ME366 Spring 2015 Professor Nathan Salowitz1

ME 366: Design of Machine Elements

Lecture 8: Failure

ME366 Spring 2015 Professor Nathan Salowitz2

Announcements & Reminders

HW 4 is posted and due Tuesday

ME366 Spring 2015 Professor Nathan Salowitz3

Agenda

Coulomb-Mohr

Brittle Failure

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Review

Tresca (maximum shear stress) Calculate principle and max shear stresses

Yield if:

von Mises (distortion energy)

=1

2

2+

2+

2 +

ME366 Spring 2015 Professor Nathan Salowitz5

Clarification

Why did I do von Mises twice Tuesday? Once with shear only, once with normal stress only?

Example

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Sty Scy in Some Materials

Grey cast iron

Sty Scy

Magnesium Alloys

Scy Sty

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Mohr Theory

Perform 3 tests

Uniaxial tension

Uniaxial compression

Pure shear

Plot Mohr's circles

Create failureenvelopes (BCD)

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Coulomb-Mohr Theory for Ductile Materials

Linearize between uniaxial tension & compression

Define a point O

Similar triangles define

Bs

Cs

Derivation

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Coulomb-Mohr Ductile Criteria

Yield when:1

3

1

Ultimate failure: 1

3

1

When Syt = Syc = Sy: 1 3

Can calculate =

+

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Summary of Ductile Failure Theories

Tresca (maximum shear stress)

Conservative

Principle stress based (remember out of plane/0 stress)

Use Ssy if you have it

von Mises (maximum distortion energy)

More accurate

Based on Sy

Coulomb-Mohr

For materials with differing tensile and compressive failure

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Brittle Materials

Normal failure

More common to have Sut Suc Brittle materials may not yield so ultimate

properties are common

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Maximum Normal Stress Theory

1 2 3 Failure when:

1 Sut Or

3 Suc

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Brittle Coulomb-Mohr

1 2 3 A B & 0

Failure predicted when:

A B 0

1 A 0 B

0 A B

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Modified Mohr

Recall: =

+similarly: =

+

Modify shear criteria yield when

A B 0

()

1 A 0 B

0 A B

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Brittle Failure Summary

Maximum normal stress

Simple

Not very accurate

Brittle Coulomb Mohr Conservative

Modified Mohr

Improved precision

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Selecting a Theory

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Example

A material is uniaxially tested to failure finding : Sut = 14 MPa and Suc = 120 MPa

The critical element of a part made of this material is in plane stress with x = 0 y = -18 MPa, and xy = 20 MPa

Is failure predicted? What would be the factor of safety

Example

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Fracture Mechanics

Cracks exist in parts from manufacture & grow during service

Damage tolerant design: Parts can function with cracks/damage up to a certain size

Design such that damage can be detected before failure and part fixed or replaced

Modeled with Linear Elastic Fracture Mechanics

ME366 Spring 2015 Professor Nathan Salowitz19

Flaw Models

Stress concentration factors

Require knowledge of geometry

Radius of curvature of stress raiser critical

Crack tip approaches 0

Concentration factor approaches infinity

Plastic deformation can compensate

Linear elastic models not applicable

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Fracture

Relatively brittle materials: fracture without yielding occurring throughout the fractured cross section.

Glass, hard steels, strong aluminum alloys

Ductile materials: yield at predictable loading states

Ductile materials will blunt sharp cracks

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Fracture Energy

Cracking is fast but not instantaneous

Time is necessary to feed energy form the stress field into crack energy

Crack growth occurs when energy release from applied loading is greater than energy for crack growth

Unstable when:

rate of change of energy release rate relative to crack length > rate of change of crack growth energy

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Crack Modes

m

Opening TearingSliding

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Mode 1 Analysis

Stress field on dx dy element at crack tip

=

2cos

21

2

3

2

=

2cos

21 +

2

3

2

=

2

2

2

3

2

= 0 plane stress

= ( + )plane stress

ME366 Spring 2015 Professor Nathan Salowitz24

Stress Intensity Factor

= Units: MPa or psi

For mode 1 crack

1 =

Rewrite:

=1

2cos

21

2

3

2

=1

2cos

21 +

2

3

2

=1

2

2

2

3

2

= 0 plane stress

= ( + )plane stress

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Stress intensity factor

=

is the stress intensity modification factor

Tabulated for basic geometries

Critical stress intensity factor / fracture toughness KIc is a material property

Dependent on temperature

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Coming Up

Fatigue

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Homework #4 Due 2/24/2015 in class

Reading

Chapter 3 & 4

Homework Assignment

3-81 (20 points)

3-100 (10 points)

3-122 (10 points)

4-17 (10 points)

4-104 (10 points)

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