Failure Theories - UPRMacademic.uprm.edu/pcaceres/Courses/INME4011/MD-5A.pdf · Failure Theories...
Transcript of Failure Theories - UPRMacademic.uprm.edu/pcaceres/Courses/INME4011/MD-5A.pdf · Failure Theories...
Failure TheoriesWhy do mechanical components fail? Mechanical components fail because the applied stresses exceeds the material’s strength (Too simple).What kind of stresses cause failure? Under any load combination, there is always a combination of normal and shearing stresses in the material.
What is the definition of Failure?Obviously fracture but in some components yielding can also be considered as failure, if yielding distorts the material in such a way that it no longer functions properly
Which stress causes the material to fail?Usually ductile materials are limited by their shear strengths. While brittle materials (ductility < 5%) are limited by their tensile strengths.
Stress at which point?
Failure Theories
Load typeUniaxialBiaxial
Pure Shear
Material PropertyDuctile Brittle
Application of StressStatic
Dynamic
Static Loading Maximum Normal Stress
Modified MohrYield strength
Maximum shear stressDistortion energy
Dynamic LoadingGoodman
GerberSoderberg
Static Failure TheoriesThe idea behind the various classical failure theories is that whatever is responsible for failure in the standard tensile test will also be responsible for failure under all other conditions of static loading.
Ductile Material Brittle MaterialCharacteristic Failure Stress
Yield Stress Ultimate Stress
Important Theories
1. Maximum Shear Stress2. Maximum Octahedral Shear Stress
1. Maximum Normal Stress
2. Modified Mohr.
Ductile MaterialsMaximum Shear Stress Theory
Failure occurs when the maximum shear stress in the part exceeds the shear stress in a tensile test specimen (of the same material) at yield.Hence in a tensile test,
2maxyS
=τ
For a general state of stresses
2231
maxyS
=−
=σστ
This leads to an hexagonal failure envelop. A stress system in the interior of the envelop is considered SAFE
for design purposes, the failure relation can be modified to include a factor of safety (n):
31 σσ −= yS
n
The Maximum Shear Stress Theory for Ductile Materials is also known as the Tresca Theory.
Several cases can be analyzed in plane stress problems:
Case 1: 021 ≥≥σσIn this case σ3=0
y
y
S
S
≥
==−
=
1
131max 222σ
σσστ
Yielding condition
Case 2: 31 0 σσ ≥≥
y
y
S
S
≥−
=−
=
31
31max 22
σσ
σστ
Distortion Energy TheoryBased on the consideration of angular distortion of stressed elements.
The theory states that failure occurs when the distortion strain energy in the material exceeds the distortion strain energy in a tensile test specimen (of the same material) at yield.
ResilienceResilience is the capacity of a material to absorb
energy when it is deformed elastically and then, upon unloading, to have this energy recovered.
∫=y dUr
εεσ
0Modulus of resilience Ur
If it is in a linear elastic region,
EEU yy
yyyr 221
21 2σσ
σεσ =⎟⎟⎠
⎞⎜⎜⎝
⎛==
For general 3-D stresses: ( )
( )( )13322123
22
21
332211
221
21
σσσσσσνσσσ
εσεσεσ
++−++=
++=
Eu
u
Applying Hooke’s Law
There are two components in this energy a mean component and deviatoric component. 33
321 zyxM
σσσσσσσ++
=++
=
MDMDMD σσσσσσσσσ −=−=−= 3,32,21,1
The energy due to the mean stress (it gives a volumetric change but not a distortion:
( )( )
( )[ ] ( )13322123
22
21
2
222
2226
2121321
221
σσσσσσσσσννσ
σσσσσσνσσσ
+++++−
=−=
++−++=
EEu
Eu
MMean
MMMMMMMMMMean
( )13322123
22
213
1 σσσσσσσσσν−−−++
+=−=
Euuu MeanD
Compare the distortion energy of a tensile test with the distortion energy of the material.
( )13322123
22
21
2
31
31 σσσσσσσσσνν
−−−+++
==+
=E
uSE
u DyTensile
1323
21
13322123
22
21
σσσσ
σσσσσσσσσ
−+=
−−−++=
y
y
S
S
Plane Stress
Von Mises effective stress : Defined as the uniaxial tensile stress that creates the same distortion energy as any actual combination of applied stresses.
( ) ( ) ( ) ( )
222
222222
3
26
xyyxyxVM
zxyzxyxzzyyxVM
τσσσσσ
τττσσσσσσσ
+−+=
+++−+−+−=
2D
This simplifies the approach since we can use the following failure criterion
VM
y
yVM
Sn
S
σ
σ
=
≥
Case of Pure Shear
yy
Max
yxyVM
SS
S
577.03
3
==
≥=
τ
τσ
Several theories have been developed to describe the failure of brittle materials, such as:
�Maximum Normal Stress Theory
�Coulomb-Mohr Theory
�Modified-Mohr Theory
Brittle Materials
Maximum Normal Stress TheoryFailure occurs when one of the three principal stresses reaches a permissible strength (TS). 21 σσ >Failure is predicted to occur when σ1=St and σ2<-Sc
Where St and Sc are the tensile and compressive strengthFor a biaxial state of stresses
σ1
σ2
St
St-Sc
-Sc
Coulomb-Mohr Theory or Internal Friction Theory (IFT)This theory is a modification of the maximum normal stress theory in the which the failure envelope is constructed by connecting the opposite corners of quadrants I and III.
The result is an hexagonal failure envelop.Similar to the maximum shear stress theory but also accounts for the uneven material properties of brittle material
Mohr’s TheoryThe theory predicts that a material will fail if a stress
state is on the envelope that is tangent to the three Mohr’s circles corresponding to:
a. uni-axial ultimate stress in tension, b. uni-axial ultimate stress in compression, and c. pure shear.
Modified Mohr’s Theory
112 ≤−TC σ
σσσ
This theory is a modification of the Coulomb-Mohr theory and is the preferred theory for brittle materials.
Maximum Normal-Strain TheoryAlso known as the Saint-Venant’s Theory.Applies only in the elastic range.Failure is predicted to occur if yy SS ±=−±=− 1221 or νσσνσσ
Where Sy is the yield strength. For a biaxial state of stress
σ1Sy-Sy
σ2
Sy
-Sy
Maximum Strain-Energy TheoryYielding is predicted to occur when the total strain energy in a given volume is greater than or exceeds the strain energy in the samevolume corresponding to the yield strength in tension or compression. The strain energy stored per unit volume (us) during uniaxial loading is E
Su y
s ⋅=
2
2
In a biaxial state of stress
( )2122
21
2211
221
22
σσνσσ
σεσε
σ
σ
⋅⋅⋅−+=
⋅+
⋅=
Eu
u
This theory is no longer used
Example:Given the material SY , σx , σv and τxy find the safety factors for all the applicable criteria. a. Pure aluminum
MPaMPaMPaMPaS xyyxY 0 10 10 30 =−=== τσσ
10-10
MPaMPaMPa Max 10 10 10 31 =−== τσσIs Al ductile or brittle? Ductile
Use either the Maximum Shear Stress Theory (MSST) or the Distortion Theory (DT)
5.12030
)10(1030
31
==−−
=−
=σσ
ySn
MSST Theory
DT Theory
73.132.17
30
32.173003 222
===
==+−+=
MPaMPaS
n
MPa
VM
y
xyyxyxVM
σ
τσσσσσ
b. 0.2%C Carbon SteelKsiKsiKsiKsiS xyyxY 10 35 5 65 =−=−== τσσ
In the plane XY the principal stresses are -1.973Ksi and -38.03Ksi with a maximum shear stress in the XY plane of 18.03KsiIn any orientation
KsiKsiKsiKsi
Max 01.1903.38 973.1 0 321
=−=−==
τσσσ
71.1)03.38(0
65
31
=−−
=−
=σσ
ySnMSST Theory
DT Theory71.1
03.3865
03.38312
32
1
===
=−+=
MPaKsiS
n
Ksi
VM
y
VM
σ
σσσσσ
Ductile
C. Gray Cast Iron
KsiKsiKsiKsiSKsiS xyyxucut 0 10 35 120 30 ==−=== τσσ
Brittle10-35
KsiKsiKsiKsi
Max 5.2235 0 10 321
=−===
τσσσ
Use Maximum Normal Stress Theory (MNST), Internal Friction Theory (IFT), Modified Mohr Theory (MMT)
0.31030
1
===σ
utSnMNST Theory (tensile)
4.335
120
3
===σ
ucSnMNST Theory (compression)
MMT
84.1
54.01
=
=
nn
)30120()30)(120(1)35(
301203010
1_4 0 0
31
31
−=−
−−
−=
−−
≤≥
n
SSSS
nSSS
quadrant
utuc
utuc
utuc
ut
th
σσ
σσ
IFT
6.1
625.0120
3530101 31
=
=−
−=−=
nSSn ucut
σσ
13
31
_
_4 0 0
σσ
σσ
ut
ucuc
th
SSSequationline
quadrant
+−=
≤≥112 ≤−
TC σσ
σσ
Example 1The cantilever tube shown is to be made of 2014 aluminum alloy treated to obtain a specified minimum yield strength of 276MPa. We wish to select a stock size tube (according to the table below). Using a design factor of n=4. The bending load is F=1.75kN, the axial tension is P=9.0kN and the torsion is T=72N.m. What is the realized factor of safety?
Consider the critical area ( top surface).
IMc
AP
x +=σ
Maximum bending moment = 120F
I
dkNxmm
AkN
x
⎟⎠⎞
⎜⎝⎛×
+= 275.1120
9σ
Jd
J
d
JTr
zx362
72=
⎟⎠⎞
⎜⎝⎛×
==τ
( ) 2122 3 zxxVM τσσ +=
GPaGPanS y
VM 0690.04276.0
==≤σ
For the dimensions of that tube57.4
06043.0276.0
===VM
ySn
σ
Example 2:A certain force F is applied at D near the end of the 15-in lever, which is similar to a socket wrench. The bar OABC is made of AISI 1035steel, forged and heat treated so that it has a minimum (ASTM) yield strength of 81kpsi. Find the force (F) required to initiate yielding. Assume that the lever DC will not yield and that there is no stress concentration at A.Solution:
1) Find the critical section
The critical sections will be either point A or Point O. As the moment of inertia varies with r4
then point A in the 1in diameter is the weakest section.
Fd
inFd
dM
IMy
x 6.1421432
64
234 =××
=⎟⎠⎞
⎜⎝⎛
==ππ
σ
2) Determine the stresses at the critical section
Fin
inFd
dT
JTr
zx 4.76)1(1516
32
234 =
××=
⎟⎠⎞
⎜⎝⎛
==ππ
τ
3) Chose the failure criteria.
The AISI 1035 is a ductile material. Hence, we need to employ the distortion-energy theory.
lbfS
F
F
VM
y
zxxxyyxyxVM
4165.194
81000
5.19433 22222
===
=+=+−+=
σ
τστσσσσσ
Apply the MSS theory. For a point undergoing plane stress with only one non-zero normal stress and one shear stress, the two non-zero principal stresses (σA and σB) will have opposite signs (Case 2).
( ) ( )( )lbfF
FF
S
S
zxxzxx
yBA
zxxyBA
3884.7646.14281000
42
2
222
2122
2222
22
max
=×+=
+=+⎟⎠⎞
⎜⎝⎛=≥−
+⎟⎠⎞
⎜⎝⎛±==
−=
τστσσσ
τσσστ
Example 3:A round cantilever bar is subjected to torsion plus a transverse load at the free end. The bar is made of a ductile material having a yield strength of 50000psi. The transverse force (P) is 500lb and the torque is 1000lb-in applied to the free end. The bar is 5in long (L) and a safety factor of 2 is assumed. Transverse shear can be neglected. Determine the minimum diameter to avoid yielding using both MSS and DET criteria.Solution
1) Determine the critical section
The critical section occurs at the wall.
3432
64
2dPL
d
dPL
IMc
x ππσ =
⎟⎠⎞
⎜⎝⎛
==34
16
32
2dT
d
dT
JTc
xy ππτ =
⎟⎠⎞
⎜⎝⎛
==
( ) ( )
( )
( ) ⎥⎦⎤
⎢⎣⎡ +×±×=
⎥⎦⎤
⎢⎣⎡ +±=⎟
⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛±=
+⎟⎠⎞
⎜⎝⎛±⎟
⎠⎞
⎜⎝⎛=+⎟⎟
⎠
⎞⎜⎜⎝
⎛ −±⎟⎟⎠
⎞⎜⎜⎝
⎛ +=
2232,1
223
2
3
2
332,1
22
22
2,1
10005500550016
16161616
2222
d
TPLPLdd
TdPL
dPL
xyxx
xyyxyx
πσ
ππππσ
τσστσσσσ
σ
indn
Sdd
yMAX
MAX
031.1
000,252
500002
4.137152
)8.980(264502
31
3331
≥
==≤=−
=−−
=−
=
τσσ
σστ
3126450
d=σ 32
8.980d
−=σ The stresses are in the wrong order.. Rearranged to
3126450
d=σ 33
8.980d
−=σ
MSS
indnS
d
dddd
yVM
VM
025.12
5000026950
8.980264508.98026450
3
33
2
3
2
3312
32
1
≥
=≤=
⎟⎠⎞
⎜⎝⎛−⎟⎠⎞
⎜⎝⎛−⎟
⎠⎞
⎜⎝⎛−+⎟
⎠⎞
⎜⎝⎛=−+=
σ
σσσσσDET
Example 4:
In the wheel suspension of a car, the spring motion is provided by a torsion bar fastened to the arm on which the wheel is mounted. The torque in the torsion bar is created by the 2500N force acting on the wheel from the ground through a 300mm long lever arm. Because ofspace limitations, the bearing holding the torsion bar is situated 100mm from the wheel shaft. The diameter of the torsion bar is 28mm. Find the stresses in the torsion bar at the bearing by using the DET theory.
Solution
The stresses acting on a torsion bar are:
1. Torsion
2. bending
( )( )( )
MPaPad
dlengtharmF
JTc 174
028.0014.03.0250032
32
)2
)(_(44 =
×=
×==
ππτ
( ) ( )( )( )
MPaPadI
116028.0
64
64
44 ====ππ
σ
dlengthbearingFMc 014.01.025002
_×⎟
⎠⎞
⎜⎝⎛×
( ) ( )
( )
MPaMPa
xyxx
xyyxyx
4.125 4.241
0.1742
0.1162
0.116
2222
21
22
2,1
22
22
2,1
−==
+⎟⎠⎞
⎜⎝⎛±=
+⎟⎠⎞
⎜⎝⎛±⎟
⎠⎞
⎜⎝⎛=+⎟⎟
⎠
⎞⎜⎜⎝
⎛ −±⎟⎟⎠
⎞⎜⎜⎝
⎛ +=
σσ
σ
τσστσσσσ
σ
The principal stresses are:
( ) ( ) ( )( )
nS
MPa yVM
VM
≤=
−−−+=−+=
6.322
4.1254.2414.1254.241 2231
23
21
σ
σσσσσ
DET
MSS
( )
nSMPa
MPa
YMax
Max
≤=×
=−−
=−
=
8.3662
4.1832
4.1254.2412
31
τ
σστ
Example 5:The factor of safety for a machine element depends on the particular point selected for the analysis. Based upon the DET theory, determine the safety factor for points A and B.
This bar is made of AISI 1006 cold-drawn steel (Sy=280MPa) and it is loaded by the forces F=0.55kN, P=8.0kN and T=30N.m
Solution:Point A 2324
432
464
2dP
dFl
dP
d
dFl
AreaP
IMc
x ππππσ +=+
⎟⎠⎞
⎜⎝⎛
=+=
( )( )( )( )
( )( )( )
MPax 49.9502.01084
02.01.01055.032
2
3
3
3
=+=ππ
σ
( )( )
MPadT
JTr
xy 10.19020.0301616
33 ====ππ
τ
( ) ( )[ ]77.2
1.101280
1.1011.19349.953 212222
===
=+=+=
VM
y
xyxVM
Sn
MPa
σ
τσσ
Point B ( )( )( )
( )( )
( )( )( )
( )[ ]22.6
02.45280
02.4543.21347.25
43.2102.0
43
1055.0402.03016
3416
47.2502.010844
2122
2
3
33
2
3
2
==
=+=
=⎟⎠⎞
⎜⎝⎛
+=+=
===
n
MPa
MPaAV
dT
MPadP
VM
xy
x
σ
πππτ
ππσ