Thank you for actually explaining the problem solving methods and their approach in detail rather than just reading the textbook examples out loud like 90% of other channels.
Hi Nik, I understand your frustrations. I remember as an undergrad thinking the same thing: "why the heck can't they just break it down and go through the step-by-step process". There are some great channels that do this but like you said a lot just want to release content and move on, release, move on, release, move on. That's not how teaching/learning works, a good instructor will take the time to teach a concept because undoubtedly this concept is the stepping stone to the next. My videos are intended to never leave anyone behind. If it does, I know I am doing something incorrectly.
if anyone wants a tip for the second question, you can recognize that Y_cp should be lower than Y_c. Since 12.67 m is the only multiple choice option below Y_c = 12.5, you know thats the right answer. It will save you some time and is a good strategy for all questions to eliminate options that arent possible.
best video on RUclips on this topic. Thank you so much! Some of them are over an hour. Sometimes when your exam is close you don't have time. I've been looking for an easy and concise example on hydrostatics.
Always assume we are working with gauge pressure unless they ask for the absolute pressure. If the atmospheric pressure was given in the problem statement, we would consider Patm. In this case, Patm=0 gauge pressure
Thank you for the example, could you please solve example explaining absolute maximum moment of moving concentrated load sets on a beam from structural analysis? and if you could explain problem 64 from the NCEES FE civil practice exam? I would appreciate that...
Thanks for the example! I was a bit confused about the hinge being at the top vs being at the bottom. So would the force needed to just *open* the gate be acting at the very bottom in the opposite direction as Fr? Call it P. I got the distance from Fr to the hinge to be d2 = 2.67 m = 12.67 - 8/sin(53.1). The distance from the hinge to the bottom of the gate where force P acts is d1 = 5 m. So the moment balance would be Fr x d2 = P x d1, solving for P I got 393 kN
Yes!! Assuming the hinge is at the top as shown in this question. The force P that needs to act the very bottom is found by taking the moment about the hinge. What you did is exactly right after tossing it around in my head. Well done!
Thank you for the lesson and the explanation, Sir. However, I have a question, how come is the centre of pressure more than 12m? since the total distance from the free surface is only 12m? If you consider the diagonal distance from the free surface to ycp, could you please explain the justification? Thank you very much
Can you help in fluid statics problem?Gate AB in the figure is 1.5 m wide into the paper, hinged at A, and restrained by a stop at B. The water is at 20°C. Compute (a)location of centre of gravity and centre of pressure (b) the force on stop B and (b) the reactions at A if the water depth h = 2.9 m Problem-
When you translate this equation from the y-world (slant depth from the water level) to the z-world (plumb depth from the water level), you actually end up squaring the sine term, rather than cancelling it. Any given z-position relates to the y-position by: z = y*sin(theta) Multiply the y-version formula by sin(theta) and simplify, and you'll get: zp = zc + Ixxc*sin(theta)^2/(zc*A + P0*A/(rho*g)) where: zc is the depth of the centroid zp is the depth of the point at which the hydrostatic force effectively acts. The y-version will produce a division by zero problem, when theta = 0. Trivially, zp should equal zc when theta=0, which the z-version reflects.
Hello, I'm honestly not sure what you're referring to. If you can, I would love to see a picture of your work using your method. Note, this example is relevant to the FE exam. It uses a derived formula rather than using partially derived equations or equations derived in differently. This formula will always work for any slanted gate that's not a curved surface. Also, please note the distance to the center of pressure is measured as an inclined distance from the top water surface. Some books shows equations with the distance to center of pressure measured from the centroid of the gate.
would you please answer my question? as you said I looked at static section of handbook and I saw "Ixc" is equal to bh3/36 (about the centroid), and also how "h" is 5 and not 4 in solving for the moment of inertia. sorry for the confusion.
Hi it will be a rectangular cross section so Ix = bh^3/12. The h is the height of the cross section into the page. So it will be the 5 m. The 4 meters is the vertical height but not the total height of the gate. In other words h will always be the hypotenuse height.
@@directhubfeexam I'm also confused about this. How do you know to use Ix instead of Ixc? Ixc in the handbook is what this person says: bh3/36. Ix is what you used: bh3/12.
Not considered since we are working with gauge pressure. The problem statement would specify the atmospheric pressure value if that needs to be considered.
Hi if you are given the pressure at the centroid of the area you would use the second one that used Pc (pressure at the centroid). From what I've seen this is hardly ever given and the top equation is more often used.
you literally just cleared up 3+ hours of confusion in the first 8 minutes...
Thank you for actually explaining the problem solving methods and their approach in detail rather than just reading the textbook examples out loud like 90% of other channels.
Hi Nik, I understand your frustrations. I remember as an undergrad thinking the same thing: "why the heck can't they just break it down and go through the step-by-step process".
There are some great channels that do this but like you said a lot just want to release content and move on, release, move on, release, move on.
That's not how teaching/learning works, a good instructor will take the time to teach a concept because undoubtedly this concept is the stepping stone to the next.
My videos are intended to never leave anyone behind. If it does, I know I am doing something incorrectly.
ruclips.net/video/m_ZPRwbWtvM/видео.html
...
Agreed!!
Saved my butt on my final for this question. THANK YOU SO MUCH!!!!
if anyone wants a tip for the second question, you can recognize that Y_cp should be lower than Y_c. Since 12.67 m is the only multiple choice option below Y_c = 12.5, you know thats the right answer. It will save you some time and is a good strategy for all questions to eliminate options that arent possible.
Absolutely! Good tip 👍
ruclips.net/video/m_ZPRwbWtvM/видео.html
Hello Sir, I just wanted to say thank you, you're out here doing god's work, incredible explanations.
Thank you Fenix! Extremely pumped to hear this. Much appreciated 🙏🏼
Thank you for the clear explanation, I was doing a problem and the Hc, Yc and Ycp was a little confusing, but now I get the differences.
Anytime 😊
ruclips.net/video/m_ZPRwbWtvM/видео.html
.
.
.
best video on RUclips on this topic. Thank you so much! Some of them are over an hour. Sometimes when your exam is close you don't have time. I've been looking for an easy and concise example on hydrostatics.
Thank you, I am happy to hear that you found this useful. Good luck!
ruclips.net/video/m_ZPRwbWtvM/видео.html
...
you are an amazing teacher, I understand much better now!
Thank you 🙏🏼
Helps me alot I feel very sad cause our professor can't explain it like that you make it easier to me thanks alot
Hi Wael, I am glad you found this video helpful! Thank you 🙂
thankyou it really helped me out & cleared all my doubts, you are one of the best teacher out here on youtube .....thankss
Good video man, this explained the concepts well. I have a fluid mechanics exam tomorrow and this is the only concept I'm stuck on
Thank you! I'm happy to hear this helped. Ace that exam man!
Sir, thank you so much, your explanation is very clear and concise!!!
You're welcome!
Thank you sir for your perfectly clear explanation. This video is realy helpful.
You’re welcome 😊
thank you so much! very clear explanation
However you did not need to find theta you needed sin theta which you already had and that’s how you got theta. You could have stoped and used 3/4
Thank u!!!
Amazing video thank you 👍🏽
Bro? Why you used 5 as based and not 1 as based?
The width is always the base. It’s the width into the page for these fluid gate problems.
woah what an explanation.
For the first question why don't we include patm. In the FE handbook, it has this equation: Wetted side: FR = (Patm+ ρgyC sin θ)A.
Always assume we are working with gauge pressure unless they ask for the absolute pressure. If the atmospheric pressure was given in the problem statement, we would consider Patm.
In this case, Patm=0 gauge pressure
thank you for the explanation
You're welcome!
Thank you for the example, could you please solve example explaining absolute maximum moment of moving concentrated load sets on a beam from structural analysis? and if you could explain problem 64 from the NCEES FE civil practice exam? I would appreciate that...
11:15...You just went backwards, there was no need for trig. You divided the height by sine then later multiplied by sine.
Thank yu
good video thanks
Thanks for the example! I was a bit confused about the hinge being at the top vs being at the bottom. So would the force needed to just *open* the gate be acting at the very bottom in the opposite direction as Fr? Call it P. I got the distance from Fr to the hinge to be d2 = 2.67 m = 12.67 - 8/sin(53.1). The distance from the hinge to the bottom of the gate where force P acts is d1 = 5 m. So the moment balance would be Fr x d2 = P x d1, solving for P I got 393 kN
Yes!! Assuming the hinge is at the top as shown in this question. The force P that needs to act the very bottom is found by taking the moment about the hinge. What you did is exactly right after tossing it around in my head. Well done!
ruclips.net/video/m_ZPRwbWtvM/видео.html
I think your answer hier is not correct. d1=5 but d2 is 5-2.66 and d2=2.34! 736*2.34=P*5, P=344
Thank you for the lesson and the explanation, Sir. However, I have a question, how come is the centre of pressure more than 12m? since the total distance from the free surface is only 12m? If you consider the diagonal distance from the free surface to ycp, could you please explain the justification? Thank you very much
What about if there was a Concrete block with a given density but asked to calculate the volume of concrete required to keep the gate closed
Thank you
Can you help in fluid statics problem?Gate AB in the figure is 1.5 m wide into the paper, hinged at A, and restrained by a stop at
B. The water is at 20°C. Compute (a)location of centre of gravity and centre of pressure
(b) the force on stop B and (b) the reactions at A if the water depth h = 2.9 m
Problem-
You are a legend
Doesn’t the sin(53.13) in the final equation cancel out the hc/sin(53.13)……..so your left with hc anyway?
You mean we are left with F_R = (ρghc) x A ?
When you translate this equation from the y-world (slant depth from the water level) to the z-world (plumb depth from the water level), you actually end up squaring the sine term, rather than cancelling it.
Any given z-position relates to the y-position by:
z = y*sin(theta)
Multiply the y-version formula by sin(theta) and simplify, and you'll get:
zp = zc + Ixxc*sin(theta)^2/(zc*A + P0*A/(rho*g))
where:
zc is the depth of the centroid
zp is the depth of the point at which the hydrostatic force effectively acts.
The y-version will produce a division by zero problem, when theta = 0. Trivially, zp should equal zc when theta=0, which the z-version reflects.
you nailed it...
Thank you!
Thank u so much sir ❤️
Thank you for watching!
Sir am kindly asking the reason why we are not multiplying the moment of inertia by sine theta squared for finding the center of pressure?thank you.
Hello, I'm honestly not sure what you're referring to. If you can, I would love to see a picture of your work using your method. Note, this example is relevant to the FE exam. It uses a derived formula rather than using partially derived equations or equations derived in differently. This formula will always work for any slanted gate that's not a curved surface. Also, please note the distance to the center of pressure is measured as an inclined distance from the top water surface. Some books shows equations with the distance to center of pressure measured from the centroid of the gate.
would you please answer my question? as you said I looked at static section of handbook and I saw "Ixc" is equal to bh3/36 (about the centroid), and also how "h" is 5 and not 4 in solving for the moment of inertia. sorry for the confusion.
Hi it will be a rectangular cross section so Ix = bh^3/12. The h is the height of the cross section into the page. So it will be the 5 m. The 4 meters is the vertical height but not the total height of the gate. In other words h will always be the hypotenuse height.
this slanted gates are almost new concept for me. please accept my apology in advance if I dont know what Im talking about :))
@@saeedehlotfi2576 yeah it’s difficult to understand, no worries :)
@@directhubfeexam Thanks for the clarifying the concept for me. I appreciate it. Have a good one
Ixc says bh3/36 in the reference manual
We are looking at a rectangular gate cross section since it's a gate with a width (w). Therefore, the rectangle cross section Ixc controls.
@@directhubfeexam I'm also confused about this. How do you know to use Ix instead of Ixc? Ixc in the handbook is what this person says: bh3/36. Ix is what you used: bh3/12.
Please where in the FE reference handbook did you get the equations for Fr , Yc and Ycp? I can't find it.
Pg.180 FE Handbook 10.2
@@directhubfeexam thank you!
sir which software are you using for drawing purpose ? please reply
Yc is 14.142 is not?
Brother 😍😍❤❤🙌🙌👍🙋 ...
also. there's Patm
@@kylecatman7738 As a rule for the FE exam, always neglect the atmospheric pressure unless the problem statement says otherwise.
what about the Patm
Not considered since we are working with gauge pressure. The problem statement would specify the atmospheric pressure value if that needs to be considered.
There are two Ycp equations. how do we know when to use which one?
Hi if you are given the pressure at the centroid of the area you would use the second one that used Pc (pressure at the centroid). From what I've seen this is hardly ever given and the top equation is more often used.
where in the new FE handbook is the pressure distrbutions written out?
Pg. 189 Fe handbook 10.0.1.
If you know sin Theta is 3/4 why would you solve for theta
I see what you’re saying. Didn’t think too much on this. Applying that FE handbook was the first goal in mind 👍🏼
Eq in the handbook adds Patm. Why didn't you do that??
Atmosphere pressure is zero I believe. We are working with gauge pressure.
Ohh dear... Why did you findout theta and then yc; you could have been replaced YcSin with hc directly. And save a lot of efforts.
Yes sir!! That’s faster. Either way, knowing what yc,hc, and ycp means is important
clutch