PILE

Structural Layout Vignette and Multiple Choice

PILE

Postby jacky » Thu Nov 28, 2013 1:02 pm

FROM ARE STUDY REVIEW 4.0

Foundation Piles 2

Four 12 inch diameter piles (each of whose frictional capacity is 300 psf) support a column carrying 120 kips total load.



How deep must each pile be to support the load?

A. 24 ft

B. 30 ft

C. 32 ft

D. 40ft



AnswerC. 32 ft. From the diameter of each pile (r/2 = 12/2 = 6 in = 0.5 ft) the circumference is (C = 2∏r = 2x3.142x0.5) 3.142. Dividing the required frictional area by the circumference of the circular piles equals the required depth (100sf / 3.142 = 31.83 ft or 32 ft)


I confuse that the where this 100sf comes from? could be very basic.... but could not find out..
jacky
 
Posts: 8
Joined: Sat Nov 23, 2013 5:41 pm

Re: PILE

Postby Corkscrewed » Sat Nov 30, 2013 3:04 am

1) total load = 120 kips = 120,000 lbs total load
2) total frictional capacity = 4 x 300 psf = 1200 psf total frictional capacity
3) 120,000 lbs / 1200 psf = 100 sf of surface area required by all four piles working together.
4) surface area = circumference x depth, thus depth = surface area / circumference
depth = 100 sf / 3.14159 ft = 31.83 ft

alternately...

1) total load = 120 kips = 120,000 lbs total load
2) assume total load is evenly distributed among all four piles, then load/pile = total load / 4 = 120,000 lbs / 4 = 30,000 lbs / pile
3) 30,000 lbs load / 300 psf frictional capacity = 100 sf surface area required per pile
4) surface area = circumference x depth, thus depth = surface area / circumference
depth = 100 sf / 3.14159 ft = 31.83 ft
User avatar
Corkscrewed
 
Posts: 517
Joined: Sat Nov 30, 2013 2:37 am

Re: PILE

Postby jacky » Mon Dec 02, 2013 6:03 am

Thanks very easy! thank you so much... I tried to consider difficult and wrong way.. Thanks again.
jacky
 
Posts: 8
Joined: Sat Nov 23, 2013 5:41 pm


Return to SS - STRUCTURAL SYSTEMS

Who is online

Users browsing this forum: No registered users and 55 guests

cron