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 !!???
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.
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
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.
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.
Here we go again, so much better than how my Uni was teaching it. You're a ripper!
Go ahead and spread the word to your classmates. I don't mind additional publicity. 👍
@@StructuresProfH will do!
I have been spreading the word like the gospel 😎
Thank you!!
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 !!???
Hi There, why don't we calculate and add the shear force strain alongside moment and axial strain for the total deflection?
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.
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
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.
why the horizontal force is 8kn？
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.