Dr. Curran Crawford

Department of Mechanical Engineering

University of Victoria

MECH 360 Design of Mechanical Elements

Solid Mechanics Review

Lecture #2

2

Lecture Outline

•Today

–Questions from last time?

• Course structure, basic stress analysis, etc

–Review examples (Chpt 4)

–Tutorial this week

–Project announcement next week

–Assignment 1 posted today

•Next time

–Static failure theories (Chpt 5)

– Intro to gearboxes

Carabiner testing

• Where might these fail?

– Why?

– http://www.youtube.com/watch?v=13poPfa8Zso&f

eature=related

3

4

Translate forces and moments to the cross-

section of interest

5

The point of maximum stress is not always

obvious

• Frequently have to check multiple locations

• Develop engineering intuition to pick points

–Changing cross-section

–Stress-risers

–Maximum internal loads

–Points around circumference

Coordinate systems and associated

stresses are arbitrary

• However, failure modes are not

6

7

The maximum stress at a point is usually

the ultimate objective of a stress analysis

• Require both shear and normal stresses

–Principle stresses: zero shear stress

–Maximum shear: associated normal stress

–General 3D problem

•Transformation equations in 3D & 2D

–Frequently concerned with 2D problem on surface of

a part

• Mohr’s circle

–Simply graphical representation of transformation

equations

8

The elemental cube is used to define stress

at a point in the material

𝜏𝑥𝑦=𝜏𝑦𝑥 , 𝑒𝑡𝑐.

9

The 2D stress transformation problem

frequently arises

10

Even for a 2D loading problem, the 3D

problem must be considered

For element!

11

This information should not be new!

• Statics & dynamics courses

–Loading

• Mechanics of solids courses

–Stress/strain analysis

• Materials courses

–Properties of different material compositions

12

We will be using material from previous

courses to cover new topics

• Static engineering analysis

–Compute loads

–Determine where to apply formulas

•This is key: we don’t have time to analyse every detail

• Fatigue failure analysis

–Load and stress determinations are identical to the

static case

–Fluctuating components must use different failure

theories

13

We will be using material from previous

courses to cover new topics

• Analysis of specific types of mechanical

elements

–Gears

–Bearings

–Fasteners

–Welds

–Etc.

Engine Gearbox(1:10 ratio)

Prop

•Engine: 3000 Hp @ 20000 RPM

•Engine & prop rotate CW as viewed

from prop

•Gearbox bolted to engine housing

•Neglect friction losses in gearbox

Find:

1. Dir. & mag. of torque applied to engine housing by

gearbox housing

2. Dir. & mag. of torque applied to aircraft3. Why use 2 props?

16

How do we ensure our parts will perform

their required function in service?

• Failure prevention analysis

17

Failure prevention is assured by following

an adaptive analysis process

•No prescribed method will work in all cases

•Generic questions to address:

–Which location(s) on the part will fail first?

–What are the consequences of failure there?

–What is the stress there (and associated load) at

which failure will occur?

–What is the actual maximum stress there (and

associated load) that will occur during service?

– Is the failure stress (load) sufficiently higher than the

actual stress (load)?

18

A real part presents many possible failure

locations for consideration

19

Even relatively simple geometry/loading

may have more than one failure location

•Where might failure occur?

20

The consequences of failure are an

important consideration

•“Failure of a loaded member can be regarded as any

behavior that renders it unsuitable for it’s intended

function”

•Failure may include

breakage or excessive

distortion/strain

Your bike handlebar

747 control yoke

VS

21

What stress is developed at failure?

•What is a critical stress?

–Ultimate tensile stress?

–Maximum shear stress?

–Tensile or torsion yield stress?

–Something else?

•Hint: this is why you’re here!

•The type of loading affects the stress at failure

–Static or steady loading

– Impulse, impact, or shock loading

–Variable loading

•What loading creates that stress?

22

Predicting the failure stress is one of two

major keys to failure analysis

•It’s not easy!

•Standard mechanical properties derived from:

–Particular specimen geometries

–Specific loading conditions

–Controlled experimental conditions

•In service, everything changes

–Part geometry

–Type of loading

–Environment

–Surface finish

–Etc…

23

There are two possible methods to

determine the failure stress

•Experimental

–Test an actual part under actual operating

conditions

–Good for final verification, bad for design

•Analytic

–Modify standard test data for specific application

–A lot less costly!

–At least do this step as a pre-cursor to testing

24

The other key to failure analysis is

accurately predicting the real loads

•How do you determine the “real” loads?

– Instrument a model or real part

•Strain gauges, accelerometers, etc.

–Analysis

•Dynamic simulation

•Statically indeterminate?

– Solid mechanics

– FEA

•The loads are frequently modified as the part

design is modified

–Changes in mass, stiffness, etc.

25

Do you feel lucky, punk?

•We must account for analysis uncertainty

•We can define a safety factor (SF):

•The book uses N as a symbol for SF

–Also called Factor of Safety (FS, FoS)

–Margin of Safety (MoS) = FoS - 1

•“Quantity” can be stress, load, stiffness, etc.

•If you are confused, remember that N should

always be > 1

SF = N = Predicted quantity at failure

Predicted maximum in-service quantity

26

Factor of Safety example

•Consider a steel cylindrical rod in tension

–Define failure as the onset of yielding

–The factor of safety could then be defined as:

where

N = SyF =A

N = safety factor

Sy = yield st rength

F = tensile load

A = cross-sect ional area

27

The magnitude of the safety factor will

depend on the application

•In general, use a higher FS to reflect:

–Uncertainty in: material properties, loading conditions

–Criticality: potential threat to life & limb

–Design refinement: e.g. weight-critical

•Some industries have standards or established

practices

28

In the absence of standards, there are

general guidelines for selecting FS

•SF = 1.25 — 1.5

–Exceptionally reliable materials

–Controllable conditions

–Subjected to loads and stresses that can be

determined with certainty

–Almost always used when low weight is a particularly

important consideration

–Reduce uncertainly and use lower SF by more

detailed testing

29

In the absence of standards, there are

general guidelines for selecting FS

•SF = 1.5 — 2

–Well-known materials

–Reasonably constant environmental conditions

–Subject to loads and stresses that can be

determined readily

30

In the absence of standards, there are general

guidelines for selecting FS

•SF = 2 — 2.5

–Average materials

–Operated in ordinary environments

–Subjected to loads and stresses that can be

determined

31

In the absence of standards, there are

general guidelines for selecting FS

•SF = 2.5 — 3

–Less tried or brittle materials

–Average conditions of environment, load, and stress

•SF = 3 — 4

–Untried materials

–Used under average conditions of environment,

load, and stress

Or

–Better known materials

–Used in uncertain environments or subjected to

uncertain loads and stresses

We need criteria for quantifying the

maximum allowable stresses

32

33

There are essentially 3 basic failure

“classes” to be aware of

•Static loading

–Most simple case

–Ductile vs. brittle behaviour

–Ductile → brittle transition

•Dynamic failure

–Fatigue

•Material imperfections

–May occur in any type of material

–Results from micro-cracks, inclusions, flaws, etc.

creating stress concentrations

–Fracture mechanics approach

34

A large number of static loading failure

theories have been developed

•Ductile

– Maximum shear stress

– Distortion energy (Von Mises

stress)

– Total strain-energy*

•Brittle

– Maximum normal stress

– Maximum normal strain*

– Coulomb-Mohr*

– Modified Mohr*

*Not Covered in Mech 360

35

Ductile materials yield significantly before

failure; brittle materials do not

Ductile >5% elongation at failure Brittle

36

Don’t confuse strength and stress

•Strength refers to the properties of the material

•Stress refers to the stress state due to actual

loading

•Safety factor compares the two

S

σ

Testing of test articles

Analysis of actual part

SF = N = 𝑆

𝜎

37

Ductile failure is usually defined by yield

strength, not ultimate strength

•Yield strength usually much less than ultimate

strength

•Plastic deformation after yield

–Non-linear deformations

–More complicated elasto-plastic stress distributions

•Ductile failure mode is safer

–Can usually visually observe deformation

–Avoid catastrophic brittle failure

–Usually post-examining a brittle (unexpected) failure,

as ductile failures usually give warning

38

Ductile failure occurs along planes of

maximum shear stress

•Failure of ductile materials is generally

controlled by shear strength

•How does Mohr’s circle explain this picture?

39

Brittle materials fail along planes of

maximum normal stress

•Failure of brittle materials is controlled by tensile

strength

•How does Mohr’s circle explain this picture?

40

Tensile specimens have characteristic

fracture surfaces1) Crack initiation and growth by coalescence of microvoids

2) Final failure along maximum shear plane

Brittle fracture at inter or transgranular surfaces

41

Ductile failure occurs in stages

•Plastic deformation

–Necking

•Microvoid formation

–Local plastic deformations

to relieve stresses

–Microvoids join up

•Final failure along

maximum shear planes

•Note difference between

plastic deformations and

final failure in shear

42

Cup-and-cone elongation & crack

initiation at center of specimen

43

Real parts behave in a similar manner to

tensile specimens

•Material composition

doesn’t tell the whole

story

•Material may be brittle

or ductile

–Cold-working

–Heat treatment

44

The failure mechanism is independent of

the loading

•Failure in shear of ductile materials

•How does Mohr’s circle explain failure in torsion

specimen?

Material fails along slip planesShear

stress is critical

45

The failure mechanism is independent of

the loading

•Failure by normal stresses of brittle materials

•How does Mohr’s circle explain failure in torsion

specimen?

46

The failure mode can be deduced by the

pieces left behind

•Ductile failure

– One piece

– Large deformations

– Preferred mode

– Most metals

•Brittle failure

– Many pieces

– Small deformations

– E.g. Ceramics

– Temperature dependence

47

Fracture surfaces tell the story of failure to

the trained eye

Chevron pattern pointing to origin of brittle fracture

48

Fracture surfaces tell the story of failure to

the trained eye

Crack initiation site

49

Fracture surfaces tell the story of failure to

the trained eye

50

Fracture surfaces tell the story of failure to

the trained eye

Radial marks

51

Failure theories derive from a fundamental

premise relating test articles and parts

•The mechanism of failure is always the same,

for either ductile or brittle material

•Therefore:

– If a certain set of conditions is responsible for failure in

the test specimen, then when this set of conditions

occurs in a part, the part will fail

OR

–Whatever is responsible for failure in the standard

tensile test will also be responsible for failure under all

other conditions of static loading

52

Brittle materials fail due to normal stresses

•From experimental evidence:

–Brittle material in a standard tensile test fails when

maximum normal stress exceeds a critical value

–Therefore we might reasonably expect components

made of that brittle material to fail when the

maximum normal stress in the component exceeds

that same critical value

53

The Maximum Normal Stress Theory is the

most simple brittle failure theory

•Failure will occur when:

–One of the principal stresses exceeds the ultimate

tensile strength (or ultimate compressive strength) of

the material

•Reasonable correlation for brittle materials but

not for ductile materials

–Conceptually simple, but often wrong!

•For brittle materials stronger in compression

than tension, should use alternate theories:

–Coulomb-Mohr

–Modified Mohr

54

Always remember the third principal stress!

Largest → Smallest

One principal stress is zero

Define as 2 non-zero principal stresses

𝜎1; 𝜎2; 𝜎3

𝜎𝐴; 𝜎𝐵

55

Labeling A & B instead of 1, 2, 3 avoids

ambiguity

*Note that the book uses 1, 2, 3, usually with 2 as zero principal stress

Maximum Normal Stress

Theory

56

The failure envelope defines the range of

safe principle stresses

Non-conservative!(¾B )

(¾A )

Stronger in

compression

57

Ductile materials ultimately fail in shear

•From experimental evidence:

–Ductile material in a standard tensile test fails when

maximum shear stress exceeds a critical value

–Therefore we might reasonably expect components

made of that ductile material to fail when the

maximum shear stress in the component exceeds

that same critical value

•Ductile failure is defined by yield, not ultimate

strength

–Recall complex nature of final failure

–We presuppose yielding is also characterized by

shear properties (to start with)

58

The Maximum Shear Stress theory is

reasonable for ductile yielding

•Note that the failure criterion for ductile

materials is yield, not rupture: 𝜎𝑚𝑎𝑥 < 𝑆𝑦;𝑠ℎ𝑒𝑎𝑟 = 0.5𝑆𝑦

𝜎𝐵 𝜎𝐴

𝜎𝐴𝜎𝐵

Normal

ShearPure

Torsion

59

Safety factors can be viewed as % along

line from origin to failure envelope

*Or just use the formulas!

60

Mohr’s circle or the equations will yield the

same answer, if applied correctly

•You have a choice of method

–Use whichever you are more comfortable with

–The graphical method shows directions and rotations

of the element

•Not always useful, but frequently is

–We’re usually after stress magnitudes, rather than

principle directions/axes

•Just remember to check all 3 principle stresses

to find maximum shear

•Check course website for Mohr’s circle tool

End of Lecture 2

•Questions?

61

62

Deflection may be a design criteria, in

addition to stress allowables

•Axial deflection

𝛿 =𝑃𝐿

𝐴𝐸

•Torsion

𝜃 =𝑇𝑙

𝐽𝐺

•Beam deflection

wE I= d4 y

dx 4

63

Beams have some additional methods:

graphical integration

64

Beams have some additional methods:

singularity functions

Find (qualitatively):

1. FBD of shaft AB

2. Shear and bending moment diagrams

3. Draw stresses on infinitesimal elements

on shaft AB @ C, at points on the surface tangent to vertical and

horizontal planes

Motor input

Output 1 (12kW)

Output 2 (8kW)

Shaft AB transmits

20kW @ 450RPM

A B