Problem 1. Calculate the ultimate load carrying capacity of a 25 ft long and 24 inch dia pile. Which is embeded in clayey soil, which coefficient of adhesion = 0.5 and
confined compressive strength of soil = 2000 psf.
Solution:
Bearing capacity of pile,
Qult = αAsCs + NcApCp
Here, Pile length = 25 ft
Pile dia = 24 inch
α = 0.5
Confined compressive strength = 2000 psf
∴ Qult ={ 0.5 *π * (24/12) * 25 * (2000 / 2)} + {9 * (π/4) (24/12)2 * (2000/2)}
= 106814.4 lb
= 106.81 kip
Answer: 106.81 kip
Problem 2. A nine pile group consisting of 18 inch diameter concrete pile was cast in situ in a soil whose unconfined compressive strength is 1000 lb/ ft2. Each pile is 60 ft long. Using a reduction factor α = 0.5 and factor of safety 3. Calculate allowable skin friction of single pile.
Solution:
Allowable skin friction,
Qs = αAsCs
Here, α = 0.5
As= π * (18/12) * 60
Cs =1000 / 2
= 500
Qs = 0.5* π * (18/12) * 60 *500
= 70686 lb
= 70.68 kip
Answer: 70.68 kip
Problem 3. A pile of embeded length is 100 ft and diameter 24" and unconfined compressive strength is 2000 psi. Determine the ultimate capacity of pile where α = 0.5?
Solution:
Qult = αAsCs + NcApCp
Here, α = 0.5
Diameter of pile = 24" = 2'
Cu = 2000 psi
C = 2000/2
= 1000 psi
Qult = (0.5 * π * 24* 100 * 12 *1000) + (9 * π/4 * 242 * 1000)
= 49310553.5 lb
= 49310.6 kip
Answer: 49310.6 kip
Problem 4. There are nine piles in a group each having 18 inch dia, 65 ft long, α = 0.5, unconfined compressive strength of soil is 1000 lb/ ft2, cast in situ pile in a layer. What is the allowable skin friction of a single pile if factor of safety is 3?
Solution:
Allowable skin friction,
Qs = αAsCs
= {0.5 * π * (18/12) * 65 *(1000/2)}
= 76576.5 lb
= 76.58 kip
Answer: 76.58 kip
Problem 5. Concrete pile of 24" dia embeded in clay, unconfined compression strength of the soil is 4ksf and the embeded length of the pile is 30 ft. Calculate ultimate load carrying capacity of the pile if adhesive friction between clay and pile is in 30 ft. Unit weight of clay is 100 lb/ ft3 and water table exists at ground?
Solution:
Ultimate load carrying capacity,
Qult = αAsCs + NcApCp
= {0.5 * π * (24/12) * 30 * (4000/2)} + { 9* (π/4) * (24/12)2 * 4000/2}
= 245044.8 lb
= 245.04 kip
Answer: 245.04 kip
Problem 6. A pile dia 20", length = 60', Cu= 3 ksf. What will be it's ultimate capacity?
Solution:
Ultimate load carrying capacity,
Qult = αAsCs + NcApCp
= {0.5 * π * (20/12) * 60 * (3/2)} + {9 * (π /4) * (20/12)2 * (3/2) }
= 265.07 kip (Ans.)
Ques# 1----- "unconfined compressive strength" is given.
ReplyDelete