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431

6

Bending of Beams

Chapter Objectives

• Develop formulas to find the deflection and stresses in a beam madeof composite materials.

• Develop formulas for symmetric beams that are narrow or wide.• Develop formulas for nonsymmetric beams that are narrow or wide.

6.1 Introduction

To study mechanics of beams made of laminated composite materials, weneed to review the beam analysis of isotropic materials. Several conceptsapplied to beams made of isotropic materials will help in understandingbeams made of composite materials. We are limiting our study to beamswith transverse loading or applied moments.

The bending stress in an isotropic beam (Figure 6.1 and Figure 6.2) underan applied bending moment,

M

, is given by

1,2

, (6.1)

where

z

= distance from the centroid

I

= second moment of area

The bending deflections,

w

, are given by solving the differential equation

σ = MzI

1343_book.fm Page 431 Tuesday, September 27, 2005 11:53 AM

© 2006 by Taylor & Francis Group, LLC

432

Mechanics of Composite Materials, Second Edition

, (6.2)

where

E

= Young’s modulus of the beam material.

The term of is defined as the curvature

, (6.3)

FIGURE 6.1

Bending of a beam.

FIGURE 6.2

Curvature of a bended beam.

MM

z

x

O

z

Neutral axis

ρ

EId wdx

M2

2= −

d wdx

2

2

κxw

x= − ∂

∂

2

2



Bending of Beams

433

giving

. (6.4)

The formula for the bending stress is only valid for an isotropic materialbecause it assumes that the elastic moduli is uniform in the beam. In thecase of laminated materials, elastic moduli vary from layer to layer.

6.2 Symmetric Beams

To keep the introduction simple, we will discuss beams that are symmetricand have a rectangular cross-section

3

(Figure 6.3). Because the beam is sym-metric, the loads and moments are decoupled in Equation (4.29) to give

(6.5)

or

. (6.6)

Now, if bending is only taking place in the

x

-direction, then

FIGURE 6.3

Laminated beam showing the midplane and the neutral axis.

zc

Mid-plane

z

Neutralaxis

EI Mxκ =

M

M

M

Dx

y

xy

x

y

xy

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

= ⎡⎣ ⎤⎦

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

κκ

κ

κκ

κ

x

y

xy

x

y

xy

D

M

M

M

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

= ⎡⎣ ⎤⎦

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

−1



434


,

, (6.7)

that is,

(6.8a)

(6.8b)

, (6.8c)

where are the elements of the [

D

]

–1

matrix as given in Equation (4.28c).Because defining midplane curvatures (Equation 4.15),

,

, (6.9)

,

the midplane deflection

w

0

is not independent of

y

. However, if we have anarrow beam — that is, the length to width ratio (L/b) is sufficiently high,we can assume that

w

0

=

w

0

(

x

) only.

. (6.10)

Writing in the form similar to Equation (6.2) for isotropic beams,

My = 0 Mxy = 0

κκ

κ

x

y

xy

x

D

M⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

= ⎡⎣ ⎤⎦

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

−100

κx xD M= ∗11

κy xD M= ∗12

κxy xD M= ∗16

Dij∗

κxwx

= − ∂∂

20

2

κywy

= − ∂∂

20

2

κxyw

x y= − ∂

∂ ∂2

20

κx xd wdx

D M= − = ∗2

02 11



Bending of Beams

435

, (6.11)

where

b

= width of beam

E

x

= effective bending modulus of beam

I

= second moment of area with respect to the

x–y

-plane

From Equation (6.8a) and (6.11), we get

. (6.12)

Also,

(6.13)

. (6.14)

To find the strains, we have, from Equation (4.16),

(6.15a)

(6.15b)

. (6.15c)

These global strains can be transformed to the local strains in each plyusing Equation (2.95):

. (6.16)

The local stresses in each ply are obtained using Equation (2.73) as

d wdx

M bE I

x

x

20

2= −

Eh D

x = ∗

123

11

Ibh=

3

12

M M bx=

∈ =x xzκ

∈ =y yzκ

γ κxy xyz=

∈∈

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

= ⎡⎣ ⎤⎦ ⎡⎣ ⎤⎦ ⎡⎣ ⎤⎦

∈∈

−1

2

12

1

γ γk

x

y

xy

R T R

⎡⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

k



436


. (6.17)

The global stresses in each ply are then obtained using Equation (2.89) as

. (6.18)

Example 6.1

A simply supported laminated composite beam of length 0.1 m and width5 mm (Figure 6.4) made of graphite/epoxy has the following layup of [0/90/–30/30]

s

. A uniform load of 200 N/m is applied on the beam. What isthe maximum deflection of the beam? Find the local stresses at the top ofthe third ply (–30

°

) from the top. Assume that each ply is 0.125 mm thickand the properties of unidirectional graphite/epoxy are as given in Table 2.1.

Solution

The shear and bending moment diagrams for the beam are given in Figure6.5. The bending moment is maximum at the center of the beam and isgiven by

, (6.19)

where

q

= load intensity (N/m)

L

= length of the beam (m)

FIGURE 6.4

Uniformly loaded simply supported beam.

σστ γ

1

2

12

1

2

12

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

= ⎡⎣ ⎤⎦

∈∈

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

k k

Q

σσ

τ

σστ

x

y

xy k

T

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

= ⎡⎣ ⎤⎦

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

−11

2

12 kk

MqL=

2

8

5 mm

q = 200 N/m

0.1 m



Bending of Beams

437

The maximum bending moment then is

= 0.25 N-m.

Without showing the calculations because they are shown in detail inChapter 4 (see Example 4.2), we get

FIGURE 6.5

Shear (a) and bending moment (b) diagrams of a simply supported beam.

q = 200 N/m

(b)

(a)

0 0

20

−20

0.05 0.1

0.25

00 0.05 0.1

−0.25

M = ×200 0 18

2.

D⎡⎣ ⎤⎦ =× × − ××

− −1 015 10 5 494 10 4 234 105 494 1

1 1 1. . .. 00 5 243 10 1 567 104 234 10 1 567 1

1 0 1

1

− −

−

× − ×− × − ×

. .. . 00 9 055 101 1

3

− −×

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥.

Pa m-



438


.

To find the maximum deflection of the beam,

δ

, we use the isotropic beamformula:

. (6.20)

Now, in Equation (6.12),

.

Thus,

From Equation (6.13),

D⎡⎣ ⎤⎦ =× − × ×

−−

− −

1

1 3 21 009 10 9 209 10 4 557 109 20. . .. 99 10 1 926 10 2 901 10

4 557 10 2 901

3 1 2

2

× × ×× ×

− − −

−

. .. . 110 1 131 10

1

2 03

− ×

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥.

Pa m-

δ = 5384

4qLE Ix

h = ( ) ×( )−8 0 125 10 3.

= 0 001. m

DPa m

111

31 009 10

1∗ −= ×−

.

Eh Dx =

=( ) ×( )

= ×

∗

−

12

12

0 001 1 009 10

1 189 10

311

3 1. .

. 111Pa

Ibh

m

=

=×( )( )

= ×

−

−

3

3 3

13 4

12

5 10 0 001

12

4 167 10

.

. .



Bending of Beams

439

Therefore, from Equation (6.20),

.

The maximum curvature is at the middle of the beam and is given by

.

The global strains (Equation 6.15) at the top of the third ply (–30

°

) are

δ =( )( )( )

( ) ×( ) × −

5 200 0 1

384 1 189 10 4 167 10

4

11 1

.

. . 33( )

= × −5 256 10 3. m

= 5 256. mm

χχ

χ

x

y

xy

D

D

D

qL⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

=

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

∗

∗

∗

11

12

16

2

88b

=×

− ××

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

−

−

−

1 009 109 209 10

4 557 10

21

3

2

..

.

000 0 18 0 005

2××

..

=×

− ××

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

−

−

−

1 009 109 209 10

4 557 105

1

3

2

..

.00

= −⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

5 0450 46052 279

1...

m

∈∈

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

=

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

x

y

xy

x

y

xy

z

γ

κκ

κ

= −( ) −⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

0 000255 0450 46052 279

....



440


.

The global stresses (Equation 6.18) at the top of the third ply (–30

°

) then are

.

Example 6.2

In Example 6.1, the width-to-height ratio in the cross-section of the beamis

b

/

h

= 5/1 = 5. This may be considered as a narrow-beam cross-section.If the

b

/

h

ratio were large, the cross-section may be considered to be widebeam. What are the results of Example 6.1 if one considers the beam to bea wide beam?

Solution

In the case of wide beams, we consider

.

Then, from Equation (6.5),

,

=− ×

×− ×

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

−

−

−

1 261 101 151 105 696 10

3

4

4

...

mmm

σσ

τ γ

x

y

xy

x

y

xy

Q

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

= ⎡⎣ ⎤⎦

∈∈

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥

=× × − ××

1 094 10 3 246 10 5 419 103 246 10 2

11 10 10

10

. . .

. .. .. .

365 10 2 005 105 419 10 2 005 10

10 10

10 10

× − ×− × − × 33 674 10

1 261 101 151 10

10

3

4

.

..

×

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

− ××

−

−

−

55 698 10 4. ×

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥−

=− ×− ×

×

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

1 034 102 680 104 511 10

8

7

7

.

..

Pa

κ κy xy= = 0

M

M

M

D D D

D D D

D D D

x

y

xy

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

=11 12 16

12 22 26

16 26 666

00

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

κx



Bending of Beams 441

giving

(6.21)

. (6.22)

Thus, from Equation (6.9a), Equation (6.11), and Equation (6.21),

and, from Equation (6.20),

The relative difference in the value of deflection between the assumptionof a wide and narrow beam is

.

M Dx x= 11κ

M M b D bx x= = 11κ

EDhx =

=( ) ×( )

( )

= ×

12

12 1 015 10

0 001

1 218 10

113

1

3

.

.

. 111Pa

δ =( )( )( )

( ) ×( ) × −

5 200 0 1

384 1 218 10 4 167 10

4

11 1

.

. . 33

35 131 10

5 131

( )= ×

=

−.

. .

m

mm

∈ = − ×anarrow wide

narrow

δ δδ

100

= − ×5 256 5 1315 256

100. .

.

= 2 357. %



442 Mechanics of Composite Materials, Second Edition

Because there is only 2.357% difference in the maximum deflection, doesthis mean that the assumption of wide beams influences the stresses onlyby a similar amount?


.

Because κy = 0, κxy = 0,

.

The global strains (Equation 6.15) at the top of the third ply (–30°) are

.

The global stresses (Equation 6.18) at the top of the third ply (–30°) then are

κxxM

D=

11

=×

501 015 101.

= 4 9261

.m

κκ

κ

x

y

xy

m

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

=⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

4 92600

1.

∈∈

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

=

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

x

y

xy

x

y

xy

z

γ

κκ

κ

= −( )⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

0 000254 926

00

..

=− ×⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

−1 232 1000

3.mm




.

The relative differences in the stresses obtained using wide and narrowbeam assumptions are

.

σσ

τ γ

x

y

xy

x

y

xy

Q

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

= ⎡⎣ ⎤⎦

∈∈

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥

=× × − ××

1 094 10 3 246 10 5 419 103 246 10 2

11 10 10

10

. . .

. .. .. .

365 10 2 005 105 419 10 2 005 10

10 10

10 10

× − ×− × − × 33 674 10

1 232 100010

3

.

.

×

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

− ×⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥

−

⎥⎥

=− ×− ×

×

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

1 348 103 999 10

6 676 10

8

7

7

.

..

Pa

∈ =−

×ax narrow x wide

x narrowxσ

σ σσ

| |

|

100

=− × − − ×( )

− ×

1 034 10 1 348 10

1 034 10

8 8

8

. .

.

= 30 37. %

∈ =−

×ay narrow y wide

y narrowyσ

σ σσ

| |

|

100

=− × − − ×( )

××

2 680 10 3 999 10

2 680 10100

7 7

7

. .

.

= 49 22. %




.

6.3 Nonsymmetric Beams

In the case of nonsymmetric beams, the loads and moment are not decou-pled. The relationship given by Equation (4.29) is

or

.

Assuming that the preceding 6 × 6 inverse matrix is denoted by [J] — that is,

, (6.23)

then

. (6.24)

∈ =−

×axy narrow xy wide

xy narrowxyτ

τ ττ

| |

|

100

= × − ××

×4 511 10 6 676 104 511 10

1007 7

7

. ..

= 48 00. %

N

M

A B

B D

⎡

⎣⎢⎢

⎤

⎦⎥⎥

=⎡

⎣⎢⎢

⎤

⎦⎥⎥

∈⎡

⎣⎢⎢

⎤

⎦⎥⎥

0

κ

∈⎡

⎣⎢⎢

⎤

⎦⎥⎥

=⎡

⎣⎢⎢

⎤

⎦⎥⎥

⎡

⎣⎢⎢

⎤

⎦⎥⎥

−0

1

κA B

B D

N

M

A B

B DJ

⎡

⎣⎢⎢

⎤

⎦⎥⎥

= ⎡⎣ ⎤⎦

−1

∈∈

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥

=

x

y

xy

x

y

xy

J J0

0

0

11 1

γ

κκ

κ

22 13 14 15 16

21 22 23 24 25 26

31 32 33 3

J J J J

J J J J J J

J J J J 44 35 36

41 42 43 44 45 46

51 52 53 54 55 5

J J

J J J J J J

J J J J J J 66

61 62 63 64 65 66J J J J J J

N

Nx⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥

yy

xy

x

y

xy

N

M

M

M

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥




If bending is taking place only in the x-direction, then Mx is the onlynonzero component, giving

. (6.25)

The strain distribution in the beam, then, from Equation (4.16) is

(6.26a)

(6.26b)

. (6.26c)

Because the beam is unsymmetric, the neutral axis does not coincide withthe midplane. The location of the neutral axis, zn, is where ∈x = 0. FromEquation (6.26a),

,

giving

. (6.27)

∈ =x xJ M014

∈ =y xJ M024

γ xy xJ M034=

κx xJ M= 44

κy yJ M= 54

κxy xyJ M= 64

∈ =∈ +x x xz0 κ

∈ =∈ +y y yz0 κ

γ γ κxy xy xyz= +0

0 0=∈ +x n xz κ

= +J M z J Mx n x14 44

zJJn = − 14

44




Because, from Equation (4.15),

,

the deflection w0 is not independent of y. However, if we have a narrowbeam — that is, the length-to-width ratio (L/b) is sufficiently high, we canassume that w0 = w0(x) only.

, (6.28)

writing in the form

, (6.29)

whereb = width of beam

Ex = effective bending modulus of beamI = second moment of area with respect to the x–y-plane

From Equation (6.28) and Equation (6.29), we get

. (6.30)

Also,

.

κxwx

= − ∂∂

20

2

κywy

= − ∂∂

20

2

κxyw

x y= − ∂

∂ ∂2

20

κx xd wdx

J M= = −2

02 44

d wdx

M bE I

x

x

20

2= −

Eh J

x = 123

44

Ibh=

3

12

M M bx=




To find the strains, we have, from Equation (4.16),

(6.31a)

(6.31b)

. (6.31c)

These global strains can be transformed to the local strains in each plyusing Equation (2.95):

. (6.32)

The local stresses in each ply are obtained using Equation (2.73) as

. (6.33)

The global stresses in each ply are then obtained using Equation (2.89) as

. (6.34)

Example 6.3

A simply supported laminated composite beam (Figure 6.4) of length 0.1 mand width 5 mm made of graphite/epoxy has the following layup: [0/90/–30/30]2. A uniform load of 200 N/m is applied on the beam. What is themaximum deflection of the beam? Find the local stresses at the top of thethird ply (–30°) from the top. Assume that each ply is 0.125 mm thick andthe properties of unidirectional graphite/epoxy are as given in Table 2.1.

Solution

The stiffness matrix found by using Equation (4.28) and Equation (4.29) is

∈ = ∈ +x xo

xzκ

∈ = ∈ +y yo

yzκ

γ γ κxy xy xyz= +0

∈∈

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

= ⎡⎣ ⎤⎦ ⎡⎣ ⎤⎦ ⎡⎣ ⎤⎦

∈∈

−1

2

12

1

γ γk

x

y

xy

R T R

⎡⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

k

σστ γ

1

2

12

1

2

12

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

= ⎡⎣ ⎤⎦

∈∈

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

k k

Q

σσ

τ

σστ

x

y

xy k

T

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

= ⎡⎣ ⎤⎦

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

−11

2

12 kk




.

The inverse of the matrix is

.

Now, in Equation (6.30),


1 027 10 1 768 10 3 497 10 1 848 10 1 848 7 10 3. . . . .× × × − ×− 88 10 1 694 101 768 10 5 986 10 2 608 10

3 3

7 7 9

× ×× × × −

.. . . 11 848 10 1 848 10 6 267 10

3 497 10 2 6

3 3 2

10

. . .. .

× − × ×× − 008 10 2 195 10 1 694 10 6 267 10 1 848 109 7 3 2× × × × ×− . . . . 33

3 3 31 848 10 1 848 10 1 694 10 9 231 1 473 4 2− × × ×. . . . . . 334 101 848 10 1 848 10 6 267 10 1 473 4

1

3 3 2

×× − × ×

−

. . . . .3319 1 567 101 694 10 6 267 10 1 848 10 4

1

3 2 3

.. . . .

×× × ×

−

2234 10 1 567 10 1 8291 1× ×

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥

− −. . ⎥⎥

1 068 10 3 409 10 7 009 10 4 298 108 9 10 6. . . .× − × × × −− − − − 77 241 10 9 809 103 409 10 1 829 10

6 6

9 8

. .. .

× − ×− × ×

− −

− − 44 042 10 6 097 10 1 142 10 3 083 1010 6 5. . . .× − × × − ×− − − −66

10 10 87 009 10 4 042 10 5 035 10 6 339 10. . . .× × × − ×− − − −− − −

−

− × − ×× − ×

6 6 5

6

3 460 10 4 989 104 298 10 6 097 1

. .. . 00 6 339 10 1 194 10 4 335 10 1 9406 6 1 2− − − −− × × − × − ×. . . . 1107 241 10 1 142 10 3 460 10 4 355

2

6 5 6

−

− − −− × × − × −. . . . ×× × − ×− × −

− − −

−

10 2 551 10 5 480 109 809 10 3 08

2 1 3

6

. .. . 33 10 4 989 10 1 940 10 5 480 10 6 16 5 2 3× − × − × − ×− − − −. . . . 223 10 1×

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥−

h = × ×( )−8 0 125 10 3.

= 0 001. m

JPa m44

131 194 10

1= × −.-

Eh Jx =

=( ) ×( )

= ×

−

12

12

0 001 1 194 10

1 005 10

344

3 1

1

. .

. 11 .Pa

Ibh=

3

12




.

Thus, from Equation (6.20),

The maximum bending moment occurs at the middle of the beam and isgiven by

Calculating the midplane strains and curvature from Equation (6.24) gives

=×( )( )−5 10 0 001

12

3 3.

= × −4 167 10 13 4. m

δ =( )( )( )

( ) ×( ) × −

5 200 0 1

384 1 005 10 4 167 10

4

11 1

.

. . 33

36 219 10

6 219

( )= ×

=

−.

. .

m

mm

MqL

N m

max

.

.

=

= ×

=

2

2

8

200 0 18

0 25 -

MM

b

N mm

x maxmax=

=

=

0 250 005

50

..

.-




,

giving

.


.

∈∈

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥⎥

=

x

y

xy

x

y

xy

0

0

0

1 0

γ

κκ

κ

. 668 10 3 409 10 7 009 10 4 298 10 78 9 10 6× − × × × −− − − −. . . .2241 10 9 809 103 409 10 1 829 10 4

6 6

9 8

× − ×− × ×

− −

− −

.. . .0042 10 6 097 10 1 142 10 3 083 10

7

10 6 5 6× − × × − ×− − − −. . ... . . .009 10 4 042 10 5 035 10 6 339 1010 10 8 6× × × − ×− − − − −− × − ×

× − ×

− −

− −

3 460 10 4 989 104 298 10 6 097 10

6 5

6

. .. . 66 6 1 26 339 10 1 194 10 4 335 10 1 940 10− × × − × − ×− − −. . . . −−

− − −− × × − × − ×

2

6 5 67 241 10 1 142 10 3 460 10 4 355 1. . . . 00 2 551 10 5 480 109 809 10 3 083

2 1 3

6

− − −

−

× − ×− × − ×

. .. . 110 4 989 10 1 940 10 5 480 10 6 1236 5 2 3− − − −− × − × − ×. . . . ××

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢−10

00050001

⎤⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥

∈∈

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

=×

− ×−

−

−x

y

xy

0

0

0

4

4

2 149 103 048 10

γ

..

33 169 10 4. ×

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥−

κκ

κ

x

y

xy

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

= −− ×

⎡

⎣

⎢

−

5 9702 178

9 700 10 1

..

.⎢⎢⎢

⎤

⎦

⎥⎥⎥

∈∈

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

=∈∈

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

+x

y

xy

x

y

xy

x

z

γ γ

κκ

0

0

0

yy

xyκ

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

=×

− ×− ×

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

−

−

−

2 149 103 048 103 169 10

4

4

4

...

++ −( ) −− ×

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥

−

0 000255 9702 178

9 700 10 1

...

. ⎥⎥

=− ×

×− ×

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

−

−

−

1 278 102 397 107 431 10

3

4

5

...

mmm




The global stresses (Equation 6.34) at the top of the third ply (–30°) are

.

Example 6.4

In Example 6.3, the width-to-height ratio in the cross-section of the beamis b/h = 5/1 = 5. This may be considered as a narrow-beam cross-section.If the b/h ratio were large, the cross-section may be considered to be widebeam. What are the results of Example 6.3 if one considers the beam to bea wide beam?

Solution

In the case of the wide beams, we consider

.

Then, from Equation (6.24),

, (6.35)

we get

σσ

τ γ

x

y

xy

x

y

xy

Q

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

= ⎡⎣ ⎤⎦

∈∈

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥

=× × − ××

1 094 10 3 246 10 5 419 103 246 10 2

11 10 10

10

. . .

. .. .. .

365 10 2 005 105 419 10 2 005 10

10 10

10 10

× − ×− × − × 33 674 10

1 278 102 397 10710

3

4

.

..

×

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

− ××

−

−

..431 10 5×

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥−

=− ×− ×

×

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

1 280 103 431 10

6 170 10

8

7

7

.

..

Pa

κy = 0

∈∈

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥

=

x

y

xy

x

xy

J J0

0

0

11 12

0

γ

κ

κ

JJ J J J

J J J J J J

J J J J

13 14 15 16

21 22 23 24 25 26

31 32 33 34 JJ J

J J J J J J

J J J J J J

35 36

41 42 43 44 45 46

51 52 53 54 55 56

JJ J J J J J

M

61 62 63 64 65 66

000

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥

xx

yM

0

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥




(6.36a)

(6.36b)

(6.36c)

(6.36d)

(6.36e)

. (6.36f)

To find the neutral axis, ∈x = 0, we use Equation (6.36a) and Equation(6.36e) to give

(6.37)

. (6.38)

From Equation (6.9a), Equation (6.11), and Equation (6.38),

Thus, from Equation (6.20), we get

∈ = +x x yJ M J M014 15

∈ = +y x yJ M J M024 25

γ xy x yJ M J M034 35= +

κx x yJ M J M= +44 45

0 54 55= +J M J Mx y

κxy x yJ M J M= +64 65

zJ J J JJ J J Jn = − −

−14 55 15 54

44 55 45 54

M bM bJ

J J J Jbeam x x= =−55

44 55 54 45

κ

Eh

JJ J J Jx =

−( )

=( )

×

12

12

0 001

2 551 1

355

44 55 45 54

3.

. 00

1 194 10 2 551 10 4 355 10

1

1 1 2

−

− − −×( ) ×( ) − − ×( ) −. . . 44 355 10

1 071 10

2

11

.

. .

×( )= ×

−

Pa

δ =( )( )( )

( ) ×( ) × −

5 200 0 1

384 1 072 10 4 167 10

4

11 1

.

. . 33( )




From Example 6.3, the maximum bendings’ moment per unit width is

.

From Equation (6.36e),

.


.

= ×

=

−5 830 10

5 830

3.

. .

m

mm

MN m

mx max = 50-

MJJ

My xmax = − 54

55

= − − ××

( )−

−

4 355 102 551 10

502

1

..

= 8 497.N m

m-

∈∈

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥

=

×

x

y

xy

x

y

0

0

0

0

1 068 1

γ

κ

κ

. 00 3 409 10 7 009 10 4 298 10 7 2418 9 10 6− − − −− × × × − ×. . . . 110 9 809 103 409 10 1 829 10 4 042

6 6

9 8

− −

− −

− ×− × × ×

.. . . 110 6 097 10 1 142 10 3 083 10

7 009

10 6 5 6− − − −− × × − ×. . .. ×× × × − × −− − − −10 4 042 10 5 035 10 6 339 10 3 410 10 8 6. . . . 660 10 4 989 10

4 298 10 6 097 10 6

6 5

6 6

× − ×× − × −

− −

− −

.. . .3339 10 1 194 10 4 335 10 1 940 107

6 1 2 2× × − × − ×−

− − − −. . ... . . .241 10 1 142 10 3 460 10 4 355 10 26 5 6 2× × − × − ×− − − − .. .. .

551 10 5 480 109 809 10 3 083 10

1 3

6 6

× − ×− × − ×

− −

− − −− × − × − × ×− − − −4 989 10 1 940 10 5 480 10 6 123 105 2 3. . . . 11

00050

8 4970

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢

.

⎤⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥

=

×− ×− ×

−

−

−

1 534 102 078 103 463 10

4

4

... 44

5 6020

1 017

.

.−

⎡

⎣

⎢⎢⎢⎢⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥⎥⎥⎥





The global stresses (Equation 6.18) at the top of the third ply (–30°) are

The relative differences in the stresses obtained using wide and narrowbeam assumptions are

∈∈

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

=∈∈

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

+x

y

xy

x

y

xyγ γ

0

0

0

zzx

y

xy

κκ

κ

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

=×

− ×

−

−

1 534 102 078 10

4

4

..

−− ×

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

+ −( )−−3 463 10

0 000255 602

01 04.

..

. 117

1 247 102 078 109 221

3

4

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

=− ×− ×− ×

−

−

.

.

. 110 5−

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

mm

.

σσ

τ γ

x

y

xy

x

y

xy

Q

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

= ⎡⎣ ⎤⎦

∈∈

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥⎥

=× × − ××

1 094 10 3 246 10 5 419 103 246 10

11 10 10

1

. . .

. 00 10 10

10

2 365 10 2 005 105 419 10 2 005 10

. .. .

× − ×− × − × 110 10

3

3 674 10

1 247 102 078 10

.

.

.×

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

− ×− ×

−

−−

−− ×

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

=− ×− ×

4

3

8

9 221 10

1 382 104 354 1

.

.

. 006 833 10

7

7..

×

⎡

⎣

⎢⎢⎢

⎤

⎦

⎥⎥⎥

∈a

∈ =−

×ax narrow x wide

x narrowxσ

σ σσ

| |

|

100




6.4 Summary

In this chapter, we reviewed the bending of isotropic beams and thenextended the knowledge to study stresses and deflection in laminated com-posite beams. The beams could be symmetric or unsymmetric, and wide ornarrow cross-sectioned. Differences in the deflection and stress are calculatedbetween the results of a wide and a narrow beam.

Key Terms

Bending stressSecond moment of area

=− × − − ×( )

− ××

=

1 280 10 1 382 10

1 280 10100

7 97

8 8

8

. .

.

. %%

∈ =−

×

=− ×

ay narrow y wide

y narrowxσ

σ σσ

| |

|

.

100

3 431 110 4 354 10

3 431 10100

26 90

7 7

7

− − ×( )− ×

×

=

.

.

. %

∈ =−

×

=

axy narrow xy wide

xy narrowxyτ

τ ττ

| |

|

.

100

6 1770 10 6 836 106 170 10

100

10 79

7 7

7

× − ××

×

=

..

. %.




Symmetric beamsWide beamsNarrow beamsUnsymmetric beams

Exercise Set

6.1 A simply supported laminated composite beam (Figure 6.6) madeof glass/epoxy is 75 mm long and has the layup of [±30]2s. A uniformload is applied on the beam that is 5 mm in width. Assume eachply is 0.125 mm thick and the properties of glass/epoxy are fromTable 2.1.1. What is the maximum deflection of the beam? 2. Find the local stresses at the top of the laminate.

6.2 A simply supported laminated composite beam (Figure 6.6) madeof glass/epoxy is 75 mm long and has the layup of [±30]4. A uniformload is applied on the beam that is 5 mm in width. Assume eachply is 0.125 mm thick and the properties of glass/epoxy are fromTable 2.1.1. What is the maximum deflection of the beam? 2. Find the local stresses at the top of the laminate.

6.3 Calculate the bending stiffness of a narrow beam cross-ply laminate[0/90]2s. Now compare it by using the average modulus of the lam-inate. Assume that each ply is 0.125 mm thick and the properties ofglass/epoxy are from Table 2.1.

FIGURE 6.6Uniformly loaded simply supported beam.

5 mmq = 100 N/m

0.075 m




References

1. Buchanan, G.R., Mechanics of Materials, HRW Inc., New York, 1988.2. Ugural, A.C. and Fenster, S.K., Advanced Strength and Applied Elasticity, 3rd ed.

Prentice Hall, Englewood Cliffs, NJ, 1995.3. Swanson, S.R., Introduction to Design and Analysis with Advanced Composite Ma-

terials, Prentice Hall, Englewood Cliffs, NJ, 1997.



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