5 failure theories 2015 bach khoa

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Failure – A part is permanently distorted (bị bóp méo) 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 = 0.67 Su

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

• Maximum shear stress theory (Tresca 1886)

(Thuyết bền: ứng suất tiếp lớn nhất – TB3)

Yield strength of a material is used to design components made of

ductile material

 = Sy

 = Sy

 =Sy

( max )component < Sy

2

To avoid failure

max =

Sy

2 n

n = Safety factor

Design equation

2

=



SPECIAL CASES

A special planar state of stress

τxy= τ

Purely shear state of stress

τxy= τ

 

n

S y

TB

2 2

4 2

2

3       



 

n

S y

2

2 



• Distortion energy theory (von Mises-Hencky)

(Thuyết bền: Thế năng biến đổi hình dáng lớn nhất – TB4)

Hydrostatic state of stress → (Sy)h

h

h

h

t

t

Simple tension test → (Sy)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

yield point → failure

The area under the curve in the elastic region is called the Elastic Strain Energy

U = ½ ε

3D case

UT = ½ 1ε1 + ½ 2ε2 + ½ 3ε3

ε1 = 1

E

2

E

3

E

v

v

ε2 = 2

E

1

E

3

E

v

v

ε3 = 3 1

E

2

E

v

v

Stress-strain relationship

E

UT = (12

+ 22

+ 32

) - 2v (12 + 13 + 23) 2E

1

Ud = UT – Uh

Distortion strain energy = total strain energy – hydrostatic strain energy

Substitute 1 = 2 = 3 = h

Uh = (h2

+ h2

+ h2

) - 2v (hh + hh+ hh) 2E

1

Simplify and substitute 1 + 2 + 3 = 3h into the above equation

Uh = (1 – 2v) =

2E

3h2

6E

(1 – 2v)

(1 + 2 + 3)2

Ud = UT – Uh =

6E

1 + v (1 – 2)2 + (1 – 3)2 + (2 – 3)2

Subtract the hydrostatic strain energy from the total energy to

obtain the distortion energy

UT = (12

+ 22

+ 32

) - 2v (12 + 13 + 23) 2E

1

(1)

(2)

Strain energy from a tension test at the yield point

1= Sy and 2 = 3 = 0 Substitute in equation (2)

3E

1 + v

(Sy)2

Utest =

To avoid failure, Ud < Utest

(1 – 2)2 + (1 – 3)2 + (2 – 3)2

2

½

< Sy

Ud = UT – Uh =

6E

1 + v (1 – 2)2 + (1 – 3)2 + (2 – 3)2

½

2D case, 3 = 0

(12

– 12 + 22

) < Sy = 

Where  is von Mises stress

′ = Sy

n Design equation

(1 – 2)2 + (1 – 3)2 + (2 – 3)2

2

½

< Sy

Pure torsion,  = 1 = – 2

(12

– 2 1 + 22

) = Sy2

32

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)

τxy= τ

τxy= τ

 

n

Sy

TB

3 3

3 12

2 1

4       



 

n

Sy

3

3 



or

′ =

Sy

Sy

2n

• Select material: consider environment, density, availability → Sy , Su

• Choose a safety factor

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

 Type of manufacturing process

 Uncertainties related to stress analysis

 Consequence of failure; human safety and economics

 Codes and standards

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

Design Process

• Formulate the von Mises or maximum shear stress in terms of size

• Optimize for weight, size, or cost

• Select material, consider environment, density, availability → Sy , Su

• Choose a safety factor

• Use appropriate failure theory to calculate the size

′ =

Sy

Sy

2n

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

Compression test

Suc

Failure envelope

The component is safe if the state of stress falls inside the failure envelope

1 > 3 and 2 = 0

Tension test





Sut

state

Sut

Suc

Sut

Suc

Safe

Safe

Cast iron data

Modified Coulomb-Mohr theory (Thuyết Bền Morh – TB5)

1

3 or2

Sut

Sut

Suc

-Sut

I

II

III

Three design zones

3 or2

Modified Coulomb-Mohr theory

1

2

Sut

Sut

Suc

-Sut

I

II III

Zone I

1 > 0 , 2 > 0 and 1 > 2

Zone II

1 > 0 , 2 < 0 and 2 < Sut

Zone III

1 > 0 , 2 < 0 and 2 > Sut 1 ( 1

Sut

1

Suc

– ) – 2

Suc =

1

n

Design equation

1 = Sut

n Design equation

1 = Sut

n Design equation