FE Exam Review - Fluid Mechanics - Fluid Statics - Submerged Slanted Gate

[@francescahunt6476]

you literally just cleared up 3+ hours of confusion in the first 8 minutes...

[@pinkhead6857890]

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.

[@directhubfeexam]

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.

[@edutechguruengineeringlear4280]

kzbin.info/www/bejne/o5C9gYWtl7ynrK8 ...

[@svongsa]

Agreed!!

[@fenixfyre]

Hello Sir, I just wanted to say thank you, you're out here doing god's work, incredible explanations.

[@directhubfeexam]

Thank you Fenix! Extremely pumped to hear this. Much appreciated 🙏🏼

[@briancarrollii4359]

Saved my butt on my final for this question. THANK YOU SO MUCH!!!!

[@Bobodododo]

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.

[@directhubfeexam]

Absolutely! Good tip 👍

[@edutechguruengineeringlear4280]

kzbin.info/www/bejne/o5C9gYWtl7ynrK8

[@ello-isa]

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.

[@directhubfeexam]

Anytime 😊

[@edutechguruengineeringlear4280]

kzbin.info/www/bejne/o5C9gYWtl7ynrK8 . . .

[@hotnspicy474]

best video on KZbin 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.

[@directhubfeexam]

Thank you, I am happy to hear that you found this useful. Good luck!

[@edutechguruengineeringlear4280]

kzbin.info/www/bejne/o5C9gYWtl7ynrK8 ...

[@onlygaming8106]

thankyou it really helped me out & cleared all my doubts, you are one of the best teacher out here on youtube .....thankss

[@diegoguatemala5287]

So, another way of doing first question and quicker is simply by finding the average pressure times the area of the door. Evarage pressure is ( (1500 x 9.8 x 8 meters above the door) + (1500 x 9.8 x 12 meters of total distance to the floor) ) divided by 2. You then multiply that times area ( 5 x 1 )m^2. You get the resultant force because F = Pressure times Area. It also works if the door in vertically positioned at 90 degrees. And if there is no wall above the door than is just the, simply just a door ((atmosphere pressure + density x g x distance of depth))/2 where the atmosphere pressure will be 0.

[@wsl939djl]

Helps me alot I feel very sad cause our professor can't explain it like that you make it easier to me thanks alot

[@directhubfeexam]

Hi Wael, I am glad you found this video helpful! Thank you 🙂

[@WearableTundra7]

you are an amazing teacher, I understand much better now!

[@directhubfeexam]

Thank you 🙏🏼

[@deadman4167]

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

[@directhubfeexam]

Thank you! I'm happy to hear this helped. Ace that exam man!

[@sadkoner2036]

Thank you sir for your perfectly clear explanation. This video is realy helpful.

[@directhubfeexam]

You’re welcome 😊

[@Raniaska0306]

Sir, thank you so much, your explanation is very clear and concise!!!

[@directhubfeexam]

You're welcome!

[@silindilenkosi1355]

This was very helpful, thank you.

[@directhubfeexam]

@@silindilenkosi1355 you’re welcome !

[@ryanlakner8950]

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

[@directhubfeexam]

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!

[@edutechguruengineeringlear4280]

kzbin.info/www/bejne/o5C9gYWtl7ynrK8

[@sambakhtiari2796]

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

[@swirldude3636]

thank you so much! very clear explanation

[@sanaahamati6556]

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...

[@dodgedart2055]

The location of the resultant force is impossible. The depth of the location can NOT be lower then the actual height of the gate, because then what is the resultant force on? Not the gate. He had solved for the hypotenuse of the location with his shifted axis. To actually solve this, once you get to the 12.67m, you must then use the sin(53.13)=height/12.67. This gives you a height of 10.14m. This is the actual answer you are looking for. It makes sense too because the middle of the gate is at 10m, so we expect a resultant force location slightly lower than that.

[@directhubfeexam]

Good point! This sounds right. The vertical location from the free water surface to the center of pressure would be: h_cp = 10.14 m Note, in the video I’m solving for the slanted distance location using the y_cp formula in the FE Handbook. Thanks for pointing this out 👍🏼

[@priyabhowmik6227]

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.

[@directhubfeexam]

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

[@dioutoroo]

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

[@kodurushashi1719]

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-

[@Ben-2]

Amazing video thank you 👍🏽

[@runtdawg666]

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

[@mohammedmiah1497]

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

[@saeedehlotfi2576]

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.

[@directhubfeexam]

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.

[@saeedehlotfi2576]

this slanted gates are almost new concept for me. please accept my apology in advance if I dont know what Im talking about :))

[@directhubfeexam]

@@saeedehlotfi2576 yeah it’s difficult to understand, no worries :)

[@saeedehlotfi2576]

@@directhubfeexam Thanks for the clarifying the concept for me. I appreciate it. Have a good one

[@desirekatongo4462]

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.

[@directhubfeexam]

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.

[@Olayinka.D]

Please where in the FE reference handbook did you get the equations for Fr , Yc and Ycp? I can't find it.

[@directhubfeexam]

Pg.180 FE Handbook 10.2

[@Olayinka.D]

@@directhubfeexam thank you!

[@al-kabirhossain5946]

sir which software are you using for drawing purpose ? please reply

[@aimore5315]

Doesn’t the sin(53.13) in the final equation cancel out the hc/sin(53.13)……..so your left with hc anyway?

[@directhubfeexam]

You mean we are left with F_R = (ρghc) x A ?

[@carultch]

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.

[@mutalejohn5295]

thank you for the explanation

[@directhubfeexam]

You're welcome!

[@kylecatman7738]

Ixc says bh3/36 in the reference manual

[@directhubfeexam]

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.

[@abigailspiers1544]

@@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.

[@andrewgrimes1771]

where in the new FE handbook is the pressure distrbutions written out?

[@directhubfeexam]

Pg. 189 Fe handbook 10.0.1.

[@misbauddin1383]

There are two Ycp equations. how do we know when to use which one?

[@directhubfeexam]

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.

[@Whacker409]

Bro? Why you used 5 as based and not 1 as based?

[@directhubfeexam]

The width is always the base. It’s the width into the page for these fluid gate problems.

[@rejiii6689]

Eq in the handbook adds Patm. Why didn't you do that??

[@directhubfeexam]

Atmosphere pressure is zero I believe. We are working with gauge pressure.

[@osmanazeem9017]

You are a legend

[@AdityaRaj-lj5wf]

woah what an explanation.

[@pubudunuwan1751]

Thank u so much sir ❤️

[@directhubfeexam]

Thank you for watching!

[@francisvalenti2541]

Excellent

[@abdulalanazi3129]

Yc is 14.142 is not?

[@keikay2733]

good video thanks

[@FakeFayaz]

what about the Patm

[@directhubfeexam]

Not considered since we are working with gauge pressure. The problem statement would specify the atmospheric pressure value if that needs to be considered.

[@micahlyolo5484]

you nailed it...

[@directhubfeexam]

Thank you!

[@runtdawg666]

If you know sin Theta is 3/4 why would you solve for theta

[@directhubfeexam]

I see what you’re saying. Didn’t think too much on this. Applying that FE handbook was the first goal in mind 👍🏼

[@theotema2053]

Thank you

[@kylecatman7738]

also. there's Patm

[@directhubfeexam]

@@kylecatman7738 As a rule for the FE exam, always neglect the atmospheric pressure unless the problem statement says otherwise.

[@litzya6494]

Thank u!!!

[@OliverMosen-yv3si]

Thank yu

[@YouTubemainul]

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.

[@directhubfeexam]

Yes sir!! That’s faster. Either way, knowing what yc,hc, and ycp means is important

[@johnpaul2021]

clutch