Prepared by Dr.A.Vinoth Jebaraj

Simple Design

Image courtesy: NPTEL

Elementary Equations

For Direct loading or Axial loading

For transverse loading

For tangential loading or twisting

Where I and J Resistance properties of cross sectional area

I Area moment of inertia of the cross section about the axes lying on the section (i.e. xx and yy)

J Polar moment of inertia about the axis perpendicular to the section

Ductile fracture of Al-Mg-Si alloy Brittle fracture of Cast Iron

Necking

Why ductile materials fail in 45° plane? Why brittle materials fail in 0° plane?Image courtesy: Google

Types of Loading

Pure shear

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

Shear stress τ = τ cos 2θ At 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 shear stress is maximum and at 45° plane normal

stress is maximum.

Eccentric load on bolts

Eccentric load on column

Eccentric load on crane hook

Eccentric loading

Eccentric load on hydraulic punching press

Image courtesy: Google

Eccentric Loading

If the line of action of a load is not passing through the Centroid ofthe machine component, then that is knows as eccentric load.

There are different kinds of stresses will be induced duringeccentric loading

For eccentric axial load,

Direct stress and bending stress

To find out the magnitude of resultant stress, these combination of stresses have to be super imposed.

For eccentric plane load,

Direct shear and torsional shear 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.

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?

Ductile fracture Brittle fracture

Shear plane Normal plane

Purpose of Tensile test

1 2

3 4

Image courtesy: YouTube

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]

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

According to this theory, the failure or yielding occurs at a point in amember when the maximum principal or normal stress in a bi-axialstress system reaches the limiting strength of the material in a simpletension test.

This theory is based on failure in tension or compression and ignoresthe possibility of failure due to shearing stress, therefore it is not usedfor ductile materials.

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

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

σ1

σ2

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.

Image courtesy: Shigley

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 von Mises stress, σ′, namedafter Dr. R. von Mises, 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,

What is VonMises Stress?

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 (Saint Venant’s Theory)

According to this theory, the failure or yielding occurs at a point in a memberwhen the maximum principal strain in a multi axial stress system reaches thelimiting value of strain (strain at yield point) as determined from a simpletension test.

The strain in the direction of σ1 [ε1] =

According to this theory of failure, σ1 could be increased to a valuesomewhat higher than σy without causing yielding if the second normal stressσ2 is a tensile stress. But if σ2 is a compressive stress the maximum value of σ1that could be applied without causing yielding would be somewhat smallerthan σy.

This theory is not applicable if the failure in elastic behavior is by yielding. Itis applicable when the conditions are such that failure occurs by brittlefracture.

Maximum Strain Energy Theory (Haigh’s Theory)

According to this theory, the failure or yielding occurs at a point in a memberwhen the strain energy per unit volume in a biaxial stress system reaches thelimiting strain energy (strain energy at yield point) per unit volume asdetermined from the simple tension test.

Stress Tensor

To define a stress at any point in a member subjected to multi axial

loading, an infinitesimally small cube around a point is assumed to

indicate the stress components in three mutually perpendicular

planes.

Planar Assumptions

All real world structures are three dimensional.

For planar to be valid both the geometry and the loads must be constant across the thickness.

When using plane strain, we assume that the depth is infinite. Thus the effects from end conditions may be ignored.

Plane Stress

All stresses act on the one plane – normally the XYplane.

Due to Poisson effect there will be strain in the Zdirection. But We assume that there is no stress inthe Z – direction.

σx, τxz, τyz will all be zero.

All strains act on the one plane – normally the XYplane. And hence there is no strain in the z-direction.

σz will not equal to zero. Stress induced to preventdisplacement in z – direction.

εx, εxz, εyz will all be zero.

Plane Strain

A thin planar structure with constant thickness and loading within the plane of thestructure (xy plane).

A long structure with uniform cross section and transverse loading along its length (z –direction).

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 at

the 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 (static load)

When the stress reaches the yield point, then there will be a local plasticdeformation near the discontinuity which will lead to redistribution of stressesnear the stress concentration zone.

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

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 theductile materials.

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

BucklingCorrosion

Creep

Fatigue failure due to cyclic load

FractureRupture

WearYielding

Cup and cone ductile fracture Brittle fracture

Why ductile material fails in a brittle fashion?

Region indicating slow growth of crack with a fine fibrous

appearanceRegion of sudden fracture with a coarse granular appearance

Crack initiation Crack propagation Fracture

Factor of Safety

For Ductile Materials

For Brittle Materials

For Variable loading

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.

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

Low cycle fatigue:

Any fatigue failure when the number of stress cycles are

less than 1000, is called low cycle fatigue.

Examples: Failure of studs on truck wheels, failure of set screws for locating

gears on shafts, short lived components like missiles.

High cycle fatigue:

Any fatigue failure when the number of stress cycles are

more than 1000, is called high cycle fatigue.

Examples: Failure of springs, ball bearings and gears that are subjected to

fluctuating stresses.

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

Surface finish factor Ka: It takes into account the reduction in

endurance limit due to variation in the surface finish between the

specimen and the actual component.

Size factor Kb : It takes into account the reduction in endurance limit

due to increase in the size of the component.

Reliability factor Kc : It depends upon the reliability that is used in the

design of the component. The greater the likelihood that a part will

survive, the more is the reliability and lower is the reliability factor.

Macro observation of the metal Surface

Micro observation of the metal Surface

Surface roughness plays major role in the fatigue life of machinecomponents.

Macro and micro observation of the polished metal Surface

Polished surface with minimum roughness increases the life of ametal due to the absence of stress raisers observed in the asreceived conditions of a 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.

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 theductile materials.

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

Keyways Screw threads Stepped shaft

Effect of stress raisers Geometrical irregularities

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)

Fluctuating torsional shear stress

The endurance limit of a component subjected totorsional shear loading is obtained from endurance limitin reversed bending using theories of failures.

Maximum shear stress theory Distortion energy theory 푺풔풆 = 0.5 푺풆 푺풔풆 = 0.577 푺풆

Endurance limit in axial reversed loading is lower thanthe endurance limit in reversed bending rotating beamtest.

For axial loading, (푺풆)axial = 0.8 (푺풆) Bending

Consider an elastic system loaded by a falling weight ‘W’

W = Falling Weight (N)

h = height through which the weight falls (mm)

δ = displacement of the point of load application (mm)

L = length of the bar (mm)

A = Cross sectional area of the bar mm2

P = impact force which produces deflection δ (N)

E = Modulus of elasticity of bar material (N/mm2)

σi = impact stress in the bar

Impact Loading

Energy released by the falling weight = potential energy = W (h + δ)

Energy absorbed by the system = strain energy = Average load x deflection = ퟏퟐ

푷 휹

Equating the above two equations,

ퟏퟐ

푷 휹= 푾 풉 + 휹

Also, P = 흈풊푨 풂풏풅 휹풍

= 휺 = 흈풊푬

OR 휹 = 흈풊풍푬

Substituting the above values, we get

흈풊ퟐ푨풍ퟐ푬 − 흈풊

푾풍푬 −푾풉 = ퟎ

The above equation is a quadratic equation. Solving the equation and using the positive sign for getting maximum value

흈풊 = 푾푨

ퟏ + ퟏ + ퟐ풉푨푬푾풍

P = W ퟏ + ퟏ + ퟐ풉푨푬푾풍

Where , 푷푾

= shock factor which indicates the magnification of the load W into the impact force P during impact.

Why Environment is Important in Design?

Titanic failure

IMPACT LOADING

Design for Strength (Based on permissible shear stress)Design for Strength (Based on permissible shear stress)

Equivalent torque 푻풆 = 푴ퟐ + 푻ퟐ 흅

ퟏퟔ 흉풎풂풙 풅ퟑ

Equivalent torque 푻풆 = 푴ퟐ + 푻ퟐ = 흅

ퟏퟔ 흉풎풂풙 풅ퟑ

Design for Rigidity (Based on permissible angle of twist)Design for Rigidity (Based on permissible angle of twist)

Torsional equation

푻푱 =

푪 휽풍 =

흉풓

Torsional equation

푻푱 =

푪 휽풍 =

흉풓

ASME code for shaft design

According to ASME code,

Permissible shear stress τmax = 0.30 σyt [Or] τmax = 0.18 σut(whichever is minimum)

If keyways are present, the permissible shear stress is to bereduced by 25%.

Also, Acc. To ASME code, bending and twisting moments are to bemultiplied by factors kb and kt respectively to account for shockand fatigue in operating condition.

Kb = Combined shock and fatigue factor applied to bending moment

Kt = Combined shock and fatigue factor applied to twisting moment

Equivalent torque Te = (푲풃푴풃)ퟐ+(푲풕푴풕)ퟐ

Critical speed or Whirling speed of shafts(Natural frequency of vibration)

The speed at which the rotating shaft becomes dynamically unstable and start to

vibrate violently in transverse direction.

Reason for vibration:

Mass is not uniformly distributed about its geometric axis

Deflection due to self weight, gear forces, belt or chain tensions

Avoiding Resonance ?

Shafts can be made very rigid with high critical speed which is far away from therunning speed

Shaft passes quickly through the critical speed.

A hollow shaft is subjected to a maximum torque of 1.5 kN-m and a maximum bending

moment of 3 kN-m. It is subjected, at the same time, to an axial thrust load of 10 kN.

Assume that the load is applied gradually and the ratio of the inner diameter to the

outer diameter is 0.5. If the outer diameter of the shaft is 80 mm, find the shear stress

induced in the shaft. Consider the length of the shaft is 1m.

Combined bending, torsion and axial loading in Shaft

Mechanical device that permanently joins two rotating shafts to each other.

(Joining shafts of two separately built units)

Engine Output shaft hydraulic pump

Electric motormachine tool gear box

Oldham’s coupling Connecting two parallel shafts when they are at a small distanceapart.

Hooke’s coupling Connecting two shafts having intersecting axes.

Rigid & Flexible coupling connecting two shafts having collinear axes.

Cannot tolerate misalignments

Simple and inexpensive

Motion should be free fromshocks and vibration

Can tolerate misalignments Lateral,Angular (5°) & Axial (5mm)

Flexible elements absorb shocks andvibration

Comparatively costlier due toadditional parts

Types of couplingClamp coupling (or) Split muff coupling Sleeve coupling

Flange coupling (Unprotected & Protected )

Rigid Flange CouplingRigid Flange CouplingBolts fitted in reamed and ground holes Bolts fitted in reamed and ground holes

Torque transmitted by the coupling

푴풕 = 푷 × 푫ퟐ × 푵

P = Force acting on each boltD = Pitch diameter of boltsN = Number of bolts

Direct shear stress 흉 = 푷흅ퟒ 풅ퟏퟐ

d1 = nominal diameter of the bolt

흉 = 푴풕

흅푫푵풅ퟏퟐ

Bolts fitted in large clearance holes Bolts fitted in large clearance holes

For uniformly distributed pressure

Friction radius 푹풇 = ퟐퟑ

푹풐ퟑ 푹풊

ퟑ

푹풐ퟐ 푹풊

ퟐ

Ro = outer radius of flangeRi = inner radius of recess

Torque transmitted by the coupling푴풕 = 흁 × 푷풊 × 푵 × 푹풇

Bushed pin flexible coupling

Step I: Shaft diameter (d)Step II: Dimensions of flanges

풅풉 = ퟐ풅풍풉 = ퟏ.ퟓ풅D = 3d to 4dt = 0.5dt1 = 0.25d

푴풕 = ퟏퟐ 흅풅풉

ퟐ풕흉

Step III: Diameter of pins (d1)

Diameter of pin d1 = ퟎ.ퟓ 풅푵

Shear stress induced in the pin 흉 = ퟖ 푴풕흅 풅ퟏퟐ 푫 푵

Bending stress induced in the pin 흈풃 = ퟑퟐ 푴풃흅 풅ퟏퟑ

Step IV: Dimensions of Bushes

푴풕 = 푷 ×푫ퟐ × 푵

푴풕 = ퟏퟐ 풑풎푫풃풍풃푫푵

푴풕 = ퟏퟐ 푫풃

ퟐ푫푵

풍풃 = 푫풃

Step V: Dimensions of keys

Length of the key L = 풍풉

Shear stress 흉 = ퟐ푴풕풅풃풍

Crushing stress 흈풄 = ퟒ푴풕풅풉풍

Design of Flexible couplingDesign of Flexible coupling

Keys Keyway (or) keyseat &

key is a machine element used to connect a rotating machine element to a shaft.

Through this connection the key prevents relative rotation between the two partsand allows torque to be transmitted through.

The whole system is called a keyed joint. Commonly keyed components includegears, pulleys, and couplings.

Types of keys

Torque Applied

Reaction Torque

Shaft

Gear

Key

Resisting Tangential force

Stress analysis of a key

A key has two failure mechanisms.

•It can be sheared off.

• It can be crushed due to compressive bearing forces.

Note: Two parallel keys can be used either 90° or 180° apart from each other if the shaftconnection needs to be more robust.

Key with gear Key with pulley

Design of Helical Springs Flexible machine element Absorb energy regains its original shape after

removing load

Used in suspension systems of Automobiles and railway wagons to withstandsudden impact load.

(Compression coil helical spring)

(Tension coil spring)

Purpose of Design: To resist stretchingPurpose of Design: To resist compression

Helical Spring Nomenclature

Stresses in Helical Spring

Axially loaded helical spring

[Using force – couple method, axial applied load is replaced to the other point without

changing its effect.]

Applied force (F) Induces direct shear stress & Twisting moment (T) Induces torsional shear stressTherefore, Resultant shear stress = Direct shear stress+ torsional shear stress

Stress distribution in helical springs

Torque (T) acting on the spring

If d is the diameter of the coil wire and polar moment of inertia,

Torsional shear stress in the spring wire

Direct shear stress in the spring wire due to force F is

Resultant shear stress

Therefore, resultant shear stress the spring wire is

Curvature Effect Shear strain [γi ] inside > shear strain [γo ] outside

A spring with smaller diameter will experience more difference of shear strainbetween outside surface and inside surface compared to its larger counter part.

The above phenomenon is termed as curvature effect.

To take care of the curvature effect, the earlier equation for maximum shearstress in the spring wire is modified as,

Where, KW is Wahl correction factor, which takes care of both curvature effect andshear stress correction factor and is expressed as

Angle of twist θ = 푻 풍푮 푱

Where, T = Torque ( ) ; θ = Angle of twist; l = length of the bar; J = Polar moment of inertia; G = Modulus of rigidity

Angle of twist θ = [푭푫ퟐ ] 흅푫푵

푮 [ 흅ퟑퟐ 풅ퟒ]

θ = ퟏퟔ 푭 푫ퟐ푵푮 풅ퟒ

The axial deflection ‘δ’ of the spring, for small values of θ,

δ = θ (Length of the bracket)

δ = θ (푫ퟐ

)Therefore,

Axial deflection δ = ퟖ 푷 푫ퟑ푵푮 풅ퟒ

Therefore, rate of spring OR Stiffness k = = 푮풅ퟒ

ퟖ푫ퟑ푵

Strain energy stored in the spring E = ퟏퟐ

푷휹

Springs in series and parallel

풌 =풌ퟏ풌ퟐ풌ퟏ + 풌ퟐ

풇풐풓 풔풆풓풊풆풔 ; 풌 = 풌ퟏ + 풌ퟐ [풇풐풓 풑풂풓풂풍풍풆풍]

Types of Springs

Leaf springHelical compression spring

Helical tension spring Belleville spring

Springs under variable loading

Subjected to millions of stress cycles during its

lifetime.

Fluctuating force in a springchanges its magnitude from Pmax toPmin.

Mean force Pm = 푷풎풂풙 푷풎풊풏ퟐ

Amplitude force Pa = 푷풎풂풙 푷풎풊풏ퟐ

Valve spring of an IC engine

Fatigue failure of spring

Therefore,

Mean shear stress 흉풎 = 푲풔 ퟖ푷풎푫흅풅ퟑ

풘풉풆풓풆,푲풔 = ퟏ + ퟎ.ퟓ푪

Where Ks = shear correction factor

Amplitude shear stress 흉풂 = 푲 ퟖ푷풂푫흅풅ퟑ

풘풉풆풓풆, 푲 = ퟒ푪 ퟏퟒ푪 ퟒ

+ ퟎ.ퟔퟏퟓ푪

K = Wahl stress factor, which takes into consideration the effect of direct shear stress as well as stress concentration due to curvature.

Pulsating stress cyclePulsating stress cycle

In general, helical springs are subjected to pulsating shear stresses.

Fatigue diagram for spring designFatigue diagram for spring design

흉풎 − 흉풂흈풔풚

+ ퟐ흉풂흉풆

= ퟏ풏

흉풎 − 흉풂흈풔풚

+ ퟐ흉풂흉풆

= ퟏ풏

This equation is used in the design of springssubjected to fluctuating stresses.

Concentric springsConcentric springs

Benefits of concentric spring design

Fail safe system

Spring vibrations are eliminated

Increased load carrying capacity

The design of concentric springs is based on the following assumptions:

The springs are made of same material.

Maximum torsional shear stresses induced in outer and inner springs are equal.

Both springs are having same free length and deflected by the same amount.

In concentric spring,흉ퟏ = 흉ퟐ

푲ퟏ ퟖ푷ퟏ푫ퟏ흅풅ퟏ

ퟑ = 푲ퟐ ퟖ푷ퟐ푫ퟐ흅풅ퟐ

ퟑ

For the time being, neglect the effect of Wahl factor and assume K1 = K2

푷ퟏ푫ퟏ풅ퟏ

ퟑ = 푷ퟐ푫ퟐ풅ퟐ

ퟑ

Since the deflections of the two springs are equal,휹ퟏ = 휹ퟐ

[ퟖ 푷ퟏ 푫ퟏퟑ푵ퟏ

푮 풅ퟏퟒ] = [ퟖ 푷ퟐ푫ퟐ

ퟑ푵ퟐ

푮 풅ퟐퟒ]

[푷ퟏ 푫ퟏퟑ푵ퟏ

풅ퟏퟒ] = [푷ퟐ푫ퟐ

ퟑ푵ퟐ

풅ퟐퟒ]

When both springs are completely compressed, their adjacent coils touch each other.

Therefore,

Solid length of the outer spring = solid length of the inner spring

푵ퟏ풅ퟏ = 푵ퟐ풅ퟐ

[푷ퟏ 푫ퟏퟑ푵ퟏ풅ퟏ풅ퟏퟓ

] = [푷ퟐ푫ퟐퟑ푵ퟐ풅ퟐ

풅ퟐퟓ]

[푷ퟏ 푫ퟏퟑ

풅ퟏퟓ] = [푷ퟐ푫ퟐ

ퟑ

풅ퟐퟓ]

Therefore,푫ퟏ

ퟐ

풅ퟏퟐ=푫ퟐ

ퟐ

풅ퟐퟐ

푫ퟏ

풅ퟏ=

푫ퟐ

풅ퟐ= 푪

푷ퟏ

풅ퟏퟐ=푷ퟐ

풅ퟐퟐ

푷ퟏ푷ퟐ

= 흅풅ퟏퟐ

흅풅ퟐퟐ= 푨ퟏ

푨ퟐ

The load shared by each spring is proportional to the cross section area of wire.

Concentric springsConcentric springs

Surge in springsSurge in springs

Condition for resonance in springs,

Natural frequency of spring = Frequency of external periodic force

Wave of successive compressions of coils that travels from one end to other end and

back vibratory motion (surge)

Time required for the wave to travel from one end to other end = Time interval between

load applications

Surge is the main cause of failure in valve springs.

How to avoid surging?

Free Length: Axial length of an unloadedhelical compression spring.

Free length = compressed length + δ= Solid length + total axial gap + δ

Compressed Length: Axial length of thespring which is subjected to maximumcompressive force. There should be some gap orclearance between the adjacent coils.

Total gap = (N – 1) × gap between two coils

Solid Length: Axial length of the spring whichis so compressed that the adjacent coils toucheach other.

Solid length = N× d

Where N = total number of coils

Design of Leaf springDesign of Leaf spring

For the purpose of analysis, Leaves are divided into two groups.

Master leaf along with graduated length leaves

Extra full length leaves

P P

2P

Act as a structural member and carry lateral loads, brake torque, driving torque etc., in addition to shocks.

A simple cantilever type leaf spring is shown.

For case 1(uniform width)

For case 2 (non-uniform width)

Non uniform width leaf is a better design than a uniform width leaf.

Graduated length leaves as triangular plate

Portion of load P taken by graduated length leaves

Length of the cantilever

Bending stress in graduated length leaves

흈풃 품 = 푴풃푰풚 = ퟔ푷품푳

풏품풃풕ퟐ

Deflection

휹품 = 푷품푳ퟑ

ퟐ푬푰=

ퟔ푷품푳ퟑ

푬풏품풃풕ퟑ

Extra full length leaves as rectangular plate

Portion of load P taken by extra full length leaves

Bending stress in graduated length leaves

흈풃 풇 = 푴풃푰풚 = ퟔ푷풇푳

풏풇풃풕ퟐ

Deflection

휹풇 = 푷풇푳ퟑ

ퟑ푬푰= ퟒ푷풇푳

ퟑ

푬풏풇풃풕ퟑ

Since the deflection of full length leaves is equal to the deflection of graduated length leaves,휹품 = 휹풇

ퟔ푷품푳ퟑ

푬풏품풃풕ퟑ= ퟒ푷풇푳

ퟑ

푬풏풇풃풕ퟑ

푷품푷풇

= ퟐ풏품ퟑ풏풇

푷품 + 푷풇 = 푷Therefore,

푷풇 = ퟑ풏풇푷ퟑ풏풇 ퟐ풏품

and 푷품 = ퟐ풏품푷ퟑ풏풇 ퟐ풏품

흈풃품 = ퟏퟐ푷푳ퟑ풏풇 ퟐ풏품 풃풕ퟐ

and 흈풃풇 = ퟏퟖ푷푳ퟑ풏풇 ퟐ풏품 풃풕ퟐ

휹 = ퟏퟐ푷푳ퟑ

푬풃풕ퟑ(ퟑ풏풇 + ퟐ풏품)

Nipping of leaf springs

Stresses in extra full length leaves are greater than the graduated length leaves.

One of the methods of equalising the stresses in different leaves is to pre-stress the spring. It is achieved by different radii of curvature is known as ‘nipping’.

Belleville spring (Coned disc spring)

Useful where very large force is desired for small deflection ofthe spring

Used in plate clutches and brakes, relief valves and gun recoilmechanisms.

Disc type flywheel Rim type flywheel

Power Smoothening

Mechanical System

Power InputPower output + losses

Power Input

Power Output + Losses

Time

speedSpeed of the machinegradually increases

Power Output + Losses

Power Input

Time

speedSpeed of the machinegradually decreases

Both are unsteady system

Output Power

Input Power

Time

Output Power

Input Power

Time

Generator IC Engine

Output power is constant

Motor Punching machine

Output power is variable

Output PowerInput Power

Ideal steady operation

Fluctuation of Energy

Turning moment diagram for a single cylinder double acting steam engine.

Turning moment diagram for a four stroke internal combustion engine.

Maximum Fluctuation of Energy

Coefficient of Fluctuation of Energy (CE)

Ratio of the maximum fluctuation of energy to the work done per cycle.

Energy Stored in a Flywheel

Mean kinetic energy of the flywheel

As the speed of the flywheel changes from ω1 to ω2,

The maximum fluctuation of energy,

STRESSES IN FLYWHEEL RIM

Tensile stress due to centrifugal force

Tensile bending stress [caused by restraint of the arms]

Shrinkage stresses [due to unequal rate of cooling]

Tensile stress due to centrifugal force

Tensile bending stress [caused by restraint of the arms]

If the arms of a flywheel do not stretch at all and are placed very close together, thencentrifugal force will not set up stress in the rim σt will be zero.

If the arms are stretched enough to allow free expansion of the rim due to centrifugalaction, there will be no restraint due to the arms σb will be zero.

Arms of a flywheel stretch about 3/4th of the amount necessary for free expansion[Lanza]

STRESSES IN FLYWHEEL ARMS

Tensile stress due to centrifugal force acting on the rim.

Bending stress due to the torque transmitted from the rim to the shaft

Welded joints are permanent fasteners which areobtained by the fusion of edges of the two parts to bejoined together, with or without the application ofpressure and a filler material.

The heat required for the fusion of the material maybe obtained by burning of gas or by an electric arc.

Components held by mechanical forces - Riveted joints

Components held by molecular forces - Welded joints

Important types of welded joints are

1. Lap joint or fillet joint, and 2. Butt joint.

h

h

h

A steel plate, 100mm wide and 10mm thick is welded to another steel plate by means

of double parallel fillet welds. The plates are subjected to a static tensile force of 50 kN.

Determine the required length of the welds if the permissible shear stress in the weld is

94 N/mm2 .

Double parallel fillet welded Joints

L

Throat section subjected to shear stress

Single transverse double parallel fillet welded Joints

A plate, 75mm wide and 10 mm thick, is joined with another steel plate by means of

single transverse and double parallel fillet welds. The joint is subjected to a maximum

tensile force of 55 kN. The permissible tensile and shear stresses in the weld material are

70 and 50N/mm2 . Determine the required length of each parallel fillet weld.

Axially loaded unsymmetrical welded Joints

P1

P2

Under Equilibrium, sum of the horizontal forces acting is equal to zero.

P = P1 + P2

Under equilibrium, the moment of the forces about the C.G. is equal to zero.P1×a = P2×b

P1 = 0.707hLaτ ; P2 = 0.707hLbτ La×a = Lb×b

A 200×150×10 mm steel angle is to be welded to a steel plate by the fillet welds along

the edges of the 200 mm leg. The angle is subjected to a static load of 200 kN. The line

of action of the load is the intersection of the centroidal plane of the angle and the

plane of the weld. Find the lengths of the weld at the top and bottom, if the allowable

shear stress for the weld material is 75 MPa.

Q. A welded connection subjected to an eccentric force of 7.5 kN is shown.

Determine the size of the welds if the permissible shear stress for the weld is 100

N/mm2. Assume static conditions.

50

150

50

7.5 KN

Eccentric load in the plane of welds

Resultant shear stress = Primary shear stress + secondary shear stress(Direct shear) (Torsional shear)

P

A shaft of rectangular cross section is welded to a support by means of fillet welds.

Determine the size of the welds, if the permissible shear stress in the weld is limited to

75 N/mm2 .

Resultant stress = Direct shear stress + Bending stress

A circular shaft of diameter 50mm is welded to a support by means of a fillet welds.

Determine the size of the weld, if the permissible shear stress in the weld is limited to

100 N/mm2 .

Resultant stress = Direct shear stress + Bending stress

Consider an elemental section of area dA. It

is located at an angle θ with x-axis and

subtends an angle dθ.

Welded joint subjected to torsional moment

Torsional shear stress

흉 = 푴풕푱

×r

Polar moment of inertiaJ = Ixx + Iyy = πtr3 + πtr3

A welded connection shown in fig is subjected to an eccentric force of 60 kN in the

plane of welds. Determine the size of the welds, if the permissible shear stress for the

weld is 100 N/mm2. Assume static conditions.

Butt welded Joints

Advantages of welded joints over riveted joints

High joint efficiency

Lighter weight

Smooth appearance

Ease in alteration and addition

Less expensive

Ease in joining at difficult locations

Threaded Fasteners

Threaded fasteners are separable joints, heldtogether by means of bolts, nuts and washers.

Threads are machined by cutting helicalgroove on the cylindrical surface and hole.

High clamping force Small tightening force Simplemanufacturing process Self locking characteristics

Bolted Joint: for relatively small thickness components and where there is enough space toaccommodate parts.

Screw Joint: fixed into a threaded hole in one of the component being assembled not in anut.

Stud Joint: cylindrical rod threaded at both ends. One end of stud is screwed to nut and theother end is screwed into connecting components.

Nomenclature of the Threads

Bolt of uniform strength

Same stress level at different cross-sections of the bolt.

Reduced shank diameter is preferred over a bolt with an axial hole.

Eccentric load perpendicular to axis of bolt

Shear force on each bolt P1’ = P2’ = 푷푵풐 풐풇 풃풐풍풕풔

Each bolt is stretched by an amount ‘δ’ which is proportional to its vertical distance from the point ‘C’.

푷ퟏ 휶 풍ퟏ and 푷ퟐ 휶 풍ퟐ

푷ퟏ = 푪 풍ퟏ and 푷ퟐ = 푪 풍ퟐ

P e = 2푷ퟏ 풍ퟏ + ퟐ푷ퟐ 풍ퟐ

푷ퟏ = 푷풆풍ퟏퟐ (풍ퟏퟐ 풍ퟐퟐ)

and 푷ퟐ = 푷풆풍ퟐퟐ (풍ퟏퟐ 풍ퟐퟐ)

Shear stress on each bolt 흉 = 푷ퟏ

푨

Tensile stress in the bolt 흈풕 = 푷ퟏ

푨

Bolt denoted by ‘1’ are subjected to maximum force.

Eccentric load on circular base

P

P

Elastic analysis of bolted jointsP

Preload: When the nut is initially tightened, the bolt is subjected to

an initial tension, which is called preload (Pi).

Under the action of preload, the bolt is elongated by an amount δb

and the two parts are compressed by an amount of δc.

Stiffness of the bolt Kb = ∆푷∆휹

Stiffness of the component Kc = 푷 ∆푷∆휹

Dividing the equations,

The resultant load of the bolt Pb = Pi + ΔP

∆푷 = 푷 푲풃

푲풄 + 푲풃

Note:푷풊휹풃

= 푷풃 풎풂풙휹풃 휹풄

A rivet is a short cylindrical bar with a head integral to

it. The cylindrical portion of the rivet is called shank or

body and lower portion of shank is known as tail.

Method of Riveting

Rivet Parts

Caulking and Fullering: To make the joints leak proof or fluid tight.

Caulking tool closes the surface asperities and cracks on the contacting surfaces between

two plates and also between the rivet and the plates, resulting in leak proof joints.

Fullering is similar to caulking except the shape of the tool. The blows of the fullering tool

result in simultaneous pressure on the entire edge of the plate.

Pitch [P]. It is the distance from the centre of one rivet to the centre of the next rivetmeasured parallel to the seam.

Back pitch [Pb]. It is the perpendicular distance between the centre lines of the successiverows.

Diagonal pitch [Pd]. It is the distance between the centres of the rivets in adjacent rows ofzig-zag riveted joint.

Triple Riveted Lap Joint

Double riveted double strap (equal) butt jointSingle riveted double strap butt joint

Double riveted double strap (unequal) butt joint

Failures of a Riveted Joint

Crushing of a RivetShearing off a rivet in double cover butt joint

Shearing off a rivet in single cover butt joint

Shearing off a rivet in lap joint Tearing of the plate across therows of rivets

Tearing of the plate at an edge

Strength equations for riveted joints

In analysis of riveted joints, mainly three types of failure areconsidered. They are as follows:Shear failure of the rivet:

Shear strength of the rivet Ps = 흅ퟒ

d2. τ

Tensile failure of the plate between the rivets:Tensile strength of the plate Pt =(p – d).t. σt

Crushing failure of the plate:Crushing strength of plate Pc = d. t.σc

Efficiency of the joint = 푳풐풘풆풔풕 풐풇 푷풔,푷풕 풂풏풅 푷풄푺풕풓풆풏품풕풉 풐풇 풕풉풆 풔풐풍풊풅 풑풍풂풕풆

= 푳풐풘풆풔풕 풐풇 푷풔,푷풕 풂풏풅 푷풄푷

Where, Strength of the solid plate ‘P’ = p. t. σt

A double riveted double cover butt joint in plates 20 mm thick is made with 25 mm diameter rivets at

100 mm pitch. The permissible stresses are : σt = 120 MPa; τ = 100 MPa; σc = 150 MPa. Find the

efficiency of joint, taking the strength of the rivet in double shear as twice than that of single shear.

Tearing resistance of the plate

Tearing resistance of the plate per pitch length Pt =( p – d ) t × σt = (100 – 25) 20 × 120 = 180 000 N

Shearing resistance of the rivets

Since the joint is double riveted butt joint, therefore the strength of two rivets in double shear istaken.

Ps = n × 2 × π/4 × d 2 × τ = 2 × 2 × π/4 (25)2 ×100 = 196 375 N

Crushing resistance of the rivets

Since the joint is double riveted, therefore the strength of two rivets is taken. We know that crushing resistance of the rivets,

Pc = n × d × t × σc = 2 × 25 × 20 × 150 = 150 000 N

Efficiency of the joint

The strength of the unriveted or solid plate, P = p × t × σt = 100 × 20 × 120 = 240 000 N

Efficiency of the joint = 푳풐풘풆풔풕 풐풇 푷풔,푷풕 풂풏풅 푷풄푺풕풓풆풏품풕풉 풐풇 풕풉풆 풔풐풍풊풅 풑풍풂풕풆

= 0.625 or 62.5%

1. Thickness of the vessel

t= 풑 푫ퟐ흈풕휼

+ 푪풐풓풓풐풔풊풐풏 풂풍풍풐풘풂풏풄풆2. Diameter of rivets (If t > 8mm then use Unwin’s formula)

풅 = ퟔ 풕3. Pitch of the rivets (According to IBR)

pmin = 2dpmax =Ct + 41.28

pt = 0.2p + 1.15d (distance between outer and middle row)

pt=0.165p+0.67d (distance between middle and inner row)

Margin m = 1.5d4. Thickness of straps

t1= 0.625t [ 풑 풅풑 ퟐ풅

]

5. Efficiency of the joint

Tensile strength of plate per pitch length Pt = (p – d) t흈풕

Shear strength of rivets per pitch length Ps = 1.875n (흅ퟒ

풅ퟐ흉)

Crushing strength Pc = n.d.t.흈풄

Tensile strength of the solid plate per pitch length P = p.t. 흈풕

Efficiency of the joint = 푳풐풘풆풔풕 풐풇 푷풔,푷풕 풂풏풅 푷풄푺풕풓풆풏품풕풉 풐풇 풕풉풆 풔풐풍풊풅 풑풍풂풕풆

Eccentrically loaded riveted Joints in shear

Direct load ‘P’ at C.G results in primary shear forces P1’, P2’, P3’, P4’ (Reactionforces).

P1’ = P2’ = P3’ =P4’ = 푷푵풐.풐풇 풓풊풗풆풕풔

The moment at C.G results in secondary shear forces P1’’, P2’’, P3’’, P4’’

P.e = P1’’.r1 + P2’’.r2+ P3’’.r3+ P4’’.r4

Secondary shear force at any rivet is proportional to its distance from the C.G.

P1’’=C. r1 ; P2’’=C. r2 ; P3’’=C.r3 ; P4’’= C.r4

푪 = 푷.풆

풓ퟏퟐ + 풓ퟐퟐ + 풓ퟑퟐ + 풓ퟒퟐ

P1’’= 푷.풆.풓ퟏ풓ퟏퟐ 풓ퟐퟐ 풓ퟑퟐ 풓ퟒퟐ

Design of Connecting RodDesign of Connecting Rod

Big End

Small End

Shank

Castle NutBolt

Transmits reciprocating motion of the piston into the rotary motion of the crank shaft

Buckling of Connecting RodBuckling of Connecting Rod Buckling in the plane of motion (Ends are Hinged) (n = 1)

Buckling in the plane perpendicular to the plane of motion (Ends are fixed) (n = 4)

Connecting rod is four times stronger for buckling about the YY axis as compared tothe buckling about the XX axis.

Therefore, for equal resistant to buckling in both the planes, Ixx = 4 Iyy

B = 4t

H = 5t

t

t

t

X X

Y

Y

Proportions for the cross section of connecting rod

Ixx = 3.2 Iyy

Width of ‘B’ is kept constantthroughout the length of theconnecting rod

Height ‘H’ varies from the big end tosmall end

At the middle section,H= 5t

At the small end,H1 = 0.75 H to 0.9 H

At the big end,H2 = 1.1H to 1.25H

Cross section for Connecting RodCross section for Connecting Rod

θ

φ

P

PS

PC

PISTON

CONNECTING ROD

CRANK

P = force acting on the piston due to gas pressure (N)

Ps = side thrust on the cylinder wall (N)

Pc = force acting on the connecting rod (N)

φ = angle of inclination of connecting rod with line of stroke

θ = angle of inclination of crank from TDC position

PC

PS

P

φ

P = Pc 풄풐풔 ∅

푷풄 = 푷

퐜퐨퐬∅

At φ = 3.3°, maximum load occursshortly after the TDC

Step I: Force acting on the connecting rod is equal to the maximum force acting on the piston due to gas pressure

푷풄 = 흅푫ퟐ

ퟒ 풑풎풂풙

Step II: Critical buckling load

푷풄풓 = 푷풄 푭푶푺 [FOS = 5 OR 6]

Step III: Calculate dimensions by applying Rankine’s formula

Pcr = critical buckling load (N)

σc = compressive yield stress (N/mm2) [σc = 330 N/mm2]

A = cross sectional area of connecting rod (mm2) [A = 11t2]

a = constant depending upon material and end fixity coefficient [a = 1/7500]

L = length of the connecting rod (mm)

Kxx = radius of gyration (mm) [Kxx = 1.78t]

Different failures in Connecting RodDifferent failures in Connecting Rod

Side buckling Front rear buckling

Catastrophic buckling Plastic torque Fatigue failure

Converts reciprocating motion of the piston into rotary motion through the connectingrod.

It Consists of crank pin, crank web and shaft.

Plain carbon steels 40C8, 45C8 and 50C4, Alloy steels 16Ni3Cr2, 35Ni5Cr2 and40Ni10Cr3Mo6

Multi throw crank shaft

Overhung crank

Has one crank web and twobearings

Used in medium size engines andlarge size horizontal engines

Has two crank webs and three bearings

More popular in automotive engines

Used in radial aircraft engines, andmarine engines

High Strength

Minimum mass

Effective sealingHigh wear resistance

Heat dissipation

capacity

High Speed with minimum

noise

Resistance to distortion

Piston: To receive the impulse from the expanding gas and to transmit the energy

to the crank shaft through the connecting rod.

Head

Rings

Skirt Piston pin

Cast iron Piston Aluminium Piston

Moderately rated engines with Piston speed below 6 m/s

High temperature strength

Highly rated engines with greater Pistonspeed

α Aluminium = 2.5 times α cast iron

K Aluminium = Nearly 4 times K cast iron

Keeps down the maximum temperaturedifference between centre and edges of thepiston head

Aluminium alloys are three times lighter than

Cast iron.

Aluminium alloys are three times lighter than

Cast iron.

Piston Head (or) Crown

To withstand the straining action due to the pressure of explosion inside the enginecylinder

Quick dissipation of heat to the cylinder walls

According to Grashoff’s formula,

Thickness of the piston Head

p = Maximum Gas pressure, N/mm2

D = cylinder bore or outside diameter of piston (mm)

σt = Permissible bending (tensile) stress for the material of the piston, N/mm2

For , Grey cast iron 35 to 40 MPa

Nickel cast iron 50 to 90 MPa

Aluminium alloy60 to 100 MPa

Treating the piston head as a Flat circular plate,

Piston rings

Compression rings (or) Pressure rings

Oil control rings

Acts as a seal between the piston and cylinder bore

Imparts radial pressure

Transfer the heat from the piston to cylinder liner

Absorb fluctuations of piston due to side thrust

Provide proper lubrication to the liner

Radial thickness (t1) of the ring

The axial thickness (t2) of the rings may be taken as 0.7 t1 to t1.

The minimum axial thickness (t2) may also be obtained from the following empirical relation

Piston Barrel

Maximum thickness of the piston barrel

b = Radial depth of piston ringgroove which is taken as 0.4 mmlarger than the radial thickness of thepiston ring (t1)

Piston skirt Acts as a bearing for the side thrust from the cylinder wall.

For low speed engines,

Bearing pressure on the piston barrel due to side thrust does not exceed 0.25 N/mm2.

For high speed engines,

Bearing pressure on the piston barrel due to side thrust does not exceed 0.5 N/mm2.

Piston pinFull floating type Semi floating type

Design check for bending:

Piston failuresDamage From Running Unmixed Fuel

Damage From Debris Getting Through the Air Filter

Damage From Detonation

Damage From Heat Seizure

DESIGN OF KNUCKLE JOINT

Knuckle Pin

Collar

Taper Pin

ForkEye

The modes of failure are :

1.Shear failure of pin.

2.Crushing of pin against rod.

3.Tensile failure of flat end bar.

4.Tensile failure of fork and eye

5. Shearing out of pin from fork and eye

A knuckle joint is used to connect the two rods which areunder the tensile load, when there is requirement of smallamount of flexibility or angular movement is necessary. There isalways axial or linear line of action of load.

Failure of the solid rod in tension

Failure of the knuckle pin in shear

Failure of the knuckle pin in bending

Failure of the single eye or rod end in shearing

Failure of the single eye or rod end in crushing

Failure of the single eye or rod end in tension

Failure of the forked end in tension

Failure of the forked end in shear

Failure of the forked end in crushing

Q. Design a knuckle joint to transmit 150 kN. The design stresses may be taken as 75 MPa intension, 60 MPa in shear and 150 MPa in compression.