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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)
2
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
2
3
Lecture 4 Failure under static load
• Failure of ductile material (Ch. 5 p. 239 – 253)
• Failure of brittle material (Ch. 5 p.254 - 261)
4
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)
3
5
Mechanical properties of some material are listed in Appendix C (Norton).
6
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
4
7
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
8
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
5
9
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
= +
10
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)
6
11
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
22
21 σσσσσσσσσ −−−++>yS
von Mises stressoreffective stressor equivalent stressσ`
12
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
7
13
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
14
3D visualization of the distortion energy failure theory.
8
15
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 :
16
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)
9
17
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 =
18
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)
10
19
Which theory should we choose ?
20
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
11
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
22
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 +−=′
12
23
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)
24
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
13
25
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
26
So the answer to the question “What is the maximum P that the beam can support without permanent yielding ?”
kNP 7.14<
14
27
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
28
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
15
29
Failure theories for brittle materials
• Even materials:
b1) Maximum Normal-Stress Theory
• Uneven materials:
b2) The Coulomb-Mohr Theory
b3) The Modified-Mohr Theory
30
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
16
31
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)
32
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)
17
33
Which theory should we use for uneven materials?
34
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
18
35
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
36
SummaryLecture 1Static loading analysis
Lecture 3Stress analysis ofstatically loaded parts
Lecture 4Failure analysis of stressed parts
19
37
Type of loadings
Statically loaded parts Dynamically loaded parts
Lecture 5 : Fatigue failure