# Sheet Piles Into Clay

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### #1 IRSAgent

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Posted 06 April 2011 - 05:43 PM

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?

### #2 MA_PE

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Posted 06 April 2011 - 06:09 PM

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.

### #3 IRSAgent

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Posted 06 April 2011 - 08:32 PM

QUOTE (MA_PE @ Apr 6 2011, 02:09 PM) <{POST_SNAPBACK}>
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, 06 April 2011 - 08:32 PM.

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