admission in india for students

1. Admission in India
By:
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2. Failure Theories – Static Loads
Static load – a stationary load that is gradually applied having an unchanging
magnitude and direction
Failure – A part is permanently distorted and will not function properly.
A part has been separated into two or more pieces.
Material Strength
Sy = Yield strength in tension, Syt = Syc
Sys = Yield strength in shear
Su = Ultimate strength in tension, Sut
Suc = Ultimate strength in compression
Sus = Ultimate strength in shear = .67 Su
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Ken Youssefi Mechanical Engineering Dept., SJSU 2

3. Ductile and Brittle Materials
A ductile material deforms significantly before fracturing. Ductility is measured
by % elongation at the fracture point. Materials with 5% or more elongation are
considered ductile.
Brittle material yields very little before fracturing, the
yield strength is approximately the same as the
ultimate strength in tension. The ultimate strength in
compression is much larger than the ultimate
strength in tension.
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Ken Youssefi Mechanical Engineering Dept., SJSU 3

4. Failure Theories – Ductile Materials
Yield strength of a material is used to design components made of
ductile material
• Maximum shear stress theory (Tresca 1886)
(max )component > ( )obtained from a tension test at the yield point Failure

=
 = Sy
Sy
2
 = Sy
admission.edhole.com Design equation
 =Sy
To avoid failure
(max )component <
Sy
2
max =
Sy
2 n
n = Safety
factor
Ken Youssefi Mechanical Engineering Dept., SJSU 4

5. Failure Theories – Ductile Materials
• Distortion energy theory (von Mises-Hencky)
Hydrostatic state of stress → (Sy)h
h
h
h
Simple tension test → (Sy)t
t
t
(Sy)t (Sy)h >>
Distortion contributes to
failure much more than
change in volume.
(total strain energy) – (strain energy due to hydrostatic stress) = strain energy
due to angular distortion > strain energy obtained from a tension test at the
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Ken Youssefi Mechanical Engineering Dept., SJSU 5

6. Failure Theories – Ductile Materials
The area under the curve in the elastic region is called the Elastic Strain Energy.
Strain
energy
U = ½ ε
3D case
UT = ½ 1ε1 + ½ 2ε2 + ½ 3ε3
Stress-strain relationship
ε1 =
1
E
2
E
3
E
v v
ε2 =
2
E
1
E
3
E
v v
ε3 =
3 1
v v
E
2
E
E
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UT = (1
2 + 2
2 + 3
2) - 2v (12 + 13 + 23)
2E
Ken Youssefi Mechanical Engineering Dept., SJSU 6

7. Failure Theories – Ductile Materials
Distortion strain energy = total strain energy – hydrostatic strain energy
Ud = UT – Uh
1 (1)
UT = (1
2 + 2
2 + 3
2) - 2v (12 + 13 + 23)
2E
Substitute 1 = 2 = 3 = h
1
Uh = (h
2 + h
2 + h
2) - 2v (hh + hh+ hh)
2E
Simplify and substitute 1 + 2 + 3 = 3h into the above equation
3h
2
Uh = (1 – 2v) =
2E
( (1 – 2v) 1 + 2 + 3)2
6E
Subtract the hydrostatic strain energy from the total energy to
obtain the distortion energy
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Ud = UT – Uh =
1 + v
6E
(1 – 2)2 + (1 – 3)2 + (2 – 3)2
Ken Youssefi Mechanical Engineering Dept., SJSU 7

8. Failure Theories – Ductile Materials
Strain energy from a tension test at the yield point
1= Sy and 2 = 3 = 0 Substitute in equation (2)
(1 – 2)2 + (1 – 3)2 + (2 – 3)2 (2)
1 + v
3E
(Sy)
2
1 + v
6E
Utest =
Ud = UT – Uh =
To avoid failure, Ud < Utest
(1 – 2)2 + (1 – 3)2 + (2 – 3)2
2
½
< Sy
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Ken Youssefi Mechanical Engineering Dept., SJSU 8

9. Failure Theories – Ductile Materials
(1 – 2)2 + (1 – 3)2 + (2 – 3)2
½
2D case, 3 = 0
(1
2
2 – 12 + 2
2) < Sy = 
½
< Sy
Where  is von Mises stress
′ =
Sy
n
Design equation
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Ken Youssefi Mechanical Engineering Dept., SJSU 9

10. Failure Theories – Ductile Materials
Pure torsion,  = 1 = – 2
(1
2 – 2 1 + 2
2
2) = Sy
32
2
= Sy
Sys = Sy / √ 3 → Sys = .577 Sy
Relationship between yield strength in
tension and shear
If y = 0, then 1, 2 = x/2 ± [(x)/2]2 + (xy)2
the design equation can be written in terms of the dominant
component stresses (due to bending and torsion)
(x)
2
+ 3(xy)
2
1/2
=
Sy
n
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Ken Youssefi Mechanical Engineering Dept., SJSU 10

11. Design Process
Distortion energy theory Maximum shear stress theory
′ =
Sy
n
max =
Sy
2n
• Select material: consider environment, density, availability → Sy , Su
• Choose a safety factor
n Size Weight Cost
The selection of an appropriate safety factor should be based
on the following:
 Degree of uncertainty about loading (type, magnitude and direction)
 Degree of uncertainty about material strength
 Uncertainties related to stress analysis
 Consequence of failure; human safety and economics
 Type of manufacturing process
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 Codes and standards
Ken Youssefi Mechanical Engineering Dept., SJSU 11

12. Design Process
 Use n = 1.2 to 1.5 for reliable materials subjected to
loads that can be determined with certainty.
 Use n = 1.5 to 2.5 for average materials subjected to
loads that can be determined. Also, human safety and
economics are not an issue.
 Use n = 3.0 to 4.0 for well known materials subjected to
uncertain loads.
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Ken Youssefi Mechanical Engineering Dept., SJSU 12

13. Design Process
• Select material, consider environment, density, availability → Sy , Su
• Choose a safety factor
• Formulate the von Mises or maximum shear stress in terms of size.
• Use appropriate failure theory to calculate the size.
′ =
Sy
n
max =
• Optimize for weight, size, or cost.
Sy
2n
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Ken Youssefi Mechanical Engineering Dept., SJSU 13

14. Failure Theories – Brittle Materials
One of the characteristics of a brittle material is that the ultimate strength
in compression is much larger than ultimate strength in tension.
Suc >> Sut
Mohr’s circles for compression and tension tests.
Suc
Compression test

Sut
Tension test

3 Stress 1
state
Failure envelope
The component is safe if the state of stress falls inside the failure envelope.
1 > 3 and 2 = 0
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Ken Youssefi Mechanical Engineering Dept., SJSU 14

15. Failure Theories – Brittle Materials
Modified Coulomb-Mohr theory
1
Sut
Suc
Sut
3 or 2
Safe Safe
Safe
Suc
Safe
-Sut
Cast iron data
1
3 or 2
Sut
Sut
I
II
III
Suc
-Sut
Three design zones
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Ken Youssefi Mechanical Engineering Dept., SJSU 15

16. Failure Theories – Brittle Materials
1
3
Sut
Sut
I
II
III
Suc
-Sut
Zone I
1 > 0 , 2 > 0 and 1 > 2
Zone II
1 =
Sut
n
Design equation
1 > 0 , 2 < 0 and 2 < Sut
Zone III
1 =
Sut
n
Design equation
1 > 0 , 2 < 0 and 2 > Sut 1 (
1
Sut
1
Suc
– ) –
2
Suc
=
1
n
Design equation
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Ken Youssefi Mechanical Engineering Dept., SJSU 16