Negative FOS for slope stability in planar failure slopes

[IronStallion]

When I calculate the factor of the safety for a slope with a discontinuity which meets all the criteria for planar slope failure it comes out as -0.69 what does this mean ?

 

[Lomarandil]

Which method are you using to analyze the slope?

 

[GeoPaveTraffic]

It means something is seriously wrong with your model. There is no such think as a negative FOS.

Mike Lambert

 

[IronStallion]

The slope I'm analysing is hypothetical. I'm using the factor of safety equation which includes the presence of a critical tension crack at the slope crest(with the tension crack full of water). Does it mean the slope is completely unstable?

 

[Lomarandil]

Without knowing more about your method, I agree with GeoPave... something is wrong.

The FOS should be the ratio of resisting forces to driving forces -- neither quantity should be negative, so the FOS should not be negative.

 

[killswitchengage]

maybe you entered a value negative somewhere by mistake . check the cohesion and angle of friction as well as the unit weight

 

[IronStallion]

Im the doing the FOS calculations on paper. Well the thing is the top value for the calculation came out negative because the tension crack was 15m and was assumed to be full which made W U and V quite large. Im guessing the slope failed at a much lower depth of water in the tension crack ?

 

[GeoPaveTraffic]

A 15m tension crack is very unlikely. However, even that shouldn't make the driving forces negative since the water pressure is positive driving force.

Check all of you normal forces. Are they all positive?

Mike Lambert

 

[Okiryu]

Are you using the method of slices? Depending on the location of the center of your slip circle, some angles at the base of the slice will be negative. These affects the weight of the slice, making it negative. Perhaps you have more "negative" weights than "positive" weights that at the end will give you a negative FOS. Check again your calcs...

 

[killswitchengage]

If his using planar failure , an infinite Fellenius Slope analysis would suffice. I still maintain the idea that he used a negative value somewhere. The driving forces in this case are the PWP in the crack which lets assume is hydrostatic and the bouyant weight of a typical slice . OP should post his work here

 

[marc2016]

I would say check your force diagrams to ensure they are complete and resolve well. Full water in tension cracks sometimes throw the assessment resulting in negative values. I would reduce the water in the crack to 80% full and see what you get...

 

[fattdad]

check the software's assumptions on coordinates.

f-d

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