Principal Stresses & Directions using a Casio fx-115es plus

[@ThePanini4]

It's crazy how clear this all was! Thanks!

[@TheBomPE]

I'm glad it was clear! Thanks for watching!

[@dannemeth2753]

@@TheBomPE I have to second this! I'm in shock at how well this cleared up the topic for me as well as making it very simple to solve using just a calculator. Thank you very much!

[@hariadiyuliantosuyanto813]

@@TheBomPE thank you so much,I've been enlightened by your excellent explanation, especially the normalization to Z=1 part

[@knowledgewarrior777]

Best video on 3D stress states so far🙏

[@TheBomPE]

Thanks! I'm glad you liked it!

[@ronaldkejenge6347]

You are exceptional Prof. Thanks for the clear explanations, Long live. You just saved me from the cosine confusion, i hope my Prof. will be flexible with this awesome alternative

[@TheBomPE]

Glad I could help! Thanks for the encouragement!

[@ronaldkejenge6347]

@@TheBomPE My Pleasure Prof.

[@RafaFeher]

What a beautiful explanation. Thanks from Brazil!

[@TheBomPE]

Thanks! I'm glad you liked it! In case you haven't seen them yet and might be interested, here are some of my playlists: ENGR122 (Statics & Engr Econ Intros): kzbin.info/aero/PL1IHA35xY5H52IKu6TVfFW-BDqAt_aZyg ENGR220 (Statics & Mech of Mat): kzbin.info/aero/PL1IHA35xY5H5sjfjibqn_XFFxk3-pFiaX MEMT203 (Dynamics): kzbin.info/aero/PL1IHA35xY5H6G64khh8fcNkjVJDGMqrHo MEEN361 (Adv. Mech of Mat): kzbin.info/aero/PL1IHA35xY5H5AJpRrM2lkF7Qu2WnbQLvS MEEN462 (Machine Design): kzbin.info/aero/PL1IHA35xY5H5KqySx6n09jaJLUukbvJvB (MEEN 361 & 462 are taught from Shigley's Mechanical Engineering Design) Thanks for watching!

[@josepacheco2460]

Wow! Thanks from Mexico!

[@TheBomPE]

Glad you liked it! Thanks for watching!

[@purplevideos]

Thank you! used this in preparing for my Finals!

[@TheBomPE]

best of luck!

[@ismetsemihuzuner9471]

Professor, you are the best!

[@TheBomPE]

You are very kind! All the best to you!

[@gameislifegl2480]

ı loved your method and also your way of telling. bravo

[@TheBomPE]

Glad you liked it!

[@dogansahutoglu2073]

time to use this calculator and get ready for PE exams

[@TheBomPE]

Good luck on your exams!

[@dogansahutoglu2073]

@@TheBomPE thanks alot passed with 80% average, I nee dto download the calculator emulator? any website name?

[@3508singh]

Thank you very much sir, Very grateful!

[@TheBomPE]

I'm glad I could help! Thanks for watching!

[@zEre326]

so easy to understand, thank you so much!!

[@TheBomPE]

I'm glad it was clear! Thanks for watching!

[@dswyl7628]

Thank you so much that was really helpful! I was so frustrated

[@TheBomPE]

I'm glad I could help! Thanks for watching!

[@AmaVelempini]

Woow very instructive. Thank you sir.

[@revathi369]

You are the best

[@TheBomPE]

You are vey kind. Thanks for watching!

[@eimansiraj6670]

Thank you very much This lesson is a life saver for me

[@TheBomPE]

I'm glad it was helpful! Thanks for watching!

[@gameislifegl2480]

Thank you so much

[@TheBomPE]

Glad I could help!

[@neelmadane2195]

Well explained. Helped me a lot. Thanks :)

[@TheBomPE]

I'm glad it helped! Thanks for watching!

[@vsivakrishna9647]

loved your explanation. Thank you so much.

[@TheBomPE]

Thanks! I'm glad you liked it! In case you haven't seen them yet and might be interested, here are some of my playlists: ENGR122 (Statics & Engr Econ Intros): kzbin.info/aero/PL1IHA35xY5H52IKu6TVfFW-BDqAt_aZyg ENGR220 (Statics & Mech of Mat): kzbin.info/aero/PL1IHA35xY5H5sjfjibqn_XFFxk3-pFiaX MEMT203 (Dynamics): kzbin.info/aero/PL1IHA35xY5H6G64khh8fcNkjVJDGMqrHo MEEN361 (Adv. Mech of Mat): kzbin.info/aero/PL1IHA35xY5H5AJpRrM2lkF7Qu2WnbQLvS MEEN462 (Machine Design): kzbin.info/aero/PL1IHA35xY5H5KqySx6n09jaJLUukbvJvB (MEEN 361 & 462 are taught from Shigley's Mechanical Engineering Design) Thanks for watching!

[@DeathScythe_ua]

That Helped alot, Thank You Very Much ♥

[@TheBomPE]

Glad to hear it! Thanks for the positive feedback!

[@tarunnaik8484]

Thank you. It was very helpful.

[@TheBomPE]

Glad I could help!

[@michaelswanbom6023]

If you found this video useful, consider helping me upgrade the old tablet PC I use to create these videos! Thanks! www.gofundme.com/help-replace-my-2011-tablet-pc

[@meshalnafea5200]

Great, Thank you very much.

[@TheBomPE]

I'm glad it helped! Thanks for watching!

[@meshalnafea5200]

@@TheBomPE your channel is giving great help, thank you wholeheartedly. I wish you nothing but your dreams become right, and visit us here in KSA hehehe, your student from Riyadh.

[@rockmechanicsuncen7045]

Thank you sir

[@TheBomPE]

Glad I could help! Thanks for watching!

[@ahahkah1360]

Very nice and good 👍

[@GeoCalifornian]

You are doing God’s work explaining some of his Intelligent Engineering Design. /In the Newtonian World

[@scienceandengineering6999]

Awesome explanation thanks alot

[@TheBomPE]

I'm glad you liked it! Thanks for watching!

[@zairimohamed6241]

thanks So match

[@TheBomPE]

Glad I could help! Thanks for watching!

[@justicemadu1668]

This helped a lot

[@TheBomPE]

I'm glad it helped! Thanks for watching!

[@meghabohot3758]

how did you get 37.084 in 2nd column 2nd row, when solving matrix for finding direction??? I don't get it.. :( @15:00

[@behzaddastjerdy]

Thank you so much for the excellent explanation. I've got a query to ask. What if we have the three princiapl stress (magnitudes and orientations), and we want to calculate the stress tensor??? Actually my quesiton is in my field of study which is the 3D geostress which is totally similar to what we have in this example you talked about.

[@gabrielcamacho9330]

amazing explanation it helped me a lot thank you:)

[@osamahjarrah6931]

I would like to know why the directions in the end are always opposite in the sign , Im I doing something wrong here , cause i got exactly the same answers but they are opposite in signs .

[@TheBomPE]

two solutions where the 3 direction cosines have the same magnitude but where all three have opposite signs between the two solutions are actually equivalent to each other. this is because the two solutions represent two possible "slope" directions along a straight line.

[@dhatrivegesana1934]

Thank you nice explaination

[@TheBomPE]

Glad it helped! Thanks for watching!

[@sebas733]

thanks sir

[@TheBomPE]

glad you found it useful!

[@franksaruhan]

Why did you multiplied left side by -1?

[@najwaalawamy2772]

thank you

[@TheBomPE]

You're welcome! Glad it helped!

[@Adek90]

The explanation is very good ... but I wonder about its correctness. I have implemented your approach and calculated the 9 directional cosine components for the case you are analyzing here. I have completely independently implemented the directional cosine procedure from Appendix B of the book: "Advanced Mechanics of Materials and Applied Elasticity" Sixth Edition by Ansel C. Ugural Saul K. Fenster. The results are almost the same but not identical. Well, the 3 directional cosines for the first principal direction are the same in value but opposite directions. In other cases, the solutions coincide. For another tensor example, all the results that I got with your method are correct in terms of value, but all of them should be in the opposite direction according to the second procedure. I am a bit confused, because probably the definition of a tensor with respect to the 3 axes should be unambiguous? If you want, I can send you both procedures (the file is written in MAthCAD, so its printout looks like I wrote it in Word - you will understand everything) - just give an e-mail if you are interested.

[@TheBomPE]

If a valid solution is found for a set of 3 direction cosines, an alternate valid solution is a set of 3 direction cosines that are all negated relative to the first set. Both of the solutions describe the same line orientation, which is the purpose of finding direction cosines.

[@roseriya3640]

It helps a lot thanku

[@TheBomPE]

I'm glad it helped! Thanks for watching!

[@sohailjanjua123]

Hi I like your video

[@TheBomPE]

I'm glad you like it! It's my first one posted! You should look at all of the playlists I have put together for all of my courses! Thanks for watching!

[@timothy4247]

Sir, Suppose only Principal stress is asked then, the 3 principal stresses that you have found are the answers or just the positive one?

[@TheBomPE]

There are always 3 principal stresses for every material element. They represent stresses along three mutually orthogonal axes.

[@abhishekpg9615]

How would we know that we have linearly dependent equations without getting into echelon form? I am running into problem while trying to solve the following set of equation represent ed by matrix 3 -3 sqrt(2) -3 3 -sqrt(2) Sqrt(2) -sqrt(2) 6

[@TheBomPE]

If you multiply your first line by -1 you get your second line. That means those 2 lines do not represent independent equations. Also, the 3rd line can't be true if the 1st two lines are true ... you'll be able to prove that by multiplying the 3rd line by 3/sqrt(2).

[@abhishekpg9615]

@@TheBomPE Thanks for the fast reply. So is there a way to solve it by putting it into calculator. Calculator shows math error if i follow the same procedure as you have shown. After gauss elimination i think get it like 3 -3 sqrt(2) 0 0 0 0 0 (16/3) From this z has to be zero. And i can choose any value of y or x. Any specific way to solve this in a calculator?

[@TheBomPE]

@@abhishekpg9615 your system is not solvable. you should check whatever theory led to those equations.

[@sahebasadaf1410]

Woww

[@oneofyk]

Running into an issue where I'm getting incorrect principal directions. The tensor is sigma xx = 110, sigma xy = 60, sigma yy = -86, sigma xz and yz = 0, and sigma zz = 55. The I1, I2 and I3 are 126.91, 55, and -102.91 (MPa) respectively. I'm getting 0,0,1 for all principal directions - which is incorrect. Well... the I2 is correct, but (for example) I1 has l, m, n of .963 and .271 and 0. Any idea what could be the issue? Your method has worked for other questions Edit: seem to be getting incorrect answers whenever sigma xz, yz = 0 Maybe I'm just doing it wrong tho

[@TheBomPE]

When you have two of your shear values =0, I call this situation "plane stress plus." I would do this problem by doing a Mohr circle for the plane that has the shear, then plot the "extra" normal stress value along with that circle to make a Mohr 3 circle diagram. One of your principal directions will be along the direction of the "extra" normal stress (that stress is already a principal stress), and the other two principal axes can be found using your first Mohr circle (a single axis rotation about the "extra" stress axis). It does not surprise me that the technique demonstrated in this video breaks down for this specific case you mention. I think it would take an explanation bigger than I want to try in a KZbin comment though.

[@oneofyk]

@@TheBomPE Thank you for your detailed reply. By "two shear values" do you mean zx, yz = 0, or just xz and zx =0. (just as examples) Sorry just starting this topic and my theory isnt up there yet.

[@TheBomPE]

If any two shear values are 0 then what I wrote would hold. Remember that the assignment of names to the axes (i.e. x, y, z) is essentially arbitrary.

[@oneofyk]

@@TheBomPE makes sense. Thank you for taking the time to reply

[@ronaldkejenge6347]

Hello Prof. how can I calculate the cosine angle using the calculated vectors

[@TheBomPE]

The elements of an eigenvector are the 3 direction cosines that describe the line along which the associated principal stress is oriented. If you want the actual angles you can take the inverse cosine of each element of the eigenvector.

[@ronaldkejenge6347]

@@TheBomPE thanks for your timely response Prof.

[@leeruy5160]

God save you

[@TheBomPE]

He has, my friend.

[@mohamedkhireldin8885]

Why equations are not mutually independent ?

[@mohamedkhireldin8885]

What is the method which you used for solving?

[@TheBomPE]

The vector of direction cosines is a unit vector, so if two of its three components are known, it determines the 3rd component.

[@mohamedkhireldin8885]

Thanks for reply, But how you eliminate the third row and replaced it by (0,0,1)?

[@mohamedkhireldin8885]

@@TheBomPE Please, tell me this numerical technique that used to get direction cosines.

[@mohamedkhireldin8885]

@@TheBomPE From where are we know two components of The vector of direction cosines? In your solution you get X,Y and Z values. which I understand that X,Y and Z values are the components of the unit normal isn't it my dear ?

[@steveokocha9860]

You are the best

[@TheBomPE]

Thanks for the encouragement, but I'm just a sinner redeemed by the best. Thanks for watching!

[@farooqadam7998]

Thank you very much

[@TheBomPE]

I'm glad you found it useful! Thanks for watching!

[@justicemadu1668]

Wow, thank you very much

[@semihtuncer12]

Thank you so much ..

[@TheBomPE]

Glad you liked it! Thanks for watching!

[@dabkepromax123]

Thank you so much

[@TheBomPE]

I'm glad you found it useful! Check out videos relevant to my other courses too! Thanks for watching! ENGR220- Statics & Mechanics of Materials - kzbin.info/aero/PL1IHA35xY5H5sjfjibqn_XFFxk3-pFiaX MEMT203- Dynamics - kzbin.info/aero/PL1IHA35xY5H6G64khh8fcNkjVJDGMqrHo MEEN361- First 6 chapters of Shigley's Mech Engr design - kzbin.info/aero/PL1IHA35xY5H5AJpRrM2lkF7Qu2WnbQLvS MEEN462- chapters 7+ in Shigley - kzbin.info/aero/PL1IHA35xY5H5KqySx6n09jaJLUukbvJvB

[@Linhesw]

Thank you so much!!