7th chap MOMS

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MECHANICS OF

MATERIALSFerdinand P. Beer

E. Russell Johnston, Jr.

John T. DeWolf

CHAPTER

7


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MECHANICS OF MATERIALS Beer Johnston DeWolf

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Introduction

The most general state of stress at a point may

be represented by 6 components,

),,:(Note

stressesshearing,,

stressesnormal,,

xzzxzyyzyxxy

zxyzxy

zyx

Same state of stress is represented by adifferent set of components if axes are rotated.

The first part of the chapter is concerned with

how the components of stress are transformed

under a rotation of the coordinate axes. The

second part of the chapter is devoted to a

similar analysis of the transformation of the

components of strain.
http://localhost/var/www/apps/conversion/tmp/scratch_4/contents.ppt


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Introduction

Plane Stress- state of stress in which two faces of

the cubic element are free of stress. For theillustrated example, the state of stress is defined by

.0,, andxy zyzxzyx

State of plane stress occurs in a thin plate subjected

to forces acting in the mid-plane of the plate.

MECHANICS OF MATERIA S


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Transformation of Plane Stress

sinsincossin

coscossincos0

cossinsinsin

sincoscoscos0

AA

AAAF

AA

AAAF

xyy

xyxyxy

xyy

xyxxx

Consider the conditions for equilibrium of a

prismatic element with faces perpendicular to

thex,y, and xaxes.

MECHANICS OF MATERIALS


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Example 7.01

For the state of plane stress shown,

determine (a) the principal planes,

(b) the principal stresses, (c) the

maximum shearing stress and thecorresponding normal stress.

SOLUTION:

Find the element orientation for the principalstresses from

yx

xyp

2

2tan

Determine the principal stresses from

22

minmax,22 xy

yxyx

Calculate the maximum shearing stress with

2

2

max2

xyyx

2

yx



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Example 7.01

SOLUTION:

Find the element orientation for the principalstresses from

1.233,1.532

333.11050

40222tan

p

yx

xyp

6.116,6.26p

Determine the principal stresses from

22

22

minmax,

403020

22

xy

yxyx

MPa30

MPa70

min

max

MPa10

MPa40MPa50

x

xyx



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Example 7.01

MPa10

MPa40MPa50

x

xyx

2

1050

2

yx

ave

The corresponding normal stress is

MPa20

Calculate the maximum shearing stress with

22

22

max

4030

2

xy

yx

MPa50max

45 ps

6.71,4.18s



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Mohrs Circle for Plane Stress

Mohr's circlea useful graphical technique for finding principal stresses and

strains in materials. Mohr's circle also tells you the principal angles

(orientations) of the principal stresses without your having to plug an angleinto stress transformation equations.

Starting with a stress or strain element in the XY plane, construct a grid with a

normal stress on the horizontal axis and a shear stress on the vertical. (Positive

shear stress plots at the bottom.) Then just follow these steps:

Plot the vertical face coordinates V(xx, xy). Plot the horizontal coordinates H(yy, xy).

You use the opposite sign of the shear stress from Step 1 because the

shear stresses on the horizontal faces are creating a couple that

balances (or acts in the opposite direction of) the shear stresses on the

vertical faces. Draw a diameter line connecting Points V (from Step 1) and H (from Step

2).

Sketch the circle around the diameter from Step 3.

The circle should pass through Points V and H as shown here.



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Mohrs Circle for Plane Stress



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Example 7.02

For the state of plane stress shown,

(a) construct Mohrs circle, determine

(b) the principal planes, (c) the

principal stresses, (d) the maximumshearing stress and the corresponding

normal stress.

SOLUTION:

Construction of Mohrs circle

MPa504030

MPa40MPa302050

MPa202

1050

2

22

CXR

FXCF

yxave



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Example 7.02

Principal planes and stresses

5020max CAOCOAMPa70max

5020min BCOCOB

MPa30min

1.532

30

402tan

p

pCP

FX

6.26p



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Example 7.02

Maximum shear stress

45ps

6.71s

Rmax

MPa50max

ave

MPa20



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MECHANICS OF MATERIALS Beer Johnston DeWolfPractice Problems

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Practice Problems:

7.1-7.16 and 7.31-7.38

7th chap MOMS

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