Assume that the footing in problem 1 is of a 6 ft radius. Calculate the ultimate bearing capacity for the assumptions of general and local shear failures. GENERAL guit - 33 834 LbHE gult = 4,744 Ubt LOCAL F=200,000 lb problem 1 SAND W=5% * = 32 ° G = 2.65 e,= 0.4 6 ft Sult =D,177 t 6 ft 10 ft Somontho Q109 ft CLAY 14 ft C = 0.4 Y sar= 115 pcf C = 0.05 C, = 0.2 ft :/day eo= 0.9 OCR =1.1 ROCK
Assume that the footing in problem 1 is of a 6 ft radius. Calculate the ultimate bearing capacity for the assumptions of general and local shear failures. GENERAL guit - 33 834 LbHE gult = 4,744 Ubt LOCAL F=200,000 lb problem 1 SAND W=5% * = 32 ° G = 2.65 e,= 0.4 6 ft Sult =D,177 t 6 ft 10 ft Somontho Q109 ft CLAY 14 ft C = 0.4 Y sar= 115 pcf C = 0.05 C, = 0.2 ft :/day eo= 0.9 OCR =1.1 ROCK
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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![Assume that the footing in problem 1 is of a 6 ft radius. Calculate the ultimate bearing
capacity for the assumptions of general and local shear failures.
GENERAL
guit
33,834 LEYHE
LOCAL
gult = 4,744 lo jet
F=200,000 lb
problem 1
SAND W=5%
$ = 32 °
G = 2.65
6 ft
Sult =D,177 dt
59months =Q109
e,= 0.4
6 ft
10 ft
CLAY
14 ft
eo = 0.9
Y sar= 115 pcf C = 0.05
OCR =1.1
C = 0.4
C, = 0.2 ft :/day
ROCK](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb6fe8eaf-5ab3-4d02-a2f2-a9d462e4c47c%2F4d320ab4-1c59-44fa-b210-1c19b65915a6%2Fu5v7hdg_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Assume that the footing in problem 1 is of a 6 ft radius. Calculate the ultimate bearing
capacity for the assumptions of general and local shear failures.
GENERAL
guit
33,834 LEYHE
LOCAL
gult = 4,744 lo jet
F=200,000 lb
problem 1
SAND W=5%
$ = 32 °
G = 2.65
6 ft
Sult =D,177 dt
59months =Q109
e,= 0.4
6 ft
10 ft
CLAY
14 ft
eo = 0.9
Y sar= 115 pcf C = 0.05
OCR =1.1
C = 0.4
C, = 0.2 ft :/day
ROCK
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