-4. Design a square foundation (find the minimum width) that meets the following two conditions: 1) The foundation should carry a vertical loading P=800 kN, with a factor of safety 2, and 2) The ultimate settlement should be less than 0.3 m. Assume the general shear failure. 1m 0.7 m 4m CLAY ROCK y = 17.2 kN/m = 32° Y = 19 kN/m eo=0.8 C = 0.3 B=1,41 m
-4. Design a square foundation (find the minimum width) that meets the following two conditions: 1) The foundation should carry a vertical loading P=800 kN, with a factor of safety 2, and 2) The ultimate settlement should be less than 0.3 m. Assume the general shear failure. 1m 0.7 m 4m CLAY ROCK y = 17.2 kN/m = 32° Y = 19 kN/m eo=0.8 C = 0.3 B=1,41 m
Principles of Foundation Engineering (MindTap Course List)
8th Edition
ISBN:9781305081550
Author:Braja M. Das
Publisher:Braja M. Das
Chapter4: Shallow Foundations: Ultimate Bearing Capacity
Section: Chapter Questions
Problem 4.2P: A square column foundation has to carry a gross allowable load of 1805 kN (FS = 3). Given: Df = 1.5...
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Please solve this problem with step by step solutions and clear calculations so I can understand the theory and concept, Thank you
![4. Design a square foundation (find the minimum width) that meets the following two
conditions: 1) The foundation should carry a vertical loading P=800 KN, with a factor of safety 2,
and 2) The ultimate settlement should be less than 0.3 m. Assume the general shear failure.
1m
0.7 m
4 m
CLAY
ROCK
Y = 17.2 kN/m
$ = 32°
Y
eo= 0.8
Co = 0.3
19 kN/m
В=1,41 т](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F78a1cdca-9a2c-4803-aedc-ba4e0e310e87%2F8ccdb097-59f8-4f2c-bee0-ed41c175ca64%2Fprdbgj_processed.jpeg&w=3840&q=75)
Transcribed Image Text:4. Design a square foundation (find the minimum width) that meets the following two
conditions: 1) The foundation should carry a vertical loading P=800 KN, with a factor of safety 2,
and 2) The ultimate settlement should be less than 0.3 m. Assume the general shear failure.
1m
0.7 m
4 m
CLAY
ROCK
Y = 17.2 kN/m
$ = 32°
Y
eo= 0.8
Co = 0.3
19 kN/m
В=1,41 т
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