Dr. A. Vinoth JebarajVIT University, Vellore.

Design For Static Loading

Design For Variable Loading

Design For Static Loading

Design For Variable Loading

Shear failure @ 45° plane Brittle fracture @ 0° plane

NeckingDuctile material Brittle material

Ductile fracture Brittle fracture

Shear plane Normal plane

Uniaxial loading

The plane perpendicular to the line of action of the load is aprincipal plane. [Because, It is having the maximum stress valueand shear stress in this plane is zero.]

The plane which is at an angle of 90° will have no normal andtangential stress.

Mohr’s circle for uniaxial loading

Purpose of Tensile test

1 2

3 4

Image Courtesy: https://www.youtube.com/watch?v=D8U4G5kcpcM

Simple Tension Test

In simple tension test, all six quantities reaches its criticalvalues simultaneously (at a single instant).

Any one of the following will cause failure.

• Principal normal stress yield stress σmax = σy or σu

• Principal shear stress yield shear stress τmax = σy /2

• Principal strain energy strain energy at yield point Utotal = ½ [σy εy]

• Principal strain strain at yield point εmax = σy /E (or) σu /E

• Distortion energy distortion energy at yield pointUdistortion = ퟏ 흁

ퟑ푬[σy

2]

Real life examples for Combined loading

Torsion and bending

Thrust and torsional shear

Axial, bending and Torsion

Tensile and direct shear

Side thrust from cylinder wall, force due to piston

Crank Shaft

Connecting rod

Coupling

Propeller shaft

Lifting Jack

Principal stress < Yield stress [safe]but, Shear stress exceeds its limit.

Why failure theories?

Types of Loading

Pure shear

Normal stress σn = τ sin 2θAt θ = 45° σn = σmax = τ

Shear stress τ = τ cos 2θAt θ = 0°, τ max = τ

Under pure shear, ductile materials will fail in 0° plane and brittle materials will fail

in 45° plane. Because, at 0° plane induces maximum shear stress and 45° plane gives

maximum normal stress.

Theories of Failure

Predicting failure in the members subjected to uniaxial stress is verysimple and straightforward. Because all failure criterions are reachingthe critical limit at an instant.

But, in multi axial loading the prediction of failure is muchcomplicated. Because, predicting the cause of failure i.e. whichquantity of failure criterion is causing failure is difficult to find.

Thus, theories were formulated to predict this issue, which are knownas failure theories.

Maximum Principal or Normal Stress Theory (Rankine’s Theory)

This theory is based on failure in tension or compression and ignores thepossibility of failure due to shearing stress, therefore it is not used for ductilematerials.

For Brittle materials which are relatively strong in shear but weak in tensionor compression, this theory is generally used.

Max principal stress [σ1] ≥ [σy] yield stress(In a multi axial loading) (In a simple tension test)

Maximum Shear Stress Theory

σ1

σ2

Maximum Distortion Energy Theory (Hencky and Von Mises Theory)

According to this theory, the failure or yielding occurs at a point in a member whenthe distortion strain energy (shear strain energy) per unit volume in a biaxial stresssystem reaches the limiting distortion energy (distortion energy per unit volume) asdetermined from a simple tension test.

Total strain energy U = Uv + Ud Ud = U - Uv

Ud = (1+µ) / 6E [(σ1 - σ2) 2 + (σ2 - σ3)2 + (σ3 – σ1)2]

For uniaxial tension test

Ud = (1+µ) / 6E [(σ1 2 + σ1)2] Ud = (1+µ) / 3E [σy

2]

For triaxial loading, the distortion energy

[When σ1 reaches σy]

Thus, the left side of the Equation is a single, equivalent, or effectivestress for the entire general state of stress given by σ1, σ2, and σ3.This effective stress is usually called the VonMises stress, σ′, namedafter Dr. R. VonMises, who contributed to the theory.

=

퐓퐨퐭퐚퐥 퐬퐭퐫퐚퐢퐧 퐄퐧퐞퐫퐠퐲 퐔 = ퟏퟐ

훔ퟏ훜ퟏ + ퟏퟐ

훔ퟐ훜ퟐ + ퟏퟐ

훔ퟑ훜ퟑ

Where ε1, ε2, ε3 are strain three principal directions

∈ퟏ= ퟏ푬 [ 흈ퟏ − 흁 흈ퟐ + 흈ퟑ ]

∈ퟐ= ퟏ푬 [ 흈ퟐ − 흁 흈ퟏ + 흈ퟑ ]

∈ퟑ= ퟏ푬 [ 흈ퟑ − 흁 흈ퟏ + 흈ퟐ ]

푼 = ퟏퟐ푬

[(흈ퟏퟐ+ 흈ퟐퟐ + 흈ퟑퟐ ) – 2μ (흈ퟏ흈ퟐ + 흈ퟐ흈ퟑ + 흈ퟑ흈ퟏ) ]

Substituting the above equations,

VonMises Stress component

Total strain energy U = Uv + Ud

Therefore, the corresponding stresses are resolved into threecomponents

훔ퟏ = 훔ퟏ퐝 + 훔퐕 ; 훔ퟐ = 훔ퟐ퐝 + 훔퐕 ; 훔ퟑ = 훔ퟑ퐝 + 훔퐕

∈ퟏ퐝 + ∈ퟐ퐝 + ∈ퟑ퐝= ퟎ

∈ퟏ풅= ퟏ푬 [흈ퟏ풅 − 흁 흈ퟐ풅 + 흈ퟑ풅 ]

∈ퟑ풅= ퟏ푬 [흈ퟑ풅 − 흁 흈ퟏ풅 + 흈ퟐ풅 ]

∈ퟐ풅= ퟏ푬 [흈ퟐ풅 − 흁 흈ퟏ풅 + 흈ퟑ풅 ]

ퟏ − ퟐ흁 ( 흈ퟏ풅 + 흈ퟐ풅 + 흈ퟑ풅 ) = 0 ퟏ − ퟐ흁 ≠ ퟎ

Therefore, ( 흈ퟏ풅 + 흈ퟐ풅 + 흈ퟑ풅 ) = 0

흈ퟏ + 흈ퟐ + 흈ퟑ = ퟑ 흈푽

Strain energy for volume change Uv = 3 흈푽흐푽ퟐ

Volumetric Strain ∈푽= ퟏ푬

[흈푽 − 흁 [흈푽 + 흈푽]

∈푽 = (ퟏ ퟐ흁)흈푽푬

Uv = ퟑ(ퟏ ퟐ흁 )흈푽 ퟐ

ퟐ푬

Uv = ퟏ ퟐ흁 흈ퟏ 흈ퟐ 흈ퟑ ퟐ ퟔ푬

Ud = U - Uv

Ud = (ퟏ 흁)ퟔ푬

[ 흈ퟏ − 흈ퟐ ퟐ + 흈ퟐ − 흈ퟑ ퟐ+ 흈ퟑ − 흈ퟏ ퟐ]

In simple tension test, when yielding starts 흈ퟏ = 흈풚 풂풏풅 흈ퟐ = 흈ퟑ = ퟎ

Ud = (ퟏ 흁)ퟑ푬

흈풚 ퟐ

Ud = (ퟏ 흁)ퟔ푬

[ 흈ퟏ − 흈ퟐ ퟐ + 흈ퟐ − 흈ퟑ ퟐ+ 흈ퟑ − 흈ퟏ ퟐ]

Distortion strain energy in triaxial loading

Distortion strain energy in uniaxial loading

Therefore, Failure criterion is,

(ퟏ 흁)ퟑ푬

흈풚 ퟐ = (ퟏ 흁)ퟔ푬

[ 흈ퟏ − 흈ퟐ ퟐ + 흈ퟐ − 흈ퟑ ퟐ+ 흈ퟑ − 흈ퟏ ퟐ]

흈풚 ퟐ = ퟏퟐ

[ 흈ퟏ − 흈ퟐ ퟐ + 흈ퟐ − 흈ퟑ ퟐ+ 흈ퟑ − 흈ퟏ ퟐ]

Maximum Principal Strain Theory (St. Venant’s Principle)

The strain in the direction of σ1 [ε1] =

Max principal strain [ɛ1] ≥ [ɛy] Strain at yield point(In a multi axial loading) (In a simple tension test)

According to this theory of failure, σ1 could be increased to a

value somewhat higher than σy without causing yielding if the

second normal stress σ2 is a tensile stress. But if σ2 is a

compressive stress the maximum value of σ1 that could be applied

without causing yielding would be somewhat smaller than σy.

This theory is not applicable if the failure in elastic behavior is

by yielding. It is applicable when the conditions are such that

failure occurs by brittle fracture.

Total strain energy [UTotal] ≥ [Uy] Strain energy at yield point(In a multi axial loading) (In a simple tension test)

Maximum Strain Energy Theory (Haigh’s Theory)

According to this theory, the failure or yielding occurs at a point in a member

when the strain energy per unit volume in a biaxial stress system reaches the

limiting strain energy (strain energy at yield point) per unit volume as

determined from the simple tension test.

Historical failures

Liberty ship failure De Havilland Comet failure

Boston molasses tank failureAloha airplane failure

Stress Concentration!

De Havilland Comet was the World first commercial jet liner, with a cruising

speed of 490 mph at altitudes up to 40,000 ft. One year after its introduction,

a Comet tore apart in mid-flight near Calcutta, India. Another Comet fell

into the sea near Elba, in January of 1954. Then three months later, a third

Comet crashed near Naples, in Italy.

Image Courtesy: http://lessonslearned.faa.gov/ll_main.cfm?TabID=1&LLID=28&LLTypeID=2

Stress Concentration

Reasons for stress concentration

Variation in properties of materials

Load application

Abrupt changes in cross section

Discontinuities in the component

Machining scratches

Stress concentration: Localization of high stresses due to the irregularities present in the component and abrupt changes of the cross section

Stress concentration in brittle materials

Brittle materials do not yield locally and there is no readjustment of stresses atthe discontinuities. (due to inability of plastic deformation)

When the magnitude of stress reaches the ultimate strength of the material, acrack will nucleate and increases the stress concentration at the crack.

Therefore, stress concentration factors have to be used in the design of brittlematerials.

Stress concentration in ductile materials (fluctuating load)

Due to fluctuating load the component may fail due to fatigue. stressconcentration will leads to the reduction in endurance limit of the ductilematerials.

Therefore stress concentration factors have to be used in the design of machinecomponents made of ductile materials.

Stress concentration in ductile materials (static load)

When the stress reaches the yield point, then there will be a local plastic deformationnear the discontinuity which will lead to redistribution of stresses near the stressconcentration zone.

There is no remarkable damage to the machine component. This redistribution ofstresses will be restricted to very small area.

Cup and cone ductile fracture Brittle fracture

Why ductile material fails in a brittle fashion?

Image courtesy: http://www.reliasoft.com/newsletter/v8i2/fatigue.htm

Fluctuating stresses

σmax = max stress ; σmin = min stress ; σa = stress amplitude σmean = mean stress

The stresses induced in a machine component due to dynamic load(change in magnitude with respect to time) is known as fluctuatingstresses.

Variable loading

• Change in magnitude of the applied load

Example: Punching machine

• Change in direction of the loadExample: Connecting rod

• Change in point of applicationExample: Rotating shaft

Types of loading• Fully Reversed loading

• Repeated loading

Fatigue failure( Time delayed fracture under cyclic loading)

Fatigue failure begins with a crack at some point in the material .

Regions of discontinuities (oil holes, keyways and screw threads)

Regions of irregularities in machining operations (scratches on thesurface, stamp mark, inspection marks)

Internal cracks due to defects in materials like blow holes

These regions are subjected to stress concentration due to crack,then due to fluctuating load the crack spreads.

Region indicating slow growth of crack with a fine fibrous

appearanceRegion of sudden fracture with a coarse granular appearance

Crack initiation Crack propagation Fracture

Design of machine components for fluctuating load

FatigueMean stress

Number of cycles

Stress amplitude

Stress concentration

Residual stresses

Corrosion & creep

Endurance limit or fatigue limit of a material is defined as the maximum amplitude ofcompletely reversed stress that the standard specimen can sustain for an unlimited numberof cycles without fatigue failure.

106 cycles are considered as a sufficient number of cycles to define the endurance limit.

Fatigue life: The total number of stress cycles that the standard specimen can completeduring the test before appearance of the first fatigue crack.

S-N Curve

Fatigue test specimen

Effect of stress concentration on fatigue life

Real-World Allowable Cyclic Stress = ka * kb * kc * kd * ke * kf * EL

Size factor, surface finish factor, load factor, reliability factor, temperature factor,impact factor

Macro observation of the metal Surface

Micro observation of the metal Surface

Micro observation of the polished metal Surface

The graph shows that the

endurance limit is very low

in the corrosive

environment.

Because the corroded

surface will induce crack

in the component surface

which will reduce the life

drastically.

Notch sensitivity factor (q)

In case of dynamic loading, if stress concentration present in the material, then it willreduce the endurance limit.

The actual reduction in the endurance limit of a material due to stress concentrationunder dynamic loading is varied by the theoretical values predicted using theoretical stressconcentration factor.

Therefore two separate stress concentration factors are used . i.e. Kt and Kf.

kf is the fatigue stress concentration factor

kf = Endurance limit of the notch free specimen / Endurance limit of the notchedspecimen

Notch sensitivity [q] : Susceptibility of a material to succumb to the damaging effects ofstress raising notches in fatigue loading.

q = Increase of actual stress over nominal stress / Increase of theoretical stress overnominal stress

Notch sensitivity (q) for different materials

σo = nominal stress obtained by the elementary equations

Actual stress due to fatigue loading = Kf σ0

Theoretical stress = Kt σ0

Increase of actual stress over nominal stress = (Kf σ0 - σ0)

Increase of theoretical stress over nominal stress = (Kt σ0 - σ0)

q =

Kf = 1 + q (Kt – 1)

When the material has no sensitivity to notches,q = 0 and Kf = 1

When the material is fully sensitive to notches,q = 1 and Kf = Kt

Design for Variable loading

Modified Goodman diagram?

Modified Goodman line

According to Soderberg line,

풏=

흉풚+

흉풆푲풔풖풓푲풔풛[For shear stress]

According to Soderberg line,

ퟏ풏

= 흈풎흈풚

+ 흈풗푲풇

흈풆푲풔풖풓푲풔풛 [퐅퐨퐫 퐧퐨퐫퐦퐚퐥 퐬퐭퐫퐞퐬퐬 (퐟퐨퐫 퐚퐱퐢퐚퐥 & 퐛퐞퐧퐝퐢퐧퐠)]

ퟏ풏

= 흉풎흉풚

+ 흉풗푲풇흉풆푲풔풖풓푲풔풛

[For shear stress]

According to Goodman line,

풏=

흉풖+

흉풆푲풔풖풓푲풔풛[For shear stress]

According to Goodman line,

ퟏ풏

= 흈풎흈풖

+ 흈풗푲풇

흈풆푲풔풖풓푲풔풛 [퐟퐨퐫 퐧퐨퐫퐦퐚퐥 퐬퐭퐫퐞퐬퐬(퐅퐨퐫 퐚퐱퐢퐚퐥 & 퐛퐞퐧퐝퐢퐧퐠)]

ퟏ풏

= 흉풎흉풖

+ 흉풗푲풇흉풆푲풔풖풓푲풔풛

[For shear stress]

Combined variable loadingCombined variable loading

ퟏ풏 =

흈풎흈 +

흈풗푲풇

흈 푲 푲

Equivalent normal stress = 흈풎 +흈풆푲풔풖풓푲풔풛

ퟏ풏 =

흈풎흈풚

+ 흈풗푲풇

흈풆푲풔풖풓푲풔풛Multiplying throughout by 흈풚 we get,

흈풚풏 =

흈풎 흈풚흈풚

+ 흈풗푲풇흈풚흈풆푲풔풖풓푲풔풛

Equivalent normal stress = 흈풎 + 흈풗푲풇흈풚흈풆푲풔풖풓푲풔풛

According to Soderberg line, (for normal stresses)According to Soderberg line, (for normal stresses)

ퟏ풏 =

흉풎흉 +

흉풗푲풇

흉 푲 푲

Equivalent shear stress = 흉풎 +흉풆푲풔풖풓푲풔풛

ퟏ풏 =

흉풎흉풚

+ 흉풗푲풇

흉풆푲풔풖풓푲풔풛Multiplying throughout by 흉풚 we get,

흉풚풏 =

흉풎 흉풚흉풚

+ 흉풗푲풇흉풚

흉풆푲풔풖풓푲풔풛

Equivalent shear stress = 흉풎 + 흉풗푲풇흉풚흉풆푲풔풖풓푲풔풛

According to Soderberg line, (for shear stresses)According to Soderberg line, (for shear stresses)