SOLUTIONS + LECTURE SLIDES
,CHAPTER 2. PRINCIPLES OF MIX AND STRUCTURAL DESIGN AND CONSTRUCTION OF ASPHALT
PAVEMENT
Problem 5: Determine the vertical and radial stresses at nine points for a 9,000 lb point load on a
homogeneous, isotropic, linear elastic, semi-infinite space. Consider a Poisson’s ratio of 0.3.
Point z, inch r, inch Point z, inch r, inch Point z, inch r, inch
1 0 0 4 6 0 7 12 0
2 0 6 5 6 6 8 12 6
3 0 12 6 6 12 9 12 12
Solution:
P= 9,000 lbs; v ? r ?
Point 1: z= 0, r= 0
Use Boussinesq’s Method; assume u= 0.3
R2 r 2 z 2 = 0
3Pz 3
v
2 R5
P(1 u) 3r 2 z (1 2u)R
r
2 R2 R
3
R z
Point 2: z= 0, r=6
R 2 36
3(9000)(0)
v 0
2 R5
9000(1 0.3) (1 2 * 0.3)(6)
r 0
2 (36) 20.69 psi
6 0
Point 3: z= 0, r= 12; R 2 144 ; R=12
v 0
9000(1 0.3) (1 2 * 0.3) *12
r 0 5.17 psi
2 (144) 12 0
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,Point 4: z= 6 in., r= 0
R 2 6 2 36; R 6
3 3
3Pz 3(9000)(6)
v 5 119.37 psi
2R 2 (6)5
2
P(1 u) 3r z (1 2u) R 9000(1 0.3) (1 2 * 0.3)(6) psi
0
r
2 R2 R3 R z 2 (6) 2 6 6 10.35
Point 5: z=6, r=6
R 2 26 26 72; R 8.48
3(9000)(6)3
v 21.17 psi
2 (8.48)5
9000(1 0.3) 3(62 )(6) (1 2 * 0.3)(8.48)
r 21.39 psi
3
2 (8.48)2 8.48 8.48 6
Point 6: z= 6, r= 12
R 2 36 144 180
R 13.42
Point 7: z= 12, r= 0
R 2 122 144
R=12
3(9000)(12)3
v 29.84 psi
2 (12)5
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, 9000(1 0.3) (1 2 * 0.3)(12)
0 2.59 psi
r
2 (12)2 12 12
Point 8: z =12, r=6
R2 = 144+36 = 180 R=13.42
3(9000)(12)3
v 17.06 psi
2 (13.42)5
9000(1 0.3) 3(6)2 (12) (1 2 * 0.3)(13.42)
3.31psi
r
2 (13.42)2 (13.42)
3
13.42 12
Point 9: z =12, r=12
R2 = 144+144 = 288 R=16.97
3(9000)(12)3
v 5.28 psi
2 (16.97)5
6. If the deflection at the center of a rigid plate of radius 6 inch is found out to be 0.03 inch from a load of
9,000 lb on a subgrade with Poisson’s ratio of 0.35, what is the estimated modulus of the subgrade?
Solution:
�1−µ �𝑃
2 (1−0.35)(9,000)
∆= =
2𝐸𝑎 2𝐸6
(1 − 0.352)(9,000)
0.03 =
12𝐸
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,E=21,937.5 psi
7. Use any layered elastic analysis program to compute the vertical and radial stresses and strains directly
below the load at a depth of 149 mm in a full depth asphalt pavement with a thickness of 150 mm, and a
modulus of 3,500 MPa and Poisson’s ratio of 0.35, for the following loading conditions. The subgrade has a
modulus of 100 MPa and a Poisson’s ratio of 0.4. In each case half of a standard 18,000 lb axle (only the
main load bearing axles, not including the steering axle) has been indicated. Can you sketch the
axle/wheel configuration of the entire vehicles?
a) Loads of 20 kN, with coordinates in cm (x, y): (0,0) (33, 0); (0,122) (133,122); tire
pressure of 690 kPa
b) Loads of 20 kN, with coordinates in inch (x, y): (0,0) (33,0); (0,122) (33,122);
(0,244) (33,244); tire pressure of 690 kPa
Solution: See the following outputs from EVERSTRESS
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,a
Layered Elastic Analysis by EverStress for Windows
Title: Chapter 2 Problem 7a
No of Layers: No of Loads:
2 4 No of X-Y Evaluation Points: 1
Layer Poisson's Thickness Moduli(1)
* Ratio (cm) (MPa)
1 0.35 15 3500
2 0.4 * 100
Load No X-Position Y-Position Load Pressure Radius
* (cm) (cm) (N) (kPa) (cm)
1 0 0 20000 690 9.605
2 33 0 20000 690 9.605
3 0 122 20000 690 9.605
4 33 122 20000 690 9.605
Location No: 1
X-Position (cm): .000
Y-Position (cm): .000
Normal
Stresses
Z-Position Layer Sxx Syy Szz Syz Sxz Sxy
(cm) * (kPa) (kPa) (kPa) (kPa) (kPa) (kPa)
14.99 1 843.06 983.73 -70.27 0.9 7.29 -4.98
Normal Strains
and Deflections
Z-Position Layer Exx Eyy Ezz Ux Uy Uz
(cm) * (10^-6) (10^-6) (10^-6) (microns) (microns) (microns)
14.99 1 149.53 203.79 -202.76 -17.654 -4.876 476.542
Line
Principal
Stresses and
Strains
Z-Position Layer S1 S2 S3 E1 E2 E3
(cm) * (kPa) (kPa) (kPa) (10^-6) (10^-6) (10^-6)
14.99 1 -70.33 842.94 983.9 -202.78 149.48 203.85
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, b)
Title: Chapter 2 Problem 7-b
No of Loads: 6
No of Layers: 2 No of X-Y Evaluation Points: 1
Layer Poisson's Thickness Moduli(1)
* Ratio (cm) (MPa)
1 0.35 15 3500
2 0.4 * 100
Load No X-Position Y-Position Load Pressure Radius
* (cm) (cm) (N) (kPa) (cm)
1 0 0 20000 690 9.605
2 33 0 20000 690 9.605
3 0 122 20000 690 9.605
4 33 122 20000 690 9.605
5 0 244 20000 690 9.605
6 33 244 20000 690 9.605
X-Position Y-Position
(cm): (cm):
Location No: 1 .000 .000
cNormal Stresses
Z-Position Layer Sxx Syy Szz Syz Sxz Sxy
(cm) * (kPa) (kPa) (kPa) (kPa) (kPa) (kPa)
14.99 1 841.78 979.93 -70.24 1.02 7.3 -5.15
0 1 -1223.82 -1339.27 -690 0 0 6.09
cNormal Strains and Deflections
Z-Position Layer Exx Eyy Ezz Ux Uy Uz
(cm) * (10^-6) (10^-6) (10^-6) (microns) (microns) (microns)
14.99 1 149.54 202.83 -202.24 -17.657 -4.922 519.524
0 1 -146.74 -191.27 59.17 18.379 10.161 533.205
cPrincipal Stresses and Strains
Z-Position Layer S1 S2 S3 E1 E2 E3
(cm) * (kPa) (kPa) (kPa) (10^-6) (10^-6) (10^-6)
14.99 1 -70.3 841.64 980.12 -202.26 149.49 202.9
0 1 -1339.59 -1223.5 -690 -191.39 -146.61 59.17
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,8. Use a layered elastic analysis program to determine the radial stresses at the bottom of the surface
layer directly under any load for a pavement with three layers as follows:
Modulus Poisson Thickness
Layer (psi) Ratio (in)
1 435113 0.35 10.63
2 21755.7 0.4 20.08 Full
Friction
3 7251.9 0.4 Infinite between
all layers
Consider three different cases of loads, as follows:
a) Single axle with dual tires:
X Y Load Pressure
Tire# (in) (in) (lb) (psi)
1 0 0 5000 100
2 13.5 0 5000 100
b) Tandem axle with dual tires
X Y Load Pressure
Tire# (in) (in) (lb) (psi)
1 0 0 5000 100
2 13.5 0 5000 100
3 13.5 54 5000 100
4 0 54 5000 100
c) Tridem axle with dual tires
X Y Load Pressure
Tire# (in) (in) (lb) (psi)
1 0 0 5000 100
2 13.5 0 5000 100
3 13.5 54 5000 100
4 0 54 5000 100
5 0 108 5000 100
6 13.5 108 5000 100
Solution: See the following outputs from EVERSTRESS
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,a)
Layered Elastic Analysis by EverStress for Windows
Title: Chapter 2 Problem 8-B
No of X-Y
No of Loads: Evaluation Points: 1
No of Layers: 3 2
Layer Poisson's Thickness Moduli(1)
* Ratio (in) (ksi)
1 0.35 10.63 435
2 0.4 20.08 21.75
3 0.4 * 7.25
Load No X-Position Y-Position Load Pressure Radius
* (in) (in) (lbf) (psi) (in)
1 0 0 5000 100 3.99
2 13.5 0 5000 100 3.99
X-Position Y-Position (in):
Location No: 1 (in): .000 .000
Line
Normal Stresses
Z-Position Layer Sxx Syy Szz Syz Sxz Sxy
(in) * (psi) (psi) (psi) (psi) (psi) (psi)
10.599 1 52.73 63.37 -5.5 0 0.97 0
0 1 -125.86 -134.74 -100 0 0 0
Line
Normal Strains and
Deflections
Z-Position Layer Exx Eyy Ezz Ux Uy Uz
(in) * (10^-6) (10^-6) (10^-6) (mils) (mils) (mils)
10.599 1 74.66 107.67 -106.06 -0.47 0 14.749
0 1 -100.46 -128.01 -20.21 0.478 0 15.706
Line
Principal Stresses and Strains
Z-Position Layer S1 S2 S3 E1 E2 E3
(10^- (10^- (10^-
(in) * (psi) (psi) (psi) 6) 6) 6)
-
10.599 1 -5.52 52.75 63.37 106.11 74.71 107.67
- -
0 1 -134.74 -125.86 -100 128.01 100.46 -20.21
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,CHAPTER 2. PRINCIPLES OF MIX AND STRUCTURAL DESIGN AND CONSTRUCTION OF ASPHALT
PAVEMENT
Problem 5: Determine the vertical and radial stresses at nine points for a 9,000 lb point load on a
homogeneous, isotropic, linear elastic, semi-infinite space. Consider a Poisson’s ratio of 0.3.
Point z, inch r, inch Point z, inch r, inch Point z, inch r, inch
1 0 0 4 6 0 7 12 0
2 0 6 5 6 6 8 12 6
3 0 12 6 6 12 9 12 12
Solution:
P= 9,000 lbs; v ? r ?
Point 1: z= 0, r= 0
Use Boussinesq’s Method; assume u= 0.3
R2 r 2 z 2 = 0
3Pz 3
v
2 R5
P(1 u) 3r 2 z (1 2u)R
r
2 R2 R
3
R z
Point 2: z= 0, r=6
R 2 36
3(9000)(0)
v 0
2 R5
9000(1 0.3) (1 2 * 0.3)(6)
r 0
2 (36) 20.69 psi
6 0
Point 3: z= 0, r= 12; R 2 144 ; R=12
v 0
9000(1 0.3) (1 2 * 0.3) *12
r 0 5.17 psi
2 (144) 12 0
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,Point 4: z= 6 in., r= 0
R 2 6 2 36; R 6
3 3
3Pz 3(9000)(6)
v 5 119.37 psi
2R 2 (6)5
2
P(1 u) 3r z (1 2u) R 9000(1 0.3) (1 2 * 0.3)(6) psi
0
r
2 R2 R3 R z 2 (6) 2 6 6 10.35
Point 5: z=6, r=6
R 2 26 26 72; R 8.48
3(9000)(6)3
v 21.17 psi
2 (8.48)5
9000(1 0.3) 3(62 )(6) (1 2 * 0.3)(8.48)
r 21.39 psi
3
2 (8.48)2 8.48 8.48 6
Point 6: z= 6, r= 12
R 2 36 144 180
R 13.42
Point 7: z= 12, r= 0
R 2 122 144
R=12
3(9000)(12)3
v 29.84 psi
2 (12)5
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, 9000(1 0.3) (1 2 * 0.3)(12)
0 2.59 psi
r
2 (12)2 12 12
Point 8: z =12, r=6
R2 = 144+36 = 180 R=13.42
3(9000)(12)3
v 17.06 psi
2 (13.42)5
9000(1 0.3) 3(6)2 (12) (1 2 * 0.3)(13.42)
3.31psi
r
2 (13.42)2 (13.42)
3
13.42 12
Point 9: z =12, r=12
R2 = 144+144 = 288 R=16.97
3(9000)(12)3
v 5.28 psi
2 (16.97)5
6. If the deflection at the center of a rigid plate of radius 6 inch is found out to be 0.03 inch from a load of
9,000 lb on a subgrade with Poisson’s ratio of 0.35, what is the estimated modulus of the subgrade?
Solution:
�1−µ �𝑃
2 (1−0.35)(9,000)
∆= =
2𝐸𝑎 2𝐸6
(1 − 0.352)(9,000)
0.03 =
12𝐸
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,E=21,937.5 psi
7. Use any layered elastic analysis program to compute the vertical and radial stresses and strains directly
below the load at a depth of 149 mm in a full depth asphalt pavement with a thickness of 150 mm, and a
modulus of 3,500 MPa and Poisson’s ratio of 0.35, for the following loading conditions. The subgrade has a
modulus of 100 MPa and a Poisson’s ratio of 0.4. In each case half of a standard 18,000 lb axle (only the
main load bearing axles, not including the steering axle) has been indicated. Can you sketch the
axle/wheel configuration of the entire vehicles?
a) Loads of 20 kN, with coordinates in cm (x, y): (0,0) (33, 0); (0,122) (133,122); tire
pressure of 690 kPa
b) Loads of 20 kN, with coordinates in inch (x, y): (0,0) (33,0); (0,122) (33,122);
(0,244) (33,244); tire pressure of 690 kPa
Solution: See the following outputs from EVERSTRESS
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,a
Layered Elastic Analysis by EverStress for Windows
Title: Chapter 2 Problem 7a
No of Layers: No of Loads:
2 4 No of X-Y Evaluation Points: 1
Layer Poisson's Thickness Moduli(1)
* Ratio (cm) (MPa)
1 0.35 15 3500
2 0.4 * 100
Load No X-Position Y-Position Load Pressure Radius
* (cm) (cm) (N) (kPa) (cm)
1 0 0 20000 690 9.605
2 33 0 20000 690 9.605
3 0 122 20000 690 9.605
4 33 122 20000 690 9.605
Location No: 1
X-Position (cm): .000
Y-Position (cm): .000
Normal
Stresses
Z-Position Layer Sxx Syy Szz Syz Sxz Sxy
(cm) * (kPa) (kPa) (kPa) (kPa) (kPa) (kPa)
14.99 1 843.06 983.73 -70.27 0.9 7.29 -4.98
Normal Strains
and Deflections
Z-Position Layer Exx Eyy Ezz Ux Uy Uz
(cm) * (10^-6) (10^-6) (10^-6) (microns) (microns) (microns)
14.99 1 149.53 203.79 -202.76 -17.654 -4.876 476.542
Line
Principal
Stresses and
Strains
Z-Position Layer S1 S2 S3 E1 E2 E3
(cm) * (kPa) (kPa) (kPa) (10^-6) (10^-6) (10^-6)
14.99 1 -70.33 842.94 983.9 -202.78 149.48 203.85
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, b)
Title: Chapter 2 Problem 7-b
No of Loads: 6
No of Layers: 2 No of X-Y Evaluation Points: 1
Layer Poisson's Thickness Moduli(1)
* Ratio (cm) (MPa)
1 0.35 15 3500
2 0.4 * 100
Load No X-Position Y-Position Load Pressure Radius
* (cm) (cm) (N) (kPa) (cm)
1 0 0 20000 690 9.605
2 33 0 20000 690 9.605
3 0 122 20000 690 9.605
4 33 122 20000 690 9.605
5 0 244 20000 690 9.605
6 33 244 20000 690 9.605
X-Position Y-Position
(cm): (cm):
Location No: 1 .000 .000
cNormal Stresses
Z-Position Layer Sxx Syy Szz Syz Sxz Sxy
(cm) * (kPa) (kPa) (kPa) (kPa) (kPa) (kPa)
14.99 1 841.78 979.93 -70.24 1.02 7.3 -5.15
0 1 -1223.82 -1339.27 -690 0 0 6.09
cNormal Strains and Deflections
Z-Position Layer Exx Eyy Ezz Ux Uy Uz
(cm) * (10^-6) (10^-6) (10^-6) (microns) (microns) (microns)
14.99 1 149.54 202.83 -202.24 -17.657 -4.922 519.524
0 1 -146.74 -191.27 59.17 18.379 10.161 533.205
cPrincipal Stresses and Strains
Z-Position Layer S1 S2 S3 E1 E2 E3
(cm) * (kPa) (kPa) (kPa) (10^-6) (10^-6) (10^-6)
14.99 1 -70.3 841.64 980.12 -202.26 149.49 202.9
0 1 -1339.59 -1223.5 -690 -191.39 -146.61 59.17
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,8. Use a layered elastic analysis program to determine the radial stresses at the bottom of the surface
layer directly under any load for a pavement with three layers as follows:
Modulus Poisson Thickness
Layer (psi) Ratio (in)
1 435113 0.35 10.63
2 21755.7 0.4 20.08 Full
Friction
3 7251.9 0.4 Infinite between
all layers
Consider three different cases of loads, as follows:
a) Single axle with dual tires:
X Y Load Pressure
Tire# (in) (in) (lb) (psi)
1 0 0 5000 100
2 13.5 0 5000 100
b) Tandem axle with dual tires
X Y Load Pressure
Tire# (in) (in) (lb) (psi)
1 0 0 5000 100
2 13.5 0 5000 100
3 13.5 54 5000 100
4 0 54 5000 100
c) Tridem axle with dual tires
X Y Load Pressure
Tire# (in) (in) (lb) (psi)
1 0 0 5000 100
2 13.5 0 5000 100
3 13.5 54 5000 100
4 0 54 5000 100
5 0 108 5000 100
6 13.5 108 5000 100
Solution: See the following outputs from EVERSTRESS
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,a)
Layered Elastic Analysis by EverStress for Windows
Title: Chapter 2 Problem 8-B
No of X-Y
No of Loads: Evaluation Points: 1
No of Layers: 3 2
Layer Poisson's Thickness Moduli(1)
* Ratio (in) (ksi)
1 0.35 10.63 435
2 0.4 20.08 21.75
3 0.4 * 7.25
Load No X-Position Y-Position Load Pressure Radius
* (in) (in) (lbf) (psi) (in)
1 0 0 5000 100 3.99
2 13.5 0 5000 100 3.99
X-Position Y-Position (in):
Location No: 1 (in): .000 .000
Line
Normal Stresses
Z-Position Layer Sxx Syy Szz Syz Sxz Sxy
(in) * (psi) (psi) (psi) (psi) (psi) (psi)
10.599 1 52.73 63.37 -5.5 0 0.97 0
0 1 -125.86 -134.74 -100 0 0 0
Line
Normal Strains and
Deflections
Z-Position Layer Exx Eyy Ezz Ux Uy Uz
(in) * (10^-6) (10^-6) (10^-6) (mils) (mils) (mils)
10.599 1 74.66 107.67 -106.06 -0.47 0 14.749
0 1 -100.46 -128.01 -20.21 0.478 0 15.706
Line
Principal Stresses and Strains
Z-Position Layer S1 S2 S3 E1 E2 E3
(10^- (10^- (10^-
(in) * (psi) (psi) (psi) 6) 6) 6)
-
10.599 1 -5.52 52.75 63.37 106.11 74.71 107.67
- -
0 1 -134.74 -125.86 -100 128.01 100.46 -20.21
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, THOSE WERE PREVIEW PAGES
TO DOWNLOAD THE FULL PDF
CLICK ON THE L.I.N.K
ON THE NEXT PAGE
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