Failures Resulting From Static Loading (1)

BITS PilaniPilani Campus

VINAYAK KALLURI

BITS PilaniPilani Campus

�

BITS Pilani, Pilani Campus

� The static load is a stationary force or couple applied to amember.

� Stationary means, the load should be unchanging in

�magnitude

� point or points of application

� direction

� or in any other manner

� A stationary force may cause axial tension orcompression, shear load or bending moment or atorsional load or any combination of these

Introduction

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Failure means a part has had:

– separated into two or more pieces

– become permanently distorted and thus ruined its geometry

– its reliability downgraded

– its function compromised

Introduction

4

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

of permanent distortion or separation


Static strength data of various materials may be availablein two forms

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

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

The first type of data is always desirable but difficult toproduce

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

Introduction

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5

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� There is no general theory that is inviolable and unique;hence we call them “failure theories”

� Rather it is a choice based on the design requirementand nature of material, whether ductile or brittle

Ductile materials are designed based on yield criteria

�Maximum shear stress (MSS) 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

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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

Brittle materials

Failure Theories

9/5/2012 7ucut

f

SS

and

≠

< 05.0strain Fracture ε

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� 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”

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

theory predicts yielding when

Maximum-Shear-Stress (MSS) Theory

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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:

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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

Sor

n

S yy =−= 31max2

σστ

n

S y

A ≥σ

n

S y

BA ≥−σσ

n

S y

B

−≤σ



9/5/2012 10unmarked lines are cases for σB ≥ σA

n

S

n

S y

A

y

BA

BA

≥≥−

===

≥≥

σσσ

σσσσσ

σσ

,

0,,

,0

31

321

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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The 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 AISI 1035

CD steel using a design factor of 2. Based on Maximum shear

stress theory, What diameter should be used for the shaft?

Problem


Table A–20; Page: 1040


Distortion-Energy (DE) Theory

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�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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ijσ pijδ'

ijij orS σ

ijijij Sp += δσ[email protected]



The strain energy per unit volume for simple tension

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

Substituting for the principal strains gives

εσ2

1=u

][2

1332211 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

[ ]133221

2

3

2

2

2

1 2226

21σσσσσσσσσ

ν+++++

−=

Euv

[ ]


321, σσσσ andforngsubstitutiby av


Then the distortion energy is

[ ]

[ ]( ) ( ) ( )

−+−+−+=

+++++−

−

++−++=−=

23

1

2226

21

)(22

1

2

13

2

32

2

21

133221

2

3

2

2

2

1

133221

2

3

2

2

2

1

σσσσσσν

σσσσσσσσσν

σσσσσσνσσσ

Eu

E

Euuu

d

vd

2

321

3

1,

0

,

yd

y

SE

uenergydistortionthe

andS

yieldattesttensionsimpletheFor

ν

σσσ

+=

===



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

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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′ = Sy

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( )21

22'

BBAA σσσσσ +−=



Using xyz components of three-dimensional stress, the von

Mises stress can be written as

and for plane stress

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( ) ( ) ( ) ( )[ ]21

222222' 62

1zxyzxyxzzyyx τττσσσσσσσ +++−+−+−=

[ ]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,

SHEAR YIELD STRENGH:

ysy

y

y

xyyxy

SSyieldAt

SS

S

577.0,

577.03

3 2

=

=== ττ

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.9/5/2012 21

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represent the belt tensions. The shaft is to be made of AISI 1035

CD steel using a design factor of 2. Based on von-Mises theory,

What diameter should be used for the shaft?

Problem


Table A–20; Page: 1040

[ ]21

222' 3 xyyyxx τσσσσσ ++−=

d= 22.04 = 25 mm


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 σ1 ≥ σ2 ≥ σ3; This theory then predicts that failure occurs

whenever

σ1 ≥ Sut or σ3 ≤ −Suc

Maximum-Normal-Stress Theory

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

respectively, given as positive quantities.

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For plane stress, with σA ≥ σB,

σA ≥ Sut or σB ≤ −Suc

Maximum-Normal-Stress Theory

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represent the belt tensions. The shaft is to be made of ASTM grade

25 cast iron using a design factor of 2. Based on maximum normal

stress theory, What diameter should be used for the shaft?

Problem


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


Ductile/Brittle Coulomb-Mohr Theory

9/5/201229

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Gray Cast iron materials, for which

To 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.

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,

tuuc SoftimesS 43 −≅


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 to

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

StA ≥σ

nSS c

B

t

A 1≥−

σσ

n

ScB ≤σ




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?

Problem


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


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:

(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 clevis pin shown in the

figure is 12 mm in diameter

and has the dimensions a =

12 mm and b = 18 mm. The

pin is machined from AISI

1018 hot-rolled steel and is

to be loaded to no more

than 4.4 kN. Determine the

factor of safety based on

Distortion Energy (DE)

theory equivalent diagram

shown in figure c.

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 to a support

at the origin of the reference system. In actuality, the support may be

an inertia which we wish to rotate, but for the purposes of a strength

analysis we can consider this to be a statics problem. The material of

the shaft AB is ASTM grade 40 cast iron. Using the maximum

normal stress theory, find the factor of safety based on the stress at

point A.

Problem

F

Failures Resulting From Static Loading (1)

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