SOLID MECHANICS II

( BDA 3033 )

CHAPTER 6

Theories of Elastic

Failures

TEST 2: 14 DEC, 3-4.30PM, F2

ATAS

ASSIGNMENT 2 SUBMIT

TODAY

LAST TOPIC TODAY

GROUP PROJECT TODAY.

BASED ON LAB EXPT.

CHAPTER VI THEORIES OF FAILURE

Maximum Normal Stress Theory

Maximum Normal Strain Theory

Maximum Shear Stress Theory (Tresca Yield

Criterion)

Strain Energy Theory

Maximum Distortion Energy Theory

Out of these five theories of failure, the maximum normal

stress theory and maximum normal strain theory are only

applicable for brittle materials, and the remaining three

theories are applicable for ductile materials.

TOPICS

Introduction

Maximum Normal Stress Theory (Rankine

Theory)

Maximum Normal Strain Theory (St-Venant

Theory)

Maximum Shear Stress Theory (Tresca Yield

Criterion)

Strain Energy Theory (Haigh Theory)

Maximum Distortion Energy Theory (Von

Misses)

INTRODUCTION ( CONTD..)

Stress-Analysis is performed on a component to determine

The required “size or geometry” (design)

an allowable load (service)

cause of failure (forensic)

For all of these, a limit stress or allowable stress value for the component material is required.

Hence, a Failure-Theory is needed to define the onset or criterion of failure

INTRODUCTION (CONTD..)

FAILURE •Occurs if a component can no longer function as intended. •Failure Modes:

• yielding: a process of global permanent plastic deformation. Change in the geometry of the object.

• low stiffness: excessive elastic deflection.

• fracture: a process in which cracks grow to the extent that the component breaks apart.

•buckling: the loss of stable equilibrium. Compressive loading can lead to bucking in columns.

INTRODUCTION (CONTD..)

FAILURE

PREDICTION

• The failure of a statically loaded member in uni-axial

tension or compression is relatively easy to predict.

• One can simply compare the stress incurred with the

strength of the material.

• When the loading conditions are Complex (i.e. biaxial

loading, sheer stresses) then we must use some method

to compare multiple stresses to a single strength value.

• These methods are known failure theories

INTRODUCTION (CONTD..)

NEED FOR FAILURE THEORIES

To design structural components and

calculate margin of safety.

To guide in materials development.

To determine weak and strong directions.

MAXIMUM NORMAL STRESS THEORY

• this theory postulates, that failure will occur in the

structural component if the maximum normal stress in

that component reaches the ultimate strength, u

obtained from the tensile test of a specimen of the

same material.

• Thus, the structural component will be safe as long as

the absolute values of the principle stresses 1 and 2

are both less than u:

1 = U and 2 = U

• This theory deals with brittle materials only.

• The maximum normal stress theory can be expresses

graphically as shown in the figure. If the point

obtained by plotting the values 1 and 2 of the

principle stress fall within the square area shown in

the figure, the structural component is safe.

• If it falls outside that area, the component will fail.

1

2

u

u

-u

-u

This theory alson known as Saint-Venant’s Theory

According to this theory, a given structural component is

safe as long as the maximum value of the normal strain in

that component remains smaller than the value u of the

strain at which a tensile test specimen of the same material

will fail.

As shown in the figure, the strain is maximum along one of

the principle axes of stress if the deformation is elastic and

the material homogenous and isotropic.

Thus denoting by 1 and 2 the values of the normal strain

along the principle axes in plane of stress, we write

1 = u and 2 = u

MAXIMUM NORMAL STRAIN THEORY

MAXIMUM NORMAL STRAIN THEORY

(CONT)

Making use of the generalized Hooke’s Law, we could

express these relations in term of the principle stresses 1

and 2 and the ultimate strength U of the material.

We would find that, according to the maximum normal

strain theory, the structural component is safe as long as

the point obtained by plotting 1 and 2 falls within the

area shown in the figure where is Poisson’s ration for the

given material.

1

2

U

U

-U

-U

1

U

1

U

MAXIMUM SHEARING STRESS THEORY

• This theory is based on the observation that

yield in ductile materials is caused by slippage

of the material along oblique surfaces and is

due primarily to shearing stress.

• A given structural component is safe as long as

the maximum value max of the shearing stress

in that component remains smaller than the

corresponding value of the shearing stress in a

tensile test specimen of the same material as

the specimen starts to yield.

• For a 3D complex stress system, the max shear

stress is given by:

max = ½ (1-2)

• On the other hand, in the 1D stress system as

obtained in the tensile test, at the yield limit,

1= Y and 2=0, therefore:-

• max= ½ Y

MAXIMUM SHEARING STRESS THEORY

(CONT.)

Thus,

Graphically, the maximum shear stress criterion

requires that the two principal stresses be within

the green zone as shown in the figure.

signs same have and

signs opposite have and -

-2

1

2

1

-2

1 and

2

1

21

Y2

Y1

21Y21

21Y

21maxYmax

MAXIMUM DISTORTION ENERGY THEORY

This theory is based on the determination of the distortion

energy in a given material, i.e. of the energy associated with

changes in shape in that material (as opposed to the energy

associated with changes in volume in the same material).

A given structural component is safe as long as the maximum

value of the distortion energy per unit volume in that

material remains smaller than the distortion energy per unit

volume required to cause yield in a tensile test specimen of

the same material.

The distortion energy per unit volume in an isotropic

material under plane stress is:

-

G6

1U

2

221

2

1d

MAXIMUM DISTORTION ENERGY THEORY (CONT)

In the particular case of a tensile test specimen that is

starting to yield, we have:-

This equation represents a principal stress ellipse as

illustrated in the figure

Von Mises criterion also gives a reasonable estimation of

fatigue failure, especially in cases of repeated tensile and

tensile-shear loading

2

221

2

1

2

Y

2

221

2

1

2

Y

2

Y

Yd

2Y1

G6

1

G6

,Thus

G6Uand

0,

PROBLEM 1

The solid shaft shown in Figure has a radius of 0.5 cm

and is made of steel having a yield stress of 360 MPa.

Determine if the loadings cause the shaft to fail

according to Tresca and von mises theories.

32.5 Nm 15 kN

1 cm

SOLUTION

Calculating the stresses caused by axial force and torque

The Principal stresses

6.286

6.95

1.1915.95

)5.165(2

0191

2

0191

22

5.165

5.02

5.025.3

19110.195.0

15

2

1

2

2

2

2

2,1

4

22

xy

yxyx

xy

x

MPaJ

Tc

MPacm

kNA

P

6.286

6.95

1.1915.95

)5.165(2

0191

2

0191

22

5.165

5.02

5.025.3

19110.195.0

15

2

1

2

2

2

2

2,1

4

22

xy

yxyx

xy

x

MPaJ

Tc

MPacm

kNA

P

6.286

6.95

1.1915.95

)5.165(2

0191

2

0191

22

5.165

5.02

5.025.3

19110.195.0

15

2

1

2

2

2

2

2,1

4

22

xy

yxyx

xy

x

MPaJ

Tc

MPacm

kNA

P

191.1MPa

165.5 MPa

SOLUTION ( CONTD..)

Applying Maximum Shear stress theory

So shear failure occurs

Applying Maximum distortion theory

No Failure

3602.382

360)6.286(6.95

21

y

1296009.118677

360)6.286()6.286)(6.95(6.95 222

2

y

2

221

2

1

PROBLEM 2

The state of plane stress shown

occurs at a critical point of a steel

machine component. As a result

of several tensile tests, it has

been found that the tensile yield

strength is Y=250 MPa for the

grade of steel used. Determine

the factor of safety with respect to

yield, using:

(a) the maximum shearing stress

theory

(b) the maximum distortion

energy theory

80 MPa

25 MPa

40 MPa

SOLUTION:

92.1MPa65

MPa125F.S

-:is yield respect toh safety wit offactor tha,Therefore

MPa65

-:circle smohr'or

formula usingby determined becan stress shearing maximum The

MPa125250MPa2

1

2

1

is yieldat stress shearing ingcorrespond the,MPa250 Since

theorystress shearing Maximum .)a(

MPa45andMPa85

circle, smohr'or

formula usingby determined becan and stresses principle The

25MPa and ,MPa40 80MPa,Given

max

Y

max

YY

Y

21

21

xyyx

SOLUTION (CONT)

19.2S.F

S.F

2503.114

S.F

25045458585

S.F

CriterionEnergy Distortion Maximum .)b(

2

22

2

Y2

221

2

1

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Lab 1 Lab 3

Lab 2 Lab 4

Lab 5