0:09

We're going to look at statically indeterminate structures in this module.

And so our learning outcome is to actually solve a statically indeterminate structure

problem for axial loading.

And so here is our example.

0:25

Please realize that as I go through today's module,

this is going to be a little bit longer video module than normal.

Because for continuity, I want to get all the way through the problem.

If I cut it up in to different modules, it just wouldn't flow as well.

And so, we'll go through it a step at a time.

Bar DE here is aluminum.

I give you the cross sectional area, the modulus of elasticity, and

what the yield stress is for aluminum.

And I do the same thing for bar BC, which is steel.

0:58

I have the cross sectional area, the modulus of elasticity, and

the yield stress.

Now we're going to consider bar ABDF to be rigid.

Both the aluminum and the steel bars are going to be deformable.

And we're also going to neglect the weight of the bars, and

that's because they're going to be negligible.

The weights are going to be negligible compared to the forces that they

are supporting.

And so we're asked to find the axial stress in the aluminum and steel bars and

the deflection at point A.

1:28

So, we will also work, just for simplicity, I'm going to work in

units of kilonewtons per millimeters squared and GPa, gigapascals,

because one gigapascal is equal to one kilonewton per millimeter squared.

You can change those after you get your final answer, but

it just helps to use consistent units.

So my question to you is, what should we do first?

And what we're going to do first

is to look at the static equilibrium of our system.

And so how would we develop the static equilibrium equation or

equations for our system?

2:06

And what you should say is what we've been doing from my very first course,

which is to draw a good free body diagram.

And so I'd like you to draw a free body diagram for this structure.

And the body of interest now is going to be our rigid bar ABDF.

2:30

So here's my free body diagram again.

You'll notice that this is what we call a statically indeterminate problem,

because we have three independent equilibrium equations,

as we've known since we started the course and my earlier courses.

But we have four unknowns.

Fx, Fy, the force in D, and the force in B.

The axial forces.

We don't have any y components at D or B, because the aluminum and

the steel bars are two force members.

And so here are my unknowns.

So I'm not so happy, four equations, excuse me, three equations, four unknowns,

statically indeterminate.

And so, let's start by at least writing the best equilibrium equation we can to

start to solve the problem.

And so, what would that equation be?

3:20

And what you should say is, well, let's go ahead and sum moments about point F,

because I'm really not interested in Fx and Fy, at least in the problem statement.

I'm interested in the axial stresses in the steel and the aluminum bar.

And so by summing moments about F, I won't have to worry about these unknowns and

I'll just Include the unknowns that I'm interested in.

So go ahead and do that moment equation on your own and come on back.

3:53

And so here is my equation of equilibrium.

One equation, two unknowns.

We're going to need an additional equation to solve these statically

indeterminate problems.

And so, what we use is what's called the deformation equation, which looks at

the geometry of the deformation of the members in the structure.

And it's all called the compatibility, can be called the compatibility equation.

Because it enforces compatibility between the equilibrium and

the deformation that the structure undergoes.

And so, we're going to assume small deformations, and so

therefore, small angles.

And here his my rigid bar in its original position.

And then I'm going to exaggerate, so I can see the deflections.

But here it is, as I move to the right.

And so I have a deflection at D and I have a deflection at B.

And I want to go ahead for my deformation or deflection equation or

compatibility equation to relate those deformations.

And so I'm going to have, by similar triangles now,

the deformation of D is to 300 as

the deformation at B is to 600.

And so what I find out is that two

times the deformation at D is equal to deformation at B.

And so, that's my deformation or compatibility equation.

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what's happening to our steel and our aluminum bars?

And so we'll go ahead and

use that idealized elastoplastic material assumption.

And since I don't know if these bars are in the elastic or

the plastic region, I'm going to have to make an assumption.

And so, what I'm going to do is I'm going to assume

that we're in the elastic region, the linear elastic region.

And if we're in the linear elastic region for prismatic bars,

we know that the deformation is equal to PL over AE.

We developed that in the last class.

In the linear elastic region for this prismatic bar,

delta is given by this expression.

And so I've got two times the force,

the axial force, and D is going to be D.

And its length is 250 and we're going to divide

that by its area, which was 5,000 times

its modulus of elasticity, which is 70.

And that has to equal the deformation at B.

Which is equal to the force B times the length of BC, which is 250.

Divided by the cross-sectional area of BC, which is 1300.

And the modulus of elasticity, which is 200.

And if we go ahead and solve then for B and D,

we end up with B equaling 1.49D.

And this is going to be equation two,

because now you see that I have two equations.

The equilibrium equation and this equation with two unknowns,

the force in member BEC and the force in member DE.

7:37

Okay, so here's those two forces.

Okay, solving simultaneously,

remember we're still assuming that both members are in the linear elastic region.

We gotta check that to make sure that that's true.

And so let's first check the aluminum bar to see if it's in the linear elastic

region.

And so I've got sigma DE.

Which is the stress in DE is going to be equal to normal force,

which is D over the cross sectional area.

The normal force in DE we found, if we use this assumption, is 446 kilonewtons.

And we're going to divide by its cross sectional area, which was 5000.

And so that gives me a stress of

0.0892 GPa, gigapascals.

We find that 0.08 92

gigapascals is indeed less than the aluminum's

yield stress, which is 0.28 GPa.

And so this is good.

We'll have a smiley face here, because this means that my

9:00

linear elastic assumption for the aluminum bar is correct, and it holds.

I want you to do the same thing now on your own for the steel bar,

find out what the stress is for this assumption in the steel bar.

9:14

Okay, so you develop again sigma in the steel bar

is equal to the force in the steel bar, the axial force divided by the area.

And this is the stress.

But, we see that this stress now is greater than the yield stress, so that

steel bar is not going to be able to hold that much stress and so that's not okay.

The steel bar has yielded, so we're going to have to go back and

rework what we've done.

And so here's our equilibrium equation.

We know aluminum is in the linear elastic region, but the steel has yielded and

we're going to say that it's in the perfectly plastic region for

the elastoplastic assumption.

And so now we can find the axial stress in the steel bar, because we know that

it's going to have a stress that's equal to the yield stress or .25 gigapascals.

And so, we've got .25 gigapascals, or

250 megapascals, is all that the steel is going to be able to hold.

10:16

Okay, so there's my equilibrium equation.

I found the stress in my steel bar.

Now I want to see how much force that steel bar is

actually going to be able to support, given that it has yielded.

And so we'll use this stress definition again, and I'd like you to go ahead and

calculate what that force is that can be held in the steel bar.

10:43

And what you should say is, okay, the force B = sigma times A.

And sigma is the yield stress, A is the cross-sectional area.

So what we see now is that the steel bar can only hold 325 kilonewtons.

When we had the linear assumption, back here,

we said it would hold 664 kilonewtons.

But it couldn't hold that much.

11:09

And so, it's 325 kilonewtons.

I can put that into my equilibrium equation and see now that D,

the force that's got to be held by the aluminum now, is more.

It's going to have to be 675 kilonewtons.

And so I find that my stress in my aluminum bar has now gone up,

it's going to be 0.135 GPa, or 135 MPa, and it's going to be in compression,

right, because we're pushing on it to the right.

But we know that the yield stress of aluminum is 0.28 GPa.

So we're still in the linear elastic region.

And so we're good to go.

And so that's the answer for our aluminum bar.

So we now have found both the actual stress in the aluminum and the steel bars.

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What we want to do finally is to calculate the deflection at point A.

In calculating the deflection we must use the aluminum bar, because that's the only

one that our deflection calculation, delta equals PL over AE, will hold.

And so there's our PL over AE, and so the deflection in the aluminum then,

deflection in the aluminum, is going to be the force in the aluminum bar,

which is 675, times its length, which is 250,

divided by its area, which is 5,000 and E which is 70.

And so it ends up coming out to be 0.482 millimeters.

That's the deflection of the aluminum bar.

And so knowing that, we can now come in and

we can look at the deflection at point A.

We know what the deflection of D is, we just found it.

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And all we do again is use similar triangles, so

I've got the deflection at D over 300 is equal to

the deflection of A by similar triangles to 750.

And so the deflection at A then is going to be,

if you calculate that out,

put in your value of 0.482 here,

the deflection of A ends up being 1.205 millimeters.

And so we've solved our problem.

13:49

Okay, so that's a good statically indeterminate problem.

You should try some on your own.

Find a reference where you can go through, either online or

in a textbook, and practice will make perfect, but they're rather long problems.

But they're straightforward, if you go step by step.

And this is a very typical example.

So that's it for today's module.

See you next time.

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