Development of Equation of Equilibrium and Boundary Conditions Using Variational Approach

3 Development of Equation of Equilibrium and Boundary Condi tions Using Variational Approach 3.1 3.1.1 Bending Theory of Plates Total Potential Energy The total potential energy of the system Π reads: Π = Ub − Vb (68) where Ub is the bending energy stored in the plate, and Vb is the work of external forces. Bending Energy 1 Ub = 2 =− Z 1 2 Mαβ καβ dS SZ S (69) Mαβ w,αβ dS where the geometrical relation καβ = −w,αβ has been used. Work of External Forces Plate Loading Lateral load: q (x) = q (xα ) (70) This is distributed load measured in [N/m2 ] or [lb/in2 ] force per unit area of the middle surface of the plate. q ( x) The distributed load contains concentrated load P as a special case: P (x0 , y0 ) = P0 δ (x − x0 ) δ (y − y0 ) (71) where δ is the Dirac delta function, [x0, y0 ] is the coordinate of the application of the concentrated force, and P0 is the load intensity. Dirac δ -function x0 x NOTE The shearing loads on the lateral surface of ice are normally not consid ered in the theory of thin plates. Load Classification • Load applied at the horizontal surfaces. transverse load • Load applied at the lateral surfaces. edge force edge moment Loads are assumed to be applied to the middle plane of the plate NOTE Other type of loading such as shear or in-plane tension or compression do not deflect laterally the plate and therefore are not considered in the bending theory. in-plane tension or compression in-plane shear Potential Energy due to Lateral Load q work on transverse deflection: Z q w dS Lateral (transverse) load does (72) S This is also called a work of external forces. Potential Energy due to Edge Moment The conjugate kinematic variable associated with the edge moment is the edge rotation dw/dxn . dl t n edge moment Γ We apply only the normal bending moment Mnn : Z ¯ nn dw dl − M dxn Γ (73) where the minus sign is included because positive bending moment results in a negative rotation and negative moment produces positive rotation. Mnn < 0 0 dw >0 dxn w At the edge, Mtt = 0 and Mtn = 0. Mnn > 0 a dw <0 dxn xn Potential Energy due to Edge Forces Z V̄n w dl (74) Γ Q2 Q1 x2 x1 Vn Potential Energy due to All External Forces Now, the work of external forces reads: Z Z Z dw q w dS − M̄nn dl + V̄n w dl (75) Vb = dxn S Γ Γ 3.1.2 First Variation of the Total Potential Energy The total potential energy reads: Z 1 Π=− Mαβ w,αβ dS 2 S Z Z Z dw ¯ − q w dS + Mnn dl − V¯n w dl dxn S Γ Γ First variation of the total potential energy δΠ is expressed: Z δΠ = − Mαβ δw,αβ dS µ ¶ Z ZS Z dw ¯ − q δw dS + Mnn δ dl − V¯n δw dl dxn S Γ Γ (76) (77) We shall transform now the first integral with the help of the Gauss theorem. First note that from the rule of the product differentiation: Mαβ δw,αβ = (Mαβ δw,α ) ,β −Mαβ ,β δw,α then Z Mαβ δw,αβ dS = S Z S (Mαβ δw,α ) ,β dS − Z Mαβ ,β δw,α dS (78) (79) S Now, the first integral on the right hand side of the above equation transforms to the line integral: Z Z Z Mαβ δw,αβ dS = Mαβ δw,α nβ dl − Mαβ ,β δw,α dS (80) S Γ S The integrand of the second integral on the right hand side of the above equation transform to: (81) Mαβ ,β δw,α = (Mαβ ,β δw) ,α −Mαβ ,αβ δw which results in: Z Z Mαβ δw,αβ dS = Mαβ δw,α nβ dl S ΓZ Z − (Mαβ ,β δw) ,α dS + Mαβ ,αβ δw dS S (82) S upon which the application of the Gauss rule gives: Z Z Mαβ δw,αβ dS = Mαβ δw,α nβ dl S ΓZ Z − Mαβ ,β δw nα dl + Mαβ ,αβ δw dS (83) S Γ We can return now to the expression for δΠ and substitute there the transformed first integral: Z δΠ = (−Mαβ ,αβ −q) δw dS (84) SZ Z + Mαβ ,β δw nα dl − V¯n δw dl ZΓ ZΓ − Mαβ δw,α nβ dl + M̄nn δw,n dl ³ Γ Γ ´ dw where δw,n = δ dx . It is seen that integrals involving the prescribed forces M̄nn n and V̄n are written in a local coordinate system xγ {xn , xt } while the remaining two integrals over the contour Γ are written in the global coordinate system xα . In order to make comparison, we have to decide on one coordinate system. We choose the local system. Consider the first integral: Z (Mαβ ,β nα ) δw dl (85) Γ The term in the parenthesis is a scalar quantity and thus remain unchanged with respect to the rotation of coordinate system. In the local system xγ , the line integral becomes: Z (Mγδ ,δ nγ ) δw dl (86) Γ where γ = 1 is the normal direction n, and γ = 2 is the tangential direction t. The coordinates of the unit normal vector is the local system are nγ {1, 0}. Hence, Z Z (Mγδ ,δ nγ ) δw dl = (M1δ ,δ n1 + M2δ ,δ n2 ) δw dl (87) Γ Γ Z = M1δ ,δ δw dl Γ Furthermore, the integrand reads: M1δ ,δ = M11 ,1 +M12 ,2 ∂Mnn ∂Mnt ∂M11 ∂M12 = + = + ∂x1 ∂x2 ∂xn ∂xt (88) and we call it the shear force in the normal direction n and denote: Qn ≡ Mnδ ,δ δ = {1, 0} or {n, t} (89) Now, we can combine two line integrals in the equation of first variation of the total potential energy: Z ¡ ¢ Qn − V̄n δw dl (90) Γ How the remaining integral is transformed? Z Z (Mαβ nβ ) δw,α dl = (Mγδ nδ ) δw,γ dl Γ (91) Γ Because it is a scalar quantity, we simply switch indices from global system (α and β) to local (γ and δ). As before nδ {1, 0} so after summing with respect to δ, we have: Z Z (Mγ1 n1 + Mγ2 n2 ) δw,γ dl = Mγ1 δw,γ dl (92) Γ ZΓ = Mγn δw,γ dl ZΓ = (Mnn δw,n +Mtn δw,t ) dl Γ The first term can be absorbed with the line integral representing potential energy of bending moment: Z ¡ ¢ ¯ nn δw,n dl Mnn − M − (93) Γ There remains though one integral which does not fit to anything. Since the boundary term must be equilibrated, it is suspected that this term might belong to the shearing force term, at least partially: Z Mtn δw,t dl transverse term (94) Γ In order to compare this term with the shearing force term, we have to make this term comparable as far as the kinematic quantity describing variation is concerned. One integral involves δw and the other one δw,t . Note that ∂w,t = ∂ (δw) /∂xt is the derivative of the function δw in the tangential direction, i.e. direction along the curve Γ. This means that we can integrate by parts along Γ. Thus, Mtn δw,t = (Mtn δw) ,t −Mtn ,t δw (95) Z Mtn δw,t dl = Γ Z Γ (Mtn δw) ,t dl − Z Mtn ,t δw dl (96) Γ The first term in the right hand is equal to the value of the integrand calculated at the beginning and end of the integration path: Z (Mtn δw) ,t dl = Mtn δw|end (97) beginning Γ Consider now two cases. • The contour Γ is a smooth closed curve, so the value at the beginning is equal to the value at the end: Mtn δw|end − Mtn δw|beginning = 0 (98) The term does not give any contribution. direction of integration Γ end begining • The contour Γ is piece-wise linear or composed of a finite number, k, of smooth curves with discontinuity. Therefore, the integration should be made in a piece-wise manner. Thus, the continuation of the beginning and end of each should be added: X Mtn δw|end (99) beginning k 3.1.3 Equilibrium Equation and Boundary Conditions Now, we can write the final expression for the first variation of δΠ: Z δΠ = (−Mαβ ,αβ −q) δw dS SZ Z ¡ ¢ ¡ ¢ ¯ nn δw,n dl + Qn − V̄n δw dl − Mnn − M Γ ÃΓ ! Z X end − Mtn δw|beginning − Mtn ,t δw dl = Γ k Z (100) (−Mαβ ,αβ −q) δw dS Z ¡ ¢ ¡ ¢ ¯ nn δw,n dl + Vn − V̄n δw dl − Mnn − M Γ Γ X end − Mtn δw|beginning SZ k where Vn = Qn + Mtn ,t is the effective shear force. In order to make the functional Π stationary under arbitrary variation of the displacement field δw, there must hold: Equation of Equilibrium Mαβ ,αβ +q = 0 on S (101) Boundary Conditions 3.1.4 ¯ nn = 0 Mnn − M Vn − V̄n = 0 or or δw,n = 0 δw = 0 Mnt = 0 or δw = 0 on Γ on Γ at corner points of the contour Γ Specification of Equation for Rectangular Plate Consider a rectangular plate. (102) a x→t b y→n Boundary Conditions For edges parallel to x-axis, the normal direction is the ydirection. Myy − M̄yy = 0 or Vy − V̄y = 0 or ∂w =0 ∂y w=0 (103) where ∂Myx ∂y ∂Mxy Vy = Qy + ∂x Vx = Qx + (104) For edges parallel to y-axis, the normal direction is the x-direction. Mxx − M̄xx = 0 or Vx − V̄x = 0 or ∂w =0 ∂x w=0 (105) where ∂Mxy ∂x ∂Myx Vx = Qx + ∂y Vy = Qy + Interpretation of Corner Points (106) 1 segment [ k − 1] 2 segment [ k ] direction of integration 3 4 Boundary condition reads: X [1] [2] [4] [3] [2] Mtn δw|end − Mtn δw[1] + Mtn δw[4] − Mtn δw[3] beginning = Mtn δw (107) where thus X δw[3] = δw[2] (108) ³ ´ [1] [3] [2] [4] [1] Mtn δw|end = −M δw + M − M δw[2] + Mtn δw[4] tn tn tn beginning (109) Consider the right angle. x [ k − 1] y [k] for the k − 1 side ³ n = x, t = y for the k side ´ Mtn δw|end beginning (110) n = y, t = x at the right angle = (Mxy − Myx ) δw (111) Interpretation of Corner Forces Plane stress: τ xy = τ yx symmetry (112) σy surface element σx dS τ xy τ yx Let us place the surface element at the corner. x y z τ xy Edge [ k −1] τ yx Edge [k] M xy M yx The shearing stresses produce twisting moments which are in the opposite direction: [k−1] [k] Mxy = −Myx Therefore, the boundary condition at the corner becomes: ³ ´ [k−1] [k] Mtn δw|end = M − M δw = 2 Mxy δw = 0 xy yx beginning Fcorner = 2 Mxy (113) (114) (115) For the Entire Plate 2 M xy 2 Mxy 2 Mxy 2 M xy Interpretation of the Effective Shear Vx x y dy z Qx dy M xy M xy + Equilibrium reads: ∂M xy ∂y µ ¶ ∂Mxy dy − Mxy Qx dy + Mxy + ∂y µ ¶ ∂Mxy = Qx + dy ∂y = Vx dy Vx = Qx + ∂Mxy ∂y dy (116) (117) 3.2 3.2.1 Bending-Membrane Theory of Plates Total Potential Energy The total potential energy of the system Π reads: Π = Ub + Um − Vb − Vm (118) where Ub is the bending strain energy, Um is the membrane strain energy, Vb is the potential energy of external loading causing flexural response, and Vm is the potential energy of external loading causing membrane response. Membrane Strain Energy The membrane strain energy reads: Z 1 Um = Nαβ ε◦αβ dS 2 S (119) where 1 1 (uα ,β +uβ ,α ) + w,α w,β 2 2 Potential Energy of External Forces Evaluation of Boundary Terms ε◦αβ = Normal in-plane loading, Nnn Z N̄nn un dl (120) (121) Γ where un is normal in-plane displacement. Shear in-plane loading, Ntt Z N̄tn ut dl (122) Γ where ut is shear component of the displacement vector. Potential Energy of External Forces Z Z Vm = N̄nn un dl + N̄tn ut dl Γ 3.2.2 (123) Γ First Variation of the Total Potential Energy The first variation of the total potential energy reads: δΠ = (δUb − δVb ) + (δUm − δVm ) (124) The first parenthesis represent the terms considered already in the bending theory of plates. All we have to do is to evaluate the term in the second parenthesis. Here, the first variation of the membrane energy reads: Z δUm = Nαβ δλαβ dS (125) S where 1 1 (δuα ,β +δuβ ,α ) + (δw,α w,β +δw,β w,α ) 2 2 Because of the symmetry of the tensor of membrane forces: δλαβ = Nαβ = Nβα (126) (127) by using the characteristics of dummy indices we obtain: Nαβ δuβ ,α = Nβα δuβ ,α = Nαβ δuα ,β (128) Now, the first variation of the membrane strain energy reads: ∙ ¸À Z ¿ 1 1 δUm = Nαβ (δuα ,β +δuβ ,α ) + (δw,α w,β +δw,β w,α ) dS (129) 2 2 ZS = (Nαβ δuα ,β +Nαβ w,β δw,α ) dS S Note that the displacement vector has now three components: {uα , w} (130) so that there are three independent variations: {δuα , δw} (131) We expect those to end up with three independent equations of equilibrium. The first term of δUm reads: Z Z Z Nαβ δuα ,β dS = (Nαβ δuα ) ,β dS − Nαβ ,β δuα dS (132) S S S Z Z = Nαβ δuα nβ dl − Nαβ ,β δuα dS ZΓ ZS = Nγδ δuγ nδ dl − Nαβ ,β δuα dS ZΓ Z S = Nγ1 δuγ dl − Nαβ ,β δuα dS ZΓ ZS = Nγn δuγ dl − Nαβ ,β δuα dS S ZΓ Z = (Nnn δun + Ntn δut ) dl − Nαβ ,β δuα dS Γ S The second term of δUm reads: Z Z Z Nαβ w,β δw,α dS = (Nαβ w,β δw) ,α dS − (Nαβ w,β ) ,α δw dS (133) S ZS Z S = Nαβ w,β δw nα dl − (Nαβ w,β ) ,α δw dS ZΓ ZS = Nγδ w,δ δw nγ dl − (Nαβ w,β ) ,α δw dS ZΓ Z S = N1δ w,δ δw dl − (Nαβ w,β ) ,α δw dS S ZΓ Z = (N11 w,1 +N12 w,2 ) δw dl − (Nαβ w,β ) ,α δw dS ZΓ ZS = (Nnn w,n +Nnt w,t ) δw dl − (Nαβ w,β ) ,α δw dS S Γ Now, the variation of external work reads: Z Z N̄nn δun dl + N̄tn δut dl δVm = Γ 3.2.3 (134) Γ Equilibrium Equation and Boundary Conditions The contribution of the term (δUm − δVm ) then becomes: Z Z δ (Um − Vm ) = (Nnn δun + Ntn δut ) dl − Nαβ ,β δuα dS (135) Γ S Z Z + (Nnn w,n +Nnt w,t ) δw dl − (Nαβ w,β ) ,α δw dS S ZΓ Z ¯nn δun dl − N ¯tn δut dl − N Γ Γ Z Z Z ¡ ¢ ¡ ¢ ¯ ¯tn δut dl = − Nαβ ,β δuα dS + Nnn − Nnn δun dl + Ntn − N Γ Γ ZS Z − (Nαβ w,β ) ,α δw dS + (Nnn w,n +Nnt w,t ) δw dl S Γ The first three integrals involve independent variations of uα , i.e. δuα or {δun , δut }. This gives us two independent equations of equilibrium in the plane of the plate: Equation of Equilibrium I Nαβ ,β = 0 on S (136) and two additional boundary conditions: Boundary Conditions I ¯nn = 0 Nnn − N ¯tn = 0 Ntn − N or or δun = 0 δut = 0 on Γ on Γ (137) The remaining two integrals involve variation in the out-of-plane displacement δw and thus should be combined with the equation of equilibrium and boundary conditions governing the flexural response. The terms involving surface integral should be added to the equation of equilibrium: Equation of Equilibrium II Mαβ ,αβ + (Nαβ w,β ) ,α +q = 0 (138) on S where the second term in the left hand is the new term arising from the finite rotation. The term with the line integral should be added to the corresponding term involving variation δw: Z ¡ ¢ Vn + Nnn w,n +Nnt w,t −V̄n δw dl = 0 (139) Γ The generalized boundary conditions reads: Boundary Conditions II- (A) Vn + Nnn w,n +Nnt w,t −V̄n = 0 or δw = 0 on Γ (140) where the second and third terms in the left hand side of the first equation are the new terms arising from the finite rotation. If the boundaries of the plate are kept undeformed w,t = 0 (simply supported or clamped plate), then the boundary condition is satisfied: Vn + Nnn w,n −V̄n = 0 or δw = 0 on Γ (141) Physically, the additional terms represent the contribution of the axial force to the vertical equilibrium. Using the in-plane equilibrium, Nαβ ,β = 0, the out-of-plane equilibrium can be transformed to the form: Mαβ ,αβ +Nαβ ,α w,β +Nαβ w,αβ +q = 0 on S (142) Equation of Equilibrium II’ Mαβ ,αβ +Nαβ w,αβ +q = 0 on S (143) which is called as the von Karman equation. Note that Nαβ is related through the Hook’s law with the gradient of the in-plane displacement uα , i.e. Nαβ = Nαβ (uα ). Therefore, the new term Nαβ w,αβ represents in fact coupling between in-plane and out-out-plane deformation. To make derivation complete, the final boundary conditions which do not changed from the bending theory of plate are presented: Boundary Conditions II- (B) ¯ nn = 0 Mnn − M or δw,n = 0 Mnt = 0 or δw = 0 on Γ at corner points of the contour Γ (144)