Statically Indeterminate Beam by Superposition Example 1 (Part 1/2) - Mechanics of Materials
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- čas přidán 3. 09. 2013
- This video demonstrates how to calculate the reactions and draw shear and moment diagrams of a statically indeterminate beam by using the method of superposition (aka force method).
Part 1 - Apply the principle of superposition to calculate reactions.
Part 2 - Draw the shear and moment diagrams of the beam
Life saver!! Got a final tomorrow and been struggling with this all day.
Going through this was an enormous help; it really cleared up my confusion on indeterminacy.
This exact example is in my notes which i did not understand, thanks man. I also used your videos 2 years ago to get a mark of 75% in my exam, there was one question in the exam not covered by your videos which i struggled with massively.
Mann so are the best. Revision never felt so good. Thanks a lot
i really like that you flash through equations rather than spend time writing them down.
OMG YASS QUEEN! I was just going to comment this. I cannot stand when youtubers make us sit through drawings.
@@aaronwilliams4509 don’t say yass queen
You are the sole reason I pass these tests
Quickly rockin your channel all the way to finals, you are so good at explaining!
+A. Kessler thanks! Good luck with finals.
Great work. Perfect for a quick revision.
Cheers mate helped me with my assignment a bunch!!
Wow, thanks man! Very helpful
Thank you so much! You just saved my life :)
you are just awesome!! I can't say just how helpful your videos are!!
Mark Zeidan thanks for the kind words.
you gave me the solution of my problem by your awesome method. thanks very much
+KONAN BLE JUSTIN awwwww yeaaah!
Hey, I really enjoy your videos! Thanks so much for all the work. I have one question though.
You say at the beginning that this problem is statically indeterminate, but why?
When you solve for a statically determinate beam with sum of forces in y, and sum of moments about B you also obtain that the vertical reaction at B is 15 kN.
Also, can't you say that the horizontal force is negligible, because the deflection is so small that the horizontal component of the applied forces is 0 because
sin(very small angle) ~ 0
Anyway, thanks again for all your help, you're most of the reason imma pass my Mechanics of Materials exam tomorrow :D
Stay awesome!
I Love This!
love your explaination
Very helpful and very well explained!
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my god. this made so much sense
god thank you it was amazing.can you tell me how can i solve an indeterminate structure by using virtual work method?
I didn't follow the rationale for using shear equation [v(x)] for deflection at B. what should clue us in to use it? awesome video, thanks!
how do we calculate the formula in a problem where we have a couple at any point on a cantilever beam?
i wish u were my dad
Say what?!
i view you as a father figure
+Hhehehe Hhhehehehehe "No, I am your father" - Darth V.
+structurefree 😮 i knew it....
Thank you so so so much
Why at time time 7:30 did you use a different equation then at time 7:50 when both are point loads along the beam? Does it have to do with the support? The one at 7:30 starts with v(x) and the one at 7:50 is the delta equation.
Can You Perhaps do an example of Statically determinate to the second degree , Would be really helpful, Thanks
you're a boss
Excellent as usual...:)
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Can we use this superposition method for Statically Indeterminate Frames ?
thx in advance :)
In my university, we have a course named Advanced Engineering Mechanics in which they cover mostly 3D rigid body mechanics and advanced topics of Strength of Materials.
cool story bro
Thank you very much for your help :) before I find your video I couldn’t understood anything :)
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I'm solving for a statically indeterminate shim (cylindrical disc fixed at center) stack
This is really helpful.
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@@dellpi3911 thanx bro, could be a good reference, I've developed eqns by myself till now. simulations will definitely be helpful.
But say something don't post just the link, that's sus you know. I thought it was self-promotion
👍
how do you know when to use the equation of elastic curvature in these problems?
+Jack Heatly the equations of the elastic curve are just one approach to calculate deflections. it's up to you to choose the most convenient method of calculating deflections (e.g. moment area theorems, double integration, virtual work, castigliano's theorem, etc.)
how to approach for above beam when there is no concentrated load at point c and a moment acting at c...????
What about Castigliano method for statically indeterminate frames ?
how to solve a problem using that method
have you ever thought about doing a frame table matrix formulation for this problem? if I were doing a test, Am I allowed to bring it with me?
+Structural Analysis Not really. I don't know if you can bring it into a test, you should probably ask your instructor.
Thanks sir
6:47 how did you get those formulas at the bottom?
Which book u have referred?
hey could you do question 7.2 from mechanics of engineering materials 2nd addition benham, crawford, armstrong using superposition I cant seem to get it right
I have the ability to express the information in matrix form and I wonder if you would like to see how I do it, so maybe you make a video with ? . Thx
+Structural Analysis Thanks. I enjoy when people share what they learn with me!
Good day to structure-free, I had a situation of a shaft, simply supported by two bearings each for both sides (near to ends), with a single concentrated force in between this bearing which are (R1 _ R2 _ F1 _ R3 _ R4) , continuous with no discontinuity // I got messed up with the suitable calculations for determining the deflection, shear, moment as there are many methods out there from books and online // May I ask what type of methodology or maybe a suitable flow to obtain all these values for this situation? Or Is there any limitations where the Method of Superposition has its limitation when having a continuous beam with more than 2 degrees of indeterminacy? Thanyou to structure-free
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Hi sir! Did you solve the same problem using double integration method? Anyways this was very helpful!!
no, he solved the primary beam with actual loading and then with unit value of load by using formulas from tables, other wise it would be very lenghty to solve this with double integration or conjugate beam methods :)
legend.
Hi, I just want to ask why is the number of equilibrium eqns is 3? :) thank you so much! Btw, great video and lecture! :)
Sum of forces in the x direction, sum of forces in the y direction and sum of moments about any point
If you had let's say 4 redundant loads, does it just mean more equations?
The question is how far do you take it? Do you need to draw up every possible scenario for each redundant point?
Hey I was wondering how different would these be if you were to use the equation PL/EA?
+Peter Truong PL/EA is for normal stresses P/A these deformations are for bending stresses (Mc/I) and shear stresses (VQ/It) so the answer is very different
What to do if EI is not constant?
Thanks for the great videos. Would it be possible to tell me in detail how you got the bellow equations please?
1- dx = L = wL4 / 8EI , 2- V(x) = ((Px2)/(6EI))*(3L - X) , 3- dx = L = PL3 / 3EI
I used here d instead of delta please. Thanks in advance.
+AbdulMalek Makhdom the equations for beam deflections were from a textbook. Most textbooks and engineering handbooks have equations for deflections for common loadings. You can also search and find on the internet. Thank you for the kind words.
+structurefree HI there. I have the table in the text book that says what the individual equations are. I don't see one for equation 2 though. It seems this is an equation which comes from the fact that the force does not act at B? How did you get this equation? Specifically why did it change from WL^3/3EI to ((Px2)/(6EI))*(3L - X) ??
plus Slope(c)=-45 and Deflection,Y(c)=-108,solved by a 5x5 matrix.....
What is the difference between delta b0p and delta 1, other than direction? More specifically, how do you know whether to use v(x) or delta x=L for each one?
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hello sir what if the pointload is applied just before the point where you want to find the magnitude of deflection
do i then say 3L + X?
and if you don't mind can you please provide the link to the Beam-deflection table/chart that you used.
images.app.goo.gl/GjA7fQ6SU49YGAe19
at 7:55 why is L 6 and not 9 since the force is acting at 9?
you replace b with a force because that is the easiest to substitute. It could also be done at point A, but would be more laborious.
why did you choose By to be the redundant? why not Ay or Ax?
shylildude I chose By so that the "determinate" structure would be a cantilever, which is familiar and also has equations from various handbooks. You can choose Ay or M_A, but you would have to use your preferred method to calculate deflections for the determinate structure. Ax is known from sum of the forces in the horizontal direction, so it is not available as an option.
I didn really get how did arrive to Delta(BOP) equation? it is really confusing.. can anyone help pls. thanks
My professor is not letting us use the tables lol...
im so sorry
why is the deflection at 6m zero?
shylildude since there is a vertical support there restraining the beam.
Thank you so much for the video.
I have a question about the deflection equation of By. How do we get that? Could you help me?
I couldn't see such formula on the tables.
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@@dellpi3911 Thanks so much.
the best
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why is the middle value of the equivalency equation a V function and the other two delta???
~ 7:33
How did you find out that we needed 3 equilibrium equations
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where the shear & bending dia ?
What's the difference between force method and method of superposition? Thanks in advance! :)
+Roila Mae Robles the force method applies the principle of superposition to breakdown statically indeterminate problems.
thank you so much!
Can you using the same question but for force method.?
the method of superposition is another name for the force method.
structurefree ouhh thank you!!
This seems statically determinate. The reaction Ax is clearly zero, and then you could use equilibrium equations for the moment and y-forces to find reactions.
There are four unknowns and three equilibrium equations in this case. So it is statically indeterminate.
There is slightly mistake in case of 2nd deflection . There is moment at the redundant point too .
Was given a similar problem for redundant load and was taught none of this by my prof...... I’m so screwed in the real world
why was it necessary to do all that work to solve for By? couldn't you have just used the M about A to solve for By?
why is the number of reactions 4? Shouldn't it be 6?
3 at the fixed end and 1 at the roller
@tw0million @structurefree i also don't get why he use the v(x) either... my lecture use the Compatibility Conditions then find the equation for the redundant that he took out. Next step he just use the v(x) equation to find the what is the displacement
EDIT: I know why now... the reason he use elastic curve equation v(x) for graph 2 because in the formula sheet there is a formula for a concentrated load P acting at the end of the cantilevered beam v= - Px^2/6EI (3L-x).
where the other 2 (first and last) is acting @ 6m (it's not in the middle) so you can't use the v(x) equation if that make sense lol
I use lowercase v to represent a function for the deflected shape or elastic curve. In the compatibility equation, I am looking at the superposition of deflection at point B. In general, I use a capital V for internal shear force.
Since there is a support there, the deflection would be zero, which is why he added the sum of the 3 deflection pieces as being equal to zero.
Can you show the energy method to determine this question next time? Thank you!
Where would the method of superposition not work?
If you were given a loading function that weren't in your table of reference functions, which is very uncommon.
Another example you might see, is if the support weren't an idealized infinitely rigid support, but were instead a spring with a finite stiffness.
Stud
You made an error its 18 kN, I used method of forces and Morhs integral. 2x6 + 6 does not equal 15, the reaction should balance the loads 😅
Cant you just use sum of moments to find By, and sum of forces in the Y direction to find Ay, at the beginning to find the reaction forces? I did and got the right answer in like 30 seconds.
unless you were just using a simple example to illustrate your point. good video though
not clear enough
😭
I don not know why V (6 m) = 0
0 = Subscript[V, ow] (6 m) + Subscript[V, op] (6 m) +
Subscript[V, 1] (6 m)
Is not there moment in fix support A?
Why don not add moment in A
because at v(6m) it was roller support present orignally...and if its there ....then at v(6m) ther will be no deflection possible....
Before I watch....I got an upward reaction of 15KN at 6m, upward of 3KN at 0m, and a 0KNm moment at 0m.
before you work for khan academy...
Why you gotta yell though.