Procedure for Analysis

Professor A. Volinsky EML3500 Mechanics of Solids 1

Ch 10. Strain Transformation

Six independent stress components, which act on the faces of an element, and

six corresponding strain components. (Three normal and three shear stresses

and strains). For simplification simpler strain states are assumed in engineering.

Plane strain has two normal and one shear components, acting on an element

with a specific orientation. Plane Strain ≠ Plane Stress due to Poisson’s effect.

Plane Strain ≠ Plane Stress.


Sign ConventionPositive normal strains cause elongation along x and y axis.

Positive shear strains make AOB angle less than 90°.

Counterclockwise rotations are positive. Since stress is

proportional to strain in the elastic regime (Hooke’s law), the strains have the same sign convention as stresses.


Translation of Plane Strain• Plane strain - deformations of the material

take place in parallel planes and are the

same in each of those planes.

• Example: Consider a long bar subjected

to uniformly distributed transverse loads.

State of plane stress exists in any

transverse section not located too close to

the ends of the bar.

• Plane strain occurs in a plate subjected

along its edges to a uniformly distributed

load and restrained from expanding or

contracting laterally by smooth, rigid and

fixed supports

( )0

:strain of components

x === zyzxzxyy γγεγεε


Normal and Shear Strains

dx=dx’ cos θ

dy=dx’ sin θ δx’=εxdx cos θ + εydy sin θ + γxy sin θ cos θ


Normal and Shear Strains

δy’= - εxdx sin θ + εydy sin θ - γxy dy sin θ

γ x’y’ = -2(εx - εy)sin θ cos θ + γ xy(cos2 θ - sin2 θ)


Plane Strain Equations• State of strain at the point Q results in

different strain components with respect

to the xy and x’y’ reference frames.

( )

( ) ( )

( )yxOBxy

xyyxOB

xyyx

εεεγ

γεεεε

θθγθεθεθε

+−=

++=°=

++=

2

45

cossinsincos

21

22

θγ

θεεγ

θγ

θεεεε

ε

θγ

θεεεε

ε

2cos2

2sin22

2sin2

2cos22

2sin2

2cos22

xyyxyx

xyyxyxy

xyyxyxx

+−

−=

−−

−+

=

+−

++

=

′′

′

′

• Applying the trigonometric relations

used for the transformation of stress,


Principal Strains


Example 10.1


Continued


Important Points

Whole for the Stress, Half for the Strain!


Mohr’s Circle. Plane Stress• The equations for the transformation of

plane strain are of the same form as the

equations for the transformation of plane

stress - Mohr’s circle techniques apply.

• Abscissa for the center C and radius R ,

22

222

+

−=

+=

xyyxyxave R

γεεεεε

• Principal axes of strain and principal strains,

RR aveave

yx

xyp

−=+=

−=

εεεε

εε

γθ

minmax

2tan

( ) 22max 2 xyyxR γεεγ +−==

• Maximum in-plane shearing strain,


Procedure for Analysis


Continued


Example 10.4


Continued


3D Strain• Previously demonstrated that three principal

axes exist such that the perpendicular

element faces are free of shearing stresses.

• By Hooke’s Law, it follows that the

shearing strains are zero as well and that

the principal planes of stress are also the

principal planes of strain.

• Rotation about the principal axes may be

represented by Mohr’s circles.


Mohr’s Circle Strain• For the case of plane strain where the x and y

axes are in the plane of strain,

- the z axis is also a principal axis

- the corresponding principal normal strain

is represented by the point Z = 0 or the

origin.

• If the points A and B lie on opposite sides

of the origin, the maximum shearing strain

is the maximum in-plane shearing strain, D

and E.

• If the points A and B lie on the same side of

the origin, the maximum shearing strain is

out of the plane of strain and is represented

by the points D’ and E’.


3D Strain• Consider the case of plane stress,

0=== zbyax σσσσσ

• Corresponding normal strains,

( ) ( )babac

bab

baa

E

EE

EE

εεν

νσσ

νε

σσνε

σνσε

+−

−=+−=

+−=

−=

1

• If B is located between A and C on the

Mohr-circle diagram, the maximum

shearing strain is equal to the diameter CA.

• Strain perpendicular to the plane of stress

is not zero.


Important Points


Example 10.7


Continued


Strain Measurements: Rosettes

• Strain gages indicate normal strain through

changes in resistance.

( )yxOBxy εεεγ +−= 2

• With a 45o rosette, εx and εy are measured

directly. γxy is obtained indirectly with,

3332

32

3

2222

22

2

1112

12

1

cossinsincos

cossinsincos

cossinsincos

θθγθεθεε

θθγθεθεε

θθγθεθεε

xyyx

xyyx

xyyx

++=

++=

++=

• Normal and shearing strains may be

obtained from normal strains in any three

directions,


Generalized Hooke’s Law

σx causes εx’ = σx/E σy causes εx

’’ = -νσy/Eσz causes εx

’’’ = -νσz/E


Bulk and Shear ModulusG=E/2(1+ν), derivation on page 547

Dilatation – volumetric strain: e=δV/dV=εx + εy + εz

Normal strains cause volume change, shear strains only cause shape change.

Generalized Hooke’s Law: e=(1-2ν)/E*(σx + σy + σz)

Hydrostatic stress for uniform pressure p of a

liquid. Shear stresses not present, only principal

stresses equal to –p.

p/e=-E/3(1-2ν)

Bulk modulus k=E/(1-2ν)

For most metals ν=1/3, so k=E. From the bulk

modulus formula νmax=0.5


Important Points


Ch10 Review

Theories of failure:

review on your own.

Ch11, review on your

own.

Procedure for Analysis

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