Plane Strain

Each strain is acting independent of one another

Due to normal strain x Due to normal strain y

x =

Due to shear strain

1

dxdu1

dv1

dv2

x

y

du2

dv3

x

du3

dy

x

Negligible

y

General Equations of Plane Strain Transformation

Sign Conventions

(1) Normal strains are + ve, if they cause elongations along x and y axes, respectivelty.

(2) Normal strains are - ve, if they cause shortening along x and y axes, respectively

(3) Shear strains are + ve, if the interior angle AOB becomes less than 900.

(4) Shear strains are – ve, if the interior angle AOB becomes greater than 900.

2

x

y

dx

dy

du = x dx

dv = y dy

O A

B

x

yx

y

dy

dy

dx

dy dx

O

From the figure

Problem

Using the above orientations of axes, determine the strains along xoy axes due to

, defined w.r.t. xoy axes.

Effect of normal strain x, along x axis

Effect of normal strain y, along x axis

3

x

yx

y

dy

dy

dx

dydx

dv1du1

du

du=xdx

x

yx

y

dy

dx

dy dx

du1 =dv sin = x dysin

dv1 = y dycos

dv = ydy

Effect of shear strain xy, along x - axis

[Assume that dx remains fixed in position, and the shear strain xy is represented by the

change in angle of dy]

+du1

=

(I)

+dv1

=

= (the angle of shear distortion along x axis)

4

x

x

y

dy

dx

dxxy

dv1=xydysindu1=xydycos

xy dy

By rotating the angle through 900, in the clockwise direction, the rotation of elemental

length dy can be obtained.

(-) = rotation of the right angle xoy

(II)

From Equation I,

(III)

From Equation II,

(IV)

can be obtained by introducing (90+) for in

5

=

=

=

To find the principle strain

i.e.,

6

2p

=

=

7

= p

= p

Similarity Between Stress and Strain Transformation EquationsStresses at a point Strains at a point

8

Principle Stresses Principle Strains

for p1,

Maximum in-plane shear stress

9

Principal stress plane and maximum

shear stress planes are inclined at 450.

Consequently,

twice the values of these

angles will be inclined at 900.

As a result

Maximum in-plane Shear Strain

Principal strain plane and maximum (in-

plane) shear stress planes are inclined at

450 to one another.

Consequently, twice the values of these

angles will be inclined at 900.

As a result

10

Mohr’s Circle

For Plane Stress For Plane Strain

2

1

Material Property Relationships

When only shear stresses are acting

i.e.,

i.e., (A)

When a body is subjected to normal stresses the body under goes only

change in volume.

Volume change =

(volume change)/unit volume =

=

p/e = bulk modulus = K =

=

Theories of FailureIn the design of structural members, it becomes important to place an upper limit on the

state of stress that defines the material's failure.

Ductile Materials Brittle Materials

Stress Stress

Y - Yield stress (steel)

ult – Ultimate stress ult – Ultimate stress

- Not used since strain is very

high at this stress level

0.1% Proof stress

(stress at o.1% elongation)

(Aluminum)

Strain Strain

y – Yield strain (0.15 to 0.2 ult – Ultimate strain (0.2% to

for mild steel) 0.3% for CAST IRON)

ult – Ultimate strain (20 to 25%

for mild steel)

The material behaviour – either ductile or brittle – does not remain a constant one for

any material. It is dependent on:

Temperature

Rate of loading

Chemical environment

Forming/shaping methods

In order to apply the theories of failure:

(i) The state of stress in a structure, at a point where the maximum stresses

are expected - - are determined first.

(ii) Thereafter, the principal stresses and maximum shear stresses are

determined -

Failure Theories

For ductile materials For brittle materials

1. Maximum shear stress theory 1. Maximum normal stress theory

Proposed by Tresca

2. Maximum distortion energy 2. Mohr’s failure criterion

theory – Proposed first by Huber - Proposed by Otto Mohr

and refined later by Von Mises and

Hencky

- Huber–Mises–Hencky theory

1. Maximum Shear Stress Theory

“Failure (by yielding) will occur in a material (at a point) when the maximum shear

stress in the material is equal to the maximum shear stress that will occur when the

material is subjected to an axial tensile test’.

P = (Y) A

For a two-dimensional stress system,

(I)

Under simple tension test,

(II)

Using in Eqn. I

450

Thin mild steel strip Luder’s lines

Governing criteria (III)

Considering a three-dimensional element (with two-dimensional stress state)

Arranging the stresses in the order of decreasing magnitudes,

(i) Case (a):

Hence

Failure will occur first in the plane.

i.e., (IV)

Failure in shear will occur, when the maximum principal stress is equal to Y.

(ii) Case (b):

1

3

2

1

3 = 0

2

V (a)

Failure will occur in the plane containing stresses

Generalizing this for a plane-stress failure wherein act along x-y axes and 3 acts

along z-axis, (zero stress), one can rewrite Equation (V (a)) as

V (b)

Failure envelope or Yield loci

2. Maximum Distortion Energy Theory

Y

Y

Y

Y

2

1

1, -2

+ve -ve

-1, +2

-ve +ve

Failure occurs in (1, 2) plane

Failure occurs in (1, 2) plane

(+ve)

(-ve)

(-ve)

(+ve)

2 = Y, 3 =0-Failure in this plane (1, 3)

1 = Y (3 =0)-Failure in this plane (1 , 3)

-Failure occurs in plane (1, 3)

“ Failure (by yielding) will occur when the shear or distortion energy in the material

(at a point) reaches the equivalent value that will occur when a material is subjected to

uniaxial tensile test”.

Let us say that the principal stresses in an element, at a point, is given by

Total strain energy stored in the given system = Total volumetric strain energy + Total

distortion strain energy

ut = uv + ud (VI)

=

+

1

3

2

1

2

3

1

3

2

1

3

11

21

31

2

(VII)

Also

= 0 (VIII)

Using the earlier stress-strain relationships

Considering the volumetric strains due to

= 0 (IX) [Since according to Eqn. (VIII)]

Equation (IX) states that no volumetric change occurs in the material due to the stresses

(but it does produce a change of shape). Due to the three stresses

,

(Total strain energy)

Hence strain energy per unit volume

=

Considering only [the mean stresses and strains due to ( ) and (

)],

(X)

Considering the normal (or principal) stresses and strains,

(XI)

Since ,

(XII)

When the specimen is under uniaxial tension,

From eqn. (XII),

(XIII)

For a general state of stress,

= (XIV)

From Eqns. (XIII) and (XIV), equating the distortional energies due to an uniaxial state

of stress and that due to a multiaxial state of stresses,

(XV)

For a two-dimensional state of stresses,

Hence equation reduce to

i.e.,

(XVI)

This is an equation to an inclined ellipse.

Y

2

Plot of Eqn. (XVI) gives the failure envelope or yield loci for a system subjected to a

two-dimensional state of stress.

Brittle Materials

Applicable to cast iron that tends to fail suddenly by fracture, without any warning.

1. Maximum normal stress Theory:

In a tension (or compression) test, brittle fracture occurs when the normal stress reaches

the ultimate stress ult.

Y Y

Y

1

Maximum distortion-energy theory Maximum

shear stress theory

450

450

450

In a torsion test, brittle fracture occurs due to a maximum tensile stress (in a plane 450 to

the shear direction) when it reaches the ultimate stress ult.

Failure criteria or failure loci:

Statement

When the maximum principal stress in the material reaches a limiting value

that is equal to the ultimate normal stress the material can sustain, failure by fracture

occurs.

CompressionTensionTorsional shear

ult

ult

1

2

-Eg. Chalk: under tension, under bending and under torsion.

2. Mohr’s Failure Criterion

For materials (brittle) that have different fracture properties in tension and compression,

this criterion holds good.

-Specially for metals

-For nonmetals like concrete (Rock, concrete, soils) another theory

is applicable (we will briefly deal with this later)

Three tests done to determine failure criteria –

- Tension test that gives (ult)t

- Compression test that gives (ult)c

- Torsion test that gives ult

Circle B1 = ult

2 = 03 = -ult

Circle C1 = (ult)t

2 = 03 = 0

(ult)c

A

BC

Mohr’s circle for each test

Mohr’s failure criteria

Failure occurs when the absolute value of either one of the principal stresses reaches a

value greater than (ult)t or (ult)c or in general, if the stress at a point is defined by the

stress coordinate (1, 2), which is plotted on the boundary or outside the shaded area.

Circle A1 = 02 = 03 = -(ult)c

ult

Failure envelope

1

2

(ult)t

(ult)t

(ult)c

(ult)c

x (1, 2) material has failed

Material is under limiting condition

(11, 2

1)