0:00

[NOISE] Welcome

Â back to Mechanics of Materials Part II.

Â Well we're nearing the end of the course, we have one more topic we're

Â going to cover and that's statically indeterminate torsional members.

Â And so the learning outcome today is to solve a statically indeterminate torsion

Â problem.

Â If you go back to my Mechanics and Materials Part I course,

Â we saw the statically indeterminate problem for axial loading, and

Â you should probably go back and review that, because this is going to be the same

Â procedure in general an analogous procedure for torsion.

Â I'll also again mention since this is one of the final problems it's rather long and

Â the module may be a little longer than usual, but stick with it and

Â make sure you understand each step.

Â 0:48

So again let's relate it to a real world problem.

Â We've got a simple model of a torsion bar here.

Â For a tracked vehicle, I've got the model shown below.

Â This time I'm going to model both ends as being fixed, and

Â I don't have a hollow section here.

Â And, I'm applying an eight foot kip torque to the center.

Â One thing I'm going to do is, I'm going to change the G that I give you for steel,

Â instead of 11.6 I'm going to make it 11.0.

Â But you could use 11.6 and redo the problem it doesn't matter.

Â That's the numbers that I'm going to go ahead and work with though.

Â Okay, so, we want to determine the maximum shear stress in each section.

Â And so, how might you start?

Â 2:11

And so now that I have my free-body diagram, what do I do?

Â Well, I'm going to have to come up with the static equilibrium equations.

Â I'm really not concerned forces in the x direction or the y direction,

Â I'm only concerned about the torques or the moments, and so

Â I'm going to have one equilibrium equation.

Â How many unknowns do I have?

Â Okay, I have two unknowns.

Â I have my monel torque and the torque in the steel.

Â So I have these two unknowns.

Â Which means I have an unhappy face and

Â I'm going to have to generate another relationship.

Â So let's first start though,

Â by writing that equilibrium equation to start the solve the problem.

Â The one for the torques or the moment.

Â And if you do that on your own, come on back, see how you did.

Â 2:56

And so we sum moments about the z direction if I

Â say that the z direction is in this direction, and

Â when I do that I get the monel torque and

Â the steel torque have to balance the eight foot kip applied torque,

Â or that converts to 96 inch kips.

Â I'll work with kips and inches.

Â And so, with that,

Â that'll become equation one, that I'll need to soften the two unknowns.

Â 3:37

Well, if you recall back to the way we approach these indeterminate problems in

Â my first course, in Mechanics and Materials, we can

Â use a deformation equation, or what's also called a compatibility equation.

Â Which, it makes sure that we look at the deformation of the geometry, or

Â the geometry of the deformation of my members and

Â make sure that there's a compatibility between equilibrium and the deformation.

Â And so, here's my structure.

Â 4:13

And so here I'm going to have a fixed end on the right-hand side.

Â I'm going to have an angle of twist of C with respect to A.

Â And then I'm going to have an angle of twist, of D with respect to C,

Â that angle of twist is going to be in this direction because of my applied torque.

Â Then I'm going to have to have an angular twist of C with respect to D,

Â excuse me, D with respect to C in the other direction,

Â to bring it back to a total of zero displacement.

Â because I got zero displacement on either end.

Â So I've got the angle of twist of C

Â with respect to A minus the angle of twist of D with respect to A has to equal zero.

Â So here's my equilibrium equation, equation one,

Â here's my deformation equation but they're in terms of different unknowns so

Â I need to have them in terms of the same unknowns.

Â To do that,

Â let's start by saying we're going to use the elastoplastic assumption and

Â assume that the steel and the monel shafts remain in the linear elastic region.

Â And we'll assume that the steel has a torsional yield strength of 18 ksi and

Â the monel has a torsional yield strength of 25 ksi.

Â Here's a plot of an idealized elasto plastic material.

Â And we're saying okay let's assume we're continuing to operate for

Â all these situations in the linear elastic range.

Â Can we find a relationship between the torque and the angle of twist?

Â And the answer is yes.

Â We've done that before.

Â The angle of twist if we assume linear elastic is equal to the torque,

Â the length of this section divided by the module of this rigidity

Â times the polar moment of inertia of this section.

Â And, so let's go ahead and do that for my deformation equation and

Â I've got the angle of twist, the C with respect to A, which is my steel section.

Â So, that's T of Steel, times its length, which is 11 feet,

Â and we're going to convert everything to inches,

Â that's 12 inches per foot, and we're going to divide then by G.

Â I said at the beginning I'm going to change this to,

Â instead of 11.6, I'm going to say 11.0 times 10 to the third or

Â 11,000 Ksi times J, solid circular cylinder.

Â So J is pi over 2 times the radius,

Â which is 3 to the fourth, 3 inches to the fourth.

Â 7:16

Times its length which is four feet convert to inches,

Â 12 inches per foot, divided by G for

Â monel is given is 9.5 times 10 to the third or

Â 9,500 Ksi and J is pi over two times its radius,

Â which is half of 4, or 2 inches, to the fourth.

Â 7:43

If I boil that down, run all those numbers I end up with

Â the equation 2.13 torque

Â monel is equal to torque steel.

Â And so I have a now, an equation, a second equation,

Â I'll call this EQN 2 for the relationship between Tmonel land Tsteel.

Â So we're in good shape now.

Â Two equations, two unknowns, happy face.

Â 8:18

So there's my equilibrium equation.

Â There's the deformation equation expressed in terms of T and

Â in torques and assuming that we're working in the linear elastic region,

Â If you solve those simultaneously you get Tmonel is 30.7 in-k.

Â T steel is 65.3 inch kips.

Â So, now we need to calculate the stresses.

Â How would we do that?

Â 8:58

And so I've got sheer stress monel is equal to the torque monel,

Â which is 30.7 inch kips, times Ro.

Â And we've got Ro for the monel section is

Â going to be 2 inches, divided by,

Â J thatÂ´s going to be pi over 2 times 2 inches to the fourth.

Â And so, the sheer stress in the monel, ends up equaling

Â 2.443 Ksi.

Â 9:49

We want the maximum so that's why we're going to the outer radius in both cases.

Â That's going to be equal to the torque.

Â Which is 65.3 inch kips times

Â Rho in this case for the steel section

Â is 3 inches divided by J which is pi

Â over 2 times 3 inches to the fourth or

Â the max sheer stress in the steel

Â section is 1.539 Ksi.

Â Okay, so there's the calculation of the stresses that we came up with but

Â we need to make sure that we stayed in the elastic region because

Â that's the only way we could use this second equation.

Â And so we can see however that, if we check the stresses for

Â the linearly elastic assumption, that 2.44,

Â 10:52

Ksi, that's rounded to 3 significant figures,

Â is much less than the torsional yield strength for monel which is 25 Ksi and

Â 1.54 Ksi rounded to three significant figures again,

Â is less than, much less than, 18 Ksi.

Â So both sections do indeed remain in the linear elastic range,

Â and so I have my solution.

Â 11:23

And we've completed the worksheet, and so now you know how to do

Â an indeterminate structure for a torsional problem.

Â And so you're in great shape.

Â We've gone through all the sections of the course, and you should have a really

Â good handle on both thin walled pressure vessels and torsional structures.

Â And we'll wrap up next lesson, or next module.

Â [MUSIC]

Â