I just passed my FE exam this week. Huge thanks to you man. I watched, went through and did almost all your practice problems. I am was happy to be able to come back and post this comment knowing that you helped a student youve never seen before. For people currently studying, i was really worried about harder sections like Structural. I would not get too worried about it and only focus on basic concepts (shear and moment) as you can get the majority of those questions with minimal effort. Instead focus on easier sections. I had some really basic math and transportation questions that i didnt know how to do well because I didnt spend that much time studying those and was worried about structural. You can pass the fe by just mastering the easy questions.
Congrats on the FE! can u please tell me if mark's videos are enough? and any extra tips ? it's been a while since i last took these subjects. it's overwhelimng :(
I just passed my FE last week, I wanna thank you Mark for your valuable videos they helped me a lot, I recommend anyone who is preparing for the exam to watch these videos at least once.
For anyone unfamiliar with Question 7 (#7) dam analysis like myself, Pg. 263 of the 10.3 Handbook covers the equations used under "Horizontal Stress Profiles and Forces"
I thought structural analysis was gonna be super duper difficult, but after watching this lesson and following along, I feel much confident about this section of the FE exam. I will be taking it soon. Wish me Luck! Thanks Matt
Thank you so much Mark. I passed the FE exam today on my first attempt. Your videos help a lot. I watched all your review session and bought an exam sample on NCEES site.
I just passed my FE, your videos were of great help in the journey! And one of your question appeared exactly on the exam (indeterminate structure finding out the reaction appeared) Which I saw in your video just a day before the exam. Thank you so much Matt!
I made problem #1 way more challenging than I had to after watching Mark's method. I used the sum of moments twice, sum of forces in x and y, and two equation two unknowns solve.
For #1, I was able to remove one step and obtain an answer of 16 by using the following equation for the sum of moments. This is in comparison to breaking EC into an x and y component like Mark did: Sum of the moment at point D = 20 kips (8 ft) + EC (10 ft) = 0 ==> EC = -16 kips
Thank you for videos. There is one question ...Question-4: AB-AC and CD are zero members ( also you mentioned) , but you wrote on final table AD and AB. I think it should be AD and BD. Anway, the result not changing...
Hello Mark! In question 4, the member BD is in compression, so its length is decreasing. I am not sure, but I think you should substract its contribution to deflection in the Unit Load Method. Great videos! Thank you!
My understanding is that both the virtual force and the actual force caused by external loads would be in compression (negative), so the product (little f times big F) ends up positive. It is not discussed directly in the video, but I believe the final calculated deflection is correct.
When you look at the equation from the handbook, it's 3m+r = 3j+c. This equation is more complicated because it's checking for both stability and static determinancy. What Mark did is take only the left-hand of the equation, which is for static determinancy, and set both sides equal, obtaining 3m = r.
I tied the problem that can be seen circled at 53:50 with a load of 70KN. I am not fully sure if what I did was right, but I used δ =PL/AE from the Mechanics of Materials section like you had said: (70KN*4000mm*(1000N/kN))/(((50mm)^2)*(200,000*10^6 N/m^2)*((1m^2)/(1000mm)^2) = 0.56 mm.
Unfortunately, for this problem since it is horizontal deflection, you can't just use PL/AE, you really have to go through the principle of virtual work for trusses. Sorry, no short cut on this one.
Hey Mark, not sure if you're still making video's, but was hoping you can do a video talking about zero force members. I am still stuck on how those are identified and would be extremely grateful if you can do that!
for #1, why wasn't the y component for BC considered when summing moments about pt D? did you just move Fbc along it's line of action to pt D, and that's why it was neglected?
Question 4: In the final table member in tension are positive and in compression negative ? so we would end up with 792-280= 512= fFL ? and the rest the same I get a deformation of 1.02 mm ? Thank you Matt you are the best
Hey Mark, love the videos, I was just wondering on Q7 so the concrete unit weight is 150 pcf and you multiply it by 26 while the other side is triangle , you made it sound to me like there is just distributed force o the rectangular side when you calculated the W, i know you calculated both but u used the same W of the rectangular
In Q9, why did you take Mba into account when summing the moments about point B? I thought that Mba was AT point B so it wouldn't affect the sum of moments about point B. I just did 9(10) - 15Ay + 20 = 0 and Ay came out to be 7.3, which still gets me to the right answer, but I want to make sure I know how to do it correctly on the FE.
for question #7 towards the end when your calculating for overturning moment you multiply the weight of water (18k) by 8'? why do you multiply by 8'? that's the width of WC1? I'm just a little confused on that. Is it because the weight of that water is acting on the whole width of WC1?
Hi Mark, really appreciate your videos, it's super helpful! For Q7, when you calculate the concrete weight, why did you use width 8' multiply stress rather than using cross section area? Generally stress=Force/Area. Could you explain this? Thank you!
I have the FE next week and I'm freaking out. I did the practice exam and the problems in this video are 100 times harder than the practice FE exam problems. I'm really praying they wonn't be asking such problems because although If I know how to solve them they will just take sooooo much time.
By is up on BC and down on AB. If the arrow is shown in the correct direction, I use the signs in the equations of equilibrium to assign positive or negative. I do not use absolute sign conventions like is used in vector mechanics with ijk unit vectors. If you learned vector mechanics for statics, it's a shift in thinking, and an easier one to implement (IMHO).
On Question 4, it specifies horizontal displacement. How do I solve for that? Or do we assume deformation only occurs along the load line? I would have thought since there is compression at DB that vertical displacement would occur as well. Or am I lost in the weeds?
First of all, your videos are great! and I like your jokes haha. Second, in Q5 you shouldn't take the minor moment of inertia I, and calculate Pcr with it? And in this case, Iy of the weak axis does not become Ix to the strong axis? If yes, then the Pcr for the strong axis should be calculated with 4 m4 (40,000 mm4) as well, and the result would be the same because you still have the restraints in the weak axis (i.e. L and K would remain the same). What I mean is that maybe is not necessary to confirm the Pcr in the other direction of the moment of inertia, since if the column buckles, it would do it in the minor of Ix or Iy. Independently of the length of the column, and if there are no eccentricities, it always tends to buckle in the direction of the minor moment of inertia, is this correct?
Hey Mark, love the videos, I was just wondering on Q7 where you got the force of the water pushing on the concrete. Why did you multiply by the height of water instead of the width like you did finding the force of the concrete?
Hey @Tom Jenkins, thanks for the comment. Both the lateral force of the water and the weight of the concrete is dependent on their height. The water is a lateral pressure acting on the dam. The concrete is a vertical pressure acting on the dam. I hope this helps!
Question 1: Why not keep EC whole? Sum moments about B: (10)(EC)=(20)(8), EC=16. This solution is coincidentally close to your answer, but doesn't seem to work for other forces/dimensions. Generally curious.
You could take the left side, but either way you need to cut it apart. The reason your answer is close, is because 10 is the approximate perpendicular distance from point B to EC. If you knew the perpendicular distance from any point to EC, you could figure out the EC force in one step. Varignon's principle is helpful with trusses and along the same lines. See also ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-LbpJf4p7mWc.html
For this problem (and in general for similar problems), you want to use the moment of inertia about the centroid, or Ixc, in the chart. I hope that helps!
hi, please, can you tell me? at Q4 to get fi we get dimensional ratios for length 4m because it the largest or the smallest long (to get fi for each member we divide the length of member by the length of the longest member or by the length of the shortest member)
Hello Mark, thank you so much for the great videos. One question, for P#5, how do we know which inertia formula to use? The Ixc =(( bh^3)/12) or the Ix = ((bh^3)/3) Like how do we know when to divide by 12 or 3 for Lx and Ly. Thank you.
Hi! I hope I can help but I believe he leaves off the c because we assume we are taking it about the center of the area (c). We are using Ixc technically.
I think you mean 6f? Reactions only show up when you cut the members apart... If you don't cut it apart, that L-shaped member only has two reactions -> making it unstable. If you cut it apart, you add three reactions, but you also require 3 more reactions for stability between the two members, so the stability doesn't change and it's still unstable.
For the determinacy problem, if the number of reactions equals 3 x the number of beams, and there are not collapsible mechanisms or other forms of instability, it is statically determinate. This is essentially comes from the 3m + r = 3j +c equation in the manual. I used 3m = r (by counting all the internal forces at the hinges).
@@MarkMattsonPE Hi Mark. I am SO confused on this question. I was told that for Beams to use r = 3 + C also how come on c and f you didn't use the "frame" equation 3m + r = 3j = c ? what is the difference? i feel like im missing something. and these are such "easy questions on the FE i dont want to miss it :(
Just wanted clarification: If you keep your delta equation negative like the FE states, your reactions at By = -22.5 kips, Ay and Cy = 29.25 kips. Sum of forces still = 36 kips, keeping your beam in equilibrium. Although 29.25 kips is not a choice, is this still a correct answer?
By has to be acting up, it is not negative. The signs can be very tricky. That means the rest of the force needs to get split between Ay and Cy as shown in the video.
I did leave the negative sign off, intentionally. The manual has a negative for the uniform load deflection indicating it goes down. The point load deflection is positive because it goes up. For this problem, we know they have to be equal to each other. If you kept the negative sign you could say the total deflection is 0 since the sum of deflections, or Dbb + dbb = 0. The negative would allow them to cancel out (or to be equal to each other if you add Dbb to the other side).
In the truss deflection problem, BD is 1 KN, but it's in compression since it's pushing toward the joint. In the video, I mistakenly labeled that force AB in the FBD. Sorry for any confusion.
Both are shown in the correct direction on the FBD. AD is shown away from the joint indicating tension, and BD is shown toward the joint indicating compression. Since they are shown in the correct direction, the sign is positive when solved using the equations of equilibrium, indicating the assumed tension/compression is correct. AD is tension and BD is compression. You only get different signs if you assume all forces at the joint are in the same direction, for example, in tension. To do that for this problem, BD would be shown down (assumed pulling away from the joint). Then, when solved for, BD would be negative, indicating the arrow is in the wrong direction and in compression. In that case, tension will be (+) and compression (-).
