R&DE (Engineers), DRDO

Theory of Plates

Ramadas Chennamsetti [email protected] Introduction R&DE (Engineers), DRDO
“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”

Plates are initially flat structural elements

Ramadas Chennamsetti 2 [email protected] Introduction R&DE (Engineers), DRDO
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 [email protected] Introduction R&DE (Engineers), DRDO
Plate stretching

z Nx t Nx x

Uniform stretching of the plate => u o q

Axial deformation due to transverse load

Net deformation = Algebraic sum of uniform stretching and axial deformationRamadas due toChennamsetti bending load 4 [email protected] Introduction R&DE (Engineers), DRDO
For plates - 1 ≥ t ≥ 1 10 b 2000
Thin & thick plates –
Thin plate => t < 20 b b = smallest side
Thick plate => t > 20 b t w ≤
Small deflections – 5
Thin plate theory – Kirchoff’s Classical Plate Theory (KCPT)
Thick plate theory – Reissner – Mindlin Plate Theory (MPT)

Ramadas Chennamsetti 5 [email protected] 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
Governing equations are derived based on undeformed geometry
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 [email protected] KCPT - Assumptions R&DE (Engineers), DRDO
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 [email protected] Sign convention R&DE (Engineers), DRDO
Following sign convention will be followed

z z

ψψψ Mz z

My ψψψ y y y Mx ψψψ x x x

Positive moments Positive rotations

Ramadas Chennamsetti 8 [email protected] Bending deformations R&DE (Engineers), DRDO z Bending takes place in both planes y

D

x Rotation => ψ A y A’

z B B’ A

x B

View ‘D’ [email protected] Ramadas Chennamsetti 9 Bending deformations R&DE (Engineers), DRDO
Deformation in ‘x’ direction = − ψ ( ) u(x, y, z) z y x, y
Deformation in ‘y’ direction = − ψ ( ) v( x, y, z) z x x, y

Vertical deformation w = w ( x , y )

Ramadas Chennamsetti 10 [email protected] Strains R&DE (Engineers), DRDO
Assumption – out of plane shear strain - negligible ∂u ∂w γ = + = 0 xz ∂z ∂x ∂w ∂w => γ = −ψ + = 0 => ψ = xz y ∂x y ∂x ∂v ∂w γ = + = 0 yz ∂z ∂y ∂w ∂w => γ = −ψ + = 0 => ψ = yz x ∂y x ∂y Ramadas Chennamsetti 11 [email protected] Strains R&DE (Engineers), DRDO
Non-zero strains ∂u ∂ψ ∂ 2 w ε = = − z y = − z xx ∂x ∂x ∂x 2 ∂v ∂ψ ∂ 2 w ε = = − z x = − z yy ∂y ∂y ∂y 2 ∂u ∂v ∂ 2 w γ = + = −2 z xy ∂y ∂x ∂x∂y

Ramadas Chennamsetti 12 [email protected] Stresses R&DE (Engineers), DRDO
Thin plate – out of plane shear strains vanish – out of plane shear stresses also vanish τ = τ = xz ,0 yz 0 z z τ τ yz xz y x τ τ yz xz
Out of plane normal stress is also assumed to be zero – logical – thin structure – plane stress conditions

Ramadas Chennamsetti 13 [email protected] Stresses R&DE (Engineers), DRDO
Non-zero stress components

z σ xx

y σ yy σ yy τ xy τ xy x σ xx

σ σ τ All three stress components, xx , yy , xy – in-plane

Ramadas Chennamsetti 14 [email protected] Curvatures R&DE (Engineers), DRDO
Curvature – reciprocal of radius of bending
Rate of change of slope

z

∂w − dx w ∂ R dx w* x x ψ x y dx

2 ∂w ∂ψ ∂ w slope = rotation = ψ = - curvatue = y = − = κ y ∂x ∂x ∂x 2 xx

Ramadas Chennamsetti 15 [email protected] Curvatures R&DE (Engineers), DRDO
Similarly bending in yz plane introduces a curvature ∂ 2 κ = − w yy ∂ y 2
Twisting of plate ∂ 2 w κ = −2 xy ∂x∂y

Ramadas Chennamsetti 16 [email protected] Constitutive law R&DE (Engineers), DRDO
Linear elastic isotropic – Hookean material
Three stress and strain components σ σ Writing all three equations in ε = xx − υ yy matrix form xx ε  −υ σ  E E xx  1 0  xx ε = 1 −υ σ  σ σ  yy  1 0  yy  yy xx E  ε = − υ γ   0 0 1(2 +υ)τ  yy E E  xy    xy  τ τ σ   υ ε  xx 1 0  xx γ = xy = xy ()+ υ   E   xy 2 1 σ = υ 1 0 ε  yy 1−υ2  −υ yy G E τ  1 γ   xy  0 0  xy 

  17 [email protected] Ramadas Chennamsetti 2 Constitutive law R&DE (Engineers), DRDO
Express strains in terms of curvatures  ∂ 2 w      2 σ  1 υ 0  ∂ x  xx   2   Ez  ∂ w  σ  = − υ 1 0    yy 1 − υ 2  + υ  ∂ y 2  τ  1    xy   0 0  ∂ 2  2   w   2   ∂ x ∂ y  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 [email protected] Equilibrium equations R&DE (Engineers), DRDO
Equilibrium of an infinitesimal element z

y

dx dy σ xx y t τ xy x x τ xy dy ττ** yzyz σ σ* Forces acting on an infinitesimal yy dx yy τ* τ* element dx dy dz xz xy τ* xy

σ* 19 [email protected] Ramadas Chennamsetti xx Equilibrium equations R&DE (Engineers), DRDO
Equilibrium in ‘x’ direction σ * + τ * + τ * xx dydz zx dxdy yx dzdx − σ − τ − τ = xx dydz zx dxdy yx dzdx 0  ∂ σ  σ * =  σ + xx dx  xx  xx ∂ x   ∂ τ  τ * =  τ + zx dz  zx  zx ∂ z   ∂ τ  τ * =  τ + yx  yx  yx dy   ∂ y  Ramadas Chennamsetti 20 [email protected] Equilibrium equations R&DE (Engineers), DRDO
Substitute – the following equation is obtained ∂τ ∂τ ∂τ xx + xy + xz = 0 - (1) ∂x ∂y ∂z Similarly take equilibrium in ‘y’ and ‘z’ directions ∂τ ∂τ ∂τ xy + yy + yz = 0 - (2) ∂x ∂y ∂z ∂τ ∂τ ∂τ xz + yz + zz = 0 - (3) ∂x ∂y ∂z Ramadas Chennamsetti 21 [email protected] Shear stresses R&DE (Engineers), DRDO τ
From equation (1) – shear stress xz can be computed

Use stress deflection/curvature relations

∂  Ez  ∂ 2 w ∂ 2 w  ∂  Ez ∂ 2 w  ∂τ −  +υ  + − + xz =  2  2 2    0 ∂x  1−υ  ∂x ∂y  ∂y  1+υ ∂x∂y  ∂z ∂τ  Ez  ∂ 3w ∂ 3w ∂ 3w  xz =  +υ + ()−υ   2  3 2 1 2  ∂z 1−υ  ∂x ∂x∂y ∂x∂y  ∂τ  Ez  ∂ xz =   ()∇ 2 w ∂z 1−υ 2  ∂x Integrate this across thickness to get shear stress Ramadas Chennamsetti 22 [email protected] Shear stresses R&DE (Engineers), DRDO
Integrate from mid plane to top surface of z plate z = h/2 z

0 h 2 Ez  ∂(∇ 2 w) E ∂(∇ 2 w)h 2 dτ =   dz = z dz ∫ xz ∫ −υ 2 ∂ −υ 2 ∂ ∫ τ 1  x 1 x xz 0 z ∂()∇ 2  2  h 2 −τ = E w  z xz −υ 2 ∂   1 x  2  z ∂()∇ 2  2  => −τ = E w  h − 2  xz 2  z  2()1−υ ∂x  4  ∂()∇ 2  2  => τ = E w  2 − h  xz 2  z  Parabolic variation 2()1−υ ∂x  4  Ramadas Chennamsetti 23 [email protected] Moments R&DE (Engineers), DRDO
Moment wrt ‘y’ axis = σ dF xx xx dz = = σ Neutral plane dM y zdF xx z xx dz z  ∂ 2 ∂ 2  σ = − Ez  w + υ w  xx 2  2 2  1 − υ  ∂x ∂x  dz Ez 2  ∂ 2 w ∂ 2 w  z => dM = −  + υ  y − υ 2  ∂ 2 ∂ 2  σ 1  x y  xx y + dx E  ∂ 2 w ∂ 2 w  h 2 M = −  + υ  z 2 dz dy y 1 − υ 2  ∂x 2 ∂y 2  ∫ x   − h 2 3  ∂ 2 ∂ 2  = − Eh  w + υ w  M y 2  2 2  12 ()1 − υ  ∂x ∂y  Ramadas Chennamsetti 24 [email protected] Moments R&DE (Engineers), DRDO
Moment wrt ‘x’ axis 3  ∂ 2 ∂ 2  = − Eh υ w + w  M x 2  2 2  12 1( −υ )  ∂x ∂y  Neutral plane z Twisting moment due to shear τ stress xy dz τ Eh 3 ∂2w z xy M = − ()1−υ xy −υ 2 ∂ ∂ y 12 1( ) x y dx dy Eh 3 x D = Compare with section 12 1( − υ 2 )

modulus of beam 25 [email protected] Ramadas Chennamsetti 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 z Q * = Q + x dx x x ∂x q ∂Q * = + y Q y Q y dy Q* ∂y * y Q x Q x Substitute these and simplify Qy y dx dy ∂Q ∂Q x + y = − q x ∂x ∂y

Ramadas Chennamsetti 26 [email protected] Shear forces R&DE (Engineers), DRDO
Shear forces across thickness can be computed by integrating shear stress across thickness + h 2 = τ Q y ∫ yz dz per unit width − h 2 ∂ ∇ 2  2  τ = E ( w )  2 − h  yz 2  z  2 1( − υ ) ∂ y  4  + E ∂ (∇ 2 w ) h 2  h 2  Q =  z 2 −  dz y − υ 2 ∂ ∫   2 1( ) y − h 2  4  Eh 3 ∂ (∇ 2 w ) ∂ (∇ 2 w ) Q = − = − D y 12 1( − υ 2 ) ∂ y ∂ y Ramadas Chennamsetti 27 [email protected] Governing equation R&DE (Engineers), DRDO

Similar expression for Qx ∂ (∇ 2 w ) Q = − D x ∂ x

Substitute Q x and Q y in the following expression – vertical equilibrium ∂Q ∂Q x + y = − q ∂x ∂y ∂ ∂ ∂ ∂  2   2   − D ()∇ w  +  − D ()∇ w  = − q ∂x  ∂x  ∂y  ∂y  ∂ 2 ∂ 2 q => ()∇ 2 w + ()∇ 2 w = ∂x 2 ∂y 2 D  ∂ 2 ∂ 2  =>  + ()∇ 2 = q  2 2  w ∂x ∂y D   28 [email protected] Ramadas Chennamsetti Governing equation R&DE (Engineers), DRDO
Governing equation  ∂ 2 ∂ 2   +  ∇ 2 = q  2 2  w  ∂ x ∂ y  D q => ∇ 2 (∇ 2 w )= D => ∇ 4 = q D ∂ 4 w ∂ 4 w ∂ 4 w q + 2 + = ∂x 4 ∂x 2 ∂y 2 ∂y 4 D Bi-harmonic equation

CompareRamadas with Chennamsetti beam equation 29 [email protected] Boundary conditions R&DE (Engineers), DRDO
Well posed problem – Governing equations and boundary conditions
Three basic boundary conditions
Simply supported
Clamped and
Free edge
Vertical deflection and their derivatives

Ramadas Chennamsetti 30 [email protected] Simply supported R&DE (Engineers), DRDO
Simply supported – for eg beam => vertical deflection = 0 and moment = 0 Simply supported For edge, x = const x = const w(x, y)= 0  ∂ 2 ∂ 2  = −  w +υ w  = M y D 2 2  0  ∂x ∂y  y w = 0 implies second derivative in the direction tangent to this line is zero x ∂ 2 w Simply supported = y = const 2 0 ∂ x Ramadas Chennamsetti 31 [email protected] Simply supported R&DE (Engineers), DRDO
Simply supported condition along edge y = const w = 0  ∂ 2 ∂ 2  = − υ w + w  M x D 2 2   ∂x ∂y 

w = 0 implies second derivative in the direction tangent to this line is zero ∂2 w = 2 0 Ramadas∂y Chennamsetti 32 [email protected] Clamped R&DE (Engineers), DRDO
Deflection and slope in normal directions vanish w = 0 ∂w x = constant = 0 ∂x w = 0 ∂w y = constant = 0 ∂y

Ramadas Chennamsetti 33 [email protected] Free edges R&DE (Engineers), DRDO
Free edge – Free from any external loads – Natural boundary conditions
Bending moment and Shear force vanish
Bending moment  ∂ 2 ∂ 2  = −  w +υ w  = = M y D 2 2  0 at x const  ∂x ∂y   ∂ 2 ∂ 2  = − υ w + w  = = M x D 2 2  0 at y const  ∂x ∂y 

Ramadas Chennamsetti 34 [email protected] Free edges R&DE (Engineers), DRDO

Moment M xy and shear forces

Mxy The net force acting on dy the face ∂ ' M xy Mxy = − − + Qx M xy M xy dy ∂y ∂ M xy ∂ M + dy ' M xy M xy ∂y => Q = − xy x ∂y

∂ Total shear force, M M xy xy M + dy xy ∂ dx ’ y Vx = Q x + Q x dy 35 [email protected] Ramadas Chennamsetti Free edges R&DE (Engineers), DRDO
Total shear force, = + ' Vx Qx Qx ∂ ∂  ∂2  => = − (∇2 )− ( −υ)  w  Vx D w D 1  2  ∂x ∂x  ∂x  ∂3 ∂3  => = − w + ()−υ w Vx D 2   ∂x3 ∂x∂y 2  ∂3 ∂3  = − w + ()−υ w Vy D 2   ∂y3 ∂x2∂y  Ramadas Chennamsetti 36 [email protected] Free edges R&DE (Engineers), DRDO

The forces, V x and V y => reduced, or Kirchoff’s or effective shear forces
In case of a free edge,

∂3 ∂3  = − w + ()−υ w = = Vx D 2  0 at x const  ∂x3 ∂x∂y 2  ∂3 ∂3  = − w + ()−υ w = = Vy D 2  0 at y const  ∂y3 ∂x2∂y 

Ramadas Chennamsetti 37 [email protected] Bending of plates R&DE (Engineers), DRDO
Governing equation of plate rectangular plate bending q ∇2 (∇2w)= D w = vertical deflection = w(x, y) external loading, q = q(x, y)
Solution to this equation – product of two functions – Assume w = w(x, y)= F (x)G(y) Ramadas Chennamsetti 38 [email protected] Bending of plates R&DE (Engineers), DRDO
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 [email protected] 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 y ∞ π ()=  m x  F x ∑ Fm sin   m  a  SS x = 0 x x = a SS F1, F 2,…….F ∞ are coefficients Symmetric loading wrt x = a/2

Maximum deflection at x = a/2 40 [email protected] Ramadas Chennamsetti 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 y ∞ π ()=  n y  G y ∑Gn sin   n  b  SS x x = a x = 0, SS y = 0 SS G1, G 2,…….G ∞ are coefficients

Selection of functions based on BCs Ramadas Chennamsetti 41 [email protected] Bending of plates R&DE (Engineers), DRDO
Final solution, ∞ ∞ π π ()=  m x   n y  w x, y ∑∑ FmGn sin  sin   m n  a   b  ∞ ∞ => ( )= α β w x, y ∑∑ wmn sin m x sin n y m n mπ nπ α = , β = m a n b

Coefficients, Fm, G n and wmn computed using Fourier Series Ramadas Chennamsetti 42 [email protected] Coefficients R&DE (Engineers), DRDO
Any periodic function can be expanded into a sine or cosine function using Fourier expansion Function of one variable ∞ π ()= m x f x ∑ fm sin m a m'π x mπ x m'π x => f ()x sin = f sin sin a m a a

Integrate the above from limits 0 to a Ramadas Chennamsetti 43 [email protected] Coefficients R&DE (Engineers), DRDO
Integration

a 'π a π 'π () m x = m x m x ∫ f x sin dx fm ∫sin sin dx 0 a 0 a a a π 'π a  π π  => fm m x m x = fm x ( − ' )− x ( + ' ) ∫2sin sin dx ∫ cos m m cos m m  dx 2 0 a a 2 0  a a  π π a  x ()− ' x ()+ '  sin m m sin m m  => fm a − a  π π  2  ()m−m' ()m+ m'   a a 0

Lower limit vanishes, evaluate upper limit 44 [email protected] Ramadas Chennamsetti Coefficients R&DE (Engineers), DRDO
Upper limit π x ()− ' a 'π sin m m () m x = fm a ∫ f x sin dx π a 2 ' 0 (m − m ) a x=a a m'π x f a => ∫ f ()x sin dx = m for m = m' 0 a 2 = 0 for m ≠ m' a π = 2 () m x fm ∫ f x sin dx Ramadasa 0 Chennamsettia 45 [email protected] Coefficients R&DE (Engineers), DRDO
Function of variable ‘y’ ∞ π ()= n y g y ∑ gn sin n b n'π y nπ y n'π y => g()y sin = g sin sin b n b b b π = 2 () n y gn ∫ g y sin dy b 0 b

Ramadas Chennamsetti 46 [email protected] Coefficients R&DE (Engineers), DRDO
Function of two variables ∞ ∞ π π ()=  m x   n y  w x, y ∑∑ wmn sin  sin   m n  a   b  x=a  'π  ∞ ∞  π x=a  π   'π  => ()  m x  = n y m x  m x  ∫ w x, y sin  dx ∑∑ wmn sin   ∫ sin  sin  dx x=0  a  m n  b  x=0  a   a  x=a  'π  ∞ ∞  π  ()  m x  = a n y ∫ w x, y sin  dx ∑∑ wmn sin   x=0  a  2 m n  b  = = y b x=a  'π   'π  ∞ ∞ y b  π   'π  => ()  m x   n y  = a n y  n y  ∫ ∫ w x, y sin  sin  dxdy ∑∑ wmn ∫ sin  sin  dy y=0x= 0  a   b  2 m n y=0  b   b  y=b x=a π π => = 4 ()  m x   n y  wmn ∫ ∫ w x, y sin  sin  dxdy ab y=0x= 0  a   b 

Ramadas Chennamsetti 47 [email protected] Simply supported plate R&DE (Engineers), DRDO
Assume loading over plate ∞ ∞ π π = ()= m x n y q q x, y ∑∑ qmn sin sin m n a b

Solution ∞ ∞ π π = ()= m x n y w w x, y ∑∑ wmn sin sin m n a b Governing equation ∂ 4 w ∂ 4 w ∂ 4 w q + 2 + = ∂ 4 ∂ 2 ∂ 2 ∂ 4

x x y y D 48 [email protected] Ramadas Chennamsetti Simply supported plate R&DE (Engineers), DRDO
Differentiating vertical displacement

∂ 4w ∞ ∞  mπ 4  mπ x   nπ y  =   w sin  sin   ∂ 4 ∑∑ mn x m n  a   a   b  ∂4w ∞ ∞  nπ 4  mπ x   nπ y  =   w sin  sin   ∂ 4 ∑∑ mn y m n  b   a   b  ∂ 4w ∞ ∞  mπ 2  nπ 2  mπ x   nπ y  2 = 2     w sin  sin   ∂ 2∂ 2 ∑∑ mn x y m n  a   b   a   b 

Ramadas Chennamsetti 49 [email protected] Simply supported plate R&DE (Engineers), DRDO
Plug in governing equation  π 4  π 2  π 2  π 4  m + m n + n = qmn   2      wmn  a   a   b   b   D

=> = qmn wmn  2 2 2 π 4  m + n  D  2 2   a b  1 ∞ ∞ q  mπ x   nπ y  w = w()x, y = ∑∑ mn sin  sin   π 4 2 2 2 D m n    a   b   m + n   2 2 

a b 50 [email protected] Ramadas Chennamsetti UDL R&DE (Engineers), DRDO
Uniformly distributed load qo Computation of coefficients

∞ ∞ π π ( ) =  m x   n y  q x, y ∑ ∑ q mn sin   sin   m n  a   b 

q(x, y) = q o a b π π = 4 () m x n y qmn ∫∫ q x, y sin sin dxdy ab 0 0 a b

Ramadas Chennamsetti 51 [email protected] UDL R&DE (Engineers), DRDO

Coefficients qmn a b π π = 4 m x m y qmn ∫ ∫ qo sin sin dxdy ab 0 0 a b a π b π = 4qo m x m y qmn ∫ sin dx ∫ sin dy ab 0 a 0 b 4q  4ab  16 q => q = o   = o mn ab  mn π 2  mn π 2

Ramadas Chennamsetti 52 [email protected] UDL R&DE (Engineers), DRDO
Vertical deflection  mπ x   nπ y  q sin  sin   1 ∞ ∞ mn  a   b  w = w()x, y = ∑∑ π 4 2 2 2 D m n  m   n     +     a   b    mπ x   nπ y  sin  sin   16 q ∞ ∞  a   b  => w()x, y = o ∑∑ π 6 2 2 2 D m n  m   n     +     a   b  

  [email protected] Ramadas Chennamsetti 53 Patch load R&DE (Engineers), DRDO
A patch load applied over an area u x v

Centroid at (x o, y o) ∞ ∞  mπ x   nπ y  q()x, y = q sin   sin   ∑ ∑ mn q v m n  a   b  o ∞ ∞ π π u => =  m x   n y  qo ∑ ∑ qmn sin   sin   m n  a   b  Patch load u v x + y + o 2 o 2 π π = 4qo m x m y qmn ∫ ∫ sin sin dxdy ab u v a b x − y −

o o [email protected] 2 Ramadas2 Chennamsetti 54 Patch load R&DE (Engineers), DRDO
Evaluating integrals

u v x + y + o 2o 2 π π = 4qo m x m y qmn ∫ ∫ sin sin dxdy ab u v a b x − y − o 2o 2

+u xo 2 mπ x 2a  mπ x   mπ u  sin dx = sin  o sin   ∫ π u a m  a   2a  x − o 2 16 q  mπ x   mπ u   nπ y   mπ v  => q = o sin  o sin  sin  o sin   mn mn π 2  a   2a   b   2b 

Ramadas Chennamsetti 55 [email protected] Patch load R&DE (Engineers), DRDO
Deflection  mπ x   nπ y  S sin  sin   16 q ∞ ∞ mn  a   b  w()x, y = o π 6 ∑∑ 2 2 D m n  m   n   mn   +     a   b   where,  mπ x   mπ u   nπ y   mπ v  S = sin  o sin  sin  o sin   mn  a   2a   b   2b 

Ramadas Chennamsetti 56 [email protected] Point load R&DE (Engineers), DRDO
Point load Assume the point load acts over P q = P an infinitesimal area u x v o uv

Corresponding UDL P (x o, y o) q = o uv From the earlier analysis 16 q mπ x nπ y mπ u nπ v q = o sin o sin o sin sin mn mn π 2 a b 2a 2b 16 P mπ x nπ y mπ u nπ v => q = sin o sin o sin sin mn mn uv π 2 a b 2a 2b Ramadas Chennamsetti 57 [email protected] Point load R&DE (Engineers), DRDO
Simplifying

 mπ u  nπ v   sin  sin  4P mπ x nπ y => q = sin o sin o  2a  2b  mn ab a b  mπ u  mπ v      2a  2b  4P mπ x nπ y => q = sin o sin o mn ab a b

Ramadas Chennamsetti 58 [email protected] Point load R&DE (Engineers), DRDO
Deflection due to point load π π ' m x m y ∞ ∞ S sin sin 4P mn w(x, y)= ∑∑ a b π 4 2 2 2 abD m m  m   n     +     a   b   mπ x mπ y S ' = sin o sin o mn a b

Ramadas Chennamsetti 59 [email protected] Bending & in-plane loading R&DE (Engineers), DRDO
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 [email protected] Bending & in-plane loading R&DE (Engineers), DRDO
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: Nx, N y, N xy and Nyx

Transverse forces / moments: Mx, M y, M xy , M yx , Qx and Qy
Small deformation and large deformation

Ramadas Chennamsetti 61 [email protected] In-plane forces R&DE (Engineers), DRDO
Infinitesimal element => dA = dx dy Q * N * N * x θ * xy x x θ * x

dx θ z x θ Nx x Q x x In-plane loading Transverse/out-of-plane loading

For small angles => Sin θ ≈ θ and Cos θ ≈ 1

Ramadas Chennamsetti 62 [email protected] In-plane forces R&DE (Engineers), DRDO
Force equilibrium in x-direction  ∂N  − N dy cos θ +  N + x dx dy cos θ * − N dx cos θ + x x  x ∂x  x xy y  ∂N   ∂Q   + xy  θ * − + x θ *  N xy dy dx cos y Qx dx dy sin x  ∂y   ∂x   ∂Q  + θ −  + y  θ * + θ = Qxdy sin x Qy dy dx sin y Qydx sin y 0  ∂y  ∂N ∂N => x + xy = 0 ∂x ∂y

If there are noRamadas in-plane Chennamsetti forces – equation vanishes 63 [email protected] In-plane forces R&DE (Engineers), DRDO
Force equilibrium in y-direction ∂N ∂N xy + y = 0 ∂x ∂y
Angle is not equal to zero, but, small sin θ ≈ θ and cos θ ≈ 1 ∂w ∂w θ = , θ = x ∂x y ∂y

Ramadas Chennamsetti 64 [email protected] In-plane forces R&DE (Engineers), DRDO
Force equilibrium in x-direction  ∂N  − N dy +  N + x dx dy − N dx + x  x ∂x  xy  ∂N   ∂Q   ∂w ∂ 2 w   + xy  − + x  +   N xy dy dx Qx dx dy  2 dx   ∂y   ∂x   ∂x ∂x  ∂  ∂Q   ∂ ∂ 2  ∂ + w −  + y   w + w  + w = Qx dy Qy dy dx  2 dy  Qy dx 0 ∂x  ∂y   ∂y ∂y  ∂y Neglect higher order term s ∂N ∂N No change in in- => x + xy = 0 ∂ ∂ plane equilibrium x y 65 [email protected] Ramadas Chennamsetti equations In-plane forces R&DE (Engineers), DRDO
Similar expression for force equilibrium in y-direction
Force equilibrium in z-direction

Contribution from in -plane normal forces, N x and N y and shear force, N xy

Contribution from shear force, Q x and Q y
Contribution from externally applied load, q

Ramadas Chennamsetti 66 [email protected] Z-direction R&DE (Engineers), DRDO ∂w  ∂N   ∂w ∂  ∂w   − + + x  +  N x dy sin  N x dx dy sin   dx  ∂x  ∂x   ∂x ∂x  ∂x   ∂w  ∂N   ∂w ∂  ∂w   − +  + y   +    N y dx sin  N y dy dx sin   dy  ∂y  ∂y   ∂y ∂y  ∂y   ∂w  ∂N   ∂w ∂  ∂w   − +  + xy   +  N xy dx sin  N xy dy dx sin   dy  ∂x  ∂y   ∂x ∂y  ∂x   ∂w  ∂N   ∂w ∂  ∂w   − +  + xy   +    N xy dy sin  N xy dx dy sin   dx  ∂y  ∂x   ∂y ∂x  ∂y   ∂w  ∂Q   ∂w ∂  ∂w   − + + x  +  Q x dy cos  Q x dx dy cos   dx  ∂x  ∂x   ∂x ∂x  ∂x   ∂  ∂Q   ∂ ∂  ∂   − w +  + y   w +  w   + = Q y dx cos Q y dy  cos   dy  qdxdy 0 ∂y  ∂Ramadasy  Chennamsetti ∂y ∂y  ∂y   67 [email protected] Z-direction R&DE (Engineers), DRDO
If no in-plane forces acting ∂Q ∂Q x + y = −q ∂x ∂y

Presence of in-plane forces

∂ 2 w ∂ 2 w ∂ 2 w ∂Q ∂Q N + 2N + N + x + y + q = 0 x ∂x 2 xy ∂x∂y y ∂y 2 ∂x ∂y

Ramadas Chennamsetti 68 [email protected] Z-direction R&DE (Engineers), DRDO

Substituting Qx and Qy ∂ ∂ Q = −D (∇2w) , Q = −D (∇2w) x ∂x y ∂y ∂ 2w ∂ 2w ∂ 2w N + 2N + N x ∂x 2 xy ∂x∂y y ∂y 2 ∂ ∂ ∂ ∂  2   2  +  − D ()∇ w  +  − D ()∇ w  + q = 0 ∂x  ∂x  ∂y  ∂y   ∂ 2 ∂ 2 ∂ 2  => ∇ 4 = 1  w + w + w +  w  N x 2 2N xy N y 2 q D  ∂x ∂x∂y ∂y  Ramadas Chennamsetti 69 [email protected] Plate buckling R&DE (Engineers), DRDO
Buckling of thin plate  ∂ 2 ∂ 2 ∂2  ∇4 = 1 + w + w + w w q N x 2N xy N y  D  ∂x2 ∂y∂x ∂y 2 

Assume, q = 0, in-plane load, N x = N 1, rest zero ∂ 4 w ∂ 4 w ∂ 4 w N ∂ 2 w + 2 + = x ∂x 4 ∂x 2∂y 2 ∂y 4 D ∂x 2 Assume all four edges are simply supported – Navier solution to plate bending Ramadas Chennamsetti 70 [email protected] Plate buckling R&DE (Engineers), DRDO
Plate deflection ∞ ∞ = ()= α β w w x, y ∑∑ wmn sin m x sin n y m n mπ nπ α = , β = Substitute in m a n b governing equation N α 4 + 2α 2 β 2 + β 4 − 1 α 2 = 0 Characteristic m m n n D m equation

=> ()α 2 + β 2 2 = N1 α 2 m n m

Ramadas ChennamsettiD 71 [email protected] Plate buckling R&DE (Engineers), DRDO
From characteristic equation

()α 2 + β 2 2 = N 1 α 2 m n D m 2 N  mπ  2  mπ  2  nπ  2  => 1   =   +    D  a   a   b   2 π 2 2  2 2  => = D a  m  +  n  N 1 2      m  a   b   Dπ 2  m c  2 => N = + n 2 1 b 2  c m  c = a / b - Aspect ratio Ramadas Chennamsetti 72 [email protected] Plate buckling R&DE (Engineers), DRDO
Critical load – smallest value 2
Increase in N 1 with n – Minimum value of n is equal to one – buckled shape in y- direction – single half sine wave Dπ 2 m c 2 N = + 1 b2  c m

N1 is a function of variable ‘m’ – for minimum value of ‘m’, differentiate N 1 wrt ‘m’

Ramadas Chennamsetti 73 [email protected] Plate buckling R&DE (Engineers), DRDO
Differentiate dN 2π 2 D  m c  1 c  1 =  +  −  = 0 dm b2  c m  c m2   1 c  =>  −  = 0  c m2  a => m = c = Whole number b π 2 ∴ = 4D N1cr

2 [email protected] Ramadas Chennamsettib 74 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 π 2 2 = D = m + c  N1 K K   b2  c m

Plotting ‘K’ vs aspect ratio = c = a/b for various integer values of ‘m’

Ramadas Chennamsetti 75 [email protected] Plate buckling R&DE (Engineers), DRDO
K vs aspect ratio

1.414 2.449

Ramadas Chennamsetti 76 [email protected] Plate buckling R&DE (Engineers), DRDO
From figure, value of ‘K’ is same as intersection points of ‘m’ and ‘m+1’

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  m c 2  m +1 c 2  +  =  +  Aspect ratio less than  c m   c m +1 √2 => m = 1 √ => c2 = m()m +1 Aspect ratio from 2 to √6, m = 2 c = m()m +1 C>4 => K ≈ 4 Ramadas Chennamsetti 77 [email protected] 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 Flexural modulus 2 Et 3 200 ×10 9 ×(1×10 −2 ) D = = =17 96. kNm 12 ()1−υ 2 12 ()1− 3.0 2 a 3.2 c = = = 3.2 => m = 2 b 1 2 2 => = m + c  =  2 + 3.2  = K     .4 0786

 c m  3.2 2  [email protected] Ramadas Chennamsetti 78 Plate buckling R&DE (Engineers), DRDO
Minimum critical load Dπ 2 Dπ 2 N = K = .4 086 × = 73 72. MN / m 1cr b2 b2 = × = N N1cr b 73 .72 MN

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 ,3.2 m ,1 K .7 4790 Minimum value of m = ,2 K = .4 0786 ‘K’ is considered for estimation of N

cr [email protected] m = ,3 KRamadas= .4 2890 Chennamsetti 79 Plate and column buckling R&DE (Engineers), DRDO
Uniaxial load for plate buckling Dπ 2 N = K 1crp b2 N Dπ 2 π 2 Et 3 σ = 1crp = K = K 1crp t b2t b2t 12 (1−υ 2 ) π 2 => σ = K E 1crp 12 ()1−υ 2 ()b 2 t π 2 => σ = E = K 1crp C1 , C1 ()b 2 12 ()1−υ 2

t 80 [email protected] Ramadas Chennamsetti Plate and column buckling R&DE (Engineers), DRDO
Buckling of a column π 2 EI P = C cr 2 l 2 π 2 EAk 2 π 2 EA => P = C = C cr 2 l 2 2 (l )2 k π 2 => σ = Pcr = E crc C 2 2 Column A ()l k π 2 E σ = C Plate crp 1 ()b 2 Ramadast Chennamsetti 81 [email protected] Plate and column buckling R&DE (Engineers), DRDO
Critical stress for plate depends on thinness ratio = t/b – not on the length
depends on width
More thinner plate – lesser bucking load
Critical stress in column depends on slenderness ratio
Longer columns – lower critical load

Ramadas Chennamsetti 82 [email protected] Strain energy R&DE (Engineers), DRDO
In thin plate theory, out-of-plane shear τ τ τ stresses vanish => xz , yz and zz
Stress components contributing to strain σ σ τ energy => xx , yy and xy
Strain energy, Linear elastic material = 1 (σ ε +σ ε +τ γ ) U ∫ xx xx yy yy xy xy dV 2 V Ramadas Chennamsetti 83 [email protected] Strain energy R&DE (Engineers), DRDO
Strain energy,

= 1 {σ σ τ }{ε ε γ }T U ∫ xx yy xy xx yy xy dV 2 V  ∂2w        2 σ  1 υ 0 ε  1 υ 0  ∂x  xx   xx   2   Ez   Ez  ∂ w  σ  = − υ 1 0 ε  = − υ 1 0   yy −υ 2  +υ  yy −υ 2  +υ  ∂ 2 τ  1 1 γ  1 1  y   xy  0 0  xy  0 0  ∂2  2   2  w  2   ∂x∂y => {}σ = []C {}ε

Ramadas Chennamsetti 84 [email protected] Strain energy R&DE (Engineers), DRDO
Strain energy, 1 1 U = ∫{}{}σ T ε dV = ∫{}ε T []C {}ε dV 2 V 2 V  ∂2w ∂ 2w  ∂2w   +υ    2 2  2  ∂x ∂y  ∂x     ∂ 2 ∂2  ∂ 2 => = E + υ w + w  w 2 U 2 ∫  2 2  2 z dV 2()1−υ   ∂x ∂y  ∂x  V   2   ∂2w   + 2()1−υ     ∂x∂y    85 [email protected] Ramadas Chennamsetti  Strain energy R&DE (Engineers), DRDO
Infinitesimal volume, dV = dxdydz
Carry out integration over thickness => dz

+ h 2 h3 z 2dz = ∫ 12 − h 2 Eh 3 D = 12 ()1−υ 2

Ramadas Chennamsetti 86 [email protected] Strain energy R&DE (Engineers), DRDO
Simplify

2  ∂ 2 w ∂ 2 w    +    ∂ 2 ∂ 2  D  x x   U =  dxdy ∫  2 2 2 2  2 A  ∂ w ∂ w  ∂ w   − ()− υ  −     2 1 2 2   ∂x ∂x  ∂x∂y     

Strain energy – Finite Element Method – Total potential approach

Ramadas Chennamsetti 87 [email protected] R&DE (Engineers), DRDO

Ramadas Chennamsetti 88 [email protected]