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# Sheet Piles Into Clay

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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?

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MA_PE    1,882

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.

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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