The center of a rectangular area at the ground surface has Cartesian coordinates (0, 0), and one corner has coordinates (7, 18). All dimensions are in meters. The area carries a uniform pressure of 135 kPa. Estimate the stresses at a depth of 15 m below the ground surface at each of the following locations: (0, 0), (0, 18), (7, 0), (7, 18), and (12, 28). a. Obtain the values by the Boussinesq method b. Compare the results with those of the 2:1 method c. Comment on the results Hint: In the Boussinesq method, the equations attached can be used.
The center of a rectangular area at the ground surface has Cartesian coordinates (0, 0), and one corner has coordinates (7, 18). All dimensions are in meters. The area carries a uniform pressure of 135 kPa. Estimate the stresses at a depth of 15 m below the ground surface at each of the following locations: (0, 0), (0, 18), (7, 0), (7, 18), and (12, 28). a. Obtain the values by the Boussinesq method b. Compare the results with those of the 2:1 method c. Comment on the results Hint: In the Boussinesq method, the equations attached can be used.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Question
The center of a rectangular area at the ground surface has Cartesian coordinates (0, 0), and one corner has coordinates (7, 18). All dimensions are in meters. The area carries a uniform pressure of 135 kPa. Estimate the stresses at a depth of 15 m below the ground surface at each of the following locations: (0, 0), (0, 18), (7, 0), (7, 18), and (12, 28).
a. Obtain the values by the Boussinesq method
b. Compare the results with those of the 2:1 method
c. Comment on the results
Hint: In the Boussinesq method, the equations attached can be used.
![►Line load P:
2P
z3
2
π (r² +2²)2
► Corner of a uniformly loaded rectangular area:
qo
mn
2 [√m² +² +1 (m²+1 +²41)
2T
2
2
mn
= qol with I=
0₂ =
O₂ =
z
1
+ tan-¹
1
= [√m²
+
2π /m² +n²+1 m² +1
2
2
2
mn
√m² +n² +1,
F1)
1
n²+1
+ tan-¹
mn
F
√m² + n²+1,
2
2
where: qo: surface stress; m = x/z; n = y/z; x and y: length and width of the
uniformly loaded area](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F40578013-3e17-4421-b4af-d1f1eb4da799%2Fc876a339-f1c1-471d-8ee9-4fcc07911574%2Fquic5u5_processed.jpeg&w=3840&q=75)
Transcribed Image Text:►Line load P:
2P
z3
2
π (r² +2²)2
► Corner of a uniformly loaded rectangular area:
qo
mn
2 [√m² +² +1 (m²+1 +²41)
2T
2
2
mn
= qol with I=
0₂ =
O₂ =
z
1
+ tan-¹
1
= [√m²
+
2π /m² +n²+1 m² +1
2
2
2
mn
√m² +n² +1,
F1)
1
n²+1
+ tan-¹
mn
F
√m² + n²+1,
2
2
where: qo: surface stress; m = x/z; n = y/z; x and y: length and width of the
uniformly loaded area
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