FE Mechanics of Materials Review Session 2022

[@patrickmoseley8865]

Hey Mark! I watched these a year ago when I took my FE and you did such a great job that I’m coming back for a refresher before I take my PE in 14 days. You are an amazing instructor. Fun to watch, easy to understand, and inspiring. Thank you sincerely.

[@misganawzewdie721]

What a wonderful method of teaching, i just passed my FE. Your videos helped me a lot, thank you.

[@jasonlewis460]

Yo Mark, do more videos. Your our only hope. Everyone is charging 100 per hour

[@mrdjf2010]

These videos are fantastic!! I wish I had taken the FE while I was still in school, but some of my peers have recommended these videos and they’re really helping me refresh on all the material!

[@kobekeise7427]

I passed my FE Exam on my first try, thanks to you and these videos. Thank YOU!!!

[@mateenfoster4595]

this man more goated than Lebron James

[@Mikelafication]

My first statics class where I learned shear and moment diagrams was in 2016. I'm out of college and grad school and just started my first job and I'm trying to pass the FE. This single video has helped me learn shear and moment diagrams when taking this class 3 times in undergrad did not. Thank you!

[@aliwow7874]

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.

[@bridgetgardner3055]

I wasn't sure which video to leave this comment on, but I enjoyed the mechanics of materials review the most so here it is! I just found out I passed the FE, and I wanted to thank you for all your review lectures. These were incredible reviews and they were the most helpful study material for my exam. Thank you for all the effort you put into these videos as well as for your encouraging words throughout all of them! It made a world of difference.

[@MarkMattsonPE]

Thanks so much! Keep up the great work!

[@purplerambutan8312]

I just passed the FE exam 2nd take! Would really recommend Mark's videos for concept training! I passed after 15 YEARS since I graduated from college! Glory to God my Lord Jesus!

[@MarkMattsonPE]

God bless!

[@rubenjerves4539]

Hello Mark, on question 12 you should calculate M as P*e, where e would be the eccentricity from the neutral axis of the square section and P the axial load. In this case, e would be equal to 45 cm. Then you can determine the stress in point A as the algebraic sum of bending and axial stresses. For instance, imagine if I wanted to calculate the stress on the opposite side of the section, as you mentioned in point B. Here, the moment would be the same, and bending stress would be adding to the axial stress; both in compression. Please correct me if I am wrong. By the way, your videos are very helpful; thank you!

[@l.valencia]

This is how i solved it first before watching Mark's explanation. I wondered why I got 40.5 Mpa at point A.

[@fabulosospucas]

I really appreciate your videos, I have my fe exam in 10 days and this has really helped.

[@michaeljaradat5185]

I just took and passed my Fe on 6/5/24 and question 6 from this review was a problem on the exam I took

[@fouzianoutash6880]

Oh my God, you are great, even though, l am a second language and my English is not great, but I fully understand your explanations, especially the last three questions. You helped me a lot. You made me anxious and hopeful for the exam. May God bless you always Thank you so much.

[@1crida1]

On Question 12 (#12), it is asking for the maximum combined stress. Why then would the maximum stress not be = - axial stress - bending stress?

[@halshehri-s3g]

Thank you so much sir, you made a great work that you shared your knowledge to us, very optimistic and very kindly man. And actually, I am watching your videos for refresh my memory in my major, and this the best review I have been seen. You don't know how I appreciated you for this professional and honest explanations, and love to help others. Thanks dear

[@paolacastillo6616]

Mark THANK YOU is not enough!

[@thomassbahtu1503]

Hi Mark I appreciate the conceptual explanation of Mohr's circle. Now I know the evolution of principles stresses.

[@carlossosa1158]

I wish my School Tied Trig Back into Everything like you do its crazy how much it helps

[@lucasvansickle]

Question 12, Pg. 13 of15: Why did you use I=(bh^4)/12 versus I=(bh^3)/12? I graduated in 2008 from FSU. Your videos are fantastic and they are really bringing it all back to me! Thank you!

[@MarkMattsonPE]

Since b=h (it's a square), I simplified bh^3/12 to h^4/12. Does that help?

[@Telco12]

Thank you. you made my mech mats final go alot smoother

[@1crida1]

On Question 14 (#14), the reference handbook provides the formulas for the principal stresses on page 132 if you don't want to draw a Mohr's Circle.

[@eugeniojimenez7268]

muchas gracias amigo I hope your channel keeps growing. Great Videos

[@1crida1]

After learning the problems, I used the print-out Mark provides as a timed practice test. Here are the answers to each question if anyone else is doing something similar: 1) B 2) B 3) B 4) C 5) D 6) C 7) B 8) D 9) D 10) C 11) B 12) B 13) N/A 14) C

[@jackohare8523]

You are the greatest of all time.

[@amberhall1119]

For problem number 7 how does the force P pulling to the left impact the system? Will the bolt only feel the force pulling to the right?

[@AlexisAlexander646]

You probably already took your test, but for future testgoers, I believe the explanation for this is that P itself is an axial force, and in the system shown, it only works provided that they are both in tension. P also has to be equal to each on both sides as a static system. Therefore, for practical purposes, P is only considered as one value when using equations and with the Free-Body Diagram.

[@tsegatedla5673]

For #2 it would be much easier to solve using the reference manual, using formula 9/128 w l^2

[@RicardoCarranza-ys3zd]

Thank you Mark! These videos ara helping me alot

[@donwilliams7768]

Problem #12. You illustrated the area to be 100 mm^2. How are you deriving that? 10cm x 10cm = 100 cm^2 = 0.01m^2.

[@naviitaramro3474]

For problem 3 why do we only solve the first and last one?

[@kathleenmil402]

i have the same question? why not solve for the area that is below 0? even if it's negative wouldn't that still be the maximum stress?

[@AliKhiat-m8j]

Any link to reference handbook?

[@construtorapoliedro5730]

Hi Mark! Just a question.. Question N7, Don´t we need to sum de other force P in the left part, so the Force P at the bolt will be ((15/2)kn + (15/2)kn))? The principle of adding (sum) opposite vectors. Tks!

[@altope9999]

i was thinking the same thing too. why did we not consider the force from the block in the other direction.

[@twigsfloat2773]

This is so great! Thank you so much!

[@arashshafigh]

problem 1 area 2 would be 6.3*4=25.2 isn't it the maximum moment?

[@TrucNguyen-xk3ud]

Hi there, I was wondering the same thing but then I remember that to find the moment @ B (10m from A to the right), we do: (10.2kip-f) - (25.2kip-f)= -15kip-ft. Then we +15kip-ft (which is area 3) to get 0 kip-ft @ the end of the beam (so the whole system is balanced).

[@C90072]

You probably dont need this anymore but for anyone who comes across: When you take the area of the shear curve you are saying the change in moment from the applied load to B is -25.2 kN-m. But the moment at the location of the load is not zero, it is 10.2 kN-m. Therefore, the moment at B is 10.2-25.2 = -15 kN-m which is the maximum moment.

[@cg2822]

Awesome job! Can you do a PE review?

[@Jonathan-fh6jj]

How is the area 30 & 20 on problem #10 if they’re 3x3 and 2x2

[@christiancorona7465]

thought the same thing for a while then realized he does 30^2. so technically he has 30x30. took me 10 minutes to figure it out haha

[@tiaraadams9875]

He's converting 3cmx3cm to 30mmx30mm to get A=900mm^2; same process for the aluminum section.

[@ColterOre]

how did he get the areas on question #10

[@WhitneyWalkerHomestead]

I am wondering the same thing

[@jgyegyiri]

@@WhitneyWalkerHomestead length * width. for steel, 250 cm * 3 cm

[@jgyegyiri]

its a rectangle

[@huseyinsanal]

Mr. Mark Mattson, thanks for the video. Do you think the questions on the real FE Civil exam would be as easy as that?

[@ssabitha]

Hi Sir, in Question 3, how much will be the bending moment of A3 , the area looks bigger than others, i don't know if i am right or wrong. please help me with that.... Thankyou!!

[@AtsouAgbemenou]

I had the same question. The prompt didn’t specify positive bending stress so shouldn’t we consider the negative bending stress too?in that case C is the answer

[@kathleenmil402]

@@AtsouAgbemenou ok i was thinking the same. has anyone provided a response to this?

[@rmaxelliott]

same question here!!!!

[@ivanzamora3333]

Hi Mark, thanks for your videos. For number 6, why is the weight of the beam taken as a punctual load instead of a distributed load?

[@jcopeland9210]

Mark, for the last question... if you take the tan inverse of y/x and then do R cos theta and add it to the center, why do the answers not come out the same and why does that not work here?

[@michaeltilly6208]

Q: Is there a difference between the Shear Modulus (G), and the Modulus of Rigidity (G)? Or is the reference manual just trying to confuse me by calling G two different names.

[@MarkMattsonPE]

I'd have to look in the manual, but I use the terms interchangeably.

[@jewbagel5969]

amazing video thank you

[@michaeltilly6208]

Very helpful! Thankyou Sir!

[@angelarinaldi1130]

In question 9, you used J = pi * r^4 /2, when in a previous problem you used J = pi * a^4 / 2 (which is also stated in the FE equation sheet). Was that an error or am I missing something? Thanks!

[@patriciagonzalez4001]

a=r, in the equation in the manual, they are referring to “a” as the radius

[@JAG_158]

⁠@@patriciagonzalez4001when would I use pi*r^4 / 32 since it’s a shaft?

[@austinburke7116]

Would you be able to have a small Q&A live stream or do it during your scheduled live streams. Or would you prefer to do that in the comments.

[@MarkMattsonPE]

Right now, my focus is on getting review problems and the lives streams posted. Feel free to put questions in the comments and I'll try to get them answered. Thanks!

[@ahmetcenk-i7z]

what is the lbs ? It is not pound-mass or pound-force, so what is it ?

[@timliebrockpuzzles]

In regards to question 1: Why did you ignore the point where the shear crossed zero on the downward slope, but considered it on question 2?

[@kathleenmil402]

i was wondering the same thing

[@miriamlriviera]

These videos are the best!! Thank you so much Professor!!!! One question if anyone has any insight. For Q12, using my TI-36x Pro, I didn't get 1337 using sys-solv, I got 2382. Anyone else? Does anyone know what I might have done wrong? Thank you!!

[@brookemclaughlin6942]

Appreciate your videos, quick question: how do you know when to use g or not when calculating the self weight in the moment/ force equations? The problems in the statics video used g but it wasn't used in question 6 here. Thanks!

[@brookemclaughlin6942]

Never mind I answered my question, weight vs. mass!

[@dylanmccoy8530]

Do you know if the problems on the exam are asked this way or are they worded weird? I've heard that the questions on the test are worded strangely.

[@Personalaccount194]

I love you Mark

[@joeluciani5823]

In question 3, why isn't the negative moment considered?

[@MarkMattsonPE]

If you want to calculate the entire VM diagram, you would add all the areas from left to right (including the negative areas). I left it out as a shortcut, but you could just go from left to right and get the same answer.

[@joeluciani5823]

@@MarkMattsonPE right but isn't 16×5 greater than ½×14×9.33? Or does it depend on how you break the diagram off into sections

[@MarkMattsonPE]

From left to right, you'd do all the areas to get to the same point. From left to right you'd have +1/2×14k×9.33' - 1/2×4k×2.67' - 4k×3' - 16k×5' = -32 kip-ft. From right to left you'd have -8k×4' = -32 kip-ft (you add the extra negative sign going right to left). Either way, you get the same answer. You just have to include all of the areas. Does that help?

[@justinzendzion8700]

@@MarkMattsonPE I have the same question, Labeling the smaller parts from left to right from A1-A5, the rectangle A4 would have a moment of -80Kip-Feet. A1 had a moment of 65.333 Kip-feet. Why would 65 be the answer and not 80. The explanation in the comment above says 32, while in the video, you say the answer is 65. What is the correct answer to Question 3?

[@justinzendzion8700]

I think I see now, you did the math on the right side of the second Max/Min point which is easier than on the left of that point, both of which will equal 32 Kips +-. While the first point, the math on the left is easier being 65.3, while calculating to right of that point would give you the same answer but negative, which means that point and the problem will have a Maximum moment of 65.33.

[@alexarmendariz8263]

P/A equation 15000N/100mm^2 shouldn't the area be 100cm^2 and not 100mm^2?

[@MarkMattsonPE]

It's 15000N/(100mm)^2. The location of the second parentheses makes all the difference. I know that a N/mm^2 = 1 MPa, so I converted everything to N and mm.

[@joshharris7902]

Hello question for you! For example 2 when we are doing the moment about point A (By x 20'-2kip/ft x 10'(10'/2)) when you multiply the moment arm of 10'/2, will we always divide by 2 when presented with this type of force is presented? For example if it were 5kips/ft of force it will be 5kips/ft x 10'(10'/2)

[@MarkMattsonPE]

A uniformly distributed load with magnitude w and length L, has a total force of wL and it acts a distance of L/2 from either end (at its center). So, if w = 5 klf and L=10', it would be a total force of 50 kips and act at a distance of 5' from either end of the load. Does that help?

[@ranchk1223]

you da best

[@Jime1248]

Hello, thank you for the videos as they are a huge help. For #12 , you stated the axial force was P/A but then squared the cross sectional area. Was that done on purpose or was that an accident?

[@jojo-vc6qe]

the cross-section was given 10 cm x 10 cm but he decided to use mm which is now 100 mm x 100 mm aka (100 mm)^2

[@Mikelafication]

VQ/It is on pg 160

[@ruthmoges]

Thank you

[@DjamicA1]

Hi Mark, Why didn’t we use 120 cm on problem 12?

[@MarkMattsonPE]

It is not needed for the moment or the vertical force calculations. Bending stress and axial stress look at the cross-section perpendicular or normal to the 120 cm dimension.

[@DjamicA1]

@@MarkMattsonPE Thank you!

[@DjamicA1]

@@MarkMattsonPE Thank you for all those videos. I passed.

[@Danger17334]

very helpful !!!

[@eugeniojimenez7268]

nice videos

[@IndianWarriors]

Awesome

[@rafia2266]

thank you