S ECTION A3 T HIN P LATE T HEORY
7.2 Plates subjected to bending and twisting
In general, the bending moments applied to the plate will not be in planes perpendicular to its edges. Such bending moments, however, may be resolved in the normal manner into tangential and perpendicular components, as shown in Fig. 7.4. The perpendicular components are seen to be Mxand Myas before, while the tangential components Mxy and Myx(again these are moments per unit length) produce twisting of the plate about axes parallel to the x and y axes. The system of suffixes and the sign convention for these twisting moments must be clearly understood to avoid confusion. Mxyis a twisting moment intensity in a vertical x plane parallel to the y axis, while Myx is a twisting moment intensity in a vertical y plane parallel to the x axis. Note that the first suffix gives the direction of the axis of the twisting moment. We also define positive twisting moments as being clockwise when viewed along their axes in directions parallel to the positive directions of the corresponding x or y axis. In Fig. 7.4, therefore, all moment intensities are positive.
7.2 Plates subjected to bending and twisting 223
Fig. 7.4Plate subjected to bending and twisting.
Fig. 7.5(a) Plate subjected to bending and twisting; (b) tangential and normal moments on an arbitrary plane.
Since the twisting moments are tangential moments or torques they are resisted by a system of horizontal shear stressesτxy, as shown in Fig. 7.6. From a consideration of complementary shear stresses (see Fig. 7.6) Mxy= −Myx, so that we may represent a general moment application to the plate in terms of Mx, My and Mxy as shown in Fig. 7.5(a). These moments produce tangential and normal moments, Mt and Mn, on an arbitrarily chosen diagonal plane FD. We may express these moment intensities (in an analogous fashion to the complex stress systems of Section 1.6) in terms of Mx, My and Mxy. Thus, for equilibrium of the triangular element ABC of Fig. 7.5(b) in a plane perpendicular to AC
MnAC=MxAB cosα+MyBC sinα−MxyAB sinα−MxyBC cosα giving
Mn=Mxcos2α+Mysin2α−Mxysin 2α (7.10) Similarly for equilibrium in a plane parallel to CA
MtAC=MxAB sinα−MyBC cosα+MxyAB cosα−MxyBC sinα or
Mt = (Mx−My)
2 sin 2α+Mxycos 2α (7.11)
(compare Eqs (7.10) and (7.11) with Eqs (1.8) and (1.9)). We observe from Eq. (7.11) that there are two values ofα, differing by 90◦and given by
tan 2α= − 2Mxy Mx−My
for which Mt=0, leaving normal moments of intensity Mnon two mutually perpen- dicular planes. These moments are termed principal moments and their corresponding curvatures principal curvatures. For a plate subjected to pure bending and twisting in which Mx, My and Mxyare invariable throughout the plate, the principal moments are the algebraically greatest and least moments in the plate. It follows that there are no shear stresses on these planes and that the corresponding direct stresses, for a given value of z and moment intensity, are the algebraically greatest and least values of direct stress in the plate.
Let us now return to the loaded plate of Fig. 7.5(a). We have established, in Eqs (7.7) and (7.8), the relationships between the bending moment intensities Mx and My and the deflection w of the plate. The next step is to relate the twisting moment Mxyto w.
From the principle of superposition we may consider Mxyacting separately from Mx and My. As stated previously Mxyis resisted by a system of horizontal complementary shear stresses on the vertical faces of sections taken throughout the thickness of the plate parallel to the x and y axes. Consider an element of the plate formed by such sections, as shown in Fig. 7.6. The complementary shear stresses on a lamina of the element a distance z below the neutral plane are, in accordance with the sign convention of Section 1.2,τxy. Therefore, on the face ABCD
Mxyδy= − t/2
−t/2
τxyδyz dz
Fig. 7.6Complementary shear stresses due to twisting momentsMxy.
7.2 Plates subjected to bending and twisting 225
and on the face ADFE
Mxyδx= − t/2
−t/2
τxyδxz dz giving
Mxy= − t/2
−t/2
τxyz dz or in terms of the shear strainγxyand modulus of rigidity G
Mxy= −G t/2
−t/2
γxyz dz (7.12)
Referring to Eqs (1.20), the shear strainγxyis given by γxy= ∂v
∂x + ∂u
∂y
We require, of course, to express γxy in terms of the deflection w of the plate; this may be accomplished as follows. An element taken through the thickness of the plate will suffer rotations equal to∂w/∂x and ∂w/∂y in the xz and yz planes respectively.
Considering the rotation of such an element in the xz plane, as shown in Fig. 7.7, we see that the displacement u in the x direction of a point a distance z below the neutral plane is
u= −∂w
∂xz Similarly, the displacementvin the y direction is
v= −∂w
∂yz
Fig. 7.7Determination of shear strainγxy.
Hence, substituting for u andvin the expression forγxywe have γxy= −2z∂2w
∂x∂y (7.13)
whence from Eq. (7.12)
Mxy=G t/2
−t/2
2z2∂2w
∂x∂ydz or
Mxy= Gt3 6
∂2w
∂x∂y
Replacing G by the expression E/2(1+ν) established in Eq. (1.50) gives Mxy= Et3
12(1+ν)
∂2w
∂x∂y
Multiplying the numerator and denominator of this equation by the factor (1−ν) yields Mxy=D(1−ν)∂2w
∂x∂y (7.14)
Equations (7.7), (7.8) and (7.14) relate the bending and twisting moments to the plate deflection and are analogous to the bending moment-curvature relationship for a simple beam.