ed batch 2
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8/13/2019 ED BATCH 2
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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=__
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