Lect03_Strain Transformation.ppt

8/16/2019 Lect03_Strain Transformation.ppt

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Dr. Wan Mohd Sabki Wan Omar

Lecture 3:

Strain Transformation

EAT 115/4EAT 115/4

Strength of MaterialsStrength of Materials

1


2/22

Dr. Wan Mohd SabkiDr. Wan Mohd Sabki

Lecture 1 !o"bined LoadingsLecture 1 !o"bined Loadings

EAT115/Strength of MaterialsEAT115/Strength of Materials

Plane Strain

General state of strain at a point in a body isrepresented by 3 normal strains and 3

shear strains .

The normal and shear strain components will vary

according to the orientation of the element.

( ) z y x ε ε ε ( ) z y x γ γ γ

2


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General Equations of Plane-Strain Transformation

Strain#transfor"ation e$uationsStrain#transfor"ation e$uations

θ γ

θ ε ε γ

θ γ

θ ε ε ε ε

ε

θ γ

θ ε ε ε ε

ε

2cos2

2sin22

2sin

2

2cos

22

2sin2

2cos22

''

'

'

xy y x y x

xy y x y x

y

xy y x y x

x

+

−−=

−−

−+

=

+−

++

=

3


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General Equations of Plane-Strain Transformation

%rinci&al Strains%rinci&al Strains

Ma'i"u" (n#%lane Shear StrainMa'i"u" (n#%lane Shear Straina!imum in-plane shear strain and associated

average normal strain are as follow"

22

2,1222

2tan

+

+±

+=

−= xy y x y x

y x

xy

p

γ ε ε ε ε ε

ε ε

γ θ

2 ,

222

2tan

22

plane-inmax y x

avg

xy y x

xy

y x

S

ε ε ε

γ ε ε γ

γ

ε ε θ

+=

+

−=

−−=

4


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E!ample #$.3 % differential element of material at a point is sub&ected to a state of plane strain

defined by which tends to distort theelement. 'etermine the ma!imum in-plane shear strain at the point and the

associated orientation of the element.

Solution"

(oo)ing at the orientation of the element*

+or ma!imum in-plane shear strain*

( ) ( ) ( )666 1080,10200,10350 −−− ==−= xy y x γ ε ε

°°=⇒

−−−=

−−=

311and9.40

80

2003502tan

s

xy

y x

s

θ

γ

ε ε θ

( ) (Ans) 10556

222

6

planeinmax

22

planeinmax

−=⇒

+

−=

γ

γ ε ε γ xy y x

5


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ohr,s irclePlane Strain

/e can also solve problems involving thetransformation of strain using ohr,s circle.

0t has a center on the 1 a!is at point 21avg* $ and a

radius R .

22

22 and

2

we!e

+

+=

+= xy y x y xavg R

γ ε ε ε ε ε

( ) 22

''2

'2

R y x

avg x =

+−

γ ε ε

6


7/22Dr. Wan Mohd SabkiDr. Wan Mohd Sabki



E!ample #$.4The state of plane strain at a point is represented by the components

'etermine the ma!imum in-plane shear strains and the orientation of an element.

Solution"

+rom the coordinates of point E * we have

To orient the element*

we can determine the cloc)wise angle*

( ) ( ) ( )666 10120,10150,10250 −−− =−== xy y x γ ε ε

( )( )

( ) ( )

( )6

6

planeinmax''

6 planeinmax''

1050

10418

108.2082

−

−

−

=

=

=

avg

y x

y x

ε

γ

γ

( )

(Ans) ".36

35.82902

1

1

°=

°−°=

s

s

θ

θ

7





%bsolute a!imum Shear Strain

Absolute maximum shear strain is found from thecircle having the largest radius.

0t occurs on the element oriented 546 about the a!is

from the element shown in its original position.

2

minmax

minmaxmaxa#s

ε ε ε

ε ε γ

+=

−=

avg

8





%bsolute a!imum Shear Strain

%lane Strain%lane Strain+or plane strain we have*

This value represents the absolute maximum shear

strain for the material.

( ) maxmax''maxa#s ε γ γ == z x ( ) minmaxmax''maxa#s ε ε γ γ −== y x

9





E!ample #$.7The state of plane strain at a point is represented by the strain components*

'etermine the ma!imum in-plane shear strain and the absolute ma!imum shear

strain.

Solution"

+rom the strain components* the centre of the circle is on the

1 a!is at

Since * the reference point has coordinates

( ) ( ) ( )666 10150,10200,10400 −−− === xy y x γ ε ε

( ) ( )66 10100102

200400 −− −−=+−

=avg ε

( )610"52

−= xyγ

( ) ( )( )66 10"5,10400 −−− A

Thus the radius of the circle is

( ) ( ) ( )9622 1030910"5100400 −− =

+−= R

10





omputing the in-plane principal strains* we have

Solution"

+rom the circle* the ma!imum in-plane shear strain is

( ) ( ) ( )( )( ) ( )66min

66

max

1040910309100

1020910309100

−−

−−

−=−−=

=+−=

ε

ε

( )[ ]( ) ( ) (Ans) 1061810409209 66minmax planeinmax−− =−−=−= ε ε γ

( ) ( ) 10409,0, 10209 6minint6

max

−− −=== ε ε ε +rom the above results* we have

Thus the ohr,s circle is as follows*

11





Strain 8osettes

9ormal strain in a tension-test specimen can bemeasured using an electrical-resistance strain

gauge.

The strain-transformation equation for each gauge is

as follow"

cc xyc yc xc

bb xyb yb xb

aa xya ya xa

θ θ γ θ ε θ ε ε



cossinsincos

cossinsincos

cossinsincos

22

22

22

++=

++=

++=

12





E!ample #$.:The state of strain at point A on the brac)et is measure using the strain rosette as

shown. 'ue to the loadings* the readings from the gauge give 1a ;





aterial-Property 8elationships

)enerali*ed +ooke,s La-)enerali*ed +ooke,s La-+or a tria!ial state of stress* the general form for

@oo)e,s law is as follow"

They are valid only for a linear–elastic materials.

@oo)e,s law for shear stress and shear strain is

written as

( )[ ] ( )[ ] ( )[ ] y x z z z x y y z y x x v E v E v E σ σ σ ε σ σ σ ε σ σ σ ε +−=+−=+−=

1

,

1

,

1

xz xz yz yz xy xyGGGτ γ τ γ τ γ

1

1

1===

14





aterial-Property 8elationships

elationshi& (noling E0elationshi& (noling E0 v v 0 and )0 and )

Dilatation and ulk ModulusDilatation and ulk Modulus

Dilatation* or volumetric strain* is caused only by

normal strain* not shear strain.

Bulk modulus is a measure of the stiffness of a

volume of material.

Plastic yielding occurs at v ; $.4.

( )v

E G

+=

12

( )v E k 213 −

=

15





E!ample #$.#$The copper bar is sub&ected to a uniform loading along its edges. 0f it has a ; 3$$

mm* b ; 4$$ mm* and t ; =$ mm before load is applied* find its new length* width*and thic)ness after application of the load. Ta)e

Solution"+rom the loading we have

The associated normal strains are determined from the generaliAed @oo)e,s law*

34.0 , $%a120 == cucu v E

0,80,&%a500,&%a800 ==−== z x y x σ τ σ σ

( ) ( ) ( ) 000850.0 , 00643.0 , 00808.0 −=+−=−=+−==+−= y x z z z x y

y z y x

x E

v

E E

v

E E

v

E σ σ

σ ε σ σ

σ ε σ σ

σ ε

The new bar length* width* and thic)ness are therefore

( )

( )( )

( )( ) (Ans) mm98.1920000850.020'

(Ans) mm68.495000643.050'

(Ans) mm4.30230000808.0300'

=−+==−+==+=

t

b

a

16





Theories of +ailure

+ailure for ductile material is by yielding * while brittle material is by fracture.

Ductile MaterialsDuctile Materials

+or yielding of a ductile material * it occurs along the

contact planes.

Maximum-shear-stress theory or Tresca yield

criterion is used to predict the failure stress of aductile material sub&ected to any type of loading.

17





Theories of +ailure

Ductile MaterialsDuctile Materials/ith reference from plane stress* the ma!imum-

shear-stress theory for plane stress can be

e!pressed for any two in-plane principal stresses.

} sinsoppositeae,

sinssameae,

2121

21

2

1

σ σ σ σ σ

σ σ σ σ

σ σ

Y

Y

Y

=−

=

=

18





Theories of +ailure

Ductile MaterialsDuctile MaterialsThe energy per unit volume of material is called the

strain-energy density .

Bielding in a ductile material occurs when the

distortion energy per unit volume of the material

equals or e!ceeds the distortion energy per unit

volume.

0t is called the maximum-distortion-energy theory .

22

221

2

1 Y σ σ σ σ σ =+−

19





Theories of +ailure

rittle Materialsrittle MaterialsMaximum-normal stress theory states that a brittle

material will fail when ma!imum principal stress is

equal to the ultimate normal stress.

lt2

lt1

σ σ

σ σ

=

=

20





E!ample #$.#5The solid shaft has a radius of $.4 cm and is made of steel having a yield stress of σ

= 3


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Solution"

Since the principal stresses have opposite signs* the absolute ma!imum shear

stress will occur in the plane*

( )

3602.382

3606.2866.95

21

>

≤−−

≤− Y σ σ σ

( )

( )( ) ( )[ ]1296009.1186""

3606.2866.2866.956.95

222

22

221

2

1

≤

≤−−−−

≤+− Y σ σ σ σ σ

?sing ma!imum-distortion-energy theory*

Thus* shear failure of the material will occur according to this theory.

?sing this theory* failure will not occur.

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Lect03_Strain Transformation.ppt

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