Module 7: Solve a thin-walled pressure vessel problem

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Del curso dictado por Georgia Institute of Technology

Mechanics of Materials II: Thin-Walled Pressure Vessels and Torsion

280 calificaciones

Georgia Institute of Technology

Mechanics of Materials II: Thin-Walled Pressure Vessels and Torsion

280 calificaciones

This course explores the analysis and design of thin-walled pressure vessels and engineering structures subjected to torsion. ------------------------------------------------ The copyright of all content and materials in this course are owned by either the Georgia Tech Research Corporation or Dr. Wayne Whiteman. By participating in the course or using the content or materials, whether in whole or in part, you agree that you may download and use any content and/or material in this course for your own personal, non-commercial use only in a manner consistent with a student of any academic course. Any other use of the content and materials, including use by other academic universities or entities, is prohibited without express written permission of the Georgia Tech Research Corporation. Interested parties may contact Dr. Wayne Whiteman directly for information regarding the procedure to obtain a non-exclusive license.

De la lección

Thin-Walled Pressure Vessels

In this section, we will learn how to analyze and design think-walled pressure vessels.

Module 7: Solve a thin-walled pressure vessel problem6:15

Module 8: Solve a thin-walled pressure vessel problem (continued)7:37

Conoce a los instructores

Dr. Wayne Whiteman, PE
Senior Academic Professional
Woodruff School of Mechanical Engineering

[MUSIC]

Welcome to Module 7 of Mechanics and Materials II.

Today's an exciting part of the course.

We start to actually solve thin-walled pressure vessel problems.

We're going to look at some real world thin wall pressure vessels, and let's

start with this real world example, which is a pressure vessel I showed before.

And it's got a diameter of roughly 2.5 inches.

The wall thickness I'm going to say is about one-sixteenth of an inch.

We´re going to assume that the maximum internal pressure

that´s expected is going to be 200 PSI, or pounds per square inch.

And we don´t want this thing to fail, and so we're going to go ahead and

find the factor of safety with respect to yielding.

And if you don't remember factor of safety, again, go back,

review the modules from my first part one course in mechanics of materials.

First of all, we need to make sure we can treat this example,

this real-world steel cylinder as a thin wall pressure vessel,

and so we're going to look at the ratio of D to t.

In this case D is 2.5 inches, the diameter,

the thickness is one-sixteenth of an inch, and so you find that ratio is equal to 40.

If you recall back from one of my earlier modules

as long as the ratio is greater than or equal to 20, we can go ahead and

treat the problem as a thin walled pressure vessel.

So we're okay.

Treat as thin walled

pressure vessel.

And so when we look at the factor of safety,

that's going to be in regard to the,

whoops that's going to be in regard to the highest stress.

What is the highest stress we're going to experience?

What you should say,

the highest stress we're going to experience is the hoop stress.

Let's go ahead and calculate the hoop stress in this vessel.

We found that the hoop stress was equal to PD over 2t.

The pressure in this case is 200 pounds per square inch.

The diameter is 2.5 inches.

And the thickness is, so it's two times the thickness,

which is one-sixteenth of an inch.

If you multiply that out, you find that the hoop stress,

that largest expected hoop stress will be 4,000 PSI, or pounds per square inch.

My question to you is this is steel.

Go ahead and do some research on your own and find out what's a good value for

the yield stress in steel in terms of PSI.

And what you'll find in most references is that the yield stress for

steel, Is about,

for most steels around 36,000 psi.

So you can see it's definitely stronger than

the stresses that we're going to experience.

How much of a factor of a safety do we have?

Well, factor of safety was defined

as the failure stress over the actual

stress being experienced.

In this case, the failure stress is also the same as the strength of the material.

Another way of expressing it.

Over the max computed actual stress is the max computed stress that we've found,

so this is just a couple of ways of expressing the factor of safety.

In this case the failure stress or the strength of the material is

36,000 pounds per square inch, PSI.

And the actual stress or

the max computed stress that we're going to experience is 4,000 PSI,

and so our factor of safety then is equal to 9.

And so my question to you then is, is that a good factor of safety for

a steel cylinder like this, pressure vessel?

And if you recall back to my first course in mechanics and materials,

you see that the typical values for factor of safety were for

buildings greater than or equal to two.

Automobiles greater than or equal to three, spacecraft or aircraft,

1.2 to 2.5 or larger, boilers and pressure vessels greater then 8.5.

If they were to fail, a pressure vessel, that's a pretty catastrophic failure,

so you want that factor of safety to be quite large, and so

since ours was 9, it's above 8.5 and we're in pretty good shape.

Again, other factors of safety lifting equipment or hooks, again,

if it fails it's a catastrophic failure, so you want it to be 8 or 9 or higher.

And bolts, 8.5 or higher.

In general you want to use, again, a higher factor of safety for

brittle materials to avoid catastrophic failure.

You also want to use,

you can use a lower factor of safety when the material properties are well known.

And as a review you also want a higher factor of safety for uncertain

environments where you don't know what kind of stresses you're experiencing.

And so in this case we're in pretty good shape.

Now that's a pretty good design, and so

we'll come back next time and do another real world engineering example.

[MUSIC]

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