Vibrations of Structures Module V: Vibrations of Plates Lesson 37: Dynamics of Plates: Variational Formulation Contents: 1. Introduction. Mathematical Modeling Keywords: Plate model, Variational formulation, Divergence theorem, Boundary conditions
Dynamics of Plates: Formulation The Variational Consider a plate of arbitrary shape in the x-y-plane, as shown in Fig. 1. Let w(x, y, t) represent the transverse displacement field along the z-axis B y ˆn = (cos α, sin α) T α A ds s x Figure 1: Representation of unit tangent and normal vectors at a domain boundary direction. The kinetic energy of a plate element can be represented by T = 1 h/ ρw,t + ρz (w,xt + w,yt) ] dz dx dy h/ = 1 ρhw,t + I(w,xt + w,yt)] dx dy, (1) where I = ρh 3 /1 is the moment of inertia per unit area of the plate. The potential energy of the plate is given by the strain energy stored in the plate when it undergoes deformation. From linear theory of elasticity, the strain
energy in this case can be written as Using the expressions and σ xx = V = 1 h/ (σ xx ɛ xx + σ yy ɛ yy + σ xy ɛ xy ) dz dy dx. h/ E 1 ν ɛ xx + νɛ yy ], σ yy = E 1 ν νɛ xx + ɛ yy ], and σ xy = E 1 + ν ɛ xy ɛ xx = zw,xx (x, y, t), ɛ yy = zw,yy (x, y, t), ɛ xy = zw,xy (x, y, t), one obtains V = D A (w,xx + w,yy ) + (1 ν)(w,xy w,xx w,yy )] dx dy, () where D = Eh 3 /1(1 ν )]. From Hamilton s principle or δ (T V) dt = 0, δt dt δv 1 dt δv dt = 0, (3) where the action integral is partitioned for convenience using the definitions V 1 = D ( w) dx dy, and V = D(1 ν) (w,xy w,xx w,yy ) dx dy. Consider the first term in (3). Using the standard arguments of the vari- 3
ational formulation of dynamics, we can rewrite the first term as δt dt = = = = ρhw,t δw,t + I(w,xt δw,xt + w,yt δw,yt )] dx dy dt t 1 ρhw,t δw + I(w,xt δw,x + w,yt δw,y )] t t1 dx dy + ρhw,tt δw I(w,xtt δw,x + w,ytt δw,y )] dx dy dt t 1 t ρhw,tt δw I{(w,xtt δw),x w,xxtt δw + (w,ytt δw),y w,yytt δw}] dx dy. Now, applying the Gauss divergence theorem to the second term, we have δt = where = t 1 t A ρhw,tt δw I ( δw w,tt ) + I( w,tt ) δw] dx dy dt. v(x, y) dx dy = I( w,tt ) ˆn δw ds + B v ˆn ds (4) ] (ρhw,tt + I w) δw dx dy dt ] dt, (5) Iw,ntt δw ds + (ρhw,tt + I w) δw dx dy w,ntt := w,tt ˆn = w,xtt cos α + w,ytt sin α. 4
The integrand of the second term in (3) can be expressed as δv 1 = D w ( δw) dx dy = D ( w δw) ( w) ( δw)] dx dy = D ( w δw) (δw w) + w δw] dx dy = D ( w δw) ˆn (δw w) ˆn] ds + D 4 w δw dx dy (using (4)) = D w δw,n w,n δw] ds + D 4 w δw dx dy. (6) The integrand of the third term in (3) can be written as δv = D(1 ν) = D(1 ν) w,xy δw,xy w,xx δw,yy w,yy δw,xx ] dx dy (w,xy δw,y ),x w,xxy δw,y + (w,xy δw,x ),y w,xyy δw,x (w,xx δw,y ),y + w,xxy δw,y (w,yy δw,x ),x + w,xyy δw,x ] dx dy = D(1 ν) F dx dy, (7) where F = (w,xy δw,y w,yy δw,x ), (w,xy δw,x w,xx δw,y )] T Using the divergence theorem (4) in (7), we get δv = D(1 ν) = D(1 ν) F ˆn ds (cos α w,xy sin α w,xx ) δw,y + (sin α w,xy cos α w,yy ) δw,x ] ds. (8) 5
Using the definitions x = cos α n sin α s, and y = sin α n + cos α s. Using the above transformations one can rewrite (8) as δv = D(1 ν) (cos α w,xy sin α w,xx ) sin α + (sin α w,xy cos α w,yy ) cos α δw,n ds + (cos α w,xy sin α w,xx ) cos α ] (sin α w,xy cos α w,yy ) sin α] δw,s ] ds. (9) Integrating by parts the second contour integral, and using the condition that the boundary term is zero over a closed contour, we have δv = D(1 ν) ( cos α sin α w,xy sin α w,xx cos α w,yy ) δw,n (cos α sin α)w,xy cos α sin α(w,xx w,yy ),s δw ds. (10) Substituting (5), (6) and (10) into (3) yields D w,n + D(1 ν){cos α sin α(w,yy w,xx ) + (cos α sin α)w,xy },s + Iw,ntt ] δw ds + D w + (1 ν)( cos α sin α w,xy cos α w,yy sin α w,xx )] δw,n ds + ρhw,tt + I w,tt D 4 w ] δw dx dy] dt = 0. The equation of motion is obtained from the last integral as ρhw,tt I w,tt + D 4 w = 0. 6
y ˆt = (0, 1) T b ˆn = (1, 0) T a x Figure : Unit boundary normal and tangent vectors for a rectangular plate The boundary conditions are obtained as w + (1 ν)(n x n y w,xy n x w,yy n y w,xx) ] = 0 or w,n = 0, (x, y) B, (11) and Iw,ntt D w,n + (1 ν){n x n y (w,yy w,xx ) + (n x n y)w,xy },s ] = 0 or w = 0, (x, y) B. (1) Consider a rectangular Kirchhoff plate (setting I = 0), as shown in Fig.. The boundary conditions (11) and (1) at x = a yield the following possible simple cases. 7
ˆt = ( sin φ, cos φ) T ˆn = (cos φ, sin φ) T φ Figure 3: Unit boundary normal and tangent vectors for a circular plate Simply-supported edge: D w (1 ν)w,yy ] x=a = 0 and w x=a = 0. Clamped edge: w x=a = 0 and w,x x=a = 0. Free edge: D w (1 ν)w,yy ] x=a = 0 and D( w),x + (1 ν)w,xyy ] x=a = 0. Using the transformations x = cos φ r sin φ r φ, and y = sin φ r + cos φ r φ the Lagrangian for a circular Kirchhoff plate may be written as L = 1 ρhw,t D{ (w,rr + w,r /r + w,φφ /r ) A +(1 ν) ( w,rr (w,r /r + w,φφ /r ) (w,φ /r w,rφ /r) )}] r dφ dr. 8
The boundary conditions (11) and (1) can be expressed in the polar coordinates at the periphery of a circular plate (see Fig. 3) for three simple cases as follows. Simply-supported edge: w (1 ν) 1 ( w,r + 1 )] r r w,φφ Clamped edge: = 0 and w = 0. w = 0 and w,r = 0. Free edge: and w (1 ν) 1 r ( w),r + (1 ν) 1 r ( w,r + 1 )] r w,φφ ] ( ) 1 r w,φφ,r = 0 = 0. 9