第二版 2002 弹性力学 第十二章 1
Chapter 12 Bending of Thin Plates,
Classical Solutions
第十二章 薄板弯曲问题。经典解答。
第二版 2002 弹性力学 第十二章 2
Section 12.1 Introduction and Assumptions
引言和假定
A plate is a body bounded by two closely
spaced parallel planes and one or more
prismatical surfaces normal to the
planes.
第二版 2002 弹性力学 第十二章 3
? Plate faces--two closely spaced parallel
planes
? Plate edges--prismatical surfaces normal to
the plate faces.
第二版 2002 弹性力学 第十二章 4
? Plate thickness--the distance between the
two plate faces,It is denoted by t.
? Thin plate-- t<<a t <<b
t<min(a,b)/8
? Thick plate
第二版 2002 弹性力学 第十二章 5
? Plate middle plane--
The plane parallel to the faces of the plate
and bisecting the thickness t is called the
middle plane of the plate,
第二版 2002 弹性力学 第十二章 6
? Coordinate system--x and y are in the
middle plane and z axis is perpendicular to
the middle plane, The system is a right
hand system.
第二版 2002 弹性力学 第十二章 7
Loads
1,Longitudinal load in the middle plane--All
the external forces are parallel to the faces
of the plate and distributed uniformly over
the thickness.--------plane stress problem.
2.Transverse load ----They are perpendicular
to the middle plane---plate bending
problem.
第二版 2002 弹性力学 第十二章 8
? The surface force component Z on the lower
face of the plate and the body force
component Z are transmitted to the upper
face of the plate along their lines of action,
that is
q(x,y)=Z)z=-t/2+Z)z=+t/2+ ?-t/2t/2 Zdz
Z,Z and q are considered positive when
they act in the positive direction of z.
q(x,y) is total transverse load per unit area.
第二版 2002 弹性力学 第十二章 9
? Deflection--the displacement of a point on the
middle plane in the direction of z,w(x,y,0),is
called the deflection of the point,
? Small deflections--the deflection is much smaller
than the thickness,W(x,y,0)<t/5
? Only small deflections are considered here,
第二版 2002 弹性力学 第十二章 10
Basic Assumptions
? Assumption stated in Sec.1.3
1,The body is continuous,perfectly elastic,
homogeneous and isotropic,
2.The displacements and strains are small.
------The deflection of the plate is small.
? Thin plates
第二版 2002 弹性力学 第十二章 11
Assumption 1.--?z rzx and rzy are neglected,
since they are small.
? The three physical equations are not valid
again and have to be abandoned.
?z=[?z- ??x- ??y]/E
rzx=?zx/G
rzy=?zy/G
第二版 2002 弹性力学 第十二章 12
?z=?w/?z ≈ 0-------w=w(x,y)
? This equation shows that all the points
aligned on a normal to the middle plane will
have the same displacement in the
transverse direction and that this
displacement is just the deflection of the
plate.
第二版 2002 弹性力学 第十二章 13
rzx= rzy ≈ 0
? Initially straight lines which are normal to
the middle plane remain straight and normal
to the deformed middle plane.
第二版 2002 弹性力学 第十二章 14
Assumption 2.--The strains due to ?z are
neglected,since ?z is small.
? The three physical equations will be the same
as those in the plane stress problems:
?x=[?x- ??y- ??z]/E
?y=[?y- ??x- ??z]/E
rxy=?xy/G
? The other physical equations are abandoned.
第二版 2002 弹性力学 第十二章 15
Assumption 3.--u)z=0 and v)z=0 are neglected.
The longitudinal displacements of all the points
in the middle plane may be neglected
? ( ?x= ?y= rxy)z=0 ≈ 0
The middle plane of the plate remains
unstrained,although the plane becomes
curved after bending.
第二版 2002 弹性力学 第十二章 16
12.2 Differential Equation for
Bending of Thin Plates
? Basic unknown function w(x,y)
? Fifteen equations for spatial problems------
one equation in term of w(x,y) for plate
bending problem.
第二版 2002 弹性力学 第十二章 17
A.express u and v in terms of w
? rzx= ?u/?z+ ?w(x,y)/?x=0 u=-z ?w /?x+f1(x,y)
rzy= ?v/?z+ ?w(x,y)/?y=0 v=-z ?w/?y+f2(x,y)
? u)z=0= v)z=0=0 f1= f2=0
? u=-z ?w /?x
v=-z ?w/?y
第二版 2002 弹性力学 第十二章 18
B.express strain components in
terms of w
? u=-z ?w /?x v=-z ?w/?y
? ?x=?u/?x= -z ?2w /?x2
?y=?v/?y= -z ?2w/?y2
rxy=?u/?y+?v/?x= -2z?2w/?x?y
? ?z rzx and rzy are neglected
第二版 2002 弹性力学 第十二章 19
C.express stress components ?x?y?xy
in terms of w--use physical equation
? ?x=?u/?x= -z ?2w/?x2
?y=?v/?y= -z ?2w/?y2
rxy=?u/?y+?v/?x= -2z?2w/?x?y
?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/?x?y
第二版 2002 弹性力学 第十二章 20
D.express strain components ?zx?zy
in terms of w--use equations of equilibrium
? ??x/?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)/?x
??zy/?z=-(??xy/?x+ ??y/?y)
= Ez/(1-? 2) ?(?2w/?x2+?2w/?y2)/?y
第二版 2002 弹性力学 第十二章 21
? ??zx/?z =-(??x/?x+??yx/?y)
= Ez/(1-? 2) ?(?2w/?x2+?2w/?y2)/?x
??zy/?z=-(??xy/?x+ ??y/?y)
= Ez/(1-? 2) ?(?2w/?x2+?2w/?y2)/?y
? ?zx =0.5Ez2/(1-? 2) ?(?2w)/?x+F1(x,y)
?zy=0.5Ez2/(1-? 2) ?(?2w)/?y+F2(x,y)
?zx)z=? t/2=0 ?zy)z=? t/2=0
F1(x,y)= 0.125Et2/(1-? 2) ?(?2w)/?x
F2(x,y)= 0.125Et2/(1-? 2) ?(?2w)/?y
? ?zx =0.5E(z2-t2/4)/(1-? 2) ?(?2w)/?x
?zy=0.5E(z2 -t2/4)/( 1-? 2) ?(?2w)/?y parabolic
第二版 2002 弹性力学 第十二章 22
E.express stress components ?z
in terms of w--use equations of equilibrium
? ??xz/?x +??yz/?y+ ??z/?z+Z=0
q(x,y)=Z)z=-t/2+Z)z=+t/2+?-t/2t/2 Zdz
??z/?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=0 F3(x,y) = -E t3/[24(1-? 2)](?4w)
? ?z=-0.5E (z3/3-zt2/4)/(1-? 2)?4w-Et3/[24(1-? 2)](?4w)
第二版 2002 弹性力学 第十二章 23
F,The differential Equation of deflection w
---B.C,on the upper face
? ?z=-0.5E (z3/3-zt2/4)/(1-? 2)?4w-Et3/[24(1-? 2)](?4w)
?z)z=-t/2= -q
? D?4w(x,y)= q(x,y)
? D=Et3/[12(1-? 2)]---flexural rigidity of the plate.
D has a dimension of [force][length]
第二版 2002 弹性力学 第十二章 24
? D?4w(x,y)= q(x,y)
1,Fifteen equations for spatial problems
become one equation in term of w(x,y) for
plate bendinf problem.
2,Boundary conditions on z= ?t/2 are
satisfied.
3,Boundary conditions on edges of plate
have to be satisfied.
第二版 2002 弹性力学 第十二章 25
12.3 Stress Resultants and Stress Couples
---Internal Forces
? In most cases,it is impossible to satisfy the
pointwise stress boundary conditions on the
edges of plate,We can satisfy only the boundary
conditions in terms of the stress resultants and
stress couples,Hence,before discussing the
boundary conditions,we proceed to investigate
the stress resultants and stress couples on the
transverse sections of a plate,
第二版 2002 弹性力学 第十二章 26
An elementary parallelepiped
第二版 2002 弹性力学 第十二章 27
A,on the transverse section normal to x axis
? ?x= -Ez/(1-? 2)(?2w/?x2+? ?2w/?y2)
?xy= -Ez/(1+?) ?2w/?x?y
linear variation along z direction,odd function of z
? ?xz =0.5E(z2-t2/4)/(1-? 2) ?(?2w)/?x
parabolic in z direction,even function of z
第二版 2002 弹性力学 第十二章 28
? ?x= -Ez/(1-? 2)(?2w/?x2+? ?2w/?y2)
? Mx=?-t/2t/2 z?x 1 dz= -D(?2w/?x2+? ?2w/?y2)
? ?x= -Ez/(1-? 2)(?2w/?x2+? ?2w/?y2) =z Mx/I
I=t3/12
linear variation along z direction,odd function of z
第二版 2002 弹性力学 第十二章 29
第二版 2002 弹性力学 第十二章 30
? ?xy= -Ez/(1+?) ?2w/?x?y
Mxy=?-t/2t/2 z ?xy 1 dz= -Et3/[12 /(1+?)] ?2w/?x?y
=-D(1- ?) ?2w/?x?y
? ?xy= -Ez/(1+?) ?2w/?x?y =z Mxy/I
I=t3/12
linear variation along z direction,odd function of z
第二版 2002 弹性力学 第十二章 31
? ?xz =0.5E(z2-t2/4)/(1-? 2) ?(?2w)/?x
parabolic in z direction,even function of z
Qx=?-t/2t/2 ?xz 1 dz= -D ?(?2w)/?x
? ?xz =0.5E(z2-t2/4)/(1-? 2) ?(?2w)/?x
=6Qx (t2/4-z2)/t3
第二版 2002 弹性力学 第十二章 32
B,on the transverse section normal to y axis
?y=E/(1-? 2)(?y+? ?x)=-Ez/(1-? 2)(?2w/?y2+? ?2w/?x2)
?yx= E/[2(1+?)] rxy = -Ez/(1+?) ?2w/?x?y
linear variation along z direction,odd function of z
?yz=0.5E(z2 -t2/4)/( 1-? 2) ?(?2w)/?y
parabolic in z direction,even function of z
第二版 2002 弹性力学 第十二章 33
? ?y= -Ez/(1-? 2)(?2w/?y2+? ?2w/?x2)
? My=?-t/2t/2 z?y 1 dz= -D(?2w/?y2+? ?2w/?x2)
? ?y= -Ez/(1-? 2)(?2w/?y2+? ?2w/?x2) =z My/I
I=t3/12
linear variation along z direction,odd function of z
第二版 2002 弹性力学 第十二章 34
? ?yx= -Ez/(1+?) ?2w/?x?y
Myx=?-t/2t/2 z ?yx 1 dz= -Et3/[12 /(1+?)] ?2w/?x?y
=-D(1- ?) ?2w/?x?y
?yx= ?xy Myx= Mxy
? ?xy= -Ez/(1+?) ?2w/?x?y =z Mxy/I
I=t3/12
linear variation along z direction,odd function of z
第二版 2002 弹性力学 第十二章 35
? ?yz =0.5E(z2-t2/4)/(1-? 2) ?(?2w)/?y
parabolic in z direction,even function of z
Qy=?-t/2t/2 ?yz 1 dz= -D ?(?2w)/?y
? ?yz =0.5E(z2-t2/4)/(1-? 2) ?(?2w)/?y
=6Qy (t2/4-z2)/t3
第二版 2002 弹性力学 第十二章 36
? Mx My----bending moment per unit width.[force]
Mxy=Myx--twisting moment per unit width,[force]
A positive moment corresponds to a positive
stress component in the positive half of the plate.
Mx My and Mxy are of order of qa2
? Qx Qy---transverse shearing force per unit width
[force][length]-1
The positive direction of Qx is the same as ?xz
The positive direction of Qy is the same as ?yz
Qx and Qy are of order of qa
第二版 2002 弹性力学 第十二章 37
Express positive internal forces on the
middle plane
第二版 2002 弹性力学 第十二章 38
? ?x?y and ?yx are of order of q(a/t)2
Max My and Mxy are of order of qa2
? ?xz and ?yz are of order of qa/t
Qx and Qy are of order of qa
? ?z--transverse normal stress----order of q
? (?x ?y ?yx)>>( ?xz ?yz )>>?z
第二版 2002 弹性力学 第十二章 39
?Fz=0,Plate,?Qx/?x+?Qy/?y+q=0 (1)
Beam,dQx/dx+q=0 (1a)
第二版 2002 弹性力学 第十二章 40
?My=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 2002 弹性力学 第十二章 41
Plate,Substituting Eqs(2)(3) into Eq(1) and
noting that Mxy=Myx,we have:
?2Mx/?x2+2?2Myx/ ?x?y+?2My/?y2+q=0 (4)
Beam,Substituting Eqs(2a) into Eq(1a),we
have:
d2Mx/dx2+q=0 (4a)
第二版 2002 弹性力学 第十二章 42
Plate,?2Mx/?x2+2?2Myx/ ?x?y+?2My/?y2+q=0 (4)
Mx= -D(?2w/?x2+? ?2w/?y2)
My= -D(?2w/?y2+? ?2w/?x2) (5)
Myx= -Et3/[12 /(1+?)] ?2w/?x?y=-D(1- ?) ?2w/?x?y
Substitution of Eqs(5) into Eq(4) yields
D?4w(x,y)= q(x,y)
Beam,d2Mx/dx2+q=0 (4a)
Mx= -EId2w/dx2 (5a)
Substitution of Eqs(5a) into Eq(4a) yields
EId4w(x)/dx4= q(x)
第二版 2002 弹性力学 第十二章 43
12.4 Boundary Conditions
第二版 2002 弹性力学 第十二章 44
A,At the clamped edge OA (x=0)
? The deflection must be zero.
W(0,y)=0
? The slope of the middle plane must be zero.
[?w(x,y)/ ?x]x=0=0
第二版 2002 弹性力学 第十二章 45
B,At the simply supported edge OC
(y=0)
? The deflection must be zero.
W(x,0)=0
? The bending moment must be zero.
(My)y=0= -D(?2w/?y2+? ?2w/?x2)y=0=0
W(x,0)=0 ?2w/?x2)y=0=0
(?2w/?y2)y=0=0
第二版 2002 弹性力学 第十二章 46
C,At the free edge AB (y=b)
? There must be no bending moments.
(My)y=b= -D(?2w/?y2+? ?2w/?x2)y=b=0
? There must be no twisting moments.
Myx )y=b = -D(1- ?)[ ?2w/?x?y ]y=b=0
? There must be no transverse shearing forces.
Qy )y=b = -D [?(?2w)/?y ]y=b =0
第二版 2002 弹性力学 第十二章 47
D,At the free edge BC (x=a)
? There must be no bending moments.
(Mx)x=a= -D(?2w/?x2+? ?2w/?y2)x=a=0
? There must be no twisting moments.
Mxy )x=a = -D(1- ?)[ ?2w/?x?y ]x=a=0
? There must be no transverse shearing forces.
Qx )x=a = -D [?(?2w)/?y ]x=a =0
第二版 2002 弹性力学 第十二章 48
E,Twisting moments are replaced by a
statically equivalent shearing force,
第二版 2002 弹性力学 第十二章 49
C,At the free edge AB (y=b)
? There must be no bending moments.
(My)y=b= -D(?2w/?y2+? ?2w/?x2)y=b=0
? The total distributed shearing force must be zero.
Vy )y=b=[Qy+ ?Myx /?x]y=b
= -D [?3w/?y3+2(1-?)?3w/ ?x2?y]y=b =0
? The equivalent forces ? Myx /?x have the same
positive direction as shearing forces Qy
第二版 2002 弹性力学 第十二章 50
D,At the free edge BC (x=a)
? There must be no bending moments.
(Mx)x=a= -D(?2w/?x2+? ?2w/?y2)x=a=0
? The total distributed shearing force must be zero.
Vx )x=a=[Qx+ ? Mxy /?y]x=a
= -D [?3w/?x3+2(1-?)?3w/ ?y2?x]x=a =0
? The equivalent forces ? Mxy /?y have the same
positive direction as shearing forces Qx
第二版 2002 弹性力学 第十二章 51
F,At the corner point B
? The total concentrated shearing force is
RB=Myx)B+Mxy)B=2Mxy)B
= -2D(1- ?)[ ?2w/?x?y ]B=0
第二版 2002 弹性力学 第十二章 52
12.5 A simple solution for elliptical plates
? Elliptical plates with clamped edges
subjected to uniform transverse loads.
第二版 2002 弹性力学 第十二章 53
? Assume,w(x,y)=m(x2/a2+y2/b2-1)2 (1)
? Boundary conditions are satisfied.
W=0 ?w/?n=?w/?x cos(n x)+?w/?y cos(n y)=0
? Substituting Eq (1) into
D(?4/?x4+2?4/?x2?y2 +?4/?y4)w=q
yields
m=q0/[D(24/a4+16/(a2 b2)+24/b4)]
第二版 2002 弹性力学 第十二章 54
补充题：四边简支的薄板，受均布荷载 q,
b>>a
第二版 2002 弹性力学 第十二章 55
? D?4w(x,y)= q (1)
x=0,w(0,y)=0 (?2w/?x2)x=0=0
x=a,w(a,y)=0 (?2w/?x2)x=a=0
y=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 2002 弹性力学 第十二章 56
? Dw’’’’(x)= q (1)
x=0,w(0)=0 w’’(0)=0
x=a,w(a)=0 w’’(a)=0 (2)
w(x)=qx4/(24D)+Ax3+Bx2+Cx+E
w’’(x)= qx2/(2D)+6Ax+2B
w(0)=0 D=0 w’’(0)=0 B=0
w(a)=0 qa4/(24D)+Aa3+Ca=0 C=qa3/(24D)
w’’(a)=0 qa2/(2D)+6Aa=0 A= -qa/(12D)
w(x)=qx4/(24D)-qax3/(12D)+qa3x/(24D)
w’’(x)= qx2/(2D) -qa x /(2D)
M(x)=-D w’’(x)=- qx2/2+qa x /2 同梁
第二版 2002 弹性力学 第十二章 57
补充题 1：四边简支的薄板，
荷载 q=q0 sin(?x/a) sin(?y/b),
第二版 2002 弹性力学 第十二章 58
D?4w(x,y)= q0 sin(?x/a) sin(?y/b) (1)
D[?4w/?x4 +2?4w/?x2?y2 +?4w/?y4]=q0
sin(?x/a)sin(?y/b)
x=0,w(0,y)=0 (?2w/?x2)x=0=0
x=a,w(a,y)=0 (?2w/?x2)x=a=0
y=0,w(x,0)=0 (?2w/?y2)y=0=0 (2)
y=b,w(x,b)=0 (?2w/?y2)y=b=0
W(x,y)=A sin(?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) / [D?4(1/a2+1/b2)2 ] (4)
第二版 2002 弹性力学 第十二章 59
补充题 2：四边简支的薄板，
q=qmn sin(m?x/a) sin(n?y/b) mn为正整数
第二版 2002 弹性力学 第十二章 60
D?4w(x,y)= qmn sin(m?x/a) sin(n?y/b) (1)
D[?4w/?x4 +2?4w/?x2?y2 +?4w/?y4]=qmn
sin(m?x/a)sin(n?y/b)
x=0,w(0,y)=0 (?2w/?x2)x=0=0
x=a,w(a,y)=0 (?2w/?x2)x=a=0
y=0,w(x,0)=0 (?2w/?y2)y=0=0 (2)
y=b,w(x,b)=0 (?2w/?y2)y=b=0
W(x,y)=A sin(m?x/a) sin(n?y/b) （ 3)
将 (3)代入 (1)得 A,A代入 (3)得 (4)：
W(x,y)=qmn sin(m?x/a) sin(n?y/b)/[D?4(m2/a2+n2/b2)2 ]
（ 4)
第二版 2002 弹性力学 第十二章 61
补充题 3：四边简支的薄板，
q= ?m=1,2,3..,?n=1,2,3..,qmn sin(m?x/a)
sin(n?y/b) mn为正整数
第二版 2002 弹性力学 第十二章 62
D?4w(x,y)
= ?m=1,2,3..,?n=1,2,3..,qmn sin(m?x/a) sin(n?y/b) (1)
D[?4w/?x4 +2?4w/?x2?y2 +?4w/?y4]
= ?m=1,2,3..,?n=1,2,3..,qmn sin(m?x/a) sin(n?y/b) (1)
x=0,w(0,y)=0 (?2w/?x2)x=0=0
x=a,w(a,y)=0 (?2w/?x2)x=a=0
y=0,w(x,0)=0 (?2w/?y2)y=0=0 (2)
y=b,w(x,b)=0 (?2w/?y2)y=b=0
W(x,y)= ?m=1,2,3..,?n=1,2,3..,qmn sin(m?x/a) sin(n?y/b) /
[D?4(m2/a2+n2/b2)2 ] (3)
第二版 2002 弹性力学 第十二章 63
12.6(解法 1) Navier‘s Solution by Double
Trigonometric Series.-四边简支，受任意荷载
第二版 2002 弹性力学 第十二章 64
D?4w(x,y)= q(x,y) (1)
x=0,w(0,y)=0 (?2w/?x2)x=0=0
x=a,w(a,y)=0 (?2w/?x2)x=a=0
y=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 2002 弹性力学 第十二章 65
? q(x,y)=?m=1,2.?n=1,2qmnsin(m?x/a) sin( n?y/b)
? Multiplying both sides of this equation by
sin(i?x/a) sin( j?y/b),where i and j are arbitrary
integers,integrating the equation with respect to x
from 0 to a and with respect to y from 0 to b,we
have
? ?0a ?0bq(x,y) sin(i?x/a) sin(j?y/b)dxdy
=?m=1,2.?n=1,2 ?0a ?0b qmnsin(m?x/a) sin( n?y/b)
sin(i?x/a) sin(j?y/b) dxdy
第二版 2002 弹性力学 第十二章 66
? ?0a ?0bq(x,y) sin(i?x/a) sin(j?y/b)dxdy
=?m=1,2.?n=1,2 ?0a ?0b qmnsin(m?x/a) sin( n?y/b)
sin(i?x/a) sin(j?y/b) dxdy
? ?0a sin(m?x/a) sin(i?x/a) dx= 0,(m≠ i)
a/2,(m=i)
? ?0b sin( n?y/b)sin(j?y/b) dy= 0,(n≠j)
b/2 (n=j)
? qij=4/ab?0a ?0bq(?,?)sin(i??/a)sin(j??/b)d?d?
? q(x,y)=?m=1,2.?n=1,2qmnsin(m?x/a) sin( n?y/b)
=4/ab?m=1,2.?n=1,2 ?0a ?0bq(?,?) sin(m??/a)
sin(n??/b) d?d? sin(m?x/a) sin( n?y/b)
第二版 2002 弹性力学 第十二章 67
D?4w(x,y)
= ?m=1,2,3..,?n=1,2,3..,qmn sin(m?x/a) sin(n?y/b) (1)
D[?4w/?x4 +2?4w/?x2?y2 +?4w/?y4]
= ?m=1,2,3..,?n=1,2,3..,qmn sin(m?x/a) sin(n?y/b) (1)
qmn=4/ab?0a ?0bq(?,?)sin(m??/a)sin(n??/b)d?d?
x=0,w(0,y)=0 (?2w/?x2)x=0=0
x=a,w(a,y)=0 (?2w/?x2)x=a=0
y=0,w(x,0)=0 (?2w/?y2)y=0=0 (2)
y=b,w(x,b)=0 (?2w/?y2)y=b=0
W(x,y)= ?m=1,2,3..,?n=1,2,3..,qmn sin(m?x/a) sin(n?y/b) /
[D?4(m2/a2+n2/b2)2 ] (3)
第二版 2002 弹性力学 第十二章 68
12.6(解法 2) Navier‘s Solution by Double
Trigonometric Series.-四边简支，受任意荷载,
第二版 2002 弹性力学 第十二章 69
D?4w(x,y)= q(x,y) (1)
x=0,w(0,y)=0 (?2w/?x2)x=0=0
x=a,w(a,y)=0 (?2w/?x2)x=a=0
y=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 2002 弹性力学 第十二章 70
W(x,y)=?m=1,2.?n=1,2… Amnsin(m?x/a) sin( n?y/b) (3)
? Boundary conditions (2) are satisfied automatically
? Substituting Eq,(3) into Eq,(1)
D(?4/?x4+2?4/?x2?y2 +?4/?y4)w=q (1)
yields
?4D?m=1,2.?n=1,2… (m2/a2+n2/b2)2
Amnsin(m?x/a) sin( n?y/b) =q(x,y)
Amn
第二版 2002 弹性力学 第十二章 71
? q(x,y)=?m=1,2.?n=1,2qmnsin(m?x/a) sin( n?y/b)
? Multiplying both sides of this equation by
sin(i?x/a) sin( j?y/b),where i and j are arbitrary
integers,integrating the equation with respect to x
from 0 to a and with respect to y from 0 to b,we
have
? ?0a ?0bq(x,y) sin(i?x/a) sin(j?y/b)dxdy
=?m=1,2.?n=1,2 ?0a ?0bqmnsin(m?x/a) sin( n?y/b)
sin(i?x/a) sin(j?y/b) dxdy
第二版 2002 弹性力学 第十二章 72
? ?0a ?0bq(x,y) sin(i?x/a) sin(j?y/b)dxdy
=?m=1,2.?n=1,2 ?0a ?0b qmnsin(m?x/a) sin( n?y/b)
sin(i?x/a) sin(j?y/b) dxdy
? ?0a sin(m?x/a) sin(i?x/a) dx= 0,(m≠ i)
a/2,(m=i)
? ?0b sin( n?y/b)sin(j?y/b) dy= 0,(n≠j)
b/2 (n=j)
? qij=4/ab?0a ?0bq(?,?)sin(i??/a)sin(j??/b)d?d?
? q(x,y)=?m=1,2.?n=1,2qmnsin(m?x/a) sin( n?y/b)
=4/ab?m=1,2.?n=1,2 ?0a ?0bq(?,?) sin(m??/a)
sin(n??/b) d?d? sin(m?x/a) sin( n?y/b)
第二版 2002 弹性力学 第十二章 73
?4D?m=1,2.?n=1,2… (m2/a2+n2/b2)2
Amnsin(m?x/a) sin( n?y/b) =q(x,y)
q(x,y)=?m=1,2.?n=1,2qmnsin(m?x/a) sin( n?y/b)
?4D?m=1,2.?n=1,2… (m2/a2+n2/b2)2
Amnsin(m?x/a) sin( n?y/b)
= ?m=1,2.?n=1,2qmnsin(m?x/a) sin( n?y/b)
比较系数得
Amn =qmn / [?4D(m2/a2+n2/b2)2 ]=
[ 4/(ab)?m=1,2.?n=1,2 ?0a ?0bq(?,?) sin(m??/a)
sin(n??/b) d?d? ] / [?4D(m2/a2+n2/b2)2 ]
第二版 2002 弹性力学 第十二章 74
q(x,y)=q0
Amn=[ 4/(ab)?m=1,2.?n=1,2 ?0a ?0bq(?,?) sin(m??/a) ]
sin(n??/b) d?d? ] / [?4D(m2/a2+n2/b2)2 ]
=q0ab(1-cos m?)(1-cos n?)/(?2mn)
=16q0/ [?6Dmn(m2/a2+n2/b2)2 ]
(m=1,3,5,…;n=1,3,5…)
? w=16q0/ [?6D] ?m=1,3,5.?n=1,3,5 sin(m?x/a) sin(
n?y/b) /[mn(m2/a2+n2/b2)2 ]
第二版 2002 弹性力学 第十二章 75
q(x,y)=q0 a=b
? W(a/2 a/2) = 0.25 [16q0a4/ (?6D)] 1 term
0.2467 [16q0a4/ (?6D)] 2 terms
0.2470 [16q0a4/ (?6D)] 3 terms
0.2469 [16q0a4/ (?6D)] 4 terms----
is exact to three digits.
? Mx(a/2,a/2)= 0.325 [16q0a2/ ?4] 1 term
0.2857[16q0a2/ ?4] 2 terms
0.2936 [16q0a2/ ?4] 3 terms
0.2920 [16q0a2/ ?4] 4 terms----is
exact to three digits.
第二版 2002 弹性力学 第十二章 76
12.7 Levy's Solution by Single
Trigonometric Series.
第二版 2002 弹性力学 第十二章 77
? D?4w(x,y)= q(x,y) (1)
x=0,w(0,y)=0 (?2w/?x2)x=0=0
x=a,w(a,y)=0 (?2w/?x2)x=a=0 (2)
y=0,specified later
y=b,specified later
第二版 2002 弹性力学 第十二章 78
Assume W(x,y)=?m=1,2.Ym(y) sin(m?x/a) (3)
? Boundary conditions (2) are satisfied automatically
? Substituting Eq,(3) into Eq,(1)
D(?4/?x4+2?4/?x2?y2 +?4/?y4)w=q (1)
yields
?m=1,2 [ym(4)-2(m?/a)2Ym(2)+(m?/a)4Ym] sin(m?x/a) =q/D
第二版 2002 弹性力学 第十二章 79
D?m=1,2 [ym(4)-2(m?/a)2Ym(2)+(m?/a)4Ym] sin(m?x/a) =q
q=2/a ?m=1,2 [?0aq(? y)sin(m??/a)d?] sin(m?x/a)
D?m=1,2 [ym(4)-2(m?/a)2Ym(2)+(m?/a)4Ym] sin(m?x/a)
=2/a ?m=1,2 [?0aq(? y)sin(m??/a)d?] sin(m?x/a)
ym(4)-2(m?/a)2Ym(2)+(m?/a)4Ym
=2/(Da) [?0aq(? y)sin(m??/a)d?]
第二版 2002 弹性力学 第十二章 80
ym(4)-2(m?/a)2Ym(2)+(m?/a)4Ym
=2/(Da) [?0aq(? y)sin(m??/a)d?]
Ym=Amch(m?y/a)+Bm (m?y/a) sh(m?y/a)+
Cmsh(m?y/a)+Dm (m?y/a) ch(m?y/a)+fm(y)
where fm(y) is a particular solution.
W= ?m[Amch(m?y/a)+Bm (m?y/a) sh(m?y/a)+
Cmsh(m?y/a)+Dm (m?y/a) ch(m?y/a)+fm(y)] sin(m?x/a)
where Am BmCm Dm are arbitrary constants and can be
determined by the boundary conditions on the edges
y=?b/2
第二版 2002 弹性力学 第十二章 81
Rectangular plate without free edges--only with
simply supported edges or clamped edges
? ?4 [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=0
? Dw is independent of ?
第二版 2002 弹性力学 第十二章 82
? Mx= - [?2(Dw)/?x2+??2(Dw/?y2)]
My= - [(?2(Dw/?y2+??2(Dw/?x2)]
? ?=0 Table
Mx= - [?2(Dw)/?x2]
My= - [(?2(Dw/?y2]
? ?≠ 0
Mx’= - [?2(Dw)/?x2+??2(Dw/?y2)]
= Mx +? My
My’= - [(?2(Dw/?y2+??2(Dw/?x2)]
= My +? Mx
第二版 2002 弹性力学 第十二章 83
12.8 Bending of Circle Plate
? Cylindrical coordinates (r,?,z)
第二版 2002 弹性力学 第十二章 84
? D?4w=D[?/(r?r)+?2/(r2??2)+?2/?r2 ]2 w =q(r,?)
? Mr=-D{?2w/?r2 +?[?w/(r?r)+?2?/(r2??2)] }
?r=zMr /I I= t3/12
? M?==-D{?w/(r?r)+?2w/(r2??2) +??2w/?r2 }
??= zM? /I
? Mr? = M?r =-D(1-?)[?2w/(r?r??)-?w/(r2??)]
?r?= zMr? /I
? Qr=-D ?(?2w )/?r
?rz= 6Qr(t2/4-z2)/t3
? Q? =-D ?(?2w )/(r??)
??z= 6Q?(t2/4-z2)/t3
第二版 2002 弹性力学 第十二章 85
Boundary conditions
? a clamped edge r=a:
w(a,?)=0,?w/?r)r=a=0
? a simply supported edge r=a
w(a,?)=0,
Mr )r=a =-D{?2w/?r2 +?[?w/(r?r)+?2?/(r2??2)] } )r=a=0
? a free edge r=a
Mr )r=a =0 Vr )r=a =[Qr+ ?Mr ? /(r??)]r=a =0
第二版 2002 弹性力学 第十二章 86
12.9 Axisymmetrical Bending of Circle Plates
? If q(r,?) is q(r) on a circle plate,
w(r,?) will be w(r) and
D?4w =D[?/(r?r)+?2/(r2??2)+?2/?r2 ]2 w =q(r,?)
will be
D[?/(r?r)+?2/?r2 ]2 w (r)=q(r)
The solution of this ordinary differential equation is
w(r)=C1 lnr+C2 r2 lnr+C3 r2 +C4+w1
where w1 is any particular solution and C1 C2 C3 C4
are arbitrary constants.
第二版 2002 弹性力学 第十二章 87
? Mr=-D{?2w/?r2 +?[?w/(r?r)+?2?/(r2??2)] }
=-D[d2w/dr2 +?dw/(rdr)]
? M?=-D{?w/(r?r)+?2w/(r2??2) +??2w/?r2 }
=-D{dw/(rdr)+?d2w/dr2 }
? Mr? = M?r =-D(1-?)[?2w/(r?r??)-?w/(r2??)] =0
? Qr=-D ?(?2w )/?r=-Dd[d2w/dr2 +dw/(rdr)]/dr
? Q? =-D ?(?2w )/(r??)=0
第二版 2002 弹性力学 第十二章 88
w(r)=C1 lnr+C2 r2 lnr+C3 r2 +C4+w1
dw/dr= C1 /r+ C2(2r lnr+r)+2 C3r+dw1/dr
dw2/dr2=- C1 /r2+ C2(2 lnr+3)+2 C3+d2w1/dr2
? Mr =-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 2002 弹性力学 第十二章 89
A,annular plate a<r<b----a circular plate with a
central hole
? There will be two Bcs at r=a,two Bcs at r=b,
so the four arbitrary constants can be
determined.
第二版 2002 弹性力学 第十二章 90
B,solid plate r?a (no centre hole)
? Qr=-D{ C2(4/r)- P
dw1/(r2dr)+d2w1/(rdr2)+d3w1/dr3}
? Two Bcs at r=a Qr
? r=0,1,since w is finite,C1 =0
2(1).No concentrated force at r=0,Since Qr is
finite,C2 =0.
2(2),Concentrated force P at r=0,
?Fz=0,2?rQr+P=0 2?r(-4DC2/r) +P=0
C2=P/(8D?)
第二版 2002 弹性力学 第十二章 91
Example 1,a circular clamped plate without
hole is subjected to a uniform load q0
w(r)=C1 lnr+C2 r2 lnr+C3 r2 +C4+w1
? particular solution:
D[?/(r?r)+?2/?r2 ]2 w (r)=q0
assume,w1 =mr4 w1 =q0r4 /(64D)
? r=0 C1 =C2 =0
? r=a w=0 dw/dr=0
C3=-20q0a2 /(32D) C4=q0a4 /(64D)
? w(r)= q0 (r2 -a2 )2 /(64D)
第二版 2002 弹性力学 第十二章 92
Exa.2,a circular clamped plate without hole
subjected to a concentrated load P at the center
r=0
w(r)=C1 lnr+C2 r2 lnr+C3 r2 +C4+w1
? particular solution,w1 =0
? r=0 1,since w is finite,C1 =0 2,C2=P/(8D?)
? r=a w=0 dw/dr=0
C3=-P(2lna+1)/(16D?) C4=Pa2 /(16D?)
? w(r)=P(2r2 ln(r/a)-r2+a2 )/(16D?)
第二版 2002 弹性力学 第十二章 93
习题作业（英文书）
? 12.12.4
? 12.12.6
? 13.6.1
? 下星期的今天交作业
第二版 2002 弹性力学 第十二章 94
Chapter 13 Bending of Thin Plates--Finite-
Difference Method
Sec 7.1 Finite-difference formulas
? f(x+dx,y0 ) ≈ f(x,y0)+?f(x,y0)/?x dx+0.5
?2f(x,y0)/?x2 dx2
? x=x0 dx=h x+dx=x1 f(x1,y0 )=f1 f(x0,y0 )=f0
f1= f0 +(?f/?x)0 h+0.5 (?2f/?x2)0 h2 (1)
? x=x0 dx=-h x+dx=x3 f(x3,y0 )=f3 f(x0,y0 )=f0
f3= f0-(?f/?x)0 h+0.5 (?2f/?x2)0 h2 (2)
?
第二版 2002 弹性力学 第十二章 95
Sec 7.1 Finite-difference formulas
? f(x+dx,y0 ) ≈ f(x,y0)+?f(x,y0)/?x dx+0.5
?2f(x,y0)/?x2 dx2
? x=x0 dx=h x+dx=x1 f(x1,y0 )=f1 f(x0,y0 )=f0
f1= f0 +(?f/?x)0 h+0.5 (?2f/?x2)0 h2 (1)
? x=x0 dx=-h x+dx=x3 f(x3,y0 )=f3 f(x0,y0 )=f0
f3= f0(?f/?x)0 h+0.5 (?2f/?x2)0 h2 (2)
? (?f/?x)0 =(f1 -f3 )/(2h)
(?2f/?x2)0 =(f1 +f3 -2f0 )/h2
第二版 2002 弹性力学 第十二章 96
? x
12
8 4 5
11 3 0 1 9
7 2 6
10
y
第二版 2002 弹性力学 第十二章 97
? Fundamental finite-differences formulas
(?f/?x)0 =(f1 -f3 )/(2h)
(?2f/?x2)0 =(f1 +f3 -2f0 )/h2
similiarly,(?f/?y)0 =(f2 -f4 )/(2h)
(?2f/?y2)0 =(f2 +f4 -2f0 )/h2
? (?2f/?x?y)0 =[?(?f/?y) / ?x]0=[(?f/?y)1-(?f/?y)3 ]/(2h)
=[(f6 -f5 )/2h-(f7 -f8 ) /2h]/2h
= [(f6+f8 )-(f5+ f7)] /(4h2)
(?4f/?x4)0 =[6f0-4(f1+f3 )+(f9+f11)] /h4
(?4f/?y4)0 =[6f0-4(f2+f4 )+(f10+f12)] /h4
第二版 2002 弹性力学 第十二章 98
Sec.13.1 Finite-difference representations
D(?4/?x4+2?4/?x2?y2 +?4/?y4)w=q (1)
D[(?4w/?x4 )0 +2(?4w/?x2?y2)0 +(?4 w/?y4)0=q0 (2)
20w0-8(w1+w2+w3+w4)+2(w5+w6+w7+w8)
+(w9+w10+w11+w12)=h4q0/D (3)
20本点 -8邻点 +2角点 +1远点 = h4q0/D
n个内点建立 n个方程。
第二版 2002 弹性力学 第十二章 99
Example,P294(E)
第二版 2002 弹性力学 第十二章 100
13.2 Plate without free edges
? Let AB represents a fixed edge A x
3--inside point
1--outside point 3 0 1
(?w/?x)0 =(w1 -w3 )/(2h) =0 h h
w1 =w3 B
? Let AB reoresents a simply supported edge
(?2w/?x2)0 =(w1 +w3 -2w0 )/h2=0
w1 =-w3
Chapter 12 Bending of Thin Plates,
Classical Solutions
第十二章 薄板弯曲问题。经典解答。
第二版 2002 弹性力学 第十二章 2
Section 12.1 Introduction and Assumptions
引言和假定
A plate is a body bounded by two closely
spaced parallel planes and one or more
prismatical surfaces normal to the
planes.
第二版 2002 弹性力学 第十二章 3
? Plate faces--two closely spaced parallel
planes
? Plate edges--prismatical surfaces normal to
the plate faces.
第二版 2002 弹性力学 第十二章 4
? Plate thickness--the distance between the
two plate faces,It is denoted by t.
? Thin plate-- t<<a t <<b
t<min(a,b)/8
? Thick plate
第二版 2002 弹性力学 第十二章 5
? Plate middle plane--
The plane parallel to the faces of the plate
and bisecting the thickness t is called the
middle plane of the plate,
第二版 2002 弹性力学 第十二章 6
? Coordinate system--x and y are in the
middle plane and z axis is perpendicular to
the middle plane, The system is a right
hand system.
第二版 2002 弹性力学 第十二章 7
Loads
1,Longitudinal load in the middle plane--All
the external forces are parallel to the faces
of the plate and distributed uniformly over
the thickness.--------plane stress problem.
2.Transverse load ----They are perpendicular
to the middle plane---plate bending
problem.
第二版 2002 弹性力学 第十二章 8
? The surface force component Z on the lower
face of the plate and the body force
component Z are transmitted to the upper
face of the plate along their lines of action,
that is
q(x,y)=Z)z=-t/2+Z)z=+t/2+ ?-t/2t/2 Zdz
Z,Z and q are considered positive when
they act in the positive direction of z.
q(x,y) is total transverse load per unit area.
第二版 2002 弹性力学 第十二章 9
? Deflection--the displacement of a point on the
middle plane in the direction of z,w(x,y,0),is
called the deflection of the point,
? Small deflections--the deflection is much smaller
than the thickness,W(x,y,0)<t/5
? Only small deflections are considered here,
第二版 2002 弹性力学 第十二章 10
Basic Assumptions
? Assumption stated in Sec.1.3
1,The body is continuous,perfectly elastic,
homogeneous and isotropic,
2.The displacements and strains are small.
------The deflection of the plate is small.
? Thin plates
第二版 2002 弹性力学 第十二章 11
Assumption 1.--?z rzx and rzy are neglected,
since they are small.
? The three physical equations are not valid
again and have to be abandoned.
?z=[?z- ??x- ??y]/E
rzx=?zx/G
rzy=?zy/G
第二版 2002 弹性力学 第十二章 12
?z=?w/?z ≈ 0-------w=w(x,y)
? This equation shows that all the points
aligned on a normal to the middle plane will
have the same displacement in the
transverse direction and that this
displacement is just the deflection of the
plate.
第二版 2002 弹性力学 第十二章 13
rzx= rzy ≈ 0
? Initially straight lines which are normal to
the middle plane remain straight and normal
to the deformed middle plane.
第二版 2002 弹性力学 第十二章 14
Assumption 2.--The strains due to ?z are
neglected,since ?z is small.
? The three physical equations will be the same
as those in the plane stress problems:
?x=[?x- ??y- ??z]/E
?y=[?y- ??x- ??z]/E
rxy=?xy/G
? The other physical equations are abandoned.
第二版 2002 弹性力学 第十二章 15
Assumption 3.--u)z=0 and v)z=0 are neglected.
The longitudinal displacements of all the points
in the middle plane may be neglected
? ( ?x= ?y= rxy)z=0 ≈ 0
The middle plane of the plate remains
unstrained,although the plane becomes
curved after bending.
第二版 2002 弹性力学 第十二章 16
12.2 Differential Equation for
Bending of Thin Plates
? Basic unknown function w(x,y)
? Fifteen equations for spatial problems------
one equation in term of w(x,y) for plate
bending problem.
第二版 2002 弹性力学 第十二章 17
A.express u and v in terms of w
? rzx= ?u/?z+ ?w(x,y)/?x=0 u=-z ?w /?x+f1(x,y)
rzy= ?v/?z+ ?w(x,y)/?y=0 v=-z ?w/?y+f2(x,y)
? u)z=0= v)z=0=0 f1= f2=0
? u=-z ?w /?x
v=-z ?w/?y
第二版 2002 弹性力学 第十二章 18
B.express strain components in
terms of w
? u=-z ?w /?x v=-z ?w/?y
? ?x=?u/?x= -z ?2w /?x2
?y=?v/?y= -z ?2w/?y2
rxy=?u/?y+?v/?x= -2z?2w/?x?y
? ?z rzx and rzy are neglected
第二版 2002 弹性力学 第十二章 19
C.express stress components ?x?y?xy
in terms of w--use physical equation
? ?x=?u/?x= -z ?2w/?x2
?y=?v/?y= -z ?2w/?y2
rxy=?u/?y+?v/?x= -2z?2w/?x?y
?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/?x?y
第二版 2002 弹性力学 第十二章 20
D.express strain components ?zx?zy
in terms of w--use equations of equilibrium
? ??x/?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)/?x
??zy/?z=-(??xy/?x+ ??y/?y)
= Ez/(1-? 2) ?(?2w/?x2+?2w/?y2)/?y
第二版 2002 弹性力学 第十二章 21
? ??zx/?z =-(??x/?x+??yx/?y)
= Ez/(1-? 2) ?(?2w/?x2+?2w/?y2)/?x
??zy/?z=-(??xy/?x+ ??y/?y)
= Ez/(1-? 2) ?(?2w/?x2+?2w/?y2)/?y
? ?zx =0.5Ez2/(1-? 2) ?(?2w)/?x+F1(x,y)
?zy=0.5Ez2/(1-? 2) ?(?2w)/?y+F2(x,y)
?zx)z=? t/2=0 ?zy)z=? t/2=0
F1(x,y)= 0.125Et2/(1-? 2) ?(?2w)/?x
F2(x,y)= 0.125Et2/(1-? 2) ?(?2w)/?y
? ?zx =0.5E(z2-t2/4)/(1-? 2) ?(?2w)/?x
?zy=0.5E(z2 -t2/4)/( 1-? 2) ?(?2w)/?y parabolic
第二版 2002 弹性力学 第十二章 22
E.express stress components ?z
in terms of w--use equations of equilibrium
? ??xz/?x +??yz/?y+ ??z/?z+Z=0
q(x,y)=Z)z=-t/2+Z)z=+t/2+?-t/2t/2 Zdz
??z/?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=0 F3(x,y) = -E t3/[24(1-? 2)](?4w)
? ?z=-0.5E (z3/3-zt2/4)/(1-? 2)?4w-Et3/[24(1-? 2)](?4w)
第二版 2002 弹性力学 第十二章 23
F,The differential Equation of deflection w
---B.C,on the upper face
? ?z=-0.5E (z3/3-zt2/4)/(1-? 2)?4w-Et3/[24(1-? 2)](?4w)
?z)z=-t/2= -q
? D?4w(x,y)= q(x,y)
? D=Et3/[12(1-? 2)]---flexural rigidity of the plate.
D has a dimension of [force][length]
第二版 2002 弹性力学 第十二章 24
? D?4w(x,y)= q(x,y)
1,Fifteen equations for spatial problems
become one equation in term of w(x,y) for
plate bendinf problem.
2,Boundary conditions on z= ?t/2 are
satisfied.
3,Boundary conditions on edges of plate
have to be satisfied.
第二版 2002 弹性力学 第十二章 25
12.3 Stress Resultants and Stress Couples
---Internal Forces
? In most cases,it is impossible to satisfy the
pointwise stress boundary conditions on the
edges of plate,We can satisfy only the boundary
conditions in terms of the stress resultants and
stress couples,Hence,before discussing the
boundary conditions,we proceed to investigate
the stress resultants and stress couples on the
transverse sections of a plate,
第二版 2002 弹性力学 第十二章 26
An elementary parallelepiped
第二版 2002 弹性力学 第十二章 27
A,on the transverse section normal to x axis
? ?x= -Ez/(1-? 2)(?2w/?x2+? ?2w/?y2)
?xy= -Ez/(1+?) ?2w/?x?y
linear variation along z direction,odd function of z
? ?xz =0.5E(z2-t2/4)/(1-? 2) ?(?2w)/?x
parabolic in z direction,even function of z
第二版 2002 弹性力学 第十二章 28
? ?x= -Ez/(1-? 2)(?2w/?x2+? ?2w/?y2)
? Mx=?-t/2t/2 z?x 1 dz= -D(?2w/?x2+? ?2w/?y2)
? ?x= -Ez/(1-? 2)(?2w/?x2+? ?2w/?y2) =z Mx/I
I=t3/12
linear variation along z direction,odd function of z
第二版 2002 弹性力学 第十二章 29
第二版 2002 弹性力学 第十二章 30
? ?xy= -Ez/(1+?) ?2w/?x?y
Mxy=?-t/2t/2 z ?xy 1 dz= -Et3/[12 /(1+?)] ?2w/?x?y
=-D(1- ?) ?2w/?x?y
? ?xy= -Ez/(1+?) ?2w/?x?y =z Mxy/I
I=t3/12
linear variation along z direction,odd function of z
第二版 2002 弹性力学 第十二章 31
? ?xz =0.5E(z2-t2/4)/(1-? 2) ?(?2w)/?x
parabolic in z direction,even function of z
Qx=?-t/2t/2 ?xz 1 dz= -D ?(?2w)/?x
? ?xz =0.5E(z2-t2/4)/(1-? 2) ?(?2w)/?x
=6Qx (t2/4-z2)/t3
第二版 2002 弹性力学 第十二章 32
B,on the transverse section normal to y axis
?y=E/(1-? 2)(?y+? ?x)=-Ez/(1-? 2)(?2w/?y2+? ?2w/?x2)
?yx= E/[2(1+?)] rxy = -Ez/(1+?) ?2w/?x?y
linear variation along z direction,odd function of z
?yz=0.5E(z2 -t2/4)/( 1-? 2) ?(?2w)/?y
parabolic in z direction,even function of z
第二版 2002 弹性力学 第十二章 33
? ?y= -Ez/(1-? 2)(?2w/?y2+? ?2w/?x2)
? My=?-t/2t/2 z?y 1 dz= -D(?2w/?y2+? ?2w/?x2)
? ?y= -Ez/(1-? 2)(?2w/?y2+? ?2w/?x2) =z My/I
I=t3/12
linear variation along z direction,odd function of z
第二版 2002 弹性力学 第十二章 34
? ?yx= -Ez/(1+?) ?2w/?x?y
Myx=?-t/2t/2 z ?yx 1 dz= -Et3/[12 /(1+?)] ?2w/?x?y
=-D(1- ?) ?2w/?x?y
?yx= ?xy Myx= Mxy
? ?xy= -Ez/(1+?) ?2w/?x?y =z Mxy/I
I=t3/12
linear variation along z direction,odd function of z
第二版 2002 弹性力学 第十二章 35
? ?yz =0.5E(z2-t2/4)/(1-? 2) ?(?2w)/?y
parabolic in z direction,even function of z
Qy=?-t/2t/2 ?yz 1 dz= -D ?(?2w)/?y
? ?yz =0.5E(z2-t2/4)/(1-? 2) ?(?2w)/?y
=6Qy (t2/4-z2)/t3
第二版 2002 弹性力学 第十二章 36
? Mx My----bending moment per unit width.[force]
Mxy=Myx--twisting moment per unit width,[force]
A positive moment corresponds to a positive
stress component in the positive half of the plate.
Mx My and Mxy are of order of qa2
? Qx Qy---transverse shearing force per unit width
[force][length]-1
The positive direction of Qx is the same as ?xz
The positive direction of Qy is the same as ?yz
Qx and Qy are of order of qa
第二版 2002 弹性力学 第十二章 37
Express positive internal forces on the
middle plane
第二版 2002 弹性力学 第十二章 38
? ?x?y and ?yx are of order of q(a/t)2
Max My and Mxy are of order of qa2
? ?xz and ?yz are of order of qa/t
Qx and Qy are of order of qa
? ?z--transverse normal stress----order of q
? (?x ?y ?yx)>>( ?xz ?yz )>>?z
第二版 2002 弹性力学 第十二章 39
?Fz=0,Plate,?Qx/?x+?Qy/?y+q=0 (1)
Beam,dQx/dx+q=0 (1a)
第二版 2002 弹性力学 第十二章 40
?My=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 2002 弹性力学 第十二章 41
Plate,Substituting Eqs(2)(3) into Eq(1) and
noting that Mxy=Myx,we have:
?2Mx/?x2+2?2Myx/ ?x?y+?2My/?y2+q=0 (4)
Beam,Substituting Eqs(2a) into Eq(1a),we
have:
d2Mx/dx2+q=0 (4a)
第二版 2002 弹性力学 第十二章 42
Plate,?2Mx/?x2+2?2Myx/ ?x?y+?2My/?y2+q=0 (4)
Mx= -D(?2w/?x2+? ?2w/?y2)
My= -D(?2w/?y2+? ?2w/?x2) (5)
Myx= -Et3/[12 /(1+?)] ?2w/?x?y=-D(1- ?) ?2w/?x?y
Substitution of Eqs(5) into Eq(4) yields
D?4w(x,y)= q(x,y)
Beam,d2Mx/dx2+q=0 (4a)
Mx= -EId2w/dx2 (5a)
Substitution of Eqs(5a) into Eq(4a) yields
EId4w(x)/dx4= q(x)
第二版 2002 弹性力学 第十二章 43
12.4 Boundary Conditions
第二版 2002 弹性力学 第十二章 44
A,At the clamped edge OA (x=0)
? The deflection must be zero.
W(0,y)=0
? The slope of the middle plane must be zero.
[?w(x,y)/ ?x]x=0=0
第二版 2002 弹性力学 第十二章 45
B,At the simply supported edge OC
(y=0)
? The deflection must be zero.
W(x,0)=0
? The bending moment must be zero.
(My)y=0= -D(?2w/?y2+? ?2w/?x2)y=0=0
W(x,0)=0 ?2w/?x2)y=0=0
(?2w/?y2)y=0=0
第二版 2002 弹性力学 第十二章 46
C,At the free edge AB (y=b)
? There must be no bending moments.
(My)y=b= -D(?2w/?y2+? ?2w/?x2)y=b=0
? There must be no twisting moments.
Myx )y=b = -D(1- ?)[ ?2w/?x?y ]y=b=0
? There must be no transverse shearing forces.
Qy )y=b = -D [?(?2w)/?y ]y=b =0
第二版 2002 弹性力学 第十二章 47
D,At the free edge BC (x=a)
? There must be no bending moments.
(Mx)x=a= -D(?2w/?x2+? ?2w/?y2)x=a=0
? There must be no twisting moments.
Mxy )x=a = -D(1- ?)[ ?2w/?x?y ]x=a=0
? There must be no transverse shearing forces.
Qx )x=a = -D [?(?2w)/?y ]x=a =0
第二版 2002 弹性力学 第十二章 48
E,Twisting moments are replaced by a
statically equivalent shearing force,
第二版 2002 弹性力学 第十二章 49
C,At the free edge AB (y=b)
? There must be no bending moments.
(My)y=b= -D(?2w/?y2+? ?2w/?x2)y=b=0
? The total distributed shearing force must be zero.
Vy )y=b=[Qy+ ?Myx /?x]y=b
= -D [?3w/?y3+2(1-?)?3w/ ?x2?y]y=b =0
? The equivalent forces ? Myx /?x have the same
positive direction as shearing forces Qy
第二版 2002 弹性力学 第十二章 50
D,At the free edge BC (x=a)
? There must be no bending moments.
(Mx)x=a= -D(?2w/?x2+? ?2w/?y2)x=a=0
? The total distributed shearing force must be zero.
Vx )x=a=[Qx+ ? Mxy /?y]x=a
= -D [?3w/?x3+2(1-?)?3w/ ?y2?x]x=a =0
? The equivalent forces ? Mxy /?y have the same
positive direction as shearing forces Qx
第二版 2002 弹性力学 第十二章 51
F,At the corner point B
? The total concentrated shearing force is
RB=Myx)B+Mxy)B=2Mxy)B
= -2D(1- ?)[ ?2w/?x?y ]B=0
第二版 2002 弹性力学 第十二章 52
12.5 A simple solution for elliptical plates
? Elliptical plates with clamped edges
subjected to uniform transverse loads.
第二版 2002 弹性力学 第十二章 53
? Assume,w(x,y)=m(x2/a2+y2/b2-1)2 (1)
? Boundary conditions are satisfied.
W=0 ?w/?n=?w/?x cos(n x)+?w/?y cos(n y)=0
? Substituting Eq (1) into
D(?4/?x4+2?4/?x2?y2 +?4/?y4)w=q
yields
m=q0/[D(24/a4+16/(a2 b2)+24/b4)]
第二版 2002 弹性力学 第十二章 54
补充题：四边简支的薄板，受均布荷载 q,
b>>a
第二版 2002 弹性力学 第十二章 55
? D?4w(x,y)= q (1)
x=0,w(0,y)=0 (?2w/?x2)x=0=0
x=a,w(a,y)=0 (?2w/?x2)x=a=0
y=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 2002 弹性力学 第十二章 56
? Dw’’’’(x)= q (1)
x=0,w(0)=0 w’’(0)=0
x=a,w(a)=0 w’’(a)=0 (2)
w(x)=qx4/(24D)+Ax3+Bx2+Cx+E
w’’(x)= qx2/(2D)+6Ax+2B
w(0)=0 D=0 w’’(0)=0 B=0
w(a)=0 qa4/(24D)+Aa3+Ca=0 C=qa3/(24D)
w’’(a)=0 qa2/(2D)+6Aa=0 A= -qa/(12D)
w(x)=qx4/(24D)-qax3/(12D)+qa3x/(24D)
w’’(x)= qx2/(2D) -qa x /(2D)
M(x)=-D w’’(x)=- qx2/2+qa x /2 同梁
第二版 2002 弹性力学 第十二章 57
补充题 1：四边简支的薄板，
荷载 q=q0 sin(?x/a) sin(?y/b),
第二版 2002 弹性力学 第十二章 58
D?4w(x,y)= q0 sin(?x/a) sin(?y/b) (1)
D[?4w/?x4 +2?4w/?x2?y2 +?4w/?y4]=q0
sin(?x/a)sin(?y/b)
x=0,w(0,y)=0 (?2w/?x2)x=0=0
x=a,w(a,y)=0 (?2w/?x2)x=a=0
y=0,w(x,0)=0 (?2w/?y2)y=0=0 (2)
y=b,w(x,b)=0 (?2w/?y2)y=b=0
W(x,y)=A sin(?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) / [D?4(1/a2+1/b2)2 ] (4)
第二版 2002 弹性力学 第十二章 59
补充题 2：四边简支的薄板，
q=qmn sin(m?x/a) sin(n?y/b) mn为正整数
第二版 2002 弹性力学 第十二章 60
D?4w(x,y)= qmn sin(m?x/a) sin(n?y/b) (1)
D[?4w/?x4 +2?4w/?x2?y2 +?4w/?y4]=qmn
sin(m?x/a)sin(n?y/b)
x=0,w(0,y)=0 (?2w/?x2)x=0=0
x=a,w(a,y)=0 (?2w/?x2)x=a=0
y=0,w(x,0)=0 (?2w/?y2)y=0=0 (2)
y=b,w(x,b)=0 (?2w/?y2)y=b=0
W(x,y)=A sin(m?x/a) sin(n?y/b) （ 3)
将 (3)代入 (1)得 A,A代入 (3)得 (4)：
W(x,y)=qmn sin(m?x/a) sin(n?y/b)/[D?4(m2/a2+n2/b2)2 ]
（ 4)
第二版 2002 弹性力学 第十二章 61
补充题 3：四边简支的薄板，
q= ?m=1,2,3..,?n=1,2,3..,qmn sin(m?x/a)
sin(n?y/b) mn为正整数
第二版 2002 弹性力学 第十二章 62
D?4w(x,y)
= ?m=1,2,3..,?n=1,2,3..,qmn sin(m?x/a) sin(n?y/b) (1)
D[?4w/?x4 +2?4w/?x2?y2 +?4w/?y4]
= ?m=1,2,3..,?n=1,2,3..,qmn sin(m?x/a) sin(n?y/b) (1)
x=0,w(0,y)=0 (?2w/?x2)x=0=0
x=a,w(a,y)=0 (?2w/?x2)x=a=0
y=0,w(x,0)=0 (?2w/?y2)y=0=0 (2)
y=b,w(x,b)=0 (?2w/?y2)y=b=0
W(x,y)= ?m=1,2,3..,?n=1,2,3..,qmn sin(m?x/a) sin(n?y/b) /
[D?4(m2/a2+n2/b2)2 ] (3)
第二版 2002 弹性力学 第十二章 63
12.6(解法 1) Navier‘s Solution by Double
Trigonometric Series.-四边简支，受任意荷载
第二版 2002 弹性力学 第十二章 64
D?4w(x,y)= q(x,y) (1)
x=0,w(0,y)=0 (?2w/?x2)x=0=0
x=a,w(a,y)=0 (?2w/?x2)x=a=0
y=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 2002 弹性力学 第十二章 65
? q(x,y)=?m=1,2.?n=1,2qmnsin(m?x/a) sin( n?y/b)
? Multiplying both sides of this equation by
sin(i?x/a) sin( j?y/b),where i and j are arbitrary
integers,integrating the equation with respect to x
from 0 to a and with respect to y from 0 to b,we
have
? ?0a ?0bq(x,y) sin(i?x/a) sin(j?y/b)dxdy
=?m=1,2.?n=1,2 ?0a ?0b qmnsin(m?x/a) sin( n?y/b)
sin(i?x/a) sin(j?y/b) dxdy
第二版 2002 弹性力学 第十二章 66
? ?0a ?0bq(x,y) sin(i?x/a) sin(j?y/b)dxdy
=?m=1,2.?n=1,2 ?0a ?0b qmnsin(m?x/a) sin( n?y/b)
sin(i?x/a) sin(j?y/b) dxdy
? ?0a sin(m?x/a) sin(i?x/a) dx= 0,(m≠ i)
a/2,(m=i)
? ?0b sin( n?y/b)sin(j?y/b) dy= 0,(n≠j)
b/2 (n=j)
? qij=4/ab?0a ?0bq(?,?)sin(i??/a)sin(j??/b)d?d?
? q(x,y)=?m=1,2.?n=1,2qmnsin(m?x/a) sin( n?y/b)
=4/ab?m=1,2.?n=1,2 ?0a ?0bq(?,?) sin(m??/a)
sin(n??/b) d?d? sin(m?x/a) sin( n?y/b)
第二版 2002 弹性力学 第十二章 67
D?4w(x,y)
= ?m=1,2,3..,?n=1,2,3..,qmn sin(m?x/a) sin(n?y/b) (1)
D[?4w/?x4 +2?4w/?x2?y2 +?4w/?y4]
= ?m=1,2,3..,?n=1,2,3..,qmn sin(m?x/a) sin(n?y/b) (1)
qmn=4/ab?0a ?0bq(?,?)sin(m??/a)sin(n??/b)d?d?
x=0,w(0,y)=0 (?2w/?x2)x=0=0
x=a,w(a,y)=0 (?2w/?x2)x=a=0
y=0,w(x,0)=0 (?2w/?y2)y=0=0 (2)
y=b,w(x,b)=0 (?2w/?y2)y=b=0
W(x,y)= ?m=1,2,3..,?n=1,2,3..,qmn sin(m?x/a) sin(n?y/b) /
[D?4(m2/a2+n2/b2)2 ] (3)
第二版 2002 弹性力学 第十二章 68
12.6(解法 2) Navier‘s Solution by Double
Trigonometric Series.-四边简支，受任意荷载,
第二版 2002 弹性力学 第十二章 69
D?4w(x,y)= q(x,y) (1)
x=0,w(0,y)=0 (?2w/?x2)x=0=0
x=a,w(a,y)=0 (?2w/?x2)x=a=0
y=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 2002 弹性力学 第十二章 70
W(x,y)=?m=1,2.?n=1,2… Amnsin(m?x/a) sin( n?y/b) (3)
? Boundary conditions (2) are satisfied automatically
? Substituting Eq,(3) into Eq,(1)
D(?4/?x4+2?4/?x2?y2 +?4/?y4)w=q (1)
yields
?4D?m=1,2.?n=1,2… (m2/a2+n2/b2)2
Amnsin(m?x/a) sin( n?y/b) =q(x,y)
Amn
第二版 2002 弹性力学 第十二章 71
? q(x,y)=?m=1,2.?n=1,2qmnsin(m?x/a) sin( n?y/b)
? Multiplying both sides of this equation by
sin(i?x/a) sin( j?y/b),where i and j are arbitrary
integers,integrating the equation with respect to x
from 0 to a and with respect to y from 0 to b,we
have
? ?0a ?0bq(x,y) sin(i?x/a) sin(j?y/b)dxdy
=?m=1,2.?n=1,2 ?0a ?0bqmnsin(m?x/a) sin( n?y/b)
sin(i?x/a) sin(j?y/b) dxdy
第二版 2002 弹性力学 第十二章 72
? ?0a ?0bq(x,y) sin(i?x/a) sin(j?y/b)dxdy
=?m=1,2.?n=1,2 ?0a ?0b qmnsin(m?x/a) sin( n?y/b)
sin(i?x/a) sin(j?y/b) dxdy
? ?0a sin(m?x/a) sin(i?x/a) dx= 0,(m≠ i)
a/2,(m=i)
? ?0b sin( n?y/b)sin(j?y/b) dy= 0,(n≠j)
b/2 (n=j)
? qij=4/ab?0a ?0bq(?,?)sin(i??/a)sin(j??/b)d?d?
? q(x,y)=?m=1,2.?n=1,2qmnsin(m?x/a) sin( n?y/b)
=4/ab?m=1,2.?n=1,2 ?0a ?0bq(?,?) sin(m??/a)
sin(n??/b) d?d? sin(m?x/a) sin( n?y/b)
第二版 2002 弹性力学 第十二章 73
?4D?m=1,2.?n=1,2… (m2/a2+n2/b2)2
Amnsin(m?x/a) sin( n?y/b) =q(x,y)
q(x,y)=?m=1,2.?n=1,2qmnsin(m?x/a) sin( n?y/b)
?4D?m=1,2.?n=1,2… (m2/a2+n2/b2)2
Amnsin(m?x/a) sin( n?y/b)
= ?m=1,2.?n=1,2qmnsin(m?x/a) sin( n?y/b)
比较系数得
Amn =qmn / [?4D(m2/a2+n2/b2)2 ]=
[ 4/(ab)?m=1,2.?n=1,2 ?0a ?0bq(?,?) sin(m??/a)
sin(n??/b) d?d? ] / [?4D(m2/a2+n2/b2)2 ]
第二版 2002 弹性力学 第十二章 74
q(x,y)=q0
Amn=[ 4/(ab)?m=1,2.?n=1,2 ?0a ?0bq(?,?) sin(m??/a) ]
sin(n??/b) d?d? ] / [?4D(m2/a2+n2/b2)2 ]
=q0ab(1-cos m?)(1-cos n?)/(?2mn)
=16q0/ [?6Dmn(m2/a2+n2/b2)2 ]
(m=1,3,5,…;n=1,3,5…)
? w=16q0/ [?6D] ?m=1,3,5.?n=1,3,5 sin(m?x/a) sin(
n?y/b) /[mn(m2/a2+n2/b2)2 ]
第二版 2002 弹性力学 第十二章 75
q(x,y)=q0 a=b
? W(a/2 a/2) = 0.25 [16q0a4/ (?6D)] 1 term
0.2467 [16q0a4/ (?6D)] 2 terms
0.2470 [16q0a4/ (?6D)] 3 terms
0.2469 [16q0a4/ (?6D)] 4 terms----
is exact to three digits.
? Mx(a/2,a/2)= 0.325 [16q0a2/ ?4] 1 term
0.2857[16q0a2/ ?4] 2 terms
0.2936 [16q0a2/ ?4] 3 terms
0.2920 [16q0a2/ ?4] 4 terms----is
exact to three digits.
第二版 2002 弹性力学 第十二章 76
12.7 Levy's Solution by Single
Trigonometric Series.
第二版 2002 弹性力学 第十二章 77
? D?4w(x,y)= q(x,y) (1)
x=0,w(0,y)=0 (?2w/?x2)x=0=0
x=a,w(a,y)=0 (?2w/?x2)x=a=0 (2)
y=0,specified later
y=b,specified later
第二版 2002 弹性力学 第十二章 78
Assume W(x,y)=?m=1,2.Ym(y) sin(m?x/a) (3)
? Boundary conditions (2) are satisfied automatically
? Substituting Eq,(3) into Eq,(1)
D(?4/?x4+2?4/?x2?y2 +?4/?y4)w=q (1)
yields
?m=1,2 [ym(4)-2(m?/a)2Ym(2)+(m?/a)4Ym] sin(m?x/a) =q/D
第二版 2002 弹性力学 第十二章 79
D?m=1,2 [ym(4)-2(m?/a)2Ym(2)+(m?/a)4Ym] sin(m?x/a) =q
q=2/a ?m=1,2 [?0aq(? y)sin(m??/a)d?] sin(m?x/a)
D?m=1,2 [ym(4)-2(m?/a)2Ym(2)+(m?/a)4Ym] sin(m?x/a)
=2/a ?m=1,2 [?0aq(? y)sin(m??/a)d?] sin(m?x/a)
ym(4)-2(m?/a)2Ym(2)+(m?/a)4Ym
=2/(Da) [?0aq(? y)sin(m??/a)d?]
第二版 2002 弹性力学 第十二章 80
ym(4)-2(m?/a)2Ym(2)+(m?/a)4Ym
=2/(Da) [?0aq(? y)sin(m??/a)d?]
Ym=Amch(m?y/a)+Bm (m?y/a) sh(m?y/a)+
Cmsh(m?y/a)+Dm (m?y/a) ch(m?y/a)+fm(y)
where fm(y) is a particular solution.
W= ?m[Amch(m?y/a)+Bm (m?y/a) sh(m?y/a)+
Cmsh(m?y/a)+Dm (m?y/a) ch(m?y/a)+fm(y)] sin(m?x/a)
where Am BmCm Dm are arbitrary constants and can be
determined by the boundary conditions on the edges
y=?b/2
第二版 2002 弹性力学 第十二章 81
Rectangular plate without free edges--only with
simply supported edges or clamped edges
? ?4 [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=0
? Dw is independent of ?
第二版 2002 弹性力学 第十二章 82
? Mx= - [?2(Dw)/?x2+??2(Dw/?y2)]
My= - [(?2(Dw/?y2+??2(Dw/?x2)]
? ?=0 Table
Mx= - [?2(Dw)/?x2]
My= - [(?2(Dw/?y2]
? ?≠ 0
Mx’= - [?2(Dw)/?x2+??2(Dw/?y2)]
= Mx +? My
My’= - [(?2(Dw/?y2+??2(Dw/?x2)]
= My +? Mx
第二版 2002 弹性力学 第十二章 83
12.8 Bending of Circle Plate
? Cylindrical coordinates (r,?,z)
第二版 2002 弹性力学 第十二章 84
? D?4w=D[?/(r?r)+?2/(r2??2)+?2/?r2 ]2 w =q(r,?)
? Mr=-D{?2w/?r2 +?[?w/(r?r)+?2?/(r2??2)] }
?r=zMr /I I= t3/12
? M?==-D{?w/(r?r)+?2w/(r2??2) +??2w/?r2 }
??= zM? /I
? Mr? = M?r =-D(1-?)[?2w/(r?r??)-?w/(r2??)]
?r?= zMr? /I
? Qr=-D ?(?2w )/?r
?rz= 6Qr(t2/4-z2)/t3
? Q? =-D ?(?2w )/(r??)
??z= 6Q?(t2/4-z2)/t3
第二版 2002 弹性力学 第十二章 85
Boundary conditions
? a clamped edge r=a:
w(a,?)=0,?w/?r)r=a=0
? a simply supported edge r=a
w(a,?)=0,
Mr )r=a =-D{?2w/?r2 +?[?w/(r?r)+?2?/(r2??2)] } )r=a=0
? a free edge r=a
Mr )r=a =0 Vr )r=a =[Qr+ ?Mr ? /(r??)]r=a =0
第二版 2002 弹性力学 第十二章 86
12.9 Axisymmetrical Bending of Circle Plates
? If q(r,?) is q(r) on a circle plate,
w(r,?) will be w(r) and
D?4w =D[?/(r?r)+?2/(r2??2)+?2/?r2 ]2 w =q(r,?)
will be
D[?/(r?r)+?2/?r2 ]2 w (r)=q(r)
The solution of this ordinary differential equation is
w(r)=C1 lnr+C2 r2 lnr+C3 r2 +C4+w1
where w1 is any particular solution and C1 C2 C3 C4
are arbitrary constants.
第二版 2002 弹性力学 第十二章 87
? Mr=-D{?2w/?r2 +?[?w/(r?r)+?2?/(r2??2)] }
=-D[d2w/dr2 +?dw/(rdr)]
? M?=-D{?w/(r?r)+?2w/(r2??2) +??2w/?r2 }
=-D{dw/(rdr)+?d2w/dr2 }
? Mr? = M?r =-D(1-?)[?2w/(r?r??)-?w/(r2??)] =0
? Qr=-D ?(?2w )/?r=-Dd[d2w/dr2 +dw/(rdr)]/dr
? Q? =-D ?(?2w )/(r??)=0
第二版 2002 弹性力学 第十二章 88
w(r)=C1 lnr+C2 r2 lnr+C3 r2 +C4+w1
dw/dr= C1 /r+ C2(2r lnr+r)+2 C3r+dw1/dr
dw2/dr2=- C1 /r2+ C2(2 lnr+3)+2 C3+d2w1/dr2
? Mr =-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 2002 弹性力学 第十二章 89
A,annular plate a<r<b----a circular plate with a
central hole
? There will be two Bcs at r=a,two Bcs at r=b,
so the four arbitrary constants can be
determined.
第二版 2002 弹性力学 第十二章 90
B,solid plate r?a (no centre hole)
? Qr=-D{ C2(4/r)- P
dw1/(r2dr)+d2w1/(rdr2)+d3w1/dr3}
? Two Bcs at r=a Qr
? r=0,1,since w is finite,C1 =0
2(1).No concentrated force at r=0,Since Qr is
finite,C2 =0.
2(2),Concentrated force P at r=0,
?Fz=0,2?rQr+P=0 2?r(-4DC2/r) +P=0
C2=P/(8D?)
第二版 2002 弹性力学 第十二章 91
Example 1,a circular clamped plate without
hole is subjected to a uniform load q0
w(r)=C1 lnr+C2 r2 lnr+C3 r2 +C4+w1
? particular solution:
D[?/(r?r)+?2/?r2 ]2 w (r)=q0
assume,w1 =mr4 w1 =q0r4 /(64D)
? r=0 C1 =C2 =0
? r=a w=0 dw/dr=0
C3=-20q0a2 /(32D) C4=q0a4 /(64D)
? w(r)= q0 (r2 -a2 )2 /(64D)
第二版 2002 弹性力学 第十二章 92
Exa.2,a circular clamped plate without hole
subjected to a concentrated load P at the center
r=0
w(r)=C1 lnr+C2 r2 lnr+C3 r2 +C4+w1
? particular solution,w1 =0
? r=0 1,since w is finite,C1 =0 2,C2=P/(8D?)
? r=a w=0 dw/dr=0
C3=-P(2lna+1)/(16D?) C4=Pa2 /(16D?)
? w(r)=P(2r2 ln(r/a)-r2+a2 )/(16D?)
第二版 2002 弹性力学 第十二章 93
习题作业（英文书）
? 12.12.4
? 12.12.6
? 13.6.1
? 下星期的今天交作业
第二版 2002 弹性力学 第十二章 94
Chapter 13 Bending of Thin Plates--Finite-
Difference Method
Sec 7.1 Finite-difference formulas
? f(x+dx,y0 ) ≈ f(x,y0)+?f(x,y0)/?x dx+0.5
?2f(x,y0)/?x2 dx2
? x=x0 dx=h x+dx=x1 f(x1,y0 )=f1 f(x0,y0 )=f0
f1= f0 +(?f/?x)0 h+0.5 (?2f/?x2)0 h2 (1)
? x=x0 dx=-h x+dx=x3 f(x3,y0 )=f3 f(x0,y0 )=f0
f3= f0-(?f/?x)0 h+0.5 (?2f/?x2)0 h2 (2)
?
第二版 2002 弹性力学 第十二章 95
Sec 7.1 Finite-difference formulas
? f(x+dx,y0 ) ≈ f(x,y0)+?f(x,y0)/?x dx+0.5
?2f(x,y0)/?x2 dx2
? x=x0 dx=h x+dx=x1 f(x1,y0 )=f1 f(x0,y0 )=f0
f1= f0 +(?f/?x)0 h+0.5 (?2f/?x2)0 h2 (1)
? x=x0 dx=-h x+dx=x3 f(x3,y0 )=f3 f(x0,y0 )=f0
f3= f0(?f/?x)0 h+0.5 (?2f/?x2)0 h2 (2)
? (?f/?x)0 =(f1 -f3 )/(2h)
(?2f/?x2)0 =(f1 +f3 -2f0 )/h2
第二版 2002 弹性力学 第十二章 96
? x
12
8 4 5
11 3 0 1 9
7 2 6
10
y
第二版 2002 弹性力学 第十二章 97
? Fundamental finite-differences formulas
(?f/?x)0 =(f1 -f3 )/(2h)
(?2f/?x2)0 =(f1 +f3 -2f0 )/h2
similiarly,(?f/?y)0 =(f2 -f4 )/(2h)
(?2f/?y2)0 =(f2 +f4 -2f0 )/h2
? (?2f/?x?y)0 =[?(?f/?y) / ?x]0=[(?f/?y)1-(?f/?y)3 ]/(2h)
=[(f6 -f5 )/2h-(f7 -f8 ) /2h]/2h
= [(f6+f8 )-(f5+ f7)] /(4h2)
(?4f/?x4)0 =[6f0-4(f1+f3 )+(f9+f11)] /h4
(?4f/?y4)0 =[6f0-4(f2+f4 )+(f10+f12)] /h4
第二版 2002 弹性力学 第十二章 98
Sec.13.1 Finite-difference representations
D(?4/?x4+2?4/?x2?y2 +?4/?y4)w=q (1)
D[(?4w/?x4 )0 +2(?4w/?x2?y2)0 +(?4 w/?y4)0=q0 (2)
20w0-8(w1+w2+w3+w4)+2(w5+w6+w7+w8)
+(w9+w10+w11+w12)=h4q0/D (3)
20本点 -8邻点 +2角点 +1远点 = h4q0/D
n个内点建立 n个方程。
第二版 2002 弹性力学 第十二章 99
Example,P294(E)
第二版 2002 弹性力学 第十二章 100
13.2 Plate without free edges
? Let AB represents a fixed edge A x
3--inside point
1--outside point 3 0 1
(?w/?x)0 =(w1 -w3 )/(2h) =0 h h
w1 =w3 B
? Let AB reoresents a simply supported edge
(?2w/?x2)0 =(w1 +w3 -2w0 )/h2=0
w1 =-w3
