1.0 Strain Analysis

8/21/2019 1.0 Strain Analysis

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CHAPTER 1

(week 1-3)Strain Analysis



• Some common engineering problems such as adam subjected to water loading, a tunnel underexternal pressure, a pipe under internalpressure, and a cylindrical roller bearingcompressed by force in a diametral plane, havesignificant strain only in a plane; that is, thestrain in one direction is much less than thestrain in the two other orthogonal directions. Ifsmall enough, the smallest strain can be ignoredand the part is said to experience plane strain.

• Assume that the negligible strain is oriented in the z -direction. To reduce the 3D strain matrix to the 2D plane stress matrix, remove allcomponents with z subscripts.

Plain Strain

http://www.efunda.com/formulae/solid_mechanics/mat_mechanics/strain.cfm










Plain Strain



GENERAL EQUATION OF PLANE STRAINTRANSFORMATION

Normal strain ex and e y arepositive if cause elongationalong x and y axis

Shear strain gxy is positive

if the interior angle AOBbecome smaller than 900. q0 will be positive using the

right-hand fingers.i.e counterclockwise



Normal and Shear Strains

• In Fig a :

• Positive ex occurline d

x elongated e

xd x , which causeline dx’ toelongated ex d x cos

q.

q

q

sin

cos

'

'

dxdy

dxdx



• Positive e y occur line d y elongated e y d y , which

cause line d y ’ to elongatede yd y sin q.

• Assuming dx fix inposition.

• Shear strain g xy changesin angle between dx anddy. dy displaced g xy dy tothe right.

• Dx’ elon ate d cos q

Normal and Shear Strains (cont.)



• Adding all the elongations

qqgqeqee

qqgqqeqqee

e

cossinsincos

dx

cos)sindx(sin)sindx(cos)cosdx(

dx

x

xy

2

y

2

xx

'

'

xy

'

y

'

x

x

'

'

x

'

'

'

qgqeqe cosdysindycosdxxyyxx '


q q

sincos

'

'

dxdydxdx






qg

qee

ee

e 2sin2

2cos22

xyyxyx

x '

qg

qee

ee

e 2sin2

2cos22

xyyxyx

y '

Using trigonometric identities:

qg

q

ee

g2cos

22sin

22

xyyxyx ''



• Principal Strains

• Direction axis of principlestrain:

• Max in Plane shear strain

• Ave shear strain

• Direction axis of shearstrain

2

xy

2

yx planeinmax,

222

g

eeg

2

yx

ave

eee

)(2tan

yx

xy

p ee

gq

g

eeq

xy

yx

s2tan

2

xy

2

yxyx

2,1222

g

ee

eee

























Construction of the Mohr’s Circle

PROCEDURE FOR ANALYSIS

The procedure for drawing Mohr's circle for strainfollows the one established for stress.Construction of the Circle.• Establish a coordinate system such that the

abscissa represents the normal strain e, withpositive to the right, and the ordinate represents

half the value of the shear strain, g/2, with positivedownward .

• Using the positive sign convention for ex, e y, gxy asshown in the Fig. , determine the center of the circleC, which is located on the e axis at a distancee avg= (e x + e y)/2 from the origin.

• Plot the reference point A having coordinatesA(e x, g xy/2). This point represents the case forwhich the x' axis coincides with the x axis. Henceq = q.

• Connect point A with the center C of the circle andfrom the shaded triangle determine the radius R ofthe circle.

• Once R has been determined, sketch the circle.



• The principal strains e1 and e2 are determinedfrom the circle as the coordinates of points Band D, that is where g/2 = 0, Fig. a.

• The orientation of the plane on which e1 acts canbe determined from the circle by calculating 2qP1,using trigonometry. Here this angle is measuredcounterclockwise from the radial reference lineCA to line CB, Fig. a. Remember that the rotationof qP1 must be in this same direction, from the

element's reference axis x to the x' axis, Fig. b.*

• When e1 and e2 are indicated as being positive asin Fig. a, the element in Fig. b will elongate in thex' and y' directions as shown by the dashedoutline.

Principal Strain



• The average normal strain and half themaximum in-plane shear strain are determinedfrom the circle as the coordinates of points Eand F, Fig. a.

• The orientation of the plane on which gmax in plane and eavg act can be determined from the circleby calculating 2qs1, using trigonometry. Herethis angle is measured clockwise from the radialreference line CA to line CE, Fig. a. Rememberthat the rotation of qs1, must be in this same

direction, from the element's reference axis xto the x' axis, Fig. c.

Maximum In Plane Shear Strain



• The normal and shear strain components ex’, and

g x’y’ for a plane specified at an angle q, Fig. d, canbe obtained from the circle using trigonometry todetermine the coordinates of point P, Fig.a.

• To locate P, the known angle q of the x' axis ismeasured on the circle as 2q. This measurement ismade from the radial reference line CA to the

radial line CP. Remember that measurements for2q on the circle must be in the same direction as q for the x' axis!

• If the value of e y, is required, it can bedetermined by calculating the e coordinate ofpoint Q in Fig. a. The line CQ lies 180o away from

CP and thus represents a rotation of 900

of the x'axis.

Strains on Arbitrary Plane































Strain Rosettes

• The axes of the threegauges are arranged at theangles of qa, qb, qc.

• If the reading of ea, eb, ec

taken, ex, e y, gxy can bedefined.

• Value of ex, e y, gxy aredetermined by solvingthese equations.

ccxyc

2

yc

2

xc

b bxy b

2

y b

2

x b

aaxya

2

ya

2

xa

cossinsincos

cossinsincos

cossinsincos

qqgqeqee

qqgqeqee

qqgqeqee



45o or Rectangular Rosette

0

0

0

90

45

0

c

b

a

q

q

q

The equation become:

ca bxy

cy

ax

2 eeeg

ee

ee

Example of 45o strain rosette



600 Strain Rosette

c bxy

ac by

ax

3

2

223

1

eeg

eeee

ee

0

c

0

b

0

a

120

60

0

q

q

q

The equation become:



Example









Stress Strain Relationship

• If a material subject to triaxial

stress (sx, s y, sz), associatednormal stress(ex, e y, ez)developed in the material.

• When sx is applied in x-direction, the element elongated with e

x

in x direction.

• Application on s y cause theelement to contract with astrain e” x in the x direction.

• Application Of sz cause the

element to contract with astrain e’’’ x in the x direction.

z x

y x

x x

s

e

s

e

s

e

'''

''

'



• The same result can bedeveloped for the normalstrain in the y and zdirection.

• Final results can bewritten as…..

yxzz

zxyy

zyxx

1

1

1

sss

e

sss

e

sss

e

Stress Strain Relationship (cont.)



Applying only shear stress, y to the element.

If to apply shear stress,t y to the element. txy will only cause deformation to gxy.

txy will not cause deformation to g yz.and gxz

t yz and t xz will only cause deformation to

g yz and gxz respectively. Hooke Law for shear stress and shear

strain written as:

xz xz

yz yz

xy xy

G

G

G

t g

t g

t g

1

1

1

Element subjected to normalstresses only

Shear stress applied to theelements




• Modulus of elasticity, E isrelated to shear modulus, G.

• Dilatation (the change in

volume per unit volume or‘volumetric strain’, e .

• Bulk Modulus (volumemodulus of elasticity), k .

12

E G

zyxE21 sssue


213

E

k





















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1.0 Strain Analysis