Mechanics of composite lamina and laminate(Classical Laminate Theory)

1

For a linear isotropic material in a 3D stress state, the stress–strain relationships as per Hooke’s law at a point in an x–y–z orthogonal system in matrix form are :

Shear Modulus G = E/2(1+v)

Compliance matrix [S]

Stiffness matrix [C]

Strain energy per unit volume is,

For Composite material which are non-isotropic, generalized Hooke’s law presents 36 constants in stiffness matrix relating stresses & strains in an 1-2-3 orthogonal system

1 2

4

3

1

=

Stiffness for various types of materials

2

The stress strain relation in eq. 4 can be expressed aswhere σ4 = τ23 , σ5 = τ31 , σ6 = τ12 and ε4 = γ 23 , ε5 = γ 31 , ε6 = γ 12

So the strain energy equation can be expressed as

which is same as

Partial differential of SE equation gives and

So Cij = Cji and thus the 36 constants can be reduced to 21 constants for Anisotropic material

Other cases of material symmetry

# of constants

Monoclinic 13

Orthotropic 9

Transverse Orthotropic 5

Isotropic 2

3

Case of 2D angle lamina

Stresses in the global x-y and local 1-2 directions in the angle lamina are related to each other through the angle of the lamina, θ as,

For a 2D laminate all the laminae are not in same directions and some laminae are at an angle θ to the global x-y direction

where [T], the transformation matrix is defined as

c = Cos(θ) and s = Sin(θ)

4a

In terms of the strains, Strain transformations relation

where

is Reuter matrix [28]

So,

So,The stress-strain relations can be written in terms of transformed reduced stiffness matrix Q ij

5

4b

5

Thus the normal, shear stresses & strains are coupled for a generally orthotropic angular lamina and varies with fiber angle.

For thin lamina with no out of plane loads, the plane stress condition gives

where• E1 = longitudinal Young’s modulus (in direction 1)• E2 = transverse Young’s modulus (in direction 2)• ν12 = major Poisson’s ratio while νij = ratio of –ve

of normal strain in j to normal strain in i when

load is applied in i dir.• G12 = in-plane shear modulus (in plane 1–2)

5c

5b

Analysis of Failure theories of composite lamina

6

Theory Criteria Remarks

Maximum stress theory (failure occurs if max normal or shear stress in local axes exceeds ultimate strength of lamina)

Similar to isotropic materials and each stress component is considered separate to others

Maximum strain theory [St. Venant](failure occurs if max normal or shear strain in local axes exceeds ultimate strain of lamina)

Similar to max stress theory but due to poisson’s ratio effect, gives different results

Tsai hill theory[Von Mises distortion energy theory]

(failure occurs when when the distortion energy is greater than the failure distortion energy of the material)

• Considers the interaction among the three strength parameters σ1, σ2, τ12

• More conservative than above 2 for applicable stress

• Does not give mode of failure• Close to experimental results

Tsai-Wu theory [Beltrami strain energy] • Distinguishes between tensile and compressive strengths of lamina

• Gives better applicable stress than all other theories

• Close to experimental results like Tsai-Hill theoryH6 = 0,

Stress strain analysis of composite laminates

7

[0/–45/902/60/0] [0/–45/60]s[0/–45/902/60/0]

Stacking notations

Assumptions•each lamina is orthotropic, homogeneous and elastic•Shear strains in xz and yz are zero•Plane stress : σz = τxz = τyz = 0•Displacements are small compared to lamina thickness and No inter-lamina slip

Resultant forces and moments on a laminate

Laminate strain displacement relations

8

Relationship between displacements ‘w’ through the thickness of a plate to mid-plane displacements u, v and curvatures α

u0, v0, w0 are displacements at mid-plane and u, v, and w are displacement at any point in the thickness

u = u0 − zxα where α = u = u0 − z

Similarly in y-z plane v = v0 - z

εx =

γ xy =

=

εy =

Mid plane strain

Mid plane curvature

So,

5d

9

The strain-displacement equations can be written in matrix form as

Stresses change over at ply since Qij depend upon material and ply orientation

Global stresses, strain can be transformed to local system 1-2 and so can be used to check failure criteria

Next, we will find the mid-plane strains and curvatures when the loads are known

6

Laminate mid-plane strains and curvatures

10

Mid plane

Ply positioning inside a laminate

Consider a system of n laminae in a laminate each having thickness of tk with ‘h’ being total thickness of laminate. The z coordinate of various plies are,

Top surface Top surfaceBottom surface Bottom surface

6b

11

If,•Nx, Ny = normal force per unit length•Nxy = shear force per unit length•Mx, My = bending moments per unit length•Mxy = twisting moments per unit length

Integrating stresses global stresses in each lamina

In matrix form,

12

So Eq 6 for relation of global stresses and mid-plane strains and curvatures can be written as,

Since mid-plane strains and curvatures are independent of z coord. and [Qk] stiffness matrix is constant for each ply. Hence,

7a

7b

13

Considering ,

8a

8b

Extensional stiffness matrix

Bending stiffness matrix

Coupling stiffness matrix

9

14

Relates in-plane forces to in-plane

strains

Relates bending moment to plate

curvatures

Couples force and moments to mid-plane strains,

curvatures

10

15

Ref: Similar to Ex 4.3 from Mechanics of composite materials, A.K Kaw

A [0/30/–45] glass/epoxy laminate is subjected to a load of Nx = Ny = 1000 N/m., Mx = 50 N-mUsing the properties of unidirectional glass/epoxy below and assuming that each lamina is 5 mm thick, aim is to find

•Mid-plane strains and curvatures•Global and local stresses on top surface of 30° ply•Percentage of load, Nx, taken by each ply

Example Problem

Property Symbol Unit ValueLongitudinal elastic modulus E1 GPa 38.6

Transverse elastic modulus E2 Gpa 8.27

Major Poisson’s ratio ν12 0.26

Shear modulus G12 Gpa 4.14

ν21 = ν12 x E2/E1

= 0.26 X 8.27/38.6 = 0.0557

Steps

16

1. Find the value of the reduced stiffness matrix [Q] for each ply using its four elastic moduli, E1, E2, ν12, and G12 in Equation (5b).

2. Find the value of the transformed reduced stiffness matrix for each ply using the [Q] matrix calculated in step 1 and the angle of the ply in Equation (5c)

3. Knowing the thickness, tk, of each ply, find the coordinate of the top and bottom surface, hi, i = 1…, n, of each ply, using Equation (6b).

4. Use the matrices from step 2 and the location of each ply from step 3 to find the three stiffness matrices [A], [B], and [D] from Equation (9).

5. Substitute the stiffness matrix values found in step 4 and the applied forces and moments in Equation (10).

6. Solve the six simultaneous equations (10) to find the mid-plane strains and curvatures.

7. Now that the location of each ply is known, find the global strains in each ply using Equation (5d).

8. For finding the global stresses, use the stress–strain Equation (5).9. For finding the local strains, use the transformation Equation (4b).10.For finding the local stresses, use the transformation Equation (4a).

Solution

17

For the 0 deg ply,

= 39.17 GPa

= 2.182 GPa

= 8.392 GPa

= 4.14 GPa

39.17 2.18 02.18 8.39 00 0 4.14

For the 30 deg ply

26.38 7.58 9.54

7.58 11.07 3.78 9.54 3.78 9.12

For the 45 deg ply

17.11 9.38 -6.65 9.38 17.11 -6.65

-6.65 -6.65 10.27

Aij = 39.17 2.18 02.18 8.39 00 0 4.14

x(-0.0025)-(-0.0075)x109

+ x(0.0025)-(-0.0025)x109

26.38 7.58 9.54 7.58 11.07 3.78 9.54 3.78 9.12

+

17.11 9.38 -6.65 9.38 17.11 -6.65

-6.65 -6.65 10.27

x(0.0075)-(0.0025)x109

Aij = 41.33 9.57 1.459.57 18.29 -1.431.45 -1.43 11.77

x 106 Pa-m

18

Bij = 1/2 39.17 2.18 02.18 8.39 00 0 4.14

x(-0.0025)2-(-0.0075)2x109 +

x(0.0025)2-(-0.0025)2x109

26.38 7.58 9.54 7.58 11.07

3.78 9.54 3.78 9.12

+

17.11 9.38 -6.65 9.38 17.11 -6.65

-6.65 -6.65 10.27

x(0.0075)2-(0.0025)2x109

1/2

+ 1/2

Bij =

-9.79 -0.55 -1.66-0.55 2.1 -1.66-1.66 -1.66 3.6

x 105 Pa-m2

Dij = 1/3 39.17 2.18 02.18 8.39 00 0 4.14

x(-0.0025)3-(-0.0075)3x109 +

x(0.0025)3-(-0.0025)3x109

26.38 7.58 9.54 7.58 11.07

3.78 9.54 3.78 9.12

+

17.11 9.38 -6.65 9.38 17.11 -6.65

-6.65 -6.65 10.27

x(0.0075)3-(0.0025)3x109

1/3

+ 1/3

7.88 1.61 -0.794.89 3.53 -0.85-0.79 -0.85 2.03

Dij =

x 103 Pa-m3

19

1.45 -1.43 11.77

41.33 9.57 1.459.57 18.29 -1.43

-9.79 -0.55 -1.66-0.55 2.1 -1.66-1.66 -1.66 3.6

x 106

x 105

-9.79 -0.55 -1.66-0.55 2.1 -1.66-1.66 -1.66 3.6

x 105

7.88 1.61 -0.794.89 3.53 -0.85-0.79 -0.85 2.03

x 103

=

-17.290.7

39.9

-56.3-

4415.2-2936.5

m/m

1/m

x 106

30 0

top

-17.290.739.9

+ (-0.025)-56.3-4415.2

-2936.5

=-17.06

101.847.24

x 10-6

30 0

top30 deg ply

26.38 7.58 9.54

7.58 11.07 3.78 9.54 3.78 9.12

= x 109

-17.06

101.847.24

x 10-6 =77.2

65.29

117.6 x 104

Pa

Mid-plane strain, curvature

Global strains

Global stress

The global strains and stress in the 30° ply at the top surface

A

B

20

The local strains and stress in the 30° ply at the top surface

=0.75 0.25 0.866

0.25 0.75 -0.866-0.433 0.433 0.5

-17.06

101.847.24

x 10-6 =53.56

31.1775.08

x 10-6

0.75 0.25 0.866

0.25 0.75 -0.866-0.433 0.433 0.5

=77.2

65.29

117.6 x 104

Pa=

143.9

50.950.14

x 104 Pa

Local stress

Global strains

This process can be repeated for middle and bottom surface for 00 and -450 plies

C

D