MIT2.810Fall2016 Homework5Solutions
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MIT2.810ManufacturingProcessesandSystemsHomework5SolutionsSheetMetalForming
RevisedOctober7,2015
Problem1.SpringbackRankthemetalsinTable2.2(Kalpakjian,alleditions)intermsofspringback,listingthosewiththelargestspringbackfirst.Answer:SpringbackislargerformaterialswithlargervaluesofY/E.
Materialsrankedintermsoftheirspringbackbehavior E[GPa] Y[MPa] Y/ETi 105 862 8.2x10-3Mg 43 218 5.1x10-3Steel 195 965 5.0x10-3Cu 128 588 4.6x10-3Al 74 293 4.0x10-3Ni 197 653 3.3x10-3Mo 345 1075 3.1x10-3W 375 620 1.7x10-3Pb 14 14 1.0x10-3
Note:ThevaluesinthisTableareaveragedandthereforecandeviateconsiderably.Example:Mildsteelcanhaveayieldstrengthofapproximately350MPa,whichleadstoaY/Eratioof1.53x10-3,avalueevensmallerthantheoneforTungsten(W).
MIT2.810Fall2016 Homework5Solutions
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Problem2.ConsiderthemanufactureofthebracketshowninFigure1fromsheetsteel(availablein4in.widestrips) using a manual brake press or presses. Assume UTS = 70,710 psi. We are interested inestimatingtheunitcosttomakethesebrackets.PleaseuseinformationfromBoothroyd’s“DesignforSheetMetalworking”andthefollowingassumptions:
1. Thesheetmaterialcosts$0.25/lb.2. Thecostofapressis:Presscost[$]=2[$/lb]×forcecapacity[lb].3. All operations take the same length of time, with no breakdowns or unusual delays
betweensteps.4. Laborcosts$30/hr.5. Equipmenthasalifeof10years,tools—of1year.
(a) Describe the sequence operations needed to make this bracket, including materials, tools,
machinesandoperators.(b) Estimatethesizeandcostoftheequipmentneededtomakethispart.(c) Estimatethetimerequiredtomakeoneofthese(ignorewaitingtime).(d) Estimate the cost per bracket (materials, labor, and equipment) assuming 106 parts are
produced.Stateanyassumptions,beyondthosegiven,explicitly.(e) Estimatetheminimumbendradiusforthepart.(f) If thepart is formed to90±,makea roughestimateof the resultingangledue to the spring
back.Use1.5x(yourminimumbendradius).Howwouldthischangeifthebracketmaterialwasaluminumortitanium(roughanswerisOK)?Howwouldyoucompensateforspringback(listatleast3ways)?
Figure1:Sheetsteelbracket.
MIT2.810Fall2016 Homework5Solutions
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Answer:
(a) Sequenceofoperations:1. Cuttingandholepunching:
Material:sheetmetalstrip,4”wideMachine:stampingpressTool:compounddieset(withcuttingandpunchingdies)(asinFig16.11,Kalpakjian,7thed)Operator:1
2. Bending:Material:sheetmetalstripforeachbracketMachine:pressbrakeTool:V-dieOperator:1
Note:AnalternativeforStep1canbeaprogressivediewith2stations,punchingandcut-off.Inthiscase,thesheetmetalstripwouldneedtobesuppliedcontinuouslyasacoil.)
(b) Machineforcecapacities:
L=perimeterofcutorbend(varies)T=thickness=0.1inUTS=ultimatetensilestrength=70,710psiW=V-dieopening≈2in
1. Cutting:maximumshearingforce(equation16.1inKalpakjian,7thed.)
𝐹! = 0.7 𝑇𝐿(𝑈𝑇𝑆) = 0.7 0.1 𝑖𝑛 4 𝑖𝑛 70,710 𝑝𝑠𝑖 =
= 19799 𝑙𝑏𝑠 4.448𝑁𝑙𝑏
= 88 𝑘𝑁
2. Holepunching:maximumshearingforce
𝐹!,!!"# = 0.7 𝑇(𝜋𝑑)(𝑈𝑇𝑆) = 0.7 0.1 𝑖𝑛 𝜋 0.25 𝑖𝑛 70,710 𝑝𝑠𝑖 = 3887 𝑙𝑏𝑠 = 17.3 𝑘𝑁 𝐹!,!"!#$ = 17.3 𝑘𝑁 × 4 = 15548 𝑙𝑏𝑠 = 69 𝑘𝑁
3. Bending:maximumbendingforce(equation16.8inKalpakjian,7thed.)
𝐹! =𝑈𝑇𝑆 𝐿𝑇!
𝑊=
70,710 𝑝𝑠𝑖 (1 𝑖𝑛) (0.1 𝑖𝑛)!
(2 𝑖𝑛)= 354 𝑙𝑏𝑠 = 1573 𝑁
Presscosts:1. Stampingpress:
𝐶!"#"$%& ≈$2𝑙𝑏
𝑙𝑏𝑠 𝑓𝑜𝑟𝑐𝑒 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 = $2 19799 + 15548 lbs ≈ $70,700
2. Pressbrake:𝐶!"#$%& ≈ $2 354 lbs ≈ $708
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Diecosts:1. Cuttingand4-holepunchdieset:
Usablearea:𝐴! = 4 𝑖𝑛 2.5 !"!"
1 𝑖𝑛 2.5 !"!"
= 25 𝑐𝑚!
Diesetmaterialcost:𝐶!" = $120 + 0.36𝐴! ≈ $130 Timefordiemaking:Basetime:𝑀!" = 23 + 0.03𝑏𝑤 = 23 + 0.03 4 𝑖𝑛 2.5 !"
!"1 𝑖𝑛 2.5 !"
!"≈ 24 ℎ𝑟
Standardpunchshapetime:𝑀!" = 𝐾𝑁! + 0.4𝑁! = 2 4 + 0.4(1) = 8.4 ℎ𝑟 Totalcostforblankingandpunchingdieset:𝐶!"#$ = $130 + 24 + 8.4 ℎ𝑟 $"#
!!= $1102
2. BendingV-die:
Diesetmaterialcost:𝐶!" = $120 + 0.36𝐴! ≈ $130(asabove)Timefordiemaking:Basetime:𝑀!" = 18 + 0.023𝑏𝑤 0.9 + 0.02𝐷 =
= 18 + 0.023 4 𝑖𝑛 2.5𝑐𝑚𝑖𝑛
1 𝑖𝑛 2.5𝑐𝑚𝑖𝑛
0.9 + 0.02 2 𝑖𝑛 2.5𝑐𝑚𝑖𝑛
≈ 19 ℎ𝑟
Bendlengthtime:𝑀!" = 0.68𝐿! + 5.8 𝑁! = 0.68 1 𝑖𝑛 2.5 !"!"
+ 5.8(1) = 7.5 ℎ𝑟TotalcostforV-die:𝐶!"#$ = $130 + 19 + 7.5 ℎ𝑟 $"#
!!= $925
(c) Timeestimateperpart:
UsingthecycletimefromBoothroyd(equation9.22),t=3.8+0.11(L+W)secondsforloadingandremovingapartmanuallyfromabrakepress.Thetimeforthestampingpressstepwillbelesssinceithasautomatedpartejection.Assuming2operatorsareoperatingthe2machinessimultaneously,thetimeforonepartwillbelimitedbythelongersteptime(bending):𝑡 = 3.8 + 0.11 𝐿 +𝑊 = 3.8 + 0.11 4 + 1 𝑖𝑛 = 4.35 𝑠 = 1.21×10!!ℎ𝑟perpart
(d) Costperbracket:
Materialcosts:𝐶!"#$ = $0.25/𝑙𝑏 0.15 𝑙𝑏𝑠 = $0.038/𝑝𝑎𝑟𝑡Laborcosts:𝐶!"#$% = $30/ℎ𝑟
𝐶!"#$ =𝐶!"#$%𝑁
+ 𝐶!"# =𝐶!"#"$%& + 𝐶!"#$%& + 𝐶!"#$ + 𝐶!"#$
𝑁+ (𝐶!"#$% + 𝐶!"#$) =
=$70,700 + $708 + $1,102 + $925
𝑁+
$30ℎ𝑟
1.21×10!! ℎ𝑟 + $0.038 =
=$73,435
𝑁+ $0.074
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If10!partsareproduced,thecostperbracketis:
𝐶!"#$ =$73,43510!
+ $0.074 ≈ $0.15
(e) Theminimumbendradiusistheradiusatwhichacrackfirstappearsattheouterfibersofasheetbeingbent(Sec.16.5,Kalpakjian,7thed).Engineeringstrainduringbending:
𝑒 =1
2𝑅𝑇 + 1
=𝑇
2𝑅 + 𝑇
2𝑅 =𝑇𝑒− 𝑇
𝑅!"# = 0.5𝑇
𝑒!"#− 𝑇
Here,T=0.1inandemax=maximumstrainatfracture=ductilityofthematerial=!!!!!!!
=
0.22forlow-carbonsteel(Boothroyd,p.403).Thus,𝑅!"# = 0.5 !.! !"
!.!!− 0.1 𝑖𝑛 = 0.18in.
(f) Fromslide64ofthesheetmetalforminglecture,theequationforspringbackradiusis:1𝑅!−1𝑅!
= 3𝑌𝐸𝑇
− 4𝑅!!𝑌𝐸𝑇
!
whereRi=1.5(0.18in)=0.27in,T=0.1inFromthediagramonthesameslide,Y/Esteel=1/909Y/EAl=1/212Y/ETi=1/188
Thenthefinalradiiwouldbe: !!!= − !
!!− 3 !
!"+ 4𝑅!!
!!"
!
Steel:𝑅! =!
0.27 − 3!
!".!+ 4 0.27 ! !
!".!
! !!= 0.2724in
Aluminum:𝑅! =!
0.27 − 3!
!".!+ 4 0.27 ! !
!".!
! !!= 0.2807in
Titanium:𝑅! =!
0.27 − 3!
!".!+ 4 0.27 ! !
!".!
! !!= 0.2822in
Estimatethefinalangleafterspringback,assumingitisinitiallybentto𝛼! = 90°:Iftheradiiaremeasuredtothecenterlineofthemetalsheet,thenthelengthofthecenterlinearcis𝑙 = 𝑅!𝛼! = 𝑅!𝛼!(Figure2).
MIT2.810Fall2016 Homework5Solutions
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Figure2.Diagramofspringbackangles.
Steel:𝛼! =!!!!𝛼! =
!.!"!.!"!#
90° = 89.2°,𝜃! = 180° − 𝛼! = 90.8°
Aluminum:𝛼! =!.!"!.!"#$
90° = 86.6°,𝜃! = 180° − 𝛼! = 93.4°
Titanium:𝛼! =!.!"!.!"!!
90° = 86.1°,𝜃! = 180° − 𝛼! = 93.9°SpringbackislargestforTitanium.
Compensationmethods:1. Over-bending2. Bottoming3. Stretchbending
