-
8/22/2019 Case Study 3 ME403
1/5
DepartmentofMechanical&AerospaceEngineering
ME403EngineeringMaterialsSelection
CaseStudy3
GroupMembers: KamalShayed
VernonLim
JonathanTee
GaryLai
TohHanWei
-
8/22/2019 Case Study 3 ME403
2/5
1. Toevaluatetheshapeefficiencyfactor,!!!
forastiffness-limiteddesign
forabeamloadedinbending,weneedtoassumeasquare-shapedcross-section
astheneutralreferencesection.
1(a)
!"#$%&!"!#$%!"!"#!!"!"#!"#!"#"!"$%"!"#$%&', !!=
!!
!
12
!"#$%&!"!#$%!"!"#!!"!"#$%&', ! = !4 5! ! 4! ! =
92.25!!!!"#$!"!"#$%&',! = ! 5! ! 4! ! = 9!!!
!!"#!""#!"#$!%!"#$%&,!!! = !!!
=
!
!!
=
12!
!!
!=
12!
!!
!!!=
12(92.25!!!)
(9!!!)!= 4.362
1(b)
! =! 10!
!
12+ 2
4! !!
12+ 4!! 4.5!
!= 246!
!
! = 10! ! + 2 4! ! = 18!!
!!!
=
12(246!!)
(18!!)!= 9
.
111
1(c)
-
8/22/2019 Case Study 3 ME403
3/5
Sincefromobservation,wecantellthatthesection(c)isasymmetrical
expansionofsection(b),wehave:
! = 2246!!= 492!
!
! = 218!!= 36!
!
!!!=
12(492!!)
(36!!)!=
4.
55
2. Similarly,toevaluatetheshapeefficiencyfactor,!!
!forastrength-limited
designforabeamloadedinbending,weneedtoassumeasquare-shapedcross-
sectionastheneutralreferencesection.
!"#$%&'!"#$%$&!!!!!"#$%&!"#"!"$%"!"#$%&',!!=
!!
!!"#,!
=
!!
!!
6
!!"#!""#$#!%$&!"#$%&,!!
!=
!!
!!,!
=
!"
!!!
=
!
!!
=
6!
!!"#
!!!
2(a)UsingresultsforIandA,derivedfrompart1(a):
!!
!=
6(92.25!!!)
(5!)(9!!!)!!
= 2.313
2(b)
UsingresultsforIandA,derivedfrompart1(b):
!!
!=
6(246!!)
(5!)(18!!)!!
= 3.866
2(c)
UsingresultsforIandA,derivedfrompart1(c):
!!
!=
6(492!!)
(5!)(36!!)!!
= 2.733
3(a)
Forastrength-limiteddesignforabendingmoment,M=104Nm,wewill
comparethemassperunitlengthforthevariousmaterials.Thefigurebelow
showstheuseofa4-segmentcharttoevaluatetheperformanceofthevarious
shapedmaterials.
(i)
MildSteelwithashapefactorof10,Strengthof200Mpaanddensityof7900kg/m3isrepresentedbytheredline.Theresultyieldsanm/L
ofabout8.(ii)
6061gradealuminiumalloywithashapefactorof3,Strengthof
200Mpaanddensityof2700kg/m3isrepresentedbytheblueline.
Theresultyieldsanm/Lofabout5.
(iii)
Titaniumalloywithashapefactorof10,Strengthof480Mpaanddensityof4420kg/m3isrepresentedbythegreenline.Theresult
yieldsanm/Lofabout2.
Basedonmassperunitlengthasthebasisofperformanceorefficientuseof
resources,wecanconcludethattitaniumisthebestchoicefollowedby6061
aluminiumalloyandlastlymildsteel.
-
8/22/2019 Case Study 3 ME403
4/5
3(b)Objective:minimizemassperunitlength!
!= !" ! =
!
!"
!"#$%#&!"#$%&",!! = !!!!!"#!"#$%&'#,!
!
!=
6!
!!!
! =!!
!!!!
6
! =
!!
!(!
!")!!
6
!! =
!!
!(!
!")!!
6!!
!
!= (6!!)
!![
!
!!
!!!
!!
]
SinceMfisaconstant,thematerialindexis!!
!!!
!!
!,andmaximizingthiswould
minimizethemassperunitlength.
(i) Formildsteel, !!!!!!
!
!= 200.94
-
8/22/2019 Case Study 3 ME403
5/5
(ii) For6061gradealuminiumalloy, !!!!!!!
!= 263.47
(iii) Fortitaniumalloy, !!!!!!!
!= 643.78
Therefore,theresultfrompart(a)areconsistentwiththeresultsofpart(b),
sincetitaniumwasfoundtohavethehighestmaterialindex,followedby
aluminiumalloyandlastlymildsteel.
