Video

I am surprised that it is not in the playlist, also. These were created so long ago (last time I taught Statics), that I do not recall how many of the inclass examples were also posted. Entirely likely that it simply was not done. Or, it sits somewhere else and just wasn't added to the playlist.

Think of it this way ... walls only have significant strength and stiffness in their plane. Out of plane, they typically behave as a simply supported beam (a 1-ft wide strip that runs up and down). They transfer the forces perpendicular to their surface (such as wind and seismic effects) to the "top" and "bottom" of that strip. Sometimes, the engineer designs the wall for the strip to be horizontal rather than vertical. Either way, the wall transfer those load effects over to perimeter elements (could be a column, beam, or wall that is perpendicular to the wall of interest; could be a floor or roof system). The reason is that in-plane behavior is very stiff involving shear and normal strains that are very stiff ... think in terms of EA/L like in an axial stiffness situation as opposed to EI/L^2 or L^3 for bending situations. Out of plane wall behavior is like bending of a beam. It is more complex than just EI/L (the latter being beam bending behavior but the wall is really structural plate behavior). If that is more theory than you like, consider it this way. Glue a bunch of long sticks together and stand the whole thing up like a wall. Clamp ahold of the "Wall" at its base. Try to push it in-plane. Compare that to what happens when you push it out of plane. In the extreme, think of the wall as a sheet of paper. That sheet of paper has almost not out of plane stiffness.

​@@dougschmucker8212Hey Doug, that makes sense but my question was along the lines of why is it always assumed as though the out of plane walls are one way slabs transferring loads to diaphragm and footings, and not as two way slabs with loads being transferred to the in plane walls as well?

@@SudheendraHerkal736 Two basic reasons: engineering design is about what is good enough as opposed to what theoretically might be, and what is simplest and easiest to do that gets the job done. Regular walls are typically designed and constructed as one-way systems and most often they behave that way as well. Maybe even more importantly: too time consuming to work through and verify the two-way slab design process even if automated in a computer ... Note that most of the computer solutions are nothing more than a way to code up the hand based solutions. It is rarely if ever a true FEM solid modeling of a three dimensional slab. We wouldn't even know what to do with those stress calc's because of the complex nature of reinforcement and matrix interactions. Instead, we always end up with approximate methods whether by hand or by computer tools. We can be more efficient with our time and end up in a safe place using one-way methods. Then, if we need to refine what we are doing, then we can do something more precise. And, finally, if you start adding openings, then you are going to end up with something way different than just a "two-way" slab wherein you add so much reinforcement that you might as well have trust treated it as a series of one way strips. Obviously, people have found all kinds of different ways to do things, but these are many of the reasons why we start with a very simplistic model and only add complexity as it is genuinely needed.

@@dougschmucker8212 Hey Doug, that makes sense! Thank you!

It doesn’t. All depends on bracing patterns. If bracing prevents lower modes, then the higher occurs.

Try the next mini lesson, video V1 3 Derivation of the Shear Stress Model

Depends what you mean by a vertical slider. Rollers on a vertical plane? Then rotation is constrained and horizontal displacement is constrained. But vertical displacements are free?

😉Yes, but of course those would come from my day job. Copyrighted and such.

Entirely welcome!

What is going on here is slightly different. The protocol being used is to first "guess" or speculate as to the answer. Then, apply a specific methodology which in turn provides an improved (and perhaps exact) answer. I believe that your question refers to the exact fixed end moments. This first step "0. Guess" assumes something simpler that is not a statically indeterminate system so that the guess is easy to obtain. In later parts to the work, you find the details of how one gets to those answers that you seem to reference. By making the system determinate or using some other assumption and then comparing those results with those of a more exact approach, you end up discovering key aspects of either the solution technique or the actual performance of the system. It also often reveals to the novice the typical mistake that is made. For instance, assuming the moment is zero in the middle of the beam (a hinge) does permit a way to estimate the end moment (PL/4). The real moment at the end will be half of that (PL/8). Indeterminate beams are stiffer than determinate beams (all other aspects being the same). As such they will develop smaller internal forces than the comparable determinate system. And that is a key understanding.

Yes

Yes and no. Certainly, yes but with different numbers. You might check my matrix channel; not entirely clear just what is or is not included in that one.

@@dougschmucker8212 what is your matrix channel?

@@colinhennessey5076 Looks like I am mistaken. I do not have a matrix playlist. I used to use many of the videos in the first structural analysis course in the second (matrix). But, not quite seeing the specific ones that I thought might be there. I am no longer the primary instructor of that second course, so don't really have motivation for putting those examples on a free channel.

Hi, I am working on a project having both tension and T and C bracing, we can connect if u are still interested.

You are welcome

You are welcome!

Thank you, Peter!

It is an example, that is all

Noted, Thanks

Thanks. It is a traditional view point not covered much anymore in the textbooks.

All depends on the sign convention used for moment. The US sign convention assigns positive bending moment to positive curvature. The European sign convention assigns positive bending moment to hogging (negative) curvature. ps. sort of sounds like you searched the internet for an answer rather than learning the principle. hope instead it was the latter instead of the former.

A fixed end would resist translation in two directions and rotation about the third (if in 2-D).

you are welcome!

first transform from local to global coordinates if not already done so. then map the local degrees of freedom into the global from which that tells you where to populate the values.

Just imagine a plate on the cross section where the force will be applied at some certain distance

Ya welcome!

you can find me at the university of utah

With respect to point C, the shear center, no. That is actually the issue at hand. The 900 N is the shear force. It needs that couple in the flanges to equilibrate the couple caused by the 900 N with respect to the shear center. These asymmetric shapes induce this type of effect.

can we have a detailed solution video for the same problems.?

Thanks, Savannah. The distributed load case would be quite similar. There, I might approximate the situation as a simple beam ... now the moment diagram is a concave downwards parabola and the maximum positive moment would be wL^2/8 in the middle (zero at the ends). We know for sure that the real maximum positive moment is less than that. (Turns out to be wL^2/24). The real moment diagram will be that from the simple case super-imposed on a moment diagram that has two equal and negative moments on the end. The key question is just big those might be. We could do the same thing as the video: insert a hinge in the middle and consider what then happens. The end moment would become wL^2/8 (wL/2 multiplied by L/4). So now we have a largest negative value bound for the end moment: It has to be between -wL^2/8 and 0. The largest moment in the middle has to be between 0 and wL^2/8. It is weird that both non-zero bounds are the same. And, the range is maybe too large to be useful, but it starts to point us in the right direction. The real answers are -wL^2/12 and wL^2/24. Note the algebraic difference of wL^2/8. That is not happenstance. It is the incremental change in the moment from the end to the mid-point due to the uniformly distributed load. Happens every time. End conditions don't change the incremental value. End conditions just change where we start the moment. It is much the same approach for any symmetrical load pattern. For asymmetric load patterns (such as off-center loads), I might insert the hinge under the point load whereever it is to find estimates. Because we are doing something extreme (inserting a hinge), there is a reasonable chance that we are creating bounding conditions on the response values.