0:00

[SOUND] Hi, and welcome to Module 4

of Mechanics of Materials Part 1.

Today's learning outcome is to determine the maximum normal and

shear stresses on inclined planes for the case of uniaxial loading.

0:31

And I have my tensile force on the left, P.

I've got an inclined plane cut with a normal and

a shear force and corresponding normal and shear stresses.

And so the first thing I want to do is, I want you to relate

the relationship, or find the relationship between the transverse area and

this nontransverse area, A.

So the transverse area we're going to call A t, the nontransverse area is A.

1:06

Okay now that you've done that, let's do it together.

I'm going to go ahead and use similar triangles.

And so I've got my A t, which is a transverse area, here.

And this is just the two dimensional view.

I have my non-transverse cross section A here.

I'll relate that to a similar triangle with a hypotenuse of 1, so

this side become cosine of theta.

This side become sine of theta.

And so now by similar triangles, I have A is related to A transpose

the same as 1 is related to cosine theta, and so

the relationship you should have come up with is A = A transpose over cosine theta.

2:00

Okay, here's our structural member with the axial force P and the incline plane.

Revealing internal forces, we found the relationship

between the cross sectional area and the transverse area.

Now we know by definition, from the previous module that sigma,

the normal stress, is equal to the normal force N over the cross sectional area.

Or N is equal to sigma times A, or sigma, in this case,

A is equal to A transverse area divided by cosine theta.

And so we'll put that in up here.

I want you now to do on your own the same thing for the shear stress and

the relationship between the sheer stress and the sheer force V.

2:49

Okay, so here is the relationship, sheer stress is equal to V over A.

So V is equal to the sheer stress times A or

the sheer stress times A transverse divided by cosine theta.

And we'll put that up there.

3:04

Okay, so here is my cross section, with P and

N and V, defined in terms of the normal stress and the shear stress.

Now I'm going to apply the equations of equilibrium to relate these normal and

shear stresses to my external force P.

And I'll begin by summing forces perpendicular to the cut.

3:29

And so I have my end force and

then by geometry I know that this angle is theta,

so I have -P cosine theta = 0,

or P cosine theta = N.

But we say that N is equal to sigma A t over cosine theta.

3:57

And so we find that, in terms of P,

sigma is equal to P over A t

times cosine squared theta.

And let's go ahead and circle that.

That's an important result that relates the external force P to the normal stress.

I want you to do the same thing to find the relationship between the shear stress

and the shear force V on your own.

4:27

Okay, coming back.

Now I'm going to sum forces and

set the equilibrium parallel to the cross-sectional face.

And so I have, since I defined up and to the left positive,

I have P sine of theta, which is the parallel

component of this P force, minus V because it's in the opposite direction.

So P sine theta is equal to V or tau a sub t over cosine theta,

or here is our relationship between tau and P.

4:59

By a trig identity, I know that sine theta,

cosine theta is the same as one half sine 2 theta, and so tau ends up being

P over 2 times the transverse area times the sine of 2 theta.

5:44

It starts at a max, goes down like this.

So this is 0 degrees, this is 90 degrees, this is 180 degrees.

The cosine is a max at 0 degrees and 180 degrees.

And so sigma max occurs at 0 degrees and 180 degrees on our cross section.

And I want you to do the same thing for tau max, when does tau max occur?

And if I look at a sine curve, so here's a sine curve,

the maximums occur at now 90 degrees and 270 degrees.

I'm looking just at the maximum value.

I don't care about the sine being either positive or negative.

And so at 90 degrees is the max, so

theta is going to be equal to one half of 90 degrees or

45 degrees, or one half of 270 degrees, or 135 degrees.

And I want you to also note that the sheer stress changes

sign when we go greater than zero is our theta is at 90 degrees.

And so here I show theta less than 90 degrees, but

if I go with theta more than 90 degrees,

the sheer force and the sheer stresses changes direction, okay?

The last thing I want to note, and

this is an important result, is that sigma max,

the magnitude of sigma max, what is the magnitude of sigma max equal to?

Okay, so the magnitude of sigma max,

that means that cosine squared theta is going to be equal to 1 is P over A sub T.

7:39

And here again, sine 2 theta being a maximum value of 1,

tau max would be P over 2 A t,

and so sigma max is equal to 2 times the absolute value of tau max.

8:13

Okay given those concepts I now want you to do a worksheet on your own.

Here's a simple model of an engineering structure.

We're going to look at bar BE.

It's going to be a round bar in this case, a round,

steel bar of 50 millimeter diameter.

We're going to go ahead neglect the weight of the individual members and

wheels in our analysis.

If I make that assumption what do we call member

BE then based on my previous courses?

Okay that would be a two force member.

And so for that we're going to do a transverse cut of that bar,

find the normal stress, and do a nontransverse cut at 30 degrees, and

find the sheer stress.

And I've put the solution in the module material and I'll see you next time.

[MUSIC]