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Dr. P. Buyung Kosasih,Spring 2008

Stress

Strain

Yield Fracture

Fracture

Load

Path to static failure of machine components

Ductile material Brittle material

Discussed last week (w3)

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Name some of ductile and brittle materials Commonly used in engineering applications

DUCTILE BRITTLE (Sy close to Su)

• Steel

• Plastics

• Rubbers

• Cast iron

• Hardened tool steel

• Concrete

• Woods

•Ceramics

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Lecture 4 Failure under static load

• Failure of ductile material (Ch. 5 p. 239 – 253)

• Failure of brittle material (Ch. 5 p.254 - 261)

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Most common test to determine material strength is Tensile test.

What do we know from tensile test ?

• Yield tensile stress (Sy)

• Ultimate tensile stress (Su)

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Mechanical properties of some material are listed in Appendix C (Norton).

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How the yield tensile stress related to failure of part ?

FF

σxσx x

y

σxσx

σy

τxy

τyxτxy

τyx

x

y

σy

Stress element ofTensile test specimen(uni-axial test)

2D stress element 3D stress element

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The answer is :

Using• For ductile materials

a1) The von Mises-Hencky theory(distortion energy theory)

a2) The maximum Shear-Stress theorya3) Maximum Normal-Stress theory

• For brittle materialsb1) Maximum Normal-Stress theoryb2) The Coulomb-Mohr theoryb3) The Modified-Mohr theory

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a1) The von Mises-Hencky or Distortion-Energy Theory

The total strain energy storedper unit volume in the material

U = (½) σ ε (J/m3)

Extending to 3D stress state

U = ½ (σ1ε1 + σ2ε2 + σ3ε3)

In terms of principle normal stresses:

( )[ ]31322123

22

21 2

21 σσσσσσνσσσ ++−++=E

U

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Component of strain energy = hydrostatic + distortion components

U = Uh + Ud

σh

σh

σh

σ1d

σ2d

σ3d

σ1

σ2

σ3

It is shown that the hydrostatic stress is (page 243)

σh = (σ1+ σ2 + σ3) / 3

= +

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Where Uh is due to hydrostatic loading i.e. equal principal stresses(volume change without changing shape → no shear)

[ ]31322123

22

213

1 σσσσσσσσσυ−−−++

+=

−=

EU

UUU

d

hd

[ ]

[ ])(26

21

)(221

31322123

22

21

222

σσσσσσσσσυ

σσσσσσυσσσ

+++++−

=

++−++=

EU

EU

h

hhhhhhhhhh

Ud is due to (angular) distortion (shape change → shear)

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From tensile test, the uni-axial stress state at yield gives σ1 = Sy ,

σ2 = σ3 = 0. So,

2@ 3

1yyieldd S

EU υ+

=

Comparing this with general stress state, so to prevent failure

dyieldd UU >@

or

)31322123

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21 σσσσσσσσσ −−−++>yS

von Mises stressoreffective stressor equivalent stressσ`

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The von Mises (effective) stress, σ`, can also be expressed in terms ofapplied stresses.

( ) ( ) ( ) ( )2

6 222222zxyzxyxzzyyx τττσσσσσσ

σ+++−+−+−

>′

For 2D case:

2331

21 σσσσσ +−=′

222 3 xyyxyx τσσσσσ +−+=′

Factor of safety : σ ′= /ySN

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2D visualization of the distortion energy failure theory.

σ1/Sy

σ3/Sy

0 0.5 1.0 1.5-1.5 -1.0 -0.5

0

0.5

1.0

1.5

-1.5

-1.0

-0.5

Safe region(Distortion -Energy theory) σ′/Sy = 1

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3D visualization of the distortion energy failure theory.

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a2) The maximum shear stress theory

The theory states thatfailure occurs when the maximum shear stressin a part exceeds the shear stress in a tensile specimen at yield (one-half of the tensile yield strength).

Sys = 0.5 Sy

i.e. part fails when

τmax > Sys

max/τysSN =Factor of safety :

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2D visualization of the maximum shear stress failure theory.

σ1/Sy

σ3/Sy

0 0.5 1.0 1.5-1.5 -1.0 -0.5

0

0.5

1.0

1.5

-1.5

-1.0

-0.5

Safe region(Maximum Shear Stress theory)

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a3) Maximum normal stress theory

The theory states that failure will occur when the normal stress in thespecimen reaches some limit on normal strength such as tensileyield strength or ultimate tensile strength.

i.e. part fails when

{σ1,σ2,σ3} > Sy

Factor of safety : },,max{/ 321 σσσySN =

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2D visualization of the three (ductile) failure theories.

σ1/Sy

σ3/Sy

0 0.5 1.0 1.5-1.5 -1.0 -0.5

0

0.5

1.0

1.5

-1.5

-1.0

-0.5

Safe region(Maximum Normal Stress theory)

Safe region(Distortion -Energy theory)

Safe region(Maximum Shear Stress theory)

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Which theory should we choose ?

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Application:

r = 50 mm

I = πr4/4

= 4.9 x10-5 m4

A = 7.9x10-3 m2

P (N)

2 m

For the loaded cantilever beam made of a material with tensile yield strength of 30 MPa. What is the maximum P that the beam can support without permanent yield ?

M

x

V

x

Mmax = 2P N.m

P N

The most critical spot1

2

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21

2041 P

σy = 0

σy = 0

τyx

τxy

τyx

x

y

σx = M r/I

= 2041 P (Pa)

τxy = 0

Element 1

PPP 20412

020412

02041 2

1 =⎟⎠⎞

⎜⎝⎛ −

++

=σ

02

020412

02041 2

2 =⎟⎠⎞

⎜⎝⎛ −

−+

=PPσ

P10202

21max =

−=

σστ

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Based on the distortion energy theory

P2041=′σ

ForP2041=′σ

kNPPx

Sy

7.1420411030 6

<>

′> σ

The von Mises stress at point 1 is

2331

21 σσσσσ +−=′

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Based on the maximum shear stress theory

Sys = 0.5 Sy = 15 Mpa

For τmax = 1020 P < Sys = 15x106

P < 14.7 kN

Based on the maximum normal stress theory

P < 14.7 kN(verify yourself)

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σy = 0

σy = 0

τyx

τxy

τyx

x

y

σx = 0

τxy = 4V/3A = 168.8 P (Pa)

Element 2

( ) PP 8.1688.1682

002

00 22

1 =+⎟⎠⎞

⎜⎝⎛ −

++

=σ

( ) PP 8.1688.1682

002

00 22

2 −=+⎟⎠⎞

⎜⎝⎛ −

−+

=σ

P8.1682

21max =

−=

σστ

σx = 0

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Based on the distortion energy theory

P4.292=′σ

ForP4.292=′σ

kNPPx

Sy

3084.2921030 6

<>

′> σ

The von Mises stress at point 2 is

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So the answer to the question “What is the maximum P that the beam can support without permanent yielding ?”

kNP 7.14<

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Failure mechanisms of brittle materials

On tension fracture is due to normal stress alone

On compression fracture is due to combination of normal stress andshear stress.

Ductile materialsdo not fracture on compression

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Characteristics of brittle materials• Failure mechanisms.In tension : due to normal stress aloneIn compression : due to combination of normal and shear stress

• Their yield strength (Sy) and ultimate strength (Su) are almost identical. So failure of brittle materials is normally associated with fracture rather than yield. And strength refers to Su.

• Sut may be equal or not to Suc. When Suc > Sut, the materials are said to be uneven materials, or else they are known as even materials.

• Their shear strength (Sus) can be greater than their tensile strength(Sut) unlike ductile materials where Sus ≅ 0.5 Sut

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Failure theories for brittle materials

• Even materials:

b1) Maximum Normal-Stress Theory

• Uneven materials:

b2) The Coulomb-Mohr Theory

b3) The Modified-Mohr Theory

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b1) Maximum Normal-Stress Theory for even materials

σ2/Sut or σ2/Suc

0 0.5 1.0 1.5-1.5 -1.0 -0.5

0

0.5

1.0

1.5

-1.5

-1.0

-0.5

Safe region(Maximum Normal Stress theory)

σ1/Sut or σ1/Suc

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b2) The Coulomb-Mohr theory for uneven materials

-0.5 0 0.5 1.0… -1.5 -1.0

-0.5

0

0.5

1.0

…

-1.5

-1.0

σ2/Sut

σ1/Sut

Safe region(Coulomb-Mohrtheory)

(-Suc/Sut, -Suc/Sut)

(Sut/Sut, Sut/Sut)

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b3) Modified-Mohr theory for uneven materials

-0.5 0 0.5 1.0… -1.5 -1.0

-0.5

0

0.5

1.0

…

-1.5

-1.0

σ2/Sut

σ1/Sut

Safe region(Modified-Mohrtheory)

(-Suc/Sut, -Suc/Sut)

(Sut/Sut, Sut/Sut)

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Which theory should we use for uneven materials?

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Calculation of factor of safety

-0.5 0 0.5 1.0… -1.5 -1.0

-0.5

0

0.5

1.0

…

-1.5

-1.0

σ2/Sut

σ1/Sut

Safe region(Modified-Mohrtheory)

1/σucSN =

1/σutSN =

1/σutSN =

)( 311 σσσ +−=

utuc

ucut

SSSS

N

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Modified-Mohr effective stress by Dowling(for general 3D case and usually programmed)

σ~utSN =

⎥⎥⎦

⎤

⎢⎢⎣

⎡+

−−

+−= )(2

21

21211 σσσσuc

ucut

SSS

C

⎥⎥⎦

⎤

⎢⎢⎣

⎡+

−

−+−= )(

221

32322 σσσσuc

ucut

SSS

C

⎥⎥⎦

⎤

⎢⎢⎣

⎡+

−

−+−= )(

221

13133 σσσσuc

ucut

SSS

C

),,,,,max(~321321 σσσσ CCC=

0max0~ <= ifσ

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SummaryLecture 1Static loading analysis

Lecture 3Stress analysis ofstatically loaded parts

Lecture 4Failure analysis of stressed parts

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Type of loadings

Statically loaded parts Dynamically loaded parts

Lecture 5 : Fatigue failure