R&DE (Engineers), DRDO rd_mech@yahoo.co.in Theory of Plates Ramadas Chennamsetti Introduction R&DE (Engineers), DRDO Plates are initially flat structural elements Ramadas Chennamsetti 2 rd_mech@yahoo.co.in “When a body is bounded by surfaces, flat in geometry, whose lateral dimensions are large compared to the separation between the surfaces is called a PLATE” Introduction Plates are subjected to transverse loads – loads normal to its mid-surface Transverse loads supported by combined bending and shear action Plates may be subjected to in-plane loading also => uniform stress distribution => membrane Membrane action – in-plane loading or pronounced curvature & slope Plate bending – plate’s mid-surface doesn’t experience appreciable stretching or contraction In-plane loads cause stretching and/or contraction of mid-surface Ramadas Chennamsetti 3 rd_mech@yahoo.co.in R&DE (Engineers), DRDO Introduction R&DE (Engineers), DRDO Plate stretching z Nx t Nx x Axial deformation due to transverse load Net deformation = Algebraic sum of uniform stretching and axial deformation due to bending load Ramadas Chennamsetti 4 rd_mech@yahoo.co.in Uniform stretching of the plate => uo q Introduction R&DE (Engineers), DRDO 1 t 1 ≥ ≥ 10 b 2000 For plates Thin & thick plates – t < 20b b = smallest side t > 20b t w ≤ 5 Small deflections – Thin plate theory – Kirchoff’s Classical Plate Theory (KCPT) Thick plate theory – Reissner – Mindlin Plate Theory (MPT) Ramadas Chennamsetti 5 rd_mech@yahoo.co.in Thin plate => Thick plate => KCPT - Assumptions R&DE (Engineers), DRDO Assumptions – Thickness is much smaller than the other physical dimensions vertical deflection w(x, y, z) = w(x, y) Displacements u, v & w are small compared to plate thickness In plane strains are small compared to unity – consider only linear strains Normal stresses in transverse direction are small compared with other stresses – neglected Ramadas Chennamsetti 6 rd_mech@yahoo.co.in Governing equations are derived based on undeformed geometry KCPT - Assumptions Material – linear elastic – Hooke’s law holds good Middle surface remains unstrained during bending – neutral surface Normals to the middle surface before deformation remain normal to the same surface after deformation => doesn’t imply shear across section is zero – transverse shear strain makes a negligible contribution to deflections. Transverse shear strains are negligible Rotary inertia is neglected Ramadas Chennamsetti 7 rd_mech@yahoo.co.in R&DE (Engineers), DRDO Sign convention R&DE (Engineers), DRDO Following sign convention will be followed z z ψz Mz ψy y y Mx x ψx x Positive moments Positive rotations Ramadas Chennamsetti 8 rd_mech@yahoo.co.in My Bending deformations R&DE (Engineers), DRDO z Bending takes place in both planes y D x z ’ BB A x B View ‘D’ Ramadas Chennamsetti 9 rd_mech@yahoo.co.in A’ Rotation => ψy A Bending deformations R&DE (Engineers), DRDO Deformation in ‘x’ direction u ( x , y , z ) = − zψ y (x , y ) v ( x, y , z ) = − zψ x (x , y ) Vertical deformation w = w (x, y) Ramadas Chennamsetti 10 rd_mech@yahoo.co.in Deformation in ‘y’ direction Strains R&DE (Engineers), DRDO => γ xz => γ yz ∂z ∂x ∂w ∂w = −ψ y + = 0 => ψ y = ∂x ∂x ∂v ∂w γ yz = + =0 ∂z ∂y ∂w ∂w = −ψ x + = 0 => ψ x = ∂y ∂y Ramadas Chennamsetti 11 rd_mech@yahoo.co.in Assumption – out of plane shear strain negligible ∂u ∂w γ xz = + =0 Strains R&DE (Engineers), DRDO Non-zero strains ε yy ∂u ∂ w y = = −z = −z 2 ∂x ∂x ∂x 2 ∂v ∂ψ x ∂ w = = −z = −z ∂y ∂y ∂y 2 γ xy 2 ∂u ∂v ∂ w = + = −2 z ∂y ∂x ∂x∂y 2 Ramadas Chennamsetti 12 rd_mech@yahoo.co.in ε xx ∂ψ Stresses R&DE (Engineers), DRDO Thin plate – out of plane shear strains vanish – out of plane shear stresses also vanish τ yz = 0 τxz τyz τyz z y x τxz Out of plane normal stress is also assumed to be zero – logical – thin structure – plane stress conditions Ramadas Chennamsetti 13 rd_mech@yahoo.co.in z τ xz = 0, Stresses R&DE (Engineers), DRDO Non-zero stress components z σxx y σyy σyy x τxy σxx All three stress components, σxx, σyy, τxy – in-plane Ramadas Chennamsetti 14 rd_mech@yahoo.co.in τxy Curvatures R&DE (Engineers), DRDO Curvature – reciprocal of radius of bending Rate of change of slope z x dx w* ∂w dx ∂x x ψy dx slope = rotation = ψ y = - ∂w ∂x ∂ψ 2 w ∂ y curvatue = =− = κ xx 2 ∂x ∂x Ramadas Chennamsetti 15 rd_mech@yahoo.co.in w R − Curvatures R&DE (Engineers), DRDO Similarly bending in yz plane introduces a curvature κ ∂ w = − 2 ∂y 2 yy κ xy ∂ w = −2 ∂x∂y 2 Ramadas Chennamsetti 16 rd_mech@yahoo.co.in Twisting of plate Constitutive law R&DE (Engineers), DRDO Linear elastic isotropic – Hookean material Three stress and strain components ε yy = γ xy = τ xy G E σ yy E = −υ −υ τ xy E σ yy E σ xx E 2 (1 + υ ) εxx 0 σxx 1 −υ 1 σ ε υ 1 0 = − yy yy E γ τ υ 0 0 2 ( 1 ) + xy xy σxx υ εxx 1 0 E σ υ = 1 0 yy ε yy 2 τ 1−υ 0 0 1−υ γ xy17 xy 2 Ramadas Chennamsetti rd_mech@yahoo.co.in ε xx = σ xx Writing all three equations in matrix form Constitutive law R&DE (Engineers), DRDO Express strains in terms of curvatures Ez = − 1−υ 2 1 υ 0 υ 1 0 Variation of stresses across thickness is linear Basis – thin plates – plane section remain plane after bending – variation of axial deflection is linear across thickness – strains also vary linearly Ramadas Chennamsetti 18 rd_mech@yahoo.co.in σ xx σ yy τ xy ∂ 2w ∂x 2 0 ∂ 2w 0 2 1 + υ ∂y ∂ 2w 2 2 ∂x∂y Equilibrium equations R&DE (Engineers), DRDO Equilibrium of an infinitesimal element z y dx σxx t τxy x Forces acting on an infinitesimal element dx dy dz y τxy x dy ττ**yz yz σyy σ*yy dx τ*xz τ*xy τ*xy Ramadas Chennamsetti σ*xx 19 rd_mech@yahoo.co.in dy Equilibrium equations R&DE (Engineers), DRDO Equilibrium in ‘x’ direction −σ xx dydz dydz σ * xx τ * zx τ * yx +τ −τ zx * zx dxdy dxdy +τ −τ yx * yx dzdx dzdx ∂ σ xx dx xx + ∂x ∂ τ zx = τ zx + dz ∂z ∂ τ yx = τ yx + dy ∂y = 0 = σ Ramadas Chennamsetti 20 rd_mech@yahoo.co.in σ * xx Equilibrium equations R&DE (Engineers), DRDO Substitute – the following equation is obtained ∂τ xx ∂τ xy ∂τ xz ∂x + ∂y + ∂z = 0 - (1) ∂τ xy ∂x + ∂τ yy ∂y + ∂τ yz ∂z = 0 - (2) ∂τ xz ∂τ yz ∂τ zz + + = 0 - (3) ∂x ∂y ∂z Ramadas Chennamsetti 21 rd_mech@yahoo.co.in Similarly take equilibrium in ‘y’ and ‘z’ directions Shear stresses R&DE (Engineers), DRDO From equation (1) – shear stress τxz can be computed Use stress deflection/curvature relations ∂2w ∂ 2 w ∂ 2 + υ 2 + ∂ y ∂y ∂x ∂τ xz Ez = 2 ∂z 1−υ Ez ∂ 2 w ∂τ xz =0 − + 1 + υ ∂x∂y ∂z 3 ∂ 3w ∂ 3w ∂ w + (1 − υ ) 3 + υ 2 ∂x∂y ∂x∂y 2 ∂ x ∂τ xz Ez ∂ 2 = ∇ w 2 ∂z 1 − υ ∂x ( ) Integrate this across thickness to get shear stress Ramadas Chennamsetti 22 rd_mech@yahoo.co.in ∂ Ez − ∂x 1 − υ 2 Shear stresses R&DE (Engineers), DRDO Integrate from mid plane to top surface of z plate z = h/2 z ∫τ dτ xz xz Ez = ∫ 2 1 − υ 0 − τ xz => −τ xz ( ) ( 2 E ∂ ∇2w ∂ ∇ w dz = 1 − υ 2 ∂x ∂x ( ) h 2 ∫ z dz z h 2 E ∂ ∇ w z2 = 2 1−υ ∂x 2 z 2 ( ) h ( ) ∂ ∇2w E = 2 1−υ 2 ∂x => τ xz = ) ( E 2 1−υ 2 ( 4 − z ∂ ∇2w 2 h2 z − Parabolic variation ∂x 4 23 ) ) 2 Ramadas Chennamsetti 2 rd_mech@yahoo.co.in h 2 0 Moments R&DE (Engineers), DRDO Moment wrt ‘y’ axis dF xx = σ dM z σ xx dz => dM z σxx dx dy y M y x M y = zdF xx = z σ xx dz ∂ 2w ∂ 2w +υ 2 2 ∂x ∂x ∂ 2w Ez 2 ∂ 2 w =− +υ 2 2 ∂y 2 1 − υ ∂x Ez =− 1−υ 2 y E =− 1−υ 2 y dz +h 2 ∂ 2w ∂ 2w 2 υ + z dz 2 2 ∫ ∂y −h 2 ∂x Eh 3 =− 12 1 − υ 2 ( Ramadas Chennamsetti ) ∂ 2w ∂ 2w +υ 2 2 ∂y ∂x 24 rd_mech@yahoo.co.in Neutral plane xx Moments R&DE (Engineers), DRDO Moment wrt ‘x’ axis Neutral plane ∂2w ∂2w Eh 3 υ Mx =− + 2 2 2 12 (1 − υ ) ∂ x ∂y z dz τxy z M xy dx y dy x Compare with section modulus of beam Eh 3 ∂2w (1 − υ ) =− 2 12(1 − υ ) ∂x∂y Eh 3 D = 12 ( 1 − υ Ramadas Chennamsetti 2 ) 25 rd_mech@yahoo.co.in Twisting moment due to shear stress τxy Shear forces R&DE (Engineers), DRDO Vertical equilibrium of plate * Q x dy + Q y* dx + qdxdy − Q x dy − Q y dx = 0 ∂Q x Q = Qx + dx ∂x ∂Q y * Qy = Qy + dy ∂y q Q* Q *y x Qx Qy y Substitute these and simplify dx dy x ∂Q y ∂Q x + = −q ∂x ∂y Ramadas Chennamsetti 26 rd_mech@yahoo.co.in * x z Shear forces R&DE (Engineers), DRDO Shear forces across thickness can be computed by integrating shear stress across +h 2 thickness Q y = ∫τ yz dz per unit width −h 2 Q Q y y yz ∂ (∇ 2 w ) E = 2 (1 − υ 2 ) ∂y 2 h2 z − 4 +h 2 2 h2 z − dz ∫ 4 −h 2 Eh 3 ∂ (∇ 2 w ) ∂ (∇ 2 w ) = − = −D 2 12 (1 − υ ) ∂y ∂y Ramadas Chennamsetti 27 rd_mech@yahoo.co.in τ ∂ (∇ 2 w ) E = 2 (1 − υ 2 ) ∂y Governing equation R&DE (Engineers), DRDO Similar expression for Qx ∂ (∇ 2 w ) Qx = −D ∂x Substitute Qx and Qy in the following expression – vertical equilibrium ∂Q x ∂Q y + = −q ∂x ∂y ∂ ∂ ∂ 2 2 D w D w − ∇ + − ∇ ∂x ∂y ∂y 2 ∂2 ∂ q 2 2 => ∇ w + ∇ w = ∂y 2 D ∂x 2 ( ) ( ( ) ∂2 ∂2 => 2 + ∂y 2 ∂x ( ( ) = − q ) ) 2 q ∇ w = D Ramadas Chennamsetti 28 rd_mech@yahoo.co.in ∂ ∂x Governing equation R&DE (Engineers), DRDO Governing equation ∂2 ∂2 + 2 2 ∂ x ∂ y 2 q ∇ w = D q 2 2 => ∇ ∇ w = D q 4 => ∇ = D ) ∂ w ∂ w ∂ w q +2 + = 4 2 2 4 ∂x ∂x ∂y ∂y D 4 4 4 Bi-harmonic equation Compare with beam equation Ramadas Chennamsetti 29 rd_mech@yahoo.co.in ( Boundary conditions R&DE (Engineers), DRDO Well posed problem – Governing equations and boundary conditions Three basic boundary conditions Vertical deflection and their derivatives Ramadas Chennamsetti 30 rd_mech@yahoo.co.in Simply supported Clamped and Free edge Simply supported R&DE (Engineers), DRDO Simply supported – for eg beam => vertical deflection = 0 and moment = 0 w( x, y ) = 0 ∂2w ∂2w M y = − D 2 + υ 2 = 0 ∂y ∂x y w = 0 implies second derivative in the direction tangent to this line is zero ∂ 2w = 0 2 ∂x Ramadas Chennamsetti x Simply supported y = const 31 rd_mech@yahoo.co.in Simply supported x = const For edge, x = const Simply supported R&DE (Engineers), DRDO Simply supported condition along edge y = const w=0 w = 0 implies second derivative in the direction tangent to this line is zero ∂ w =0 2 ∂yChennamsetti Ramadas 2 32 rd_mech@yahoo.co.in ∂2w ∂2w M x = − Dυ 2 + 2 ∂y ∂x Clamped R&DE (Engineers), DRDO Deflection and slope in normal directions vanish x = constant w=0 ∂w =0 ∂y y = constant Ramadas Chennamsetti 33 rd_mech@yahoo.co.in w=0 ∂w =0 ∂x Free edges R&DE (Engineers), DRDO ∂2w ∂2w M y = − D 2 + υ 2 = 0 at x = const ∂y ∂x ∂2w ∂2w M x = − Dυ 2 + 2 = 0 at y = const ∂y ∂x Ramadas Chennamsetti 34 rd_mech@yahoo.co.in Free edge – Free from any external loads – Natural boundary conditions Bending moment and Shear force vanish Bending moment Free edges R&DE (Engineers), DRDO Moment Mxy and shear forces Mxy The net force acting on the face Q = − M xy − Mxy ' x dy M xy + Mxy Mxy dy M xy + ∂M xy ∂y ∂M xy ∂y dy ∂M xy => Qx' = − + M xy ∂y ∂M xy ∂y Total shear force, dy dx Ramadas Chennamsetti Vx = Qx + Qx’ 35 rd_mech@yahoo.co.in dy Free edges R&DE (Engineers), DRDO Total shear force, Vx = Qx + Qx' ∂ 2 ∂ ∂ w => Vx = − D ∇ w − D(1 − υ ) 2 ∂x ∂x ∂x ) ∂3w ∂3w => Vx = − D 3 + (2 − υ ) 2 ∂x∂y ∂x 3 3 ∂ w ∂w V y = − D 3 + (2 − υ ) 2 ∂x ∂y ∂y Ramadas Chennamsetti 36 rd_mech@yahoo.co.in ( 2 Free edges R&DE (Engineers), DRDO The forces, Vx and Vy => reduced, or Kirchoff’s or effective shear forces In case of a free edge, ∂3w ∂3w V y = − D 3 + (2 − υ ) 2 = 0 at y = const ∂x ∂y ∂y Ramadas Chennamsetti 37 rd_mech@yahoo.co.in ∂3w ∂3w Vx = − D 3 + (2 − υ ) = 0 at x = const 2 ∂x∂y ∂x Bending of plates R&DE (Engineers), DRDO Governing equation of plate rectangular plate bending ( ) q ∇ ∇ w = D 2 2 w = vertical deflection = w(x, y) Solution to this equation – product of two functions – Assume w = w( x , y ) = F ( x )G ( y ) Ramadas Chennamsetti 38 rd_mech@yahoo.co.in external loading, q = q(x, y) Bending of plates Choice of functions – algebraic, trigonometric, hyperbolic etc or combination of these function Selection of a function – depends on boundary conditions Simply supported edges – trigonometric function – Navier solution Deflection of a plate can be written as sum of infinite trigonometric functions Ramadas Chennamsetti 39 rd_mech@yahoo.co.in R&DE (Engineers), DRDO Bending of plates R&DE (Engineers), DRDO Edges, x = 0 and x = a simply supported Vertical deflection vanish w(x=0, y)=0, w(x=a, y)=0 Possible form of solution ∞ SS x=0 x SS x=a F1, F2,…….F∞ are coefficients Symmetric loading wrt x = a/2 Maximum deflection at x = a/2 Ramadas Chennamsetti 40 rd_mech@yahoo.co.in mπ x F ( x ) = ∑ Fm sin a m y Bending of plates R&DE (Engineers), DRDO Other edges simply supported Vertical deflection vanish SS y=b w(x, y=0)=0, w(x, y=b)=0 Possible form of solution ∞ G1, G2,…….G∞ are coefficients SS x = 0, y=0 x SS x=a SS Selection of functions based on BCs Ramadas Chennamsetti 41 rd_mech@yahoo.co.in nπ y G ( y ) = ∑ Gn sin b n y Bending of plates R&DE (Engineers), DRDO Final solution, mπ x nπ y w( x, y ) = ∑∑ FmGn sin sin a b m n ∞ ∞ ∞ ∞ m n mπ nπ , βn = αm = a b Coefficients, Fm, Gn and wmn computed using Fourier Series Ramadas Chennamsetti 42 rd_mech@yahoo.co.in => w( x, y ) = ∑∑ wmn sin α m x sin β n y Coefficients R&DE (Engineers), DRDO Any periodic function can be expanded into a sine or cosine function using Fourier expansion Function of one variable f (x ) = ∑ m mπ x f m sin a m 'π x mπ x m 'π x sin => f (x )sin = f m sin a a a Integrate the above from limits 0 to a Ramadas Chennamsetti 43 rd_mech@yahoo.co.in ∞ Coefficients R&DE (Engineers), DRDO Integration m'πx mπx m'πx ∫0 f (x)sin a dx = fm ∫0 sin a sin a dx a fm f m πx mπx m πx πx ' => ∫ 2 sin sin dx = ∫ cos m − m − cos m + m' 2 0 a a 2 0 a a a ( a ' ( ) ) ( πx πx ' ' m m sin m m sin − + fm a a => − π 2 π m − m' m + m' a a ( ) ( ) ( ) a 0 Lower limit vanishes, evaluate upper limit Ramadas Chennamsetti ) dx 44 rd_mech@yahoo.co.in a Coefficients R&DE (Engineers), DRDO a ∫ 0 m 'π x fm f (x )sin dx = a 2 a => ∫ 0 sin πx π ( m−m ) a fma m 'π x f (x )sin dx = a 2 =0 ( m−m ) a ' ' x=a for m = m ' for m ≠ m 2 mπ x f m = ∫ f (x )sin dx a a 0 Ramadas Chennamsetti ' a 45 rd_mech@yahoo.co.in Upper limit Coefficients R&DE (Engineers), DRDO Function of variable ‘y’ nπ y g ( y ) = ∑ g n sin b n ∞ nπ y nπ y nπ y => g ( y )sin = g n sin sin b b b b 2 nπ y g n = ∫ g ( y )sin dy b0 b ' Ramadas Chennamsetti 46 rd_mech@yahoo.co.in ' Coefficients R&DE (Engineers), DRDO Function of two variables mπ ( ) w x, y = ∑ ∑ wmn sin ∞ ∞ m n x nπ sin a b y x=a x=a ' ∞ ∞ m 'π x nπ y mπ x m π x dx = ∑ ∑ wmn sin dx => ∫ w( x, y )sin ∫ sin sin b x =0 a a m n a x =0 x=a m 'π x a ∞ ∞ nπ y ( ) w x , y sin dx = w sin ∑ ∑ mn ∫x =0 a 2 m n b y =b x = a y =b => wmn 4 = ab y =b x = a m π x nπ y ( ) w x y , sin sin dxdy ∫y =0 x∫=0 a b Ramadas Chennamsetti 47 rd_mech@yahoo.co.in ' m 'π x n 'π y a ∞ ∞ nπ y n π y sin dxdy = ∑ ∑ wmn ∫ sin dy => ∫ ∫ w( x, y )sin sin 2 m n b b a b y =0 x =0 y =0 Simply supported plate R&DE (Engineers), DRDO Assume loading over plate mπ x nπ y q = q ( x, y ) = ∑ ∑ qmn sin sin a b m n ∞ ∞ Solution mπ x nπ y w = w( x, y ) = ∑∑ wmn sin sin a b m n ∞ Governing equation ∂ w ∂ w ∂ w q +2 + = 4 2 2 4 ∂x ∂x ∂y ∂y D 4 4 Ramadas Chennamsetti 4 48 rd_mech@yahoo.co.in ∞ Simply supported plate R&DE (Engineers), DRDO Differentiating vertical displacement ∂ w mπ mπ = ∑∑ wmn sin 4 ∂x a m n a 4 ∞ ∞ 4 x nπ y sin b ∂ w nπ mπ x nπ y = ∑∑ wmn sin sin 4 ∂y a b m n b ∞ ∞ 4 ∂ w mπ nπ mπ 2 2 2 = 2∑∑ wmn sin ∂x ∂y a m n a b 4 ∞ ∞ 2 2 Ramadas Chennamsetti x nπ y sin b 49 rd_mech@yahoo.co.in 4 Simply supported plate R&DE (Engineers), DRDO mπ 4 mπ 2 nπ 2 nπ 4 qmn + 2 + wmn = D a b b a qmn => wmn = 2 2 2 n 4 m Dπ 2 + 2 b a 1 w = w( x, y ) = 4 Dπ mπ x nπ y sin sin ∑∑ 2 2 2 a b m n m n 2 + 2 a b 50 Ramadas Chennamsetti ∞ ∞ qmn rd_mech@yahoo.co.in Plug in governing equation UDL R&DE (Engineers), DRDO Uniformly distributed load qo Computation of coefficients ∞ ∑∑q m mn n mπ x nπ y sin sin a b q(x, y) = qo 4 mπ x nπ y = q( x, y )sin sin dxdy ∫ ∫ ab 0 0 a b a b qmn Ramadas Chennamsetti 51 rd_mech@yahoo.co.in q (x , y ) = ∞ UDL R&DE (Engineers), DRDO Coefficients qmn q mn mπ x mπ y ∫0 ∫0 q o sin a sin b dxdy 4 qo mπ x mπ y = sin dx ∫ sin dy ∫ a b ab 0 0 a q mn => q mn b 4qo = ab 4 ab 2 mn π Ramadas Chennamsetti 16 q o = 2 mn π 52 rd_mech@yahoo.co.in 4 = ab a b UDL R&DE (Engineers), DRDO m π x nπ y q mn sin sin ∞ ∞ 1 a b w = w( x , y ) = 4 ∑ ∑ 2 2 2 π D m n m n + a b m π x nπ y sin sin ∞ ∞ 16 qo a b => w(x, y ) = 6 ∑ ∑ 2 2 2 π D m n m n + a b 53 Ramadas Chennamsetti rd_mech@yahoo.co.in Vertical deflection Patch load R&DE (Engineers), DRDO A patch load applied over an area u x v Centroid at (xo, yo) mπ x nπ y q ( x , y ) = ∑ ∑ q mn sin sin a b m n ∞ ∞ m π x nπ y => q o = ∑ ∑ q mn sin sin a b m n q mn ∞ 4qo = ab u v xo + yo + 2 2 ∫ xo − qo v u Patch load mπ x mπ y ∫ v sin a sin b dxdy u yo − 2 Ramadas 2 Chennamsetti 54 rd_mech@yahoo.co.in ∞ Patch load R&DE (Engineers), DRDO Evaluating integrals xo + u 2 o 2 ∫ xo − u v yo + 2 2 mπ x mπ y ∫ v sin a sin b dxdy u yo − 2 2 mπ x 2a mπ xo mπ u ∫ u sin a dx = mπ sin a sin 2a x − => qmn rd_mech@yahoo.co.in qmn 4 qo = ab xo + 16 qo mπ xo mπ u nπ yo mπ v = sin sin sin sin 2 mn π a 2a b 2b Ramadas Chennamsetti 55 Patch load R&DE (Engineers), DRDO Deflection m π x nπ y S mn sin sin ∞ ∞ 16 qo a b w( x , y ) = 6 ∑ ∑ π D m n m 2 n 2 mn + a b S mn m π xo = sin a m π u nπ y o sin sin 2a b Ramadas Chennamsetti mπ v sin 2b 56 rd_mech@yahoo.co.in where, Point load R&DE (Engineers), DRDO Point load Assume the point load acts over an infinitesimal area u x v Corresponding UDL P qo = uv qo = P uv P (xo, yo) mπ xo nπ yo 16 qo mπ u nπ v sin sin qmn = sin sin 2 mn π a b 2a 2b mπ xo nπ y o 16 P mπ u nπ v sin sin sin sin => qmn = 2 mn uvπ a b 2a 2b57 Ramadas Chennamsetti rd_mech@yahoo.co.in From the earlier analysis Point load R&DE (Engineers), DRDO Simplifying Ramadas Chennamsetti 58 rd_mech@yahoo.co.in => qmn mπ u nπ v sin sin mπ xo nπ yo 4P 2a 2b = sin sin ab a b mπ u mπ v 2a 2b 4P mπ xo nπ yo => qmn = sin sin ab a b Point load R&DE (Engineers), DRDO Deflection due to point load mπ x mπ y sin sin ∞ ∞ S 4P a b w( x, y ) = 4 ∑∑ 2 2 2 π abD m m m n + a b mπ xo mπ yo ' S mn = sin sin a b Ramadas Chennamsetti 59 rd_mech@yahoo.co.in ' mn Bending & in-plane loading Plates are subjected to in-plane loading also – in addition to lateral / transverse loads In-plane loading – tensile or compressive Large in-plane compressive loads – Buckling takes place Buckling – non-linear phenomenon – disproportionate increase of displacement with load Critical load – ability to resist axial load ceases – change in deformation shape Ramadas Chennamsetti 60 rd_mech@yahoo.co.in R&DE (Engineers), DRDO Bending & in-plane loading R&DE (Engineers), DRDO In-plane forces: Nx, Ny, Nxy and Nyx Transverse forces / moments: Mx, My, Mxy, Myx, Qx and Qy Small deformation and large deformation Ramadas Chennamsetti 61 rd_mech@yahoo.co.in Thin walled members – cross-sections like ‘I’, ‘L’, ‘H’, ‘C’ etc – undergo buckling – thin plates of small widths Combined loading of a rectangular plate – loads In-plane forces R&DE (Engineers), DRDO Infinitesimal element => dA = dx dy Nxy * Qx* Nx* θx* θx* dx z θx Nx θx x In-plane loading Transverse/out-of-plane loading For small angles => Sinθ ≈ θ and Cosθ ≈ 1 Ramadas Chennamsetti 62 rd_mech@yahoo.co.in Qx In-plane forces R&DE (Engineers), DRDO ∂N x − N x dy cos θ x + N x + dx dy cos θ x* − N xy dx cos θ y + ∂x ∂N xy ∂Qx * N xy + dy dx cos θ y − Qx + dx dy sin θ x* ∂y ∂x ∂Q y + Qx dy sin θ x − Q y + dy dx sin θ y* + Q y dx sin θ y = 0 ∂y ∂N x ∂N xy => + =0 ∂x ∂y If there are no in-plane forces – equation vanishes Ramadas Chennamsetti 63 rd_mech@yahoo.co.in Force equilibrium in x-direction In-plane forces R&DE (Engineers), DRDO Force equilibrium in y-direction ∂N xy ∂N y + =0 ∂x ∂y Angle is not equal to zero, but, small ∂w θx = , ∂x ∂w θy = ∂y Ramadas Chennamsetti 64 rd_mech@yahoo.co.in sin θ ≈ θ and cos θ ≈ 1 In-plane forces R&DE (Engineers), DRDO ∂N x dx dy − N xy dx + − N x dy + N x + ∂x ∂N xy ∂Q x ∂ w ∂ 2 w N xy + dy dx − Qx + dx dy + 2 dx ∂y ∂x ∂x ∂ x ∂ Q y ∂w ∂ 2 w ∂w ∂w dy dx =0 + Qx dy − Q y + + 2 dy + Q y dx ∂y ∂x ∂y ∂y ∂ y Neglect higher order term s No change in in∂N x ∂N xy => + = 0 plane equilibrium ∂x ∂y Ramadas Chennamsetti equations 65 rd_mech@yahoo.co.in Force equilibrium in x-direction In-plane forces R&DE (Engineers), DRDO Contribution from in-plane normal forces, Nx and Ny and shear force, Nxy Contribution from shear force, Qx and Qy Contribution from externally applied load, q Ramadas Chennamsetti 66 rd_mech@yahoo.co.in Similar expression for force equilibrium in y-direction Force equilibrium in z-direction Z-direction R&DE (Engineers), DRDO ∂w ∂ ∂w ∂w ∂N x − N x dy sin + Nx + dx dy sin + dx ∂x ∂x ∂x ∂x ∂x ∂ N xy ∂w ∂ ∂w ∂w dx − N xy dy sin + N xy + + dx dy sin ∂y ∂x ∂y ∂x ∂y ∂w ∂ ∂w ∂w ∂Q x − Q x dy cos + Qx + + dx dy cos dx ∂x ∂x ∂x ∂x ∂x ∂Q y ∂w ∂ ∂w ∂w dy + qdxdy = 0 − Q y dx cos + Q y + dy cos + 67 ∂y ∂Ramadas y Chennamsetti ∂y ∂y ∂y rd_mech@yahoo.co.in ∂N y ∂w ∂ ∂w ∂w dy dy dx sin − N y dx sin +Ny + + ∂y ∂y ∂y ∂y ∂y ∂ N xy ∂w ∂ ∂w ∂w − N xy dx sin + N xy + dy dx sin + dy ∂x ∂y ∂x ∂y ∂x Z-direction R&DE (Engineers), DRDO If no in-plane forces acting ∂Q x ∂Q y + = −q ∂x ∂y ∂ 2w ∂ 2w ∂ 2 w ∂Q x ∂Q y Nx + 2 N xy + Ny + + +q=0 2 2 ∂x ∂x∂y ∂y ∂x ∂y Ramadas Chennamsetti 68 rd_mech@yahoo.co.in Presence of in-plane forces Z-direction R&DE (Engineers), DRDO Substituting Qx and Qy ∂ 2 ∂ 2 Qx = − D ∇ w , Qy = − D ∇ w ∂x ∂y ( ) ( ) ∂ ∂ + − D ∇2w ∂x ∂x ( ∂ ∂ 2 + − D ∇ w ∂y ∂y ) ( ) + q = 0 2 2 2 1 ∂ ∂ w ∂ w w 4 => ∇ w = N x 2 + 2 N xy + N y 2 + q D ∂x ∂ x∂ y ∂y Ramadas Chennamsetti 69 rd_mech@yahoo.co.in ∂2w ∂2w ∂2w N x 2 + 2 N xy + Ny 2 ∂x ∂ x∂ y ∂y Plate buckling R&DE (Engineers), DRDO Buckling of thin plate 2 2 2 1 ∂ w ∂ w ∂ w 4 ∇ w = q + N x 2 + 2 N xy + Ny 2 D ∂x ∂y∂x ∂y Assume, q = 0, in-plane load, Nx = N1, rest zero ∂ w ∂ w ∂ w Nx ∂ w +2 2 2 + = 4 4 2 ∂x ∂x ∂y ∂y D ∂x 4 4 2 Assume all four edges are simply supported – Navier solution to plate bending Ramadas Chennamsetti 70 rd_mech@yahoo.co.in 4 Plate buckling R&DE (Engineers), DRDO Plate deflection ∞ ∞ m n w = w( x, y ) = ∑∑ wmn sin α m x sin β n y N1 2 α + 2α β + β − α m = 0 D N1 2 2 2 2 => α m + β n = αm D Ramadas Chennamsetti 4 m 2 m ( 2 n 4 n Substitute in governing equation Characteristic equation ) 71 rd_mech@yahoo.co.in mπ nπ αm = , βn = a b Plate buckling R&DE (Engineers), DRDO From characteristic equation 2 m + β n2 ) 2 = N1 2 αm D m π N1 mπ nπ => = + D a b a 2 2 Dπ a => N 1 = m2 2 2 2 m n + b a 2 Dπ 2 m c 2 n => N 1 = + 2 b c m c = a / b - Aspect ratio Ramadas Chennamsetti 2 2 2 2 72 rd_mech@yahoo.co.in (α Plate buckling R&DE (Engineers), DRDO Critical load – smallest value Increase in N1 with n2 – Minimum value of n is equal to one – buckled shape in ydirection – single half sine wave 2 N1 is a function of variable ‘m’ – for minimum value of ‘m’, differentiate N1 wrt ‘m’ Ramadas Chennamsetti 73 rd_mech@yahoo.co.in Dπ m c N1 = 2 + b c m 2 Plate buckling R&DE (Engineers), DRDO Differentiate dN1 2π D m c 1 c = + − 2 = 0 2 dm b c m c m 1 c => − 2 = 0 c m a => m = c = b 2 4 Dπ ∴ N1cr = 2 b Ramadas Chennamsetti Whole number 74 rd_mech@yahoo.co.in 2 Plate buckling R&DE (Engineers), DRDO Number of half sine waves can’t be a whole number – it should be an integer Equation for critical load for n = 1 m c K = + c m 2 Plotting ‘K’ vs aspect ratio = c = a/b for various integer values of ‘m’ Ramadas Chennamsetti 75 rd_mech@yahoo.co.in Dπ N1 = K 2 b 2 Plate buckling R&DE (Engineers), DRDO 1.414 2.449 Ramadas Chennamsetti 76 rd_mech@yahoo.co.in K vs aspect ratio Plate buckling R&DE (Engineers), DRDO From figure, value of ‘K’ is same as intersection points of ‘m’ and ‘m+1’ c m c m +1 + + = m +1 c m c 2 => c = m(m + 1) 2 2 c = m(m + 1) Ramadas Chennamsetti Aspect ratio less than √2 => m = 1 Aspect ratio from √2 to √6, m = 2 C>4 => K ≈ 4 77 rd_mech@yahoo.co.in N1 critical load at m – when load is increased, buckled form changes from ‘m’ to ‘m+1’ Curves for m = 1 and At transition from ‘m’ to ‘m+1’ m = 2 meet at c = √2 Plate buckling R&DE (Engineers), DRDO Estimation of buckling load t = 1 cm, a = 2.3 m, b = 1 m, E = 200 GPa, υ = 0.30 ( ) −2 2 Et 200 ×10 × 1×10 D= = = 17.96 kNm 2 2 12 1 − υ 12 1 − 0.3 a 2.3 c= = = 2.3 => m = 2 b 1 3 ( 9 ) ( 2 ) 2 m c 2 2.3 => K = + = + = 4.0786 2.3 2 c m Ramadas Chennamsetti 78 rd_mech@yahoo.co.in Flexural modulus Plate buckling R&DE (Engineers), DRDO In the above expression, for a given width and elastic properties of plate, critical load depends on ‘K’. In turn ‘K’ depends on ‘m’ for a given aspect ratio, c c = 2.3, m = 1, K = 7.4790 m = 2, K = 4.0786 Minimum value of ‘K’ is considered for estimation of Ncr Chennamsetti m = 3, K Ramadas = 4.2890 79 rd_mech@yahoo.co.in Minimum critical load 2 2 Dπ Dπ N1cr = K 2 = 4.086 × 2 = 73.72 MN / m b b N = N1cr × b = 73.72 MN Plate and column buckling R&DE (Engineers), DRDO Uniaxial load for plate buckling t ( K π 2E => σ 1crp = 12 1 − υ 2 b 2 t K π 2E => σ 1crp = C1 , C1 = 2 12 1 − υ 2 b t ( ) )( ) ( ) Ramadas Chennamsetti ( ) 80 rd_mech@yahoo.co.in σ 1crp = N1crp Dπ 2 N1crp = K 2 b Dπ 2 π 2 Et 3 =K 2 =K 2 bt b t 12 1 − υ 2 Plate and column buckling R&DE (Engineers), DRDO Buckling of a column => Pcr = C 2 l2 π 2 EAk 2 l 2 = C2 π 2 EA (l k ) 2 Pcr π 2E => σ crc = = C2 Column 2 A l k π 2E Plate σ crp = C1 2 b t Chennamsetti Ramadas ( ) ( ) 81 rd_mech@yahoo.co.in Pcr = C 2 π 2 EI Plate and column buckling R&DE (Engineers), DRDO Critical stress for plate depends on thinness ratio = t/b – not on the length More thinner plate – lesser bucking load Critical stress in column depends on slenderness ratio Longer columns – lower critical load Ramadas Chennamsetti 82 rd_mech@yahoo.co.in depends on width Strain energy R&DE (Engineers), DRDO Linear elastic material 1 U = ∫ (σ xxε xx + σ yyε yy + τ xyγ xy ) dV 2V Ramadas Chennamsetti 83 rd_mech@yahoo.co.in In thin plate theory, out-of-plane shear stresses vanish => τxz, τyz and τzz Stress components contributing to strain energy => σxx, σyy and τxy Strain energy, Strain energy R&DE (Engineers), DRDO Strain energy, ε yy γ xy } dV T ∂2w 2 ∂x σ xx 1 0 1 0 υ ε υ xx ∂ 2 w Ez Ez υ 1 υ 1 0 ε yy = − 0 σ yy = − 2 2 2 − ∂ y 1 1 − υ υ 1 + υ 1 + υ τ γ 0 0 0 0 2 xy xy 2 2 2 ∂ w ∂x∂y => {σ } = [C ]{ε } Ramadas Chennamsetti 84 rd_mech@yahoo.co.in 1 U = ∫ {σ xx σ yy τ xy }{ε xx 2V Strain energy R&DE (Engineers), DRDO Strain energy, ∂ 2 w ∂2w ∂2w 2 + υ 2 2 ∂y ∂x ∂x 2 2 2 ∂ w ∂ w ∂ w 2 E => U = + υ 2 + 2 2 z dV 2 ∫ 2 1 − υ V ∂x ∂y ∂x 2 2 ∂ w + 2(1 − υ ) ∂x∂y ( ) Ramadas Chennamsetti 85 rd_mech@yahoo.co.in 1 1 T T U = ∫ {σ } {ε }dV = ∫ {ε } [C ]{ε }dV 2V 2V Strain energy R&DE (Engineers), DRDO Infinitesimal volume, dV = dxdydz Carry out integration over thickness => dz h + 2 3 h ∫h z dz = 12 − 2 3 Eh D= 2 12 1 − υ ( ) Ramadas Chennamsetti 86 rd_mech@yahoo.co.in 2 Strain energy R&DE (Engineers), DRDO Simplify dxdy 2 Strain energy – Finite Element Method – Total potential approach Ramadas Chennamsetti 87 rd_mech@yahoo.co.in ∂ 2 w ∂ 2 w 2 + 2 2 ∂x D ∂ x U = ∫ ∂ 2w ∂ 2w ∂ 2w 2 A − 2 (1 − υ ) ∂ x 2 ∂ x 2 − ∂ x ∂ y Ramadas Chennamsetti 88 rd_mech@yahoo.co.in R&DE (Engineers), DRDO