Deriving the equation of motion and for an Euler-Bernoulli beam and solving for the response. Download notes for THIS video HERE: bit.ly/37O9x7N Download notes for my other videos: bit.ly/37OH9lX
´great video. i had 4 classes of these and neither me or my colleagues understood anything of this subject. i've learn more in 30 minutes than 8 hours. great job, thank you
Hello sir! First, thanks for that great video. I have one question: if the beam has a circular cross-section, what method should be used to solve the problem? Is it something we need to solve in polar coordinates?
The current method makes no assumption as to the shape of the cross-section. Does not require polar coordinates. Changing the shape of the cross-sections should only affect the value of I (the moment of area).
This is a consequence of the Euler-Bernoulli assumption and is a result of the shear effects being ignored. When shear effect are considered, then the rotational inertia effects play an important role. I probably should have mentioned this in the video.
Yes, however, the 'i' is absorbed into the constant (d4 in this case) - i.e. the constants, in general, can be complex numbers. I probably should have mentioned this in the video.
Great video, my only question is, is the force f(x,t) pointing in the +z direction because thats the direction the force is being applied? Or because when the beam moves laterally the forces inertia is moving up? When I make f(x,t) point in the -z in my derivation I get EIw_x_x_x_x' + pAw_t_t+ f(x,t) = 0. Which makes sense to me. Thank you and I look forward to more videos on this topic
You have exactly the same equation as I do with the exception of the sign on the "f(x,t)" which makes sense because you have defined your force, f, to be in the opposite direction from mine. Both are equivalent and both are correct.
Hello sir, At the time stamp 29:20 when the edge is pinned, w(x,t) [displacement] is zero, but why is it that W(x) [space dependent part] is equal to 0?
w(x,t) = W(x)T(t). So, w(0,t) = W(0)T(t) = 0. That means either W(0) = 0 or T(t) = 0. However, if T(t) = 0, then w(x, t) would be 0 all the time (which is a trivial solution). So, we must go with W(0) = 0.
Hello, first of all thank you for this great video. I need your help about my project which is the free vibration of a cantilever(clamped-free end) beam with initial displacement. as you know to plot displacement=time graph of the tip end point of the beam, i need to have final equation for the graph. Could anyone please help me? thanks
At 31:50 you say there is a different kind of boundary condition (b.c.) if there is a mass attached to the end of the beam. What kind of b.c. would that be?
This would be a "force" boundary condition. Instead of that boundary being a free edge, there would instead be an inertial load (both translational and rotational) at the edge. This will result in there being a non-zero shear stress and moment at that boundary. This is common, for example, when modeling water towers or a tip-tank mounted on a wing.
@@Freeball99 Thanks for prompt reply. A colleague and I are trying to model the damped oscillations of a polo mallet. So for example if there was a 10 kg mass at the end of a 2 m beam, the shear stress would be 10*9.8 = 98 N and moment would be 98*2 = 196 N.m? Or does the shear stress need to be calculated by dividing force by the area of the beam? Any corresponding modification required for the moment?
@@kenlouie5453 No this is not correct. Remember, we are talking about the dynamic case here - so inertial loads must be included. Ignore gravity for now and imagine that all of this is taking place on a frictionless horizontal surface. Imagine for now that the mallet is a cantilevered beam with a tip mass. The tip is displaced a small amount and then released and allowed to vibrate. As the tip of the beam translates and rotates, it is resisted by the mass. We are trying to apply Newton's 2nd Law, namely, F = ma and M = Jθ_ddot at the boundary. The F is the shear force (not stress). So, the boundary conditions would look like (ignoring gravity): @ x = 0 or L: EIw,xxx = m·w_ddot EIw,xx = J·w,x_ddot where w_ddot = translational acceleration and w,x_ddot = rotational acceleration at boundary. In each case, I might be off by a -ve sign depending on which boundary you are applying it to and depending on how you've defined your axes. In order to include gravitational effects, just add (or subtract depending on axes) mg to the right hand side of the equation. Also, if you are going to include gravity, then you should include the weight of the beam too (by applying it as an external load). Again, my advice is to ignore gravity as it changes very little about the nature of this problem and only confuses the issue. Hope this makes sense, it's a little hard to write equations in these comments.
I have background in fracture mechanics, but I am not familiar with breathing cracks. Just googling it now...looks interesting. Definitely a topic for a more advanced class and definitely not introductory material. I will read up on it, but am unlikely to write about it in this forum. Thanks for the question - always happy to learn.
In this problem, I have not assumed any gravity is present. I have assumed that the external load is some general function so if you wanted to include the weight of the beam, you could include it in the external load.
This is a fairly common technique when dealing with "higher-order" terms. These terms are negligible compared with the other terms. One way to view this is to divide everything through by dx. Then only the 3rd and 4th terms still have dx remaining. However, dx is very small (it is infinitesimally small), so anything multiplied by it will be very small too when compared with the 1st and 2nd term and can thus be neglected...and so we can throw away these terms with very little loss of accuracy.
Euler-Bernoulli assumption applies to slender beams (i.e. beam of high aspect ratio) and says that the cross-sections of the beam remain undeformed and normal to the beam's elastic axis at all times - so there are a couple of things there: 1) that the beam's cross-section doesn't deform and 2) it remains perpendicular to the elastic axis at all times. Consistent with this, one can also say that the bending of the beam is due to normal stresses only and that shear stresses can be ignored. Also consistent with this is that the rotatory kinetic of the beam can be ignored. Again, this applies to slender beams only (like sailplane wings, helicopter rotors, wind turbine blades, etc.). For the case of shorter, stubbier beams, the Euler-Bernoulli assumption falls away and usually Timoshenko beam theory is used.
I'm just following tradition. Typically in the texts, the transverse displacements are referred to as v & w (as can be seen in the figure that I borrowed).
