upper Limit

45/55 Blend

100

4.8 Laboratory measurement of the specific gravity and absorption of two coarse aggregate sizes are: Aggregate A: Bulk dry specific gravity = 2.81; absorption = 0.4% Aggregate B: Bulk dry specific gravity = 2.44; absorption = 5.2 % What is the average specific gravity of a blend of 50% aggregate A and 50% aggregate B by weight and what is its average absorption of the mixture?

According to the equations in the book (page 86)

Combined specific gravity G =

1 a

100 G A

+

b

100 G B

)

+ ....

Combined absorption = a Absorption A + b Absortpion B + ….. Where a, b, and c are the percentages of aggregates A, B, and C , respectively, used in the blend.

Combined specific gravity G =

1 50 50 + 100 × 2.81 100 × 2.44

=

Combined absorption = a Absorption A + b Absortpion B = 50 * 0.4/100 + 50 * 5.2/100 = 0.2 + 2.6 = 2.8

1 0.178 + .205

=

1 0.383

= 2.61

Chapter 5 5.1 Discuss the influence of oxidation on mixture molecular structure and viscosity. Solution

Asphalt is composed of hydrocarbons and heteroatoms. Heteroatoms distribution is responsible for asymmetric charge and polarity in the molecules. The higher the polarity is, the higher the interaction/association among molecules tends to be. The reaction of gaseous oxygen (from air) and asphalt increases the surface area and temperature promoting oxidation. The oxidation process changes asphalt chemical structure and increases polarity. Those changes promote more molecular association and lead to a more brittle asphalt structure. Oxidation is a slow process because the driving forces for molecular orientation are small. Polar molecules formed during oxidation process tend to orient themselves to the thermodynamic stable state or equilibrium forming a better packing system and the whole asphalt become stiffer. Oxidation is a permanent hardening process that generally causes an increase in viscosity and a decrease in penetration. Oxidation rate depends on: asphalt composition and pavement condition (temperature, compaction, traffic, asphalt film thickness, etc).

5.2 Use a book on rheology to describe the behavior of a Bingham fluid and a thixotropic fluid. Solution

Binghan fluid Binghan fluid is a type of non-Newtonian fluid. This specific fluid type has a flow curve, shear stress (τ) versus rate of shear ( γ ) represented by a straight line with an intercept in the shear stress axis (yield point: τy) (Figure (a) on following page). The yield point represents the stress magnitude that needs to be exceeded before flow starts. For stress values lower than the yield stress, the fluid structure has sufficient rigidity to resist such stress, there is no change in its volume, and the shape will deform elastically (like a solid). For stress values higher than the yield point, the fluid can not resist (its structure disintegrates) and there is an irreversible change in its volume (like a Newtonian fluid with shear stress equals to τ- τy). Examples of this type of fluid are: slurries, drilling muds, greases, oil paints, toothpaste, sludges, and lipsticks masses. Thixotropic fluids Thixotropic fluid is a type of non-Newtonian fluid that does not have a one-one relation between shear stress ( τ) and rate of shear ( γ ). Those fluids present structural changes and viscosity decreases during shear, and they are also time-dependent. There is a shear stress diminution for the same shear rate. The flow curve is represented by a

hysteresis loop (Figure (b) shown below). For those materials, its structure can be reformed if they are allowed to rest. The potential for having the material’s structure reformed is called thixotropy’s magnitude. The broken structure is called “sol” and the reformed one is called “gel”. The transformation between “sol” and “gel” depends on the material’s nature and temperature. Examples of this type of fluid are: meat products, cosmetics, and pharmaceuticals.

τ

τ

Thixotropy Magnitude

τy

γ

γ

(a) Binghan fluid (b) Thixotropic fluid Flow curves for different types of fluids Sources:

♦

Scharamm, G., A Pratical Approach to Rheology and Rheometry, Gebrueder Haake Gmbh, Karlsruhe, Federal Republic of Germany, 1994. Tanner, R.I., Engineering Rheology, Oxford University Press Inc., New ♦ York, 2nd Edition, 2000.

5.3 Use the LTPPBind software to determine the required span of pavement temperatures and the appropriate PG binder grades in each of the following locations.

♦ ♦ ♦

Houston, Texas (Bush Intercontinental Airport) Anchorage, Alaska (Anchorage International Airport) New York, New York (New York JF Kennedy Airport)

The LTPPBind software can be accesses through the book Web site Solution

After opening the LTPPBind software:

♦

Go to Select Station / Name Location and select desired states;

State/Province selection

♦

Go to Tabular Report (tool bar) and check station ID, county/district, and station name for desired locations;

Weather station details: ID, county/district, and station name For the three desired stations, the following information was collected: Houston, Texas (Bush Intercontinental Airport) Station ID: TX4300 County/District: Harris Weather Station: Houston int’cntnl ap

Anchorage, Alaska (Anchorage International Airport) Station ID: AK0280 County/District: Cook Inlet Weather Station: Anchorage intl ap New York, New York (New York JF Kennedy Airport) Station ID: NY5803 County/District: Queens Weather Station: New York j f kennedy

♦

Go to File / Preferences and select the target rut depth (mm);

Target rut depth selection

♦

Go to Select Station / Name Location and write the name of the desired station; On the main screen, click on the red square on the map location ♦ correspondent to the desired station. The weather station information should appear on the both of the main screen;

Specific station selection

♦

Go to PG (tool bar) and check the span of pavement temperatures and appropriate PG binder grade for each location (see earlier Figures).

(a) 50 % reliability (b) 98% reliability Span of pavement temperature and PG binder grade: Houston, Texas (Bush Intercontinental Airport)

(a) 50 % reliability (b) 98% reliability Span of pavement temperature and PG binder grade: Anchorage, Alaska (Anchorage Intercontinental Airport)

(a) 50 % reliability (b) 98% reliability Span of pavement temperature and PG binder grade: New York, New York (New York JF Kennedy Airport) The summary of the required span of pavement temperatures and the appropriate PG binder grades in each of the following locations are on the following Table. Pavement temperatures and PG binder grades for the desired locations Pavement Temperature (ºC) Station ID Reliability (%) Binder PG High Low 50 64.3 -0.7 70-10 TX4300 98 65.0 -6.9 70-10 50 35.0 -15.7 40-16 AK0280 98 35.1 -22.5 40-28 50 52.0 -10.0 52-10 NY5803 98 54.4 -15.6 58-16 Comment: To solve this question it was assumed a target rut depth of 12.5mm, no adjustment for traffic was used, the calculations were based on a surface layer, and the results were taken just for the desired location (although the software also gives the average results for five stations closest to the desired location). 5.4 An engineer wants to determine if a certain asphalt would be graded as PG 5828. At what temperatures should he or she run the following tests?

♦ ♦ ♦

DSR for rutting analysis DSR for fatigue cracking analysis BBR

Solution

♦

DSR for rutting analysis: DSR is used to measure the binder viscoelastic properties. The test is conducted at the maximum average seven-day pavement temperature to assess the binder resistance to rutting. For a binder graded as PG 58-28, this test should be run at 58 ºC.

♦

DSR for fatigue cracking analysis: DSR is also used to measure binder resistance to fatigue cracking. The test temperature is taken as 0.5 × (seven-day average maximum pavement temperature + minimum pavement temperature) + 4. For a binder graded as PG 58-28, this test should be run at 19 ºC.

♦

BBR: BBR is used to assess low-temperature cracking. This test is conducted at a temperature 10 ºC higher than the pavement’s lowest temperature. For a binder graded as PG 58-28, this test should be run at -18 ºC.

5.5 What is the PG grade of the asphalt whose results are shown in Table 5.14? Show all calculations and comparisons with Superpave requirements. Table 5.14 Asphalt Cement Test Results – Problem 5.5 Test Results Original Properties Flash point temperature, ºC 278 Viscosity at 135ºC 0.490 Pa.s Dynamic shear rheometer at 82ºC G* = 0.82 kPa, δ = 68º at 76ºC G* = 1.00 kPa, δ = 64º at 70ºC G* = 1.80 kPa, δ = 60º Rolling Thin Film Oven-Aged Binder Dynamic shear rheometer at 82ºC G* = 1.60 kPa, δ = 65º at 76ºC G* = 2.20 kPa, δ = 62º at 70ºC G* = 3.50 kPa, δ = 58º Rolling Thin Film Oven and PAV-Aged Binder Dynamic shear rheometer 34ºC G* = 2500 kPa, δ = 60º 31ºC G* = 3700 kPa, δ = 58º 28ºC G* = 4850 kPa, δ = 56º Bending beam rheometer -6ºC S = 255 MPa, m = 0.329 -12ºC S = 290 MPa, m = 0.305 -18ºC S = 318 MPa, m = 0.277

Solution

The Superpave PG grading system controls rutting based on the rheological parameters G* and δ, as follows: (1) G*/sin δ ≥ 1.0 kPa for unaged asphalt, and (2) G*/sinδ ≥ 2.2 kPa for short-term aged asphalt (RTFO-aged asphalt). The G*/sin δ values for the unaged asphalt in this question are: (1) G*/sinδ = 0.88 kPa for unaged asphalt tested at 82 ºC, (2) G*/sinδ = 1.11 kPa for unaged asphalt tested at 76 ºC, and (3) G*/sin δ = 2.08 kPa for unaged asphalt tested at 70 ºC. The G*/sin δ values for the RTFO-aged asphalt in this question are: (1) G*/sinδ = 1.77 kPa for RTFO-aged asphalt tested at 82 ºC, (2) G*/sinδ = 2.49 kPa for RTFO-aged asphalt tested at 76 ºC, and (3) G*/sinδ = 4.13 kPa for RTFO-aged asphalt tested at 70 ºC. It can be concluded that this asphalt binder does satisfy the Superpave requirements in its original conditions when it was tested at 76 ºC and 70 ºC; but it does not satisfy the Superpave requirements in its original condition when is tested at 82 ºC. For the tests conducted using the short-term aged (RTFO-aged) asphalt, it can be concluded that this short-term aged asphalt binder does satisfy the Superpave requirements when tested at 76 ºC and 70 ºC; but the short-term aged asphalt binder does not satisfy the Superpave requirements when is tested at 82 ºC. Fatigue cracking is controlled by the following requirement: G*×sinδ ≤ 5000 kPa (in long-term aged asphalt samples using the PAV). The G*×sinδ values for the long-term aged asphalt in this question are: (1) G*×sinδ = 2165.06 kPa for long-term aged asphalt tested at 34 ºC, (2) G*×sinδ = 3137.78 kPa for long-term aged asphalt tested at 31 ºC, and (3) G*×sinδ = 4020.83 kPa for long-term aged asphalt tested at 28 ºC. The results for G*×sinδ obtained from the DSR show that the long-term aged asphalt binder under analysis satisfy the Superpave requirement for the three temperatures tested (34 ºC, 31 ºC, and 28 ºC). Low-temperature cracking is controlled by limiting the stiffness (S) and the m-value obtained from the BBR test in long-term aged samples (PAV), as follows: (1) stiffness (S) ≤ 300 MPa (in long-term aged samples), and (2) m-value ≥ 0.3 (in the same conditions). The BBR values for the long-term aged asphalt in this question are: (1) S = 255 MPa and m = 0.329 for long-term aged asphalt tested at -6 ºC, (2) S = 290 MPa and m = 0.305 for long-term aged asphalt tested at -12 ºC, and (3) S = 318 MPa and m = 0.277 for long-term aged asphalt tested at -18 ºC. The BBR test conducted using the longterm aged asphalt binder does satisfy the Superpave requirements (for both: stiffness and m-value) when tested at -6 ºC and -12 ºC. The BBR test conducted using the long-term aged asphalt binder does not satisfy the Superpave requirements (for both: stiffness and m-value) when tested at -18 ºC. The summary of the Superpave test results and requirements for the asphalt binder whose results are showed in Table 5.14 are summarized in the following Table. DSR test to assess binder resistance to permanent deformation is conducted at the maximum average seven-day pavement temperature for unaged binder and RTFO-aged binder. Based on the results presented in the following Table, we can conclude that the binder in question can have a maximum average seven-day pavement temperature equal to 76 ºC or 70 ºC. The resistance to low-temperature cracking is assessed using the BBR. BBR test is conducted in a binder that is aged using both RTFO and PAV. BBR test is conducted at a temperature 10 ºC higher than the pavement’s lowest temperature. Based on the results presented in the following Table, we can conclude that the binder in question can have

the pavement’s lowest temperature equal to -22 ºC or -16 ºC. The DSR test is also used to measure the binder resistance to fatigue cracking. Test is conducted after aging the binder in both the RTFO and PAV. The test temperature is taken as 0.5 ×(seven-day average maximum pavement temperature + minimum pavement temperature) + 4. Based on the results presented in the following Table , we can conclude that the binder in question can have the pavement’s lowest temperature equal to -16 ºC, -22 ºC or -28 ºC. Considering that this specific binder needs to satisfy all the Superpave requirements and that we want a binder PG grad that covers the largest temperature span, this binder can be graded as a PG 76-22. Summary of the Superpave test results and requirements Test

Results

Superpave Requirements

Status

≤ 3 Pas

SATISFY

Original Properties Flash point temperature, ºC Viscosity at 135 ºC

278 0.490 Pa.s

Dynamic shear rheometer at 82ºC at 76ºC at 70ºC

G* (kPa) 0.82 1.00 1.80

Rolling Thin Film Oven-Aged Binder Dynamic shear rheometer G* (kPa) at 82ºC 1.60 at 76ºC 2.20 at 70ºC 3.50

δ (º) 68 64 60

δ (º) 65 62 58

G*/sinδ 0.88 1.11 2.08

G*/sinδ ≥ 1.0kPa

G*/sinδ 1.77 2.49 4.13

G*/sinδ ≥ 2.2kPa

G*xsinδ 2165.06 3137.78 4020.83

G*xsinδ ≤ 5000 kPa

NOT SATISFY SATISFY SATISFY

NOT SATISFY SATISFY SATISFY

Rolling Thin Film Oven and PAV-Aged Binder Dynamic shear rheometer at 34ºC at 31ºC at 28ºC Bending beam rheometer -6ºC -12ºC -18ºC

G* (kPa) 2500 3700 4850

δ (º)

S=255 MPa S=290MPa S=318MPa

m=0.329 m=0.305 m=0.277

60 58 56

S (t=60 sec) ≤ 300 MPa m(t=60sec) ≥ 0.3

SATISFY SATISFY SATISFY

SATISFY SATISFY NOT SATISFY

5.6 An asphalt mixture has been compacted with the Marshall hammer, using 50 blows. The following data was obtained in the laboratory:

Aggregate Blend Aggregate saturated surface dry (SSD) weight = 459.34 gm Weight of flask and water = 2345.67 gm Weight of flask, water, sample = 2640.35 gm Aggregate weight after being dried in oven = 454.12 gm Asphalt Mixture Weight of dry-compacted asphalt mixture in air = 3600.0 gm Weight of SSD-compacted asphalt mixture in air = 3724.2 gm Weight of compacted mixture in water = 2200.86 gm Theoretical maximum specific gravity = 2.50 Asphalt binder percent per weight of mix = 5.0% Specific gravity of asphalt binder = 1.00 Calculate the bulk dry-specific gravity of aggregate, bulk-specific gravity of the asphalt mixture, the void in mineral aggregate of the mix, and the percent of air voids in the compacted mix. Solution

Bulk dry-specific gravity of aggregate (Gagg-sb) bulk (dry) specific gravity = G agg − sb =

W D

=

W SSD − W sub

weight of oven dry aggregate volume of aggregate plus volume of impermeable voids and permeable voids * 454.12

459.34 − (2640.35 − 2345.67 )

= 2.758

* Volume is determined by displacement in water with saturated surface dry (SSD) aggregate. Bulk-specific gravity of the asphalt mixture (Gmb) G mb

=

W D W SSD

− W Sub

=

3600.00 3724.20 − 2200.86

= 2.363

Void in mineral aggregate (VMA) of the mix



VMA = 100 1 −



G mb (1 − Pb )  G sb

2.363(1 − 0.05 )   = 100 1 −  = 18.606 % 2 . 758   

Percent of air voids in the compacted mix (VTM)



VTM =  1 −



G mb 

2.363   × 100 =  1 −  = 5.48 % G mm   2.500 

5.7 A uniaxial creep test was conducted on an axial mix at 40ºC. The data for this test is available on the book’s Web site under the name “mix creep data.xls”. Use the data to calculate flow-time value, flow-time slope, and flow time intercept. Solution

Creep compliance was calculated using: l i − l0 D( t ) =

ε σ

=

LVDT gauge length F specimen area

where, li is the LVDT axial measurement for a specific time, l0 is the first LVDT axial measurement (at t = 0), and F is the axial load. For this exercise, only axial LVDT measurements were considered. Radial LVDT measurements were not use d. Creep compliance ( D(t)) versus time was plotted in the following Figure. Primary, secondary and tertiary zones can be identified. Flow-time slope and flow-time intercept can be obtained graphically from the log compliance versus log time plot shown. The regression constants (slope and intercept) can be find using the relationship: log D( t ) = m log( t ) + log( a )

where, a is the flow time intercept and m is the flow time slope. Flow-time slope (m) is equal to 0.1795 and flow-time intercept (a) is equal to 5.84×10-9. To find the flow time slope and flow time intercept a regression analysis was done for the secondary region. Flow-time value can be obtained graphically from the log rate of change of creep compliance versus log time plot shown in the following Figure. Flow-time value is marked by the time at which the rate of change in compliance is minimum. For this exercise flow-time is equal to 1,000 sec. The rate of change of compliance was calculated for intervals of 50 sec.

1.0E-06

1.0E-07 ) t ( D e c n 1.0E-08 a i l p m o C

1.0E-09

Secondary Zone

Primary Zone

Tertiary Zone

1.0E-10 0.01

0.1

1

10

100

1000

10000

Time (sec)

Creep compliance versus time -6.0 -6.5 ) -7.0 t ( D e c -7.5 n a i l p m-8.0 o C g o L -8.5

y = 0.1795x - 8.2337 R2 = 0.9559

-9.0 -9.5 -2.0

-1.0

0.0

1.0

2.0

Log Time

Log creep compliance versus log time

3.0

4.0

-9.0

e c n a -9.5 i l p m o C f o e -10.0 g n a h C f o e t a -10.5 R g o L

-11.0 1.0

1.5

2.0

2.5

3.0

Log Tim e (sec)

Log rate of change of compliance versus log time

3.5

4.0

Chapter 6 6.1

False set refers to rapid development of rigidity in the concrete without generating much heat. By further mixing without adding water, the plasticity of the concrete can be recovered. False set occurs as a result of hydration of dehydrated gypsum and forming a rigid crystalline matrix. The fluidity of the concrete can be recovered by further mixing as there is small fraction of rigid crystals. Flash set refers to rapid development of rigidity or premature stiffening which subjected with generating significant heat. The plasticity of the concrete cannot be recovered. The flash set occurs as a result of the intermediate reaction between tricalcium aluminate (C3A) and the water which produces calcium aluminates hydrate (CAH). Gypsum is used to decrease the possibility of the flash set by it reaction with dissolved aluminate to produce ettringite (C-A-S-H).

6.2

Sulfate attack occurs as a result of the reaction of sulfates from soil and seawater with the free calcium hydroxide and the aluminates in the cement. Sulfate attack causes expansion of concrete or increasing in volume which induces cracks. Using cement that is low in C3A increases the concrete resistance to sulfate attack. Cement t ype II and type V are used for moderate sulfate resistance and high sulfate resistance respectively. Also, the use of fly ash and blast furnace slag as shown in table 6.2 decreases the potential of sulfate attack.

6.3

It is believed that the concrete pavement fails in flexure and thus the modulus of rupture is the accurate measure of the concrete strength. The modulus of rapture is required input in the design process after 28 days. In order to assure the desired performance of the plain concrete pavement, laboratory mix design based on flexural strength is needed.

6.4

Concrete shrinkage is a reduction in volume caused by the loss of water. The water loss occurs early in the life of the pavement (plastic shrinkage) or during the subsequent drying (drying shrinkage). The volume reduction is restrained by the friction between the concrete pavement layer and the underlying layer. Due to this friction, stresses develop in the concrete pavements inducing cracks at the surface. These cracks called shrinkage cracks. Plain concrete experience more shrinkage cracks than the reinforced concrete. The reinforcement controls the location and the width of the cracks. There is another type of shrinkage which is associated with the size of the hydration products of cement. Some

of these products occupy less space than the original materials (autogenously shrinkage). Concrete contraction under cooling temperature can also cause Shrinkage. 6.5

Use Equations 6.3 and 6.5 to compute tensile strength and the flexural strength, respectively: = 3.10 MPa (449 lb/in2) = 4.18 MPa (606 lb/in2)

6.6

The diffusion in concrete is governed by Fick’s second law which is given in Equation 6.8. The solution of this equation is as follows:

Where: = chloride concentration at time t = distance for the surface = initial concentration of chloride prior to the application of surface chloride = chloride concentration at the surface = diffusion coefficient = error function, given in Table 6.3 for In order to simplify the solution, assume that the concrete is exposed to a constant concentration of chloride ( which is not change with time, in addition the initial concentration ( =0 The solution to Equation 6.8 becomes:

From table 6.3 at

= 0.4,

= 0.3710. Therefore, time (sec) is obtained by solving:

. Thus the time

= 52.36 years

6.7 a.

Flexural strength and versus compressive strength

7 6 a P M , 5 h t g n e r 4 t S l a r 3 u x e l F 2 1 15

20

25

30

35

40

45

Compressive Strength, MPa

b. Tensile strength versus compressive strength

50

55

60

65

4.5 4 a P M3.5 , h t 3 g n e r t S 2.5 e l i s n e 2 T 1.5

1 15

20

25

30

35

40

45

50

55

60

65

Compressive Strength, MPa c.

Modulus of elasticity versus compressive strength

45

a 40 P G , y t i c 35 i t s a l E f o 30 s u l u d o M25

ρ=2320 ρ=2360 ρ=2400 ρ=2440 ρ=2480

20 15

20

25

30

35

40

45

50

Compressive Strength, MPa

55

60

65

Chapter 7 7.1:

Input P= 20 kN, r = 0.25 m and z =0.10 m and the layer properties specified into Equ. 7.2: σ z = -6.75 kPa σ r = -32.57 kPa σ θ = -4.71 kPa τ zr = 16.87 kPa

Compute strains from Equ. 7.3: ε z =

1 (− 6.75 − 0.35(− 32.57 − 4.71)) = 42 10-6 150000

ε r =

1 (− 32.57 − 0.35(− 6.75 − 4.71)) = -190 10-6 150000 1

ε θ =

150000

γ zr =

(− 4.71 − 0.35(− 32.57 − 6.75)) = 60 10-6

2 16.87 (1 + 0.35) 150000

= 303 10-6

Compute surface deflection from Equ. 7.6: w=

20 (1 − 0.35 2 ) π 150000 0.25

= 0.149 mm

7.2:

Use Equ. 2.1 to compute the tire imprint radius of curvature a as 0.130 m Input data into Equ. 7.7:

σ z

σ r

  0.25 3 = 750− 1 +  = -226.2 kPa 2 2 3/ 2 (0.13 + 0.25 )  

= σ θ =

750 

 − − (1 + 2 0.35) +  = -1.08 kPa 2 2 3/ 2 2 2 2  (0.13 + 0.25 )  0.13 + 0.25 2 (1 + 0.35)0.25

0.25 3

τ zr = 0 w=

2 (1 − 0.35 2 ) 150000

750 0.13 = 0.00114 m (1.1 mm)

7.3:

Input radial offsets r of 0.1, 0.25, 0.50, 0.75 and 1.0 m into Equ. 7.6: w=

50 (1 − 0.45 2 ) π 135000 r

= 0.09402 10-3/ r = 0.94 10-3, 0.376 10-3, 0.188 10-3, 0.125 10-3 and

0.094 10-3 m. 7.4:

The strains in the top layer contribute little in the overall surface deflection of this twolayer system. Therefore, the Odemark’s method of equivalent thicknesses can be applied in computing the strains in the lower layer. These strains are translated into vertical deformations and added to compute surface deflection. The equivalent layer thickness is computed from Equ. 7.10 as: 1

he

1300  3 = 0.9   0.4 = 0.74 m 150  

Vertical strains below the layer interface are computed at intervals of 0.0 5 m using Equ. 7.3a and the stress value given by Equ. 7.7a and 7.7b. The calculations are shown in the following table. The resulting vertical defection at the surface is 0.38 mm.

E mu

kPa

1,300,000 0.5

Load radius pressure

kN m kPa

40 0.15 565.8842421

z

sigm a z

sigma r

150,000

sigma th

s z - mu(s r + s th)

ez

0.74 0.79 0.84 0.89

-33.17 -29.28 -26.03 -23.28

69.24 63.22 57.82 53.00

69.24 63.22 57.82 53.00

-102.41 -92.50 -83.85 -76.29

-6.8273E-04 -6.1664E-04 -5.5901E-04 -5.0858E-04

0.94 0.99 1.04 1.09

-20.95 -18.94 -17.21 -15.70

48.70 44.84 41.39 38.29

48.70 44.84 41.39 38.29

-69.64 -63.78 -58.60 -53.99

-4.6428E-04 -4.2522E-04 -3.9066E-04 -3.5996E-04

1.14 1.19 1.24 1.29 1.34 1.39 1.44 1.49 1.54 1.59 1.64 1.69

-14.38 -13.22 -12.20 -11.29 -10.47 -9.74 -9.09 -8.49 -7.96 -7.47 -7.03 -6.62

35.51 33.00 30.73 28.68 26.81 25.12 23.57 22.16 20.87 19.68 18.59 17.59

35.51 33.00 30.73 28.68 26.81 25.12 23.57 22.16 20.87 19.68 18.59 17.59

-49.89 -46.22 -42.93 -39.96 -37.29 -34.86 -32.66 -30.66 -28.83 -27.15 -25.62 -24.21

-3.3260E-04 -3.0813E-04 -2.8618E-04 -2.6641E-04 -2.4857E-04 -2.3241E-04 -2.1774E-04 -2.0438E-04 -1.9219E-04 -1.8103E-04 -1.7080E-04 -1.6139E-04

1.74 1.79 1.84 1.89

-6.25 -5.91 -5.59 -5.30

16.66 15.80 15.01 14.27

16.66 15.80 15.01 14.27

-22.91 -21.71 -20.60 -19.57

-1.5273E-04 -1.4474E-04 -1.3734E-04 -1.3050E-04

1.94 1.99

-5.04 -4.79

13.58 12.95

13.58 12.95

-18.62 -17.73

-1.2414E-04 -1.1823E-04

-3.8083E-04

7.5:

Using the Burmister nomograph (Fig. 7.1) for a/h = 0.15/0.4= 0.375 and E 2 /E 1 = 0.115 gives an F w value of 0.28, which substituted into Equ. 7-9 with a contact stress of 565 kPa (=40/(π 0.152)) gives: w=

1.5 565 0.15 150000

0.28 = 0.237 10-3 m or 0.237 mm.

Using layered analysis software (EVERETSRESS) two types of analyses were condu cted with respect to the interface between the two layers, namely one with full friction (100%) and the other with no friction (0%). The output of the computer runs is shown next. The Resulting vertical deflection values computed are 0.243 10-3 m and 0.262 10-3 m,

respectively. These values compare favorably with the one obtained from the Burmister nomograph. 100% friction between layers No of Layers: 2 Layer * 1 2

No of Loads: 1 No of X-Y Evaluation Points: 1

Poisson's Ratio (cm) .50 .50

Thickness (MPa)

Moduli(1)

40.000 1300.00 * 150.00

Load No X-Position * (cm) (cm)

Y-Position (N) (kPa)

Load (cm)

Pressure

Radius

1 .00 .00 40000.0 565.00 15.012 Location No: 1 X-Position (cm): .000 Y-Position (cm): .000 Normal Stresses Z-Positio n Layer Sxx Syy Szz Syz Sxz (cm) * (kPa) (kPa) (kPa) (kPa) (kPa) (kPa) .000 1 -692.56 -692.56 -565.00 .00 .00 Normal Strains and Deflections Z-Position Layer Exx Eyy Ezz Ux (cm) * (10^-6) (10^-6) (10^-6) (microns)

Sxy

.00 Uy Uz (microns)

(microns)

.000 1 -50.03 -50.03 97.06 .000 .000 243.068 Principal Stresses and Strains Z-Position Layer S1 S2 S3 E1 E2 E3 (cm) * (kPa) (kPa) (kPa) (10^-6) (10^-6) (10^-6) .000

1

-692.56 -692.56 -565.00 -50.03

-50.03

97.06

0% friction between layers Layer * 1 2

Poisson's Ratio (cm) .50 .50

Thickness (MPa)

Moduli(1)

40.000 1300.00 * 150.00



Load (cm)

Pressure

Radius

1 .00 .00 40000.0 565.00 15.012 Location No: 1 X-Position (cm): .000 Y-Position (cm): .000 Normal Stresses Z-Position Layer Sxx Syy Szz Syz Sxz (cm) * (kPa) (kPa) (kPa) (kPa) (kPa) (kPa) .000 1 -706.13 -706.13 -565.00 .00 .00 Normal Strains and Deflections Z-Position Layer Exx Eyy Ezz Ux (cm) * (10^-6) (10^-6) (10^-6) (microns)

Sxy

.00 Uy Uz (microns)

.000 1 -55.26 -55.26 107.47 .000 .000 262.026 Principal Stresses and Strains Z-Position Layer S1 S2 S3 E1 E2 E3 (cm) * (kPa) (kPa) (kPa) (10^-6) (10^-6) (10^-6)

(microns)

.000

1

-706.13 -706.13 -565.00 -55.26

-55.26

107.47

7.6 and 7.7:

The EVERSTRESS output for this problem is shown next. The plots of normal strains and vertical deflection follow.

Layered Elastic Analysis by EverStress for W indows Title: problem 7.10 No of Layers: 3 No of Loads: 4 No of X-Y Evaluation Points: 8 Layer * 1 2 3

Poisson's Ratio (cm) .35 .40 .50

Thickness (MPa)

Moduli(1)

25.000 1350.00 40.000 500.00 * 125.00



1 2 3 4

23561.9 23561.9 23561.9 23561.9

.00 40.00 400.00 440.00

Location No: 1

.00 .00 .00 .00

X-Position (cm):

Normal Stresses Z-Position Layer (cm) * (kPa) 24.999 1 .000 1

Sxx (kPa)

Load (cm)

750.00 750.00 750.00 750.00

.000

Syy (kPa)

99.84 134.61 -110.78 -724.64 -759.38 -750.00

Pressure

10.000 10.000 10.000 10.000

Y-Position (cm):

Szz (kPa) .00 .00

Syz (kPa) 15.09 .00

Normal Strains and Deflections Z-Position Layer Exx Eyy Ezz Ux (cm) * (10^-6) (10^-6) (10^-6) (microns) 24.999 1 .000 1

67.78 102.54 -142.84 -145.45 -180.19 -170.81

Principal Stresses and Strains Z-Position Layer S1 (cm) * (kPa) (kPa) 24.999 1 .000 1

Location No: 2

S2 (kPa)

Radius

-7.138 12.434

.000

Sxz (kPa)

Sxy

.00 .00

Uy Uz (microns) .000 .000

237.139 296.857

S3 E1 E2 E3 (10^-6) (10^-6) (10^-6)

-111.85 100.92 134.61 -143.91 68.85 102.54 -759.38 -750.00 -724.64 -180.19 -170.81 -145.45

X-Position (cm):

Normal Stresses Z-Positio n Layer

Sxx

5.000

Y-Position (cm):

Syy

Szz

Syz

.000

Sxz

Sxy

(microns)

(cm)

*

24.999 1 .000 1

(kPa)

(kPa)

(kPa)

(kPa)

86.32 132.96 -108.52 -725.39 -765.31 -750.00

(kPa)

.00 .00

2.08 .00


57.61 104.25 -137.24 -144.46 -184.39 -169.08


Location No: 3

-3.932 5.174


X-Position (cm): 10.000

Sxx (kPa)

Syy (kPa)

Y-Position (cm):

Szz (kPa)

55.95 122.79 -98.51 -346.86 -507.79 -375.00

35.15 101.99 -119.31 -28.06 -188.99 -56.20


Location No: 4

.00 .00

Syz (kPa)

S2 (kPa)


Sxx (kPa)

Syy (kPa)

-1.587 -1.963

Principal Stresses and Strains Z-Position Layer S1

S2

Sxz (kPa)

Sxy

.00 .00

.000 .000

(microns)

242.003 270.759

S3 E1 E2 E3 (10^-6) (10^-6) (10^-6)

Y-Position (cm):

Szz (kPa)

26.85 111.86 -88.02 -27.47 -188.72 .00

13.71 98.72 -101.16 28.58 -132.67 56.05

.000

Uy Uz (microns)

.00 .00

Syz (kPa)

-.398 .537

S3

101.99 -28.06

.000

Sxz (kPa)

-5.11 .00


240.742 295.231

-5.23 .00

-98.68 56.13 122.79 -119.49 35.33 -507.79 -375.00 -346.86 -188.99 -56.20

Normal Stresses Z-Positio n Layer (cm) * (kPa) 24.999 1 .000 1

(microns)

S3 E1 E2 E3 (10^-6) (10^-6) (10^-6)


.00 .00

-108.54 86.35 132.96 -137.26 57.63 104.25 -765.31 -750.00 -725.39 -184.39 -169.08 -144.46


S2 (kPa)

(kPa)

.00 .00


E1

Sxy

242.119 248.168

E2

E3

(microns)

(cm)

*

24.999 1 .000 1

Location No: 5

(kPa)

-88.25 27.07 -188.72 -27.47

(kPa)

(10^-6) (10^-6) (10^-6)

111.86 -101.39 .00 -132.67

13.93 28.58

X-Position (cm): 20.000 Y-Position (cm):

Normal Stresses Z-Position Layer (cm) * (kPa) 24.999 1 .000 1

(kPa)

Sxx (kPa)

Syy (kPa)

15.12 107.30 -83.69 -38.48 -177.00 .00

Szz (kPa) .00 .00

Syz (kPa) -.08 .00


5.08 97.26 -93.73 17.38 -121.13 55.87


Location No: 6

-83.69 15.12 -177.00 -38.48

Sxx (kPa)

Syy (kPa)

26.84 111.86 -88.02 -27.37 -188.73 .00

13.70 98.72 -101.16 28.66 -132.70 56.02


Location No: 7

-88.24 27.06 -188.73 -27.37

.000 .000

5.08 17.38

Y-Position (cm):

Szz (kPa) .00 .00

Sxx (kPa)

.402 2.633

Syy (kPa)

55.95 122.79 -98.51 28.35 -245.31 .00

Sxy

.00 .00

(microns)

242.215 244.632

Syz (kPa) 4.95 .00

97.26 55.87

.000

Sxz (kPa)

Sxy

.00 .00


242.738 248.790

S3 E1 E2 E3 (10^-6) (10^-6) (10^-6)

111.86 -101.38 .00 -132.70



S2 (kPa)

Sxz (kPa)

Uy Uz (microns)


.000

S3 E1 E2 E3 (10^-6) (10^-6) (10^-6)

107.30 -93.73 .00 -121.13


cNormal Stresses Z-Positio n Layer (cm) * (kPa) 24.999 1 .000 1

S2 (kPa)

.002 1.583

98.72 56.05

13.92 28.66

Y-Position (cm):

Szz (kPa) .00 .00

Syz (kPa) 5.07 .00

98.72 56.02

.000

Sxz (kPa) .00 .00

Sxy

(microns)


35.15 101.99 -119.31 84.60 -189.06 56.25


-98.67 -245.31

56.11 .00

S2 (kPa)

1.589 5.141


(microns)

243.242 271.916

S3 E1 E2 E3 (10^-6) (10^-6) (10^-6)

122.79 -119.47 28.35 -189.06

35.31 56.25

101.99 84.60

150 100 50 6 -

0 1 n i a r t S

0

exx 0

10

20

30

40

eyy

-50

ezz

-100 -150 -200 Offset x (cm)

Uz 350 300 6 -

0 250 1 ) m200 ( n o i t 150 c e l f e 100 D

Uz

50 0 0

10

20 Offset x (cm)

30

40

7.8:

Input all specified data into EVERSTRESS. Assume full friction between the layers. Initial layer moduli were obtained from Figure 7.8 in the book The modulus of the asphalt concrete was calculated from the creep compliance, by setting time t = 0. The computer output is given below. The solution converged after 5 iterations to within 3.4%. The computed Exx, Eyy and Ezz strains are 168.71,116.02 and -242.41 x10-6, resp. No of Layers: 3 No of Loads: 2 No of X-Y Evaluation Points: 1 Layer * 1 2 3

Poisson's Ratio (cm) .45 .45 .45

Thickness (MPa) (MPa)

Moduli(1) (MPa) *

Multiplier

15.200 2950.00 * * * 45.700 413.00 254.22 380.00 .500 * 103.00 60.92 100.00 .300



1 2

19948.6 689.00 9.600 19948.6 689.00 9.600

.00 .00

Moduli(2)

.00 35.00

Load (cm)

Pressure

Radius

No of Iteration: 5Maximum Error in Modulus: 3.4

Location No: 1

X-Position (cm):

Normal Stresses Z-Position Layer (cm) * (kPa) 15.199 1 38.050 2 60.900 3

Sxx (kPa)

.000

Syy (kPa)

730.78 623.58 -105.64 10.21 5.43 -37.10 -.60 -1.53 -15.59

Y-Position (cm):

Szz (kPa)

Syz (kPa)

12.54 10.42 2.32

.00 .00 .00

Normal Strains and Deflections Z-Position Layer Exx Eyy Ezz Ux (cm) * (10^-6) (10^-6) (10^-6) (microns) 15.199 1 38.050 2 60.900 3

168.71 116.02 -242.41 96.24 68.97 -173.64 116.65 94.44 -240.17

Principal Stresses and Strains Z-Position Layer S1 (cm) * (kPa) (kPa) 15.199 1 38.050 2 60.900 3

S2 (kPa)

.000 .000 .000

.000

Sxz (kPa)

Sxy

.00 .00 .00

Uy Uz (microns)

-11.261 376.234 -12.953 322.644 -17.971 284.506

S3 E1 E2 E3 (10^-6) (10^-6) (10^-6)

-105.85 623.79 730.78 -242.51 116.13 168.71 -39.52 7.85 10.21 -187.43 82.75 96.24 -15.96 -1.16 -.60 -249.01 103.28 116.65

(microns)

Power

7.9:

The 1.2 meters length of the load influence was subdivided into 0.01 cm increments, stress increments in radial coordinates were computed for each increment and combined into an increase in stress stress function. This was multiplied by the creep compliance versus time function to yield the strain increase. Examples of the calculations are shown on the following table. The radial strain versus offset distance is plotted in the following figure. er

1000 er

800

) s n o r c i m (

600 400 200

n i a r t s

0 -1

-0 .8 -0 -0 .6 -0 -0.4 -0-200 .2 0

0.2

-400 radial offset (m)

0.4

0.6

0.8

1

radial stress r

t

sigma z

sig ma r

sigma theta

s r - mu(s z + s theta)

0

-0.6

0

-0.53

-7.54

-0.70

-6.99

1

-0.59

0.0006

-0.58

-7.91

-0.72

-7.33

2

-0.58

0.0012

-0.63

-8.30

-0.73

3

-0.57

0.0018

-0.68

-8.72

-0.75

4

-0.56

0.0024

-0.74

-9.16

5

-0.55

0.003

-0.80

6

-0.54

0.0036

7

-0.53

0.0042

8

-0.52

9

stress increase

*10^6 *10^6

radial strain

SUM

1/E

er microns

er

2.08E+00

-1.46E+01

-1.46E+01

-0.34

2.10E+00

-7.18E-01

-1.53E+01

-7.69

-0.36

2.12E+00

-7.68E-01

-1.60E+01

-8.08

-0.38

2.13E+00

-8.21E-01

-1.69E+01

-0.76

-8.49

-0.41

2.15E+00

-8.79E-01

-1.77E+01

-9.63

-0.78

-8.92

-0.43

2.17E+00

-9.42E-01

-1.87E+01

-0.87

-10.13

-0.79

-9.38

-0.46

2.18E+00

-1.01E+00

-1.97E+01

-0.95

-10.67

-0.81

-9.87

-0.49

2.20E+00

-1.08E+00

-2.08E+01

0.0048

-1.04

-11.24

-0.82

-10.40

-0.52

2.22E+00

-1.16E+00

-2.19E+01

-0.51

0.0054

-1.14

-11.85

-0.84

-10.95

-0.56

2.24E+00

-1.25E+00

-2.32E+01

10

-0.5

0.006

-1.25

-12.50

-0.86

-11.55

-0.59

2.25E+00

-1.34E+00

-2.45E+01

11

-0.49

0.0066

-1.37

-13.19

-0.87

-12.18

-0.63

2.27E+00

-1.44E+00

-2.60E+01

12

-0.48

0.0072

-1.50

-13.94

-0.89

-12.86

-0.68

2.29E+00

-1.55E+00

-2.75E+01

13

-0.47

0.0078

-1.65

-14.73

-0.91

-13.58

-0.72

2.31E+00

-1.67E+00

-2.92E+01

14

-0.46

0.0084

-1.82

-15.59

-0.92

-14.35

-0.77

2.32E+00

-1.79E+00

-3.10E+01

15

-0.45

0.009

-2.01

-16.51

-0.94

-15.18

-0.82

2.34E+00

-1.93E+00

-3.29E+01

16

-0.44

0.0096

-2.23

-17.49

-0.96

-16.06

-0.88

2.36E+00

-2.08E+00

-3.50E+01

17

-0.43

0.0102

-2.47

-18.55

-0.97

-17.00

-0.94

2.38E+00

-2.24E+00

-3.72E+01

18

-0.42

0.0108

-2.74

-19.69

-0.99

-18.01

-1.01

2.40E+00

-2.41E+00

-3.96E+01

19

-0.41

0.0114

-3.05

-20.91

-1.00

-19.08

-1.08

2.41E+00

-2.60E+00

-4.22E+01

20

-0.4

0.012

-3.40

-22.22

-1.02

-20.24

-1.15

2.43E+00

-2.80E+00

-4.50E+01

21

-0.39

0.0126

-3.80

-23.64

-1.03

-21.47

-1.23

2.45E+00

-3.02E+00

-4.81E+01

22

-0.38

0.0132

-4.25

-25.17

-1.04

-22.79

-1.32

2.47E+00

-3.26E+00

-5.13E+01

23

-0.37

0.0138

-4.77

-26.82

-1.05

-24.20

-1.41

2.49E+00

-3.51E+00

-5.48E+01

24

-0.36

0.0144

-5.36

-28.60

-1.06

-25.71

-1.51

2.50E+00

-3.77E+00

-5.86E+01

25

-0.35

0.015

-6.04

-30.51

-1.07

-27.32

-1.61

2.52E+00

-4.06E+00

-6.26E+01

26

-0.34

0.0156

-6.82

-32.58

-1.07

-29.03

-1.72

2.54E+00

-4.36E+00

-6.70E+01

27

-0.33

0.0162

-7.72

-34.81

-1.07

-30.86

-1.83

2.56E+00

-4.68E+00

-7.17E+01

28

-0.32

0.0168

-8.77

-37.22

-1.06

-32.80

-1.94

2.58E+00

-5.00E+00

-7.67E+01

29

-0.31

0.0174

-9.98

-39.82

-1.05

-34.86

-2.06

2.60E+00

-5.34E+00

-8.20E+01

30

-0.3

0.018

-11.39

-42.62

-1.03

-37.03

-2.17

2.62E+00

-5.68E+00

-8.77E+01

Chapter 8 8.1

A 28 cm thick isolated slab, with dimensions of 3.6x4.2 m, is resting on a liquid subgrade with a modulus of subgrade subgrade reaction of 60 MPa/m. It is subjected subjected to a decrease in temperature of -12 °C at its upper surface and an increase of +7°C on its lower surface. Determine and plot the stresses versus slab depth at mid slab, as well as at mid-span of the two free boundaries. Additional information given for the P ortland concrete: E = = 28 GPa, µ = = 0.15 and at = 9.0 10-6/°C.

Chapter 8 8.1

A 28 cm thick isolated slab, with dimensions of 3.6x4.2 m, is resting on a liquid subgrade with a modulus of subgrade subgrade reaction of 60 MPa/m. It is subjected subjected to a decrease in temperature of -12 °C at its upper surface and an increase of +7°C on its lower surface. Determine and plot the stresses versus slab depth at mid slab, as well as at mid-span of the two free boundaries. Additional information given for the P ortland concrete: E = = 28 GPa, µ = = 0.15 and at = 9.0 10-6/°C.

Solution

The temperature distribution is disaggregated into a uniform increase of (-12+7)/2=-2.5

°C and a pure bending inducing temperature T a of (-12-7)/2=8.5 °C. Consider stresses only from pure bending. Equation 8-21 gives the radius of relative stiffness: stiffness:

 28000 0.28 3  =  12(1 − 0.15 2 )60   

1/ 4

=

0.967 m

This gives normalized slab dimensions of 3.6/0.967= 3.72 and 4.2/0.967= 4.34 in the x and y directions, respectively.

Accordingly, Figure 8-6 yields C x and C y values of 0.48

and 0.69, respectively. The stresses are computed using Equations 8-20:

At location 1 (mid-span):

σ y =

2810 6 9 10 −6 8.5

(1 − 0.15 ) 2

(0.48 + 0.15 0.69) = 1,253 kPa

σ x

=

2810 6 9 10 −6 8.5

(1 − 0.15 ) 2

(0.69 + 0.15 0.48) =

1,636kPa

The slab has a concave shape with outer fiber stresses that are tensile at the top and compressive at the bottom.

At location 2:

At this location, there is obviously no stress in the x direction, hence no contribution from it to the stress in the y direction. Accordingly: σ y =

2810 6 9 10 −6 8.5

(1 − 0.15 ) 2

(0.48) =

1030 kPa

σ x = 0

At location 3:

The reverse is true for location 3, hence: σ y = 0

σ x

=

2810 6 9 10 −6 8.5

(1 − 0.15 ) 2

(0.69) =

1,484 kPa

8.2

Compute the tensile stresses generated by subgrade friction in a 8.0 m long concrete slab. How high is this stress compared to the tensile strength of concrete? Given that the 28-

′

day compressive strength of the concrete f c is 20 MPa. Also, given γ for the concrete equal to 22.5 kN/m3. Solution

Refer to the free-body diagram shown in Fig. 8.8. Use Equ. 8.23 to compute stresses due to subgrade friction: σ =

L

2

y f = σ =

8 2

22.5 1.5 = 135 kPa

Use Equ. 6.3 to compute the tensile strength of the concrete as: 2,486 MPa. Hence, the stress generated due to subgrade friction amount to about 5% of the tensile strength of the concrete.

8.3

Consider the rigid pavement layout shown in Figure 8-22. The slabs are 0.20 m thick and were poured in two halves by two separate passes of a slip-form paver, (i.e., construction joint indicated by dotted line). Compute the necessary area of tiebar steel across the construction joint and the average bond stress between the tiebars and the concrete. The allowable stress of the steel f r is given as 200 MPa and the length of the tiebars is 1.00 m.

Solution

Refer to Fig. 8.22 The distance between the construction joint and the free slab face is 8.4 m. Substitute into Equ. 8.25 gives: Ar =

L ′ h γ f f r

=

8.4 0.2 22.5 1.5 200000

= 0.0028 m2/m width (2.8 cm2/m)

Select 2 tiebars per meter width gives a bar diameter of (2.8/2x4/π)0.5= 1.34 cm with a circumference of 4.2 cm. The corresponding bond stress is computed from Equ. 8.26:

u=

L ′ h γ f nΣ o (t / 2 )

=

8.4 0.20 22.5 1.5 2 0.042 (1 / 2 )

= 1,350 kPa.

8.4

Compute the maximum tensile stress on a 0.25 m thick slab of a JPCP m under a corner point load of 40 kN.

Solution

Use Equ. 8.29: σ c =

3P h2

=

3 40 0.25 2

= 1,820 kPa

8.5

Determine the maximum tensile stress and the corner deflection unde r a circular load of 0.15 m radius carrying 700 kPa pressure, given a slab thickness of 0.22 m, a modulus of subgrade reaction of 60 MPa/m, a concrete modulus of 28 GPa and a Poisson’s ratio of 0.15. Solution

Compute the radius of relative stiffness using Equ 8.21: 1/ 4

 28000 0.22 3  =  12(1 − 0.15 2 )60  = 0.8 m   And the load using Equ. 2.1 as 49.5 kN Use Equ. 8.30 and 8.31. The corner stress is:

3 49.5   0.15 2   1 −  σ c = 0.22 2   0.8 



0.6

  = 1,684 kPa  

The corner deflection ∆c is:

∆c =

49.5 60000 0.8 2

  0.15 2   = 0.0011 m or 1.1 mm 1.1 − 0.88    0.8 

8.6

A 3.0 cm diameter dowel bar is transferring a vertical load of 4500 N across a 0.35 cm wide joint. Compute the dowel bar deflection at the edge of the joint and the corresponding concrete bearing stresses. Can the concrete handle this stress? Given, K c

′

of 120,000 MPa/m, E r of 200,000 MPa and f c 30 MPa

Solution

Concrete bearing strength from Equ. 8.36:

 4 − 0.3937 d  ′  4 − 0.3937 3   f c =   30000 = 28,081 kPa 3 3    

f b = 

Deflection and bearing stress are calculated from Equ. 8.37 and 8.39, reps. Compute first the moment of inertia of the dowel bar I r as 3.974 10-8 m4. Compute β :

 K c d   β =  E I 4  r r  y 0

=

0.25

Pt 3

4 β E r I r

  120000 0.03  =  −8  4 200000 3 . 974 10  

(2 + β z ) =

0.25

= 18.34 m-1

4.5 3

6

4 18.34 200 10 3.974 10

−8

(2 + 18.34

0.0035) = 4.73 10-5 m

The corresponding bearing concrete stress at the joint face is computed as:

σ = K c y 0 = 120 10 6 4.73 10 −5 = 5,676 kPa which is lower than the bearing strength

computed.

8.7

Consider a JDRCP, consisting of slabs 25 cm thick and 3.6 m wide, resting on a subgrade with a modulus of subgrade reaction of 50 MPa/m. An axle load consisting of 2 identical tires 1.8 meters apart, each carrying 44 kN, is located at the edge of the joint, 0.30 m from the edge of the slab. Load across the joint is carried by 30 mm diameter dowel bars placed at 0.3 m center-to-center distances, as shown in Figure 8-15. Compute the load carried by each dowel bar. Given, E for the Portland concrete of 28,000 MPa and a Poisson’s ratio of 0.15. Assume that the tires apply point loads, the load transfer across the slabs is 50/50 and the distribution of load varies linearly with the distance from each tire load location. Solution

Compute the radius of relative stiffness using Equ 8.21:

 28000 0.253  =  12(1 − 0.15 2 )50   

1/ 4

= 0.925 m

Therefore the influence of each load is 1.8x0.925=1.67 m or 5.58 dowel bars away. The force diagram is shown on the following page. Using simple geometry allows computing the fraction of the loads x and y corresponding to each dowel bar location. Equilibrium of forces for the left-hand-side load gives:

 

x1 +

4.58 5.58

+

3.58 5.58

+

2.58 5.58

+

1.58 5.58

+

0.58 

 = 22

5.58 

which results in a value for x of 6.64 kN. Similarly, equilibrium of forces for the righthand side load gives:

 

y 1 +

4.58 5.58

+

3.58

+

5.58

2.58 5.58

+

1.58 5.58

+

0.58 5.58

+

3.5 4.5

+

2.5 4.5

+

1.5 4.5

+

0.5 

 = 22

4.5 

which results in a value for y of 4.33 kN. Stresses from both wheels can be computed through superposition.

22 kN

1.8 m

22 kN

1.8x0.93=1.6 0.25 m 0.3 0.3 x

1.67 m

k k

1.35 m y

8.8

Design the amount of rebar reinforcement required for a CRCP 0.25 m thick slab subjected to a temperature difference between pou ring and the coldest winter day of – 40°C and estimate the anticipated average spacing of the transverse concrete cracks. Given a coefficient of thermal expansion for the concrete or 9.0 10-6/°C, a tensile concrete strength of 3.5 MPa, a steel elastic modulus of 200000 MPa, a concrete elastic

modulus of 28000 MPa, a steel elastic limit of 340 MPa, and an allowable bonding stress between steel and concrete of 2.6 MPa.

8.9

Given the stiffness matrix of the plate element described in Example 8-11, compute the slab stiffness for the slab below made up of 4 identical plate elements, laid out as shown below in Figure 8-23.

y

8

9 6 1

4

7 3

5

1

2

2

3

4 x

Solution

Note: The element stiffness was derived using element dimensions of 2α = 2.0 m and 2 β = 1.0 m as in Example 8.11. The stiffness matrix for each of these plate elements labeled as 1, 2 3 and 4 in red above was derived as part of Example 8.11. The stiffness matrix of the slab is assembled from the individual plate element stiffnesses taking into account the

common nodes of the plate elements (e.g., node 6 is shared between elements 1 and 4, node 5 is shared by all elements and so on). The following tables show the stiffness matrices for each of the elements. Their summation gives the overall stiffness of the slab.

SLAB 1 1 1 1 2 2 2 3 3 3 4 4 4 5 5 5 6 6 6 7 7 7 8 8 8 9 9 9

SLAB 2

1 11.46 29.72 -2.03 -4.30 -28.64 -2.03

1 -2.86 -19.10 2.03 1.43 9.55 0.00

-7.16 -1.07 -2.03 0.00 0.00 -2.03

0.00 0.00 2.03 0.00 0.00 0.00

1

1

1

9.55 1.43 -4.06 0.00 0.00 -4.06

2 -4.30 -28.64 2.03 11.46 29.72 2.03

2 -1.43 -9.55 0.00 2.86 19.10 2.03

-4.77 -0.72 0.00 0.00 0.00 0.00

0.00 0.00 2.03 -7.16 -1.07 2.03

0.00 0.00 0.00 0.00 0.00 4.77

1

1 1 1 2 2 2 3 3 3 4 4 4 5 5 5 6 6 6 7 7 7 8 8 8 9 9 9

SLAB 3 1 1 1 2 2 2 3 3 3 4 4 4 5 5 5 6 6 6 7 7 7 8 8 8 9 9 9

1

SLAB 4 1 1 1 2 2 2 3 3 3 4 4 4 5 5 5

1

1

1

2 0.00 0.00 4.06 9.55 1.43 4.06

3

3

5 -7.16 -1.07 2.03 0.00 0.00 2.03

0.00 0.00 0.00 -2.03 -0.72 0.00

11.46 29.72 2.03 -4.30 -28.64 2.03

2

2

2

3

3

11.46 29.72 -2.03 -4.30 -28.64 -2.03

-2.86 -19.10 2.03 1.43 9.55 0.00

9.55 1.43 -4.06 0.00 0.00 -4.06

-4.30 -28.64 2.03 11.46 29.72 2.03

-1.43 - 9.55 0.00 2.86 19.10 2.03

1

1

2

2

2

2

2

2

3

4

3

4

0.00 0.00 4.06 9.55 1.43 4.06

3

3

4

4

-7.16 - 1.07 2.03 0.00 0.00 2.03 -7.16 -1.07 -2.03 0.00 0.00 -2.03

3

3

4

0.00 0.00 - 2.03 0.00 0.00 0.00 0.00 0.00 2.03 0.00 0.00 0.00

3

3

5

5

6

6

6

0.00 0.00 -2.03 0.00 0.00 0.00

4.77 0.72 0.00 0.00 0.00 0.00

0.00 0.00 -2.03 -4.30 -4.30 -2.03

0.00 0.00 0.00 0.00 0.00 -2.03

0.00 0.00 0.00 4.77 0.72 0.00

-2.86 -19.10 -2.03 1.43 9.55 0.00

-9.55 -1.43 -4.06 0.00 0.00 -4.06

-4.30 -4.30 -4.77 11.46 29.72 -2.03

-1.43 -9.55 0.00 2.86 19.10 -4.77

0.00 0.00 4.06 -9.55 -9.55 4.06

4

5

5

5

6

6

4.77 0.72 0.00 0.00 0.00 0.00 -4.77 -0.72 0.00 0.00 0.00 0.00

0.00 0.00 - 2.03 -4.30 - 4.30 - 2.03 0.00 0.00 2.03 -7.16 -1.07 2.03

0.00 0.00 0.00 0.00 0.00 - 2.03 0.00 0.00 0.00 0.00 0.00 4.77

0.00 0.00 0.00 4.77 0.72 0.00 0.00 0.00 0.00 -2.03 -0.72 0.00

11.46 29.72 2.03 -4.30 -28.64 2.03

-2.86 -19.10 -2.03 1.43 9.55 0.00

4

4

4

5

5

5

6

-9.55 -1.43 -4.06 0.00 0.00 -4.06

6

7

7

7

8

8

8

9

9

9

7

7

7

8

8

8

9

9

9

-4.30 -4.30 -4.77 11.46 29.72 -2.03

-1.43 -9.55 0.00 2.86 19.10 -4.77

6

6

0.00 0.00 4.06 -9.55 -9.55 4.06

7

7

7

8

8

8

11.46 29.72 2.03 -4.30 -28.64 2.03

2.86 19.10 2.03 -1.43 -9.55 0.00

9.55 1.43 4.06 0.00 0.00 4.06

-4.30 -28.64 -2.03 11.46 29.72 -2.03

1.43 9.55 0.00 -2.86 -19.10 2.03

0.00 0.00 -4.06 9.55 1.43 -4.06

0.00 0.00 2.03 -7.16 -1.07 2.03

0.00 0.00 0.00 0.00 0.00 -2.03

0.00 0.00 0.00 4.77 0.72 0.00

-4.30 -4.30 -2.03 0.00 0.00 -2.03

0.00 0.00 -2.03 0.00 0.00 0.00

4.77 0.72 0.00 0.00 0.00 0.00

0.00 0.00 2.03 -7.16 -1.07 2.03

0.00 0.00 0.00 0.00 0.00 4.77

0.00 0.00 0.00 -2.03 -0.72 0.00

-7.16 -1.07 -2.03 0.00 0.00 -2.03

0.00 0.00 2.03 0.00 0.00 0.00

-4.77 -0.72 0.00 0.00 0.00 0.00

11.46 29.72 2.03 -4.30 -28.64 2.03

-2.86 -19.10 -2.03 1.43 9.55 0.00

-9.55 -1.43 -4.06 0.00 0.00 -4.06

-4.30 -4.30 -4.77 11.46 29.72 -2.03

-1.43 -9.55 0.00 2.86 19.10 -4.77

0.00 0.00 4.06 -9.55 -9.55 4.06

7

7

7

8

8

4

4

4

5

5

11.46 29.72 2 03

2.86 19.10 2 03

5

6

9.55 1.43 4 06

-4.30 -28.64 -2 03

6

1.43 9.55 0 00

6

0.00 0.00 -4 06

0.00 0.00 2 03

8

0.00 0.00 0 00

0.00 0.00 0 00

9

9

9

9

9

9

-4.30 -4.30 -2 03

0.00 0.00 -2 03

4.77 0.72 0 00

SUM 1 1 1 2 2 2 3 3 3 4 4 4 5 5 5 6 6 6 7 7 7 8 8 8 9 9 9

SLAB 3 1 1 1 2 2 2 3 3 3 4 4 4 5 5 5 6 6 6 7 7 7 8 8 8 9 9 9

1

SLAB 4 1 1 1 2 2 2 3 3 3 4 4 4 5 5 5 6 6 6 7 7 7 8 8 8 9 9 9

1

1 11.46 29.72 -2.03 -4.30 -28.64 -2.03 0.00 0.00 0.00 0.00 0.00 0.00 -7.16 -1.07 -2.03 0.00 0.00 -2.03 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00

1 -2.86 -19.10 2.03 1.43 9.55 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 2.03 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00

1

1

1

1 9.55 1.43 -4.06 0.00 0.00 -4.06 0.00 0.00 0.00 0.00 0.00 0.00 -4.77 -0.72 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00

2

1

2 -4.30 -28.64 2.03 22.91 59.44 0.00 -4.30 -28.64 -2.03 0.00 0.00 0.00 0.00 0.00 2.03 -7.16 -1.07 2.03 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00

2

2

2 -1.43 -9.55 0.00 0.00 0.00 4.06 1.43 9.55 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 4.77 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00

2

2

3

2

3

3

2

3

3

0.00 0.00 4.06 19.10 2.86 0.00 0.00 0.00 -4.06 0.00 0.00 0.00 0.00 0.00 0.00 -2.03 -0.72 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00

0.00 0.00 0.00 -4.30 -28.64 2.03 11.46 29.72 2.03 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00

0.00 0.00 0.00 -1.43 -9.55 0.00 2.86 19.10 2.03 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00

3

3

3 0.00 0.00 0.00 0.00 0.00 4.06 9.55 1.43 4.06 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00

4

4

4

5

5

5

6

6

6

7

7

7

8

8

8

11.46 29.72 2.03 -4.30 -28.64 2.03

2.86 19.10 2.03 -1.43 -9.55 0.00

9.55 1.43 4.06 0.00 0.00 4.06

-4.30 -28.64 -2.03 11.46 29.72 -2.03

1.43 9.55 0.00 -2.86 -19.10 2.03

0.00 0.00 -4.06 9.55 1.43 -4.06

0.00 0.00 2.03 -7.16 -1.07 2.03

0.00 0.00 0.00 0.00 0.00 -2.03

0.00 0.00 0.00 4.77 0.72 0.00

-4.30 -4.30 -2.03 0.00 0.00 -2.03

0.00 0.00 -2.03 0.00 0.00 0.00

4.77 0.72 0.00 0.00 0.00 0.00

0.00 0.00 2.03 -7.16 -1.07 2.03

0.00 0.00 0.00 0.00 0.00 4.77

0.00 0.00 0.00 -2.03 -0.72 0.00

-7.16 -1.07 -2.03 0.00 0.00 -2.03

0.00 0.00 2.03 0.00 0.00 0.00

-4.77 -0.72 0.00 0.00 0.00 0.00

11.46 29.72 2.03 -4.30 -28.64 2.03

-2.86 -19.10 -2.03 1.43 9.55 0.00

-9.55 -1.43 -4.06 0.00 0.00 -4.06

-4.30 -4.30 -4.77 11.46 29.72 -2.03

-1.43 -9.55 0.00 2.86 19.10 -4.77

0.00 0.00 4.06 -9.55 -9.55 4.06

7

7

7

8

8

3

4

4

4

0.00 0.00 0.00 -7.16 -1.07 2.03 0.00 0.00 2.03 4.30 28.65 0.00 -4.30 -28.64 0.00 0.00 0.00 0.00 0.00 0.00 2.03 -7.16 -1.07 2.03 0.00 0.00 0.00

0.00 0.00 0.00 0.00 0.00 -2.03 0.00 0.00 0.00 2.86 19.10 4.06 -1.43 -9.55 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 4.77 0.00 0.00 0.00

4

4 0.00 0.00 0.00 4.77 0.72 0.00 0.00 0.00 0.00 4.78 0.71 4.06 0.00 0.00 4.06 0.00 0.00 0.00 0.00 0.00 0.00 -2.03 -0.72 0.00 0.00 0.00 0.00

4

5 -7.16 -1.07 2.03 0.00 0.00 0.00 -4.30 -4.30 -2.03 -4.30 -28.64 0.00 27.21 69.82 35.81 -10.62 -57.28 4.06 -7.16 -1.07 -2.03 0.00 0.00 0.00 -7.16 -1.07 2.03

5

5

5

6

6

6

8

9

9

9

9

9

9

11.46 29.72 2.03 -4.30 -28.64 2.03

2.86 19.10 2.03 -1.43 -9.55 0.00

9.55 1.43 4.06 0.00 0.00 4.06

-4.30 -28.64 -2.03 11.46 29.72 -2.03

1.43 9.55 0.00 -2.86 -19.10 2.03

0.00 0.00 -4.06 9.55 1.43 -4.06

0.00 0.00 2.03 -7.16 -1.07 2.03

0.00 0.00 0.00 0.00 0.00 -2.03

0.00 0.00 0.00 4.77 0.72 0.00

-4.30 -4.30 -2.03 0.00 0.00 -2.03

0.00 0.00 -2.03 0.00 0.00 0.00

4.77 0.72 0.00 0.00 0.00 0.00

0.00 0.00 2.03 -7.16 -1.07 2.03

0.00 0.00 0.00 0.00 0.00 4.77

0.00 0.00 0.00 -2.03 -0.72 0.00

-7.16 -1.07 -2.03 0.00 0.00 -2.03

0.00 0.00 2.03 0.00 0.00 0.00

-4.77 -0.72 0.00 0.00 0.00 0.00

11.46 29.72 2.03 -4.30 -28.64 2.03

-2.86 -19.10 -2.03 1.43 9.55 0.00

-9.55 -1.43 -4.06 0.00 0.00 -4.06

-4.30 -4.30 -4.77 11.46 29.72 -2.03

-1.43 -9.55 0.00 2.86 19.10 -4.77

0.00 0.00 4.06 -9.55 -9.55 4.06

7

7

0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 -4.30 -4.30 -2.74 4.30 28.64 0.00 0.00 0.00 0.00 11.46 29.72 2.03 -4.30 -28.64 2.03 0.00 0.00 0.00

0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 -1.43 -9.55 0.00 2.86 19.10 -6.80 0.00 0.00 0.00 -2.86 -19.10 -2.03 1.43 9.55 0.00 0.00 0.00 0.00

5 0.00 0.00 -2.03 0.00 0.00 0.00 0.00 0.00 -2.03 1.43 9.55 0.00 -2.86 -2.86 -14.33 2.03 0.00 0.00 0.00 0.00 2.03 0.00 0.00 0.00 0.00 0.00 4.77

5 4.77 0.72 0.00 0.00 0.00 0.00 4.77 0.72 0.00 0.00 0.00 -4.06 7.52 8.83 1.43 -4.06 0.00 0.00 -4.77 -0.72 0.00 0.00 0.00 0.00 -2.03 -0.72 0.00

6

6

0.00 0.00 -2.03 -4.30 -4.30 -2.03 0.00 0.00 0.00 11.46 29.72 2.03 -12.90 -61.58 -4.77 22.91 59.44 -4.06 0.00 0.00 0.00 -7.16 -1.07 -2.03 0.00 0.00 -2.03

0.00 0.00 0.00 0.00 0.00 -2.03 0.00 0.00 0.00 -2.86 -19.10 -2.03 1.43 9.55 0.00 0.00 0.00 -2.74 0.00 0.00 0.00 0.00 0.00 2.03 0.00 0.00 0.00

6 0.00 0.00 0.00 4.77 0.72 0.00 0.00 0.00 0.00 -9.55 -1.43 -4.06 0.00 0.00 -4.06 0.00 -8.12 0.00 0.00 0.00 0.00 -4.77 -0.72 0.00 0.00 0.00 0.00

7 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 4.06 -4.78 -8.83 4.06 0.00 0.00 0.00 -9.55 -1.43 -4.06 0.00 0.00 -4.06 0.00 0.00 0.00

8 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 -4.30 -4.30 -2.03 0.00 0.00 0.00 -7.16 -1.07 2.03 -4.30 -4.30 -4.77 22.91 59.44 0.00 -4.30 -28.64 2.03

8 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 -2.03 0.00 0.00 0.00 0.00 0.00 -2.03 -1.43 -9.55 0.00 0.00 0.00 -6.80 1.43 9.55 0.00

8

9

9

0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 4.77 0.72 0.00 0.00 0.00 0.00 4.77 0.72 0.00 0.00 0.00 4.06 -19.10 -10.98 0.00 0.00 0.00 -4.06

0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 -4.30 -4.30 -2.03 0.00 0.00 -2.03 0.00 0.00 0.00 -4.30 -4.30 -4.77 11.46 29.72 -2.03

0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 -2.03 0.00 0.00 0.00 0.00 0.00 0.00 -1.43 -9.55 0.00 2.86 19.10 -4.77

9 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 4.77 0.72 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 4.06 -9.55 -9.55 4.06

SUM 1 1 1 2 2 2 3 3 3 4 4 4 5 5 5 6 6 6 7 7 7 8 8 8 9 9 9

1 11.46 29.72 -2.03 -4.30 -28.64 -2.03 0.00 0.00 0.00 0.00 0.00 0.00 -7.16 -1.07 -2.03 0.00 0.00 -2.03 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00

1 -2.86 -19.10 2.03 1.43 9.55 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 2.03 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00

1 9.55 1.43 -4.06 0.00 0.00 -4.06 0.00 0.00 0.00 0.00 0.00 0.00 -4.77 -0.72 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00

2 -4.30 -28.64 2.03 22.91 59.44 0.00 -4.30 -28.64 -2.03 0.00 0.00 0.00 0.00 0.00 2.03 -7.16 -1.07 2.03 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00

2 -1.43 -9.55 0.00 0.00 0.00 4.06 1.43 9.55 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 4.77 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00

2

3

3

0.00 0.00 4.06 19.10 2.86 0.00 0.00 0.00 -4.06 0.00 0.00 0.00 0.00 0.00 0.00 -2.03 -0.72 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00

0.00 0.00 0.00 -4.30 -28.64 2.03 11.46 29.72 2.03 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00

0.00 0.00 0.00 -1.43 -9.55 0.00 2.86 19.10 2.03 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00

3 0.00 0.00 0.00 0.00 0.00 4.06 9.55 1.43 4.06 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00

4

4

0.00 0.00 0.00 -7.16 -1.07 2.03 0.00 0.00 2.03 4.30 28.65 0.00 -4.30 -28.64 0.00 0.00 0.00 0.00 0.00 0.00 2.03 -7.16 -1.07 2.03 0.00 0.00 0.00

0.00 0.00 0.00 0.00 0.00 -2.03 0.00 0.00 0.00 2.86 19.10 4.06 -1.43 -9.55 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 4.77 0.00 0.00 0.00

4 0.00 0.00 0.00 4.77 0.72 0.00 0.00 0.00 0.00 4.78 0.71 4.06 0.00 0.00 4.06 0.00 0.00 0.00 0.00 0.00 0.00 -2.03 -0.72 0.00 0.00 0.00 0.00

5 -7.16 -1.07 2.03 0.00 0.00 0.00 -4.30 -4.30 -2.03 -4.30 -28.64 0.00 27.21 69.82 35.81 -10.62 -57.28 4.06 -7.16 -1.07 -2.03 0.00 0.00 0.00 -7.16 -1.07 2.03

5 0.00 0.00 -2.03 0.00 0.00 0.00 0.00 0.00 -2.03 1.43 9.55 0.00 -2.86 -2.86 -14.33 2.03 0.00 0.00 0.00 0.00 2.03 0.00 0.00 0.00 0.00 0.00 4.77

5 4.77 0.72 0.00 0.00 0.00 0.00 4.77 0.72 0.00 0.00 0.00 -4.06 7.52 8.83 1.43 -4.06 0.00 0.00 -4.77 -0.72 0.00 0.00 0.00 0.00 -2.03 -0.72 0.00

6

6

0.00 0.00 -2.03 -4.30 -4.30 -2.03 0.00 0.00 0.00 11.46 29.72 2.03 -12.90 -61.58 -4.77 22.91 59.44 -4.06 0.00 0.00 0.00 -7.16 -1.07 -2.03 0.00 0.00 -2.03

0.00 0.00 0.00 0.00 0.00 -2.03 0.00 0.00 0.00 -2.86 -19.10 -2.03 1.43 9.55 0.00 0.00 0.00 -2.74 0.00 0.00 0.00 0.00 0.00 2.03 0.00 0.00 0.00

6 0.00 0.00 0.00 4.77 0.72 0.00 0.00 0.00 0.00 -9.55 -1.43 -4.06 0.00 0.00 -4.06 0.00 -8.12 0.00 0.00 0.00 0.00 -4.77 -0.72 0.00 0.00 0.00 0.00

7

7

0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 -4.30 -4.30 -2.74 4.30 28.64 0.00 0.00 0.00 0.00 11.46 29.72 2.03 -4.30 -28.64 2.03 0.00 0.00 0.00

0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 -1.43 -9.55 0.00 2.86 19.10 -6.80 0.00 0.00 0.00 -2.86 -19.10 -2.03 1.43 9.55 0.00 0.00 0.00 0.00

7 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 4.06 -4.78 -8.83 4.06 0.00 0.00 0.00 -9.55 -1.43 -4.06 0.00 0.00 -4.06 0.00 0.00 0.00

8 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 -4.30 -4.30 -2.03 0.00 0.00 0.00 -7.16 -1.07 2.03 -4.30 -4.30 -4.77 22.91 59.44 0.00 -4.30 -28.64 2.03

8 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 -2.03 0.00 0.00 0.00 0.00 0.00 -2.03 -1.43 -9.55 0.00 0.00 0.00 -6.80 1.43 9.55 0.00

8

9

9

0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 4.77 0.72 0.00 0.00 0.00 0.00 4.77 0.72 0.00 0.00 0.00 4.06 -19.10 -10.98 0.00 0.00 0.00 -4.06

0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 -4.30 -4.30 -2.03 0.00 0.00 -2.03 0.00 0.00 0.00 -4.30 -4.30 -4.77 11.46 29.72 -2.03

0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 -2.03 0.00 0.00 0.00 0.00 0.00 0.00 -1.43 -9.55 0.00 2.86 19.10 -4.77

8.10

Expand the stiffness matrix of the slab above, to account for a liquid foundation, given a modulus of subgrade reaction of 40 MPa/m. Solution

The Stiffness Matrix is obtained from the one derived above (Problem 8.9) by adding 40 to the value of the diagonal elements related to the vertical displacement only.

8.11

Compute the solid foundation stiffness matrix for a plate element with dimensions 2 a=

9 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 4.77 0.72 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 0.00 4.06 -9.55 -9.55 4.06

8.10

Expand the stiffness matrix of the slab above, to account for a liquid foundation, given a modulus of subgrade reaction of 40 MPa/m. Solution

The Stiffness Matrix is obtained from the one derived above (Problem 8.9) by adding 40 to the value of the diagonal elements related to the vertical displacement only.

8.11

Compute the solid foundation stiffness matrix for a plate element with dimensions 2 a= 4.0 m and 2b=1.5 m. Given, subgrade E s of 300 MPa and µ s of 0.40. Solution

Non-diagonal components of the foundation “softness” matrix can be readily computed from Equation 8-53: 1/ 4 1 / 4.272 1 / 1.5   F i   wi   d w  (1 − 0.40 2 )  1 / 4   d 1 / 1.5 1 / 4.272  F j   j      =  1 / 1.5 d 1 / 4   F k  π 300 1 / 4.272 wk    wl  1 / 4.272 1/ 4 d    1 F l 

The diagonal components d is computed from Equ. 8.54 using MathCad  d =

1

2

∫

2 0.75 0

0.75 2  1 2 − 0 .5  ∫ ( x + y ) dy  dx = 2 0.75 2.104 = 1.403 0 

Hence, the stiffness matrix is obtained from: 0.25 0.2341 1   F i   1.403  F   1.403 0.667 0.2341 π 300  0.25  j    = 2   F 0 . 2341 0 . 667 1 . 403 0 . 25 1 0 . 4 ( ) −  k     F l  0.2341 0.25 1.403   1

−1

− 0.0779 − 0.024 − 1.023   wi   wi   1.460 w   − 0.0779 0.933 − 0.426 − 0.024   w j   j     = 1122    − 0.024 − 0.426 0.933 − 0.0779 w k  wk    wl   − 1.023 − 0.024 − 0.0779 1.460  wl 

8.12

A Falling Weight Deflectometer (FWD) load of 44 kN is applied to the right-hand side slab in the middle of a JPCP joint. The deflection measurements obtained in the right and the left of slabs are 2.08 and 1.82 mm, respectively. Compute the elastic constant of normal load transmission cw across this joint, given that the modulus of subgrade reaction k is 60 MPa/m.

Solution

Use Equ. 8.58 to compute the stiffness of the joint: c w (2.08 − 1.82 ) = 60 1.82

Therefore c w = 420 MPa/m

8.13

Compute the coefficients of force and moment transmission across a dowel in a JDRPC pavement. Given, slab thickness of 0.25 m, dowel bar diameter of 0.030 m, joint opening of 0.01 m, elastic modulus of the dowel bar steel 250,000 MPa, Poisson’s ratio of the dowel bar steel of 0.4, spring constant of concrete support K c 100,000 MPa/m. Solution:

Compute moment of inertia and area of the dowel bar: I r

Ar

= =

π d 4 64

=

π 0.03 4

π 0.03 2

4

64

= 3.98 10-8 m4

= 7.07 10-4 m2

Compute the quantity b from Equation 8-38:

 100000 0.03   β =  −8  4 250000 3 . 98 10  

0.25

= 16.57 m-1

Compute the shear modulus of steel as: Gr =

E r

2(1 + µ )

=

250000 2(1 + 0.4 )

= 89,285 MPa.

Use Equation 8-62 to compute the constant of transmitting vertical load cw: cw

=

1

  0.01 (2 + 16.57 0.01)  2 16.57 3 250000 3.98 10 −8  + 89285 7.07 10 −4  

= 41.2 MN/m

8.14

Utilizing a FEM software package, compute the stresses at the bottom of a 0.20 m thick Portland concrete slab subjected to a combined load of 100 kN and a thermal gradient consisting of a reduction at the surface of 5°C and an increase at the bottom of 5°C. The layout of the slab and the load is shown below. Given, modulus of subgrade reaction of 80 MPa/m, tensionless subgrade, slab modulus of elasticity of 28000 MPa, Poisson’s ratio of 0.15, unit weight of 23.54 kN/m3 and coefficient of linear thermal expansion at of 9.0 10-6/°C.

1 . 8 m

0.20

3.6

0.15 0.15

1.5

0.2 2.8

5.6 Solution

Using EVERFE 2.22, allows producing a variety of stress plots. Examples: Maximum tensile stresses at the bottom of the slab

Maximum compressive stresses at the bottom of the slab

Chapter 9 9.1

Synthesize a pavement profile by superimposing random elevations ranging between – 0.01 to 0.01 m to two in-phase sinusoidal waves with amplitudes of 0.03 and 0.02 and wave lengths of 3 and 5 meters, respectively. Plot the pavement profile for a distance of 32 meters using 0.25 m increments. Plot the trace of a rolling straight edge (RSE) with a base length of 4 meters, (i.e., assume that the transport wheels and the tracing wheels are small enough to neglect their dimensions).

0.10 0.08 0.06 ) m ( n o i t a v e l e

0.04 0.02 raw profile

0.00 -0.02

0

10

20

30

RSE 4 m

-0.04 -0.06 -0.08 -0.10

distance (m)

9.2

Filter the profile generated in Problem 1 using a low-pass MA filter with a base length of 1.0 m. Plot the results.

0.10 0.08 0.06 0.04 ) m 0.02 ( n o i t 0.00 a v e -0.02 l e

raw profile 0

5

10

15

20

25

MA B=1 m

30

-0.04 -0.06 -0.08 -0.10 distance (m)

Low-Pass 1.0-m base filter 9.3

Filter the profile generated in Problem 1 using a high-pass MA filter with a base length of 2.0 m. Plot the results. MA B=1 m 0.02 0.01 ) m (

0.01

n 0.00 o i t a -0.01 v e l e

MA B=1 m

-0.01 -0.02 -0.02

5

10

15 distance (m)

High-pass 1.0-m base filter

20

Note: since the random number generated are case-specific, your plots may look slightly different than those above. 9.4

Compute and plot the power spectral density (PSD) of the artificial pavement profile given below. What are the dominant wave lengths and corresponding amplitudes?

Distance (m)

Elevation (m)

Distance (m)

Elevation (m)

0

10.0000

3.2

10.0165

0.2

10.0219

3.4

10.0229

0.4

10.0264

3.6

10.0101

0.6

10.0216

3.8

9.9918

0.8

10.0163

4

9.9835

1

10.0074

4.2

9.9788

1.2

9.9901

4.4

9.9726

1.4

9.9783

4.6

9.9783

1.6

9.9818

4.8

9.9994

1.8

9.9917

5

10.0219

2

9.9936

5.2

10.0272

2.2

9.9913

5.4

10.0218

2.4

9.9994

5.6

10.0169

2.6

10.0081

5.8

10.0076

2.8

10.0077

6

9.9895

3

10.0071

6.2

9.9788

0.0003

)

2

0.0002

m ( d e r a 0.0002 u q S e d 0.0001 u t i l p m A 0.0001

0.0000 0.00

0.25

0.50

0.75

1.00

1.25

1.50

1.75

2.00

Wave Number (cy/m)

The PSD of the elevation profile given is shown above. The dominant wave numbers are 0.48 and 1.3 cy/m (i.e., dominant wavelengths of 2.08 and 0.77 m, resp.) and the corresponding amplitudes are 0.0154 m and 0.0057 m.

9.5

The pavement profile data tabulated next was obtained with an inertial profilometer at intervals of 0.1394 m. Compute its IRI using commercially available software. What is the corresponding pavement serviceability, (i.e., PSI )? Use the computer program ProVal. One of the profile input formats it accepts is ASCII. The particular format is used by the TxDOT and it is suited for importing the data given. This format includes 5 lines for handling alphanumeric information (i.e., c omments, profile sampling distance increment and other reference data). The format is fully

described under ftp://ftp.dot.state.tx.us/pub/txdot-info/cst/TMS/1000S_series/pdfs/spe1001.pd . f Distance (m)

Elevation (m)

Distance (m)

Elevation (m)

Distance (m)

Elevation (m)

0.0000

-0.02024

2.9274

-0.012725

5.8548

-0.002845

0.1394

-0.02022

3.0668

-0.011760

5.9942

-0.002718

0.2788

-0.01984

3.2062

-0.011608

6.1336

-0.002565

0.4182

-0.01910

3.3456

-0.010541

6.2730

-0.001905

0.5576

-0.01895

3.4850

-0.010058

6.4124

-0.001549

0.6970

-0.01910

3.6244

-0.010109

6.5518

-0.001219

0.8364

-0.01875

3.7638

-0.009398

0.9758

-0.01783

3.9032

-0.008915

1.1152

-0.01768

4.0426

-0.008509

1.2546

-0.01788

4.1820

-0.008255

1.3940

-0.01793

4.3214

-0.007823

1.5334

-0.01699

4.4608

-0.007264

1.6728

-0.01648

4.6002

-0.006528

1.8122

-0.01636

4.7396

-0.006426

1.9516

-0.01638

4.8790

-0.005791

2.0910

-0.01549

5.0184

-0.005232

2.2304

-0.01443

5.1578

-0.004750

2.3698

-0.01422

5.2972

-0.004293

2.5092

-0.01458

5.4366

-0.004064

2.6486

-0.01443

5.5760

-0.003810

2.7880

-0.01356

5.7154

-0.003226

The data file developed is shown on the next page. It has an extension *.DAT. Elevations are specified in mils (10-3 inches).

HEAD3,08082008,07,007,IH0037S,0010 +00.200,R1 CMET3,Book Problem 9.5,07082008 Manufacturer,mil,R,5.4882,in COMMENT COMMENT -797, -796, -781, -752, -746, -752, -738, -702, -696, -704, -706, -669, -649, -644, -645, -610, -568, -560, -574, -568, -534, -501, -463, -457, -415, -396, -398, -370, -351, -335, -325, -308, -286, -257, -253, -228, -206, -187, -169, -160, -150, -127, -112, -107, -101, -75,

The IRI value calculated with ProVal is 37.4 inches/mi (0.59 m/km) Inputting this metric IRI value into Equation 9.4 gives a PSI of 4.46, which reflects a very smooth pavement.

9.6

The following FWD measurements were obtained on a flexible pavement using a plate radius of 10.3 cm and a contact pressure of 500 kPa. Estimate the value of the subgrade modulus. The layer thicknesses and the assumed values for the Poison ratio are given next. Sensor, s

1

2

3

4

5

6

7

Offset (cm)

0

20

40

60

80

100

120

Deflection (µm)

280

235

190

175

85

60

30

Layer, k

1

2

3

Thickness (cm)

20

50

∞

Poison’s ratio, µ

0.33

0.45

0.45

The solution is shown in the following Excel-generated tables. The surface deflection calculations were carried out by Everstress. The subgrade modulus value that corresponds to the minimum deflection error is 83.6 MPa.

TRIAL

E1/ESG

E2/ESG

ESG

pa/Esg

Esg/pa

(1/m)

(m)

W1

" measured values"

W2 280

W3

W4

W5

W6

W7

235

190

175

85

60

30 25.4

estimated v alues e.g., 10 and 1 1400 and 140 with 140 MPa subgrade 1

3

1

140 0.3678571 2.7184466

335.29

149.24

84.03

54.4

39.36

30.82

2

3

2

14 3.6785714 0.2718447

279.05

109.32

66.47

48.8

38.7

31.8

26.8

3

5

1

140 0.3678571 2.7184466

258.75

138.11

82.25

55.78

40.2

31.2

25.55

4

5

2

140 0.3678571 2.7184466

212.51

103.11

66.48

48.96

38.7

31.8

26.8

5 6 7 8 9 10 11 12

10

1

140 0.3678571 2.7184466

188.82

121.81

82.4

57.29

41.76

32.15

26.01 26.22

10

10

140 0.3678571 2.7184466

102.21

53.26

42.32

37.01

32.9

29.35

100

1

140 0.3678571 2.7184466

77.27

68.31

58.99

49.95

41.9

35.08

29.48

100

10

140 0.3678571 2.7184466

48.06

41.03

36.16

32.27

29.05

26.31

23.91

100

30

140 0.3678571 2.7184466

35.57

29.68

26.97

25.08

23.49

22.03

20.67

300

1

140 0.3678571 2.7184466

52.39

49.07

45.03

40.63

36.28

32.19

28.47

300

10

140 0.3678571 2.7184466

37.85

35.04

32.32

29.69

27.25

25.01

22.97

300

30

140 0.3678571 2.7184466

28.69

26.3

24.55

23.04

21.7

20.46

19.31

S

ESG est pa/Esg

ERRORS (wm-wk)/wm 1

2

3

4

5

6

7

0.1974643 -0.364936 -0.557737 -0.689143 -0.536941 -0.486333 -0.153333 -0.003393 -0.534809 -0.650158 -0.721143 -0.544706

-0.47 -0.106667

-0.075893 -0.412298 -0.567105 -0.681257 -0.527059

-0.48 -0.148333

-0.241036 -0.561234 -0.650105 -0.720229 -0.544706

-0.47 -0.106667

-0.325643

-0.48166 -0.566316 -0.672629 -0.508706 -0.464167

-0.133

-0.634964 -0.773362 -0.777263 -0.788514 -0.612941 -0.510833

-0.126

-0.724036 -0.709319 -0.689526 -0.714571 -0.507059 -0.415333 -0.017333 -0.828357 -0.825404 -0.809684

-0.8156 -0.658235

-0.5615

-0.203

-0.872964 -0.873702 -0.858053 -0.856686 -0.723647 -0.632833

-0.311

-0.812893 -0.791191

-0.051

-0.763 -0.767829 -0.573176

-0.4635

-0.864821 -0.850894 -0.829895 -0.830343 -0.679412 -0.583167 -0.234333 -0.897536 -0.888085 -0.870789 -0.868343 -0.744706

f1

f2

f3

f4

f5

-0.659 -0.356333

f6

f7

911.4680

405.7010

228.4311

147.8835

106.9981

83.7825

69.0485

758.5825

297.1806

180.6951

132.6602

105.2039

86.4466

72.8544

703.3981

375.4447

223.5922

151.6350

109.2816

84.8155

69.4563

577.6971

280.2990

180.7223

133.0951

105.2039

86.4466

72.8544

513.2971

331.1340

224.0000

155.7398

113.5223

87.3981

70.7068

277.8524

144.7845

115.0447

100.6097

89.4369

79.7864

71.2777

210.0544

185.6971

160.3612

135.7864

113.9029

95.3631

80.1398

130.6485

111.5379

98.2990

87.7243

78.9709

71.5223

64.9981 56.1903

96.6951

80.6835

73.3165

68.1786

63.8563

59.8874

142.4194

133.3942

122.4117

110.4505

98.6252

87.5068

77.3942

102.8932

95.2544

87.8602

80.7107

74.0777

67.9883

62.4427

77.9922

71.4951

66.7379

62.6330

58.9903

55.6194

52.4932

fi/f1/wm

S2

1

0 .0 03 6

0 .0 01 9

0 .0 01 3

0 .0 00 9

0 .0 01 4

0 .0 01 5

0 .0 02 5

0 .0 00 0 0 .0 13 14 99

2

0 .0 036

0.0 017

0.0 013

0 .0 010

0 .0016

0 .0019

0 .0032

0 .0000 0.014 2237

1 05 .5 7 0 .4 87 85 07 95.09

3

0 .0036

0 .0023

0 .0017

0 .0012

0 .0018

0.0020

0.0033

0.0000 0.01 58765

92.28 0.55 80888

4

0 .0036

0 .0021

0.0016

0.0013

0.0021

0.002 5

0.004 2

0.000 0 0.0 17439 3

85.2 0 0.6 04433 9

5 0.00 36 6 0 .00 36 7 0.00 36 8 0 .00 36

0.00 27

0.00 23

0.0 017

0.0 026

0.0 028

0.0 046

0.0 001 0.0203 786

83.64 0.6157 258

0.00 22

0.00 22

0.0 021

0.0 038

0.0 048

0.0 086

0 .0 001

0.027 161

72.21 0.713 1755

0.00 38

0.00 40

0.0 037

0.0 064

0.0 076

0.0 127

0 .0 003 0.0417 086

82.06 0.627 6092

0.00 36

0.00 40

0.0 038

0.0 071

0.0 091

0.0 166

0 .0 005 0.0478 198

65.45

0.5 41566

0.78 6827

f1

f2

f3

f4

f5

f6

f7

911.4680

405.7010

228.4311

147.8835

106.9981

83.7825

69.0485

758.5825

297.1806

180.6951

132.6602

105.2039

86.4466

72.8544

703.3981

375.4447

223.5922

151.6350

109.2816

84.8155

69.4563

577.6971

280.2990

180.7223

133.0951

105.2039

86.4466

72.8544

513.2971

331.1340

224.0000

155.7398

113.5223

87.3981

70.7068

277.8524

144.7845

115.0447

100.6097

89.4369

79.7864

71.2777

210.0544

185.6971

160.3612

135.7864

113.9029

95.3631

80.1398

130.6485

111.5379

98.2990

87.7243

78.9709

71.5223

64.9981 56.1903

96.6951

80.6835

73.3165

68.1786

63.8563

59.8874

142.4194

133.3942

122.4117

110.4505

98.6252

87.5068

77.3942

102.8932

95.2544

87.8602

80.7107

74.0777

67.9883

62.4427

77.9922

71.4951

66.7379

62.6330

58.9903

55.6194

52.4932

fi/f1/wm

S2

S

ESG est pa/Esg

1

0 .0 03 6

0 .0 01 9

0 .0 01 3

0 .0 00 9

0 .0 01 4

0 .0 01 5

0 .0 02 5

0 .0 00 0 0 .0 13 14 99

2

0 .0 036

0.0 017

0.0 013

0 .0 010

0 .0016

0 .0019

0 .0032

0 .0000 0.014 2237

95.09

3

0 .0036

0 .0023

0 .0017

0 .0012

0 .0018

0.0020

0.0033

0.0000 0.01 58765

92.28 0.55 80888

4

0 .0036

0 .0021

0.0016

0.0013

0.0021

0.002 5

0.004 2

0.000 0 0.0 17439 3

85.2 0 0.6 04433 9

5 0.00 36 6 0 .00 36 7 0.00 36 8 0 .00 36 0.00 36 9 10 0.00 36 0.00 36 11 12 0.00 36

0.00 27

0.00 23

0.0 017

0.0 026

0.0 028

0.0 046

0.0 001 0.0203 786

83.64 0.6157 258

0.00 22

0.00 22

0.0 021

0.0 038

0.0 048

0.0 086

0 .0 001

0.027 161

72.21 0.713 1755

0.00 38

0.00 40

0.0 037

0.0 064

0.0 076

0.0 127

0 .0 003 0.0417 086

82.06 0.627 6092

0.00 36

0.00 40

0.0 038

0.0 071

0.0 091

0.0 166

0 .0 005 0.0478 198

65.45

0.00 36

0.00 40

0.0 040

0.0 078

0.0 103

0.0 194

0 .0 006 0.0526 037

56.77 0.907 2368

0.00 40

0.00 45

0.0 044

0.0 081

0.0 102

0.0 181

0 .0 006 0.0530 141

78.60 0.655 2261

0.00 39

0.00 45

0.0 045

0.0 085

0.0 110

0.0 202

0 .0 007 0.0561 991

63.25 0.814 2393

0.00 39

0.00 45

0.0 046

0.0 089

0.0 119

0.0 224

0 .0 008 0.0597 842

53.28 0.966 5077

Adju sted Deflec tio ns (wip )

1

2

3

4

1 05 .5 7 0 .4 87 85 07

5

0.5 41566

0.78 6827

6

7

4 44 .6 6

1 97 .9 2

1 11 .4 4

7 2. 15

5 2. 20

4 0. 87

3 3. 69 - 0. 58 80 72 0 .1 57 78 09 0 .4 13 47 24 0 .5 87 74 25 0 .3 85 89 33 0 .3 18 77 73 - 0. 12 28 46

1 .1 5

4 10 .8 2

1 60 .9 4

9 7. 86

7 1. 84

5 6. 97

4 6. 82

3 9. 46 - 0. 46 72 23 0 .3 15 13 66 0 .4 84 95 61 0 .5 89 46 14 0 .3 29 70 77 0 .2 19 72 43 - 0. 31 51 82

1 .1 6

3 92 .5 6

2 09 .5 3

1 24 .7 8

8 4. 63

6 0. 99

4 7. 33

3 8. 76 - 0. 40 19 95 0 .1 08 37 68 0 .3 43 24 04 0 .5 16 42 42 0 .2 82 48 46

3 49 .1 8

1 69 .4 2

1 09 .2 3

8 0. 45

6 3. 59

5 2. 25

4 4. 04

3 16 .0 5

2 03 .8 9

1 37 .9 2

9 5. 89

6 9. 90

5 3. 81

4 3. 54 - 0. 12 87 51 0 .1 32 39 26 0 .2 74 09 16 0 .4 52 03 99 0 .1 77 66 31 0 .1 03 11 26

- 0. 45 12

0 .5 6

1 98 .1 6

1 03 .2 6

8 2. 05

7 1. 75

6 3. 78

5 6. 90

5 0. 83 0 .2 92 29 45 0 .5 60 60 96 0 .5 68 17 35 0 .5 89 98 64 0 .2 49 59 76 0 .0 51 63 81

- 0. 69 44 5

1 .6 2

1 31 .8 3

1 16 .5 5

1 00 .6 4

8 5. 22

7 1. 49

5 9. 85

5 0. 30 0 .5 29 17 12

0. 51 30 24 0 .1 58 98 22 0 .0 02 48 73 - 0. 67 65 49

1 .5 0

1 02 .8 0

8 7. 76

7 7. 34

6 9. 02

6 2. 14

5 6. 28

5 1. 14

0. 63 28 65 0 .6 26 54 89 0 .5 92 92 46 0 .6 05 57 81 0 .2 68 98 33 0 .0 62 07 17 - 0. 70 47 41

2 .0 8

8 7. 73

7 3. 20

6 6. 52

6 1. 85

5 7. 93

5 4. 33

5 0. 98

0. 68 66 95 0 .6 88 51 47 0 .6 49 91 88 0 .6 46 54 76 0 .3 18 43 77 0 .0 94 46 61 - 0. 69 92 63

2 .3 9

9 3. 32

8 7. 40

8 0. 21

7 2. 37

6 4. 62

5 7. 34

5 0. 71 0 .6 66 72 53 0 .6 28 07 09 0 .5 77 85 62 0 .5 86 45 69 0 .2 39 74 31 0 .0 44 38 77 - 0. 69 03 56

2 .0 5

8 3. 78

7 7. 56

7 1. 54

6 5. 72

6 0. 32

5 5. 36

5 0. 84 0 .7 00 78 68 0 .6 69 95 81 0 .6 23 47 78 0 .6 24 46 97 0 .2 90 38 88 0 .0 77 35 35 - 0. 69 47 77

2 .2 9

7 5. 38

6 9. 10

6 4. 50

6 0. 54

5 7. 01

5 3. 76

5 0. 74 0 .7 30 78 54 0 .7 05 95 49 0 .6 60 51 23 0 .6 54 08 41 0 .3 29 24 04 0 .1 04 05 68 - 0. 69 11 69

2 .4 9

3.00

2.50

2.00

1.50

0 .2 11 09 - 0. 29 20 93

0 .7 7

- 0. 24 70 7 0 .2 79 05 43 0 .4 25 08 05 0 .5 40 30 16 0 .2 51 89 65 0 .1 29 14 57 - 0. 46 78 55

0 .9 1

0. 50 40 63

0. 47 02 94

Adju sted Deflec tio ns (wip )

1

2

3

4

5

6

7

4 44 .6 6

1 97 .9 2

1 11 .4 4

7 2. 15

5 2. 20

4 0. 87

3 3. 69 - 0. 58 80 72 0 .1 57 78 09 0 .4 13 47 24 0 .5 87 74 25 0 .3 85 89 33 0 .3 18 77 73 - 0. 12 28 46

1 .1 5

4 10 .8 2

1 60 .9 4

9 7. 86

7 1. 84

5 6. 97

4 6. 82

3 9. 46 - 0. 46 72 23 0 .3 15 13 66 0 .4 84 95 61 0 .5 89 46 14 0 .3 29 70 77 0 .2 19 72 43 - 0. 31 51 82

1 .1 6

3 92 .5 6

2 09 .5 3

1 24 .7 8

8 4. 63

6 0. 99

4 7. 33

3 8. 76 - 0. 40 19 95 0 .1 08 37 68 0 .3 43 24 04 0 .5 16 42 42 0 .2 82 48 46

3 49 .1 8

1 69 .4 2

1 09 .2 3

8 0. 45

6 3. 59

5 2. 25

4 4. 04

3 16 .0 5

2 03 .8 9

1 37 .9 2

9 5. 89

6 9. 90

5 3. 81

4 3. 54 - 0. 12 87 51 0 .1 32 39 26 0 .2 74 09 16 0 .4 52 03 99 0 .1 77 66 31 0 .1 03 11 26

- 0. 45 12

0 .5 6

1 98 .1 6

1 03 .2 6

8 2. 05

7 1. 75

6 3. 78

5 6. 90

5 0. 83 0 .2 92 29 45 0 .5 60 60 96 0 .5 68 17 35 0 .5 89 98 64 0 .2 49 59 76 0 .0 51 63 81

- 0. 69 44 5

1 .6 2

1 31 .8 3

1 16 .5 5

1 00 .6 4

8 5. 22

7 1. 49

5 9. 85

5 0. 30 0 .5 29 17 12

0. 51 30 24 0 .1 58 98 22 0 .0 02 48 73 - 0. 67 65 49

1 .5 0

1 02 .8 0

8 7. 76

7 7. 34

6 9. 02

6 2. 14

5 6. 28

5 1. 14

0. 63 28 65 0 .6 26 54 89 0 .5 92 92 46 0 .6 05 57 81 0 .2 68 98 33 0 .0 62 07 17 - 0. 70 47 41

2 .0 8

8 7. 73

7 3. 20

6 6. 52

6 1. 85

5 7. 93

5 4. 33

5 0. 98

0. 68 66 95 0 .6 88 51 47 0 .6 49 91 88 0 .6 46 54 76 0 .3 18 43 77 0 .0 94 46 61 - 0. 69 92 63

2 .3 9

9 3. 32

8 7. 40

8 0. 21

7 2. 37

6 4. 62

5 7. 34

5 0. 71 0 .6 66 72 53 0 .6 28 07 09 0 .5 77 85 62 0 .5 86 45 69 0 .2 39 74 31 0 .0 44 38 77 - 0. 69 03 56

2 .0 5

8 3. 78

7 7. 56

7 1. 54

6 5. 72

6 0. 32

5 5. 36

5 0. 84 0 .7 00 78 68 0 .6 69 95 81 0 .6 23 47 78 0 .6 24 46 97 0 .2 90 38 88 0 .0 77 35 35 - 0. 69 47 77

2 .2 9

7 5. 38

6 9. 10

6 4. 50

6 0. 54

5 7. 01

5 3. 76

5 0. 74 0 .7 30 78 54 0 .7 05 95 49 0 .6 60 51 23 0 .6 54 08 41 0 .3 29 24 04 0 .1 04 05 68 - 0. 69 11 69

2 .4 9

0 .2 11 09 - 0. 29 20 93

0 .7 7

- 0. 24 70 7 0 .2 79 05 43 0 .4 25 08 05 0 .5 40 30 16 0 .2 51 89 65 0 .1 29 14 57 - 0. 46 78 55

0 .9 1

0. 50 40 63

0. 47 02 94

3.00

2.50

2.00

1.50

1.00

0.50

0.00 1

2

3

4

5

6

7

8

9

10

11

12

9.7/9.8

The following FWD measurements were obtained on a Portland concrete slab 0.30 m thick under a load of 40 kN and a plate radius of 15 cm. Determine the modulus of subgrade reaction and the elastic modulus of the slab assuming a liquid foundation and a Portland concrete Poison’s ratio, µ of 0.15. For the data given in the previous question, determine the elastic moduli of the subgrade and the slab, assuming a solid foundation and a subgrade Poison’s ratio of 0.40. Sensor number

0

1

2

3

Offset (m)

0

0.3

0.6

0.9

9.7/9.8

The following FWD measurements were obtained on a Portland concrete slab 0.30 m thick under a load of 40 kN and a plate radius of 15 cm. Determine the modulus of subgrade reaction and the elastic modulus of the slab assuming a liquid foundation and a Portland concrete Poison’s ratio, µ of 0.15. For the data given in the previous question, determine the elastic moduli of the subgrade and the slab, assuming a solid foundation and a subgrade Poison’s ratio of 0.40. Sensor number

0

1

2

3

Offset (m)

0

0.3

0.6

0.9

Deflection (µm)

98

80

75

60

The normalized deflection calculations are shown below: Sensor, s

0

1

2

3

Offset (m)

0

0.3

0.6

0.9

98

80

75

60

Deflection (m)

0.000098

0.00008

0.000075

0.00006

Deflection (in)

0.00385827

0.00314961

0.00295276

0.0023622

1

0.81632653

0.76530612

0.6122449

Deflection (µm)

Normalize

Equation 9.41 gives the AREA as 28.65 in, which through Figure 9.30 gives radii of curvature equal to 30 and 21 inches for liquid and solid foundation assumptions, respectively. The estimate d i values are obtained from Figures 9.31 and 9.32. The calculations are summarized below. Estimates of di liquid predcitting k (pci)

0.122

0.11

0.09

0.07

316.20

349.25

304.80

296.33

prediciting E (psi)

3,157,715 21772.44

Estimates of di solid Predicting Es Preidicting E

0.19 0.163 33662.41 35376.53 2,366,356 16316.02

0.14 0.107 32410.4 30963.51

316.65

33,103.21 228.2466

The resulting modulus of subgrade reaction is 316.6 pci and the resulting subgrade stiffness is 33,103 psi 9.9

A distress survey conducted on 250 m2 of flexible pavement surface produced the following results. Compute the PCI.

Distress

Extent: Area/length affected

Low severity fatigue cracking

15 m

High severity longitudinal cracking

10 m

Medium Low severity rutting

8 m2

2

Compute ratios of distress extent by dividing the coverage o f each distress by 250: 15 6.00% 0.22 0.29

DV (Figs 9.35-9.37) rearrange DV

10 4.00% 0.15 0.22

8 3.20% 0.29 0.15

Compute the maximum number of deduct values m using Equ. 9.43: m = 1+

9 98

(100 − HDV ) =

m = 1+

9 98

(100 − 29) = 7.52 rounded to 7, which

allows using all 3 distresses present. Computations are tabulated below: TDV Summing distresses

0.29 0.29 0.29

0.22 0.22 0.02

0.15 0.02 0.02

The PCI is computed form Equ. 9.44 as: 100-41=59.

q 0.66 0.53 0.33

3 2 1

CDV from Fig 9.38 0.4 0.41 max 0.32

9.10

A coefficient of friction of 0.35 was measured with an ASTM E-274 tester equipped with treaded tires. The sand-patch test macrotexture of this pavement, (i.e., MPD), was 0.7 mm. Calculate the value of the normalized coefficient of friction at 60 km/h, give the IFI and plot the coefficient of friction as a function of slip speed.

The constants for the ASTM E-274 Trailer with treaded tires are sho wn in Table 9.23. Its reference slip speed is 65 km/h. Equation 9.49 gives the speed constant S p 89.7x0.7+14.2= 77 km/h. Equation 9.46 gives the normalized coefficient of friction at 60 km/h as: 65− 60

f (60) = −0.023 + 0.607 0.35 e

77

+ 0.098 0.7 = 0.273

Equ. 9.47 allows calculating the coefficient of friction at any speed as: 60 − S

f ( S ) = f (60) e

S p

60 − S

= 0.273 e

77

. The resulting IFI is (77, 0.273).

Chapter 10 10.1

According to equation (10.6) the total discharge rate for this pavement is: 14.4   2 +  = 0.105 m / day  14.4 14.4 ⋅ 3 

q = 0.223

Figure 10.3 gives a precipitation rate of 1.4in/hour(0.853 m/day) which according to equation (10.5) the infiltration rate for Portland cement concrete is calculated to be between 0.43 to 0.57 m/day. 10.2

Equation (10.9) gives the maximum water infiltration rate that can be drained through this layer as: 2

2

 h   0.4   ⇒ q = 2  = 0.015 m / hour  b   4.6  The actual speed of pore water movement is obtained by dividing the discharge speed by the effective porosity, that is 0.015/0.28 = 0.054 m/hour. According to this speed the required time for one drop of water to transverse the full length of drainage path under steady state conditions would be determined as 4.6/0.054 = 85.17 hours. q = k 

10.3

The length of the drainage path could be calculated by use of equation (10.15) as: AD =

3.8

(0.03 0.03

2

+ 0.06 2

)

0.5

= 8.5 m

Use the Equation (10.16) to compute the difference in elevation between A and D: ∆h AD =

3.8

(0.03 0.03

2

)

+ 0.06 2 = 0.57 m

Which gives a combined grade for the flow path of 0.0670, or 6.7%. This value is confirmed using Equation 10.17:

(

g f = 0.03 2 + 0 .06 2

)

0.5

= 0.0670 m

The maximum rate of flow discharge for this layer is computed from Darcy’s law as: q = 2 . 0.067 = 0.1342 m/hour 10.4

Use the Equation 10.18 to compute the gradient factor S1 as 4.5×0.044/0.35 = 0.57. Use the Figure 10.6 with a 0.57 gradient factor to obtain a time factor t/m of 0.35. Use the Equation (10.19) to solve for actual time :

0.35 =

0.2 ⋅ 0.35 ⋅ t 0.30 ⋅ 4.5 2

⇒ t = 30.38 hours or 1.27 days

According to table 10.1 this layer would be ranked as a good drainage quality.

10.5

Figure 10.11 allows obtaining the characteristics grain size p ercentiles shown in the table below: Size D85 D50 D15

Soil (mm) 4.1 0.8 0.1

Base Layer (mm) 21 13 6.5

The criteria suggested by equations 10.23 to 10.25 are tested by substituting in the values from table 10.2 as: D15, filter D85, soil D15, filter D15, soil D50, filter D50, soil

=

6.5

=

6.5

=

13

4.1

0.1

0.8

= 1.59 < 5 = 65 > 5 = 16.25 < 25

Which are all satisfied; hence, no intermediate filter layer is required b etween this base and the subgrade layer. 10.6

The heat diffusivity of the asphalt concrete and the subgrade are computed as f: k

ρ c

=

1.4 3

2.65 × 10 ⋅ 0.5 × 10

3

= 1.06 ×10 − 6 m 2 / sec or 0.0038 m 2 / hr

and : k

ρ c

=

1.1 3

2.4 × 10 ⋅ 0.85 × 10

3

= 0.54 ×10 − 6 m 2 / sec or 0.0019 m 2 / hr

∆t is small. In this problem a ρ c ∆ z 2 depth increment of 0.05 and time increment of 0.1 hour is selected that results in : A small time and Δz is selected such that the ratio

k

ρ c k

ρ c

∆t = 0.0038 ⋅ 40 = 0.152 ∆ z 2 ∆t = 0.002 ⋅ 40 = 0.08 ∆ z 2

k

for the asphalt concrete and subgrade layers, respectively. Applying Equation 10.31 is applied afterwards. The following curve shows the results of the analysis.

Depth(m) 0.00 0.05 0.10 0.15 0.20 0.25 0.30 0.35

0 25.00 12.50 12.50 12.50 12.50 12.50 12.50 12.50

0.1 25.00 14.40 12.50 12.50 12.50 12.50 12.50 12.50

0.2 25.00 15.72 12.79 12.50 12.50 12.50 12.50 12.50

0.3 25.00 16.69 13.19 12.54 12.50 12.50 12.50 12.50

0.4 25.00 17.42 13.62 12.64 12.51 12.50 12.50 12.50

0.5 25.00 17.99 14.05 12.77 12.53 12.50 12.50 12.50

0.6 25.00 18.46 14.45 12.92 12.56 12.50 12.50 12.50

Tempreature(C) 0.00

10.00

20.00

30.00

40.00

0.00

0.20

0.40

0.60 5 hr ) m ( h t 0.80 p e D

10hr 15hr 20 hr 25 hr

1.00

1.20

1.40

1.60

10.7

Using Equation 10.29 to compute the latent heat of the soil: L = 1.43 . 12.140 = 2402.4 BTU/ft3 or 89,510 kJ/m3 Use Equation 10.34 and 10.35 to obtain the fusion parameter μ and the thermal ratio α, respectively:

µ =

α =

C avg FI nL V 0 n FI

=

=

35.8 ⋅ 850 85 ⋅ 2402.4

(40 − 32)85 850

= 0.15

= 0.8

Entering the parameter into Figure 10.9 gives a value of λ of 0.84, which allows computing the depth of the frost penetration form Equation 10.33 as: z = λ

48 k FI L

= 0.84

48 ⋅ 0.43 ⋅ 850 2402.4

= 2.27 ft (0.7m)

Chapter 11 11.1

Calculate the required layer thicknesses of a new asphalt concrete pavement on a fair draining base and subgrade (the water drains out the pavement within a period of two days). It is estimated that the pavement structure becomes saturated less frequently than 5% of the time. The following data is also given:

• Estimated number of ESALs over a 15-year maximum performance period = 3 • • • •

million Subgrade resilient modulus = 25,000 lbs/in2 (172.4 MPa) Design reliability = 95% Standard error in predicting serviceability = 0.45 ∆ PSI = 2.5 (from 4.5 to 2.0)

Solution

The AASHTO 1993 design guide is used to design the pavement. A classical flexible pavement composed by an asphalt surface layer, an unbounded granular base and an unbounded granular subbase will be designed. The following table summarizes the information required for the design: Given information: Traffic Period Subgrade resilient modulus Design reliability Standard error in predicting serviceability ∆ PSI Other required information: Standard Normal Deviate, Z R Drainage coefficients for untreated granular base and subbase Asphalt concrete layer coefficient (a1) Modulus of unbounded aggregate base ( E 2) Base layer coefficient (a2)† Modulus of unbounded aggregate base ( E 3)

3 million ESALs 15 years 25,000 lbs/in2 (172.4 MPa) 95% 0.45 2.5 (from 4.5 to 2.0)

-1.645 1.25

From Table 11.1 Good water removal conditions

0.44

Recommended

60,000 psi (414 MPa)

Assumed

0.13 40,000 psi (276 MPa)

Based on E 2 Assumed

83

Subbase layer coefficient (a3) †

0.10

Based on E 3

†

layer coefficients vary according to the modulus of the layer. For detailed information and recommended values consult reference 2.

The moduli of the layer materials were assumed based on experience. In reality these values are known from tests results performed on the material that is available near the project area. The value of SN for this pavement is obtained from the performance equation (Equation 11.1):

 2.5  4.2 − 1.5   6 + 2.32 log(25000) − 8.07 log(310 ) = −1.645 0.45 + 9.36 log(SN + 1) − 0.20 + 1094 0.4 + (SN + 1)5.19 log 

From which it is found that the structural number, SN , for the whole structure is 2.69. However, as mentioned in example 11.1, there is not a unique combination of D1, D2 and D3 that satisfies the design equation: 2.69 = 0.44 D1

+ 0.14 1.25 D2 + 0.10 1.25 D3

Then, a sequential design process is conducted. Initially, the thickness of the asphalt concrete layer ( D1) is determined based on the structural number that satisfies equation 11.1 (SN 1) when it is assumed that the layer is supported on a subgrade with the modulus of the unbounded aggregate base (E 2). Then, the thickness of the base is found using the structural number calculated with the modulus of the subbase (SN 2 calculated with E 3). Finally, the thickness of the subbase is computed using the original design equation with the structural number for the whole structure (SN = 2.69). The following tables summarizes the results and presents the final design. Modulus and Layer Coefficients for the Pavement’s Layers Layer

Modulus (psi)

Layer coefficient

Asphalt concrete Base Subbase Subgrade

60,000 40,000 25,000 = M R

0.44 0.13 0.11 ---

Design Procedure to Obtain the Thicknesses of the Layers. SN

(Equation 11.1)

Equation for D (in)

Design value of D

SN1 = 1.96

D1 = 1.96/0.44 = 4.45

5 in

84

SN2 = 2.28 SN3 = 2.69

D2 = (2.28 – 5 0.44)/ (0.13 1.25) = 0.49 D3 = (2.69 – 5 0.44 -1 0.1625 )/(0.10 1.25) = 2.62

1 in 3 in

The final design satisfies the serviceability performance criterion. Note that this is only one possible design, and that the method provides the possibility to design many other possibilities (e.g a pavement with an asphalt concrete layer and only a granular base, a pavement with stabilized layers, etc.) Design engineers should analyze different possibilities, and based on their experience and taking into account minimal technical specifications, construction issues and economical considerations determine the optimal design for this particular project. 11.2

Compute the anticipated life of the pavement designed in problem 11.1 considering the combined effects of traffic and environment. The subgrade is fair draining clayey sand (i.e., designated as SC according to the UC system), having less than 10% by weight finer that 0.02mm, and a Plasticity Index of 10%. The subgrade layer is 10 ft deep; it is exposed to high moisture levels and exhibits a medium level of structural fracturing. The percent of the pavement surface subjected to swelling is estimated to be 20%, while the probability of frost is estimated at 30%. The depth of the frost penetration is 4 ft. Solution

To calculate the anticipated life of the pavement it is first necessary to compute the serviceability loss due to the combined effects of traffic and environment. From problem 11.1 it is known that the total serviceability loss for the pavement is 2.5 ( ∆PSI=2.5). That means that the combined loss due to traffic (∆PSI traffic) and environment (∆PSI envir ) must sum up 2.5. The calculation of the actual life in years to terminal serviceability is based on an iterative process. In this process different period of years are selected, and the total loss of serviceability due to the environment is calculated for each period. The available traffic serviceability loss for that number of years is computed as the difference between the total available serviceability loss (2.5) and the environmental loss. The net PSI that results from this calculation is used in Equation 11.1 to obtain the number of accumulated ESALs that are required to obtain that specific loss in serviceability due to traffic load. The number of ESALs resulting from this equation is then used to estimate the cumulative period in years at which it is expected to reach the traffic level. This step requires a traffic growth rate value. In this exercise, it is assumed that the traffic load (i.e. number of ESALs) growths at a constant annual rate of 3.5%, as shown in the following Figure. The pavement life is obtained when the period of years used in the initial calculation of ∆PSI envir equals the number of years associated with the ESALs calculated from equation 11.1 using the available loss in serviceability for traffic (∆PSI traffic).

85

) s n o i l l i m ( s L A S E d e t a l u m u c c A

3.5 3 2.5 2 1.5 1 0.5 0 1

2

3

4

5

6

7

8

9

10 11

12 13

14 15

16 17

Age (yea rs)

Accumulation of ESALs versus Time for Problem 11.2.

The serviceability loss due to the environment requires the computation of the loss due to the subgrade swelling (∆PSISW) and the loss due to frost heave ( ∆PSIFH). ∆PSISW is computed using Equation 11.2. For this particular case, the low plasticity index of the subgrade makes the value of the potential vertical rise due to swelling (V R , Figure 11.3) to be almost 0. Therefore, ∆PSISW = 0, and ∆PSI envir = ∆PSI FH.

∆PSIFH is computed using Equation 11.3. According to Figure 11.5, the maximum serviceability loss due to frost heave ( ∆PSImax) for a depth of frost penetration of 4 ft and good drainage conditions is 0.8. The frost heave rate, φ, is found to be 1.6 mm/day (Figure 11.6). Therefore, the loss of serviceability due to frost heave as a function of the period of service life in years (t ) is equal to:

∆PSI FH = ∆PSI envir = 0.01 p f ∆PSI max (1 − e −0.02φ t ) = 0.01 30 0.8 (1 − e −0.02 1.6 t ) = 0.0915 The following table presents the results for the calculation of ∆PSI envir for different periods of service life, as well as the calculation of the net PSI available for traffic, the ESALs associated with that level as serviceability (equation 11.1) and the period in years required to accumulate that amount of ESALs. For example, after 10 years PSI envir = 0.0657, therefore, ∆PSI traffic = 2.5-0.0657 = 2.43. The number of ESALs corresponding to a loss in serviceability of 2.43 is obtained from Equation 11.1:

86

log(W 18 ) = −1.645 0.45 + 9.36 log(2.69 + 1) − 0.20 +

 2.43   4.2 − 1.5 

log  0.4 +

+ 2.32 log(25000) − 8.07

1094

(2.69 + 1)5.19

That provides a value of W 18 of 2.95 millions of ESALs. This number of ESALs are reached after a period of 14.9 years.

Summary of Serviceability Loss Computations Due to Traffic and Environmental Factors. Iteration

Performance period (years)

PSI loss Due to Environment

1 2 3 4 5 6 7 8 9

8 9 10 11 12 13 14 15 16

0.05421 0.06006 0.06572 0.07121 0.07653 0.08168 0.08666 0.09149 0.09617

Net PSI available for traffic 2.5-0.0542 2.5-0.0601 2.5-0.0657 2.5-0.0712 2.5-0.0765 2.5-0.0817 2.5-0.0867 2.5-0.0915 2.5-0.0962

= = = = = = = = =

2.4458 2.4399 2.4343 2.4288 2.4235 2.4183 2.4133 2.4085 2.4038

ESALs obtained with Equation 11.1 (millions) 2.98 2.98 2.97 2.97 2.96 2.96 2.96 2.95 2.95

Period to Acc um ul ate Corresponding ESALs (years) 14.9 14.9 14.9 14.9 14.9 14.8 14.8 14.8 14.8

The data in Table 11.2.1 suggests that the overall life for this structure under the combined effect of traffic and environment is 15 years. Note that for this particular example, the loss in serviceability due to the environment is very low, and therefore, the overall life of the pavement depends mainly on the traffic conditions.

11.3

Design an asphalt concrete pavement with an untreated granular base to accommodate 2.5 million ESALs without failing in fatigue cracking or rutting. The subgrade resilient modulus is 80 MPa, and the MAAT is 15.5°C. (Note: You need to obtain the proper chart from Reference 3). Solution

The Asphalt Institute (AI) design method is used to find the thicknesses of layers. Based on the traffic level and the quality of the subgrade it is concluded that a granular base is required as part of the design. Therefore, a 10 inches (0.254 m) untreated granular base layer is selected.

87

The appropriate chart from reference 3 that corresponds to a MAAT of 15.5°C and an unbounded granular base of 10 inches is required for this design. Using this chart, it is found that a pavement subjected to 2.5 million ESALs, and having a subgrade with resilient modulus of 80 MPa requires 8 inches (0.203 m) of an asphalt concrete layer. Note that the method provides other possibilities for the pavement design. As an example, the following table presents the results when unbounded granular bases of 12 or 18 inches are selected instead of the 10 in base (note: different charts are required). For this particular case, it is observed that a change of 8 inches in the granular base produces only a change of 0.5 inches in the asphalt concrete base. The final design would be selected based on economical and technical considerations. Three Possible Designs using the Asphalt Institute Method Layer Asphalt concrete Unbounded granular base Subgrade

Design 1 8 in 10 in

Design 2 7.8 in 12 in

Design 3 7.5 in 18 in

Note: 1 in = 2.54 cm

11.4

A pavement section has accumulated a total of 20% bottom-up fatigue damage. Estimate its fatigue cracking, given that the asphalt concrete layer has a thickness of 15 cm (5.9 in). Solution

The coefficient c2’ is computed from Equation 11.13: c 2' = −2.40874 − 39.748 (1 + 5.9 )

−2.856

= −2.569

This parameter is used in combination with the accumulated bottom-up fatigue cracking (FD = 20%) in Equation 11.12 to obtain the fatigue cracking (FC ), as follows: FC =

100 1 + e − 2.569 (− 2+ log 20 )

= 14.24%

11.5

Compute and plot the plastic strain accumulated at the mid-depth of an asphalt concrete layer 0.23 m (9 in) thick at a temperature of 75°F after 1, 10, 100 and 200 load cycles.

88

Also compute the plastic deformation after 200 load cycles. The elastic vertical strain is 120 10-6. Solution

Equation 11.18 is used to compute C 1 and C 2, the coefficients that are required in equations 11.17 and 11.16: C 1

= −0.1039 9 2 + 2.4868 9 − 17.342 = −3.3767

C 2

= 0.0172 9 2 − 1.7331 9 + 27.428 = 13.2233

These values provide a value of the calibration factor that accounts for the increased in confinement with depth (k 1) at the middepth of the layer of 0.373 (Equation 11.17): k 1

= (−3.3767 + 13.2233 4.5) 0.328196 4.5 = 0.373

Then, the plastic strain is calculated as a function of the number of loading cycles as (Equation 11.16): ε p = (0.373 10 −3.4488 T 1.5606 N 0.479244 )ε v = (0.373 10 −3.4488 751.5606 N 0.479244 )120 10 −6

= 1.3437 10 −5 N 0.479244 The plot of plastic strain versus the number of loading cycles is presented in the following figure

4.0E-04 3.5E-04 3.0E-04 n i a r 2.5E-04 t s c 2.0E-04 i t s a 1.5E-04 l P

1.0E-04 5.0E-05 0.0E+00 0

200

400

600

800

1000

1200

Number of strain cycles

Plastic Strain as a Function of the Number of Strain Cycles for the Asphalt Concrete Layer

89

According to equation 11.16, after 200 cycles the plastic strain is 1.70 10-4. The total plastic deformation in the asphalt layer is equal to the plastic deformation multiplied by the total thickness of the asphalt concrete layer: 1.70 10-4 x 9 = 1.53 10-3 in (0.0389 mm).

11.6

Compute the plastic strain and the plastic deformation in a 14-in thick (35.5 cm) granular base layer with a moisture content of 6% after 1000 cycles, at a compressive strain level of 180 10-6. Treat the base as a single layer. Solution

Equations 11.21, 11.22 and 11.23 can be used to determine β and ρ and the ratio ε 0 /ε r , the parameters that are required to compute the plastic strain in the granular base layer: log β = −0.6119 − 0.017638 W c

= −0.6119 − 0.017638 × 6 = −0.7177 ,

then

β = 0.1915 1

1

β

0.1915

 − 4.89285     = 10 9  − 4.89285   1 − (10 9 ) β   1 − (10 9 )0.1915     

ρ = 10 9 

 ρ    ε 0 1   ( ρ ) = 0.15 e + 20 e  10  ε r 2  

β

β

9

= 4.406 10 3

 1   =  0.15 e (4.406 10 )  2   

3 0.1915

+ 20 e

 4.406 103     109   

0.1915

   = 21.975  

The plastic strain is computed from equation 11.20 (assuming a single layer), using the compressive strain level (180 10-6), the β G value (1.673 for granular base) and N , the number of loading cycles (1 103): ε p

 ε  = β G  0  e  ε r 

β

 ρ  −   N 

ε v

= 1.673 × 21.975 e

 4.406 103   − 3   10 

0.1915

180 10 − 6

= 1.753 10 −3

Therefore, the plastic deformation in the granular layer is: 1.753 10-3 * 14 = 2.454 10-2 in (0.62 mm).

11.7

The constant describing the master curve of the creep compliance of an asphalt concrete at -20°C is given in Table 11.6. Compute its fracture parameters and the growth of an existing 5.1 cm (2 in) deep transverse crack caused by an increase in stress at its tip of 334.7 kPa (50 lbs/in2). The tensile strength of the asphalt concrete is 4.0 MPa (580 lbs/in2).

90

Constants Defining the Creep Compliance of the Asphalt Concrete Constant Units Value 2 1/lbs/in. 2.8 10-07 D0 1/lbs/in.2 5.4 10-08 D1 1/lbs/in.2 9.5 10-08 D2 1/lbs/in.2 4.0 10-08 D3 D4 1/lbs/in.2 3.0 10-07 T0 sec 13,000 T1 sec 0.9 T2 sec 1.9 T3 sec 2.8 T4 sec 3.6 Solution Table 11.6 provides the information to characterize the creep compliance of the asphalt concrete in terms of the reduced time, ξ, using a generalized Voight-Kelvin model (Equation 11.24) D(ξ ) = D0 +

D0ξ

= 2.8 10

T 0 −7

+

N

+ ∑ Di (1 − e −ξ / T ) = i

i =1

2.8 10 −7 ξ 13,000

+ 5.4 10 −8 (1 − e −ξ / 0.9 ) + 9.5 10 −8 (1 − e −ξ / 1.9 )

+ 4.0 10 −8 (1 − e −ξ / 2.8 ) + 3.0 10 −7 (1 − e −ξ / 3.6 ) To compute the fracture parameter n, the linear part of master curve (in log-log scale) is fitted to an exponential function: D(ξ ) = D0 + D1ξ m . The parameters D0, D1, and m are obtained by minimizing the sum of square errors between the estimated and the actual compliance values: D0 = 5.8 10-7 (1/psi), D1 = 6.70 10-7 (1/psi) and m = 0.192. Equation 11.28 is used to determine n based on the m-value:

 

n = 0.8 1 +

1  1    = 0.8 1 +  = 4.967 m   0.192 

Master curve of the Creep Compliance for the Asphalt Concrete (-20°C)

91

1.00E-06 ) i s p / 1 ( e c n a i l p m o C p e e r C

1.00E-07 1

10

100

1000

10000

Time (sec)

Obtaining the Slope m of the Creep Compliance Master Curve after Fitting an Exponential Function (-20°C)

1.00E-06 ) i s p / 1 (

slope of the curve:

e c n a i l p m o C p e e r C

D

m

= D + D

-m

1.00E-07 10

100

1000

Time (sec)

The n-value and the tensile strength of the material (580 lbs/in2) are substituted in Equation 11.29 to compute the fracture parameter A:

log A = 4.389 − 2.52 log (10000 S t n ) = 4.389 − 2.52 log (10000 580 4.967 ) = −14.41 Then, −14.41 A = 10 = 3.899 10 −15

92

Equation 11.27 provides the change in the stress intensity factor:

(

)

(

)

K = σ 0.45 + 1.99C 00.56 = 50 0.45 + 1.99 2 0.56 = 169.19

Finally, Equation 11.26 gives the resulting increase in crack propagation:

∆C = A∆K n = 3.899 10 −15 169.19 4.967 = 4.56 10 −4 inches (0.0l159 mm)

93

Chapter 12 12.1

The drainage coefficient and the load transfer coefficient for the conditioned specified are obtained as 1.1 and 3.1. from the table 12.1 and 12.2, respectively. The value of the standard normal deviate for the 95% is -1.645. Substituting this values into equation 12.1 gives :  2.5  log 0.75 525 ⋅1.1( D − 1.132)  4.5 − 1.5  + (4.22 − 0.32 ⋅ 2) × log log(5,000,000) = −1.654 ⋅ 0.40 + 7.35 log( D + 1) − 0.06 + 7  0.75  1.624 ⋅10 18.42 1+ − 215.63 ⋅ 3.1 D  8.46 ( D + 1) (4,000,000 / 140)0.25   which, solved for the D gives a value of 9.85in(25 cm)to be rounded up to 10in.(26 cm).

12.2

First estimate the serviceability loss due to subgrade swelling. Use Figures 11.2 and 11.3 to compute the potential swelling rise, VR , as 0.78 in and swelling rate θ as 0.15, respectively. Substituting these values into Equation 12.2 gives: ΔPSIsw = 0.00335×0.78×40(1-e-0.15×30) = 0.103 Next estimate the serviceability loss due to subgrade frost heave. Use the figure 11.4 and 11.5 to compute the maximum loss ΔPSImax as 0.85 and the frost heave rate as 5 mm/day, respectively. Substituting these values into Equation 12.3 gives: ΔPSIFH = 0.01×50×0.85×(1-e-0.02×5×30) = 0.403 Adding the two serviceability loss components give the total serviceability loss due to environment, ΔPSISWFH = 0.506 12.3

Given the ESAL growth assumption, the accumulated ESAL versus pavement age relationship is plotted in the figure below. Select a performance period shorter than the stipulated 30 years, say 20 years, for this problem(i.e., it is assumed that at the end of this performance period the pavement section will need rehabilitation). For the selected performance period, compute the loss in serviceability due to environment ΔPSIsw = 0.00335×0.78×40(1-e-0.15×20) = 0.103 ΔPSIFH = 0.01×50×0.85×(1-e-0.02×5×20) = 0.403 ΔPSISWFH = 0.467 Subtract this value from the maximium performance period serviceability loss (4.5-2 = 2.5) to obtain the serviceability loss available for the traf fic alone, ΔPSITR, which after 20 years is : ΔPSITR = ΔPSI – ΔPSISWFH = 2.5 – 0.403 = 2.097 Compute the ESALs that will bring about this serviceability loss using Eq uation 12.1:

94

6

5

) s 4 n o i l l i m ( s L A S E 3 d e t a l u m u c c A 2

1

0 0

5

10

15

20

25

30

35

Age(y ears)

 2.097  0.75  − 1.132) 525 ⋅ 1.1(10  4.5 − 1.5  + (4.22 − 0.32 ⋅ 2) × log log(W t ) = −1.654 ⋅ 0.40 + 7.35log(10 + 1) − 0.06 + 7  0.75  1.624 ⋅ 10 18.42 1+ ⋅ − 215 . 63 3 . 1 10   (10 + 1)8.46 (4,000,000 /140)0.25   log

Solving this equation gives Wt = 4,560,163 ESALs. This level of ESAL is accumulated over a period of 25.6 years. The environmental and serviceability loss calculations are repeated for another trial period, until a time period is found over which the sum of serviceability losses due to traffic an environment add to 2.5.The results are tabulated in table below. It is evident that the pavement section remains serviceable for approximately 25 years under the combined effect of traffic and environment.

iteration 1 2 3

Performance period (years) 20 25 25.5

PSI Loss due to environment 0.403 0.492 0.494

Net PSI available for traffic 2.097 2.008 2.006

Period to accumulate corresponding ESALs (years) 25.6 25.4 25.4

95

12.4

Figure 12.2 is used to compute the tensile stress in the slab due to 22,000 lb l oad, σw as 170 lbs/in2. Table 12.5 for a limestone aggregate gives a value of 3.8×10-6/oF for the coefficient of the Portland concrete, αc,. The tensile strength of the concrete suggests a shrinkage coefficient z of 0.00045 from A Table 12.4. Selecting as a trial a No.5 steel bar gives a limiting steel stress of 61,000 lbs/in 2 from Table 12.6. Substituting the available values in Equations 12.4, 12.5, and 12.6 allows solving for the minimum percent steel reinforcement, ρ: 6.70

1.15

5   500   1 1.321 +  (1 + 0.625)2.19   +  1000   2 ⋅ 3.8  8= 5.20  1 + 170  (1 + ρ )4.6 (1 + 1000 ⋅ 0.00045)1.79    1000  6.53

 500  1 0.625 2.20 0.009321 + )  (+ 1000   0.04 = 4.91  170   (1 + ρ )4.55 1 +  1000  0.425

4.09

 65   1 500  473001 +   +  100  1000    61000 = 3.14  170  1  +  (1 + 1000 ⋅ 0.00045)0.494 (1 + ρ )2.74  1000  The calculated values for the minimum percent are 0.26, 0.39, and 0.41, respectively which yield a minimum percent steel reinforcement pmin of 0.41. The maximum percent steel reinforcement value is obtained by substituting the minimum crack spacing value equal to 3.5 ft into Equation 12.4. The corresponding maximum percent steel reinforcement value is 0.51. Since pmax is larger than pmin the design is feasible. Utilizing the Equation 12.8 and 12.9 gives the number of reinforcement bars required: Nmin = 0.01273 × 0.41 ×12×12/0.6252 = 1.92 No.5 bars Nmax = 0.01273 × 0.51 ×12×12/0.6252 = 2.39 No.5 bars Finally, select two No. 5 bars per foot-width of slab, that gives 6-in center to center bar spacing 12.5

Compute the total number of heavy trucks in the design lane over the 25 years life: 20000×365×25×0.50×0.75×0.15 = 10,265,625 Use the load frequency distribution given in table 12.8 to calculate the actual number of axle passes, by configuration as shown in table below: Ax le l oad (kN) 125-133

Single axle Ac tu al n um ber of axl e pas ses 5,954

96

115-125 107-115 97.8-107 88.8-97.8 80-88.8

13,859 28,436 60,773 100,911 222,456

Ax le l oad (kN) 213-231 195-213 178-195 160-178 142-160 125-142

Tandem axle Ac tu al n um ber of axl e pas ses 20,121 40,447 117,849 351,803 835,827 878,122

Select a trial slab of 220 mm to test the fatigue damage. For 220 mm slab Table from Ref. 17) gives ES value of 1.04 MPa and 1.21 MPa for single and tandem axles, respectively. Using Equation 12.10 and a modulus of rupture of 4.0 MPa allows computing SRF values of 1.21/4 = 0.30 and 1.03/4 = 0.26, respectively. Entering these SRF values into Figure 12.3 allows computing the number of axle passes to fatigue failure. In doing so, it was conservatively elected to use the upper limit of the load interval. Finally, the damage ratios caused by each axle load interval are computed and summed for all load intervals and axle configurations. The results tabulated in table below show that 220 mm is perhaps too thick for the traffic loads specified. Before testing a lesser slab thickness for fatigue, however, it is prudent to make sure that the 220mm slab can pass the erosion test as well. Single axle Number of passes to failure Ax le l oad (kN) 125-133 115-125 107-115 97.8-107 88.8-97.8 80-88.8

Ax le l oad (kN) 213-231 195-213 178-195 160-178 142-160 125-142

Fig ur e(12.7) 1,300,000 10,000,000

Tandem axle Number of passes to failure Fig ur e(12.7) >10,000,000

Ac tu al n um ber of axl e pas ses 5,954 13,859 28,436 60,773 100,911 222,456

Ac tu al n um ber of axl e pas ses 20,121 40,447 117,849 351,803 835,827 878,122 Total accumulated fatigue damage

Damage ratio(%) 0.46% 0.14% 0 0 0 0

Damag e rat io 0 0 0 0 0 0 0.60%

97

For a 220mm JDRCP slab with concrete shoulders on a subgrade with a 60 MPa/m reaction modulus, Tables found in reference (17) give EF values of 2.26 and 2.43 for single and tandem axles, respectively. The number of axle load passes and erosion factors are in table below:

Single axle Number of passes to failure Ax le l oad (kN) 125-133 115-125 107-115 97.8-107 88.8-97.8 80-88.8

Ax le l oad (kN) 213-231 195-213 178-195 160-178 142-160 125-142

Fig ur e(12.8) 40,000,000 100,000,000

Tandem axle Number of passes to failure Fig ur e(12.8) 50,000,000

Ac tu al n um ber of axl e pas ses 5,954 13,859 28,436 60,773 100,911 222,456

Ac tu al n um ber of axl e pas ses 20,121 40,447 117,849 351,803 835,827 878,122 Total accumulated fatigue damage

Damage ratio(%) 0.01% 0.01% 0 0 0 0

Damag e rat io 0.04% 0 0 0 0 0 0.02%

That shows the total accumulated erosion damage for this trial is 0.06%. Another slab thickness should be checked to see if some slab thickness can be economized without raising the accumulated fatigue and erosion higher than 100%.

Thickness(mm) 200 190 180

Thickness(mm) 200 190 180

Single 1.37 1.47 1.58

Single 2.37 2.44 2.5

ES Tandem 1.16 1.24 1.33

Fatigue cracking SRF Damage ratio Single Tandem Single Tandem total 0.34 0.29 9.60% 0 9.60% 0.37 0.31 49.17% 0.20% 49.37% 0.40 0.33 224% 0.80% 224.80%

Erosion ES Damage ratio Tandem Single Tandem 2.44 0.04% 0.05% 2.49 0.19% 0.06% 2.54 0.44% 0.18%

total 0.09% 0.25% 0.62%

98

According to above table the 190mm in slab could be selected. 12.6

Use the Equation 12.27 to compute the joint opening as: 12000 × 8 × 0.85 × (5 ×10-6 × (55+25) + 1.5 × 10-4) = 44.88×10-3 Use Equation 12.28 to compute the initial joint shear capacity, so: s0 = 0.05 × 10 × exp(-0.032 ×44.88) = 0.11892 Use equation 12.29 to compute the dimensionless joint transverse stiffness from aggregate interlock, Jagg jw =

 0.11892− 0.35    −    log( J agg ) = −3.19626 + 16.09737 exp − 2.7183  0.38   = −0.63192  

Which gives a Jagg value of 0.233. Substituting this value into Equation 12.30 gives: LTE agg

=

100 1 + 1.2 ⋅ 0.233 − 0.849

= 19.50%

12.7

For the month of January the LTEagg was calculated as 19.50%, which given a deflection wl of 12×10-3in allows the computing a deflection wul of 2.34×10-3in. This gives the dimensionless shear stress τ, using Equation 12.25: τ =

0.233 10

(12 − 2.34) = 0.225

Use Equation 12.33 to compute the reference shear stress. τref = 111.1 exp{-exp[0.9988exp(0.1089log 0.233)]} = 6.02

Considering that 3.8×10 = 38 is smaller than jw = 44.88 and larger and that there is only one axle configuration /load to be considered, allows computation of the loss in shear stress Δs, using Equation 12.32, as: 0.0068 1 + 6(44.88 / 10 − 3)

−1.98

 250,000  0.255     = 1.93 × 10 −5 6 6 . 02   10 

Which is to be subtracted from the dimensionless shear stress for the following month.(Equation 12.31) 12.8

During the first month, the load transfer stiffness variable J d is computed from Equation 12.34:

99

J d = J 0

=

120 ⋅1.25 2 14

= 13.4

Which allows computing of LTEdowel using Equation 12.38: LTE dowel

=

100 1 + 1.2 ⋅13.4 − 0.849

= 88.30%

Which allows computing the wul for the current month (Equation 12.18) as 0.015 × 0.883 = 0.0132in. This, in turn, allows computing the dimensionless bearing stress of the Portland concrete(Equation 12.26): σ b =

13.4 1.25

(0.015 − 0.0132) = 0.0188

Use the Equation 8.21 to compute the radius of relative stiffness:  4,000,000 ×14 3  l=  12 1 − 0.15 2 ⋅120 

(

0.25

)

= 52.84in.

Substituting this values into Equation 12.37 gives the dowel support damage at the end of the first month:  100,000  0.0188 ×120 × 52.84    = 1.135 4200   10 6 

DAM dowel = 400

Which allows computing the Jd for the following month (Equation 12.35), after computing Jd * as : Jd * = Min(118, Max[165×1.252/14-19.8120, 0.4]) = 0.4 Jd = 0.4 +(13.4-0.4)exp(-1.135) = 4.58 As a result, the LTEdowel for the following month will be: LTE dowel

=

100 1 + 1.2 ⋅ 4.58 −0.849

= 75.2%

12.10

Use the equation 12.58 to compute the site factor: SF = 35(1 + 0.5556 ⋅1000 )

1 + 20 10 6

= 0.4091

Use equation 12.26 to compute the scaling factor: SCF = -1400 + 350×7×(0.5+0)+1.36×4000-0.2×100×35+43×12-536×0.45 = 4839.8 Use the equation 12.55 to compute the extent of spalling:

100

100   35   ⋅ = 2.6 ×10 − 6 %   − ⋅ + ( ) 12 35 4839 . 8  35 + 0.01 1 + 1.005 

SPALL = 

Use Equation 12.57 to compute the present roughness IRI = 65 +0.0823×22+0.4417×2.6×10-6+1.4929×25+25.24×0.4091 = 114.46 in/mi

101

Chapter 13 13.1

Use the equation 13.4 to compute the resilient modulus of the subgrade: 0.24 P 0.24 ⋅10000 M r = = = 17142 ib / in 2 wr r 0.007 ⋅ 20 Given a contact area of π62 = 113.1 in , the contact pressure p on the FWD plate is 88.4 lb/in2. Use Equation 13.5 to solve for E p:

 1−    1  0.010 = 1.5 ⋅ 88.4 ⋅ 6 + 2    17142 1 +  16.5 3 E P   6 17142     

  2  16.5   1+     6    E P     1

That gives the value of approximately 195000lbs/in2. Entering this value along with the D value of the 16.5 in. into Figure 13.3 gives an SNEFF value of 4.2..Consideration of the structural number required to accommodate future traffic allows computation of the required overlay thickness. Considering the require structural number of overlay to accommodate future traffic allows to compute the overlay thickness as: DoL =

5.2 − 4.2 0.44

= 2.27in.

Which rounds up to 2.5in. over lay thickness. 13.2

Draining within one week characterizes the drainage as “fair” (Table 10.1), which combined with 5% of time saturation, allows estimating drainage coefficients for the base and subbase layers of 1.1(Table 2.7).Hence, the original structural number of the section post construction can be computed from Equation 2.5:

SN0= 0.44×8.67+0.14×1.1×7.1+0.11×1.1×5.9 = 5.62 The remaining life factor is computed from Equation 13.1 as:  4.5  RL = 1001 −  = 18.2%  5.5  Using a 18.2% remaining life, Figure 13.1 gives a condition factor of 0.73which allows computing the effective structural number SNEFF from Equation 13.2 as: SNEFF = 0.73×5.62 = 4.1 Given a required future structural number of 4.8, the overlay thickness can be computed from Equation 13.3 as: 102

DoL =

4.8 − 4.1 0.44

= 1.6in.

Which rounds up to 2in. over lay thickness. 13.3

Use Equation 12.1 to compute the required thickness of a Portland concrete layer to carry future traffic, DF. Compute the necessary input from the information given. For the 95% confidence, the value of the standard normal deviate ZR is -1.645. The adjusted modulus of the subgrade reaction is 13.4.

Substituting given moduli values into Equation 13.12 and solve it for d AC: 325000 = 190000 +

400000 − 190000 1 + e −0.3+ 5 log

d AC

Which gives d AC value of 87.5%.

103

Chapter 14

14.1

A company purchased a paving machine for $350,000 in 1992. What its actual replacement cost for an equivalent machine would have been in the year 2000 and subsequently 2008, assuming an average inflation rate of 3.5%? Solution

Use compound interest formula (Tab. 14.3). 350000 (1 + 0.035) 8 = $460,883 16 350000 (1 + 0.035) = $606,895

14.2

Explain why the incremental benefit-cost method is superior to the net method and the plain benefit-cost method in comparing economic alternatives. Solution

Incremental B/C explains the differential benefit gained from increases in e xpenditure and therefore removes the bias introduced by the scale of the the problem. For examples Using the Net method: B-C = $2,000,000-$1,900,000 = $100,000 would be equally desirable to B-C = $200,000-$100,000 = $100,000 Although clearly in differential terms the second alternative is preferable, since for the first alternative the extra $1,800,000 in cost returns only onl y $1,800,000 in extra profit, which makes it undesirable. 104

Similarly, the plain B/C method cannot differentiate between two alternatives of different magnitude, such as for example: B/C = $100,000/$50,000 = 2 B/C = $2,000,000/$1,000,0000 = 2 While computing B/C in differential terms:

∆B/∆C = $1,900,000/$950,000 = 2 Suggests that spending the additional $950,000 returns benefit of $1,900,000, which suggests that the more expensive alternative is worthwhile.

14.3

Land was acquired for the right-of-way right-of-way of a roadway in 1955 for for $2 million. Its estimated value in 2001 was $36 million. million. Calculate the inflation-free inflation-free rate of return of this investment, given that the inflation inflation runs at 2.5% average per annum. Carry out your calculations in terms of (a) actual $ and (b) real (i.e., inflation-free) $. Solution

(a) Use the compound rate formula from Tab. 14.3 F / P = 36 / 2 = (1 + i )

46

which gives i = 181/46 – 1 = 6.49% ignoring inflation (actual $) (b) Use Equ. 14.2 to compute compu te the inflation-free rate of return (real-1955$) i′ =

i − f

1 + f

=

0.0649 − 0.025 1 + 0.025

= 3.89%

105

14.4

Compare the agency costs for two flexible pavement treatments with the following characteristics:

• capital cost of $60,000 per 2-lane-km and an expected life of 9 years, • capital cost of $100,000 per 2-lane-km and an expected life of 13 years. Assume a discount rate of 3.5%. Solution

Compare with the EUA Cost method. Annualize costs using the Capital Recovery formula in Tab. 14.3. A = P

A = P

i (1 + i ) n

(1 + i ) n − 1

i (1 + i) n

(1 + i ) n − 1

= $60000

0.035 (1.035) 9

= $100000

1.0359 − 1

= $7,890 /yr

0.035 (1.035)13 1.03513 − 1

= $9,710 /yr

Which suggest that the first alternative is the best.

14.5

Compare the agency costs for a flexible and a rigid pavement with the following characteristics:

•

capital cost of $405,000 and an expected life of 15 years,

• capital cost of $675,000 and an expected life of 35 years. Assume a discount rate of 3.5%. Perform the calculations short-hand short-hand and through the built-in functions of a spreadsheet.

106

Solution

Compare with the EUA Cost method. Annualize costs using the Capital Recovery formula in Tab. 14.3. A = P

A = P

i (1 + i ) n

(1 + i ) n − 1 i (1 + i ) n

(1 + i ) n − 1

= $450000

= $675000

0.035 (1.035)15 1.03515 − 1 0.035 (1.035) 35 1.035 35 − 1

= $39,071 /yr

= $33,748 /yr

which makes the rigid pavement a better choice.

14.6

Compare the rolling resistance forces acting on a large passenger car operating on two different pavements with IRI roughnesses of 1 m/km and 6 m/km. Given:

• vehicle speed of 110 km/h, • radial tires, •

Tdsp of 2 mm,

• vehicle mass M of 1,800 kgr. Ignore the effect of pavement surface stiffness. Solution

Table 14-5 gives: CR1 = 1.0 b11 = 24.42 b12 = 0.097 b13 = 0.1102

107

Assuming a flexible pavement, Tab. 14.6 gives he value of CR2 as: 2

CR2 = 0.89 +0.03 1 (0.38+0.93 2) = 1.04 for IRI of 1 m/km 2

CR2 = 0.89 +0.03 6 (0.38+0.93 2) = 1.79 for IRI of 6 m/km

which, substituted into Equ. 14-5 give the rolling resistance force for roughness levels of 1 m/km and 6 m/km in the IRI scale as:

F r = 1.04 (24.42 4 + 1 (0.097 1800 + 0.1102 30.55 2 )) = 389.8 N F r = 1.79 (24.42 4 + 1 (0.097 1800 + 0.1102 30.55 2 )) = 670.9 N respectively. Hence, the difference in traction forces due solely to the difference in pavement roughness is 281 N.

14.7

What is the difference in fuel consumption for the circumstance of the previous problem? Solution

Table 14-7 suggests that Pmax for a large passenger car is 90 kW, hence Ptot at mid-power is 45 kW. Equation 14-6 suggests that the difference in the power requirement due to the calculated difference in rolling resistance force (i.e., 281 N calculated earlier) is:

∆Ptot = ∆Ptr = v ∆F r / 1000 = 30.55

281/ 1000 = 8.6 kW

which according to Equs. 14-9 and 14-10 results in a difference in fuel consumption of:

∆ IFC = ξ ∆Ptr = ξ b (1 + ehp

Ptot − Peng Pmax

)∆Ptr = 0.067 (1 + 0.25

45 90

) 8.6 = 0.65 ml/s

or 2.33 liters of fuel per hour (0.61 gallons/hr) difference in fuel consumption. At 110 km/h (68.35 miles/h), this represents a difference of 0.009 gallons/mile. Considering a

108

25 miles/gallon (0.04 gallons/mi) typical car fuel consumption, this represents a difference of 22.3%.

14.8

Compare the cost of vehicle maintenance parts and labor for a large passenger car that has an odometer reading of 100,000 km and operates on pavements with IRI of 6 m/km versus operating on a pavement with an IRI of 1 m/km. Given:

• the purchase cost of the vehicle is $20,000 and • the shop charges $60 per hour labor. Solution

Substituting the constants for the large passenger car from Table 14-10 into Equation 14-11 gives:

 100,000  PARTS = 1.0   100 , 000  

0.23

(1 + 0.230 (6- 3))=1.69 of the price of a new vehicle in

$1000s, that is 1.69 x 20 = $33.8 per 1000 km for operating on a pavement with an IRI of 6 m/km. The corresponding labor is calculated from Equation 14-13 as: LH = 77.14 PARTS

0.547

0.547

0 e = 77.14 0.00169

= 2.35 hours per 1000 km

which translates to labor cost of $141 and brings the total vehicle maintenance cost to $174.75 per 1000 km. Repeating these calculations for an IRI of 1 m/km gives: -8

13

RI = 3.0 + 5.54x10 IRI = 3.000

which when substituted into Equation 14-11 gives:

109

 100,000  PARTS = 1.0   100 , 000  

0.23

(1 + 0.23 ( 3 - 3))=1.00 of the price of a new vehicle in

$1000s, that is $20 per 1000 km and involves labor of: LH = 77.14 PARTS

0.547

0.547

0

e = 77.14 0.001

= 1.76 hours per 1000 km,

which translates to a labor cost of $106 and brings the total vehicle maintenance cost to $126 per 1000 km.

14.9

Calculate the delay and the associated cost for a two-lane lane closure of 250 m in length on a 6-lane divided freeway. Given that:

• the arrival rate is 1,300 pce/hour in the 10 minute interval to be analyzed, • no lower speed limit is posted through the work zone, • all vehicles have double occupancy and • the unit cost of travel time is $10/hour. Solution

Assuming a 50% confidence that the capacity will be equal or lower than the one computed and utilizing the constant values found in Tab. 14.15. Equ. 14-21 gives: CAPWZ = (1370 – 4.05 50) = 1167 passenger car equivalents/hr, (pce/hr)

The number of vehicles at the end of the interval is calculated from Equation 14-20 as: VEHQUE i = (1,300 - 1,167)

10 = 22.2 rounded to 23 passenger car equivalents. 60

Assuming 0 vehicles in the queue in the previous time interval, the total time delay DQUE i (vehicle-sec) is calculated as:

110

DQUE i =

(0 +23) 2

60 10 = 6900 vehicle-sec or 1.91 vehicle-hrs, which translates to 3.83

person-hrs or $38.3 for the time interval of 10 minutes analyzed.

14.10

Calculate the total hourly delay for a 1,000 meter lane closure on a 2-lane undivided roadway with a directional volume of 800 pce/hour, a work zone cycle of 700 seconds, a work-zone capacity or 1,100 pce/hour and a posted speed through the work zone of 45 km/hour (12.5 m/sec). Assume lost times at the beginning and end of the green interval of 2 seconds and 0 second, respectively. Solution

Equs. 14-26 to 14-28 give: IG =

700 12.5

EFFG =

= 56 sec

700 2

GCRATIO =

+ (56 + 2 + 0) = 408 sec 408 = 0.582 700

The volume over capacity ratio is: X =

800 1100

= 0.73

which when substituted into Equation 14-30b gives: DELAYUN = 0.38

700 (1−0.582) 2 1−0.582 0.73

= 80.81 s/pce.

Equation 14-32 gives: 111

Papagiannakis Solutions for text book

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