So we can go ahead and also solve for the bending stress here.

So the bending stresses are going to occur at points A like this,

where we're going to have stresses compressing point A and

then a neutral axis, and then stresses and tension at the bottom.

So at A we can see we're in compression and

that our bending stress is coming from this load F.

So, I can see the stresses are acting in the x-direction, the equation for

bending stress is negative MC over I.

We're going to simplify this equation to be relevant for a cylinder.

So we've done this a couple of times, where we set c equal to d over 2.

And I is Pi d to the fourth divided by 64.

And when we simplify that out we get 32M over pi d cubed.

And that's our bending stress equation.

Now because our moment is actually going to be positive and

our diameter is going to be positive, the negative sign is going to hold

which make sense because we know we're in compression here.

So we know this is going to be a negative stress.

And so really the only question we have left is,

what is the bending moment at this point A?

And because we have no other loads or

constraints between point A and the place where we're applying

the force F we can just say our moment is force times distance.

Which in this case is going to be 25 pounds of

force times the distance of 20 inches.

And that is divided by Pi times 0.05

inches cubed and we get sigma x is equal to,

let's see 40.7 KSI and

that's negative, so negative 40.7.

Okay, so we figured out our stresses, our normal stresses due

to axial loading and due to bending loading.

Let's go ahead and figure out our stress now due to shear or due to torsion,

which we know our point A is in the XZ plane at the top of the rods so

tau XZ is going to be equal to Tr over J.

And you guys have done this a few times now.

So you know that our J is going to be

Pi d to the fourth divided by 32.

And our R is going to be d over 2, so

that's going to simplify to 16T over Pi d cubed.

And so when we work through that we have been directly given a torque here.

So 100 pounds of force per inch

times 16 divided by Pi times 0.5

inches cubed is equal to 4.1 KSI.

So now, if I looked at my stresses,

I can see all of the stresses start to add up and

so what I have is sigma X due to axial is

equal to negative 1.3 KSI.

I have a sigma X due to bending which is

equal to negative 40.7 KSI, and

I have a tau Xz, which is equal to 4.1 KSI.

And actually, that is negative as well,

because the torque is in the negative direction.

So when I plug this into my effective stress,

you can see immediately I don't have any stress in the y, or in the z, so

a lot of these components are going to fall out.

And then I only have a shear stress in the Xz plane, so these components fall out.

So what I end up with is a sigma prime

is equal to 1 divided by the square

root of 2 times sigma x squared plus

sigma x squared plus 6 tau Xz squared,

and this is all to the one half.

I have parentheses there, that's not really necessary but that's okay.

So this is going to simplify down to.

I can pull out, this is going to be a 2 here, right?

So I'm going to add these two together, so I can pull out a square root of 2, and

that will simplify this equation down to sigma x squared plus 3 tau

xz squared to the one-half is equal to my effective stress.