bending of beams - dhu.edu.cnstructures.dhu.edu.cn › _upload › article › files › e1 › 14...
TRANSCRIPT
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
1343_book.fm Page 432 Tuesday, September 27, 2005 11:53 AM
© 2006 by Taylor & Francis Group, LLC
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
1343_book.fm Page 433 Tuesday, September 27, 2005 11:53 AM
© 2006 by Taylor & Francis Group, LLC
434
Mechanics of Composite Materials, Second Edition
,
, (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
1343_book.fm Page 434 Tuesday, September 27, 2005 11:53 AM
© 2006 by Taylor & Francis Group, LLC
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
1343_book.fm Page 435 Tuesday, September 27, 2005 11:53 AM
© 2006 by Taylor & Francis Group, LLC
436
Mechanics of Composite Materials, Second Edition
. (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
1343_book.fm Page 436 Tuesday, September 27, 2005 11:53 AM
© 2006 by Taylor & Francis Group, LLC
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-
1343_book.fm Page 437 Tuesday, September 27, 2005 11:53 AM
© 2006 by Taylor & Francis Group, LLC
438
Mechanics of Composite Materials, Second Edition
.
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
.
. .
1343_book.fm Page 438 Tuesday, September 27, 2005 11:53 AM
© 2006 by Taylor & Francis Group, LLC
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
....
1343_book.fm Page 439 Tuesday, September 27, 2005 11:53 AM
© 2006 by Taylor & Francis Group, LLC
440
Mechanics of Composite Materials, Second Edition
.
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
1343_book.fm Page 440 Tuesday, September 27, 2005 11:53 AM
© 2006 by Taylor & Francis Group, LLC
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. %
1343_book.fm Page 441 Tuesday, September 27, 2005 11:53 AM
© 2006 by Taylor & Francis Group, LLC
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
© 2006 by Taylor & Francis Group, LLC
Bending of Beams 443
.
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. %
1343_book.fm Page 443 Tuesday, September 27, 2005 11:53 AM
© 2006 by Taylor & Francis Group, LLC
444 Mechanics of Composite Materials, Second Edition
.
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
⎡
⎣
⎢⎢⎢⎢⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥⎥⎥⎥⎥
1343_book.fm Page 444 Tuesday, September 27, 2005 11:53 AM
© 2006 by Taylor & Francis Group, LLC
Bending of Beams 445
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
© 2006 by Taylor & Francis Group, LLC
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
© 2006 by Taylor & Francis Group, LLC
Bending of Beams 447
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
© 2006 by Taylor & Francis Group, LLC
448 Mechanics of Composite Materials, Second Edition
.
The inverse of the matrix is
.
Now, in Equation (6.30),
From Equation (6.13),
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
1343_book.fm Page 448 Tuesday, September 27, 2005 11:53 AM
© 2006 by Taylor & Francis Group, LLC
Bending of Beams 449
.
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
..
.-
1343_book.fm Page 449 Tuesday, September 27, 2005 11:53 AM
© 2006 by Taylor & Francis Group, LLC
450 Mechanics of Composite Materials, Second Edition
,
giving
.
The global strains (Equation 6.31) at the top of the third ply (–30°) are
.
∈∈
⎡
⎣
⎢⎢⎢⎢⎢⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥⎥⎥⎥⎥⎥
=
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
1343_book.fm Page 450 Tuesday, September 27, 2005 11:53 AM
© 2006 by Taylor & Francis Group, LLC
Bending of Beams 451
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
© 2006 by Taylor & Francis Group, LLC
452 Mechanics of Composite Materials, Second Edition
(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( )
1343_book.fm Page 452 Tuesday, September 27, 2005 11:53 AM
© 2006 by Taylor & Francis Group, LLC
Bending of Beams 453
From Example 6.3, the maximum bendings’ moment per unit width is
.
From Equation (6.36e),
.
From Equation (6.35),
.
= ×
=
−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
.
.−
⎡
⎣
⎢⎢⎢⎢⎢⎢⎢⎢
⎤
⎦
⎥⎥⎥⎥⎥⎥⎥⎥
1343_book.fm Page 453 Tuesday, September 27, 2005 11:53 AM
© 2006 by Taylor & Francis Group, LLC
454 Mechanics of Composite Materials, Second Edition
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°) 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
1343_book.fm Page 454 Tuesday, September 27, 2005 11:53 AM
© 2006 by Taylor & Francis Group, LLC
Bending of Beams 455
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
× − ××
×
=
..
. %.
1343_book.fm Page 455 Tuesday, September 27, 2005 11:53 AM
© 2006 by Taylor & Francis Group, LLC
456 Mechanics of Composite Materials, Second Edition
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
1343_book.fm Page 456 Tuesday, September 27, 2005 11:53 AM
© 2006 by Taylor & Francis Group, LLC
Bending of Beams 457
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.
1343_book.fm Page 457 Tuesday, September 27, 2005 11:53 AM
© 2006 by Taylor & Francis Group, LLC