Solution Manual for Metal Forming: Mechanics and Metallurgy, 4th Edition (Complete Solutions)

All Chapters Covered
h h




SOLUTION MANUAL
h

, SolutionhManualh3rdhEd.hMetalhForming:hMechanicshandhMetallurgyh
Chapterh1


Determinehthehprincipalhstresseshforhthehstresshstate
10 3 4
hijh h 3 5 2h.
4 2 7
Solution: I1h=h10+5+7=32,hI2h=h-(50+35+70)h +9h+4h +16h=h-126,h I3h =h350h-48h-40h-80
-63h=h119;h h  h–h222h-126h-119h=h0.h Ahtrialhandh errorhsolutionhgiveshh-=h13.04.
3

 Factoringhouth13.04, 2h-
8.96h +h9.16h=h0.hSolving;h h =h13.04,hh =h7.785,hh =h1.175.

1-2 Ah5-
cm.hdiameterhsolidhshafthishsimultaneouslyhsubjectedhtohanhaxialhloadhofh80hkNhandhahtorqueh
ofh400hNm.
a. Determinehthehprincipalhstresseshaththehsurfacehassuminghelastichbehavior.
b. Findhthehlargesthshearhstress.
Solution:ha.hThehshearhstress,h,hathahradius,hr,hishh=hsr/Rhwherehsishthehshearhstresshaththehsu
rfacehRhishthehradiushofhthehrod.hThehtorque,hT,hishgivenhbyhTh=h∫2πtr2drh=h(2πsh/R)∫r3dr
=hπsR3/2.hSolvinghforh=hs,hsh=h2T/(πR3)h=h2(400N)/(π0.0253)h=h16hMPahTh
ehaxialhstresshish.08MN/(π0.0252)h=h4.07hMPa
1,2h =h4.07/2h±h[(4.07/2)2h +h(16/2)2)]1/2h=h 1.029,h -0.622h MPa
b.hthehlargesthshearhstresshish(1.229h+h0.622)/2h=h0.925hMPa

Ahlonghthin-
wallhtube,hcappedhonhbothhendshishsubjectedhtohinternalhpressure.hDuringhelastichloading,hdo
eshthehtubehlengthhincrease,hdecreasehorhremainhconstant?
Solution:hLethyh=hhoophdirection,hxh=haxialhdirection,handhzh=hradialhdirection.h–
hexh=he2h=h(1/E)[h-h(h3h+h1)]h=h(1/E)[2h-h(22)]h=h(2/E)(1-2)


Sincehuh<h1/2hforhmetals,hexh=he2hishpositivehandhthehtubehlengthens.

4 Ahsolidh2-
cm.hdiameterhrodhishsubjectedhtohahtensilehforcehofh40hkN.hAnhidenticalhrodhishsubjectedhtohah
fluidhpressurehofh35hMPahandhthenhtohahtensilehforcehofh40hkN.hWhichhrodhexperienceshthehla
rgesthshearhstress?
Solution:hThehshearhstresseshinhbothharehidenticalhbecausehahhydrostatichpressurehhashnohshe
arhcomponent.

1-5 Considerhahlonghthin-
wall,h5hcmhinhdiameterhtube,hwithhahwallhthicknesshofh0.25hmmhthathishcappedhonhbothhends.h
Findhthehthreehprincipalhstresseshwhenhithishloadedhunderhahtensilehforcehofh40hNhandhanhinter
nalhpressurehofh200hkPa.
Solution:hxh=hPD/4th+hF/(πDt)h=h12.2hMPa
yh=hPD/2th=h 2.0hMPa

1

, yh=h0




2

, 1-6 Threehstrainhgaugesharehmountedhonhthehsurfacehofhahpart.hGaugehAhishparallelhtoht
hehx-axishandhgaugehChishparallelhtohthehy-
axis.hThehthirdhgage,hB,hishath30°htohgaugehA.hWhenhthehparthishloadedhthehgaugeshread
GaugehA 3000x10-6
GaugehB 3500hx10-6
GaugehC 1000hx10-6
a. Findhthehvaluehofhxy.
b. Findhthehprincipalhstrainshinhthehplanehofhthehsurface.
c. SketchhthehMohr’shcirclehdiagram.
Solution:hLeththehBhgaugehbehonhthehx’haxis,hthehAhgaugehonhthehx-axishandhthehChgaugehon
2 2
thehy-axis.hexxhexxhxhxheh yyhxyh hhxyhxxhxyh,hwherehhxxh=hcosexh=h 30h=h√3/2handhhxyh=
cosh60h=h½.hSubstitutinghthehmeasuredhstrains,h3500
h=h3000(√2/3) h–h1000(1/2) h+hxy(√3/2)(1/2)
2 2
xy
h=h (4/√3/2){3500-[3000(1000(√3/2)1/2+1000(1/2) ]}h =h2,309h(x10 )  2
2 2 -6


b.h e1,e2h =h(exh+ey)/2±h [(ex-ey)2h +h xy2] /2h=h(3000+1000)/2h±h[(3000-1000)h +
23092]1/2/2h.e1h=h3530(x10-6),he2h=h470(x10-6),he3h=h0.
c)

 x



2 1
2=60°



 y


Findhthehprincipalhstresseshinhthehparthofhproblemh1-
6hifhthehelastichmodulushofhthehparthish205hGPahandhPoissons’shratiohish0.29.
Solution:he3h=h0h=h(1/E)[0h-hh(1+2)],h 1h=h2
e1h =h (1/E)(1h -hh1);h1h =hEe1/(1-)h =h205x109(3530x10-6)/(1-.292)h=h 79h MPa
1
Showhthaththehtruehstrainhafterhelongationhmayhbehexpressedhash hhln(h )h wherehrhishthe
1hhr
1h
reductionhofharea.h hhln(h ).
1hhr
Solution:hrh =h(Ao-A1)/Aoh =1h–hA1/Aoh=h1h –h Lo/L1.hh=h ln[1/(1-r)]

Ahthinhsheethofhsteel,h1-mmhthick,hishbenthashdescribedhinhExampleh1-11.hAssuminghthathE

=hish205 GPahandhh=h0.29,hh=h2.0hmhandhthaththehneutralhaxishdoesn’thshift.
a. Findhthehstatehofhstresshonhmosthofhthehouterh surface.
b. Findhthehstatehofhstresshaththehedgehofhthehouterh surface.


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