failures resulting from static loading (1)
DESCRIPTION
okTRANSCRIPT
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
9/5/2012 [email protected]
BITS Pilani, Pilani Campus
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
Focus of this chapter is on first two cases i.e. predictability
of permanent distortion or separation
BITS Pilani, Pilani Campus
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
9/5/2012
5
BITS Pilani, Pilani Campus
� 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
9/5/2012 6
BITS Pilani, Pilani Campus
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
BITS Pilani, Pilani Campus
� 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
9/5/2012 8
y
ySor
S≥−≥
−= 31
31max
22σσ
σστ
ysy SS 5.0 by, given isshear in strength yieldThe =
BITS Pilani, Pilani Campus
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:
9/5/2012 9
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
n
Sor
n
S yy =−= 31max2
σστ
n
S y
A ≥σ
n
S y
BA ≥−σσ
n
S y
B
−≤σ
Maximum-Shear-Stress (MSS) Theory
BITS Pilani, Pilani Campus
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
Maximum-Shear-Stress (MSS) Theory
BITS Pilani, Pilani Campus
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
BITS Pilani, Pilani Campus
Table A–20; Page: 1040
BITS Pilani, Pilani Campus
Distortion-Energy (DE) Theory
9/5/2012 13
�Other names of distortional energy criterion
• The Hencky von- Mises or von-Mises theory
• The shear energy theory
• The octahedral shear stress theory
BITS Pilani, Pilani Campus
9/5/2012 14
ijσ pijδ'
ijij orS σ
ijijij Sp += δσ[email protected]
Distortion-Energy (DE) Theory
BITS Pilani, Pilani Campus
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
Distortion-Energy (DE) Theory
BITS Pilani, Pilani Campus
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
[ ]
Distortion-Energy (DE) Theory
321, σσσσ andforngsubstitutiby av
BITS Pilani, Pilani Campus
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
+=
===
Distortion-Energy (DE) Theory
BITS Pilani, Pilani Campus
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
9/5/2012 18
σ′ is a single, equivalent, or effective stress called von-Mises stress
( ) ( ) ( )yy SorS ≥≥
−+−+− '2
12
13
2
32
2
21
2σ
σσσσσσ
Distortion-Energy (DE) Theory
BITS Pilani, Pilani Campus
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
9/5/2012 19
( )21
22'
BBAA σσσσσ +−=
Distortion-Energy (DE) Theory
BITS Pilani, Pilani Campus
Using xyz components of three-dimensional stress, the von
Mises stress can be written as
and for plane stress
9/5/2012 20
( ) ( ) ( ) ( )[ ]21
222222' 62
1zxyzxyxzzyyx τττσσσσσσσ +++−+−+−=
[ ]21
222' 3 xyyyxx τσσσσσ ++−=
Distortion-Energy (DE) Theory
BITS Pilani, Pilani Campus
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
Distortion-Energy (DE) Theory
BITS Pilani, Pilani Campus
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 von-Mises theory,
What diameter should be used for the shaft?
Problem
BITS Pilani, Pilani Campus
Table A–20; Page: 1040
[ ]21
222' 3 xyyyxx τσσσσσ ++−=
d= 22.04 = 25 mm
BITS Pilani, Pilani Campus
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
9/5/2012 24
where Sut and Suc are the ultimate tensile and compressive strengths,
respectively, given as positive quantities.
BITS Pilani, Pilani Campus
For plane stress, with σA ≥ σB,
σA ≥ Sut or σB ≤ −Suc
Maximum-Normal-Stress Theory
9/5/2012 25
BITS Pilani, Pilani Campus
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 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
BITS Pilani, Pilani Campus
BITS Pilani, Pilani Campus
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
BITS Pilani, Pilani Campus
Ductile/Brittle Coulomb-Mohr Theory
9/5/201229
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 −≅
BITS Pilani, Pilani Campus
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
BITS Pilani, Pilani Campus
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 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
BITS Pilani, Pilani Campus
BITS Pilani, Pilani Campus
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
BITS Pilani, Pilani Campus
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
BITS Pilani, Pilani Campus
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
BITS Pilani, Pilani Campus
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