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!

# Help! Lindeburg sample exam morning #14

Started by
ronnnald
, Mar 02 2011 12:46 AM

2 replies to this topic

### #1

Posted 02 March 2011 - 12:46 AM

### #2

Posted 06 March 2011 - 04:53 AM

QUOTE (ronnnald @ Mar 1 2011, 04:46 PM) <{POST_SNAPBACK}>

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!

Help please....Thank you!

This type of chart doesn't work for all type of Sloping soils, you can used when certain conditions applied:

1 No open water outside of the slope

2 No surcharge or tension cracks

3 The soil is Homogeneous (for this problem the soil is sat)

4 the shear strength is from cohesion only (c=50 KPa)

5 Failure takes place on circular arc

if you check the problem you'll see that this problem has all these characteristics.

Hope this won't confused you more!

### #3

Posted 08 March 2011 - 02:36 PM

QUOTE (ronnnald @ Mar 1 2011, 07:46 PM) <{POST_SNAPBACK}>

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!

Help please....Thank you!

I do not have this problem in front of me, but if I understood your question correct, that is mean that the layer of soil you are working on, is below water level. In this case you need to find effective stress or you may find effective specific weight. See chapter 35.7 page 35-14 in CERM.

Hope it helped

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