Failure Theories, Static Loads

Ken Youssefi Mechanical Engineering Dept., SJSU 1

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 (fracture)

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


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.


Failure Theories Ductile Materials

Maximum shear stress theory (Tresca 1886)

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

ductile material

= Sy

Sy

2 =

= Sy

=Sy

(max )component > ( )obtained from a tension test at the yield point Failure

(max )component < Sy

2

To avoid failure

max = Sy

2 n n = Safety factor

Design equation



Distortion energy theory (von Mises-Hencky)

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

Hydrostatic state of stress (Sy)h

h

h

h



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

Strain

energy

U =

3D case

UT = 11 + 22 + 33

1 = 1 E

2 E

3 E

v v

2 = 2 E

1 E

3 E

v v

3 = 3 1

E

2 E

v v

Stress-strain relationship

E

UT = (12 + 2

2 + 32) - 2v (12 + 13 + 23)

2E

1



Ud = UT Uh

Distortion strain energy = total strain energy hydrostatic strain energy

Substitute 1 = 2 = 3 = h

Uh = (h2 + h

2 + h2) - 2v (hh + hh+ hh)

2E

1

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

Uh = (1 2v) = 2E

3h2

6E

(1 2v) (1 + 2 + 3)2

UT = (12 + 2

2 + 32) - 2v (12 + 13 + 23)

2E

1 (1)

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

(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

< Sy

Ud = UT Uh = 6E

1 + v (1 2)

2 + (1 3)2 + (2 3)

2 (2)



2D case, 3 = 0

= (12 12 + 2

2) < Sy Where is von Mises stress

= Sy

n Design equation

(1 2)2 + (1 3)

2 + (2 3)2

2

< Sy


Pure torsion, = 1 = 2

(12 2 1 + 2

2) = Sy2


32 = Sy

2 Sys = Sy / 3 Sys = .577 Sy

Relationship between yield strength in

tension and shear

(x)2 + 3(xy)

2 =

Sy

n

1/2

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)


Design Process

= Sy

n max =

Sy 2n

Maximum shear stress theory Distortion energy theory

consider strength, environment, cost, weight,

availability, codes and standards, manufacturing process 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

n Cost Weight Size

Select material:


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 - Static load & Ductile material

Formulate the maximum stresses in the component in terms of size: x, xy

Optimize for weight, size, or cost.

Select material Sy , Su

Choose a safety factor, n

Determine the principal stresses, maximum shear stress and von

Mises stress in terms of the size; 1, 2 , max and Use appropriate failure theory to calculate the size.

= Sy

n max =

Sy 2n

Distortion energy theory Maximum shear stress theory

Choose a cross section: round, rectangular, hollow, I-beam, C channel,


Failure Theories Brittle Materials

Characteristics of brittle materials;

Perform two tests, one in compression and one in tension, draw the

Mohrs circles for both tests.

Compression test

Suc

Failure envelope

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

1 > 2 and 3 = 0

Tension test

Sut 2 1 Stress state

2. Suc >> Sut 1. Sut Syt 3. elongation < 5%



1 Sut

Suc

Sut

Suc

Safe

Safe

Safe Safe

-Sut

Cast iron data

Modified Coulomb-Mohr theory

1

2 or 3

Sut

Sut

Suc

-Sut

I

II

III

Three design zones

2 or 3



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

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Failure Theories, Static Loads

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