Sign in to follow this  
Followers 0
IRSAgent

Sheet Piles Into Clay

3 posts in this topic

In Lindeburg's Practice Problems, Chapter 39 Problem #2:

A 35 ft long sheet pile is driven through 10 ft of clay to bedrock below. The sheet pile supports a 25 ft vertical cut through drained sand. A tie rod is located 8 ft below the surface, terminating at a deadman behind the failure plane. There is no significant water table.

What is the tensile force in the tie rod, given the following soil parameters?

Silty Sand

Internal Friciton Angle: 32 degrees

Specific Weight: 110 lbf/ft3

Cohesion = 0

Clay

Internal Friciton Angle: 0 degrees

Specific Weight: 120 lbf/ft3

Cohesion = 750 lbf/ft2

The answer they give is 8050 lbf

I can follow most everything they do to get that result, but it appears that they use a rectangular pressure distrubution within the clay layer. Am I mistaken in this understanding? Is it a rectangular pressure because phi=0, or are they somehow referring to this component as a surcharge, which equates to a uniform lateral pressure?

Share this post


Link to post
Share on other sites

Yes. The key here is that they give you the internal friction angle of clay=0. The reaction on the lower portion of the sheet pile will not engage a wedge of soil (triangle), it will provide uniform lateral resistance (rectangle).

The clues are in recognizing what is given in the problem statement.

Share this post


Link to post
Share on other sites
Yes. The key here is that they give you the internal friction angle of clay=0. The reaction on the lower portion of the sheet pile will not engage a wedge of soil (triangle), it will provide uniform lateral resistance (rectangle).

The clues are in recognizing what is given in the problem statement.

Thanks for your response. I couldn't find anywhere in the CERM where it explicitly states the shape of the distribution of pressure in clays. It makes sense, but I just wanted to confirm.

Edited by IRSAgent

Share this post


Link to post
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!


Register a new account

Sign in

Already have an account? Sign in here.


Sign In Now
Sign in to follow this  
Followers 0