Hello to you Mr. Professor and all the Viewers This is an Water and Environmental Engineering student from Afghanistan. I am in my 4th semester and in this one we study Mechanics of Soil, I`m so happy that I found your channel and lessons, these lessons are so useful and effective for me and I`m sure that I would get high level of information in this subject. THANK YOU IN ADVANCE.....
A sample of dry soil is uniformly mixed with 16% water and compacted in a cylindrical mould. The volume of wet soil is 0.001m^3 and it's mass is 1.60kg. If the relative density of the soil particles is 2.68, calculate the dry density, the void ratio and degree of saturation. Find moisture content if the volume remains unchanged. Also complete the soil diagram
Hi, I would like to know one thing about the phase diagram. My question is that, to what extent does this conceptualization of soil as a three phase diagram is correct and are there any limitations for the soil phase diagram?
We have very rare cases where another phase may exist .. that is materials that decay or dissolve and get leached out.. other than these cases the 3 phase material representation is perfect for soil mechanics...
Not exactly, Gamma sat considers that all soil voids are filled with water (that's why he says that it's used for soils below the water table. Gamma m (you said w but I guess you mean m) means is wet but not necessarily all the voids are filled with water (i.e. voids are filled with a combination of air and water). So to summarize, Gamma d (dry) is for when saturation is 0%, Gamma sat is when saturation is 100% and gamma m (moist) is when saturation is between 0 and 100. Hope this helps.
Hello sir, I had a question... Since the dry unit weight is meant to represent the per unit weight of the soil without water, shouldn't the denominator be 'V total - V of water' since water isn't in the mix anymore? So in the first example that would be: 34.54/(0.33-0.086).
If you remove the volume of the water, you are also removing that part of the void volume that is occupied by water which is inaccurate. If you recall the first slide, he showed that the total volume is made up by the volume of the solid + the volume of the voids. Water sometimes occupy some of the void volume. So to get the actual dry unit weight, you have to consider the total volume, which is includes the total void volume.
for the fill volume: e=Vv/Vs-->Vs(barrow)=(8,000-Vs(barrow))/1.2-->Vs(barrow)=3,636--> since Vs(barrow=Vs(fill)-->Vv(fill)=3,636*e(fill)=2,909. : The Vt(fill)= Vs(fill)+Vv(fill) = 3,636+2,909=6,545 Cu. Yd. for the new amount of water: we know that Vs = 3,636 Cu. Yd. --> w=Ww/Ws--> Ws=Vs Gs Gama(w)--> Ws(fill)=Ws(barrow)-->Ww(barrow)=3636*6.27*62.4*0.15 and Ww(fill)= 3636*6.27*62.4*0.18
Hello to you Mr. Professor and all the Viewers
This is an Water and Environmental Engineering student from Afghanistan.
I am in my 4th semester and in this one we study Mechanics of Soil, I`m so happy that I found your channel and lessons, these lessons are so useful and effective for me and I`m sure that I would get high level of information in this subject.
THANK YOU IN ADVANCE.....
This is a big help for students esp since we're studying online. Thank you so much sir. I make sure to watch as many ads as I can as a thank you
These lecture series are VERY good.
THANK YOU.
This is very good , our teacher recommend your videos
Thank you so much Professor!
hey i got a problem here could you help me?
Highly appreciate your efforts.
Thank you sir
Thank you so much
Thanks Prof
Thank you. Very helpful for me to face final exam of soil mechanics.
Thank you so much for this, Sir.
Thankyou professor, this helps alot
TY sir, hope i pass my exam
Did you pass your exam?
A sample of dry soil is uniformly mixed with 16% water and compacted in a cylindrical mould. The volume of wet soil is 0.001m^3 and it's mass is 1.60kg. If the relative density of the soil particles is 2.68, calculate the dry density, the void ratio and degree of saturation. Find moisture content if the volume remains unchanged. Also complete the soil diagram
Can you do this for me?
Thank you
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thank you sir
Hi, I would like to know one thing about the phase diagram. My question is that, to what extent does this conceptualization of soil as a three phase diagram is correct and are there any limitations for the soil phase diagram?
We have very rare cases where another phase may exist .. that is materials that decay or dissolve and get leached out.. other than these cases the 3 phase material representation is perfect for soil mechanics...
Can you explain more about why the weight of the borrowed soil and the filled soil is being taken as the same
Hello Mr. Is it possible for us to access the slides you are using?
thank so much sir.. for my research about weight volume relationship to my assignment sorry for bad english grammar
🔥
How do you compute for the "Air Content"?
sir can u plss let me know what is ment by e in relative density formulae
At 10:07 gamma w and gamma sat are the same then 🤷🏻♀️
Not exactly, Gamma sat considers that all soil voids are filled with water (that's why he says that it's used for soils below the water table. Gamma m (you said w but I guess you mean m) means is wet but not necessarily all the voids are filled with water (i.e. voids are filled with a combination of air and water). So to summarize, Gamma d (dry) is for when saturation is 0%, Gamma sat is when saturation is 100% and gamma m (moist) is when saturation is between 0 and 100. Hope this helps.
Hello. The dropbox link is not working, Thank you
for example #2, how do we know when to use dry unit weight?
Hello sir, I had a question... Since the dry unit weight is meant to represent the per unit weight of the soil without water, shouldn't the denominator be 'V total - V of water' since water isn't in the mix anymore? So in the first example that would be: 34.54/(0.33-0.086).
If you remove the volume of the water, you are also removing that part of the void volume that is occupied by water which is inaccurate. If you recall the first slide, he showed that the total volume is made up by the volume of the solid + the volume of the voids. Water sometimes occupy some of the void volume. So to get the actual dry unit weight, you have to consider the total volume, which is includes the total void volume.
for the fill volume: e=Vv/Vs-->Vs(barrow)=(8,000-Vs(barrow))/1.2-->Vs(barrow)=3,636--> since Vs(barrow=Vs(fill)-->Vv(fill)=3,636*e(fill)=2,909.
: The Vt(fill)= Vs(fill)+Vv(fill) = 3,636+2,909=6,545 Cu. Yd.
for the new amount of water: we know that Vs = 3,636 Cu. Yd. --> w=Ww/Ws--> Ws=Vs Gs Gama(w)--> Ws(fill)=Ws(barrow)-->Ww(barrow)=3636*6.27*62.4*0.15 and Ww(fill)= 3636*6.27*62.4*0.18
sir, how did u get 62.4pcf, please explain.
its not in metric units either
It must be 9.81 kN/m^3
Thank you so much for replying to the question. I was about to ask the same thing :D
It is the equivalent of 9.81KN/m3 (SI) in Imperial (English) system - for those who might still know the answer to this question.
hello Professor, can you share with me the textbook authors'name you use for lecturing this course? Thank you
Structural Engineers are wacky guys indeed🤣
Error 404 on dropbox
John Malkovich, is that you?
Thank you