L=L, Q,-Σα"c,p ΔL (10.37) L=0 atmospheric pressure = 100 kN/m² (~2000 lb/ft?). So, conservatively, we may assume that where Pa = a* = 0.4 (10.39) Clay D, Cu(1) Clay L2 Cu(2) Figure P10.7
L=L, Q,-Σα"c,p ΔL (10.37) L=0 atmospheric pressure = 100 kN/m² (~2000 lb/ft?). So, conservatively, we may assume that where Pa = a* = 0.4 (10.39) Clay D, Cu(1) Clay L2 Cu(2) Figure P10.7
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Concept explainers
Question
Figure P10.7 shows a drilled shaft without a bell. Assume the following values:
L1 = 6 m cu(1) = 50 kN/m2
L2 = 7 m cu(2) = 75 kN/m2
Ds = 1.5 m
Determine:
a. The net ultimate point bearing capacity [use Eqs. (10.33) and (10.34)]
b. The ultimate skin friction [use Eqs. (10.37) and (10.39)]
c. The working load Qw (factor of safety = 3)
![L=L,
Q,-Σα"c,p ΔL
(10.37)
L=0
atmospheric pressure = 100 kN/m² (~2000 lb/ft?).
So, conservatively, we may assume that
where Pa =
a* = 0.4
(10.39)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff85071e9-beed-4083-845d-0dfdce09d330%2F6fabea99-5e82-453c-9322-a7f47004bc0d%2F225vl6p_processed.png&w=3840&q=75)
Transcribed Image Text:L=L,
Q,-Σα"c,p ΔL
(10.37)
L=0
atmospheric pressure = 100 kN/m² (~2000 lb/ft?).
So, conservatively, we may assume that
where Pa =
a* = 0.4
(10.39)
![Clay
D,
Cu(1)
Clay
L2
Cu(2)
Figure P10.7](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff85071e9-beed-4083-845d-0dfdce09d330%2F6fabea99-5e82-453c-9322-a7f47004bc0d%2Fqu2jz19_processed.png&w=3840&q=75)
Transcribed Image Text:Clay
D,
Cu(1)
Clay
L2
Cu(2)
Figure P10.7
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