力学课件弹性力学 第十二章板弯曲

Chapter12BendingofThinPlates.ClassicalSolutions第十二章薄板弯曲问题。经典解答。第二版20021Section12.1IntroductionandAssumptions引言和假定Aplateisabodyboundedbytwocloselyspacedparallelplanesandoneormoreprismaticalsurfacesnormaltotheplanes.第二版20022Platefaces--twocloselyspacedparallelplanesPlateedges--prismaticalsurfacesnormaltotheplatefaces.第二版20023Platethickness--thedistancebetweenthetwoplatefaces.Itisdenotedbyt.Thinplate--t<<at<<bt<min(a,b)/8Thickplate第二版20024Platemiddleplane--

Theplaneparalleltothefacesoftheplateandbisectingthethicknesstiscalledthemiddleplaneoftheplate.第二版20025Coordinatesystem--xandyareinthemiddleplaneandzaxisisperpendiculartothemiddleplane.Thesystemisarighthandsystem.第二版20026Loads1.Longitudinalloadinthemiddleplane--Alltheexternalforcesareparalleltothefacesoftheplateanddistributeduniformlyoverthethickness.--------planestressproblem.2.Transverseload----Theyareperpendiculartothemiddleplane---platebendingproblem.第二版20027ThesurfaceforcecomponentZonthelowerfaceoftheplateandthebodyforcecomponentZaretransmittedtotheupperfaceoftheplatealongtheirlinesofaction,thatisq(x,y)=Z)z=-t/2+Z)z=+t/2+-t/2t/2ZdzZ,Zandqareconsideredpositivewhentheyactinthepositivedirectionofz.q(x,y)istotaltransverseloadperunitarea.第二版20028Deflection--thedisplacementofapointonthemiddleplaneinthedirectionofz,w(x,y,0),iscalledthedeflectionofthepoint.Smalldeflections--thedeflectionismuchsmallerthanthethickness.W(x,y,0)<t/5Onlysmalldeflectionsareconsideredhere.第二版20029BasicAssumptionsAssumptionstatedinSec.1.31.Thebodyiscontinuous,perfectlyelastic,homogeneousandisotropic.2.Thedisplacementsandstrainsaresmall.------Thedeflectionoftheplateissmall.Thinplates第二版200210

Assumption1.--zrzxandrzyareneglected,

sincetheyaresmall.Thethreephysicalequationsarenotvalidagainandhavetobeabandoned.z=[z-x-y]/Erzx=zx/Grzy=zy/G第二版200211z=w/z≈0-------w=w(x,y)Thisequationshowsthatallthepointsalignedonanormaltothemiddleplanewillhavethesamedisplacementinthetransversedirectionandthatthisdisplacementisjustthedeflectionoftheplate.第二版200212

rzx=rzy

≈

0

Initiallystraightlineswhicharenormaltothemiddleplaneremainstraightandnormaltothedeformedmiddleplane.第二版200213

Assumption2.--Thestrainsduetoz

are

neglected,sincez

issmall.Thethreephysicalequationswillbethesameasthoseintheplanestressproblems:x=[x-y-z]/Ey=[y-x-z]/Erxy=xy/GTheotherphysicalequationsareabandoned.

第二版200214

Assumption3.--u)z=0andv)z=0areneglected.

Thelongitudinaldisplacementsofallthepointsinthemiddleplanemaybeneglected(x=y=rxy)z=0≈

0Themiddleplaneoftheplateremainsunstrained,althoughtheplanebecomescurvedafterbending.

第二版20021512.2DifferentialEquationforBendingofThinPlatesBasicunknownfunctionw(x,y)Fifteenequationsforspatialproblems------oneequationintermofw(x,y)forplatebendingproblem.第二版200216A.expressuandvintermsofwrzx=u/z+w(x,y)/x=0u=-zw/x+f1(x,y)rzy=v/z+w(x,y)/y=0v=-zw/y+f2(x,y)u)z=0=v)z=0=0f1=

f2=0u=-zw/xv=-zw/y第二版200217B.expressstraincomponentsintermsofwu=-zw/xv=-zw/yx=u/x=-z2w/x2

y=v/y=-z2w/y2rxy=u/y+v/x=-2z2w/xyzrzxandrzyareneglected第二版200218C.expressstresscomponentsxyxy

intermsofw--usephysicalequation

x=u/x=-z2w/x2

y=v/y=-z2w/y2rxy=u/y+v/x=-2z2w/xy

x=E/(1-2)(x+y)=-Ez/(1-2)(2w/x2+2w/y2)

y=E/(1-2)(y+x)=-Ez/(1-2)(2w/y2+2w/x2)

xy=

E/[2(1+)]rxy=

-Ez/(1+)2w/xy第二版200219D.expressstraincomponentszxzy

intermsofw--useequationsofequilibriumx/x+yx/y+zx/z+X=0(8.1.1)xy/x+y/y+zy/z+Y=0(8.1.2)zx/z=-(x/x+yx/y)=Ez/(1-2)(2w/x2+2w/y2)/xzy/z=-(xy/x+y/y)=Ez/(1-2)(2w/x2+2w/y2)/y第二版200220zx/z=-(x/x+yx/y)=Ez/(1-2)(2w/x2+2w/y2)/xzy/z=-(xy/x+y/y)=Ez/(1-2)(2w/x2+2w/y2)/yzx=0.5Ez2/(1-2)(2w)/x+F1(x,y)zy=0.5Ez2/(1-2)(2w)/y+F2(x,y)

zx)z=t/2=0zy)z=t/2=0

F1(x,y)=0.125Et2/(1-2)(2w)/x

F2(x,y)=0.125Et2/(1-2)(2w)/yzx=0.5E(z2-t2/4)/(1-2)(2w)/xzy=0.5E(z2-t2/4)/(

1-2)(2w)/yparabolic第二版200221E.expressstresscomponentsz

intermsofw--useequationsofequilibriumxz/x+yz/y+z/z+Z=0

q(x,y)=Z)z=-t/2+Z)z=+t/2+-t/2t/2Zdzz/z=-(xz/x+yz/y)=-0.5E(z2-t2/4)/(1-2)(4w)z=-0.5E(z3/3-zt2/4)/(1-2)(4w)+F3(x,y)z)z=t/2=0F3(x,y)=-Et3/[24(1-2)](4w)z=-0.5E(z3/3-zt2/4)/(1-2)4w-Et3/[24(1-2)](4w)第二版200222F.ThedifferentialEquationofdeflectionw---B.C.ontheupperface

z=-0.5E(z3/3-zt2/4)/(1-2)4w-Et3/[24(1-2)](4w)z)z=-t/2=-qD4w(x,y)=q(x,y)D=Et3/[12(1-2)]---flexuralrigidityoftheplate.Dhasadimensionof[force][length]第二版200223D4w(x,y)=q(x,y)1.Fifteenequationsforspatialproblemsbecomeoneequationintermofw(x,y)forplatebendinfproblem.2.Boundaryconditionsonz=t/2aresatisfied.3.Boundaryconditionsonedgesofplatehavetobesatisfied.第二版20022412.3StressResultantsandStressCouples

---InternalForcesInmostcases,itisimpossibletosatisfythepointwisestressboundaryconditionsontheedgesofplate.Wecansatisfyonlytheboundaryconditionsintermsofthestressresultantsandstresscouples.Hence,beforediscussingtheboundaryconditions,weproceedtoinvestigatethestressresultantsandstresscouplesonthetransversesectionsofaplate.第二版200225Anelementaryparallelepiped第二版200226A.onthetransversesectionnormaltoxaxisx=-Ez/(1-2)(2w/x2+2w/y2)xy=

-Ez/(1+)2w/xylinearvariationalongzdirection,oddfunctionofzxz=0.5E(z2-t2/4)/(1-2)(2w)/x

parabolicinzdirection,evenfunctionofz第二版200227x=-Ez/(1-2)(2w/x2+2w/y2)

Mx=-t/2t/2zx1dz=-D(2w/x2+2w/y2)x=-Ez/(1-2)(2w/x2+2w/y2)=zMx/II=t3/12linearvariationalongzdirection,oddfunctionofz第二版200228第二版200229xy=

-Ez/(1+)2w/xyMxy=-t/2t/2zxy1dz=-Et3/[12/(1+)]2w/xy

=-D(1-)2w/xyxy=

-Ez/(1+)2w/xy=zMxy/II=t3/12linearvariationalongzdirection,oddfunctionofz第二版200230xz=0.5E(z2-t2/4)/(1-2)(2w)/x

parabolicinzdirection,evenfunctionofzQx=-t/2t/2

xz1dz=-D(2w)/xxz=0.5E(z2-t2/4)/(1-2)(2w)/x=6Qx(t2/4-z2)/t3

第二版200231B.onthetransversesectionnormaltoyaxisy=E/(1-2)(y+x)=-Ez/(1-2)(2w/y2+2w/x2)

yx=

E/[2(1+)]rxy=

-Ez/(1+)2w/xylinearvariationalongzdirection,oddfunctionofzyz=0.5E(z2-t2/4)/(

1-2)(2w)/yparabolicinzdirection,evenfunctionofz第二版200232y=-Ez/(1-2)(2w/y2+2w/x2)

My=-t/2t/2zy1dz=-D(2w/y2+2w/x2)y=-Ez/(1-2)(2w/y2+2w/x2)=zMy/II=t3/12linearvariationalongzdirection,oddfunctionofz第二版200233yx=

-Ez/(1+)2w/xyMyx=-t/2t/2zyx1dz=-Et3/[12/(1+)]2w/xy

=-D(1-)2w/xyyx=xyMyx=Mxyxy=

-Ez/(1+)2w/xy=zMxy/II=t3/12linearvariationalongzdirection,oddfunctionofz第二版200234yz=0.5E(z2-t2/4)/(1-2)(2w)/y

parabolicinzdirection,evenfunctionofzQy=-t/2t/2

yz1dz=-D(2w)/yyz=0.5E(z2-t2/4)/(1-2)(2w)/y=6Qy(t2/4-z2)/t3

第二版200235MxMy----bendingmomentperunitwidth.[force]Mxy=Myx--twistingmomentperunitwidth.[force]Apositivemomentcorrespondstoapositivestresscomponentinthepositivehalfoftheplate.MxMyandMxyareoforderofqa2QxQy---transverseshearingforceperunitwidth[force][length]-1

ThepositivedirectionofQxisthesameasxzThepositivedirectionofQyisthesameasyzQxand

Qyareoforderofqa第二版200236Expresspositiveinternalforcesonthemiddleplane第二版200237xyandyxareoforderofq(a/t)2Max

MyandMxyareoforderofqa2

xzandyzareoforderofqa/tQxand

Qyareoforderofqaz--transversenormalstress----orderofq(xyyx)>>(

xzyz)>>z第二版200238

Fz=0:Plate:Qx/x+Qy/y+q=0(1)

Beam:

dQx/dx+q=0(1a)

第二版200239My=0:Plate:Mx/x+Myx/y=Qx(2)

Beam:

d

Mx/dx=Qx(2a)

Mx=0:Plate:My/y+Mxy/x=Qy(3)

Beam:

0=0

第二版200240Plate:SubstitutingEqs(2)(3)intoEq(1)andnotingthatMxy=Myx,wehave:

2Mx/x2+22Myx/xy+2My/y2+q=0(4)Beam:SubstitutingEqs(2a)intoEq(1a),wehave:

d2Mx/dx2+q=0(4a)第二版200241

Plate:2Mx/x2+22Myx/xy+2My/y2+q=0(4)

Mx=-D(2w/x2+2w/y2)

My=-D(2w/y2+2w/x2)(5)

Myx=-Et3/[12/(1+)]2w/xy=-D(1-)2w/xy

SubstitutionofEqs(5)intoEq(4)yields

D4w(x,y)=q(x,y)

Beam:d2Mx/dx2+q=0(4a)Mx=-EId2w/dx2(5a)SubstitutionofEqs(5a)intoEq(4a)yields

EId4w(x)/dx4=q(x)

第二版20024212.4BoundaryConditions第二版200243A.AttheclampededgeOA(x=0)Thedeflectionmustbezero.

W(0,y)=0Theslopeofthemiddleplanemustbezero.[w(x,y)/x]x=0=0第二版200244B.AtthesimplysupportededgeOC(y=0)Thedeflectionmustbezero.

W(x,0)=0Thebendingmomentmustbezero.(My)y=0=-D(2w/y2+2w/x2)y=0=0W(x,0)=02w/x2)y=0=0

(2w/y2)y=0=0第二版200245C.AtthefreeedgeAB(y=b)Theremustbenobendingmoments.

(My)y=b=-D(2w/y2+2w/x2)y=b=0Theremustbenotwistingmoments.Myx)y=b

=-D(1-)[2w/xy]y=b=0Theremustbenotransverseshearingforces.Qy)y=b

=-D[(2w)/y]y=b=0

第二版200246D.AtthefreeedgeBC(x=a)Theremustbenobendingmoments.

(Mx)x=a=-D(2w/x2+2w/y2)x=a=0Theremustbenotwistingmoments.Mxy)x=a

=-D(1-)[2w/xy]x=a=0Theremustbenotransverseshearingforces.Qx)x=a

=-D[(2w)/y]x=a=0

第二版200247E.Twistingmomentsarereplacedbyastaticallyequivalentshearingforce.第二版200248C.AtthefreeedgeAB(y=b)Theremustbenobendingmoments.

(My)y=b=-D(2w/y2+2w/x2)y=b=0Thetotaldistributedshearingforcemustbezero.Vy)y=b=[Qy+

Myx/x]y=b

=-D[3w/y3+2(1-)3w/x2y]y=b=0

Theequivalentforces

Myx/xhavethesamepositivedirectionasshearingforcesQy

第二版200249D.AtthefreeedgeBC(x=a)Theremustbenobendingmoments.

(Mx)x=a=-D(2w/x2+2w/y2)x=a=0Thetotaldistributedshearingforcemustbezero.Vx)x=a=[Qx+

Mxy/y]x=a

=-D[3w/x3+2(1-)3w/y2x]x=a=0

Theequivalentforces

Mxy/yhavethesamepositivedirectionasshearingforcesQx

第二版200250F.AtthecornerpointBThetotalconcentratedshearingforceisRB=Myx)B+Mxy)B=2Mxy)B

=

-2D(1-)[2w/xy]B=0

第二版20025112.5AsimplesolutionforellipticalplatesEllipticalplateswithclampededgessubjectedtouniformtransverseloads.第二版200252Assume:w(x,y)=m(x2/a2+y2/b2-1)2(1)Boundaryconditionsaresatisfied.W=0w/n=w/xcos(nx)+w/ycos(ny)=0SubstitutingEq(1)intoD(4/x4+24/x2y2+4/y4)w=qyieldsm=q0/[D(24/a4+16/(a2b2)+24/b4)]第二版200253补充题：四边简支的薄板，受均布荷载q,b>>a第二版200254D4w(x,y)=q(1)x=0:w(0,y)=0(2w/x2)x=0=0x=a:w(a,y)=0(2w/x2)x=a=0y=0:w(x,0)=0(2w/y2)y=0=0(2)y=b:w(x,b)=0(2w/y2)y=b=0设：w(x,y)≈w(x)(y=b/2附近）第二版200255

Dw’’’’(x)=q(1)x=0:w(0)=0w’’(0)=0x=a:w(a)=0w’’(a)=0(2)w(x)=qx4/(24D)+Ax3+Bx2+Cx+Ew’’(x)=qx2/(2D)+6Ax+2Bw(0)=0D=0w’’(0)=0B=0w(a)=0qa4/(24D)+Aa3+Ca=0C=qa3/(24D)w’’(a)=0qa2/(2D)+6Aa=0A=-qa/(12D)w(x)=qx4/(24D)-qax3/(12D)+qa3x/(24D)w’’(x)=qx2/(2D)-qax/(2D)M(x)=-Dw’’(x)=-qx2/2+qax/2同梁第二版200256补充题1：四边简支的薄板，

荷载q=q0sin(x/a)sin(y/b).第二版200257

D4w(x,y)=q0sin(x/a)sin(y/b)(1)D[4w/x4+24w/x2y2+4w/y4]=q0sin(x/a)sin(y/b)x=0:w(0,y)=0(2w/x2)x=0=0x=a:w(a,y)=0(2w/x2)x=a=0y=0:w(x,0)=0(2w/y2)y=0=0(2)y=b:w(x,b)=0(2w/y2)y=b=0W(x,y)=Asin(x/a)sin(y/b)(3)将（3）代入（1）得A,A代入（3）得（4）：W(x,y)=q0

sin(x/a)sin(y/b)/[D4(1/a2+1/b2)2](4)第二版200258补充题2：四边简支的薄板，

q=qmnsin(mx/a)sin(ny/b)mn为正整数第二版200259D4w(x,y)=qmnsin(mx/a)sin(ny/b)(1)D[4w/x4+24w/x2y2+4w/y4]=qmnsin(mx/a)sin(ny/b)x=0:w(0,y)=0(2w/x2)x=0=0x=a:w(a,y)=0(2w/x2)x=a=0y=0:w(x,0)=0(2w/y2)y=0=0(2)y=b:w(x,b)=0(2w/y2)y=b=0W(x,y)=Asin(mx/a)sin(ny/b)（3)将(3)代入(1)得A,A代入(3)得(4)：W(x,y)=qmn

sin(mx/a)sin(ny/b)/[D4(m2/a2+n2/b2)2]（4)第二版200260补充题3：四边简支的薄板，

q=m=1,2,3...n=1,2,3...

qmnsin(mx/a)sin(ny/b)mn为正整数第二版200261

D4w(x,y)=m=1,2,3...n=1,2,3...

qmnsin(mx/a)sin(ny/b)(1)D[4w/x4+24w/x2y2+4w/y4]=m=1,2,3...n=1,2,3...

qmnsin(mx/a)sin(ny/b)(1)x=0:w(0,y)=0(2w/x2)x=0=0x=a:w(a,y)=0(2w/x2)x=a=0y=0:w(x,0)=0(2w/y2)y=0=0(2)y=b:w(x,b)=0(2w/y2)y=b=0W(x,y)=m=1,2,3...n=1,2,3...

qmn

sin(mx/a)sin(ny/b)/[D4(m2/a2+n2/b2)2](3)第二版20026212.6(解法1)Navier‘sSolutionbyDoubleTrigonometricSeries.-四边简支，受任意荷载第二版200263D4w(x,y)=q(x,y)(1)x=0:w(0,y)=0(2w/x2)x=0=0x=a:w(a,y)=0(2w/x2)x=a=0y=0:w(x,0)=0(2w/y2)y=0=0(2)y=b:w(x,b)=0(2w/y2)y=b=0第二版200264q(x,y)=m=1,2.n=1,2qmnsin(mx/a)sin(ny/b)Multiplyingbothsidesofthisequationbysin(ix/a)sin(jy/b),whereiandjarearbitraryintegers,integratingtheequationwithrespecttoxfrom0toaandwithrespecttoyfrom0tob,wehave0a0bq(x,y)sin(ix/a)sin(jy/b)dxdy=m=1,2.n=1,20a0b

qmnsin(mx/a)sin(ny/b)sin(ix/a)sin(jy/b)dxdy第二版2002650a0bq(x,y)sin(ix/a)sin(jy/b)dxdy=m=1,2.n=1,20a0b

qmnsin(mx/a)sin(ny/b)sin(ix/a)sin(jy/b)dxdy0asin(mx/a)sin(ix/a)dx=0,(m≠i)a/2,(m=i)0bsin(ny/b)sin(jy/b)dy=0,(n≠j)b/2(n=j)qij=4/ab0a0bq(,

)sin(i/a)sin(j/b)dd

q(x,y)=m=1,2.n=1,2qmnsin(mx/a)sin(ny/b)=4/abm=1,2.n=1,20a0bq(,

)sin(m/a)sin(n/b)ddsin(mx/a)sin(ny/b)第二版200266

D4w(x,y)=m=1,2,3...n=1,2,3...

qmnsin(mx/a)sin(ny/b)(1)D[4w/x4+24w/x2y2+4w/y4]=m=1,2,3...n=1,2,3...

qmnsin(mx/a)sin(ny/b)(1)qmn=4/ab0a0bq(,

)sin(m/a)sin(n/b)ddx=0:w(0,y)=0(2w/x2)x=0=0x=a:w(a,y)=0(2w/x2)x=a=0y=0:w(x,0)=0(2w/y2)y=0=0(2)y=b:w(x,b)=0(2w/y2)y=b=0W(x,y)=m=1,2,3...n=1,2,3...

qmn

sin(mx/a)sin(ny/b)/[D4(m2/a2+n2/b2)2](3)第二版20026712.6(解法2)Navier‘sSolutionbyDoubleTrigonometricSeries.-四边简支，受任意荷载.第二版200268D4w(x,y)=q(x,y)(1)x=0:w(0,y)=0(2w/x2)x=0=0x=a:w(a,y)=0(2w/x2)x=a=0y=0:w(x,0)=0(2w/y2)y=0=0(2)y=b:w(x,b)=0(2w/y2)y=b=0第二版200269W(x,y)=m=1,2.n=1,2…Amnsin(mx/a)sin(ny/b)(3)Boundaryconditions(2)aresatisfiedautomaticallySubstitutingEq.(3)intoEq.(1)D(4/x4+24/x2y2+4/y4)w=q(1)yields4Dm=1,2.n=1,2…(m2/a2+n2/b2)2

Amnsin(mx/a)sin(ny/b)=q(x,y)Amn

第二版200270q(x,y)=m=1,2.n=1,2qmnsin(mx/a)sin(ny/b)Multiplyingbothsidesofthisequationbysin(ix/a)sin(jy/b),whereiandjarearbitraryintegers,integratingtheequationwithrespecttoxfrom0toaandwithrespecttoyfrom0tob,wehave0a0bq(x,y)sin(ix/a)sin(jy/b)dxdy=m=1,2.n=1,20a0bqmnsin(mx/a)sin(ny/b)sin(ix/a)sin(jy/b)dxdy第二版2002710a0bq(x,y)sin(ix/a)sin(jy/b)dxdy=m=1,2.n=1,20a0b

qmnsin(mx/a)sin(ny/b)sin(ix/a)sin(jy/b)dxdy0asin(mx/a)sin(ix/a)dx=0,(m≠i)a/2,(m=i)0bsin(ny/b)sin(jy/b)dy=0,(n≠j)b/2(n=j)qij=4/ab0a0bq(,

)sin(i/a)sin(j/b)dd

q(x,y)=m=1,2.n=1,2qmnsin(mx/a)sin(ny/b)=4/abm=1,2.n=1,20a0bq(,

)sin(m/a)sin(n/b)ddsin(mx/a)sin(ny/b)第二版2002724Dm=1,2.n=1,2…(m2/a2+n2/b2)2

Amnsin(mx/a)sin(ny/b)=q(x,y)q(x,y)=m=1,2.n=1,2qmnsin(mx/a)sin(ny/b)4Dm=1,2.n=1,2…(m2/a2+n2/b2)2

Amnsin(mx/a)sin(ny/b)=m=1,2.n=1,2qmnsin(mx/a)sin(ny/b)比较系数得Amn=qmn/[4D(m2/a2+n2/b2)2]=[4/(ab)m=1,2.n=1,20a0bq(,

)sin(m/a)sin(n/b)dd]/[4D(m2/a2+n2/b2)2]第二版200273q(x,y)=q0Amn=[4/(ab)m=1,2.n=1,20a0bq(,

)sin(m/a)]sin(n/b)dd]/[4D(m2/a2+n2/b2)2]=q0ab(1-cosm)(1-cosn)/(2mn)=16q0/[6Dmn(m2/a2+n2/b2)2](m=1,3,5,…;n=1,3,5…)w=16q0/[6D]m=1,3,5.n=1,3,5sin(mx/a)sin(ny/b)/[mn(m2/a2+n2/b2)2]第二版200274q(x,y)=q0a=b

W(a/2a/2)=0.25[16q0a4/(6D)]1term0.2467[16q0a4/(6D)]2terms0.2470[16q0a4/(6D)]3terms0.2469[16q0a4/(6D)]4terms----isexacttothreedigits.Mx(a/2,a/2)=0.325[16q0a2/4]1term0.2857[16q0a2/4]2terms0.2936[16q0a2/4]3terms0.2920[16q0a2/4]4terms----isexacttothreedigits.第二版20027512.7Levy'sSolutionbySingleTrigonometricSeries.第二版200276D4w(x,y)=q(x,y)(1)x=0:w(0,y)=0(2w/x2)x=0=0x=a:w(a,y)=0(2w/x2)x=a=0(2)y=0:specifiedlatery=b:specifiedlater第二版200277AssumeW(x,y)=m=1,2.Ym(y)sin(mx/a)(3)Boundaryconditions(2)aresatisfiedautomaticallySubstitutingEq.(3)intoEq.(1)D(4/x4+24/x2y2+4/y4)w=q(1)yieldsm=1,2[ym(4)-2(m/a)2Ym(2)+(m/a)4Ym]sin(mx/a)=q/D

第二版200278Dm=1,2[ym(4)-2(m/a)2Ym(2)+(m/a)4Ym]sin(mx/a)=qq=2/am=1,2[0aq(y)sin(m/a)d]sin(mx/a)Dm=1,2[ym(4)-2(m/a)2Ym(2)+(m/a)4Ym]sin(mx/a)=2/am=1,2[0aq(y)sin(m/a)d]sin(mx/a)ym(4)-2(m/a)2Ym(2)+(m/a)4Ym

=2/(Da)[0aq(y)sin(m/a)d]

第二版200279ym(4)-2(m/a)2Ym(2)+(m/a)4Ym

=2/(Da)[0aq(y)sin(m/a)d]

Ym=Amch(my/a)+Bm(my/a)

sh(my/a)+Cmsh(my/a)+Dm(my/a)

ch(my/a)+fm(y)wherefm(y)isaparticularsolution.W=m[Amch(my/a)+Bm(my/a)

sh(my/a)+Cmsh(my/a)+Dm(my/a)

ch(my/a)+fm(y)]sin(mx/a)whereAmBmCmDmarearbitraryconstantsandcanbedeterminedbytheboundaryconditionsontheedgesy=b/2第二版200280Rectangularplatewithoutfreeedges--onlywithsimplysupportededgesorclampededges4[Dw(x,y)]=q(x,y)(1)Bcs:(Dw)s=0[(Dw)/x]s=0=0[(Dw)/y]s=0[2(Dw)/x2]s=0=0[2(D

w)/y2]s=0=0Dwisindependentof第二版200281Mx=-[2(Dw)/x2+2(Dw/y2)]

My=-[(2(Dw/y2+2(Dw/x2)]=0TableMx=-[2(Dw)/x2]

My=-[(2(Dw/y2]≠0Mx’=-[2(Dw)/x2+2(Dw/y2)]=Mx+My

My’

=-[(2(Dw/y2+2(Dw/x2)]=My+Mx第二版20028212.8BendingofCirclePlateCylindricalcoordinates(r,,z)第二版200283D4w

=D[/(rr)+2/(r22)+2/r2]2w

=q(r,)Mr=-D{2w/r2+[w/(rr)+2/(r22)]}r=zMr/II=t3/12M==-D{w/(rr)+2w/(r22)

+2w/r2}=

zM/IMr=Mr

=-D(1-)[2w/(rr)-w/(r2)]

r=zMr/IQr=-D(2w

)/rrz=6Qr(t2/4-z2)/t3Q=-D(2w

)/(r)z=6Q(t2/4-z2)/t3第二版200284Boundaryconditionsaclampededger=a:w(a,)=0,w/r)r=a=0

asimplysupportededger=aw(a,)=0,Mr)r=a

=-D{2w/r2+[w/(rr)+2/(r22)]})r=a=0afreeedger=aMr)r=a

=0Vr)r=a

=[Qr+Mr/(r)]r=a

=0第二版20028512.9AxisymmetricalBendingofCirclePlatesIfq(r,)isq(r)onacircleplate,

w(r,)willbew(r)andD4w

=D[/(rr)+2/(r22)+2/r2]2w

=q(r,)willbe

D[/(rr)+2/r2]2w

(r)=q(r)Thesolutionofthisordinarydifferentialequationisw(r)=C1lnr+C2r2lnr+C3r2+C4+w1wherew1isanyparticularsolutionandC1C2C3C4arearbitraryconstants.第二版200286Mr=-D{2w/r2+[w/(rr)+2/(r22)]}=-D[d2w/dr2+dw/(rdr)]M=-D{w/(rr)+2w/(r22)

+2w/r2}

=-D{dw/(rdr)+d2w/dr2}Mr=Mr

=-D(1-)[2w/(rr)-w/(r2)]=0Qr=-D(2w

)/r=-Dd[d2w/dr2+dw/(rdr)]/drQ=-D(2w

)/(r)=0

第二版200287w(r)=C1lnr+C2r2lnr+C3r2+C4+w1dw/dr=C1/r+C2(2rlnr+r)+2C3r+dw1/drdw2/dr2=-C1/r2+C2(2lnr+3)+2C3+d2w1/dr2Mr=-D[d2w/dr2+dw/(rdr)]M

=-D{dw/(rdr)+d2w/dr2}Qr=-D(2w

)/r=-Dd[d2w/dr2+dw/(rdr)]/dr=-D{C2(4/r)-dw1/(r2dr)+d2w1/(rdr2)+d3w1/dr3}第二版200288A.annularplatea<r<b----acircularplatewithacentralhole

TherewillbetwoBcsatr=a,twoBcsatr=b,sothefourarbitraryconstantscanbedetermined.第二版200289B.solidplatera(nocentrehole)Qr=-D{C2(4/r)-Pdw1/(r2dr)+d2w1/(rdr2)+d3w1/dr3}TwoBcsatr=aQrr=0:1.sincewisfinite,C1=02(1).Noconcentratedforceatr=0.SinceQrisfinite,C2=0.2(2).ConcentratedforcePatr=0.Fz=0:2rQr+P=02r(-4DC2/r)+P=0C2=P/(8D)第二版200290Example1:acircularclampedplatewithoutholeissubjectedtoauniformloadq0w(r)=C1lnr+C2r2lnr+C3r2+C4+w1particularsolution:D[/(rr)+2/r2]2w

(r)=q0assume:w1=mr4w1=q0r4/(64