Important Properties of Subgrade, Subbase, and Base Materials

Chapter 7 includes questions and practical problems that cover the key properties of subgrade, subbase, and base materials including the California bearing ratio (CBR), resilient modulus, stabilometer R-value, dry density, and moisture content. The unbound materials are used in the underlying layers of the pavement structure, which provide strength and support to the pavement, serve as drainage layers and prevent frost protection, and minimize settlements in the pavement. The underlying layers of the pavement increase the structural capacity of the pavement to carry the traffic loads during the design life of the pavement structure and distribute the traffic loading to the subgrade soil. Therefore, the properties of the unbound materials used in these layers are of high significance to pavement engineers. In this chapter, a proper understanding of these properties will be achieved through presenting practical problems on this aspect. 7.1

If the R-value (from the Hveem stabilometer test) of a soil material is 15, determine the California Bearing Ratio (CBR) of the material.

Solution:

M R = 1155 + 555   R https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0001.tif"/>

Where:

M_R = resilient modulus

R = Hveem stabilometer resistance value

⇒

M R = 1155 + 555 ( 15 ) = 9840   psi   ( 1427   kPa ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0002.tif"/>

M R = 1500 ( CBR ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0003.tif"/>

Where:

M_R = resilient modulus

CBR = California bearing ratio

⇒

CBR = M R 1500 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0004.tif"/>

CBR = 9840 1500 = 6.3 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0005.tif"/>

The computations of this problem are conducted using an MS Excel worksheet. An image of the MS Excel worksheet used is shown in Figure 7.1.212

7.2

In a triaxial test for resilient modulus of a granular material, the confining pressure was 15 psi (103.4 kPa) and the bulk stress was 47 psi (324.1 kPa), determine the deviator stress.

Solution:

Given:

θ = 47 psi, and σ ₃ = 15 psi

7.3 θ = σ 1 + 2 σ 3 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0006.tif"/>

7.4 σ d = σ 1 − σ 3 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0007.tif"/>

⇒

θ = σ d + 3 σ 3 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0008.tif"/>

Or:

σ d = θ − 3 σ 3 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0009.tif"/>

And therefore,

σ d = 47 − 3 ( 15 ) = 2   psi   ( 13.8   kPa) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0010.tif"/>

7.3

In Problem 7.2 above, determine the axial stress (σ₁).

Solution:

θ = σ 1 + 2 σ 3 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0011.tif"/> 213

⇒

σ 1 = θ − 2 σ 3 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0012.tif"/>

σ 1 = 47 − 2 ( 15 ) = 17   psi   ( 117.2   kPa) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0013.tif"/>

The computations and results of the MS Excel worksheet used for Problems 7.2 and 7.3 are shown in Figure 7.2.

7.4

In Problem 7.2 above, if the recoverable strain for the specimen is 140 microstrains, then the resilient modulus of the specimen is:

Solution:

7.5 M R = σ d ε r https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0014.tif"/>

Where:

M_R = resilient modulus

ε_r = resilient (recoverable) strain

⇒

M R = 2 140 × 10 − 6 = 14286   psi   ( 98500   kPa) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0015.tif"/>

The MS Excel worksheet used to compute the results of this problem is shown in Figure 7.3.214

7.5

In a resilient modulus test of a granular material sample, the plot shown in Figure 7.4 for the resilient modulus (M_R) versus the stress invariant (θ) is obtained. Based on this figure, answer the following questions:

Determine the nonlinear coefficient k₁ and the exponent k₂ of the granular material knowing that the relationship between M_R and θ is given by the model M R = k 1 θ k 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0016.tif"/> .

Determine the confining pressure and the deviator stress at a bulk stress of 8 psi if the axial load is 75 lb (333.8 N) and the sample diameter = 4.0 in (10.2 cm).215

Solution:

By taking any two arbitrary points (the points with a circle legend) on the curve and computing the slope as the difference between log (resilient modulus value) divided by the difference in log (stress invariant), the value of k₂ is obtained (see Figure 7.5):

Point 1 (32, 11000), Point 2 (60, 15500)

k ₂ = slope of log-log relationship

k 2 = log M R 2 − log M R 1 log θ 2 − log θ 1 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0017.tif"/>

Where:

M_R1 and M_R2 = resilient modulus of points 1 and 2, respectively.

θ₁ and θ₂ = bulk stress (stress invariant) of points 1 and 2, respectively.

⇒

k 2 = log( 15500 ) − log( 11000 ) log(60) − log(32) = 0.546 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0018.tif"/>

k₁ can be determined by substituting k₂ in the given model and using any of the two points as shown below:

7.7 M R = k 1 θ k 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0019.tif"/>

⇒

15500 = k 1 ( 60 ) 0.546 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0020.tif"/>

⇒

k 1 = 1661 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0021.tif"/>

σ 1 = Axial   Load Circular   Area https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0022.tif"/>

7.8 σ 1 = Axial   Load Circular   Contact   Area https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0023.tif"/> 216

⇒

σ 1 = 75 ( π ( 4 ) 2 4 ) = 6.0   psi   ( 41.2   kPa) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0024.tif"/>

θ = σ 1 + 2 σ 3 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0025.tif"/>

⇒

σ 3 = ( θ − σ 1 ) 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0026.tif"/>

⇒

σ 3 = ( 8 − 6 ) 2 = 1.0   psi   ( 6.9   kPa) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0027.tif"/>

σ d = σ 1 − σ 3 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0028.tif"/>

⇒

σ d = 6.0 − 1.0 = 5.0   psi   ( 34.1   kPa) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0029.tif"/>

The MS Excel worksheet used for rapid and efficient solution is shown in Figure 7.6.

217

7.6

The results from a confined resilient modulus test of a granular soil material are given in Table 7.1.

Plot the data to determine the nonlinear coefficient K₁ and the exponent K₂ of the granular material knowing that the relationship between M_R and θ is given by the model M R = k 1 θ k 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0030.tif"/> .

Solution:

The data of the resilient modulus (M_R) and the bulk stress (θ) is plotted using MS Excel and a trend regression line is added to the scatter plot of the power type as shown in Figure 7.7.

By displaying the equation and the coefficient of determination (r²) in MS Excel worksheet, the following model, which is also shown in the figure, is obtained:

M R = 1.6691 ( θ ) 0.545 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0031.tif"/>

Note that the resilient modulus (M_R) values are multiplied by 10³.218

7.7

In a triaxial compression test of a granular material specimen of 4-inch diameter to determine the resilient modulus, the relationship between M_R and the bulk stress (θ) is given as log M R = 4.00 + 0.5 log θ https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0032.tif"/> . At a deviator stress and bulk stress of 10 psi (69.0 kPa) and 55 psi (379.2 kPa), respectively, determine the following:

The axial compressive stress

The axial load

The resilient modulus

The confining pressure

The recoverable strain in the specimen

Solution:

Given:

D = 4   inches   (0 .102   m) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0033.tif"/> , log M R = 4.00 + 0.5 log θ https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0034.tif"/> , σ d = 10   psi   ( 69   kPa) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0035.tif"/> , and θ = 55   psi   ( 379.2   kPa) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0036.tif"/>

The axial compressive stress (σ₁):

θ = σ 1 + 2 σ 3 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0037.tif"/>

σ d = σ 1 − σ 3 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0038.tif"/>

⇒

θ = 2 σ 1 − σ d https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0039.tif"/>

⇒

σ 1 = θ + 2 σ d 3 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0040.tif"/> 219

Therefore:

σ 1 = 55 + 2 ( 10 ) 3 = 25   psi   ( 172.4   kPa) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0041.tif"/>

The axial load:

σ 1 = Axial   Load Circular   Contact   Area https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0042.tif"/>

⇒

Axial   Load = σ 1 ( Contact   Area ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0043.tif"/>

⇒

Axial   Load = 25 ( π ( 4 ) 2 4 ) = 314.2   lb   ( 1398   N) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0044.tif"/>

The resilient modulus (M_R):

log M R = 4.00 + 0.5 log θ https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0045.tif"/>

⇒

M R = 10 ( 4.00 + log ( 55 ) ) = 74162   psi   ( 511347   kPa ≅ 511.3   MPa ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0046.tif"/>

The confining pressure (σ₃):

θ = σ 1 + 2 σ 3 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0047.tif"/>

⇒

σ 3 = θ − σ 1 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0048.tif"/>

⇒

σ 3 = 55 − 25 2 = 15   psi   ( 103.4   kPa) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0049.tif"/>

The recoverable strain in the specimen (ε_r):

M R = σ d ε r https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0050.tif"/>

⇒

ϵ r = σ d M R https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0051.tif"/>

⇒

ϵ r = 10 74162 = 135 × 10 − 6   in/in = 135   μ ε https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0052.tif"/>

The solution of this problem is also done using the MS Excel worksheet shown in Figure 7.8.

220

7.8

The results of a triaxial test of a granular base material to determine the resilient modulus are plotted in Figure 7.9. Based on these results, determine the following:

The nonlinear coefficient k₁ of the material if the model that describes the behavior of the material is M R = k 1 θ k 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0053.tif"/> .

The exponent k₂ of the material in the model M R = k 1 θ k 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0054.tif"/> .

If the axial load applied on the specimen is 250 lb (1112.5 N) and the specimen diameter is 4 inches (10.2 cm), then compute the confining pressure and the deviator stress at a stress invariant of 50 psi (344.8 kPa).

The resilient modulus in Part c above.221

Solution:

The nonlinear coefficient k₁ of the material:

The nonlinear coefficient k₁ of the material can be determined from the intercept of the line that represents the relationship between log θ and log M_R. By extending the line as shown in Figure 7.10, the line intersects the y-axis at 3.58.

Hence,

k 1 = 10 3.58 = 3802. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0055.tif"/>

The exponent k₂ of the material:

The exponent k₂ is determined from the slope of the relationship between log θ and log M_R since the original model is a power model M R = k 1 θ k 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0056.tif"/> . Therefore, by taking two arbitrary points on the line (Point 1 and Point 2) in Figure 7.10, the slope can be computed as shown below:

Point 1 (1, 3.92)

Point 2 (1.5, 4.12)

⇒

k 2 = Slope = 4.12 − 3.92 1.5 − 1.0 = 0.40. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0057.tif"/>

In other words, the model between M_R and θ is:

M R = 3802 ( θ ) 0.40 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0058.tif"/>

If the axial load applied on the specimen is 250 lb and the specimen diameter is 4 inches, then compute the confining pressure and the deviator stress at a stress invariant of 50 psi:

σ 1 = Axial   Load Contact   Area https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0059.tif"/>

⇒

σ 1 = 250 ( π ( 4 ) 2 4 ) = 19.9   psi   ( 137.2   kPa) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0060.tif"/>

θ = σ 1 + 2 σ 3 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0061.tif"/> 222

⇒

σ 3 = θ − σ 1 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0062.tif"/>

σ 3 = 50 − 19.9 2 = 15.2   psi   ( 103.8   kPa) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0063.tif"/>

σ d = σ 1 − σ 3 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0064.tif"/>

⇒

σ d = 19.9 − 15.1 = 4.8   psi   ( 33.4   kPa) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0065.tif"/>

The resilient modulus in Part c above:

M R = 3802 ( θ ) 0.40 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0066.tif"/>

⇒

M R = 3802 ( 50 ) 0.40 = 18180   psi   ( 125353   kPa ≅ 125.4   MPa) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0067.tif"/>

The MS Excel worksheet shown in Figure 7.11 is used to solve the problem for rapid solution.223

7.9

In a confined triaxial compression test of a granular material to determine the resilient modulus, the relationship between M_R and the bulk stress (θ) is described by the k₁-k₃ ­universal model being used in the Mechanistic-Empirical Pavement Design Guide (M-EPDG) M R = k 1 ( θ ) k 2 ( σ d ) k 3 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0068.tif"/> . If the test data are given (see Table 7.2), use the multiple non-linear regression analysis techniques to:

Develop the non-linear model

Determine the material-related regression coefficients k₁, k₂, and k₃

Compute the coefficient of determination (r²) of the model

Plot the relationship between resilient modulus (M_R) and total stress (θ)

Solution:

Develop the non-linear model:

The MS Excel solver tool is utilized to develop the model and to determine the regression coefficients k₁, k₂, and k₃. In order to do that, initial values will have to be used for the three coefficients (k₁, k₂, and k₃) as shown in the Excel sheet shown in Figure 7.12. Excel solver uses numerical iterative approach to minimize the error associated with the developed model and provide optimum values for these coefficients. The error is simply defined as:224

Error = ( ( log ( M R ) ) − log ( M R-Predicted ) ) 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0069.tif"/>

Where:

M R-Predicted = k 1 ( θ ) k 2 ( σ d ) k 3 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0070.tif"/>

The sum of the errors is obtained for all test data points (see Table 7.3 and Figure 7.12).225

Determine the material-related regression coefficients k₁, k₂, and k₃:

The values of k₁, k₂, and k₃ are determined using MS Excel solver tool as shown in Figure 7.12. Numerical iterative approach is followed in the MS Excel solver tool to solve for the three regression coefficients by minimizing the sum of the errors associated with the model.

In this case,

k₁ = 3295

k₂ = 0.322

k₃ = 0.112

Compute the coefficient of determination, r² of the model.

The final model that fits the given resilient modulus data is:

M R-Predicted = 3295 ( θ ) 0.322 ( σ d ) 0.112 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0071.tif"/>

The coefficient of determination, r² of the model is computed using the following formula:

r 2 = S t − S r S t https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0072.tif"/>

Where:

S_t = total sum of squares around the mean ( y ¯ ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0073.tif"/>

S t = ∑ i = 1 n ( y i − y ¯ ) 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0074.tif"/>

S_r = sum of squares of residuals (errors) around the regression model.

y_i = given (measured) resilient modulus data point.

S r = ∑ i = 1 n ( y i − y i-Predicted ) 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0075.tif"/> 226

y_i-Predicted = calculated (predicted) resilient modulus value from the developed model.

Hence,

r 2 = 1.40 × 10 11 − 2.08 × 10 8 1.40 × 10 11 = 0.999 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0076.tif"/>

The MS Excel computations for the coefficient of determination, r², are shown in Figure 7.13.227

Plot the relationship between resilient modulus (M_R) and total stress (θ):

The relationship between the given resilient modulus values and the total (bulk) stress values is plotted using arithmetic scale as shown in Figure 7.14.

7.10

A granular material specimen of 6-inch (152.4-mm) diameter is tested in a triaxial test to determine the resilient modulus. If the axial compressive load applied on the specimen is 276 lb (1228.2 N), and the stress invariant (bulk stress) is 20 psi (137.9 kPa), compute the axial stress, the confining pressure, and the deviator stress.

Solution:

σ 1 = Axial   Load Contact   Area https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0077.tif"/>

⇒

σ 1 = 276 ( π ( 6 ) 2 4 ) = 9.76   psi   ( 67.3   kPa) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0078.tif"/>

θ = σ 1 + 2 σ 3 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0079.tif"/>

⇒

σ 3 = θ − σ 1 2 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0080.tif"/>

σ 3 = 20 − 9.76 2 = 5.12   psi   ( 35.3   kPa) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0081.tif"/> 228

σ d = σ 1 − σ 3 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0082.tif"/>

⇒

σ d = 9.76 − 5.12 = 4.64   psi   ( 32.0   kPa) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0083.tif"/>

The MS Excel worksheet used to conduct the computations of this problem is shown in Figure 7.15.

7.11

In a resilient modulus test of a subgrade soil material, the confining pressure is 12 psi (82.7 kPa), the axial stress is 30 psi (206.9 kPa), and the resilient strain is 715 microstrains, determine the resilient modulus of the material.

Solution:

σ d = σ 1 − σ 3 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0084.tif"/>

⇒

σ d = 30 − 12 = 18   psi   ( 124.1   kPa) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0085.tif"/>

M R = σ d ε r https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0086.tif"/>

⇒229

M R = 18 715 × 10 − 6 = 25175   psi   ( 173580   kPa ≅ 173.6   MPa ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0087.tif"/>

The image in Figure 7.16 shows the MS Excel worksheet computations for the results of this problem.

7.12

In a triaxial compression test of a fine-grained material to determine the resilient modulus, the results plotted in Figure 7.17 were obtained. Based on these results, determine the following:

The coefficient k₁ of the material

The coefficient k₂ of the material

The coefficient k₃ of the material

The coefficient k₄ of the material

The recoverable strain at a deviator stress of 8 psi.

Solution:

The coefficient k₁ of the material (see Figure 7.18):

k₁ = 5400 from the figure; k₁ is the distance between the intersection point of the two linear lines and the x-axis.

The coefficient k₂ of the material:

k₂ = 4 from the figure; k₂ is the distance between the intersection point of the two linear lines and the y-axis.230

The coefficient k₃ of the material:

Again, by choosing two arbitrary points on the first upper portion of the curve, the slope k₃ can be estimated from these points:

Point 1 (2.4, 7200), Point 2 (4.0, 5400)231

k 3 = ( 7200 − 5400 ) ( 4.0 − 2.4 ) = 1124 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0088.tif"/>

The coefficient k₄ of the material:

By choosing two arbitrary points on the second lower portion of the curve, the slope k ₄ can be estimated from these points:

Point 3 (5.6, 4800), Point 4 (9.6, 3180)

k 4 = ( 4800 − 3180 ) ( 9.6 − 5.6 ) = 405 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0089.tif"/>

The recoverable strain at a deviator stress of 8 psi:

From the figure; when σ _d  = 8 psi, M _R  = 3800 psi.

M R = σ d ε r https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0090.tif"/>

⇒

ϵ r = σ d M R https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0091.tif"/>

ϵ r = 8 3800 = 2105 × 10 − 6   in/in = 2105   μ ϵ https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0092.tif"/>

⇒

ε r = σ d M R https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0093.tif"/>

ε r = 8 3800 = 2105 × 10 − 6   in/in = 2105   μ ε https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0094.tif"/>

The MS Excel worksheet in Figure 7.19 shows the solution of this problem.232

7.13

The CBR test results for a sand material used for the subgrade of a highway asphalt pavement are plotted in Figure 7.20. If the diameter of the piston applying the load on the sand sample is 5.0 cm (about 2 in), determine the CBR value of the sand subgrade (see Figure 7.20).233

Solution:

At a penetration of 0.10 inch (2.54 mm), the value of load is determined. From Figure 7.21, it is equal to 2.8 kN (629.2 lb).

CBR = Pressure   at   0.10   in Standard   Pressure   at   0.10   in = 1000   psi × 100 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0095.tif"/>

Or:

CBR = Load   at   0.10   in Standard   Load   at   0.10   in = 13.2   kN × 100 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0096.tif"/>

⇒

CBR = 2.8 13.2 × 100 = 21.2 % https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0097.tif"/>

At a penetration of 0.20 inch (5.08 mm), the value of load is determined. From Figure 7.21, it is equal to 4.6 kN (1033.7 lb).

CBR = Pressure   at   0.20   in Standard   Pressure   at   0.20   in = 1500   psi × 100 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0098.tif"/>

Or:

CBR = Load   at   0.20   in Standard   Load   at   0.20   in = 20.06   kN × 100 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0099.tif"/>

⇒234

CBR = 4.4 20.06 × 100 = 21.9 % https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0100.tif"/>

Typically; if the CBR value at 0.20 in (50.08 mm) is higher than the CBR value at 0.10 in (25.4 mm), the test should be repeated. In this case, the difference is not significant.

7.14

If the relationship between the dry density and moisture content of a subgrade soil material in a standard AASHTO compaction test takes the form

γ d = − 0.4 ( MC ) 2 + 8.9 ( MC ) + 94.3 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0101.tif"/> , where γ_d is the dry density in lb/ft³ and MC is the moisture content (%), determine the following:

The maximum dry density (max γ_d)

The optimum moisture content (OMC)

Solution:

Given:

The relationship between the dry density (γ_d) and the moisture content (MC):

γ d = − 0.4 ( MC ) 2 + 8.9 ( MC ) + 94.3 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0102.tif"/>

The maximum dry density (max γ_d):

The relationship is plotted on arithmetic scale as shown in the graph in Figure 7.22.

The maximum dry density (γ_d) is determined from the graph.

The optimum moisture content (OMC):

The optimum moisture content (OMC) is also determined from the graph, which is the moisture content on the x-axis at the maximum dry density.

Another approach:

The first derivative of the relationship is obtained, and the OMC is the value of MC when the derivative is zero:

− 0.8   ( MC ) + 8.9 = 0 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0103.tif"/>

⇒

OMC = 8.9 0.8 = 11.1 % https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0104.tif"/>

The maximum dry density is the dry density at the OMC, which is determined using the relationship:

γ d = − 0.4   ( MC ) 2 + 8.9   ( MC ) + 94.3 https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0105.tif"/>

⇒

γ d = − 0.4   ( 11.1 ) 2 + 8.9   ( 11.1 ) + 94.3 = 143.8   lb/ft 3   ( 2302.6   kg/m 3 ) https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/TNF-CH007_eqn_0106.tif"/> 235 236

The relationship between the moisture content and the dry density of the soil material is shown in Figure 7.22.

The MS Excel worksheet used to solve this problem is shown in Figure 7.23.

Figure 7.1 MS Excel worksheet image for the computations of Problem 7.1. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig7_1_B.tif"/> Figure 7.2 MS Excel worksheet image for the computations of Problems 7.2 and 7.3. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig7_2_B.tif"/> Figure 7.3 MS Excel worksheet image for the computations of Problem 7.4. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig7_3_B.tif"/> Figure 7.4 Stress invariant versus resilient modulus (Problem 7.5). https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig7_4_B.tif"/> Figure 7.5 The selected two points on the curve (Problem 7.5). https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig7_5_B.tif"/> Figure 7.6 MS Excel worksheet image for the computations of Problem 7.5. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig7_6_B.tif"/> Table 7.1 Bulk Stress Versus Resilient Modulus of a Granular Soil Material M_R (10³ psi) Bulk Stress, θ (psi) 18.9 73.8 19.1 75.6 21.2 81.0 22.5 90.0 20.8 99.0 21.2 108.0 15.0 55.8 16.4 57.6 14.2 63.0 13.8 72.0 18.6 81.0 10.8 37.8 10.3 39.6 11.3 45.0 13.4 54.0 12.9 63.0 8.0 19.8 10.9 21.6 9.4 27.0 15.0 36.0 16.9 45.0 Figure 7.7 Stress invariant versus resilient modulus for a granular soil. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig7_7_B.tif"/> Figure 7.8 MS Excel worksheet image for the computations of Problem 7.7. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig7_8_B.tif"/> Figure 7.9 Bulk stress versus resilient modulus for a granular base material (log–log scale). https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig7_9_B.tif"/> Figure 7.10 The two points selected on the line (Problem 7.8). https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig7_10_B.tif"/> Figure 7.11 MS Excel worksheet image for the computations of Problem 7.8. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig7_11_B.tif"/> Table 7.2 Resilient Modulus Test Data for a Granular Material Confining Pressure, σ₃ (psi) Deviator Stress, σ_d (psi) Elastic Strain, ε_r (με) M_R (psi) Bulk Stress, θ (psi) 20 1.0 114.0 9115 73.8 2.0 225.6 18615 75.6 5.0 510.0 20120 81.0 10.0 960.0 21915 90.0 15.0 1560.0 22870 99.0 20.0 2040.0 21435 108.0 15 1.0 144.0 12315 55.8 2.0 264.0 14360 57.6 5.0 762.0 12175 63.0 10.0 1560.0 12385 72.0 15.0 1740.0 20645 81.0 20.0 1920.0 22175 90.0 10 1.0 200.4 10780 37.8 2.0 420.0 9285 39.6 5.0 960.0 11245 45.0 10.0 1608.0 12450 54.0 15.0 2520.0 12860 63.0 5 1.0 270.0 11575 19.8 2.0 396.0 7615 21.6 5.0 1152.0 8665 27.0 10.0 1440.0 15425 36.0 15.0 1920.0 16625 45.0 1 1.0 270.0 9870 19.8 2.0 396.0 12980 21.6 5.0 1152.0 9315 27.0 7.5 1440.0 14560 36.0 10.0 1920.0 17745 45.0 Figure 7.12 An Image of the MS Excel Solver and computations of Problem 7.9. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig7_12_B.tif"/> Table 7.3 The Predicted Resilient Modulus Values with the Associated Errors Confining Pressure, σ₃ (psi) Deviator Stress, σ_d (psi) Elastic Strain, ε_r (με) M_R (psi) Bulk Stress, θ (psi) Predicted M_R (psi) Error 20 1.0 114.0 9115 73.8 13162 2.5458E-02 2.0 225.6 18615 75.6 14334 1.2883E-02 5.0 510.0 20120 81.0 16238 8.6679E-03 10.0 960.0 21915 90.0 18153 6.6918E-03 15.0 1560.0 22870 99.0 19587 4.5292E-03 20.0 2040.0 21435 108.0 20802 1.6929E-04 15 1.0 144.0 12315 55.8 12029 1.0448E-04 2.0 264.0 14360 57.6 13132 1.5073E-03 5.0 762.0 12175 63.0 14975 8.0842E-03 10.0 1560.0 12385 72.0 16894 1.8181E-02 15.0 1740.0 20645 81.0 18361 2.5919E-03 20.0 1920.0 22175 90.0 19616 2.8354E-03 10 1.0 200.4 10780 37.8 10611 4.7207E-05 2.0 420.0 9285 39.6 11639 9.6332E-03 5.0 960.0 11245 45.0 13438 5.9857E-03 10.0 1608.0 12450 54.0 15399 8.5252E-03 15.0 2520.0 12860 63.0 16934 1.4284E-02 5 1.0 270.0 11575 19.8 8616 1.6436E-02 2.0 396.0 7615 21.6 9576 9.8993E-03 5.0 1152.0 8665 27.0 11400 1.4190E-02 10.0 1440.0 15425 36.0 13514 3.2978E-03 15.0 1920.0 16625 45.0 15195 1.5255E-03 1 1.0 270.0 9870 19.8 8616 3.4807E-03 2.0 396.0 12980 21.6 9576 1.7453E-02 5.0 1152.0 9315 27.0 11400 7.6925E-03 7.5 1440.0 14560 36.0 13086 2.1474E-03 10.0 1920.0 17745 45.0 14521 7.5815E-03 Sum of Errors 2.1388E-01 Figure 7.13 MS Excel computations to determine the coefficient of determination for Problem 7.9. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig7_13_B.tif"/> Figure 7.14 Bulk stress versus resilient modulus relationship for Problem 7.9. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig7_14_B.tif"/> Figure 7.15 MS Excel worksheet image for the computations of Problem 7.10. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig7_15_B.tif"/> Figure 7.16 MS Excel worksheet image for the computations of Problem 7.11. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig7_16_B.tif"/> Figure 7.17 Deviator stress versus resilient modulus for a fine-grained material. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig7_17_B.tif"/> Figure 7.18 Determination of k1 and k2 graphically using the deviator stress versus resilient modulus relationship. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig7_18_B.tif"/> Figure 7.19 MS Excel worksheet image for the computations of Problem 7.12. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig7_19_B.tif"/> Figure 7.20 A Plot for the CBR test data for a subgrade sand material. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig7_20_B.tif"/> Figure 7.21 Determination of the CBR value using the CBR test penetration versus load relationship. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig7_21_B.tif"/> Figure 7.22 Moisture content versus dry density for a subgrade soil material using a standard compaction test. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig7_22_B.tif"/> Figure 7.23 MS Excel worksheet image for the computations of Problem 7.14. https://s3-euw1-ap-pe-df-pch-content-public-p.s3.eu-west-1.amazonaws.com/9780429054297/5885f92e-b0a3-419d-837a-bad4e02024b0/content/fig7_23_B.tif"/>