then both normal and shear failure stress will be used. in this case divide normal load corrected area and find the new normall stress against failure point of shear stress.... now in this case graph of shear stress vs normal stress corrected one will be used
Yes about corrected area ! Un your curve we have to use the stresses at faillure (normal stress at the moment of faillure too). This is calculated whit the corrected area
@@cheikhoundiaye5556 your point is valid but we can take corrected area when we have equipment with vertical displacement..usually direct shear test apparatus dont have vertical displacement.thats why..and ofcourse it will be failure shear stress vs normal stress..also we make we make a failure envelop considering linear trend so c and phi are already somehow have conservative values..due to all these we have to perform triaxial
No, its not imposed to get vertical displacement ! The corrected area is the same. So the normal stres is calculated for a ratio between the initial normal force considered and the corrected area
@@cheikhoundiaye5556 in my opinion we must have vertical displacement as well..but i consider your point...and secondly tbere is no area correction recommendatio n ASTM D3080-04 thats why i didnt include this point..but i have included this point in triaxial test video
Thanks very helpful but can you send me a Geotech worksheet and excel files? and i have questions? where the shear force came from? is it from Proving ring? and the Displacement? is it horizontal displacement or vertical displacement?
Aftet lockdown I will be uploading detailed videos of sample preparations, practical performance etc and the technical demonstration videos as per ASTM and domestic practices. There many youtube channel but i don't know a specific channel.
Area correction is not recommended for small scale laboratory tests. It has not been considered in ASTM. For large scale Direct shear test, area correction should be apply.
How do you choose those values of 50 100 and 200 for Normal stress? If the answer is checking the depth of the sample, could you specify more? Because I never see an example where they calculate the normal stress for this type of problem. Thank you
In direct shear test, we don't simulate the effective depth rather than that in accordance to ASTM D3080 we need to develop a failure envelop. so its upto you, you can either adopt normal stresses of 25, 50 and 100 kpa as well. so it is not mendatory to adopt 50, 100 and 200 kpa. For more details plz read ASTM carefully
It's quite helpful.
Thanks for sharing...
Great demonstration. Thanks.
Hi which one would be vertical and horizontal displacement?
thank you so much, sir. It's quite helpful.
Excellent demonstration
Really helpful sir
In some cases load reading multiply with constant & 5
Can you explain it?
informative. thanks.
Very informative videoo sir.
Nice vedio, by d way can we fix cohesion value automatically.
Thanks for sharing sir
you are a legend sir
Hello Naqeeb i have got a data in which there is strain as well along with the normal and shear load. How to consider strain in the test?
then both normal and shear failure stress will be used. in this case divide normal load corrected area and find the new normall stress against failure point of shear stress.... now in this case graph of shear stress vs normal stress corrected one will be used
Hello! Amazing video. Can you please share the XL Sheet?
it was helpful me. thank you
thankyou
What does it mean if the cohesion intercept (y intercept) is negative?
it cant be negatv.... then you need to repeat the test
thanks a lot
thank you
what about rock direct shear test
If the trendline is not coming straight, is it consider ok or not?
Pliz reply
what do you mean??
@@GeotechwithNaqeeb i had done my test and my results shows curve graph!
Is it correct or not
I think that your normal stresses must be adapted whit the displacement. At rigour studies the impacted zone is the corrected surface !!!
can you elaborate plz...are u saying about use of corrected area??
Yes about corrected area ! Un your curve we have to use the stresses at faillure (normal stress at the moment of faillure too). This is calculated whit the corrected area
@@cheikhoundiaye5556 your point is valid but we can take corrected area when we have equipment with vertical displacement..usually direct shear test apparatus dont have vertical displacement.thats why..and ofcourse it will be failure shear stress vs normal stress..also we make we make a failure envelop considering linear trend so c and phi are already somehow have conservative values..due to all these we have to perform triaxial
No, its not imposed to get vertical displacement ! The corrected area is the same. So the normal stres is calculated for a ratio between the initial normal force considered and the corrected area
@@cheikhoundiaye5556 in my opinion we must have vertical displacement as well..but i consider your point...and secondly tbere is no area correction recommendatio n ASTM D3080-04 thats why i didnt include this point..but i have included this point in triaxial test video
Is this horizontal displacement?
yes horizontal
Thanks very helpful but can you send me a Geotech worksheet and excel files?
and i have questions? where the shear force came from? is it from Proving ring?
and the Displacement? is it horizontal displacement or vertical displacement?
@geotech with naqeeb
Yes I am now struggling on the same thing
Hi if you still here can the cohesion be negative thanks
no it cant be negative.
Could you recommend some laboratory testing vedio to see preparing sample and test conducted?
Aftet lockdown I will be uploading detailed videos of sample preparations, practical performance etc and the technical demonstration videos as per ASTM and domestic practices. There many youtube channel but i don't know a specific channel.
@@GeotechwithNaqeeb thanks for the response.
I think u forgot.sir.
Thanks
may you please share the excel sheet?
Is there any rules for normal load
you need to develop a failure envelop, so normal load can be adjusted with experience. There is no specific rule defined in ASTM.
very helpful thank you.
what about the correction of area ?
Area correction is not recommended for small scale laboratory tests. It has not been considered in ASTM. For large scale Direct shear test, area correction should be apply.
How do you choose those values of 50 100 and 200 for Normal stress? If the answer is checking the depth of the sample, could you specify more? Because I never see an example where they calculate the normal stress for this type of problem. Thank you
In direct shear test, we don't simulate the effective depth rather than that in accordance to ASTM D3080 we need to develop a failure envelop. so its upto you, you can either adopt normal stresses of 25, 50 and 100 kpa as well. so it is not mendatory to adopt 50, 100 and 200 kpa. For more details plz read ASTM carefully
If chohesion increases angle of friction also increases aa please reply me
not necessarily. it depends on soil type and particle shape etc
Thank you what happens with black soil
kindly share the excle file as well as the software used if possible CASTeR: Computer Aided Soil Test Report
email??
@@GeotechwithNaqeeb eidmuhammadzurmutay@gmail.com
may you please share the excel sheet?
email??
Please share it if u have recieved it brother
Sir Excel sheet provide Kara digye
Excel sheet download
thank you
Thanks