Hi Mark, thank you for your dedication to helping students. It is much appreciated. I passed my FE thanks to your resourceful videos and review packets. Onward and upward!
Hey Mark! Thank you so much for your videos. I passed the Exam. It was really helpful to me. I did my exam after 10 year from graduation. I happy to say inly 2 months work hard with your videos.
On prob 7, he should've written (100)^3/1000 when doing the conversion from cm^3/m^3 to kg/m^3 I believe. 10^3/1000 = 1. Thank you for the videos Mark! Exam in 2 days!
For #11 at 1:13:29, I'm wondering why you don't consider the saturated unit weight of the soil that's submerged in water? I haven't taken a soils/geotechnical class yet so I'm not sure if my concept is correct. I used formula in reference manual to get (gamma)sat and used that as (gamma)2 in the Vertical Stress Profile. Thanks!
Your question is a good one. My question is not written very well. I should have given a second unit weight, gamma 2 for the moist soil below the water table. If you use, say 130 pcf for gamma 2, you get an effective stress of about 1000 psf. The question is solved assuming the dry unit weight above and the moist unit weight below the water table are the same, which should be changed. I had a practicing engineer with an MS in geotech qa/qc these problems for me before doing the live session... but neither of us caught this. Thanks for asking! I hope the clarification helps.
Good day Mark. I have a question in no. 18. Did you make Ls an area? Based on the video, you converted the length to ft^2. I hope you will notice my question. Thank you 😊
Hi Mark, Thank you for the great videos. One question regarding question 14. The dimensions of the footing are given as 5x5ft but the thickness is unknown. The second approach for solving the question involves calculating I (Moment of Inertia), and I see that you are not using the thickness of the foundation to get the moment of inertia. However; I think the way the footing is bending, the Moment of Inertia should be calculated as width x thickness^3 / 12. and not width x length^3/12
Hey Mark for (E) the equation is uses specific weight (gamma) so I think we have to multiply by gravity (32.2 ft/s^2) since you gave a density of 115lbs/ft^2. Look at the units. Even if you square the 8ft….. the answer would be lbs/ft. Sorry I am studying back over your notes because I my test is scheduled for May 2nd!!!! Please let me know if I’m looking at this wrong
Great Video Mark! One question: On problem 18, the equation used for TFF doesnt seem dimensionally correct, C*Ls (kip/ft) + Wn*cos*tan (kip). The equation is pulled from the handbook, so I am not sure whether it is calibrated to work that way or if Wn should be divided over the length in the second part of the equation.
i think the assumption is per unit feet, or 1 feet perpendicular to the paper. i think it would be better if mark shows the per feet length value by multiplying it
I noticed that too. I solved for Tff by converting ft to inches and tons to lbs and I got the wrong answer. if the units need to be in ft for the equation to work, I wish it specified that in the manual.
Thanks for posting this! Question on #18 - How do the units work for the Tff calculation? When doing c * Ls, you converted the units to lb/ft, but it's not clear why it needs to be lb/ft instead of (say) lb/inch. Is the 30-ton weight of soil given implied to be for a 1-foot cross section?
Great question. I converted everything to kips and feet. And, yes, to get the units to work out and get that other ft to cancel out in the cohesion piece of it, we need to multiply by a 1-foot depth into the page. Thanks!
Hey Mark! Thanks to your time, I have a doubt regarding Q:11, the Handbook is showing Gama 2 as a "saturated" unit weight, but it is not given in the question the saturated value !! is it correct to use the normal unit weight instead?
He responded to this in a previous comment - the question was poorly written. They will typically give you the saturated unit weight of the submerged soil layer on a problem like this.
Yes, thank you! The 72% is correct, and you get this value when using 122/115 (not 122/100 as shown) for the last part of the relative density formula.
Hi , In Q11, there should be 2 unit weights given. one is dry unit weight and the other one is saturated unit weight of the soil. Or, we can assume the dry unit and saturated unit weight are equal.....thanks
Thats exactly what I thought, as Gama2 should be saturated unit weight instead of the other type od soil! In this case, the saturated unit weight didn't given
Can you elaborate more on question 18 and how did you get the units to work out? I know you said in the comments that you need to multiply kips/ft by 1 ft to end up with kips only, can you elaborate on that more please? Is the 30 tons per ft? as in 30 ton/ ft? If yes, how are we supposed to know? Thank you!
I should have written the 30 tons was for a 1 ft width into the problem statement. At that point, if you multiply through by 1 ft, the units all work out to the answers shown. Many geotech problems are solved on a 1-ft unit width basis. Good catch and thanks for the clarification!
Hi Mark! Thank you for your videos. Will take the exam in a month and it helped me a lot to refresh with all the concepts. For Question 13, I read somewhere that "Normal stress at failure" should be at the point where the mohr-coulomb failure meet the Mohr's circle and not the major principal stress which you used? Please can you clarify?
@ 2:04:30, why are you using a wieght of 60 tones instead of 30 like we did in Tmob force. I appreciate the help with studying and have found your content very helpful!
Nobody's said this and it's driving me wild. For question 17 the only way you get 2 cm is if you use the natural log (ln) instead of log base 10. If you use log base 10 then you get closer to 0.90cm
Hi Mark, thanks for your efforts, I have a question on number 14, why the weight of the retaining wall was not considered in calculating the vertical forces and for calculating the resisting moment?
Hi Mark, for #12, could explain the units for the final answer. This might be just a simple misunderstanding on my end, but pcf (from density) multiplied by ft^2 (from h^2) gets 1130 lb/ft
Hi @AndyZaz, I should have indicated the analysis was for a 1 ft width into the page. That extra foot into the page cancels out the last foot. Does that help?
@@MarkMattsonPE I am confused on how density 115 pcf is unit weight. or was that a wording mistake in problem #12? as well as why multiply unit weight x 1 ft width of wall. I don't understand the relation there?
It was a wording mistake. It should be unit weight. Unit weight includes the impact of gravity. The 1' of wall, is essentially because we are looking at the load per foot of wall into the page. Often, stability analyses are done per foot of wall.
Hi Mark! Finally I paaaaaaseeeed the FE civil exam from the second try 😍😍 I am soooo happy and want to thank you for all your videos I watched it several times Also I want to recommend the book of “ Islam “ which contains 800 questions it was also amazing and let me understand many additional things that not included in the yellow book So it was great knowledge from your videos, the yellow book, and Islam book after completing it I passed the exam 😍😍😍
+ (cc log po + delta p / pc ) did you make a mistake because cr log pc/po is not the same as cc can you please verify the correct way to complete this problem. I'm taking my fe in two weeks and am nervous about this test. I failed this test and trying to understand everything to be more prepared this time.
Hi Mark, I hope you are doing well. In question 14, when you calculated value for Q heel and it comes to be positive. So, why did you say it is compression? It should be tension right?
For problem 12 you bring up surcharge, is there an equation to solve for S or will that possibly be given and we have to factor that value in to solve for a vertical stress?
You don't really have to... you'll get the same answers if you don't divide P and M by 5. That said, I did that for this problem because in the Handbook equation (see pp. 264-265), B is used which defined as "width of base." I wanted to approach this question like a retaining wall so you could see how the equations work and know B is in the same direction as the overturning forces. I actually used a combined stress P/A ± M/S approach when I first solved it, but wanted to demonstrate the use of the Handbook equations in case you get a retaining wall instead.
For determining the section modulus (or moment of inertia) for a footing or retaining wall, wouldn’t you consider one face of a cross section of the footing? So you would need to know the depth of the footing or need to make an educated guess about its depth? I want to not mess this up when I the exam. Thank you so much for your incredible videos, they have been a huge help to me!
For the Tff, the friction is developed based on the normal force. The normal force is perpendicular to slope. Here's another, more detailed, example on components of a force normal to a sloped surface. ruclips.net/video/ubDV4zhnQv4/видео.html
first thanks for content , in Q no 11 i need to understand why you neglect term of water pressure in equation,, i think something wrong, with my respect if you can reply
![Felix "Unfair" | [Stray Kids : SKZ-PLAYER]](http://i.ytimg.com/vi/Oswujxm2Ag0/mqdefault.jpg)
Hey Mark! Thanks to your videos, I passed the FE!! I spent all summer watching and studying your videos. I truly appreciate the content. Thanks!
That's awesome! Thanks for letting me know. Congrats!
Sin = opposite side/hypotenuse..@1:33:16
Hi Mark, thank you for your dedication to helping students. It is much appreciated. I passed my FE thanks to your resourceful videos and review packets. Onward and upward!
Hey Mark! Thank you so much for your videos. I passed the Exam. It was really helpful to me. I did my exam after 10 year from graduation. I happy to say inly 2 months work hard with your videos.
What were your resources?
Hey Mark, your videos are really awecome! I'm sure I can pass the exam after learning all of your videos!
On prob 7, he should've written (100)^3/1000 when doing the conversion from cm^3/m^3 to kg/m^3 I believe. 10^3/1000 = 1. Thank you for the videos Mark! Exam in 2 days!
UPDATE: I PASSED!
Thank you for everything Matt! Flow nets came for my test. Just know how to read the points on a flow-net graph.
I wonder why because its not listed in the Handbook no more. Was it simple? there is an new version 10.4 out now.
For #11 at 1:13:29, I'm wondering why you don't consider the saturated unit weight of the soil that's submerged in water? I haven't taken a soils/geotechnical class yet so I'm not sure if my concept is correct. I used formula in reference manual to get (gamma)sat and used that as (gamma)2 in the Vertical Stress Profile. Thanks!
Your question is a good one. My question is not written very well. I should have given a second unit weight, gamma 2 for the moist soil below the water table. If you use, say 130 pcf for gamma 2, you get an effective stress of about 1000 psf. The question is solved assuming the dry unit weight above and the moist unit weight below the water table are the same, which should be changed.
I had a practicing engineer with an MS in geotech qa/qc these problems for me before doing the live session... but neither of us caught this. Thanks for asking! I hope the clarification helps.
I really appreciate these videos as well as your stories!!
Good day Mark. I have a question in no. 18. Did you make Ls an area? Based on the video, you converted the length to ft^2. I hope you will notice my question. Thank you 😊
I asked the same question that how was it possible. But no reply
Hi Mark, Thank you for the great videos. One question regarding question 14. The dimensions of the footing are given as 5x5ft but the thickness is unknown. The second approach for solving the question involves calculating I (Moment of Inertia), and I see that you are not using the thickness of the foundation to get the moment of inertia. However; I think the way the footing is bending, the Moment of Inertia should be calculated as width x thickness^3 / 12. and not width x length^3/12
Hey Mark for (E) the equation is uses specific weight (gamma) so I think we have to multiply by gravity (32.2 ft/s^2) since you gave a density of 115lbs/ft^2. Look at the units. Even if you square the 8ft….. the answer would be lbs/ft. Sorry I am studying back over your notes because I my test is scheduled for May 2nd!!!! Please let me know if I’m looking at this wrong
I think the same thing about multiplying the density of 115lb/ft3 by the acceleration of gravity, could you please @MarkMattsonPE review this point.
Great Video Mark! One question:
On problem 18, the equation used for TFF doesnt seem dimensionally correct, C*Ls (kip/ft) + Wn*cos*tan (kip).
The equation is pulled from the handbook, so I am not sure whether it is calibrated to work that way or if Wn should be divided over the length in the second part of the equation.
someone please clarify this.
i think the assumption is per unit feet, or 1 feet perpendicular to the paper. i think it would be better if mark shows the per feet length value by multiplying it
I noticed that too. I solved for Tff by converting ft to inches and tons to lbs and I got the wrong answer. if the units need to be in ft for the equation to work, I wish it specified that in the manual.
Thanks for posting this! Question on #18 - How do the units work for the Tff calculation? When doing c * Ls, you converted the units to lb/ft, but it's not clear why it needs to be lb/ft instead of (say) lb/inch. Is the 30-ton weight of soil given implied to be for a 1-foot cross section?
Great question. I converted everything to kips and feet. And, yes, to get the units to work out and get that other ft to cancel out in the cohesion piece of it, we need to multiply by a 1-foot depth into the page. Thanks!
@@MarkMattsonPE Hey Mark , would you not consider the "g" acceleration due to gravity on the weight part ? 30 X 2000 X 62.4 ? thank you
Do you have any plans to do a review for 2023?
Thank you Mark, very helpful content!
Can you explain why you choose to do a 1 ft section for problem 14 rather than just doing it with the 5x5 footing?
Hey Mark! Thanks to your time, I have a doubt regarding Q:11, the Handbook is showing Gama 2 as a "saturated" unit weight, but it is not given in the question the saturated value !! is it correct to use the normal unit weight instead?
I agree. I don't think gamma 2 is just another unit weight altogether, I think its gamma 1 but saturated
He responded to this in a previous comment - the question was poorly written. They will typically give you the saturated unit weight of the submerged soil layer on a problem like this.
Thanks for the videos! In Q12, the density (rho) shouldn't be multiplied by gravity acceleration to get the specific weight (gamma)?
For question #9, is it supposed to be 122/115 instead of 122/100? Thank you
Yes, thank you! The 72% is correct, and you get this value when using 122/115 (not 122/100 as shown) for the last part of the relative density formula.
Hi , In Q11, there should be 2 unit weights given. one is dry unit weight and the other one is saturated unit weight of the soil. Or, we can assume the dry unit and saturated unit weight are equal.....thanks
Thats exactly what I thought, as Gama2 should be saturated unit weight instead of the other type od soil! In this case, the saturated unit weight didn't given
Can you elaborate more on question 18 and how did you get the units to work out? I know you said in the comments that you need to multiply kips/ft by 1 ft to end up with kips only, can you elaborate on that more please? Is the 30 tons per ft? as in 30 ton/ ft? If yes, how are we supposed to know? Thank you!
I should have written the 30 tons was for a 1 ft width into the problem statement. At that point, if you multiply through by 1 ft, the units all work out to the answers shown. Many geotech problems are solved on a 1-ft unit width basis. Good catch and thanks for the clarification!
@@MarkMattsonPE Thank you! Keep these videos coming, please!
Hi mark, I am looking at problem No 11. Why are we using dry unit weight for gamma 2? Shouldn't Gamma 2 be saturated unit weight?
Im also confused. It doesnt seem specify what kind of unit weight given in question.
I also think it's not correct. The problem statement should give the saturated unit weight though.
Hi Mark! Thank you for your videos. Will take the exam in a month and it helped me a lot to refresh with all the concepts.
For Question 13, I read somewhere that "Normal stress at failure" should be at the point where the mohr-coulomb failure meet the Mohr's circle and not the major principal stress which you used? Please can you clarify?
In question 13, I get an answer of 28 degrees. Shouldn't Tangent be opposite over adjacent? Sine is opposite over hypotenuse
See 2:08:15. That error got caught and corrected later in the video. Thanks!
Thanks a lot for your efforts, very helpful lecture,
Hey Mark, at 2:04:40, are the unit compatible while adding the cohesive force and frictional force?
Having the same question.
@ 2:04:30, why are you using a wieght of 60 tones instead of 30 like we did in Tmob force. I appreciate the help with studying and have found your content very helpful!
I think he kind of skipped a step. It meant by 30tons * (2kip/tons) = 60 kips for Wm.
The 60 is because he converted 30 tons to 60 kips. I have the same question
Nobody's said this and it's driving me wild. For question 17 the only way you get 2 cm is if you use the natural log (ln) instead of log base 10. If you use log base 10 then you get closer to 0.90cm
in the last problem, Tff ends up with k/ft, due to the initial conversion, how can it cancel out?
In the second last example, c*L units are lb/ft. How was it added to weight?
Thanks so much!!
Hey Mark, not sure if you still answer these, but I was wondering, when do we use Ixc and just Ix- area moment of inertia?
Did you get your answer?
thanks, Mark
Hi Mark, thanks for your efforts, I have a question on number 14, why the weight of the retaining wall was not considered in calculating the vertical forces and for calculating the resisting moment?
Thanks for the amazing videos!!
Question: Why are the equations for the Cr and Cc values the same equation when not using a correlation?
Thanks again!
Hi Mark, for #12, could explain the units for the final answer. This might be just a simple misunderstanding on my end, but pcf (from density) multiplied by ft^2 (from h^2) gets 1130 lb/ft
Hi @AndyZaz, I should have indicated the analysis was for a 1 ft width into the page. That extra foot into the page cancels out the last foot. Does that help?
@@MarkMattsonPE I am confused on how density 115 pcf is unit weight. or was that a wording mistake in problem #12? as well as why multiply unit weight x 1 ft width of wall. I don't understand the relation there?
It was a wording mistake. It should be unit weight. Unit weight includes the impact of gravity.
The 1' of wall, is essentially because we are looking at the load per foot of wall into the page. Often, stability analyses are done per foot of wall.
Thanks!!
Hi Mark! Finally I paaaaaaseeeed the FE civil exam from the second try 😍😍
I am soooo happy and want to thank you for all your videos I watched it several times
Also I want to recommend the book of “ Islam “ which contains 800 questions it was also amazing and let me understand many additional things that not included in the yellow book
So it was great knowledge from your videos, the yellow book, and Islam book after completing it I passed the exam 😍😍😍
how can I get this book?
+ (cc log po + delta p / pc ) did you make a mistake because cr log pc/po is not the same as cc can you please verify the correct way to complete this problem. I'm taking my fe in two weeks and am nervous about this test. I failed this test and trying to understand everything to be more prepared this time.
Hi Mark, I hope you are doing well. In question 14, when you calculated value for Q heel and it comes to be positive. So, why did you say it is compression? It should be tension right?
A positive value means that we drew it correctly and it was originally drawn in compression
For problem 12 you bring up surcharge, is there an equation to solve for S or will that possibly be given and we have to factor that value in to solve for a vertical stress?
In question 14: Why did you need to find the moment (4) and force (16) for the 1'? Thanks.
You don't really have to... you'll get the same answers if you don't divide P and M by 5. That said, I did that for this problem because in the Handbook equation (see pp. 264-265), B is used which defined as "width of base." I wanted to approach this question like a retaining wall so you could see how the equations work and know B is in the same direction as the overturning forces. I actually used a combined stress P/A ± M/S approach when I first solved it, but wanted to demonstrate the use of the Handbook equations in case you get a retaining wall instead.
For determining the section modulus (or moment of inertia) for a footing or retaining wall, wouldn’t you consider one face of a cross section of the footing? So you would need to know the depth of the footing or need to make an educated guess about its depth? I want to not mess this up when I the exam. Thank you so much for your incredible videos, they have been a huge help to me!
thanks so good
[ where did you get the 21 degrees friction angle in question 18? ]
for the tangent
Problem 12 why did you use density and not unit weight?
Also, for the last question, I am not sure why the Wm for Wm*cos(30) is 60k not 30k (for the Tff). Could you please explain a bit? thank you!
For the Tff, the friction is developed based on the normal force. The normal force is perpendicular to slope. Here's another, more detailed, example on components of a force normal to a sloped surface. ruclips.net/video/ubDV4zhnQv4/видео.html
very helpful
I think #17 should maybe be 2.00 cm. It looks like the Cr and Cc values were flip flopped. Thank you for your content!
Ope. Just had to wait a few more minutes haha.
At 3:29 so funny with the ground up joke
first thanks for content , in Q no 11 i need to understand why you neglect term of water pressure in equation,, i think something wrong, with my respect if you can reply
I'm not sure I understand your question... effective vertical stress does not include the water, but the total vertical stress does.
He didn't neglect it. First we're assuming gamma1=gamma2 so he rewrote it a factor of H. (gamma2*H2)-(gammaw*H2) = (gamma2-gammaw)*H2
these units are going to be the death of me
Thank you!