Thank you Mark for all of your videos! I just passed my FE exam after several failed attempts. This time I studied with you and passed it. Your videos are incredible, I appreciate how you put them together like this.
Thanks Mark, I have a quick workaround for problem #10, instead of calculating slant distance for "Yc" and "Ycp" we can directly calculate the vertical distances "hc" and "hcp". Hc will be 50cm, however little caution needs to be taken when calculating "Ixc" and Area "A" of the gate, instead of using 16cm, use 8cm as a vertical dimension of the gate to calculate "Ixc" and "A". Just wanted to put this method as a comment as it reduces time to calculate vertical distance to the centre of pressure.
Mark, just wanted to say a quick thank you for putting these videos together. I watched them a few days prior to my exam and I am happy to say I passed. THANK YOU!
To find velocity on Question #3, a faster route, although it does not explain all the concepts you did while solving, is to equate the difference in height between the two tubes to (V^2)/2g, and then solve for V. edit: Mark did explain it right after solving the first way.
gage pressure is measured with respect to the atmospheric pressure. Overall whenever the flow is exposed to atmosphere the pressure at that point is 0.
It may be too late for your question at this point. The reason is the 100kPa at point A is gage pressure. Gage pressure 0 is atmospheric pressure. The atmospheric pressure would need to be added if the pressure at A was absolute pressure.
Hi Mark, Thank you for the lecture. On question 5, you multiplied the height of the raft (7.5"/12") by 6' by 8' feet to get the volume of wood in the raft, but since the raft is made of 7.5x7.5 timbers, there has to be some space between the timbers since 7.5 inches does not evenly divide into 6 feet. Only 9 of these timbers can fit into a 6 foot span, unless of course you are using a partial timber (0.6 the width of the others), which kind of defeats the purpose of telling the width dimension of the timbers, right?
Hi Mark, thank you for the FE review sessions! I find them extremely valuable! For Fluid Mechanics Problem 10, can you go over how you calculated Ixc ?
Ixc is the moment of inertia of the plate about its centroid. So you take the base (or width into the page) times the height cubed (length of plate) divided by 12. I hope that helps clarify. I did make a mistake when talking in the video. See the comment string from @Lay also.
For problem 3, why is it that we use .15m for the height of Ps instead of .10m? would we use .10m if the problem didn't have the first pipe sticking out on the left?
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 😍😍😍
So how do you spell gage/gauge? USGS and the Steel Deck Institute uses gage. However, ASTM and others use gauge. I normally associate gage with stream flow gages and steel deck, and gauge with meters used to measure a pressure. Merriam-Webster indicates gauge is preferred. Both are in the FE Reference Handbook. Thoughts?
Dear Mark, Im taking this video from korea and it's really wonderful! Just a question. From the 1:17:48, you explained about the applicable b and h if it is not the square type gate. So the h value will be derived from the shown value, e.g.16 cm as per problem and the b will be the value from page direction? You mentioned 12cm in y direction and 20cm in x direction and I suppose Ixc will be then 20 x 12^3/12 but the video says the other way. Appreciate if you can clarify. Thank you so much!
Hey Lay! thanks for the question from the other side of the globe :) You are totally right. It should be 20 × 12^3 / 12 if the 12 is in the direction of y and 20 in the direction of x. I flipped it when I was trying to add some clarification :( Thanks so much for pointing this out! Hopefully, the comment and response helps clarify it for you and others.
@@MarkMattsonPE Thanks for swift response! I really appreciate your lecture, always happy to see your smiley face on the right bottom of screen with stunning jokes from time to time :)
Dear Mark, I passed it thanks to you! (2months of studying with 2 hours/weekday+5hours on weekends) Could you please advise me on how to pass in PE Geotech in general? e.g.) List of books to study with, Online lecture will help, right?, sequence of study, preparation duration and etc… Thank you!
Question 5: W= γ*V, But the problems states that the density ρ=45 lb/ft3. why are you using density as Specific weight γ ?? Thanks Matt. Keep the jokes up man!
You're right... gamma is unit weight and includes gravity, whereas density or rho is mass per volume. The problem should say the "unit weight of the saturated wood is 45 pcf..." Thanks!
Grateful you asked that question, Daniel. Just spent almost 10 minutes racking my brain on why gravity wasn't used, but you and Mark cleared it all up here.
@@LeaderZ2000 Everybody brought gravity to the party in that example. Both the surrounding water and our raft, are both affected by the same gravitational field, and buoyancy comes in proportion to the weight of the water. We'd get the same answer, regardless of what planet the problem takes place upon. So whether we use specific weight, or density, we'd get the same answer, as long as we were consistent.
A correction factor (in this case, 0.9) is often to adjust average velocity and area measurements. For example, see this sample lesson/lab on finding flow in a small stream www.nps.gov/common/uploads/teachers/lessonplans/Climate%20Science%20in%20Focus%20Field%20Trip%20-%20Measuring%20Stream%20Flow.pdf
Can you explain why, in problem 12, the impulse momentum principle doesn't come into play? Why can we set Fx & Fy to 0? Doesn't our momentum change in both x & y directions because our V1 and V2 are in different directions? Why isn't it "P1A1 - P2A2*cos(α) + Fx = Qp(V2*cos(α) - V1)" and "Fy + P2A2*sin(α) = Qp(V2*sin(α) - 0)" ?
The distribution lines are are usually sized to keep the velocities relatively low in the pipes. Adding in the mass flow due to a velocity would increase the forces slightly. I probably could have defined the problem more fully by asking for the thrust due to the static pressure.
Good morning Mark I have a question on problem # 7 When you were getting Resultant 2 why did you do the average of the two pressure and why not use only the 5386 like when you got resultant 1 using 3973 only?
You actually do the average of the top and bottom pressures for each area... only for a triangle, the top pressure is zero, so you take (0+3973)/2, versus (3973+5386)/2 for the trapezoid. Does that help?
For question 3, if we know that v^2/2g represents the change in height, then why can't we use the .2m and set it equal to that equation to find our velocity? Wouldn't it be faster?
@@MarkMattsonPE How do you know whether the given ratio in that problem, is a volumetric ratio or a mass ratio? Is it convention to give the ratio by volume if not otherwise specified?
I don't understand why you used Q = Av to solve question #1 when the problem statement defines it as a channel. We can see it is open. Open-channel flow is manning's equation?
