7 stress transformations lecture thursday

7/18/2019 7 Stress Transformations Lecture Thursday

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7 - 1


2/19

Mohrs Circle for Plane Stress

7 - 2

With the physical significance of Mohrs

circle for plane stress established, it may be

applied with simple geometric considerations.

Critical values are estimated graphically or

calculated. For a known state of plane stress

plot the pointsXand Yand construct the

circle centered at C.

The principal stresses are obtained atAandB.

The direction of rotation of Oxto Oais

the same as CXto CA.

Shear angle always (angle +45O)

xyyx ,,

xy

yxyxave R

+

=

+=

yx

xyp

ave R

=

=

tan

minma!,


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7 - 3


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7 - 4


5/19

7 - 5


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7 - 6


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7 - 7


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7 - 8


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7 - 9

With Mohrs circle uni"uely defined, the state

of stress at other a!es orientations may bedepicted.

For the state of stress at an angle with

respect to thexya!es, construct a new

diameterXYat an angle with respect toXY.

#ormal and shear stresses are obtained

from the coordinatesXY.

Observe direction of rotationObserve double of the element angle

Shear angle always (angle +45O


10/19


7 - 10

Mohrs circle for centric a!ial loading$

Mohrs circle for torsional loading$

%, === xyyxA

P

A

Pxyyx

===

J

!xyyx === % %=== xyyx

J

!


11/19

Example 1

7 - 11

For the state of plane stress shown,

&a' construct Mohrs circle, determine

&b' the principal planes, &c' the

principal stresses, &d' the ma!imumshearing stress and the corresponding

normal stress.

()*+T)#$

Construction of Mohrs circle( ) ( )

( ) ( ) M-a%/%0%

M-a/%M-a0%%%

M-a%

1%%

=+==

===

=+

=+

=

CXR

"XC"

yxave


12/19

Example 1

7 - 12

-rincipal planes and stresses

%%ma! +=+== CAOCOAM-a2%ma! =

%%ma! === BCOCOB

M-a0%ma! =

=

==

1.0

0%

/%tan

p

pCP

"X

= 3.3p


13/19

Example 1

7 - 13

Ma!imum shear stress (hear angle always &angle 4/)

+= /ps

= 3.21s

R=ma!

M-a%ma! =

ave =

M-a%=


14/19

Sample Problem 7.2

7 - 14

For the state of stress shown,

determine &a' the principal planes

and the principal stresses, &b' the

stress components e!erted on the

element obtained by rotating thegiven element counterclockwise

through 0% degrees.

()*+T)#$

Construct Mohrs circle

( ) ( ) ( ) ( ) M-a/5%

M-a5%

3%1%%

=+=+=

=+

=+

=

"XC"R

yxave


15/19

Sample Problem 7.2

7 - 15

-rincipal planes and stresses

=

===

/.32

/.%

/5tan

p

pC"

X"

clockwise2.00 =p

5%

ma!

+=+== CAOCOA

5%

ma!

=== BCOCOA

M-a10ma! += M-a5min +=


16/19

Sample Problem 7.2

7 - 16

(tress components after rotation by 0%o

-ointsXand Yon Mohrs circle that

correspond to stress components on therotated element are obtained by

rotatingXYcounterclockwise through

==

+=+=====

==

3.sin

3.cos5%

3.cos5%

3./.323%15%

X#

C$OCO$

#COCO#

yx

y

x

= 3%

M-a0./1

M-a3.111

M-a/./5

=

+=+=

yx

y

x


17/19

Yield Criteria for Ductile Materials nder Plane Stress

7 - 17

Failure of a machine component

sub6ected to unia!ial stress is directly

predicted from an e"uivalent tensile test

Failure of a machine component

sub6ected to plane stress cannot be

directly predicted from the unia!ial state

of stress in a tensile test specimen

t is convenient to determine the

principal stresses and to base the failure

criteria on the corresponding bia!ial

stress state

Failure criteria are based on the

mechanism of failure. 7llows

comparison of the failure conditions for

a unia!ial stress test and bia!ial

component loading


18/19

Yield Criteria for Ductile Materials nder Plane Stress

7 - 18

Ma!imum shearing stress criteria$

(tructural component is safe as long as thema!imum shearing stress is less than the

ma!imum shearing stress in a tensile test

specimen at yield, i.e.,

For aand %with the same sign,

For aand %with opposite signs,

ma!Y

Y

=

7 stress transformations lecture thursday

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