FE Structural Design Review Session 2022
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- เผยแพร่เมื่อ 25 ก.ค. 2024
- FE Exam Review Session: Structural Design
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:
The second half of Structural Engineering!
Structural Engineering
F. Loads, load combinations, and load paths (e.g., dead, live, lateral, influence
lines and moving loads, tributary areas)
G. Design of steel components (e.g., codes and design philosophies, beams,
columns, tension members, connections)
H. Design of reinforced concrete components (e.g., codes and design
philosophies, beams, columns)
For the series see the playlist at • FE Civil Review 2022
![THE TOYS x NONT TANONT - ดอกไม้ที่รอฝน (spring) [Official MV]](http://i.ytimg.com/vi/pPa1d5cC8M4/mqdefault.jpg)
Hi Mark,
Thanks for your help and posting these great videos in youtube. I passed my FE civil today. I have watched each and every videos of FE civil from your channel. You have covered almost everything in your videos. You are doing great job. Thank you again!!
Excellent! Thanks for sharing and I wish you all the best!
I've been stressed out of my mind the past month over the exam, but small things like your concrete truck joke make the pressure a bit more bearable
massively impressed by that quick 3 dimensional cross-section with the bolt holes. Very artistic
I plan on taking my FE in April for the 2nd time, your videos have been very helpful! Thank you
You make hard concepts easy to understand such as Steel Bean design. Thank you!!
Mark, awesome videos! another thing I learned in a different course was that phi BF makes that nominal Moment equation simpler especially since that value can be found in the AISC Table. That equation would become CB[Mp-BF(Lb-Lp)]. Which I used in solving the problem and came to the same result of 383 (382.1394).
Hi Mark thank you so much I passed the FE going Onward and Upward!
Awesome videos all of them,
Thanks, I have passed FE CIVIL,
I am an international student,
6 years out of college,
Great job putting in the work, especially at 6 years out!
FE Handbook 10.3 did add the equation Mn = As fy (d - a/2) on page 275.
Great video Mark. Thank you 😊
Glad it was helpful!
Thank you so much for the amazing video, I just wanted to add the equation that we have to memorize to use the KL weak Table is coming from dividing the slenderness ratio of y to slenderness ratio of x in ,(KL/r)y/(KL/r)x .
I passed my FE last year after watching all your videos. Thankyou!! Can you do some videos for the PE Structural? pleaseee
Hi Mark, Lcy eq equation can be found in the Steel Construction Manual (15th Edition) on page 4-5.
So true and thanks for the note! The AISC Manual is full of useful information, but unfortunately (or fortunately, depending on how you look at it), not all of the information is referenced in the Handbook. I should have said "it's not in the Reference Handbook."
For those of you wondering, this is the equation where Lcy-equivalent = KLy-equivalent = KLx × rx/ry that is referenced at 47:40. I don't recall seeing it in the reference handbook though.
Thanks!
Mark
Hey mark, I'm currently studying for my FE and your videos are helping a ton! I never took a steal design class, do you have any tips or resources for learning this material?
I was completely taken aback hearing judy lius name!! she was my steel design professor at Oregon State University. Funny coincidence. She was awesome and one of my favorite professors in all of college
Prof. Liu is a superstar! She can probably correct all my mistakes without blinking!!!
I think, for the question 7 for Vs, the 8*bw*d*squre root of fc' should also be checked.
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.
Hi Ali, I am about 30 days out from my Exam. I am glad these videos helped a lot. Because I am very nervous about what is on the test. If you dont mind me asking, was the test easier than the NCEES practice test or harder? And did these videos provide coverage for all the material? Thank you if you see this message!
@@silasstapp3577 did u pass?
@@silasstapp3577 so. how did u do? plz let me know
One problem I got on the FE that I was really confused on was how to determine the amount of rebar needed for a concrete section. I know In my structures class we worked a singular problem like this, however I was given more tables than I believe is provided on the FE Manual.
thanks MARK
Thanks!
Thank you for the video sir !
Just a quick question, for question 9, you use the interaction diagram to estimate the answer, I intended to use another interaction diagram(Graph A.11) to solve it, but it didn't work, why is that?
Hi Mark appreciate for your useful videos but i have one comment that in Q1 the LL reduction is not apply if At 100 psf
Sir, can problem 3 be solved using the design compressive strength formula on page 281 from the manual? If so, may you please do a video guiding us through those steps? thanks!
You should be able to find phi Pn using the equations. You still need to calculate slenderness ratios for each axis to see which one controls and then which equation to use.
does the fe include T beam design
Hey on question 8, when solving for Ast why did you use 1in^2 having trouble figuring that out. If you have a chance pls lmk thx. Love the series btw FE on July 20th this year!.
A #9 bar has an area of 1 in^2. There is a table in the Reference Handbook that gives the steel area for each bar. See 1:16:56
Hello Mark, I just had a quick question regarding question 3, the steel column example. When doing it on my own, I tried to use Table 4-14 provided in the reference using my least KL/r to get phiFcr and ended up getting about 770 after multiplying by my Ag. I understand how you did it with Table 4-1, but how do you use Table 4-14 to get the same answer? Table 4-14 is on the page right before Table 4-1. Awesome video by the way!
Hey Cole! Thanks for the comment and the question. I'm glad the video helps. What you suggested is another great approach to the problem. I probably should have covered it in the solution as well, especially since the KLy-equivalent formula is not in the handbook (only in the AISC Manual).
To use Table 4-14, you use the largest KL/r, not the smallest. For this problem, KLx/rx = 30 ft × 12 in/ft /5.28" = 68.2 and KLy/ry = 12 ft × 12 in/ft /2.51" = 57.4. Again, the larger slenderness value controls since it's in the bottom of the Fe equation.
So, going to Table 4-14, you find for KL/r of 68, phi Fcr is 32.1 and for KL/r of 69, phi Fcr is 31.8. If you estimate phi Fcr as 32, and multiply by the gross area of the W12x58, you get 544, which is still pretty close to the selected answer of 550. The actual value if you use the AISC equations is 544.56 kips. I hope that helps!
I got same too!
@@MarkMattsonPE Thank you for the answer but still trying to figure out why to take K=1. I thought both sides are rotation and translation fixed, that's why I took K=0.5.
Thank you for the review video. For the 1st question; Is this the correct way to calculate the tirbutary area for beam B1 ? Both slabs are two-way slabs and if I'm not wrong, each slab should contain 2 trapezoidal areas and 2 triangular areas. I am gonna take my Fe in 3 weeks and this question made me a little bit confused about tirbutary areas. Thanks...
The funny looking double-headed arrows indicate one-way slabs for this problem. So the trib. areas are rectangular shapes.
For two-way slabs, what you describe would be correct.
Typically, steel deck only spans one-direction, whereas concrete slabs will span two-directions.
Thank you so much for the great and super fast explanation.
On question 7, when calculating Vs, where did the 2 come from on the numerator? Is that the number of stirrups? How did you know it was 2? Thanks.
It's a rectangular shaped stirrup and has 2 legs that both carry shear.
You are not supposed to reduce live loads greater than 100psf. That is not included in the handbook, so do we go by what you did here or by ASCE 7-16/22 on the FE exam?
Thank u for the video! In the problem with shear and required spacing u used s=d/2 but why did u choose exactly this column in the table ? It doesnt satisfie upper condition (phi*Vc)/2
Good question. I should have developed that more around 1:35:35. For spacing, here are the numbers and limits for the left column.
[phi Vc /2 = 14k] < [Vu = 25 k] < [phi Vc = 0.75 × 38 kips = 28.5 kips]
So that means we need a max spacing of d/2 or 24". For our case with d = 17.5", d/2 = 8.75", and that will control. The given spacing of 12" is too far. The spacing is limited by the Code so that shear cracks do not develop in between stirrups; and that is based on geometry. Generally, shear cracks develop at about a 45° angle, so that means, a stirrup at d/2 should intercept a crack to prevent failure. Does that help?
If you want to see a nice video of a shear failure, I like this one by Materials Lab ONLINE: th-cam.com/video/GHMCG4fUUpM/w-d-xo.html
Thank you very much.
can onyone confirm if Cb was on the test?
For problem 2, my reference manual has d_b (not d_h) for calculating A_n...
This equation was updated in version 10.1. Make sure you download the latest handbook.
Mark why you add 1/16 in number 3. It should have no 1/16 right? Check the formula in the handbook.
Hi Mark,
I am loving your videos. I am scheduled to take the FE at the end of the month and they are helping a ton.
On question 7 when you divided phiVc by 2 to determine if stirrups were required-- is this essentially saying that a safety factor of 2 is needed? If so why is the safety factor of 2 not needed with the stirrups? Since you are already over Vu without the stirrups, is the Vs calculation necessary?
Hello Mark
Q7 I agree the stirrups is not adequate but the spacing is 12 inches which is bigger than dh /2 8.75 inches not less than is this will impact the answer
Hi Mark, for question #3 I tried the Euler equation (same as question 5 in structural analysis section), but I get different results. both questions look the same, and the table and Euler equation should give same results. Could you please advise?
mark I love your videos but you're killing me with this structures section man
For the first question, Live load reduction is not allowed for heavy loads greater than 100 psf (4.7.3 of ASCE 7-16). If it supports more than 2 floors (not given) then 20% reduction is allowed. So there should be no reduction taking place, right? I do not see thus stipulation in the handbook, but it is in the code. So the answer should be the 1300 Kip-ft right?
For question 2 : For the bolt holes having a diameter of 13/16''= 3/4''+1/16''. we already had the nominal diameter of the bolt holes ? or is just because the formula said that you have to add 1/16'' ? Thanks Mattt your review is the best of the best !! ( same your jokes haha )
The formula in the Reference Handbook v 10.1 is dh + 1/16. Also, dh is db + 1/16" up to 1" diameter bolts. In total you get db +1/8 or dh +1/16 for 1" diameter bolts or smaller. This is consistent with the AISC 360 code.
@@MarkMattsonPE OH wow I was using the v.10.0.1 that's why!!! Thank you Matt
@@MarkMattsonPE the formula in FE handbook is (db + 1/16”) not (dh +1/16”) in the problem you used (dh+1/16”) I don’t know why ??
@@okjjjjjj dh = db + 1/6". That's what the FE handbook says. And the formula to reduce the area of the bolts is dh + 1/16"
@@okjjjjjj is correct. The FE handbook 10.0.01 says (db + 1/16"). This problem already gave us the bolt hole diameter so we don't need to calculate it. (dh + 1/16") doesn't really mean anything. The addition of the 1/16" is to allow the bolt to have wiggle room to enter the bolt hole. If you use the formula from the book, you'll calculate 162 kips for yielding and 164 kips for rupture. So the correct answer should be B.
Can someone let me know why the last problem is in compression?
I should have explicitly stated compression. Sorry for the confusion. In tension, you just get the strength of the steel.
Why is it 6 bars for number 8?
In question 1, the beam can not considered simply supported, because one can suppose that the beam has a rigid connections to the columns and therefore, the suposition of the maximum positive moment equals wl^2/8 cant be used because the value of the moment is lesser. That value could be wl^2/10 or eventually, lesser, considering the true structural analysis which is more than evident that it si not avaliable for this test.