ME5720Chap7E_F07

8/11/2019 ME5720Chap7E_F07

http://slidepdf.com/reader/full/me5720chap7ef07 1/33

Deflection and Buckling of

Laminates

• Transverse deflections generally much

larger than in – plane deflections

• Buckling is an instability under in –plane

compressive or shear loading

• Equations used in both cases are similar

Stress resultants and external loads acting on laminate.

(From Halpin, 1984.)

8/11/2019 ME5720Chap7E_F07


0=∑ x F

dxdy y

N dx N dxdy x

N dy N xy

xy x

x∂∂++

∂∂+

0=−− dx N dy N xy x

(7.119)

Equation (7.119) may be simplified as

0=∂

∂+

∂

∂

y

N

x

N xy x (7.120)

The summation of forces along the y – direction yields

0=∑ y F

dxdy x

N

dy N dxdy y

N

dx N

xy

xy

x

y∂

∂

++∂

∂

+0=−− dy N dx N xy y

(7.121)

0=∂

∂+

∂

∂

y

N

x

N xy x (7.122)

or

The summation of forces along the z – direction gives

0=∑ z F

y

QdxQdxdy

x

QdyQ

y

y x

x∂∂++

∂∂+

( ) 0, =+−− y xqdxQdyQ y x

(7.123)

8/11/2019 ME5720Chap7E_F07


or

( ) 0, =+

∂

∂+

∂

∂ y xq

y

Q

x

Q y x (7.124)

The summation of moments about the x axis yields

0=∑ x M

dxdyQdxdy x

M dy M dydx

y

M dx M y

xy

xy

y

y +∂

∂−−

∂

∂−−

( ) 2/2/, dydyQdxdydy y xqdydxdy y

Q x

y ++∂

∂+

02/2/ =−++∂∂+ dydyQdy M dx M dxdydy

x

Q x xy x

x

(7.125)

The summation of moments about the y axis gives

0=∑ y M

x

xy x Q y

M

x

M =

∂

∂+

∂

∂ (7.127)

Substitution of Eqs. (7.126) and (7.127) in Eq. (7.124)

yields

( ) 0,22

22

2

2

=+∂

∂+

∂∂

∂+

∂

∂ y xq

y

M

y x

M

x

M y xy x (7.128)

8/11/2019 ME5720Chap7E_F07


• These are the differential equations of equilibriumof the plate in terms of stress and moment

resultants.

• The corresponding equilibrium equations in terms

of displacements can be derived by substituting in

the laminate force – deformation equations, the

strain – displacement relations, and the curvature –

displacement equations

• The resulting equations are,

2

02

162

02

66

02

162

02

11 2 x

v A

y

u A

y x

u A

x

u A

∂

∂+

∂

∂+

∂∂

∂+

∂

∂

( ) y x

w B

x

w B

y

v A

y x

v A A

∂∂

∂−

∂

∂−

∂

∂+

∂∂

∂++

2

3

163

3

112

02

26

02

6612 3

( ) 023

3

262

3

6612 =∂

∂−

∂∂

∂+−

y

w B

y x

w B B

(7.129)

8/11/2019 ME5720Chap7E_F07


( ) 2

02

662

02

26

02

66122

02

16 x

v

A y

u

A y x

u

A A x

u

A ∂

∂

+∂

∂

+∂∂

∂

++∂

∂

3

3

162

02

22

02

262 x

w B

y

v A

y x

v A

∂

∂−

∂

∂+

∂∂

∂+

( ) 0323

3

222

3

262

3

6612 =∂

∂−

∂∂

∂−

∂∂

∂+−

y

w B

y x

w B

y x

w B B

(7.130)

( ) 22

4

66123

4

164

4

11 224 y x

w

D D y x

w

D x

w

D ∂∂

∂++

∂∂

∂+

∂

∂

y x

u B

x

u B

y

w D

y x

w D

∂∂

∂−

∂

∂−

∂

∂+

∂∂

∂+

2

03

163

03

114

4

223

4

26 34

( )3

03

163

03

262

03

6612 2 x

v B

y

u B

y x

u B B

∂

∂−

∂

∂−

∂∂

∂+−

(7.131)

( ) ( ) y xq yv B

y xv B

y xv B B ,32 3

03

222

03

262

03

6612 =∂∂−∂∂∂−∂∂∂+−

Note: when Bij = 0, the transverse displacements w are decoupled

from the in – plane displacements u and v

8/11/2019 ME5720Chap7E_F07


• For symmetric laminates with Bij = 0, Eq. (7.131)

alone becomes the governing equation for transverse

bending displacements.

• The governing partial differential equations must be

solved subject to the appropriate boundary

conditions. In the general case, when the in – plane

displacements are coupled with the transverse

displacements, the boundary conditions must reflect

this.

• Here we will restrict the discussion to bending of

symmetric laminated plates. That is, we will only

consider transverse bending displacements according

to Eq. (7.131) with all Bij = 0.

Consider the case of transverse deflection of arectangular specially orthotropic plate which is

simply supported on all edges and loaded with

a distributed load, q(x,y). For this case, all of

the Bij = 0. Also A16 = A26 = D16 = D26 = 0 and

Eq. (7.131) becomes

( ) ( ) y xq y

w D y xw D D

xw D ,22 4

4

2222

4

66124

4

11 =∂

∂+∂∂

∂++∂

∂

(7.132)

8/11/2019 ME5720Chap7E_F07


Simply supported, specially orthotropic plate with

distributed loading.

For the simply supported boundary

condition the transverse displacements and bending moments must vanish at the edges.

In order to use the bending moment

boundary conditions to solve the differential

equation for displacements, however, the

bending moments must be expressed in

terms of displacements.

2

2

122

2

111211 y

w D x

w D D D M y x x

∂

∂−∂

∂−=+= κ κ (7.133)

2

2

222

2

122212 y

w D

x

w D D D M y x y

∂

∂−

∂

∂−=+= κ κ (7.134)

8/11/2019 ME5720Chap7E_F07


8/11/2019 ME5720Chap7E_F07


Substitution of Eqs. (7.138) and (7.137) in Eq. (7.132)

( ) mnq y

w D y xw D D

xw D =

∂∂+

∂∂∂++

∂∂

4

4

2222

4

66124

4

11 22

And taking the indicated derivatives:

( ) mnmn q Db

n

b

n

a

m D D D

a

mw =

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟ ⎠

⎞⎜⎝

⎛ +⎟

⎠

⎞⎜⎝

⎛ ⎟ ⎠

⎞⎜⎝

⎛ ++⎟

⎠

⎞⎜⎝

⎛ 22

422

661211

4

22 π π π π

Note that all sine terms like and

cancel out on both sides, so that the deflection isa

xmπ sin b

ynπ sin

( )( ) ( )( )4

22

2

6612

4

11

4

4

22 nR DmnR D Dm D

qaw mn

mn+++

=π

(7.139)

where the plate aspect ratio R = a/b. The Fourier

coefficients qmn can be found for the particular assumedload distribution.

8/11/2019 ME5720Chap7E_F07


For the uniform load q(x, y) = q0, a constant, it can be

shown that the Fourier coefficients are

mn

qqmn 2

016

π = for ,...5,3,1, =nm

(7.140)and

0=mnq ,...6,4,2, =nmfor

Boundary conditions for the specially

orthotropic plate are satisfied by assumed

solution of the form

( ) ∑∑∞

=

∞

=

=1 1

sinsin,m n

mnb

yn

a

xmw y xw π π

8/11/2019 ME5720Chap7E_F07


Proof:

( ) 0,0 = yw

b

yn

a

xm π π sinsin

because ( )

00

sin =

a

mπ

( ) 0, = yaw because ( )

0sin =a

amπ

( ) 0,0 = y M x since both2

2

2

2

y

w

x

w

∂

∂+

∂

∂

contain

same for ( ) 0, = ya M x

( ) 00, = x M y

( ) 0, =b x M y

( ) 00, = xw

( ) 0, =b xw

However, for a symmetric angle ply

laminate, the differential equation is

( )22

4

66123

4

164

4

11 224 y x

w D D

y x

w D

x

w D

∂∂

∂++

∂∂

∂+

∂

∂

( ) y xq y

w D

y x

w D ,4

4

4

223

4

26 =∂

∂+

∂∂

∂+

Boundary conditions.:along x = 0 and x = a, we must have

w = 0

022

162

2

122

2

11 =∂∂

∂−

∂

∂−

∂

∂−=

y x

w D

y

w D

x

w D M x

8/11/2019 ME5720Chap7E_F07


and along y = 0 and y = b, we must have

w = 0

022

262

2

222

2

12 =∂∂

∂−

∂

∂−

∂

∂−=

y x

w D

y

w D

x

w D M y

∴a solution of the form

( ) ∑∑∞

=

∞

=

=1 1

sinsin,n m

mnb

yn

a

xmw y xw π π

will not satisfy the diff. eqns. or the boundary

conditions since all terms in the resulting

equations do not contain the same sin or cos

functions as in specially orthotropic case

Buckling Analysis

Differential element of laminate in out – of – plane position

for buckling analysis

8/11/2019 ME5720Chap7E_F07


The summation of forces in the z direction

now becomes

0=∑ z F

( ) 02,2

22

2

2

=∂

∂+

∂∂

∂+

∂

∂++

∂

∂+

∂

∂

y

w N

y x

w N

x

w N y xq

y

Q

x

Q y xy x

y x

(7.141)

Note new terms due toin-plane forces

Note same terms that appeared in

equation for tranverse displacements

Substitution of Eqs. (7.141), (7.29), and (7.30)

in Eq. (7.142) yields the equation,

2

22

2

2

2 y

M

y x

M

x

M y xy x

∂

∂+

∂∂

∂+

∂

∂

( ) 0,22

22

2

2

=+∂

∂+

∂∂

∂+

∂

∂+ y xq

y

w N

y x

w N

x

w N y xy x

(7.142)

Combining Eqs. (7.141), (7.126), and (7.127),

we find that

Note new terms due to

in-plane forces

8/11/2019 ME5720Chap7E_F07


( )22

4

66123

4

164

4

11 224 y x

w D D

y x

w D

x

w D

∂∂

∂++

∂∂

∂+

∂

∂

y x

u B

x

u B

y

w D

y x

w D

∂∂

∂−

∂

∂−

∂

∂+

∂∂

∂+

2

03

163

03

114

4

223

4

26 34

( )3

03

163

03

262

03

6612 2 x

v B

y

u B

y x

u B B

∂

∂−

∂

∂−

∂∂

∂+−

(7.143)

( )3

03

222

03

262

03

6612 32 y

v B

y x

v B

y x

v B B

∂

∂−

∂∂

∂−

∂∂

∂+−

( )2

22

2

2

2, y

w N

y x

w N

x

w N y xq y xy x

∂∂+

∂∂∂+

∂∂+=

New terms due to in-plane forces

We now consider the case of buckling of a rectangular,simply supported, specially orthotropic plate under a

single compressive axial load, N x = - N . In this case the

loads N y = N xy = q(x, y) = 0, all Bij = 0, the stiffnesses

A16 = A26 = D16 = D26 = 0 and Eq. (7.143) becomes

( )2

2

4

4

2222

4

66124

4

11 22 x

w N

y

w D

y x

w D D

x

w D

∂

∂−=

∂

∂+

∂∂

∂++

∂

∂

(7.144)

8/11/2019 ME5720Chap7E_F07


Simply supported, specially orthotropic plate under

compressive uniaxial in – plane loading

For the simply supported boundary conditiondescribed previously by Eqs. (7.135) and

(7.136) we may assume a solution of the form

( )b

yn

a

xmw y xw mn

π π sinsin, =

The mode shape for a particular buckling mode

is described by the subscripts m and n since m

is the number of half – sine waves along the xdirection and n is the number of half – sine

waves along the y direction.

(7.145)

8/11/2019 ME5720Chap7E_F07


Substitution of this solution in the governing

differential equation (7.144) leads to the equation

( )( ) ( )( )4

22

2

6612

4

11

2 22 bR DmnR D Dm Dwmn +++π

22m Nawmn=(7.146)

where again R = a/b. This equation has the trivial

solution wmn = 0, which is of no interest. For

nontrivial solutions the critical buckling load must

be

( )( ) ( )( )[ ]4

22

2

6612

4

1122

2

22 bR DmnR D Dm Dma

N cr +++= π

(7.147)

where the smallest buckling load occurs for n = 1, and the lowest

value of the load corresponding to a particular value of m can only be determined if the Dij and the dimensions a and b are known. As

shown in Fig.

n = 1 for all curves

Comparison of

predicted and

measured normalized

buckling load, N xb2,

vs. plate aspect ratio,

a/b, for [012]

graphite/epoxylaminates.

(From Hatcher and Tuttle, 1991)

8/11/2019 ME5720Chap7E_F07


8/11/2019 ME5720Chap7E_F07


Nxy

Nxy

Shear Buckling - buckling of thin panels can be caused

by shear forces as well as by compressive forces

Design with Metals

• Look up material properties in handbook – properties are same for all configurations

• Part is made by cutting a piece of materialdown to the desired size and shape

8/11/2019 ME5720Chap7E_F07


Design with Composites• Too many possible combinations (fiber,

matrix, lamina orientations, stacking

sequence, etc.) for hand book tabulation of

material properties - use computer software

or carpet plots

• Part is made by building up the material to

the desired size and shape

Micromechanics

Lamina Properties

Laminate Properties

Design

Ply Orientations

and

Stacking Sequence

8/11/2019 ME5720Chap7E_F07


Analysis and Design of Laminates

• Analysis –

Given a composite laminate and allowable ply stresses, determine the load that it willsupport, or given the laminate, loads and

properties, determine the stresses and strains(unique solution)

• Design –

Given a set of loads and other designconstraints, select the materials and laminate

configuration to withstand the loads (infinitenumber of solutions which may or may not beoptimized for a given design variable).

Design Criteria Associated Failure ModesStrength Fracture (either partial or complete)

Stiffness Excessive deformation

Stability Buckling

Hygrothermal Effects Property degradation, expansion andcontraction, residual stresses

Life, or Durability Fatigue, Creep

Weight Heavier than conventional design

Cost Not affordable

Manufacturability Impractical to build, warping due toresidual stresses

8/11/2019 ME5720Chap7E_F07


8/11/2019 ME5720Chap7E_F07


8/11/2019 ME5720Chap7E_F07


φ °90°0

66 A

Unidirectional

Quasi - isotropic

°45

Example:An existing power transmission shaft consists of a hollow composite tube, and the tube

wall is a filament wound quasi-isotropic [60/0/60]s laminate of thickness t. A new shaft

of the same wall thickness t is to be designed from the same lamina material, but the new

laminate is to have a shear stiffness greater than that of the existing shaft. Over what

range of angles θ will a [+θ/-θ/-θ]s angle-ply laminate achieve this design objective?

The shear st iffness of the new angle-ply laminate is

3

2)(

3)()( 6666

2/

2/6666

t Q

t Qdz Q A

t

t ap θ θ −+

−+== ∫

T

+θ

-θ

8/11/2019 ME5720Chap7E_F07


Recalling that the lamina stiffnesses can be expressed in terms of invariants as

θ 4cos2

341

66 U U U

Q −−

=

and that cos4θ = cos(-4θ), the new laminate stiffness can be written as

θ

θ θ

4cos2

3

2)4cos(

234cos

2)(

341

341

341

66

t U t U U

t U

U U t U

U U A ap

−−

=

⎥⎦

⎤⎢⎣

⎡−−

−+⎥

⎦

⎤⎢⎣

⎡−

−=

The shear stiffness of a quasi-isotropic laminate is

t U U

A QI 2

)( 4166

−=

Therefore the shear stiffness of the new laminate can be expressed as

θ 4cos)()( 36666 t U A A QI ap −=

8/11/2019 ME5720Chap7E_F07


The variations of ap A )( 66 and QI A )( 66 with θ are shown below, where it can be seen that

QI ap A A )()( 6666 > for angles θ in the rangeoo

5.675.22 ≤≤θ

QI A )( 66

ap A )( 66

θ

0

o

45

o

90

o

22.5

o

67.5

o

66 A

Example: Stiffness-critical design where

deflection must be limited for some reason

8/11/2019 ME5720Chap7E_F07


( ) ∑∑∞

=

∞

=

=1 1

sinsin,m n

mnb

yn

a

xmw y xw π π

Deflection of simply supported orthotropic plate

where

( )( ) ( )( )4

22

2

6612

4

11

4

4

22 nR DmnR D Dm D

qaw mn

mn+++

=π

and maximum deflection is w(a/2, b/2)

Example of strength-critical design wherestresses must be limited according to some

failure theory:

filament wound composite pressure vessel

must be designed to withstand some internal

pressure from stored gas or fluid

8/11/2019 ME5720Chap7E_F07


Procedure:

1. For an assumed laminate design, findengineering constants for the plies, E 1 , E 2 ,G12 , υ12, then calculate the Qij, for plies.

2. Using ply orientation, calculate

3. Using ply stacking sequence and plythicknesses, calculate laminate stiffnesses,

Aij , Bij , Dij

4. Form stiffness matrix and invert

ijQ

5. Find strains and curvatures from

6. Find lamina stresses from

7. Apply failure criterion to check design

8. Repeat Steps 1-7 until suitable design isfound

⎭⎬⎫

⎩⎨⎧

⎥⎦

⎤⎢⎣

⎡=

⎭⎬⎫

⎩⎨⎧ °

−

M

N

D B

B A1

κ

ε

{ } { } { }( )κ ε σ z Q k k +°=

8/11/2019 ME5720Chap7E_F07


Carpet Plots

Convenient graphical representation showing

how a given laminate property depends on the

percentages of the plies at various ply

orientations.

To generate carpet plots, the ply orientations

must be restricted to certain angles, such as.90°

,0° ,45°

±Carpet plots for [0i/ 45 j/90k ] Kevlar/epoxy laminates. From

Peters, et al., 1991 Reprinted by permission of the society for the Advancement of Material and Process engineering.

8/11/2019 ME5720Chap7E_F07


±Carpet plots for [0i/ 45 j/90k ] Kevlar/epoxy laminates. From

Peters, et al., 1991 Reprinted by permission of the society for

the Advancement of Material and Process engineering.

Ex: Kevlar/epoxy to have E x = 30 GPaSeveral possibilities, such as

(1)

⎪⎭

⎪⎬

⎫

°

°±

°

90@%0

45@%70

0@%30

(2)

⎪⎭

⎪⎬

⎫

°

°±

°

90@%65

45@%0

0@%35

8/11/2019 ME5720Chap7E_F07


8/11/2019 ME5720Chap7E_F07


Composite Laminate Analysis Software

ASCA

AdTech Systems Research, Inc., 1342 N. Fairfield Road, Beavercreek, OH 45432-2698,web address: http://www.adtechsystems.com/

CompositePro

Peak Composite Innovations LLC, 11372 W. Parkhill Drive, Littleton, CO 80127, webaddress:http://borderworlds.com/compositepro/

FiberSim

Vistagy, Inc., 486 Totten Pond Road, Waltham, MA 02451-1917, web address:http://www.vistagy.com/index.htm

HyperSizer

Collier Research, Harbour Centre, 2 Eaton Street, Suite 504, Hampton, VA 23669, webaddress:http://www.collier-research.com/

LAP

Anaglyph Ltd, Suite 33, 10 Barley Mow Passage, London W4 4PH, United Kingdom,web address: http://www.anaglyph.co.uk/

Mic-Mac Lite

Think Composites, 101 Alma Street, #703, Palo Alto, CA 94301, web address:http://www.thinkcomp.com/

The Laminator

web address: http://www.thelaminator.net/

V-Lab

Applied Research Associates, Inc., Southeast Division, 811 Spring Forest Road, Suite100, Raleigh, NC 27609, web address: http://www.ara.com/v-lab.htm

Design Guides

• Composite Materials Handbook – MIL 17

3 volumes + CD – ROM, published byTechnomic Pub. Co., 1990. Available

for check out at WSU Science andEngineering Library Reserve Desk

• Composites Engineering Handbook,P. K. Mallick, Editor, published by MarcelDekker, Inc., 1997

8/11/2019 ME5720Chap7E_F07


Design Allowables

• Allowable values of mechanical propertiessuch as strength and modulus, based on

statistical analysis of test data.

• A – basis allowable – 99% of test data

should exceed this allowable value with

95% confidence.

• B – basis allowable – 90% of test data

should exceed this allowable value with

95% confidence.

Do not forget the “secondary stresses”!!!

Example: composite linkage delamination due to

interlaminar stresses, even though load is axial

Interlaminar stresses around

edges of holes

Load

8/11/2019 ME5720Chap7E_F07


Interlaminar Stresses in

Unidirectional Composites

Example: through – the – thickness stresses at root of notch

combined with low transverse strength of unidirectional

composite cause transverse tensile failure at root of notch

Transverse

cracks inwall of

vessel due

to bending

Transverse cracking of filament wound pressure

vessel at edge of metal end cap

Filament wound

cylindrical

pressure vessel

Metal end cap

Internal

pressure

Deformed

shape

Documents

ME5720Chap7E_F07