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Slide #Chapter 16: Failure Resulting from Static Loading
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Slide #Chapter 16: Failure Resulting from Static Loading
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6-9 Modifications of the Mohr Theory forBrittle Materials
Brittle-Coulomb-Mohr (BCM) Theory
Modified I-Mohr Theory
Slide #Chapter 16: Failure Resulting from Static Loading
4
-
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Brittle Coulomb-Mohr (BCM) Theory
Same as discussed before for Coulomb-Mohr Theory for ductilematerials. For
0utS
=
A B
Slide #Chapter 16: Failure Resulting from Static Loading
5
10
0
A BA B
B A B
n
nut uc
uc
n
S S
S
=
=
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Graphically
0
10
A A B
A BA B
ut
n
n
S
S S
=
=
Slide #Chapter 16: Failure Resulting from Static Loading
6
0B A B
uc
n
S =
Figure 6-27
Biaxial fracture data ofgray cast iron comparedwith various failurecriteria.
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Slide #Chapter 16: Failure Resulting from Static Loading
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Modified I-Mohr (M1M) Theory
Observed experimental data resulted in modification of BrittleCoulomb Mohr (BCM) Theory.
Similar to the Maximum-Normal-Stress (MNS) Theory and theBrittle Coulomb-Mohr (BCM) Theory in the First and Thirdquadrants.
Slide #Chapter 16: Failure Resulting from Static Loading
8
In the fourth quadrant, it is less conservative than the BCMTheory but more conservative than the MNS theory.
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0
0 1and
A B
BA B
Slide #Chapter 16: Failure Resulting from Static Loading
9
10 1
0
and
A
BA B
A
A B
nc
= >
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Fail
Safe
Safety factor:
i. 1st and 4th Quadabove Pure Shear Line
ii. 4th Quad belowPure
1&0
0
|| >>
>>=
A
BBA
BAut
An
S
Slide #Chapter 16: Failure Resulting from Static Loading
10
ABuc
BnS
>>= 0
ear ne
nSSS
SS
uc
B
utuc
Autuc 1)(=
1&0 || >>>A
BBA
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Slide #Chapter 16: Failure Resulting from Static Loading
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Modified II-Mohr (M2M) Theory
Similar to the previous theories but different in the fourthquadrant.
Data lying in the fourth quadrant were found to be outside theextended region.
Slide #Chapter 16: Failure Resulting from Static Loading
12
ME 307 MACHINE DESIGN I12
The Modified II-Mohr (M2M) extends the region in the fourthquadrant where using a parabolic relation.
The parabola is of the form such that itis tangent to the vertical at
1A B
21B B Aa b c + + =
A B ut
S = =
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: 0
0 1and
Case A A A B
BA B
utn
S
=
Similar toprevious case(M1M)
Slide #Chapter 16: Failure Resulting from Static Loading
13
2
1 0 1
0
anCase B:
Case C:
d
BA B
A B
A
B A B
nn ut
ut ut uc
uc
n
S
S S S
S
+ + = >
=
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Modified II-Mohr
(M2M) Theory
Safety factor:
4th Quad below Pure ShearLine.
Safe
Slide #Chapter 16: Failure Resulting from Static Loading
14
ME 307 MACHINE DESIGN I14
Figure 6-28
Fourth quadrant
data for a grade 40 cast iron,compared with a modified IIMohr fracture locus.
( ),A B
Fail
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6-10 Failure of Brittle Materials Summary
All theories agree in the first and third quadrant.
In the fourth quadrant:
MNS fails.
CM Conservative.
Slide #Chapter 16: Failure Resulting from Static Loading
15
Mod. I Mohr less conservative. Mod. II Mohr matches experimental data better.
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Figure 6-29
A plot of experimentaldata points obtainedfrom tests on Cast Iron.Shown also are thegraphs of three failure
Slide #Chapter 16: Failure Resulting from Static Loading
16
usefulness for brittlematerials. Notice pointsA, B, C and D. To avoidcongestion in the firstquadrant, points havebeen plotted for
As well as for theopposite sense.
A B >
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6-11 Selection of Failure Criteria
For Ductile Materials:
The preferred criterion is the DE theory, although somedesigners also apply the MSS theory because of its simplicityand conservative Nature.
Slide #Chapter 16: Failure Resulting from Static Loading
17
For Brittle behavior:Refer to the flowchart of Figure 6-30
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Ductile Materials and Brittle Material
Slide #Chapter 16: Failure Resulting from Static Loading
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Ductile Materials and Brittle Material
Slide #Chapter 16: Failure Resulting from Static Loading 19
Figure 6-30
Failure Theoryselection flowchart.
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6-12Static or Quasi Static Loadingon a Shaft
In machinery, the general
term shaft refers to amember, usually of circularcross-section, whichsupports gears, sprockets,
Slide #Chapter 16: Failure Resulting from Static Loading
20
, , .,
which is subjected totorsion and to transverse oraxial loads acting singly orin combination.
An axle is a non-rotatingmember that supportswheels, pulleys, andcarries no torque.
Quasi static loading on a shaft means
shaft rotating at a very low speed.
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6-12 Static or Quasi Static Loadingon a Shaft
The design of a shaft involvesthe study of
Slide #Chapter 16: Failure Resulting from Static Loading
21
. a c an a ue s ress
and strength
2. Bending and TorsionalDeflection and rigidity
3. Critical Speed
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6-12 Static or Quasi Static Loadingon a Shaft
The fundamentalkinematic component ofour mechanical universeis the wheel and axle.An essential part of this
Slide #Chapter 16: Failure Resulting from Static Loading
22
revolute joint is theshaft. It is a goodexample of a static andquasi-static loading, anddynamically loadedapplication.
chapter
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6-12 Static or Quasi Static Loadingon a Shaft
The stress at an element located
on the surface of a solid roundshaft of diameter dsubjected tobending, axial loading, andtwisting is
Slide #Chapter 16: Failure Resulting from Static Loading
23
3 232 4
xM Fd d
= +
3
16
xy
T
d
=
Normal stress
Shear stress
Non-zero principal stresses
1 22
2,
2 2
x y x y
A B xy
+ = +
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6-12 Static or Quasi Static Loadingon a Shaft
( )
1/2 1/22 2 2 2
1/22 2
3
' 3
4' 8 48
A A B B x xy
M Fd Td
= + = +
= + +
Von Mises stress
Slide #Chapter 16: Failure Resulting from Static Loading
24
( )
( )
1 22 2
max
1/22 2
max 3
14
2 2
2
8 64
A Bx xy
M Fd Td
= = +
= + +
Maximum Shear
Stress Theory
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6-12 Static or Quasi Static Loadingon a Shaft
( )
1/2 1/22 2 2 2
1/22 2
3
' 3
4' 8 48
A A B B x xy
M Fd Td
= + = +
= + +
Von Mises stress (6-37)
Slide #Chapter 16: Failure Resulting from Static Loading
25
( )
( )
1 22 2max
1/22 2
max 3
1 42 2
28 64
A Bx xy
M Fd T
d
= = +
= + +
Maximum Shear
Stress Theory(6-38)
The above equations permit an estimate of and whendiameter is given, or an estimate of when an allowable valueof or is given.
' maxd d
' max
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6-12 Static or Quasi Static Loadingon a Shaft
(6-39)
For a design factor of ,the distortion energy theory of ductile failure
gives an allowable stress of
dn
' y
all
S =
Slide #Chapter 16: Failure Resulting from Static Loading
26
d
For a design factor of ,the maximum shear theory of ductile failuregives an allowable shear stress of
dn
2
sy y
ll
d d
S Sn n
= = (6-40)
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Static or Quasi Static Loading on a Shaft-bendingand Torsion
1/22 2
3
16' 4 3M T
d
= + Von Mises stress (6-41)
Under many conditions, the axial force F in Eqs. (6-37) and (6-38) iseither zero or so small that its effect may be neglected. With F = 0,
Eqs. (6-37) and (6-38) become
Slide #Chapter 16: Failure Resulting from Static Loading
27
1/22 2
max 316 M T
d
= + Maximum Shear
Stress Theory
(6-42)
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Static or Quasi Static Loading on a Shaft-bendingand Torsion
( )1/3
1/22 2
1/2
16 4 3y
nd M TS
= +
Von Mises stress
(6-43)
Substitution of the allowable stresses from Eqs. 6-39 and 6-40 we find
Slide #Chapter 16: Failure Resulting from Static Loading
28
3
4 3y
M Tn d S
= +
( )
( )
1/3
1/22 2
1/22 2
3
32
1 32
y
y
nd M T
S
M Tn d S
= +
= +
Maximum Shear
Stress Theory
(6-44)
(6-45)
(6-46)
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Example 6-5
Consider the wrench in Ex. 6-3, Fig. 6-24, as made of cast
iron, machined to dimension.The force F required tofracture this part can beregarded as the strength of
Slide #Chapter 16: Failure Resulting from Static Loading
2929
.
material is ASTM grade 30 castiron, find the force Fwith
a) Coulomb-Mohr failuremodel
b) Mod. I-Mohr failure modelc) Mod. II-Mohr failure model
Figure. 6-24
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Example 6-5(Contd)
Assumptions:1. Lever DC strong enough
and is not part of theproblem.
2. Since Grade Cast Iron is abrittle material and cast
Slide #Chapter 16: Failure Resulting from Static Loading
3030
iron, the stressconcentration factorsand are set to unity.
3. Table A-24 =
Figure. 6-24
tK
tsK
31109
KpsiKpsi
ut
uc
SS
=
=
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Example 6-5(Contd)
Assumptions:
4. Stresses on element at A
Tensile bending stress
Torsional stress.
Slide #Chapter 16: Failure Resulting from Static Loading
3131
5. The location at A is theweakest location and itgoverns the strength ofthe assembly.
Figure. 6-24
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Example 6-5 (Contd)
( ) ( )
( )
( )3332 1432
1 142.61x t t
FM M
K K FI C d
= = = =
Shear stress
Normal stress
Slide #Chapter 16: Failure Resulting from Static Loading
3232
( ) ( )( )
33
16 1516 1 76.41
xy ts tsFT r TK K F
J d
= = = =
1 22
2,
2 2
x y x y
A B xy
+ = +
Non-zero principal stresses
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Example 6-5 (Contd)
Non-zero principal stresses
( )
1 22
142.6 0 142.6 0
, 76.42 2
175.8
A B
A
F F
F
F
+ = +
=
Slide #Chapter 16: Failure Resulting from Static Loading
3333
.B =
Case II
It locates the principal stresses in
the fourth quadrant of theplane ( )
,A B
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Example 6-5
(Contd)
a) Brittle Coulomb-Mohr (BCM) Theory
10A B A B
nut uc
S S
=
Slide # 3434Chapter 16: Failure Resulting from Static Loading
34
FourthQuadrant( )
( )( )3 3
32.2175.8 131 10 109 10
FF =
Solving for F yields
167lbfF =
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Example 6-5
(Contd)
b) Modified I MohrTheory
The slope of the load line is
32.2
Slide # 3535Chapter 16: Failure Resulting from Static Loading
35
( )3175.8
131 10
A Fut
S
= =
FourthQuadrant
176lbfF =
.
175.8B A
= =
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