Linear plate bending and laminate theory

Transcription

1 Linear plate bending and laminate theor 4M: Design Tools Eindhoven Universit of Technolog Linear plate bending A plate is a bod of which the material is located in a small region around a surface in the threedimensional space. A special surface is the mid-plane. Measured from a point in the mid-plane, the distance to both surfaces of the plate is equal. The geometr of the plate and position of its points is described in an orthonormal coordinate sstem, either Cartesian (coordinates {,, }) or clindrical (coordinates {r,θ,}).. Geometr The following assumptions about the geometr are supposed to hold here : the mid-plane is planar in the undeformed state, the mid-plane coincides with the global coordinate plane =, the thickness h is uniform. Along the edge of the plate a coordinate s is used and perpendicular to this a coordinate n.

2 s θ h n r. Eternal loads The plate is loaded with forces and moments, which can be concentrated in one point or distributed over a surface or along a line. The following loads are defined and shown in the figure : force per unit of area in the mid-plane : s (,),s (,) or s r (r,θ), s t (r,θ) force per unit of area perpendicular to the mid-plane : p(, ) or p(r, θ) force per unit of length in the mid-plane : f n (s), f s (s) force per unit of length perpendicular to the mid-plane : f (s) bending moment per unit of length along the edge : m b (s) torsional moment per unit of length along the edge : m w (s) p θ r s f (s) m b (s) s f s (s) m w (s) n f n (s).3 Cartesian coordinate sstem In the Cartesian coordinate sstem, three orthogonal coordinate aes with coordinates {,, } are used to identif material and spatial points. As stated before, we assume that the mid-plane of the plate coincides with the plane = in the undeformed state. This is not a restrictive assumption, but allows for simplification of the mathematics.

3 3 s θ h n r.3. Displacements The figure shows two points P and Q in the undeformed and in the deformed state. In the undeformed state the point Q is in the mid-plane and has coordinates (,,). The out-of-plane point P has coordinates (,, ). As a result of deformation, the displacement of the mid-plane point Q in the (,, )-coordinate directions are u, v and w, respectivel. The displacement components of point P, indicated as u, u and u, can be related to those of point Q. θ θ θ P T P S R u Q v w Q u = u SR = u QRsin(θ ) u = v ST = v QT sin(θ ) u = w+sq QR = cos(θ ) QT = cos(θ ) SQ = cos(θ) u = u sin(θ )cos(θ ) u = v cos(θ )sin(θ ) u = w+ cos(θ)

4 4 No out-of-plane shear When it is assumed that there is no out-of-plane shear deformation, the so-called Kirchhoff hpotheses hold : straight line elements, initiall perpendicular to the mid-plane remain straight, straight line elements, initiall perpendicular to the mid-plane remain perpendicular to the midplane. The angles θ and θ can be replaced b the rotation angles φ and φ respectivel. This is illustrated for one coordinate direction in the figure. φ P Q P Q φ θ = φ ; θ = φ ; θ = φ Small rotation With the assumption that rotations are small, the cosine functions approimatel have value and the sine functions can be replaced b the rotation angles. These rotations can be epressed in the derivatives of the -displacement w w.r.t. the coordinates and. cos(φ) = cos(φ ) = cos(φ ) = sin(φ ) = φ = w, ; sin(φ ) = φ = w, u = u w, u = v w, u = w+ } Derivatives are denoted as e.g. ( ) = ( ), and ( ) = ( ),

5 5 Constant thickness The thickness of the plate ma change due to loading perpendicular to the mid-plane or due to contraction due to in-plane deformation. The difference between and is obviousl a function of. Terms of order higher than are neglected in this epression. It is now assumed that the thickness remains constant, which means that ζ(,) = has to hold. With the assumption, which is correct for small deformations and thin plates, the displacement components of the out-of-plane point P can be epressed in mid-plane displacements. (,) = (,) (,) = ζ(,)+η(,) + O( 3 ) h = ζ(,)( h ) ζ(,)( h ) h = ζ(,) = u (,,) = u(,) w, u (,,) = v(,) w, u (,,) = w(,)+η(,).3. Curvatures and strains The linear strain components in a point out of the mid-plane, can be epressed in the mid-plane strains and curvatures. A sign convention, as is shown in the figure, must be adopted and used consistentl. It was assumed that straight line elements, initiall perpendicular to the mid-plane, remain perpendicular to the mid-plane, so γ and γ should be ero. Whether this is true will be evaluated later. ε = u, = u, w, = ε κ ε = u, = v, w, = ε κ γ = u, + u, = u, + v, w, = γ κ ε = u, = η(,) γ = u, + u, = η, γ = u, + u, = η, ε = ε κ ε ε ε ε κ κ κ κ

6 6.3.3 Stresses It is assumed that a plane stress state eists in the plate : σ = σ = σ =. Because the plate ma be loaded with a distributed load perpendicular to its surface, the first assumption is onl approimatel true. The second and third assumptions are in consistence with the Kirchhoff hpothesis, but has to be relaed a little, due to the fact that the plate can be loaded with perpendicular edge loads. However, it will alwas be true that the stresses σ, σ and σ are much smaller than the in-plane stresses σ, σ, and σ. σ σ σ σ.3.4 Isotropic elastic material behavior For linear elastic material behavior Hooke s law relates strains to stresses. Material stiffness (C) and compliance (S) matrices can be derived for the plane stress state. ε ε γ σ σ σ = E = E ν ν ν (+ν) ν ν ( ν) σ σ σ ε ε γ ; ε = ν E (σ + σ ) short notation : ε = Sσ σ = S ε = Cε = C(ε κ ) Inconsistenc with plane stress The assumption of a plane stress state implies the shear stresses σ and σ to be ero. From Hooke s law it then immediatel follows that the shear components γ and γ are also ero. However, the

7 7 strain-displacement relations result in non-ero shear. For thin plates, the assumption will be more correct than for thick plates. ε = η η(,) = ε = ν E (σ ν + σ ) = ( ν) (w, + w, ) γ = γ = ν ( ν) (w, + w, ) ν ( ν) (w, + w, ).3.5 Cross-sectional forces and moments Cross-sectional forces and moments can be calculated b integration of the in-plane stress components over the plate thickness. A sign convention, as indicated in the figure, must be adopted and used consistentl. The stress components σ and σ are much smaller than the relevant in-plane components. Integration over the plate thickness will however lead to shear forces, which cannot be neglected. Ñ = = M D = N N N M M M h/ h/ = = h/ h/ h/ h/ d = {C(ε κ )}d = σ h/ hcε h/ h/ d = {C(ε σ κ )}d = h/ h3 Cκ h/ σ d ; D = σ d h/ N N N D N D M M M M Stiffness- and compliance matri Integration leads to the stiffness and compliance matri of the plate.

8 8 [ Ñ M ] [ Ch = Ch 3 / ][ ε κ ] [ ε κ ] [ S/h = S/h 3 ][ Ñ M ].3.6 Equilibrium equations Consider a small column cut out of the plate perpendicular to its plane. Equilibrium requirements of the cross-sectional forces and moments, lead to the equilibrium equations. The cross-sectional shear forces D and D can be eliminated b combining some of the equilibrium equations. N, + N, + s = N, + N, + s = D, + D, + p = M, + M, D = M, + M, D = M, + M, + M, + p = The equilibrium equation for the cross-sectional bending moments can be transformed into a differential equation for the displacement w. The equation is a fourth-order partial differential equation, which is referred to as a bi-potential equation. M M M Eh 3 ν = ( ν ν ) ( ν) M, + M, + M, + p = w, w, w, 4 w w w = ( + )( w + w ) = p B B = Eh 3 ( ν = plate modulus ).3.7 Orthotropic elastic material behavior A material which properties are in each point smmetric w.r.t. three orthogonal planes, is called orthotropic. The three perpendicular directions in the planes are denoted as the -, - and 3-coordinate aes and are referred to as the material coordinate directions. Linear elastic behavior of an orthotropic material is characteried b 9 independent material parameters. The constitute the components of the compliance (S) and stiffness (C) matri.

9 9 ε ε ε 33 γ γ 3 γ 3 = E ν E ν 3 E3 ν E E ν 3 E3 ν 3 E ν 3 E E3 G G 3 G 3 σ σ σ 33 σ σ 3 σ 3 ε = Sσ σ = S ε = Cε C = ν 3 ν 3 E E 3 ν 3 ν 3 +ν ν 3 ν 3 +ν E E 3 ν 3 ν 3 ν ν 3 +ν 3 ν ν 3 +ν 3 E E 3 ν ν 3 +ν 3 E E 3 ν ν 3 +ν 3 ν ν E E E E E E 3 E E 3 E E G G 3 G 3 with = ( ν ν ν 3 ν 3 ν 3 ν 3 ν ν 3 ν 3 ν ν 3 ν 3 ) E E E 3 Orthotropic plate In an orthotropic plate the third material coordinate ais (3) is assumed to be perpendicular to the plane of the plate. The - and -directions are in its plane and rotated over an angle α (counterclockwise) w.r.t. the global - and -coordinate aes. For the plane stress state the linear elastic orthotropic material behavior is characteried b 4 independent material parameters : E, E, G and ν. Due to smmetr we have ν = E E ν. α ε ε γ σ σ σ E ν E = ν E E G = ν ν σ σ σ E ν E ν E E ( ν ν )G ε ε γ

10 Using a transformation matrices T ε and T σ, for strain and stress components, respectivel, the strain and stress components in the material coordinate sstem (inde ) can be related to those in the global coordinate sstem. c = cos(α) ; s = sin(α) ε c s cs ε = s c cs γ cs cs c s σ σ σ = c s cs s c cs cs cs c s ε ε γ σ σ σ = T ε ε ε = T σ σ σ ε = T ε ε = T ε σ = T σ σ idem = T ε σ σ σ ε = T ε S σ = T ε S T σ = σ Sσ = T σ C ε = T σ C T ε = C ε ε.4 Clindrical coordinate sstem In the clindrical coordinate sstem, three orthogonal coordinate aes with coordinates {r, θ, } are used to identif material and spatial points. It is assumed that the mid-plane of the plate coincides with the plan = in the undeformed state. We onl consider aismmetric geometries, loads and deformations. For such aismmetric problems there is no dependenc of the clindrical coordinate θ. Here we take as an etra assumption that there is no tangential displacement. p(r) f θ f s m b mw f n r

11 .4. Displacements All assumptions about the deformation, described in the former section, are also applied here : no out-of-plane shear (Kirchhoff hpotheses), small rotation, constant thickness. The displacement components in radial (u r ) and aial (u ) direction of an out-of-plane point P can then be epressed in the displacement components u and w of the corresponding same and coordinates point Q in the mid-plane. φ P Q P Q φ r u r (r,θ,) = u(r,θ) w,r u (r,θ,) = w(r,θ)+η(r,θ).4. Curvatures and strains The linear strain components in a point out of the mid-plane, can be epressed in the mid-plane strains and curvatures. A sign convention, as indicated in the figure, must be adopted and used consistentl. It was assumed that straight line elements, initiall perpendicular to the mid-plane, remain perpendicular to the mid-plane, so γ r =. ε rr = u r,r = u,r w,rr = ε rr κ rr ε tt = r u r = r u r w,r = ε tt κ tt γ rt = u r,t + u t,r = ε = ε κ ε = u, = η(r,) ε r = u r, + u,r = η,r γ t = u t, + u,t =

12 ε rr ε tt κ tt κ rr r θ.4.3 Stresses A plane stress state is assumed to eist in the plate. The relevant stresses are σ rr and σ tt. Other stress components are much smaller or ero. σ tt σ rr θ r.4.4 Isotropic elastic material behavior For linear elastic material behavior Hooke s law relates strains to stresses. Material stiffness (C) and compliance (S) matrices can be derived for the plane stress state.

13 3 [ εrr ε tt [ σrr σ tt ] = [ ν E ν ] = E ν [ ν ν ][ σrr σ tt ][ εrr ε tt ] ] ; ε = ν E (σ rr + σ tt ) short notation : ε = Sσ σ = S ε = Cε = C(ε κ ).4.5 Cross-sectional forces and moments Cross-sectional forces and moments can be calculated b integration of the in-plane stress components over the plate thickness. A sign convention, as indicated in the figure, must be adopted and used consistentl. The stress component σ r is much smaller than the relevant in-plane components. Integration over the plate thickness will however lead to a shear force, which cannot be neglected. Ñ = M = D = [ Nr N t [ Mr M t h h ] h [ σrr = h σ tt ] h = h σ r d ] [ σrr σ tt h d = h C(ε κ ) d = hcε ] h d = h C(ε κ )d = h3 Cκ D N r N t M t M r r θ Stiffness- and compliance matri Integration leads to the stiffness and compliance matri of the plate. [ Ñ M ] [ Ch = Ch 3 / ][ ε κ ] [ ε κ ] [ S/h = S/h 3 ][ Ñ M ]

14 4.4.6 Equilibrium equations Consider a small column cut out of the plate perpendicular to its plane. Equilibrium requirements of the cross-sectional forces and moments, lead to the equilibrium equations. The cross-sectional shear force D can be eliminated b combining some of the equilibrium equations. N t dr M t dr (M r + M r,r dr)(r+dr)dθ N r rdθ Drdθ prdrdθ (N r + N r,r dr)(r+dr)dθ dθ r M r rdθ (D+D,r dr)(r+dr)dθ M t dr N t dr N r,r + r (N r N t )+s r = rd,r + D+rp = rm r,r + M r M t rd = rm r,rr + M r,r M t,r + rp = The equilibrium equation for the cross-sectional bending moments can be transformed into a differential equation for the displacement w. The equation is a fourth-order partial differential equation, which is referred to as a bi-potential equation. [ Mr M t ] Eh 3 [ ][ ν w,rr = ( ν ) ν r w,r rm r,rr + M r,r M t,r + rp = d 4 w dr 4 + d 3 w r dr 3 d w r dr + dw r 3 dr = p B B = ( d ] dr + r Eh 3 ( ν = plate modulus ) )( d d w dr dr + r ) dw = p dr B

15 5.4.7 Solution The differential equation has a general solution comprising four integration constants. Depending on the eternal load, a particular solution w p also remains to be specified. The integration constants have to be determined from the available boundar conditions, i.e. from the prescribed displacements and loads. The cross-sectional forces and moments can be epressed in the displacement w and/or the strain components. ( d w M r = B M t = B w = a + a r + a 3 ln(r)+a 4 r ln(r)+w p ) dw dr dr + ν r ( dw r dr + ν d w dr ) D = dm r dr + r (M r M t ) = B ( d 3 w dr 3 + d w r dr ) dw r dr.4.8 Eamples This section contains some eamples of aismmetric plate bending problems. Plate with central hole Solid plate with distributed load Solid plate with vertical point load Plate with central hole A circular plate with radius R has a central hole with radius R. The plate is clamped at its outer edge and is loaded b a distributed load q perpendicular to its plane in negative -direction. The equilibrium equation can be solved with proper boundar conditions, which are listed below. q r R R

16 6 boundar conditions ( ) dw w(r = R) = (r = R) = dr m b (r = R) = M r (r = R) = f (r = R) = D(r = R) = differential equation general solution d 4 w dr 4 + d 3 w r dr 3 d w r dr + dw r 3 dr = q B w = a + a r + a 3 ln(r)+a 4 r ln(r)+w p particulate solution w p = αr 4 substitution 4α + 48α α + 4α = q B α = q 64B solution w = a + a r + a 3 ln(r)+a 4 r ln(r) q 64B r4 The 4 constants in the general solution can be solved using the boundar conditions. For this purpose these have to be elaborated to formulate the proper equations. dw dr = a r+ a 3 r + a 4(r ln(r)+r) ( d w M r = B dr + ν ) dw r dr = B ( M t = B ν d w = B q 6B r3 { (+ν)a ( ν) r a 3 + (+ν)ln(r)a 4 +(3+ν)a 4 (3+ν) q ) dw dr 6B r dr + r { (+ν)a +( ν) r a 3 + (+ν)ln(r)a 4 +(3ν + )a 4 (3ν + ) q 6B r D = dm r dr + r (M r M t ) ( = B 4 r a 4 + q ) B r } } The equations are summaried below. The 4 constants can be solved. After solving the constants, the vertical displacement at the inner edge of the hole can be calculated.

17 7 w(r = R) = a + 4a R + a 3 ln(r)+4a 4 R ln(r) q 4B R4 = dw dr (r = R) = 4a R+ R a 3 + 4a 4 Rln(R)+a 4 R q B R3 = { M r (r = R) = B (+ν)a ( ν) R a 3+ ((+ν)ln(r)+(3+ν))a 4 (3+ν) q 6B R} = ( D(r = R) = B 4 R a 4 + q } B R = w(r = R) = qr4 64B { 9(5+ν)+48(3+ν)ln() 64(+ν)(ln()) } 5 3ν Solid plate with distributed load A solid circular plate has radius R and uniform thickness. It is clamped at its outer edge and loaded with a distributed force per unit of area q in negative -direction. The equilibrium equation can be solved with proper boundar conditions, which are listed below. Besides the obvious boundar conditions, there are also two smmetr conditions, one for the deformation and the other for the load. q r R boundar conditions differential equation general solution w(r = R) = ( dw dr ( ) dw (r = R) = dr ) (r = ) = D(r = ) = d 4 w dr 4 + d 3 w r dr 3 d w r dr + dw r 3 dr = q B w = a + a r + a 3 ln(r)+a 4 r ln(r)+w p particulate solution w p = αr 4 substitution

18 8 4α + 48α α + 4α = p B α = q 64B solution w = a + a r + a 3 ln(r)+a 4 r ln(r) q 64B r4 The 4 constants in the general solution can be solved using the boundar and smmetr conditions. For this purpose these have to be elaborated to formulate the proper equations. The two smmetr conditions result in two constants to be ero. The other two constants can be determined from the boundar conditions. After solving the constants, the vertical displacement at r = can be calculated. dw dr = a r+ a 3 r + a 4(r ln(r)+r) q 6B r3 D = dm r dr + r (M r M t ) ( = B 4 r a 4 + q ) B r boundar conditions ( ) dw (r = ) = a 3 = dr D(r = ) = a 4 = ( dw dr displacement in the center ) (r = R) = a R q 6B R3 = a = q w(r = R) = a + a R 3B R q 64B R4 = a = q 64B R4 w(r = ) = q 64B R4 Solid plate with vertical point load A solid circular plate has radius R and uniform thickness d. The plate is clamped at its edge. It is loaded in its center (r = ) b a point load F perpendicular to its plane in the positive -direction. The equilibrium equation can be solved using appropriate boundar conditions, listed below. Besides the obvious boundar conditions, there is also the smmetr condition that in the center of the plate the bending angle must be ero. In the same point the total cross-sectional force K = πd must equilibrate the eternal force F.

19 9 F r R boundar conditions differential equation general solution w(r = R) = ( dw dr ( ) dw (r = R) = dr ) (r = ) = K(r = ) = F d 4 w dr 4 + d 3 w r dr 3 d w r dr + dw r 3 dr = w = a + a r + a 3 ln(r)+a 4 r ln(r) The derivative of w and the cross-sectional load D can be written as a function of r. The smmetr condition results in one integration constant to be ero. Because the derivative of w is not defined for r =, a limit has to be taken. The cross-sectional equilibrium results in a known value for another constant. The remaining constants can be determined from the boundar conditions at r = R. After solving the constants, the vertical displacement at r = can be calculated. dw dr = a r+ a 3 r + a 4(r ln(r)+r) D = dm r dr + r (M r M t ) = B ( 4 r a 4 boundar conditions ( ) dw { lim = lim a r+ a } 3 r dr r r + a 4(r ln(r)+r) = a 3 = K(r) = πrd(r) = 8πBa 4 = F a 4 = F 8πB w(r = R) = a + a R + F 8πB R ln(r) = a = FR ( ) 6πB dw (r = R) = Ra + F dr 8πB (Rln(R)+R) = a = F (ln R+) 6πB displacement at the center FR { ( r ) ( r ) ( r w(r = ) = lim + ln r 6πB R R R) } = FR 6πB )

20 The stresses can also be calculated. The are a function of both coordinates r and. We can conclude that at the center, where the force is applied, the stresses become infinite, due to the singularit, provoked b the point load. σ rr = 3F { ( r πd 3 +(+ν)ln R)} σ tt = 3F { ( r πd 3 ν +(+ν)ln R)}

21 Laminates Laminates are plates, which are made b stacking a number of laers, also called plies or lamina s. Each pl can be fabricated from various materials, having different mechanical properties within a wide range. The properties ma be isotropic, completel anisotropic, or orthotropic. The latter occurs when a pl is a composite material, consisting of a matri which is enforced b long fibers in one direction. The material directions are denoted as (fiber, longitudinal) and (transversal). To increase the bending stiffness of the laminate, the thickness can be enlarged, b appling a laer of filler material. e.g. a foam, which is considered to be isotropic. The figure shows an eploded view of a laminate with fiber reinforced plies. Also a foam laminate is shown. The mechanical properties of the laminate are determined b its stiffness matri, which can be calculated from the properties of the plies, their thicknesses and their stacking sequence. The behavior of the laminate is based on linear plate bending theor.. Lamina strain For one pl (k) the strain components in the global coordinate sstem can be related to the strain in the mid-plane and the curvature of the mid-plane. The strain components in the material coordinate sstem indicated with superscript are calculated with the transformation matri T ε. k k α with k () = ε ε κ T ε = k ε () = T k ε,kε () c s cs s c cs cs cs c s

22 . Lamina stress Assuming linearl elastic material behavior according to Hooke s law, the stress components in the material directions can be determined, using the material stiffness matri C. Stress components in the global coordinate sstem are calculated with the transformation matri T σ. with k σ = C k k kε σ = T σ,k C k T k ε,kε = C k kε = C k κ ) (ε T σ = c s cs s c cs cs cs c s ; T σ = c s cs s c cs cs cs c s k k.3 ABD-matri The cross-sectional forces and moments can be determined b summation of the integrated stress components over each individual pl. The result is the so-called ABD-matri, which relates crosssectional forces and moments to mid-plane strains and curvatures. k k Ñ k = k M k = k σ k d = ( k k )C k ε ( k k )C k κ = A k ε + B kκ k summation over all plies k σ k d = ( k k )C k ε + 3 (3 k 3 k )C kκ = B k ε + D kκ

23 3 n Ñ = k = k=ñ Aε + Bκ [ ] [ ][ Ñ A B = B D ε M κ ] n ; = k = M k=m Bε + Dκ [ ] [ ][ a b Ñ ε = b d κ M ] The ABD-matri characteries the mechanical behavior of the laminate, as it relates the cross-sectional loads to strains and curvatures of the mid-plane. The figure shows the influence of some components of the ABD-matri. A A A 3 B B B 3 A A 3 A 33 B B B 3 B 3 B 33 B 3 D D 3 D D D 3 D 33.4 Thermal and humidit loading When the laminate is subjected to a change in temperature, thermal strains will occur due to thermal epansion. In each pl these strains will generall be different, leading to deformation. Analogousl, the laminate can be placed in a humid environment, which also causes strains caused b absorption. The thermal and humidit strains can be calculated in each pl, given its coefficients of thermal epansion (α and α ) and humid driven epansion (β and β ). To calculate stresses these strains ˆε k must be subtracted from the mechanical strains ε k.

24 4 The cross-sectional forces and moments are related to the mid-plane strains and curvatures b the ABD-matri. The forces caused b thermal and humidit epansion are simpl added. strains in pl k due to temperature change T k and absorption c k ˆε α β ˆε = α T k + β c k k ˆγ ˆε = α k T k + k β c k stresses due to thermal(absorption) and mechanical strains ε k k σ k (ε k ˆε k ) σ k = C k (ε k ˆε k) = C k (ε κ k) ˆε forces and moments in pl k : Ñ k = k k k = M k σ k d = A kε + B kκ A k ˆε k Ñ = n Ñ k k= n k σ k d = B kε + D kκ B k ˆε k M = [ Ñ [ M ε κ ] [ A B = B D ] [ a b = b d ][ ε ]{[ κ Ñ M ] [ ˆÑ ˆM ] + [ ˆÑ ˆM ] ]} k= M k.5 Stacking The fibers in the plies can be oriented randoml, however, mostl a certain orientation relative to the orientation in other plies is chosen, resulting in cross-pl, angle-pl and regular angle-pl laminates. Fiber orientation in plies above and under the mid-plane can be smmetric and anti-smmetric. The resulting mechanical behavior is related to certain components in the ABD-matri. The notation of the pl-orientations is illustrated with the net eamples : [///9/9/9/45/..] total stacking and orientation sequence is given [ 3 /9 /45/..] total stacking and orientation sequence given; inde gives number of plies [ 3 /9 /45/ 45 3 ] s smmetric laminate

25 5 cross-pl orthotropic plies material directions = global directions. A 3 = A 3 = angle-pl orthotropic plies each pl material direction is rotated over α o w.r.t. global direction. regular angle-pl orthotropic plies subsequent plies have material direction rotated alternatingl over α o and α o w.r.t. the global -ais. even number of plies A 3 = A 3 = smmetric smmetric stacking w.r.t. mid-plane B = anti-smm. anti-smmetric stacking w.r.t. mid-plane D 3 = D 3 = quasi-isotropic α k = k π n with k =,..,n (n = number of plies).5. Recommendations for laminate stacking There are no general rules for the number of plies in a laminate, their (mutual) fiber orientation and their stacking sequence. For most applications, however, there are some recommendations. Choose a smmetric laminate. Because B =, there is no coupling between forces and curvatures and between strains and moments. Minimie material direction differences in subsequent plies. Large differences lead to high inter-laminar shear stresses, which ma cause delamination. Minimie stiffness differences of subsequent plies. Large difference lead to high inter-laminar shear stresses. Avoid neighboring plies with the same orientation. The same orientation results in high inter-laminar shear stresses under thermal loading. Ensure negative inter-laminar normal stresses (pressure). Negative stresses (tensile) ma lead to delamination..6 Damage Several damage phenomena can occur in a laminate : fibre rupture fibre buckling matri cracking

26 6 fibre-matri de-adhesion interlaminar delamination The occurence of these damage phenomena can be monitored with a damage criterion..6. Damage : Tsai-Hill For anisotropic materials the ield criterion of Tsai-Hill is often used. Either tensile (T i ) or compressive (C i ) ield limits are used. The shear limit is denoted as S. The ield criterion is used in each pl of the laminate in the material coordinate sstem. Failure can also occur b eceeding the maimum strain in longitudinal (ε f l ) or transversal (ε ft ) direction. Tsai-Hill ield criterion for tensile loading ( ) σ Tl ( ) σ σ Tl Tsai-Hill ield criterion for compressive loading ( σ C l ) ( σ σ C l ) ) σ +( σ +( Tt S = ) ) ) σ +( σ +( Ct S =.6. Damage : ILSS An important failure mode in laminates is the loss of adhesion between plies. This occurs when the interlaminar shear stress eceeds a limit value : the interlaminar shear strength (ILSS). The interlaminar shear stresses (ils) are calculated as the difference between the global stress component values between two adjacent plies, indicated here as ( ) t (top) and ( ) b (bottom). ils = σ b σ t ; ils = σ b σ t ; ils = σ b σ t.7 Material parameters for some materials Material parameters for Carbon fibre, HP-PE fibre and epo are listed in the table. Material parameters for fibre-matri composites are also listed in the table. HP-PE fibres can be plasma treated (pl.tr.)to improve adhesion to the epo matri The fibre volume fraction is V f =.5. from PhD thesis T.Peijs

27 7 fibre and matri parameters E E G G 3 ν ν [GPa] [GPa] [GPa] [GPa] [-] [-] Carbon HP-PE Epo fibre/epo composite parameters V f =.5 PVOH HP-PE HP-PE Aramid pl.tr. untr. E l [GPa] E t [GPa] G lt [GPa].9.. ν lt [-].3.3 T l [MPa] T t [MPa] C l [MPa] C t [MPa] 44 S [MPa] ε f l [%] ε ft [%] ILSS [MPa] > Eamples Some laminates have been modeled and analed with Matlab. For each eample the laminate buildup and the loading is presented. The resulting stiffness matri is shown. The loading results in strains and curvatures of the mid-plane. The deformation is visualied and the pl-strains and stresses are plotted, both in global directions and in fiber directions. Random 4-pl laminate Smmetric 8-pl laminate Anti-smmetric 8-pl laminate.8. Random 4-pl laminate ============================================================ Laminate build-up (lam) - + ang El Et nutl Gl

28 Mechanical load {ld) [N N N M M M] = [ ] Stiffness matri.57e e+7 4.5e e e+4 -.7e e+7.5e+8.83e e+4-4.6e e+4 4.5e+7.83e e+7 -.7e e e e e+4 -.7e e+ 9.8e+ 5.7e e+4-4.6e e+4 9.8e+.e e+ -.7e e e+4 5.7e+ 4.73e+.43e Strains in the mid-plane (e) [ e e e ] = [ 3.8e e-4-3.e-4 ] Curvatures of the mid-plane (kr) [ k k k ] = [ 4.784e e e- ] ============================================================ [mm] [mm] σ [Pa] σ [Pa] w. [mm] [mm] ε [ ] ε [ ] 3.8. Smmetric 8-pl laminate ============================================================ Laminate build-up (lam) - + ang El Et nutl Gl

29 Mechanical load {ld) [N N N M M M] = [ ] Stiffness matri.6e+8.3e+7 -.8e+6 -.e-.6e-3.5e-4.3e e+7 -.7e+6.6e e-3-5.7e-6 -.8e+6 -.7e+6 4.9e+7.5e-4-5.7e-6-4.4e-3 -.e-.6e-3.5e-4 7.4e+.55e+.97e-.6e e-3-5.7e-6.55e+ 7.4e+ 4.58e-.5e-4-5.7e-6-4.4e-3.97e- 4.58e- 3.e Strains in the mid-plane (e) [ e e e ] = [ 9.57e e-7.444e-6 ] Curvatures of the mid-plane (kr) [ k k k ] = [.6e+.97e+ 3.9e+ ] ============================================================ [mm] [mm] σ [Pa] σ [Pa] 9 4 w [mm] [mm] ε [ ] ε [ ].8.3 Anti-smmetric 8-pl laminate ============================================================ Laminate build-up (lam) - + ang El Et nutl Gl

30 Mechanical load {ld) [N N N M M M] = [ ] Stiffness matri.6e+8.3e+7.6e- -.e-.6e e+.3e e+7.e-.6e e-3 -.7e+.6e-.e- 4.9e e+ -.7e+ -4.4e-3 -.e-.6e e+ 7.4e+.55e+ 4.8e-9.6e e-3 -.7e+.55e+ 7.4e+.9e e+ -.7e+ -4.4e-3 3.7e-8.9e-8 3.e Strains in the mid-plane (e) [ e e e ] = [ 4.859e e e-5 ] Curvatures of the mid-plane (kr) [ k k k ] = [.5e+.5e+ 3.7e+ ] ============================================================ [mm] [mm] σ [Pa] σ [Pa] 9 4 w [mm] [mm] ε [ ] ε [ ]