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Mechanics of MaterialsMAE 243 (Section 002)

Spring 2008

Aaron Kessman, substitute for Dr. K.A. Sierros

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CHAPTER 7

ANALYSIS OF

STRESS

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Overview

Introductionhavent we been analyzing stress the whole time???

7.1-2 Plane stresshow uniaxial normal stress creates a shearcomponent

Problem solving example

7.3 Principal stresses and max shear stresswill the materialbreak under loading?

Problem solving example

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Introduction

Up till now, youve been dealing mostly with the big picture:uniaxial loading, torsion, and some combined loading in 2-Dand 3-D.

The end result has been to solve for the stress/moment at agiven location on some loaded objecteither explicitly or by ashear/moment diagram.

Now well take a microscopic look at the combined stressesand the effects of those loadings on the fabric of the materialthat is being loaded.

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Introduction stresses at a point

When a body is loaded by normal and shear stresses, we can

consider any point in that body as a stress element.

The stress element can be depicted by a little square (in 2-D

or more correctly a cube in 3-D) with the stresses acting

upon it. Well just ignore 3-D for the meantime

*https://ecourses.ou.edu

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Plane Stress components and conventions

And thats what we mean by plane stress: the 2-D

representation of combined stresses on the four faces

of a stress element

Two normal stress components, sx, sy

One shear stress component txy

Which btw, txy = tyx

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Elements in plane stress, note sign conventions:

(a) three-dimensional view of an element oriented to thexyzaxes,

(b) two-dimensional view of the same element, and

(c) two-dimensional view of an element oriented to thex1y1 axes -

rotated by some angle q from original

For now well deal with plane stress, the 2-D biaxial stress

projection of the 3-D cube

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Plane Stress How do we look at stresses in rotation?

If you were to rotate that little

square stress element some angle q,what would happen?

Well, stresses arent vectors, so

they cant be resolved the same(easy) way.

We have to account for:

Magnitude Direction

AND the orientation of the area upon

which the force component acts

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Stress Transformation - equations

The stress transformation is a way to describe the effect of combined loading on

a stress element at any orientation.

From geometry and equilibrium conditions (SF = 0 and SM = 0),

)2cos()2sin(2

)2sin()2cos(22

1

x

1

1

qtqss

t

qtqssss

s

xy

yx

yx

xy

yxyx

)90()2sin()2cos(

22

11 xy

qsqtqssss

s xyyxyx

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Stress Transformation - Ramifications

Given stresses at one angle we can calculate stresses at any arbitrary angle

Even a uniaxial loading (sx) will create both perpendicular (sy) and shear (txy)loadings upon rotation

Why this is important:If any of the transformed stresses at angle qexceed the materials yield stress,

the material will fail in this direction,

even if it was loaded by lower stresses.

Sometimes the way this works out is

failure by shear, which is not obvious.

Materials are often weaker in shear.

*https://ecourses.ou.edu

)2sin()2cos(221

x qtqssss

s xyyxyx

)2cos()2sin(211

qtqss

t xyyx

yx

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Stress Transformations Example 7.2-11

)2cos()2sin(2

)2sin()2cos(22

1

x

1

1

qtqss

t

qtqssss

s

xy

yx

yx

xy

yxyx

Approach:

1. Determine sx, sy, txy, q

2. Plug3. Chug

)2sin()2cos(221

y qtqssss

s xyyxyx

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Principal Stresses and Maximum Shear Stress

If material failure is what we ultimately care about, then we

really want to know what are the maximum and minimum normal stresses

maximum shear stress

orientation (q) at which these occur

These are called the principal stresses (s1, s2) andmaximum shear stress (txy).

The equations for these can be found from the stresstransformation equations by differentiation ( ) and

some algebraic manipulation.

This is really just a more general look at the material in theprevious section.

0

q

s

d

d

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s1, s2, txy, and q - equations

2221,2

2

2/)(

)2tan(

2

xy

yx

yx

xy

p

yx

avg

yx

tss

s

ss

tq

ss

s

ss

qp = planes of principal stresses

qp = qp1, qp2, 90 apart

no shear stress acts on the principal

planes

2

2/)(

)2tan(

21

max

222max

sst

tt

t

ssq

ss

xy

xy

yx

s

yx

IP

qs = planes of max shear stress

qs = qs1, qs2, 90 apart, 45 offset qp

tmaxIP = max in-plane shear stress

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Summary

Principal stresses represent the max and min normal stresses at the point.

At the orientation at which principal stresses act, there is no acting shear stress.

At the orientation at whichmaximum in-plane shearstress acts, the average

normal stress acts in bothnormal directions (x, y)

The element acted upon bythe maximum in-plane shear

stress is oriented 45 fromthe element acted upon bythe principal stresses

*https://ecourses.ou.edu

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Principal Stresses and Max. Shear Stress - Example 7.3-18

Approach:

Determine sx, txy

Find sy(sx, txy, t0)

Find numerical rangesy cannot be = 0 because at

some angles the combined

effect will raise txy above t0.

222max xy

yx

IPtt

ss