# Lindeburg sample exam problem 14



## ronnnald (Mar 1, 2011)

Sorry didn't mean to cross post, I should have posted this question here.

In the Lindeburg sample exam, problem #14 morning, we are suppose to find the cohesion factor of safety. This should be easy enough, just use the Taylor chart to get No, c, the effective specific weight and H. In the solution, Lindeburg doesn't use the effective specific weight...Why??? This is driving me insane. The text says to use effective specific weight...take the saturated density subtract out the specific weight of water...What am I doing wrong?

Help please....Thank you!


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## geo pe (Mar 2, 2011)

I think you have to use the effective unit weight, when the question specifically tells you that the soil is submerged (see CERM Example 40.2). Also the CERM says "The effective unit weight is used when the clay is submerged". In this particular problem, soil is saturated but not submerged. Hence no deduction for water. I hope it helps!!


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## RIP - VTEnviro (Mar 2, 2011)

Mmmm...yeahhh. If you could just go ahead and find that cohesion factor, that'd be great, mmmkay? The thing is, we're putting the new cover sheets on all our TPS reports. I'll make sure you get that memo. Thanks.







Oh wait, you said Lindeburg?


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