now we just need to figure out the other components in this equation.

Let's give me some more space to work with here.

We have V is equal to P, which is 500 newtons.

And then I, I is going to be the moment of inertia of this rod or

cylinder, which is going to be pi D to the 4th, over 64.

I'm going to draw the cross section of the rod OA and

here is our point C which is acting along this neutral axis.

And if you recall B is the width of the beam at the point of interest,

which will be right here.

B Is equal to D.

And now we have to figure out now is Q, which was, A prime, y prime.

Where A, is the area above, the point at the level at the point of interest.

A is going to be this cross sectional area,

right here, which is going to be one half of pi

d squared divided by 4 or pi d squared over 8.

And y prime is going to be the distance between,

if this is our centroid of that cross sectional area here,

it's going to be the distance between the centroid and

the neutral axis, and if you, you can go ahead and

either derive the centroid if you want, or you can look it up,

and what you'll find is the centroid of the half circle is 4d over 6 pi.

If we plug all of this into our equation,

what we'll see is tau is equal to V [SOUND],

times Q which is pi d squared over 8,

times 4d over 6 pi.

[INAUDIBLE]

[SOUND]

Which is over I which will if we take

the inverse of I will be 64 divided by pi d to the 4th.

And if we take the inversive of P, we get one over d.

[SOUND] And so when we break this

down, what we end up with is,

tau is equal to 4 times 64 times v,

divided by 8 times 6 times pi d squared.

And what we can break this out into is

actually combined pi D squared and this 4 and

that will give you the inverse of the area, so

we can end up with 64 divided by 48 times V over

A which boils down to 4 3rds V over A [SOUND].

And if you go into any machine design textbook,

you'll often see this shortcut equations for cross sectional areas,

or I'm sorry, for transverse shear at the maximum point.

At point C our transverse shear is going to be maximum,

because we're right at that neutral axis right here.

this is actually one of those shortcut equations where if you're at

the neutral axis, your transverse shear in a cylinder is 4 3ds V over A.

If we go ahead and we calculate four thirds V over

A what we will find is 4 times 500 Newtons divided

by 3 times pi times 0.04 divided by 2 squared

which is going to be right around a half of megapascal.

And the distribution for this [SOUND]

stress would look something like this.

If we're right here at this cross section, what we would see is, you would

see really low stresses at to zero at the exterior, and then you'd see really

high stresses at the neutral axis and it ends up looking something like that.

There's also a short cut for a rectangle.

If you're looking at the maximum stress or transverse shear at the neutral axis,

what you'll see, again, your stress is going to be max at the neutral axis.

You end up with this almost circular shape on the transverse

shear at the neutral axis the max is 3V over 2A which you

guys could derive out on your own time for fun, [SOUND].

Okay so that's it for today, and that concludes the module on transfer shear.

I'll see you next time.

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