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[SOUND] Hi, and welcome to Module 4

Â of Mechanics of Materials Part 1.

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

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

Â 0:31

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

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

Â a shear force and corresponding normal and shear stresses.

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

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

Â this nontransverse area, A.

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

Â 1:06

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

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

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

Â And this is just the two dimensional view.

Â I have my non-transverse cross section A here.

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

Â this side become cosine of theta.

Â This side become sine of theta.

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

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

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

Â 2:00

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

Â Revealing internal forces, we found the relationship

Â between the cross sectional area and the transverse area.

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

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

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

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

Â And so we'll put that in up here.

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

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

Â 2:49

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

Â So V is equal to the sheer stress times A or

Â the sheer stress times A transverse divided by cosine theta.

Â And we'll put that up there.

Â 3:04

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

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

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

Â shear stresses to my external force P.

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

Â 3:29

And so I have my end force and

Â then by geometry I know that this angle is theta,

Â so I have -P cosine theta = 0,

Â or P cosine theta = N.

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

Â 3:57

And so we find that, in terms of P,

Â sigma is equal to P over A t

Â times cosine squared theta.

Â And let's go ahead and circle that.

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

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

Â and the shear force V on your own.

Â 4:27

Okay, coming back.

Â Now I'm going to sum forces and

Â set the equilibrium parallel to the cross-sectional face.

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

Â I have P sine of theta, which is the parallel

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

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

Â or here is our relationship between tau and P.

Â 4:59

By a trig identity, I know that sine theta,

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

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

Â 5:44

It starts at a max, goes down like this.

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

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

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

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

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

Â the maximums occur at now 90 degrees and 270 degrees.

Â I'm looking just at the maximum value.

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

Â And so at 90 degrees is the max, so

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

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

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

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

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

Â if I go with theta more than 90 degrees,

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

Â The last thing I want to note, and

Â this is an important result, is that sigma max,

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

Â Okay, so the magnitude of sigma max,

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

Â 7:39

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

Â tau max would be P over 2 A t,

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

Â 8:13

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

Â Here's a simple model of an engineering structure.

Â We're going to look at bar BE.

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

Â steel bar of 50 millimeter diameter.

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

Â wheels in our analysis.

Â If I make that assumption what do we call member

Â BE then based on my previous courses?

Â Okay that would be a two force member.

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

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

Â find the sheer stress.

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

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