Introduction to statically indeterminate problems and the principle of superposition
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- Опубликовано: 9 сен 2024
- This mechanics of materials tutorial introduces the method of superposition and explains how it can be used to solve statically indeterminate problems.
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Thanks for watching, I hope it helps!
You can't imagine how helpful this video was to me. I've never thought of solving the problems this way.
Thanks a lot!
Hey your comment slipped past me somehow without a reply, thanks for watching and letting me know!
I've taken 2 classes teaching this concept and this video actually made me finally understand it
Glad I could help! Depended on what you're learning right now, I recommend checking engineer4free.com/mechanics-of-materials or engineer4free.com/structural-analysis for all the other tuts I did on statically indeterminate structures 👌👌
Geez... this was beautifully explained! Definitely the best I could find. Earned a sub with this one.
I never understood what is principle of superposition.. Thank-you sir... Keep making more problems...
Glad I can help 🙂
Incredible video, you did in 7.5 mins what took my lecturer 2 hours
Thanks! You can check out the full playlist here: engineer4free.com/mechanics-of-materials =)
Shouldnt the sum of forces be Ra + 150kn - Rb? seeing that Rb is in the opposite direction of Ra and the force?
Thanks for this tutorial! This helped me understand what I did not catch in class. Wish I'd found your video's and website earlier - I now swear by them.
Awesome glad to hear it! Please tell some friends about the videos too =)
Sir thank you so much for the tutorial~!
Great video!
Question: @6:35 Rb*600mm-> why 600? shouldn't it be 200mm? I thought the pressure acts on 200mm and the pulling acts on 400mm... THX!
We don't even need the values of the modulus of Elasticity and the cross sectionalrea of the beam. We can have 2 equations for this problem by first using the displacement equation of B with respect A equal zero. Second equation would be just the body diagram of the whole beam. Don't need extra calculations
Why does the right section equal 0? I am confused on why when finding the deformation the whole 0.4 + 0.2 isnt used? Anything will help! Thank you!
A little late but the section has a net force of zero because the section is not accelerating. If there was a net force greater or less than zero then there would also be a net acceleration that would cause the section to move.
Thank you for the video, it helps clarify a lot of things.. Curiously though, why should we use 600mm rather then 600+1.5mm? It seems more logical to consider the entire length of the material after/under deformation rather than the rest length.. Is there any explanation for this?
The deflection 1.5mm is accounted for on the left hand side of the equation, it has been designed to work that way with original length and deflection being treated as separate quantities. Keep in mind that the length of this member really never does change because it's bound by the rigid walls. Removing the one wall and letting it elongate is a hypothetical scenario that happens to reveal useful information in the actual case where the wall is there the whole time.
@@Engineer4Free thanks for the explanation!
Where do you create or simulate these problems?
teacher i have a question
What is the importance of the statically indeterminate?
Hey Abdullah. Nearly every actual structure in real life is statically indeterminate. Most buildings, or even chairs or tables are not built exlusivily with pin connections and 2 force members. In order to design and analyze real structures, you need to learn the methods, which are beyond the level of statics. Check out the three courses I have on structures here: engineer4free.com/statics - engineer4free.com/mechanics-of-materials - engineer4free.com/structural-analysis ✌️
Shouldn’t the length be 700mm?
What is the name of this programme ?
Hey Tamer, I've got a full list of the hardware and software that I use over at engineer4free.com/tools 👌
Is the area the same in the first formula and then in the last one?
It should be different, I think
A is the Area is the cross-sectional area of the member. It's a constant 200m^2 anywhere in the member.
@@Engineer4Free I think maybe he is referring to the beams Area expanding due to the conflict of not beeing able to stretch in length?
isnt MPa = N/mm^2? not GPa
sum of Fx= 0 should be Ra-150+Rb=0 sir?
I should have actually drew on the unknown forces of Ra and Rb, but I didn’t. I assumed them to both be pointing to the right, and the right being the positive x direction, we have all three terms, Ra, Rb, 150kN, then being positive values summing to zero.