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By

GAUtam janaRDHan(09A207)

AhmadullaH M(10A202)Ashwin A(10A206)

Jayanth R(10A215)

Ramki A(10A238)

Gowtham kumar(11A434)

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While designing a component there are no factors

which are complex to evaluate with accuracy.Some

of these are

magnitude of external force variation in properties of materials

variation in changes of dimension during

castingTo overcome thesefactor of safetyis used

FoS decribes the structural capacity of system beyond

the expected loads or actual loads

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Type of load:: FoS low if the external force

acting is static .if force actiing is impact load

Degree of accuracy of Force analysis::if force

acting is uncertain FoS is high .if the force

acting determined precisely FoS is low

Material of component::component made up of

homogenous ductile material FoS is low.if

material is non-homogenous FoS is high

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SELECTION OF FOS

The selection of FoS is one of the difficult tasks facedby the designer.Some guidelines for selection of FoS

cast iron for cast iron components ultimatetensile strength is considered to be the failure

criterion.Cast iron is non-homogenous structure andthere are residual stresses.A large FoS usually3 to 5 isused

ductile materials(like steel):yield strength isconsidered to be the failure criterion.These ductilecomponents have a homogenous structure andresidual stresses can be relieved by proper heattreatment process.The recommended FoS is 1.5 to 2

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SELECTION OF FOS

For components subjected tofluctuating forcesendurance limit is considerde to be the failurecriterion.These components fail due to fatigue.Anumber of factors like stress concentration,surface

finish and size affect the endurance limit. Therecommended FoS is 1.3 to 1.5

For components like cams and followers,gears,rollingcontact bearings FoS is calculated based on endurancelimit. The recommended FoS is 1.8 to 2.5

For components like piston rods,power srewsor studs ,are designed on basis of bucklingconsiderations.The recommended FoS is 3 to 6

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CALCULATION OF FOS

General formula

FoS=failure stress/allowable stress

For ductile materials

FoS=Syt/allowable stress

For brittle materials

FoS=Sut/allowable stress

where Sytyield strength

Sutultimate strength

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DESIGN FOR AXIAL STRESSProcedure

Selection of material

Selection of FoS

Calculation of permissible stress

Calculation of dimensions

Problem

It is required to design a cotter joint to connect twosteel rods of equal diametre.Each rod is subjected to anaxial tensile force of 50kN.Design the joint and specifyits main dimensions.

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DESIGN FOR AXIAL STRESSSTEP 1:

Selection of materialon basis of strength the material of the two

rods and the cotter is selected as plaincarbon steel of Grade 30C8(Syt=400N/mm)

STEP 2:

Selection of component FoS is taken 4for cotter joint for following reasons no stressconcentration in cotter cost of cotter is less

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DESIGN FOR AXIAL STRESSSTEP 3:

Calculation of permissible stresst=Syt/FoS

=400/4=100N/mmSTEP 4:

calculation of dimensionsd= (4*p/) 0.5

=(4*50*10/*66.67)0.5

=30.90mm

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DESIGN FOR BENDING STRESSPROCEDUREStep 1:Calculation of stresses

b=Mb/Z

STEP 2:

SHEAR STRESS=T/Zp

STEP 3:

Calculation of dimension

y/FoS=(+(4*))0.5

d=___

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DESIGN FOR TORSIONAL STRESSStep 1:

=P/A

Step 2:=T/Zp

Step 3:

( +(3*))0.5=y/FoSd=__