Module 25: Stresses on any given plane using Mohr’s Circle

Loading...

来自 Georgia Institute of Technology 的课程

Mechanics of Materials I: Fundamentals of Stress & Strain and Axial Loading

466 个评分

Georgia Institute of Technology

Mechanics of Materials I: Fundamentals of Stress & Strain and Axial Loading

466 个评分

This course explores the topic of solid objects subjected to stress and strain. The methods taught in the course are used to predict the response of engineering structures to various types of loading, and to analyze the vulnerability of these structures to various failure modes. Axial loading with be the focus in this course. ------------------------------ 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.

从本节课中

Stresses on Inclined Planes

In this section, we will develop the stress transformation equations for inclined planes and introduce Mohr’s Circle for Plane Stress

Module 22: Introduction to Mohr’s Circle4:42

Module 23: Mohr’s Circle for Plane Stress5:57

Module 24: Determine Principal Stresses, Principal Planes, and Maximum Shear Stress using Mohr’s Circle9:54

Module 25: Stresses on any given plane using Mohr’s Circle6:31

Module 26: Principal Stresses and Principal Planes by solving Eigenvalue Problem5:00

与讲师见面

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

[NOISE]

[MUSIC]

This is module 25 of mechanics of materials part one.

And the learning outcome today is to find

the stress on any given plane at a point for

a given set of plane stress conditions.

And so here was our development of our Mohr's circle that we did a couple

of modules ago.

And then last module we solved this example with actual numbers and

we found the principle stresses sigmus of one and sigmus of two.

And the maximum sheer stress and

we also drew it on a properly oriented stress block.

Today we're going to look at the normal stresses and

the sheer stresses on plane AA on our stress block and so

we'll start by finding out what this angle theta sub A is

on our stress block and so we have theta sub A, is equal to,

that's going to be the opposite over the adjacent, or

the inverse tangent of three over four which ends up being

36.9 degrees on the stress block.

But we know on Mohr's circle,

that's going to be two theta sub A and

if we round off to three significant figures,

that ends up being 73.7 degrees on Mohr's Circle and

so we're going to go now from our

horizontal face counterclockwise

73.7 degrees and

this is two theta sub A.

And this is going to be our point A and so

we now have two theta sub A, we have two theta sub P.

We also, let's look at this angle here which is phi.

I'll call it phi and so we have phi which is equal to

two theta sub A minus two theta sub P or

73.7 degrees minus,

if you recall back from last lesson two theta sub P was 22.6

degrees and so, phi ends up being 51.1 degrees and

so this angle phi is 51.1 degrees.

Now, I can drop a a vertical line here and so, I know what this angle is and

now I can find, at point A, my shear stress and my normal stress at point A.

So, Sine of phi is equal to the opposite side,

which is going to be the shear stress at point A divided by the radius and

we know the radius is 130 from last time, it's also written here and

so, we've got tau sub A equals 130 times the sine

of 51 degrees, which ends up being 101 MPa.

And so, that's one of our answers.

Similarly, we can go and find the normal stress at point A.

This is the adjacent side, we have the hypotenuse.

So, sigma sub A is equal to 130, the radius,

times cosine of 51.1 degrees and

this should have been 0.1 degrees up her but

this radius, remember now, we want the normal stress from here to here.

So, we're going to have to subtract out this 20 from our center and

we find that the normal stress at point A is going to end up being 61.6 or

sigma sub a equals and this is positive.

So, it's 61.6 megapascals in tension and that's our other answer.

So, on our Mohr circle, we can plot it up here.

This state of stress or the principle stress condition is 61.6 and

the normal stress and 101 for the sheer stress and

you can notice now that for this state of plane stress,

you could do this for any plane through that stress block.

So, you could find the normal sheer stress on any plane.

Let's go ahead and draw the stress on a properly-oriented stress block.

We're moving from our horizontal face here.

So, this is my horizontal face.

I'm going to turn my block, in this case I've turned it 36.9 degrees.

This is 36.9 degrees and

on that face our stresses are 61.6

normal stress mega pascals in tension.

And we have 101 and it's positive.

So, that means it's going to be in a clockwise direction on our stress block

and so our shear stress on that face is

101 megapascals and so that's the solution.

And you should now have a pretty good grasp on how to use Mohr's Circle

to find the principal stresses, the max shear stress and

actually the normal and shear stress condition for any plane for

plane stress at a point and I'll see you next time.

[MUSIC]

探索我们的目录

免费加入并获得个性化推荐、更新和优惠。

Coursera 致力于普及全世界最好的教育，它与全球一流大学和机构合作提供在线课程。

© 2018 Coursera Inc. 保留所有权利。

通过 App Store 下载

通过 Google Play 获取