Dr.. A. Az Dr Aziz iz Ba Bazou zoune ne King Fahd University of Petroleum & Minerals
Mechanical Engineering Department ME 307 MACHINE DESIGN I 1
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ME 307 MACHINE DESIGN I Slide #
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FAILURE CRITERIA
Failure Criteria (Static Load +Ductile Material) Material) 6-6
Coulomb Mohr Theory for Ductile Materials
Not all materials have their tensile strength equal to their compressive strength. Examples are Magnesium Suc = 0.5Sut uc
ut
This section deals with materials
Suc
≠
Sut
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S sy Practical difficulties to construct 3 Mohr circles for three tests. Figure 6-20 Three Mohr circles, one for the uniaxial compression test, one for the test in , the uniaxial tension test are used to define failure by the Mohr hypothesis. The compressive and tensile strengths, respectively; they can be used for yield or ultimate strength .
ME 307 MACHINE DESIGN I Slide #
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Figure 6-20 Three Mohr circles, one for the uniaxial compression test, one for the test in pure shear, and one for the uniaxial tension test are used to define failure by the Mohr hypothesis. The compressive and tensile strengths, respectively; they can be used for yield or ultimate strength ME 307 MACHINE DESIGN I Slide #
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Coulomb Mohr Theory for Ductile Materials
σ
1
≥ σ 2 ≥ σ 3
Failure will occur if Mohr’s circle (3-D) describin a state of stress in a body grows until it becomes tangent to the failure envelope.
σ
1
St
−
σ
3
Sc
=1
σ1
> 0, σ 3 ≤ 0
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Coulomb Mohr Theory for Ductile Materials Failure will occur if σ1
−
St
σ 3
Sc
=1
σ1
> 0, σ 3 ≤ 0
Failure criteria according to Coulomb-Mohr Theory
Factor of safety σ1
St
−
σ 3
Sc
=
1
n
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Special Cases: Application of CoulombCoulomb -Mohr for Plane Stress σ A ≥ σ B (Failure will occur if ☺ Case 1:
σ
A
≥ σ B ≥
⇒
0
σ A
Case 2:
σ
≥
−
St 0 ≥ σ A
≥ σ B
σ B
Sc
⇒
σ B
1
= σ A , σ 2 = σ B , σ 3 = 0
≥ S t
⇒
0 ≥ σ σ A
Case 3:
σ
≤−
=σ
σ
,σ
=
0, σ
=σ
≥1
σ
1
=
0, σ 2
= σ A , σ 3 = σ B
Sc
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Factor of safety σ1
−
St
σ 3
Sc
n
For Coulomb-Mohr Theory we do not need the torsional shear strength circle, we can deduce it from Eq.(6-22). For pure shear τ , σ 1 = − σ 3 = τ . The torsional ield stren th occurs when τ
max
=
1
=
S sy
For yielding, Eq. (6-22) gives
S sy
S
=
S
S
yt
+
yt
yc
S
yc
ME 307 MACHINE DESIGN I Slide # 11
Example 6-2 (Textbook Page 268)
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6-7 Failure of Ductile Materials Summary
This study is limited materials and parts that are known to fail in a ductile manner.
Experimental data collected for several materials from many sources are shown in Figure 6-23.
Figure 6-23 Experimental data superposed on failure theories ME 307 MACHINE DESIGN I Slide # 14
6-7 Failure of Ductile Materials Summary
Figure 6-23 shows that either the Maximum Shear Stress Theory (MSS) or the Distortion Energy Theory (DE) is acceptable for design and analysis of materials that would fail in a ductile manner.
that you, the engineer, MUST DECIDE.
ME 307 MACHINE DESIGN I Slide # 15
6-7 Failure of Ductile Materials Summary
For Design Purposes: Maximum Shear Stress (MSS) Theory is easy quick to use conservative
If the problem is to learn why a part failed, then the Distortion Energy (DE) Theory may be best to use. Figure 6-23 shows that the locus of the DE passes closer to the central area of the data points, and thus is generally a better predictor of failure.
ME 307 MACHINE DESIGN I Slide # 16
6-7 Failure of Ductile Materials Summary
For ductile materials with S yt≠ S yc , the Mohr Theory is the best available. However, the theory requires the results from three separate modes of tests, graphical construction of the failure locus, and fitting the largest Mohr’s circle to the failure locus.
Coulomb-Mohr Theory (CM) which requires only tensile and compressive yield strengths and represents an alternative to Mohr Theory. It is easily dealt with in equation form.
ME 307 MACHINE DESIGN I Slide # 17
Example 6-3 A certain force F applied at D near the end of the 15-in lever shown in Fig. 6-24, which is quite similar to a socket wrench, results in certain stresses in the cantilevered bar OABC. The bar OABC is of AISI 1035 steel, forged minimum (ASTM) yield strength of 81 Kpsi. We presume that this component would be of no value after yielding. Thus the force F required to initiate yielding can be regarded as the strength of the component part. Find this force.
Figure. 6-24
ME 307 MACHINE DESIGN I Slide # 18
Example 6-3 (Cont.’ d) Assuming lever DC strong enough
σ x
τ zx
=
=
( d2 ) McM (π
I =
Tr
4
64 ) d
T ( d 2)
=
J
π
32
=
=
d
32 π
16T π
3
=
d =
d
32 ) M(14 F
π
(1)
3
16 (15 F )
π
=
(1)
= 142.6 F
76.4 F
Using the DE theory, we find, from Eq. (6-15) 12
(142.6 F ) + 3 × ( 76.4 F ) Equating the Von Mises stress to S , we solve for F and get y σ
'
=
(
σ
2
2
+ x 3τ
12
)
zx
F =
2
=
S y 194.5
=
81,000 194.5
2
=
= 194.5F
416 lbf
ME 307 MACHINE DESIGN I Slide # 19
Example 6-3 (Cont.’ d) In this example the strength of the material point A is S y = 81 kpsi . The strength of the assembly or component is F = 416 lbf . Using the MMS Theory, for a point undergoing plane stress with only one non-zero normal stress and one shear stress, the two nonzero principal stressesσ and σ A B will have opposite signs and hence fit case 2 for the MSS Theory. From Eq. (4-13) 12
σ
− A σ
= B
2
σ x
2
+τ
2
zx
=
(
σ
2
+ x
4τ
2
12
)
zx
case 2 for the MSS Theory, Eq. (6-5) applies and hence
(
σ
2
x+
2
4τ
12
)
zx
=
S
y 12
(142.6 F )2 + 4 ( 76.4 F ) 2
=
209 F = 81, 000
F = 388 lbf ME 307 MACHINE DESIGN I Slide # 20
Example 6-4 (Textbook Page 271) The cantilever tube shown in Fig. 6-25 is to be made of 2014 aluminum alloy treated to obtain a specified minimum yield strength of 276 MPa. We wish to select a stock-size tu e rom a e - us ng a design factor nd = 4 . The bending load is F=1.75 kN, the axial tension is P = 9.0 kN, and the torsion is T = 72 N.m. What is the realized factor of safety?
Figure 6-25 ME 307 MACHINE DESIGN I Slide # 21
Example 6-4 (Textbook Page 271) SOLUTION The normal stress for an element on the top surface of the tube at the origin is
σ x
=
P A
+
Mc
=
I
9
+
120 (1.75 )( d o 2 )
A
=
9
I
+
A
105do
I
(1)
The torsional stress at the same point is
τ zx
=
Tr J
=
72 ( do 2 )
J
=
36 ( d o )
J
(2)
ME 307 MACHINE DESIGN I Slide # 22
Example 6-4 (Cont.’d) For accuracy, we choose the DE as the design basis. The Von Mises stress is
'=
σ
(
σ
2
2
+ 3τ x
12
)
(3)
zx
On the basis of the given design factor, the goal forσ '
σ '
≤
S y
=
0.276 4
nd
=
is (4)
0.0690 GPa
Programming Using MATLAB or EXCEL , entering metric sizes from Table A-8 reveals that a 42-X 5- mm is satisfactory. The Von Mises stress is found to be σ ' = 0.06043 GPa for this size. Thus the realized factor of safety is
n=
S y σ '
=
0.276 0.06043
=
4.57
ME 307 MACHINE DESIGN I Slide # 23
