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Macro-Mechanics

ME 429 Introduction to Composite Materials



ME429: Int. to Composite Materials

Macro Mechanics – Intro

• Micromechanics allowed us to obtain average

material properties for a composite ply frommatrix and fiber properties.

• Ply mechanics considers average material

properties for a single ply with any fiberorientation angle.

• Macromechanics will put several plies together in

order to evaluate stresses and strains within each

ply when a composite laminate is subject to forceor displacement boundary conditions




Deformations in a Plate




Plate Assumptions

• Each layer is in state of plane stress: Length of

line AD is constant.-- this neglects normal strain in z direction

• Layers are perfectly bonded together, therefore

displacements are continuous across boundaries• Kirchoff-Love hypothesis: Normals to the center

line remain normal to center line after deformation

-- this neglects through the thickness shears




Stress Profile for Homogeneous Material• Continuous stress and strain (Under effect of Moment)




Stress and Strain Profiles for Laminates

(Bending)•

Continuous strain• Discontinuous stress




Displacements

• u(x,y,z)=u0(x,y)-zfx(x,y)

• v(x,y,z)=v0(x,y)-zfy(x,y)

– if plate is thin

• assuming:

– length of A-D is constant ezz~0

– fx and fx are very small

• w(x,y,z)=w0(x,y)

y

w)y,x(

xw)y,x(

0y

0x

f

f

Displacements of all points are related to the mid-plane displacement




Strain-Displacement Relations

Linear strain-displacement relationship:

x

w

z

u)z,y,x(

yw

zv)z,y,x(

x

v

y

u)z,y,x(

x

v

)z,y,x(

x

u)z,y,x(

xz

yz

xy

x

x

g

g

g

e

e




Strains and CurvaturesSubstituting displacement from the previous page and

simplifying yields:Strains in terms of midplane strains and curvatures

curvaturesz

strains

surface

mid

plate

in

strains

z

xy

y

x

xy0

y0

x0

xy

y

x

g

e

e

g

e

e




Midsurface Strains and Curvatures

• Midsurface Strains

• Curvatures

x

v

y

u)y,x(

y

v)y,x(

x

u

)y,x(

00xy

0

0y

0

0

x

0

g

e

e

yx

w2)y,x(

yw

y)y,x(

x

w

x)y,x(

0

2

xy

2

02y

y

2

0

2

xx

f

f

Due to

bending

Due to

twisting




Force and Moment Resultants

Resultant force have units of force per length of laminate (N/m)

Resultant moments have units of length times force per length of laminate

(N.m/m)




Stress Resultants for a Single Ply• Stress resultants

•

Moment Resultants

• What are the units?

dz

N

N N 2/t

2/txy

y

x

xy

y

x

dzzMM

M2/t

2/txy

y

x

xy

y

x




Geometry




Stress Resultants for a Laminate

•

Simply sum over the number of layers• Stress resultants

• Moment Resultants

k is the layer number counting from the bottom up N

N

1k

2/t

2/txy

y

x

xy

y

x

dz

N

N

N

N

1k

2/t

2/txy

y

x

xy

y

x

dzzMM

M

k

k




Plate Stiffness and Compliance

• We already know the stress strain

relationships for a single ply

k

xy

y

x

k

662616

262212

161211

k

xy

y

x

xyk k

xyk

12k k

12k

QQQ

QQQ

QQQ

Q

Q

g

e

e

e

e




Laminate Stiffness and Compliance

• Substitute plate stiffness relationships into

laminate stress and moment resultant

equations in terms of strains and curvatures

g

e

e

xy

y

x

xy0

y0

x0

662616662616

262212262212

161211161211

662616662616

262212262212

161211161211

xy

y

x

xy

y

x

DDDBBB

DDDBBBDDDBBB

BBBAAA

BBBAAA

BBBAAA

M

MM

N

N

N




A-B and D matrices

– where i,j =1,2,6

– zk is the coordinate of the top and bottom of ply surface

• 18 Constants

N

1k

1k 3

k 3

k ijij

N

1k

1k 2

k 2

k ijij

N

1k

1k k k ijij

)zz(Q3

1D

)zz(Q2

1B

)zz(QA




A-B-D Matrices (part 2)

• Coefficients Aij, Bij, Dij are functions of

thickness, orientation, stacking sequence

and material properties of each layer

• [A] =in-plane stiffness matrix

• [B] = bending-extension coupling matrix

• [D] =bending stiffness matrix

(B=0 if laminate is symmetric wrt mid-plane)




Coupling of Different Loads



Computation of Stresses

Section 6.2 pg. 142




Stresses within Plies

• Several steps involved

1) Calculate strains through the thickness from

stress and moment resultants

2) Convert strains to material coordinate system

3) Calculate stresses within each ply in materialcoordinate system

• IMPORTANT: We cannot simply transform from

exy to xy using Q because

– may be several plies with different fiber orientations

– strain can vary with thickness if is nonzero




Stresses within Plies (2)• Calculate Strains Through Thickness

ee

e

z

M

NF

o

o




Stresses within Plies (3)• Convert Strains to Material Coordinates

• Calculate Stresses within Each Ply

xy

1

12 ]R ][T][R [ ee

1212 ]Q[ e





• Given Material Properties, Stiffness matrix [Q] can be calculated.

•

Calculate [Q]• Calculate the [A], [B], [D], and [H] matrices.

• Calculate mid-surface strains, curvatures, and interlaminar shearstrains.

• Once these are known, the strains at any point through the thicknessof the plate can be determined.

•The stress at any point throughout the plate can then be determined.

Steps:





• Given Material Properties, Stiffness matrix [Q] can be

calculated using 5.23. (pg. 114)

12

2212

2121

G00

0/E/E

0/E/E

]Q[




Computation of Stresses• [Q] bar can then be calculated using 5.48 or 5.49. (pg. 123)

5.48

5.49





• The [A], [B], [D], and [H] matrices can be calculated using 6.16.

(pg. 138)





• Calculate mid-surface strains, curvatures, and interlaminar shear strains

using 6.17. (pg. 139)





•

The strains at any point through the thickness of the plate can bedetermined using 6.20. (pg. 142)

curvatureszstrainssurface

mid

platein

strains

zxy

y

x

xy0

y0

x0

xy

y

x

ge

e

ge

e





• The stress at any point throughout the plate can be determined using 6.21.

(pg. 143)

k

xy

y

x

k

662616

262212

161211

k

xy

y

x

xyk k

xyk

12k k

12k

QQQ

QQQ

QQQ

Q

Q

ge

e

e

e




Computation of StressesExample 6.2

• Compute the stresses at the bottom surface of a single-layer plate subjected to Mx = 1. The plate thickness is t = 0.635 mmand the material properties are given in example 5.1. Thefiber direction coincides with the global x-axis.





• From Example 5.1:

Example 6.2





• Shear forces = 0

• Inplane forces = 0

• Plate is symmetric

Example 6.2

Therefore:

[H] = 0

[A] = 0[B] = 0





• To determine the [D] matrix:

– Use (6.16)

Example 6.2

(6.16)Where zk = t/2, zk-1 = -t/2

[D] = [Q] (0.02134)





• Compute [d] bending compliance) from [D] to determine curvature:

Where; [d] = [D]-1

Example 6.2

6.19

=

Symmetry allows (6.17) to be separated into (6.18) and (6.19)

Using (6.19) to determine the curvatures:





• To determine strains at the bottom layer (z = -t/2):

– Use (6.20)

Example 6.2

(6.20)

Therefore:

=

curvaturesz

strains

surface

mid

plate

in

strains

z

xy

y

x

xy0

y0

x0

xy

y

x

g

e

e

g

e

e

– Midplane strains = 0



ME429: Int to Composite Materials


• Finally The stresses at the bottom

surface can be determined using 6.21

k

xy

y

x

k

662616

262212

161211

k

xy

y

x

xyk

k

xyk

12k k

12k

QQQ

QQQ

QQQ

Q

Q

g

e

e

e

eExample 6.2

=

Since:

q = 0[Q] = [Q]

Where e is a very small value

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