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Deflection of Frames using Principle of Virtual Work - Intro to Structural Analysis

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Structures with Prof. H

Structures with Prof. H

Күн бұрын

Пікірлер: 12
@silencerrh
@silencerrh Жыл бұрын
Here we go again, so much better than how my Uni was teaching it. You're a ripper!
@StructuresProfH
@StructuresProfH Жыл бұрын
Go ahead and spread the word to your classmates. I don't mind additional publicity. 👍
@silencerrh
@silencerrh Жыл бұрын
@@StructuresProfH will do!
@mitchelltracy6260
@mitchelltracy6260 Жыл бұрын
I have been spreading the word like the gospel 😎
@忘歸客
@忘歸客 Жыл бұрын
why the horizontal force is 8kn?
@StructuresProfH
@StructuresProfH Жыл бұрын
We need to a take the sum of moments equals zero about point A to find the horizontal reaction at C. Doing that, we see... -(20 kN)(1.6 m) + (Rcx)(4 m) = 0, where Rcx is the reaction at point C, assuming it points to the left. Solving for Rcx, we get Rcx = (20 kN)(1.6 m)/(4 m) = 8 kN to the left. We then apply sum of forces in the x-direction equals zero, which shows that the reaction at A in the x-direction is 8 kN to the right.
@Moxie_Valor
@Moxie_Valor Жыл бұрын
Awesome video. Thank you for posting. quick question on this: Why does the integral for the moment at member B-C only go from 0 to 1.6 instead of from 0 to 3.2? Thanks
@StructuresProfH
@StructuresProfH Жыл бұрын
Thanks for watching! The integral goes from 0 to 1.6 only because the moment from 1.6 to 3.2 is equal to zero (I was being lazy). So yes, we should integrate along the entire length from 0 to 3.2, but the portion of that integral from 1.6 to 3.2 goes away.
@clarencefigueroa8552
@clarencefigueroa8552 10 ай бұрын
Hi There, why don't we calculate and add the shear force strain alongside moment and axial strain for the total deflection?
@StructuresProfH
@StructuresProfH 10 ай бұрын
Shear deformation usually is not significant for beams that are relatively slender (i.e., length/depth ratio > 10), so it is often ignored. However, as beams become deeper (i.e., lower length/depth ratio), shear deformation may be important. It can be included by adding another integral term to the equation: integrate (K*Vi*Vr)/(A*G) over x from 0 to L, where K is a cross-section factor that describes the shear stress distribution, Vi is the shear due to the unit load, Vr is the shear due to the real applied loads, A is the cross-sectional area, and G is the shear modulus. Here are a few simple assumptions for the cross-section K factor of common shapes: Rectangular cross sections, K = 1.2 Circular cross sections, K = 1.11 I-shaped cross sections, K = 1 but the area A only includes the area of the web, not the flanges Technically, K depends on the Poisson's ratio of the material, and it can be rather involved to compute exactly. Also note that my definition of the K-factor is the inverse of what some other texts use, so just bear that in mind if you are searching for K values of typical sections. For example, if you look up Timoshenko beam theory on Wikipedia (en.wikipedia.org/wiki/Timoshenko-Ehrenfest_beam_theory), they use K*A*G whereas I'm defining K/(A*G). It still works out to the same thing if you are consistent. The Wikipedia site approximates K = 5/6 for rectangles, but this is just the inverse of 1.2 which is what I've cited above.
@gelgl5488
@gelgl5488 5 ай бұрын
when we talk about principle of virtual work , we say that external force * delta distance = ∫ internal force * deflection for any system thus when we use it for unit load method , how we use this principle for 1 unit load (virtual load) and for real load in the same time , how this combination !!???
@uzjenompajicek2023
@uzjenompajicek2023 3 ай бұрын
Thank you!!
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