67031921 failures resulting from static loading compatibility mode

I/C: Kalluri Vinayak

Introduction

• The static load is a stationary force or couple applied to a member.

• Stationary means, the load should be unchanging in…

– magnitude

– point or points of application

– direction

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– direction

– or in any other manner

• A stationary force may cause

– axial tension or compression, shear load or bending moment or a torsional load or any combination of these

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Introduction• Failure means a part has had:

– separated into two or more pieces

– become permanently distorted and thus ruined its geometry

– its reliability downgraded


– its reliability downgraded

– its function compromised


Focus of this chapter is on first two cases i.e. predictability

of permanent distortion or separation

Introduction

• Static strength data of various materials may be available in two forms

– Data obtained in tests conducted in the similar conditions as the actual service life

– Data collected in tests conducted in certain ideal conditions that may not necessary simulate the exact service conditions


exact service conditions

• The first type of data is always desirable but difficult to produce

• When second type of data is used, then the engineer must be clever enough to apply suitable factors


Failure Theories• There is no general theory that is sacrosanct and unique; hence

we call them “failure theories”

• Rather it is a choice based on the design requirement and nature of material, whether ductile or brittle

• Ductile materials are designed based on yield criteria

�Maximum shear stress (MSS) theory

�Distortion energy (DE) theory


�Distortion energy (DE) theory

�Ductile Coulomb-Mohr (DCM) theory

• Brittle materials, are designed based on fracture criteria

�Maximum normal stress (MNS) theory

�Brittle Coulomb-Mohr (BCM) theory


Failure Theories• How do we know that the given material is ductile? The

judgment is based on the fracture strain.

• Ductile materials

Fracture strain, εf ≥ 0.05 and

identifiable yield strength is often the same in

compression as in tension i.e. Syt = Syc = Sy


compression as in tension i.e. Syt = Syc = Sy

• Brittle materials

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ucut

f

SS

and

≠

< 05.0strain Fracture ε

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Maximum-Shear-Stress (MSS) Theory

• Also referred to as the Tresca or Guest theory.

• Theory predicts that “yielding begins whenever the

maximum shear stress in any element equals or

exceeds the maximum shear stress in a tension test

specimen of the same material when that specimen

begins to yield”


begins to yield”

• For a general state of stress, the maximum-shear-

stress theory predicts yielding when


y

ySor

S≥−≥

−= 31

31max

22σσ

σστ

ysy SS 5.0 by, given isshear in strength yieldThe =


• For design purposes, Equation can be modified to

incorporate a factor of safety, n.

for plane stress (one of the principal stresses is zero) and

assuming that σA ≥ σB:

n

Sor

n

S yy =−= 31max2

σστ


assuming that σA ≥ σB:

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Case 1: σA ≥ σB ≥ 0.

For this case, σ1 = σA and σ3 = 0. Equation reduces to a yield condition of

Case 2: σA ≥ 0 ≥ σB .

Here, σ1 = σA and σ3 = σB , and Equation becomes

Case 3: 0 ≥ σA ≥ σB .

For this case, σ1 = 0 and σ3 = σB and Equation gives

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n

S yA ≥σ

n

S yBA ≥−σσ

n

S yB

−≤σ


n

S

n

S y

A

y

BA

BA

≥≥−

===

≥≥

σσσ

σσσσσ

σσ

,

0,,

,0

31

321

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unmarked lines are cases for σB ≥ σA

n

S

n

S y

BA

y

BA

BA

≥−≥−

===

≥≥

σσσσ

σσσσσ

σσ

,

,0,

,0

31

321

n

S

n

S

y

B

y

B

A

BA

−≥−

≥−

=

=

=

≥≥

σ

σσ

σσ

σσ

σ

σσ

,

,

,0

,0

31

3

2

1

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Problem

A ductile hot-rolled steel bar has minimum yield

strength in tension and compression of 350MPa.

Using the maximum-shear-stress theory,

determine the factors of safety for the following

plane stress states:plane stress states:

(a) σx = 100 MPa, σy = 50 MPa

(b) σx = 100 MPa, τxy = −75 MPa

(c) σx = −50 MPa, σy = −75MPa, τxy = −50 MPa

(d) σx = 100 MPa, σy = 20 MPa, τxy = −20 MPa


Problem• The figure shows a crank loaded by a force F = 800 N which

causes twisting and bending of the 20 mm-diameter shaft fixed toa support at the origin of the reference system. In actuality, thesupport may be an inertia which we wish to rotate, but for thepurposes of a strength analysis we can consider this to be a staticsproblem. The material of the shaft AB is hot-rolled AISI 1018steel (Table A–20; Page: 1040). Using the maximum-shear-stresstheory, find the factor of safety based on the stress at point A.


F

Distortion-Energy (DE) Theory

�Other names of distortional energy criterion

• The Hencky von- Mises or von-Mises theory

• The shear energy theory

• The octahedral shear stress theory

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• The octahedral shear stress theory

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ijσ pijδ'

ijij orS σ

ijijij Sp += δσ

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• The strain energy per unit volume for simple tension

• For tri-axial stress, the strain energy per unit volume is


εσ2

1=u

][2

1332211 σεσεσε ++=u


• Substituting for the principal strains gives

][2

332211 σεσεσε ++=u

[ ])(22

1133221

2

3

2

2

2

1 σσσσσσνσσσ ++−++=E

u

( )[ ] ( )[ ] ( )[ ]

+−=+−=+−= 213313223211

11;

1.. σσνσεσσνσεσσνσε

Eand

EEei

• The strain energy for producing only volume change is

• Substituting σav = (σ1 + σ2 + σ3)/3 gives


)21(2

3 2

νσ

−=E

u avv 321, σσσσ andforngsubstitutiby av[ ]


• Substituting σav = (σ1 + σ2 + σ3)/3 gives

[ ]133221

2

3

2

2

2

1 2226

21σσσσσσσσσ

ν+++++

−=

Euv


[ ]

[ ]( ) ( ) ( ) −+−+−+

=

+++++−

−

++−++=−=

1

2226

21

)(22

1

222

133221

2

3

2

2

2

1

133221

2

3

2

2

2

1

σσσσσσν

σσσσσσσσσν

σσσσσσνσσσ

E

Euuu vd

• Then the distortion energy is


( ) ( ) ( )

−+−+−+=

23

1 133221 σσσσσσνE

ud

2

321

3

1,

0

,

yd

y

SE

uenergydistortionthe

andS

yieldattesttensionsimpletheFor

ν

σσσ

+=

===

Distortion-Energy (DE) Theory• Theory predicts that “yielding occurs when the

distortion strain energy per unit volume reaches or

exceeds the distortion strain energy per unit volume

for yield in simple tension or compression of the

same material”

• For a general state of stress, the Distortion-Energy

Theory predicts yielding when


Theory predicts yielding when

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Where “σ′ “is a single, equivalent, or effective stress called von Mises stress

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( ) ( ) ( )yy SorS ≥≥

−+−+− '2

12

13

2

32

2

21

2σ

σσσσσσ


• For plane stress, let σA and σB be the two nonzero

principal stresses, then the von Mises stress is

• The above equation is a rotated ellipse in the σA, σB

plane with σ′ = S

( )21

22'

BBAA σσσσσ +−=


plane with σ′ = Sy



• Using xyz components of three-dimensional stress,

the von Mises stress can be written as

( ) ( ) ( ) ( )[ ]21

222222' 62

1zxyzxyxzzyyx τττσσσσσσσ +++−+−+−=


• and for plane stress


[ ]21

222' 3 xyyyxx τσσσσσ ++−=

• According to DE (von Mises) criterion, substituting

the pure shear state of stress in the 2-D DE criterion,

the two normal stresses being zero,

y

y

xyyxy SS

S 577.03

3 2 === ττ

SHEAR YIELD STRENGH:



ysy

yxyyxy

SSyieldAt 577.0,

3

=

ysy SS 5.0=

According to the MSS criterion,

DE criterion predicts the shear yield strength to be 15 percent more than

that predicted by the MSS criterion. Hence MSS is more conservative.


Distortion-Energy (DE) TheoryOctahedral shear stress:

� Considering the principal directions as the coordinate

axes, a plane whose normal vector makes equal angles

with each of the principal axes (i.e. having direction

cosines equal to ) is called an octahedral plane

� The shear stress acting on the so-called ‘octahedral

planes’ which are the eight planes that are normal to the

3

1

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planes’ which are the eight planes that are normal to the

equi-inclined directions in the eight quadrants formed by

the principal stresses.

Sec. 2.4.4 (page : 36) of Arthur P. Boresi and Richard J.

Schmidt, Advanced Mechanics of Materials, 6th ed., John

Wiley & Sons, New York, 2003,

Distortion-Energy (DE) TheoryTHE OCTAHEDRAL PLANES: Octahedral shear stress theory

Failure occurs when the octahedral shear stress in the

given state of stress reaches the octahedral shear

stress in simple tension at yield.


Problem

A ductile hot-rolled steel bar has minimum yield

strength in tension and compression of 350MPa.

Using the distortion energy theory, determine

the factors of safety for the following plane

stress states:stress states:

(a) σx = 100 MPa, σy = 50 MPa

(b) σx = 100 MPa, τxy = −75 MPa

(c) σx = −50 MPa, σy = −75MPa, τxy = −50 MPa

(d) σx = 100 MPa, σy = 20 MPa, τxy = −20 MPa


Problem• The figure shows a crank loaded by a force F = 800 N which

causes twisting and bending of the 20 mm-diameter shaft fixed toa support at the origin of the reference system. In actuality, thesupport may be an inertia which we wish to rotate, but for thepurposes of a strength analysis we can consider this to be a staticsproblem. The material of the shaft AB is hot-rolled AISI 1018steel (Table A–20; Page: 1040). Using the distortion energytheory, find the factor of safety based on the stress at point A.


F

Table A–20; Page: 1040


Ductile/Brittle Coulomb-Mohr TheoryTo be applied when the material has unequal strength in tension and

compression,ycyt SS ≠

Examples: magnesium alloy materials, for which ytyc SS 5.0≅

DCM is a simplification of the Coulomb- Mohr yield criteria.

Sec. 4.5.1 (page : 126) of Arthur P.

Gray Cast iron materials, for whichtuuc SoftimesS 43−≅


OR ⇒

where either yield strength or ultimate strength can be used

Sec. 4.5.1 (page : 126) of Arthur P.

Boresi and Richard J. Schmidt,

Advanced Mechanics of Materials,

6th ed., John Wiley & Sons, New

York, 2003,

Ductile Coulomb-Mohr Theory or Internal friction theory

For design equations, incorporating the factor of safety n, divide all strengths by n.

nSS ct

131 =−σσ

for plane stress (one of the principal stresses is zero) and assuming that σA ≥ σB:

Case 1: σA ≥ σB ≥ 0. Here, σ1 = σA and σ3 = 0. Equation reduces ton

StA ≥σ

1σσ


Case 2: σA ≥ 0 ≥ σB . Here, σ1 = σA and σ3 = σB , Equation reduces to

Case 3: 0 ≥ σA ≥ σB . Here, σ1 = 0 and σ3 = σB, Equation reduces to

n

nSS c

B

t

A 1≥−

σσ

n

ScB ≤σ

Maximum-Normal-Stress Theory

• The maximum-normal-stress (MNS) theory

states that failure occurs whenever one of the

three principal stresses equals or exceeds the

strength.

• For the principal stresses for a general stress

state in the ordered form σ ≥ σ ≥ σ ; This


state in the ordered form σ1 ≥ σ2 ≥ σ3; This

theory then predicts that failure occurs

whenever

σ1 ≥ Sut or σ3 ≤ −Suc

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where Sut and Suc are the ultimate tensile and compressive strengths, respectively,

given as positive quantities.

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Maximum-Normal-Stress Theory• For plane stress, with σA ≥ σB,

σA ≥ Sut or σB ≤ −Suc


Modified Mohr Theory


Failure theory selection flowchart


ProblemThe figure shows a shaft mounted in bearings at A and D and having pulleys at

B and C. The forces shown acting on the pulley surfaces represent the belt

tensions. The shaft is to be made of ASTM grade 25 cast iron using a design

factor of 2. Based on Brittle Coulomb Mohr theory, What diameter should

be used for the shaft?

150 mm dia

1200 N


A

B

C

D

200 mm

200 mm

150 mm200 mm dia

200 N120 N

1600 N

Table A–24 ; Page :1046-1047

Mechanical Properties of Three Non-Steel (Gray cast iron,

Aluminum, Titanium Alloy) Metals

(a) Typical Properties of Gray Cast Iron


67031921 failures resulting from static loading compatibility mode

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