Solution Manual for Metal Forming: Mechanics and Metallurgy 4th Edition

All Chapters Covered
j j




SOLUTION MANUAL
j

, SolutionjManualj3rdjEd.jMetaljForming:jMechanicsjandjMetallurgyjC
hapterj1


Determinejthejprincipaljstressesjforjthejstressjstate
10 3 4
jijj j 3 5 2j.
4 2 7
Solution: I1j=j10+5+7=32,jI2j=j-(50+35+70)j+9j+4j+16j=j-126,j I3j=j350j-48j-40j-80
-63j=j119;j j  j–j222j-126j-119j=j0.j Ajtrialjandjerrorjsolutionjgivesjj-=j13.04.
3

 Factoringjoutj13.04, 2j-
8.96j +j9.16j=j0.jSolving;j j =j13.04,jj =j7.785,jj =j1.175.

1-2 Aj5-
cm.jdiameterjsolidjshaftjisjsimultaneouslyjsubjectedjtojanjaxialjloadjofj80jkNjandjajtorquejofj4
00jNm.
a. Determinejthejprincipaljstressesjatjthejsurfacejassumingjelasticjbehavior.
b. Findjthejlargestjshearjstress.
Solution:ja.jThejshearjstress,j,jatjajradius,jr,jisjj=jsr/Rjwherejsisjthejshearjstressjatjthejsurfacej
Rjisjthejradiusjofjthejrod.jThejtorque,jT,jisjgivenjbyjTj=j∫2πtr2drj=j(2πsj/R)∫r3dr
=jπsR3/2.jSolvingjforj=js,jsj=j2T/(πR3)j=j2(400N)/(π0.0253)j=j16jMPajThejax
ialjstressjisj.08MN/(π0.0252)j=j4.07jMPa
1,2j=j4.07/2j±j[(4.07/2)2j +j(16/2)2)]1/2j=j1.029,j-0.622j MPa
b.jthejlargestjshearjstressjisj(1.229j+j0.622)/2j=j0.925jMPa

Ajlongjthin-
walljtube,jcappedjonjbothjendsjisjsubjectedjtojinternaljpressure.jDuringjelasticjloading,jdoesjt
hejtubejlengthjincrease,jdecreasejorjremainjconstant?
Solution:jLetjyj=jhoopjdirection,jxj=jaxialjdirection,jandjzj=jradialjdirection.j–
jexj=je2j=j(1/E)[j-j(j3j+j1)]j=j(1/E)[2j-j(22)]j=j(2/E)(1-2)


Sincejuj<j1/2jforjmetals,jexj=je2jisjpositivejandjthejtubejlengthens.

4 Ajsolidj2-
cm.jdiameterjrodjisjsubjectedjtojajtensilejforcejofj40jkN.jAnjidenticaljrodjisjsubjectedjtojajfluid
jpressurejofj35jMPajandjthenjtojajtensilejforcejofj40jkN.jWhichjrodjexperiencesjthejlargestjshe

arjstress?
Solution:jThejshearjstressesjinjbothjarejidenticaljbecausejajhydrostaticjpressurejhasjnojshearjc
omponent.

1-5 Considerjajlongjthin-
wall,j5jcmjinjdiameterjtube,jwithjajwalljthicknessjofj0.25jmmjthatjisjcappedjonjbothjends.j Fin
djthejthreejprincipaljstressesjwhenjitjisjloadedjunderjajtensilejforcejofj40jNjandjanjinternaljpres
surejofj200jkPa.
Solution:jxj=jPD/4tj+jF/(πDt)j=j12.2jMPa
yj=jPD/2tj=j 2.0jMPa

1

, yj=j0




2

, 1-6 Threejstrainjgaugesjarejmountedjonjthejsurfacejofjajpart.jGaugejAjisjparalleljtojthejx
-axisjandjgaugejCjisjparalleljtojthejy-
axis.jThejthirdjgage,jB,jisjatj30°jtojgaugejA.jWhenjthejpartjisjloadedjthejgaugesjread
GaugejA 3000x10-6
GaugejB 3500jx10-6
GaugejC 1000jx10-6
a. Findjthejvaluejofjxy.
b. Findjthejprincipaljstrainsjinjthejplanejofjthejsurface.
c. SketchjthejMohr’sjcirclejdiagram.
Solution:jLetjthejBjgaugejbejonjthejx’jaxis,jthejAjgaugejonjthejx-axisjandjthejCjgaugejon
2 2
thejy-axis.jexxjexxjxjxjej jxyyyj jjxyjxxjxyj,jwherejjxxj=jcosexj=j 30j=j√3/2jandjjxyj=
cosj60j=j½.jSubstitutingjthejmeasuredjstrains,j3500j=
j3000(√2/3) j–j1000(1/2) j+jxy(√3/2)(1/2)
2 2
xy
j=j(4/√3/2){3500-[3000(1000(√3/2)1/2+1000(1/2) ]}j=j2,309j(x10 )  2
2 2 -6


b.j e1,e2j =j(exj+ey)/2±j[(ex-ey)2j +j xy2] /2j=j(3000+1000)/2j±j[(3000-1000)j +
2309 ] /2j.e1j=j3530(x10 ),je2j=j470(x10-6),je3j=j0.
2 1/2 -6

c)

 x



2 1
2=60°



 y


Findjthejprincipaljstressesjinjthejpartjofjproblemj1-
6jifjthejelasticjmodulusjofjthejpartjisj205jGPajandjPoissons’sjratiojisj0.29.
Solution:je3j=j0j=j(1/E)[0j-jj(1+2)],j1j=j2
e1j=j(1/E)(1j-jj1);j1j=jEe1/(1-)j =j205x109(3530x10-6)/(1-.292)j=j79jMPa
1
Showjthatjthejtruejstrainjafterjelongationjmayjbejexpressedjasj jjln(j )j wherejrjisjthe
1jjr
1j
reductionjofjarea.j jjln(j).
1jjr
Solution:jrj=j(Ao-A1)/Aoj =1j–jA1/Aoj=j1j–jLo/L1.jj=jln[1/(1-r)]

Ajthinjsheetjofjsteel,j1-mmjthick,jisjbentjasjdescribedjinjExamplej1-11.jAssumingjthatjE

=jisj205 GPajandjj=j0.29,jj=j2.0jmjandjthatjthejneutraljaxisjdoesn’tjshift.
a. Findjthejstatejofjstressjonjmostjofjthejouterjsurface.
b. Findjthejstatejofjstressjatjthejedgejofjthejouterjsurface.


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