FE Statics Review Session 2022
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- čas přidán 31. 01. 2022
- FE Exam Review Session: Statics
Problem sheets are posted below. Take a look at the problems and see if you can solve them ahead of time. See the video for solutions. Feel free to ask questions!
See for problem sheets: sites.google.com/view/markmat...
This is a question based review with questions that will cover topics that address each topic in the NCEES FE Civil exam specification.
These topics include:
A. Resultants of force systems
B. Equivalent force systems
C. Equilibrium of rigid bodies
D. Frames and trusses
E. Centroid of area
F. Area moments of inertia
G. Static friction
For the series see the playlist at • FE Civil Review 2022
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Wow, better than any course i've paid for. The books are hard to stick with and hard to stay motivated to do it yourself. Feels like I'm back in the classroom with a teacher who cares. Thanks for the great videos and thank you for giving me hope that I can pass the FE.
Do you have yours scheduled? I'm scheduled for Oct 30.
Thank you for all the videos!! I watched every video of the FE CIVIL and I passed the exam! The exam questions are similar and a bit easier than the questions in these videos. If you master the questions in the videos, you will pass the fe test. Thank you so much Mark, waiting for your PE videos.
Congrats!
I watched your entire FE review course and it proved to be an excellent preparation for the exam. This morning I found out I passed! Thank you for all the thought and effort you put into your content! I found your examples to be harder than the actual FE questions. Cheers!
correction on problem 2: you wrote Rt=11kN, Rt= 33kN. Thank you for the videos im currently prepping for the FE and this has been quite helpful!
I just love your subtle humor and you being so patient with the explanations and your insights about the industry. thanks a lot for these series mark
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.
*Appreciation Post*
Started studying for FE since FEB 2022. It was a bit discouraging to start with Lindeburg's FE PRACTICE book, since while attempting the question I always felt the urge to check the solution in b/w. But your videos are helping me to revive my concepts. No I'm aiming to practice watching your videos and then try solving all the questions on the books later on to get my confidence back.
P.S: Good Stuff! Hoping to finish all your reviews and pass FE CIVIL! Thanks!
I was having SO much trouble with #12 when solving for P. I guess I’m so used to having the surfaces be horizontal that I completely disregarded P factoring into the normal force. Once I realized that, it made a lot more sense. Thanks, Mark!
Thank you so much for the video! FE in 32 days and this video series is fantastic and is keeping me motivated to study for the FE along with my course load :)
Taking FE in mechanical in two weeks. Your videos are helpful. I am practicing your lectures. I am not going to give up- I will eventually pass the FE. Thanks, Matt.
was the 3D DIMENSIONS resultant forces F=(i,j,k) on it.
So so so so so thankful for these videos! Thank you for taking the time to make these and help us reach our goal of passing the FE!
Watched all of your videos Mark and passed on my first try! Thank you for the work you put into these videos. Helped so much.
Thank you so much for your helpful and accessible study materials! I am so sorry for your loss, and I hope you are taking care of yourself!
I can't thank you enough for the amazing videos you've shared on CZcams. As I prepare for the FE exam in the US, your teachings have been very helpful for me and your explanations are clear,. I feel more confident than ever in pursuing my engineering career. Thank you for making a difference! Now I am moving to next video.
thank you! i've always struggled with statics, and you made it super clear
Always incredible Mark Mattson,
I have passed FE CIVIL
Thank you, Mark Matson, Your videos were instrumental in helping me pass the FE exam. Grateful for your work and excited about this achievement. You are definitely the best!
Excellent job Mark - very comprehensive and fun!
Mark thank you so much for the video, I never understood when to use either Cos or Sin and you helped me so much!
Happy to hear that!
I have my FE exam next week, Thanks Mr. Mattson!
I am definitely a student of yours. Very much enjoy and love your content. Thank you so much for sharing.
Thank you from Puerto Rico! Great job! I'm very very thankful for these invaluables resources. Blessings
Dios le bendiga!
Watched all your videos and Passed my FE on my first try. Thank you 🙏👏🏻
Have you studied anything else plus these videos?
Having taken and failed the FE twice and now working towards passing it on my third go I would say statics is probably THE most useful section to be completely and utterly familiar with, knowing how free body diagrams and sum of moments work is useful on more than just the statics section and can really help you out on quite a few others.
Thank you mark for your help! I got my results and pass first try because of your videos. I made a small donation but I couldnt write a message but I wanted to say thank you for all you do for the community.
Congrats! Thanks! Keep up the good work!!!
amazing teacher!! Thank you so much for this help!
RT should be 33 KN not 11 KN
Thanks for these videos
For Question 6, I simply used a single moment equilibrium equation, still using method of joints, rather than the x and y equilibrium method you showed, which I found more difficult to understand and longer to execute. I got the same answer of 25.3 from it.
To explain, I selected the same joint, Joint C, and then did moment equilibrium at Point B, which cancelled out the intersecting axial force BC, and just left AC unknown. So, essentially I treated it as a method of sections problem in a way, and I got the same answer of 25.3. Just another method that's possible for those that may find it easier.
@MarkMattsonPE Thank you so much for these videos Mark! I'm 6 years out of college and was able to pass the FE with the help of your review sessions!
DAMN!! what took you so long to take the FE?
@@Dana__black I am 27 years out of college and trying to get FE soon:::::)))))) no jock!
For problem 6 why did you multiply BCx by (5/3) and not multiply BCy from a specific ratio?
FE in 9 days. Thanks for the videos, Mark!
howd you do?
This is amazing thank you
Hello Mark, love your videos! I have a question regarding question 7 (1:11:46) would the frame be stable and statically indeterminate? I got values of 3(2)+3 > 3(4)+0.
You are an inspiration
Hi Mark, Thanks for the videos. I'm taking my FE exam soon. Your Method of Joints (Part 1) video shows you using the sum of moments, but this video says not to use it. Would you please help clarify if/when to use the sum of moments with trusses?
Thank you
hey just one quick question , in problem no 4 why wasn't the direction of reaction ax and reaction bx the same
Method of joints. Are you gonna teach us how to roll Mr. Mattson
For problem 4, I get that B is a roller so it only has a reaction in the x direction but in this case why is it not in the y direction?
Hello Mark, I gave the exam 2 weeks ago. Did not pass. I will try again soon. Your videos are very helpful. I should watch your videos before the exam. I got some similar question you have covered in this class.
Thanks for uploading videos
I'm glad the content is relevant. Many of the basic concepts are things that will apply regardless of the specific question. If you can really master those first principles, like the equations of equilibrium, it will take you very far. Keep practicing and learning. Best!
For problem 6, why did we ignore the cable that the pulling force P is applied? when doing the summation of moments shouldn’t the length of the cable be part of the moment arms?
For Question 7:
Why is it that Py and 200 are listed as separate forces in the Y direction? When solving this question before watching the video, I assumed that Py was 200lbs no?
Hey Mark! I'm loving your videos! Thank you, for all of your content!
I have a question for problem 6. I got -25.33kn indication that AC in compression. When taking the sum of forces using the method of joints, I didn't consider positive to be up/right and negative left/down. I considered any force acting towards a point(compression) as negative and positive going away(tension). Can you please clarify if I am doing this correctly? Thanks.
It sounds like your sign convention "any force acting towards a point(compression) as negative and positive going away(tension)" is correct. There are a lot of signs involved in this equation. Check that you added AC to the other side of the equation and that you had the double negative for BCx.
For #12, why would the friction force be pointing down the inclined plane? Thanks for your videos btw! Studying for my FE exam in March :)
Hey Mr. Mattson, hope you are doing well! Thanks for the great videos!
Was curious on the last problem. I also got 1337 for P, but if we selected 1300 wouldn’t that be less than the force required to push the block up the ramp? I was conflicted between choosing between 1300 and 1400 due to the wording, but leaned to 1400 instead. If we got into this type of scenario on the FE, is it better to round to our closest answer instead of the logical answer?
Either way - thanks so much for the help!
Problem 11: when initially stated Ff= μk x N ? are we considering both 7k and 3 k forces as normal forces?
Thanks for the nice videos. would you like to share the software name that you use to edit the pdf. Thanks in advance.
Hello on question 6, don't we have to take the magnitude of Bcx and Bcy, making the answer 38.8kN, D.39 kN?
I genuinely busted out laughing when he said "hopefully nothing alcoholic cause that makes it harder to do the FE" 😂 he's not wrong!
3 - 2 = 1 so 33(1) why did you use 33(2) ??? 7-2=5 for rectangular and for triangle 2/3 (3) = 2 then 3m - 2 = 1 what did I do wrong?
Why 1/2 x22kN/mx3m =11KN? It should be 33kNM
For question 7, how come you dont have to split the distributed load on the incline plane into x and y components? why can you just get the distance from A to the center of the distributed load?
Hi! I looked into your question. He somewhat did do this by calculating the distance using trig. He found the perpendicular distance from the force acting at that point to the moment point. Balancing moments about a point is different from balancing the forces in their x and y components. When calculating the moment, we factor in all the forces acting at some perpendicular distance rotating about that point, I believe they all don't necessarily need to be x and y acting forces, as long as there is a calculable perpendicular distance from the force. Hope this helps!
Do we have to consider the tension AB for Question 5?
You can either use Ax and Ay or the tension in AB, but because I summed moments at A, neither causes a moment at A, because the line of action for all of those passes through A.
Can someone explain why for question 6 it was Bcx/Bcy and not Bcx/Hypotenuse ?
If you want to find the member force BC, you could use Bcy/Hyp or Bcx/Hyp. For this problem, I wanted to get back to sum forces in the X direction, so I used Bcx/Bcy to go from the Y-component to the X-component for the vector force. I hope that helps! These problems can stretch your mind for sure.
For question 7, the area of the triangle should be 0.5*9*2? Rt*9=0.5*9*2*9, so final answer is 18.22
Keep watching... the final answer is 18.2. Thanks!
Question on problem 12… i understand the method and how you get to your answer. But for this to work, do my assumptions on the direction of the components of force P need to be correct before entering into the system solver? On my first attempt assumed that the y component of force P goes up rather than down with respect to the plane. This yielded a force of ~1100 for P rather than the correct answer of 1337. I’m worried that I’ll make incorrect assumptions on the exam.
Hi! After looking into your question, I realized that when he switched the axis P became a force acting slightly downwards and to the right which indicates a negative Y (down) and positive X (right). I rotated my paper slightly to match his axis and I could see the way P points according to his defined axis. Hope this helps, good luck studying!
first thanks a lot for all ur effort and time, prof. Second, for the last problem, he asked for the force that will let the body start moving upward, I get 1338 N as u did solve it, but why we do not choose any force bigger than this number which is 1400 N as 1300 N is not enough to move it up. Plz clarify, thank you.
Yes I believe Matt missed the part of the question where it says "find the Force P that moves the block up" . P should be 1400N
On Question no. 3) How do we know the direction of the forces at point A and B. And why did you take Ax and Bx different direction. Tnx
Hey Ruth! Do you mean question 4 at about 36:20? If so, I just made a guess based on the way applied forces (50 kN and 20 kN) are acting and what direction I think Ax should go. If you assume both Ax and Bx are to the right, when you solve the sum of moments equation at B, you'll get Ax is negative, meaning the arrow was shown in the wrong direction and the force really goes to the left. Does that help?
Yes it does. Thank you very much.
Hello Mark. I have a question about question 12. I am okey all with entire solution. I found P value 1337 like you. But Why did you choose 1300 N. Arent we supposed to choose bigger than 1337 N. I thought 1337 N is breaking point. If we want to cause motion to begin up Shouldn't we pick up bigger number from 1337? Bye the way thank you so much for your videos they are very helpful. I appreciate for your help. Thanks.
I can't argue with your logic... I probably could have written the question better, but "most nearly" is still closest to the 1300 answer. That's really the reason I picked it. You're absolutely right in that (in theory), you'd have to go to 1400 to get the block to actually move.
So, hopefully, you don't get any questions that are poorly written on the exam :)
Q#7 - Why is the moment arm for Rt 9? How is 3ft half of the length of the triangle?
for triangle loads, the moment arm is 1/3 Length (from the heavier loaded side); not 1/2 (that's for uniform (rectangle) loads only)
As for how 3' could be half the length of the triangle, it isn't! But the moment arm is indeed 9'.
The resultant occurs at the centroid. So for the triangular distributed load (Rt), it occurs at b/3 = 9/3 = 3 (see FE Handbook 10.3 pg. 98). Then to get the moment arm, you have to make sure you are measuring the horizontal distance from A. This gives you 6 + 3 = 9.
It would only occur at the halfway point for a symmetrically (often rectangularly) distributed load.
On question 5, why is the moment arm for Py not equal to 9' and the moment arm for px not equal to 5'?
So weird, I get the same answer as Mark even when I use these moment arms.
@@amberhall1119 I believe because, the force in the cable is the same if you analyze it at point B or at the end of the cable where it shows in the diagram. The moment distances are a ratio, and as one increases the other does as well, so mathematically they will cancel out because the force is the same throughout that section of the cable. Does that make sense? I know you posted this months ago, but at first glance I had the same question before I thought about it for a sec.
hello,
In Q2 Rt O.5x22x3 = 11kn. But the answer is 33kn
u made a mistake there
Problem #10 The Ixc should be 11, not 19 which would make your final answer 110 not 120 if I am not mistaken.
you multiplied instead of added why you got 19 instead of 11.
I see where I went wrong.
Question 2, F shouldn't be 99 KN?
The rectangle gives 22kN/m × 4 m = 88 kN. The triangle gives 1/2 × 22 kN/m × 3 m = 33 kN. The total vertical force should be 121 kN.
Yes he calculated it wrong and wrote 11 instead of 33. just shows how easy it is to muck up a question if you’re not taking your time.
1:44:58 😂😂😂
The answer of question 12 will be 1400
Matt, I did it by hand as following:
1- Forces acting in the X direction:
Fx= 507.8 N, Fy= N= 1,895.15, so therefore friction force (Ff) Ff=0.35x1895.15N= 663.30N
Summary force acting in the X direction:
Fx=507.80N & Ff=653.30N
2- Component Px= Fx+Ff= 507.80+653.30= 1161.10N
3- Problems states to find the magnitude of force 'P' (Acting Horizontally) that will cause motion. We can to calculate "P" having Px= 1161.10N as following:
P= 1161.10N / Cos(15) = 1,202.06N
we picked the same answer 'C' because P has to be more then 1,202.06N to be able to cause motion but we don't get the same answer of P= 1337N ; N 2241N and in this case the answer maybe should be "D" due to P has to be more then 1337N to be able to cause motion ?
What am I missing ?
Thank you Matt you are helping a lot !!
The normal force also has a component of P (not just the weight of the block). I think your friction force is low since I don't see Py in what you wrote above. Does that help?
@@MarkMattsonPE that is what I am missing. I see why you used the elimination method using the calculator due to we have 2 unknowns for 'N'. Thanks Matt Taking My FE very soon!
Your last question is wrong. Double-check it. I used substitution AND elimination, and got P=1882.37. And N=2382.15
I may have messed you up with N for a variable and N for Newtons. The N on the left is a variable and on the right is Newtons.
I just double checked it with Wolfram... see www.wolframalpha.com/input?i=systems+of+equations+calculator&assumption=%7B%22F%22%2C+%22SolveSystemOf2EquationsCalculator%22%2C+%22equation1%22%7D+-%3E%22-.35N%2B.9659P%3D508%22&assumption=%22FSelect%22+-%3E+%7B%7B%22SolveSystemOf2EquationsCalculator%22%7D%7D&assumption=%7B%22F%22%2C+%22SolveSystemOf2EquationsCalculator%22%2C+%22equation2%22%7D+-%3E%22N-.2588P%3D1895%22 for a solver of the two equations I wrote out.
I think you used 0.55 instead of 0.35 for the coefficient of static friction