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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
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
1343_book.fm Page 433 Tuesday, September 27, 2005 11:53 AM
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?
From Equation (6.21),
.
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
1343_book.fm Page 442 Tuesday, September 27, 2005 11:53 AM
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
1343_book.fm Page 445 Tuesday, September 27, 2005 11:53 AM
446 Mechanics of Composite Materials, Second Edition
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=
1343_book.fm Page 446 Tuesday, September 27, 2005 11:53 AM
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
1343_book.fm Page 447 Tuesday, September 27, 2005 11:53 AM
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
⎡
⎣
⎢⎢⎢⎢⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥⎥⎥⎥⎥
1343_book.fm Page 451 Tuesday, September 27, 2005 11:53 AM
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
× − ××
×
=
..
. %.
1343_book.fm Page 455 Tuesday, September 27, 2005 11:53 AM
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
1343_book.fm Page 456 Tuesday, September 27, 2005 11:53 AM
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-