Exam

Monday April 18

Chapter 5: Elastic Strain Deflection Stability

Chapter 6: Failure TheoriesSafety Factors

Chapter 7: Impact

Chapter 8: Fatigue

5.1-2, 5.5-7, 5.10-12

6.1-2, 6.5-8, 6.10-12

7.1-2, 7.4

8.1-11

Chapter 5: Elastic Strain 5.1-2, Deflection 5.5-7 p184-185p186 Tab 5.1p187 Tab 5.2 Stability 5.10-12 p209-214Fig 5.27Fig 5.28Eq 5.10-13

Chapter 6: 6.1-2,6.5-8, 6.10-12 Failure Theories p253-254Fig 6.10Eq 6.5-8Safety Factors p263

Chapter 7: 7.1-2,7.4 Impact p275-284, 290-295

Chapter 8: Fatigue 8.1-11 alternating stressp304-310 readingFig 8.11Fig 8.13Tab 8.1

Mean + alternating stressp317-322Fig 8.16Fig 8.17

Tables and Charts provided for Exam:1. Table 5.1, p186 (Deflection, Stiffness)

2. Table 5.2, p187 (Torsional Deflection)

3. Appendix B1-a, p834 (Section Properties)

4. Appendix C-1, p840 (Material Properties)

5. Figure 5.27, p212 (eq. Column Length)

6. Table 8.1, p316 (Fatigue Strength Factors)

7. Figure 8.13, p314 (CS Surface Finish)

Elastic Strain, Deflection & Stability

Linearly elastic stress-strain relationship (Hooke’s Law)

strain: (uniaxial stress)

E

11

δ=ε

E…Young’s Modulus(Elasticity Modulus)

Shear strain normally can’t be measured directly.

Shear strain: (Hook’s Law) G…shear modulus of elasticity

dx

γ Gτ

=γ

( )ν+=

12EG

uniaxial: E

11

δ=ε

dy (neg.)

dz (neg.)

dx

Axial strain

also causes Lateral strain (Poisson’s Ratio)

strainaxialstrainlateral

=ν

Uniaxial Linear Strain:

Strain Φ-direction: Φ⋅⎟⎠⎞

⎜⎝⎛ ε−ε

+ε+ε

=εΦ 2cos22

2121

Shear Strain Φ-direction: Φ⋅ε−ε

=γ Φ 2sin

2221

Mohr’s Circle:

•Half shear strain

+γ/2

+ε

•Angles twicethe real angles

substitute: σ? ε, τ? γ/2substitute: σ→ ε, τ→ γ/2

Mohr Strain Circle

Deflection or stiffness, rather than stress, is controlling factor in design

• satisfying rigidity • preventing interference or disengagement of gears

Elastic stable systems: small disturbance corrected be elastic forces

Tension Bending Torsion Compression

short column

Elastic unstable systems: small disturbance can cause buckling (collapse)

Compression slender column

Elastic Strain, Deflection & Stability

Deflection in direction of Load p186

Deflection not in direction of Load

K’ …Section Property (Table 5.2)

Deflection Spring Rate

Rigidity Property Section Material

θ in radiant

Table 5.2: Section Properties for Torsional Deflection

Buckling does not vary linearly with load • it occurs suddenly and is therefore dangerous

Elastically Elastically Stable Unstable

P < 2e

2

cr LEIP π

= P > 2e

2

cr LEIP π

=

E …modulus of Elasticity I …moment of inertia (smallest value)

Axis of least I and ρ becomes neutral bending axis

Le…equivalent column length (length between zero bending)

Le=L

Both ends hinged

fixed fixed

fixed

0.707L 0.8 L

0.5 L 0.65 L

L 1.2 L

2 L 2.1 L

Elastic Instability - Buckling

J.B.Johnson Parabola Euler Buckling

P > 2

e2

2y

ycrL

E4S

SS ⎟⎟⎠

⎞⎜⎜⎝

⎛ρπ

−= ( )2e

2

crL

ESρ

π=

ρ …radius of gyration Le/ρ…slenderness ratio

Euler Column Buckling

Johnson Parabola 2

Sy

21

y

2

SE2

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛ π

Failure TheoriesPredict capability of materialsto withstand the infinite combination of non-standard loads

Failure:Behavior of a member that renders it unsuitable for its intended function.

Static loading (no impact, fatigue nor surface wear)

How much distortion is too much ?Impact loading,

Sharp notches,Low temperatures,

Can promote brittle fracture of usually ductile material

Max. Normal-Stress-Theory(W.J.M. Rankine 1802-1872, engl. scientist & educator)

+σ

+τ

Suc

Suc+σ1

Sut

+σ2

Sut

Suc Sut

Failure occurs, when greatest tensile stress

exceeds uniaxial tensile strength.

No Failure occurs within this area

No Failure occurs within these bounds

Principle Mohr circles σ1 –σ2 plot

Correlates well for brittle fracture

simplest

Max. Shear-Stress-Theory(C.A. Coulomb 1736-1806, French scientist)

+τ

Syt+σ1

Sy+σ2

Sy

Failure occurs, when maximum shear stress

exceeds shear strength in uniaxial tension test.

No Failurewithin this area

No Failurewithin

these bounds

Principle Mohr circles σ1 –σ2 plot

Correlates well for ductile yielding

oldestAlso called: Tresca Theory or Guest’s Law

+σ

2S

2y

max21 =τ=

σ−σ

Max. Distortion-Energy-Theory(Maxwell 1856 Engl., Hueber 1904 Pol., Mises 1913 & Hencky 1925 Ger./US)

Failure occurs, when distortion-energy in unit volume (arbitrary load condition)

equals distortion-energy in same volume for uniaxial yielding.

best

Also called: Max.-Octahedral-Shear Theory

with spring rateyPk =

axially loaded bar:L

EAk ⋅=

Energy stored in a spring y2PU ⋅=

k2PU

2

⋅=

EA2LPU

2

⋅⋅

=

per unit volume: ( )LAVUu

⋅==

EA2Pu 2

2

⋅=

stress:AP

=σ E2u

2σ=

stress-strain rel.:εδ

=E 2u σ⋅ε

=

3-D: 222u 332211 σ⋅ε

+σ⋅ε

+σ⋅ε

=

Max. Distortion-Energy-Theory222

u 332211 σ⋅ε+

σ⋅ε+

σ⋅ε=σ

with 3-D stress-strain rel.

with 3-D averaged stress:

[ ]ν−σ

= 21E2

3u2

avav

3321

avσ+σ+σ

=σ

( )[ ]1332212

32

22

1 2E21u σσ+σσ+σσν−σ+σ+σ=σ

avd uuu −= σDistortion energy

2yd S

E31u ν+

=y1 S=σSimple tension test:

( ) ( ) ( )⎥⎦

⎤⎢⎣

⎡ σ−σ+σ−σ+σ−σν+=

2E31u

213

232

221

d

( ) ( ) ( )2

S2

132

322

21ey

σ−σ+σ−σ+σ−σ=σ>

actual average difference

EEE321

1νδ

−νδ

−δ

=ε

EEE312

2νδ

−νδ

−δ

=ε

EEE213

3νδ

−νδ

−δ

=ε

Max. Distortion-Energy-Theory

(Juvinall p254)

( ) ( ) ( )2

S2

132

322

21ey

σ−σ+σ−σ+σ−σ=σ>

3D:

2D:

Example:

Failure Theoriesσ1 = 35 ksi

σ2 = -25 ksi

SteelSy=100 ksi

Load Line

Shear Diagonal

σ1= Sy=

σ2=

Sy=

Max. S

hear

Stress

Th.

Max. Distortio

n Energy Th.M

ax. N

orm

al S

ress

Th.

Load Point

y212

22

1 S<σσ−σ+σ

y21 S<σ−σ

ymax S<σ

1.7

1.9

2.9

also called shock, sudden or impulsive loadingImpact

Compare Time of Load Application t with Natural Period of Vibration τ

a) Rapidly moving

dashpot

■driving a nail with a hammer, automobile collisions ….

b) Suddenly applied c) Direct Impact■Combustion in

engine cylinder■ vehicles crossing

a bridge■ pile drive,■ drop forge,■ car crash

To distinguish:

km

π= 2

Carrying loads

Absorbing energy

t > 3 τStatic load “Gray area” Impact load

3 τ > t > 1/2 τ t < 1/2 τ

Design for

Impact Factor

material properties vary with loading speed Impact

Su

Sy

Sy/Su

elongation

ksi %•Works favorably

•Promotes brittle fracture

•Loading rate only approximated

In praxis:•Static propertiesknown

■ x4 for automobile suspension parts

Linear & Bending Impact

Assumptions:

•No massof structure

•No deflectionwithin impacting mass

•No damping

dynamic = static deflection curveonly multiplied by impact factor

all energy goes into structure

most severe impact

W

Wk k

hδ

Fe

W

Guide rod

δst δ h

Elastic-strain energy stored in structure= 0.5Feδ

Work of falling weight = W(h+δ)

W (h + δ) = 0.5 Fe δFe = k δ

k = W / δstFe = W δ/δst

⎟⎟⎠

⎞⎜⎜⎝

⎛++=

stst

h211δ

δδ

⎟⎟⎠

⎞⎜⎜⎝

⎛++=

ste

h211WFδ

Impact Factor

Impact Deflection …

Impact Load …

⎟⎟⎠

⎞⎜⎜⎝

⎛++=

stste

h211δ

σσImpact Stress …

Linear & Bending ImpactW

Wk k

hδ

Fe

W

Guide rod

δst δ h

Elastic-strain energy stored in structure= 0.5Feδ

Work of falling weight = W(h+δ)

W (h + δ) = 0.5 Fe δFe = k δ

k = W / δstFe = W δ/δst

v2 = 2gh → h= v2/2g

h = 0 or v=0suddenly applied load

Impact Factor = 2 (Double Safety Factor)

Impact Factor

For h >> δst

Linear & Bending ImpactW

Wk k

hδ

Fe

W

Guide rod

δst δ h

Elastic-strain energy stored in structure= 0.5Feδ

Work of falling weight = W(h+δ)

W (h + δ) = 0.5 Fe δFe = k δ

k = W / δstFe = W δ/δst

Impact Factor

For h >> δst

Vertical movement

Horizontal movement

W = k δst = g m

Impact kin. Energy … U = 0.5 m v2

As greater Stiffness and kin. Energy as greater Equivalent static force

Linear Impact of Straight Bar

σ = Fe / A k = A E / L

Impact Energy Capacity of a straight rod

• Basic relationship → Gives optimistic results

• Calculates stress lower than actual stressbecause of- Non-uniform loading & stress distribution- Mass of the impacted rod

σ

Term fatigue introduced by Poncelet (France) 1839progressive fracture is more descriptive

Fatigue

1. Minute crack at critical area of high local stress(geometric stress raiser, flaws, preexisting cracks)

2. Crack gradually enlarges(creating “beach marks”)

3. Final fracture(suddenly, when section sufficiently weakened)

Fatigue: no or only microscopic distortionstatic failure: gross distortion

• Repeated plastic deformation

• Thousands/millions of microscopic yielding(far below conventional yield or elastic point)

• Highly localized plastic yielding(holes, sharp corners, threads, keyways, scratches, corrosion)

• Strengthen vulnerable location often as effectiveas choosing a stronger material

• (If local yielding is sufficiently minutestrain-strengthen may stop the yielding)

Standard Fatigue Strength Sn’Empirical data from R.R. Moore fatigue test(Highly standardized and restricted conditions)

Rotating-beam fatigue-testing machine

Pure bending (zero traverse shear)

N cycles of tension-to-compression-to-tension

1750 rpm

various

Endurance LimitS-N curve approximation

Endurance limit

Steel Sn‘ = 0.5 x Su

Cast Iron

Aluminum

Magnesium Sn‘ = 0.35 x Su

Nickel alloys Sn‘ = 0.35…0.5 x Su

Cooper alloys Sn‘ = 0.25…0.5 x Su

Titanium Sn‘ = 0.45…0.6 x Su

@ N=106

@ N=108Sn‘ = 0.4 x Su

Endurance Limit

Reversed Bending (not rotating bending like in Moore testing)

maximum stresses only @ top and bottomhigh probability not weakest point

Rotating Bending (Moore testing)

maximum stresses on surfaceweakest point fatigue start

Fatigue strength usually slightly greaterdeliberately neglected safe side

Reversed Axial Loadingmaximum stresses entire cross section

no reserve !Fatigue strength about 10% lesseccentric loads about 20…30% less

CG=0.7…0.9gradient factor

Reversed Torsional Loading

reversed biaxial stressdistortion energy theory 58%

CL=0.58load factor

maximum stresses on surfaceshear stresses fatigue start

Fatigue StrengthJuvinall p.312

Fig. 8.12

Steel Sus=0.8 Su

Other ductile material Sus=0.7 Su

Fatigue StrengthInfluence of SurfaceSo far special “mirror polish” surface (only in laboratory ! )Minimizes 1.) surface scratches (stress concentration)

2.) differences of surface & interior material3.) residual stresses from finishing

Commercial surfaces have localized pointsof greater fatigue vulnerability.

Surface factor CS

Use only for endurance limit !

+ cast iron

Fatigue StrengthInfluence of Size

Reversed Axial Loadingmaximum stresses entire cross section

no reserve !Fatigue strength about 10% lesseccentric loads about 20…30% less

CG=0.7…0.9gradient factor

large specimen> 0.4” CG = 0.92..4” CG = 0.84..6” CG = 0.7

0.3” test specimen

Use equivalent round section !

Bending & Torsional Reversed Loading

small specimen< 0.3” CG = 1

Fatigue Strength

Sn = Sn’CLCGCS

Summery use Table 8.1 Juvinall p.316106-cycle strength (endurance limit)

Sn’… Moore endurance limit

1000-cycle strengthBending Axial Torsion

CL (load factor) 0.9Su 0.75Su 0.9Sussteel: Sus=0.8Su

other ductile metals: Sus=0.7Su

Bending Axial TorsionCL (load factor) 1 1 0.58

CG (gradient factor)< 0.4” 1 0.7…0.9 1

0.4”…2” 0.9 0.7…0.9 0.92”…4” r e d u c e - 0.1 4”…6” r e d u c e - 0.2

CS (surface factor) see Fig. 8.13

Fatigue StrengthFluctuating stress = static stress + completely reversed stress

mean + alternating

Effect of mean stress

Fatigue StrengthEffect of mean stressStatic tensile stress reduces amplitude of reversed stress

that can be superimposed

Macroscopic Yielding on first load application

Microscopic Yielding

empiric concept

Sy

Sn

Su

-Sn

σa

σm

Fatigue StrengthEffect of mean stress

Compressive mean stress does not reduce amplitudethat can be superimposed

Sy-Sy

Sy

Sn

σa

-σm(compression)

σm(tension)

Goodman lines

empiric concept

Su

Values from S-N curve (σm=0)

No macroscopic yielding

Ext

ends

infin

ite fo

r fat

igue

(o

nly

stat

ic fa

ilure

Suc

)

Juvinall p.318 Fig. 8.16constant-life fatigue diagram

Fatigue StrengthEffect of mean stress

Sy Su

Sy

-σm(compression)

σm(tension)

σa

-Sy

67.0m

a =σσ

Sn=Sn’CLCGCS=(0.5x150ksi)x1x0.9x0.9=61ksi

Sy=120 ksiSu=150 ksi

polishedfor N=106 & 103

d= ? < 2in

1000…5000lbaxial load

Alb6000

ASFxFm

m ==σσm

SF=2

Alb4000

ASFxFa ==σa

σa(N=106)=38ksi d=0.367in < 3/8 in σa(Sy)=48ksi d=0.326in <11/32in

S=0.75Su=0.75x150ksi=112ksi

Homework P 8.23Known: D=in, Su = 110ksi, Sy = 77ksi,

reversed bending, axial, torsional loading, steel, machinedsurface

Find: Sn(6x104 life cycles)

Homework P 8.23Known: D=in, Su = 110ksi, Sy = 77ksi,

reversed bending, axial, torsional loading, steel, machinedsurface

Find: Sn(6x104 life cycles)

Homework P 8.23Known: D=in, Su = 110ksi, Sy = 77ksi,

reversed bending, axial, torsional loading, steel, machinedsurface

Find: Sn(6x104 life cycles)

Homework P 8.23Known: D=in, Su = 110ksi, Sy = 77ksi,

reversed bending, axial, torsional loading, steel, machinedsurface

Find: Sn(6x104 life cycles)