[Mother Goose]

Any mechanical engineers around?

I have the following problem:

• A thin elastic (2D) beam or plate is pushed down with a force F onto a fluid. Everything is in equilibrium.
• Go ahead and ignore the fluid mechanics. The important forces are...
• We have a load p(x) which is applied between L* < x < L
• We have downward forces T1 and T2 applied at x = L* and x = L.

What I am trying to determine is the necessary boundary conditions at x = L* and x = L. In particular, what is the moment M(x) at x = L (or L*) and the shear V(x) at x = L (or L*).

For the below picture, ignore some of the stuff in the boxes. What I'm trying to verify is that at the rightmost boundary,

Moment = M(L) = 0
Shear = V(L) = T2

Can someone verify that I'm correct? I'm not exactly sure why the moment at x = L should be zero, except that it seems to make sense.

Edit: I realised now that if we're holding down the left-hand side of the plate, then there needs to be a moment there, as well.



This seems like a basic problem, but I'm afraid I'm not up to snuff in my solid mechanics...

[LoneWolf121188]

The shear at x=L is [-F - T1 + p(x=L*) + p(x=L) + T2]. For the moments, you need the physical boundary conditions. Are the ends fixed or free? Free ends always have a moment equal to 0.

Here's a very, very crappy shear and bending moment diagram. Not numerically correct, but it should get the point across. The slope of the region on the M(x) diagram after the linear region should be negative up until the inflection point, where it becomes positive after. The inflection point corresponds to the point where the V(x) graph crosses the axis. On the V(x) diagram, the forces labeled are the point loads at each point, not what the shear is equal to at that point (ie, where it says T1-p(x=L*), that means the point load at that point is equal to T1-p(x=L*), but the total shear at that point is actually -F - [T1 - p(x=L*)].

Sorry I'm a little complacent with signs...I'm an engineer, what do you expect? You don't need signs to tell you which way something bends...you just push on it and see which way it goes.

[Mother Goose]

Thanks, I think I managed to figure out most of the aspects of the problem. At the free end, you should indeed have Moment M = 0, but the shear I think is, V = T2. I'm not sure if you're quite right...

Originally Posted by LoneWolf121188

The shear at x=L is [-F - T1 + p(x=L*) + p(x=L) + T2].

In particular, I considered a small segment from x > L* to x = L. The vertical force balance on this segment is:

V(x) - T2 + Integral[ p(s), s = x to L] = 0

where V(x) is the unknown shear at x, and p(x) is the load. Then let x tend to L and so long as the load is integrable at the end (and not a singular force), then,

V -> T2

Similarly at x = L*

V(left of L*) = -T1 - T2 + Integral[ p(s), s = L* to L]
V(right of L*) = -T2 + Integral[ p(s), s = L* to L]

and V suffers a jump of T1 upon crossing L*. One comes up with the formulae for the moment doing something similar. The moment should be continuous at x = L*and I think can be computed using

M(L*) = -T2(L - L*) + Integral[ p(s), s = L* to L] * (L* - C)

where x = C is the centroid for the pressure distribution.

Sorry for all the beginner questions. All I have is a first year physics course I took eons ago.

[Mother Goose]

Okay, maybe a harder one?

Here, we have a plate that is clamped at one end, at a height h0, and left to scrape over the ground. They'll be a reactive force and a reactive moment in the clamped end, and a normal reactive force on the ground (I'm pretty sure the ground can't produce a reactive moment).

What is the equation for theta? Or maybe an easier but related question is what angle is produced at the end of the plate?

I'm still working on this one, and any help would be appreciated.

[Mother Goose]

It seems that the above problem is mostly the same as all the others. The difference here is that two things need to be taken in account. The first is that the reactive force Ff is unknown, but only its component normal to the plate is important. The second is that we have to write an equationwhich relates h0 to the deflected board. This equation is something like L - Integral[ cos theta(s), s = 0 to L] = h0.

I'm sorry to have occupied so many unanswered posts. I thought that there were more engineers around here, but this should be a happy end to this post.

[LoneWolf121188]

^^ Wouldn't this be highly dependent on the rigidity/elasticity of the material? Doing that with a rubber band would produce a very different result than if you did it with a plate of hardened steel...

Originally Posted by Mother Goose

Thanks, I think I managed to figure out most of the aspects of the problem. At the free end, you should indeed have Moment M = 0, but the shear I think is, V = T2. I'm not sure if you're quite right...



In particular, I considered a small segment from x > L* to x = L. The vertical force balance on this segment is:

V(x) - T2 + Integral[ p(s), s = x to L] = 0

where V(x) is the unknown shear at x, and p(x) is the load. Then let x tend to L and so long as the load is integrable at the end (and not a singular force), then,

V -> T2

Similarly at x = L*

V(left of L*) = -T1 - T2 + Integral[ p(s), s = L* to L]
V(right of L*) = -T2 + Integral[ p(s), s = L* to L]

and V suffers a jump of T1 upon crossing L*. One comes up with the formulae for the moment doing something similar. The moment should be continuous at x = L*and I think can be computed using

M(L*) = -T2(L - L*) + Integral[ p(s), s = L* to L] * (L* - C)

where x = C is the centroid for the pressure distribution.

Sorry for all the beginner questions. All I have is a first year physics course I took eons ago.

Interesting...mathematically, that is correct...I think it's just weird for me because I'm used to being given boundary conditions and then figuring out the shear, not figuring out what the necessary boundary conditions are...

[Mother Goose]

Originally Posted by LoneWolf121188

^^ Wouldn't this be highly dependent on the rigidity/elasticity of the material? Doing that with a rubber band would produce a very different result than if you did it with a plate of hardened steel...

Well, the Euler-Bernouilli equations only holds for linearised theory, anyways, so you're always assuming that the deflection of the plate is small (compared to its length). This should be clear by the fact you approximate the curvature of the beam by the second derivative of the deflection. So no, everything should be equally valid for the elastic band vs. the steel plate, on the assumption that you're only working with forces that deflect both a small amount.