Failure from static loading

TopicsQuiz 2/1/07Failures from static loading

ReadingChapter 5

HomeworkHW 3 due 2/1HW 4 due 2/8

What is Failure?

Failure – any change in a machine part which makes it unable to perform its intended function.(From Spotts M. F. and Shoup T. E.) We will normally use a yield failure criteriafor ductile materials. The ductile failure theories presented are based on yield.

Failure Theories [Note Theories]

Static failureDuctileBrittleStress concentration

RecallDuctile

Significant plastic deformation between yield and fracture

BrittleYield = fracture

Tensile Test

Linear Stress Strain Plot

Failure Theory Problem Statement

Given:Stress-strain data for simple uniaxial tension

Find:When failure occurs for general state of stress

Mohr’s Circle for Tensile Test

Static Ductile Failure

Two primary theories for static ductile failureVon Mises criterion

Maximum Distortion-energy TheoryMDE

Maximum Shear Stress criterionMSS

Static Ductile Failure

Max Shear Stress criterion Material yields (fails) when:

Factor of Safety:

( ) y

y

S

S

≥−

≥

31

max 2σσ

τ

( ) max31 2τσση yy SS

=−

=

or1)

2)

Safety Factors

Maximum Shear Stress Criteria

Static Ductile Failure

Von Mises criterion Let the Mises stress (σe, equivalent stress) be:

Then failure (yield) occurs when:

Factor of Safety:

Typically,Want a margin of error but not completely overdesigned

( ) ( ) ( )[ ]2312

322

2121 σσσσσσσ −+−+−=e

ye S≥σ

e

ySσ

η =

425.1 ≤≤ η

Comparison of MDE, MSS,MNS

Which theory to use?

Look at a plot of the principal stresses

σB vs. σAThe non-zero principal stresses

Failure occurs when the principal stresses lie outside the enclosed areaShape of area depends on the failure theoryData points are experimental resultsMSS

Slightly more conservativeEasier to calculate

MDEMore accurateIf not specified, use this one!

Hydrostatic Stress State Diagonal

Ductile failure theory example

Given:Bar is AISI 1020 hot-rolled steel

A DUCTILE materialF = 0.55 kNP = 8.0 kNT = 30 Nm

Find:Factor of safety (η)

Two areas of interest:A

Top – where max normal stress is seen (bending!)

BSide – where max shear stress is seen

Element A

Consider the types of loading we haveAxial?

Yes – due to PBending?

Recall that bending produces σand τ, depending on the element of interestYes – due to M (σ at A, τ at B)

Torsion?Yes – due to T

Element ACalculate stresses due to each loadAxial:

Bending (A has normal Stresses):

Normal

Shear:

Torsion:

22

4

4DP

DP

AP

x ππσ =

⎟⎟⎠

⎞⎜⎜⎝

⎛==

( )34

32

64

2DFL

D

DFL

IMy

x ππσ =

⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎠⎞

⎜⎝⎛

==

0=xyτ

( )34

16

32

2DT

D

DT

JTc

xz ππτ =

⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎠⎞

⎜⎝⎛

==

Element ALook at a stress elementSum up stresses due to all the loads

σx = 95.5 MPaτxz = 19.1 MPa

332324324

DFLPD

DFL

DP

x πππσ +

=+=

316DT

xz πτ =

Element A

Draw Mohr’s Circle with the stresses that we calculated

σx = 95.5 MPaτxz = 19.1 MPax at (σx, τxz)

(95.5, 19.1)y at (σy, τzx)

(σy, -τxz)(0, -19.1)

Find C

Find radius, R

( )0,8.470,205.950,

2=⎟

⎠⎞

⎜⎝⎛ −

=⎟⎟⎠

⎞⎜⎜⎝

⎛ += yxC

σσ

( ) ( ) 4.511.198.475.95 2222 =+−=+−= xzxx CR τσ

Out of Plane Maximum Shear for Biaxial State of StressMax. Shear:

Case 1σ1,2 > 0σ3 = 0

21

maxστ =

231

maxσσ

τ−

=23

maxσ

τ =

Case 2σ2,3 < 0σ1 = 0

Case 3σ1 > 0, σ3 < 0σ2 = 0

231

maxσσ

τ−

=

Element A

Find principal stresses (2D)σ1 = C + R

99.2 MPaσ2 = C - R

-3.63 MPaThink about 3-D Mohr’s Circle!This is Case #3…We want σ1 > σ2 > σ3Assign σ2 = 0 and σ3 = -3.63 MPa

No failure theory was given, so use MDE

Element A

Find the von Mises stress (σe)

Sy for our material = 331 MPaCalculate the factor of safety

( ) ( ) ( )[ ]

( ) ( ) ( )[ ]MPae

e

e

101

63.32.9963.3002.992121

222

231

232

221

=

++++−=

−+−+−=

σ

σ

σσσσσσσ

28.3101331

===e

ySσ

η For yield

22.38.102

331)( 31

==−

=σσ

η ySMSM Theory:

Element B

Consider the types of loading we haveAxial?

Yes – due to PBending?

Recall that bending produces σ and τ, depending on the element of interestYes – due to M (σ at A, τ at B)

Torsion?Yes – due to T

Element BCalculate stresses due to each loadAxial:

Bending (B has Shear Stresses due to bending): No Normal Stresses at B:Shear Stresses at B:

Use equation for round solid cross-section

Torsion:

22

4

4DP

DP

AP

x ππσ =

⎟⎟⎠

⎞⎜⎜⎝

⎛==

( )22 3

16

43

434

DF

DF

AV

IbVQ

xy ππτ =

⎟⎟⎠

⎞⎜⎜⎝

⎛===

( )34

16

32

2DT

D

DT

JTc

xy ππτ =

⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎠⎞

⎜⎝⎛

==

0=xσ

Element BLook at a stress elementSum up stresses due to all the loads

σx = 25.5 MPaτxy = 19.1 MPa

Note small contribution of shear stress due to bending

24DP

x πσ =

002.1.1916316

32 +=+=DT

DF

xy ππτ

Element BDraw Mohr’s Circle with the stresses that we calculated

σx = 25.5 MPaτxy = 19.1 MPax at (σx, τxy)

(25.5, 19.1)y at (σy, τyx)

(σy, -τxy)(0, -19.1)

Find C

Find radius

( )0,8.120,205.250,

2=⎟

⎠⎞

⎜⎝⎛ −

=⎟⎟⎠

⎞⎜⎜⎝

⎛ + yx σσ

( ) ( ) 96.221.198.125.25 2222 =+−=+−= xzxx CR τσ

Out of Plane Maximum Shear for Biaxial State of Stress

Case 1σ1,2 > 0σ3 = 0

21

maxστ =

231

maxσσ

τ−

=23

maxσ

τ =

Case 2σ2,3 < 0σ1 = 0

Case 3σ1 > 0, σ3 < 0σ2 = 0

Element B

Find principal stresses (2D)σ1 = C + R

35.8 MPaσ2 = C - R

-10.2 MPaThink about 3-D Mohr’s Circle!This is Case #3…We want σ1 > σ2 > σ3Assign σ2 = 0 and σ3 = -10.2 MPa

No failure theory was given, so again use MDE

Element B

Find the von Mises stress (σe)

Sy for our material = 331 MPaCalculate the factor of safety

( ) ( ) ( )[ ]

( ) ( ) ( )[ ]MPae

e

e

8.41

2.108.352.10008.352121

222

231

232

221

=

++++−=

−+−+−=

σ

σ

σσσσσσσ

91.78.41

331===

e

ySσ

η For yield

Example, concluded

We found the factors of safety relative to each element, A and B

A – 3.28B – 7.91

A is the limiting factor of safety

η = 3.3

Static Brittle Failure

Three primary theories for static brittle failureMaximum Normal Stress (MNS)Coulomb-Mohr Theory Modified-Mohr Theory

Mohr’s Circle for MNS

Static Brittle Failure

Maximum Normal Stress (MNS)Oldest failure hypothesis, attributed to RankineFailure occurs whenever one of the three principal stresses equals the yield strength

Say σ1 > σ2 > σ3 (our convention)Failure occurs when either

σ1 = St or σ3 = -Sc

Note – brittle materials have both a tensile and compressive strengthη = St / σ1 or η = -Sc / σ3

Plot of σB vs. σA

Static Brittle FailureCoulomb-Mohr Theory (AKA Internal Friction) Sut

Suc

Suc

Sut

Stress Region

Mohr’s Circle

Failure Factor of Safety

σA,B> 0

σA> 0, σB < 0

σA,B ≤ 0

utA S≥σ

1≥−uc

B

ut

A

SSσσ

ucB S≥σ

1ση utS

=

uc

B

ut

A

SSσσ

η−=

1

B

ucSσ

η =Use table, or look at load line…

The Load Line – Coulomb-Mohr

σB = rσA Equation of line from origin to point (σA, σB)Then,

Note: “Strength” of a part can be considered the stress necessary tocause failure. To find a part’s strength at onset of failure, use η = 1.

( )utuc

utucA rSS

SS−

=ησ

Modified Mohr Failure Theory

Static Brittle FailureModified Mohr Theory

Stress Region

Mohr’s Circle

Failure Factor of Safety

σA,B >0

σA,B≤ 0

σA > 0 -Sut < σB

σB < 0

σA > 0σB < -Sut

SeeEquation

A

SeeEquation

B

utA S≥σ

ucB S≥σ

A

utSσ

η =

B

ucSσ

η =

|σB| = Sut

Sut

Sut

utA S≥σA

utSσ

η =

( ) ButAutuc

utuc

SSSSS

σση

−−=

( ) 1≤−− ButAutuc

utuc

SSSSS

σσA: B:

Which to use? (C-M or Mod-M)

In general, Mod-M is more accurate

The Load Line – Modified Mohr

σB = rσA Equation of line from origin to point (σA, σB)Then,

Note: Again, “strength” of a part can be considered the stress necessary tocause failure. To find a part’s strength at onset of failure, use η = 1.

For both Coulomb-Mohr and Modified Mohr, you can use either the table equations or the load line equations.

( )( )utuc

utucA SrS

SS+−

=1

ησ

Brittle Failure example

Given: Shaft of ASTM G25 cast iron subject to loading shownFrom Table A-24

Sut = 26 kpsiSuc = 97 kpsi

Find: For a factor of safety of η = 2.8, what should the diameter of the shaft (d) be?

Brittle Failure example

First, we need to find the forces acting on the shaft

Torque on shaft from pulley at BTB = (300-50)(4) = 1000 in·lb

Torque on shaft from pulley at CTC = (360-27)(3) = 1000 in·lb

Shaft is in static equilibriumNote that shaft is free to move along the x-axis (bearings)

Draw a FBD Reaction forces at points of attachment to show constrained motion

Equilibrium

Use statics to solve for reactions forcesRAy = 222 lbRAz = 106 lbRDy = 127 lbRDz = 281 lb

OK, now we know all the forces. The problem gives us a factor of safety, but unlike our last example, we aren’t told specific places (elements) at which to look for failure!We are going to have to calculate stresses What do we need?

Axial forces, bending moments, and torquesWe need to find our moments… HOW?Shear-Moment diagrams will give us the forces and moments along the shaft.Failure will likely occur where the max values are seen

Torsion and moment diagrams

Let’s look at torsion and how it varies across the shaft

We calculated the torques at B and C to be 1000 in·lbeachPlot that along the shaft and we see that max torque occurs at B and C (and all points between)

Torsion and moment diagrams

Now let’s look at the momentsWe have a 3-D loading

How are we going to do the V-M diagrams?Look at one plane at a time

Moment in the x-yplane

From geometry you can calculate the values of the moment at B and C

Torsion and moment diagrams

Moment in the x-z planeFailure is going to occur at either B or C, since these are locations where maximum moments are seenBut we have moments in both planesTo find the max bending stresses, we must find the total maximum momentJust as we would vectorallyadd the two force components to find the force magnitude, we can vectorally add the two moment components to find the moment magnitude

22xzxy MMM +=

We found the following:MB x-y = 1780 in·lbMB x-z = 848 in·lbMC x-y = 762 in·lbMC x-z = 1690 in·lb

Calculating the magnitudes with MB = 1971.7 in·lbMC = 1853.8 in·lb

Since the overall max moment is at B, we will expect failure there, and use MB in our stress calculationsIf we had been told the location of interest, we would essentially start here.

Calculate the max moment

22xzxy MMM +=

Calculate the stresses at B

Bending stress (σ and τ)We know from experience that σ is the predominant stress, so essentially we will look for failure at an element at the top of the shaft

M = 1971Plug in known valuesσmax = (20x103)/d3

Torsional stressT = 1000τ = (5.1x103)/d3

64

24dI

dy

IMy

π

σ

=

=

=

32

24dJ

dc

JTc

π

τ

=

=

=

Mohr’s Circle

Let’s look at our stress elementNow construct Mohr’s circle

C at (10 x 103)/d3

R = (11.2 x 103)/d3

σ1 = (21.2 x 103)/d3

σ3 = (-1.2 x 103)/d3

Use Coulomb-Mohr theory for brittle failure

If making a design recommendation, you would recommend the next largest standard dimension (16th’s)

d = 1.375 in

"32.18.21

972.1

262.21

1

33

31

=

=+

=−

ddd

SS ucut ησσ

Stress concentration

A stress concentration is any geometric discontinuity in an element that is subjected to stressAside from reducing the cross-sectional area, these stress concentrations do not significantly affect static ductile failureStress concentrations DO, however, have a significant influence on brittle failure

Analytical approach to stress concentrations

σmax = ktσnom

τmax = ktsτnom

kt, kts are stress concentration (SC) factorsσnom, τnom are nominal stresses

Nominal – those stresses that are calculated before taking the SC’s into account

SC factors are given in the text on page 1006-1012Equations for the nominal stresses (taking into account geometry change due to the SC’s) are given in the same charts

Stresses at a Hole in an Infinite Plate

Hoop Stress at a Hole in an Infinite Plate

Radial Stress at a Hole in an Infinite Plate

Stress Concentration in Ductile Material

Brittle failure example

Given:ASTM 30 cast iron

Sut = 31 ksiSuc = 109 ksi

Find:How much torque before failure with and without the stress concentration?Note – asked to find failure (not given a safety factor)Use η = 1 to find onset of failure

Brittle failure exampleWithout the SC

τ = (5.1)TσA = 5.1 T, σB = -5.1 TUse Coulomb-Mohr (easier)

44

098.03232

5.0

inDJ

cJTc

===

=

=

ππ

τ

( ) 11011.2

1101091.5

10311.5

1

4

33

≤×

≤×

−−

×

≤−

−T

TTSS uc

B

ut

A σσ

T ≤ 4730 in·lb

Brittle failure exampleWith the SC

Refer to figure A-15-15, pg. 1010Picture shows us the loading and geometryEquation is given to calculate the nominal stress considering the geometry with the SCAxis and data labels tell us the quantities we need to calculate (using the figure as a guide)

Brittle failure exampleWith the SC

d = D – 2r = 1 – 2(.025) = 0.95D/d = 1/0.95 = 1.05r/d = 0.025/0.95 = 0.026kt ~ 1.8τmax = ktτnom

TdT

JTc

nom 94.5163 ===

πτ

Brittle failure exampleτmax = ktτnom

τmax = ktτnom = (1.8)(5.94T) = 10.69 TConstruct stress element and Mohr’s Circle as beforeUse Coulomb-Mohr theory

T ≤ 2258 in·lbAbout half the load that could be withstood in the absence of the SC!

TdT

JTc

nom 94.5163 ===

πτ

( ) 11043.4

11010969.10

103169.10

1

4

33

≤×

≤×

−−

×

≤−

−T

TTSS uc

B

ut

A σσ

Brittle failure exampleWhat if we consider a solid shaft (no SC’s) with a diameter of d (0.95) ?

Again, use Coulomb-Mohr

Note, this is a greater amount of torque than a shaft with larger diameter but with a SC

( ) Td

TJTc d

9.5

32

42 =

⎟⎟⎠

⎞⎜⎜⎝

⎛==

πτ

lbinT

TTSS uc

B

ut

A

⋅≤

≤×

−−

×

≤−

4090

1101099.5

10319.5

1

33

σσ

Design to avoid stress concentrations

Avoid sudden changes in cross-sectionAvoid sharp inside cornersForce-flow analogy

Imagine flow of incompressible fluid through partSudden curvature in streamlines…

High stress concentration!

Design to avoid stress concentrations