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Laminated Beam Theory
Assumptions
Derivation of ODE's from weighted residual form
Constitutive relations for thermoelastic beam
Shear stress distribution (warning: uses sign conventions from aero 306)
Transverse normal stress distribution … assign as HW?
Summary
HW solution
HW solution…maple
HW solution…maple.. mod
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Laminated Beam Theory
Assumptions
z, w
y, v
h
b
x
Chose yz plane because when we go to plate theory, z will be taken perpendicular to plate (of course we
could have taken xz…but xy is not a good choice)
Assumptions/Approximations

b and h are constant

Axial displacement v varies linearly in z-direction.

Transverse displacement w is constant in z-direction   z  0

z  0

 yz y,  h 2  0

 yz  0

f y  constant in z-direction
b g
“Weakest” Assumption:
 yz  0
For “thin” beams the contributions of
“thick” beams, the assumption
 yz to strain energy and deformation is negligible. For
 yz  0 can cause large errors. The shear modulus affects how thick a
beam can be and still be considered thin. The lower the shear modulus the smaller the thickness must be
for a beam to be considered thin. Composites tend to have a smaller ratio of shear to extensional modulus
than homogeneous metals.
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Derivation of Differential Equations of Equilibrium
Change notes… include “b” in the integrations and allow it
to vary through thickness and along length. e.g. N=
integral(stress dArea)
We will derive the equilibrium equations based on the weighted form. This will suffice for us, since we
assumed that “b” and “h” are constant. Generally, it is safer to start with the weak form or just start from
scratch with a free body (not as convenient for higher order theories)
General 3D:
F
zG
Hx
ji
I
J
K
 f i ui dV  0
j
Let’s satisfy equilibrium in a weighted sense for the cross section. We will assume that y= axial direction,
z= vertical direction, and x= through the width.
  y  yz



f

  v dxdz dy  0
y
    y z

 yz
 

    zz  y  f z   w dx dz dy  0
We are assuming no variation in the x-direction (we can relax this assumption in terms of the stresses…
variation that is caused by variation in material properties) Note that if we allow b to vary with the axial
direction, we should put it inside the derivatives.


  y  yz

b

 f y   v dz dy  0
z
 y

 yz
 

b z 
 f z   w dz dy  0
y
 z

These integrals must equal to zero regardless of the limits on the y-integration.
 integral with respect to z (which is integrand for y integration) must = 0
\
æ ¶s y ¶s yz
ö
+
+ f y ÷d v dz = 0
¶y
¶z
ø
òçè
æ¶s z ¶s yz
ö
+
+
f
d w dz = 0
z
òçè ¶z ¶y
÷
ø
This is the weighted residual form. (virtual work before integration by parts). Beam theory has certain
simplifying kinematic assumptions and assumptions about the stresses that will be imposed to simplify
these expressions. The final result will be the governing differential equations of equilibrium.
The next step is to impose the kinematic assumptions.
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Assumed Displacement Field
1.
2.
bg
bgbg bg
w  w y (ie. constant through thickness)
v
v  v y, z  v y, o  z
y, o
z
v
w
v w
 Recall: Assumed  yz  0 



z
y
z y
 Define: v( y,0) v0
w
w
 v  vo  z
or v  v o  z
where   
y
y
Note that positive rotation

is clockwise.
Strains
v vo
2 w

z 2
y y
y
w
z 
0
 yz  0
z
y 
  yo  zk
where k  
2 w
y 2
Substitute assumed displacement into the weighted residual equations
b
b
zFG
H
F
zG
H
 y
y

 yz
z
 z  yz

z
y
Ib
g
J
K
Ibwgdz  0
f J
K
 f y v o  z dz  0
z
Note: v o , , w do not vary with z and they are arbitrary and independent.
Therefore,
æ¶s y
òçè ¶y
æ¶s y
òçè ¶y
æ¶s z
òçè ¶z
+
+
¶s yz
ö
+ f y ÷ dz = 0
¶z
ø
¶s yz
+
ö
+ f y ÷ zdz = 0
¶z
ø
¶s yz
ö
+ f z ÷ dz = 0
¶y
ø
These are equilibrium equations for force in the y-direction, moment about the x-axis, and
force in the z-direction.
We will consider one equation at a time and impose the simplifying assumptions about the stresses.
(1)
(2)
(3)
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  y  yz


 f y  dz  0
y
z

h 2


 y dz   yz dz   f y dz  0

y
z
h
Equation (1)
2
 



z
Define:
z
h
dz   yz
2
|
h
z
=0 since we assumed
2
N   y dz ; f y 
N
 fy  0
y
Equation (2)

But
 yz
(valid since h is constant)
b g
 yz y,  h 2  0
z
f y dz
  y  yz

  y  z  f y  z dz  0
Moment about x-axis due to residual in y-equilibrium
Since we assumed "h" is constant, this can be written as
 yz

z y dz  
zdz   f y z dz  0

y
z

If we assume fy = constant or an even function in z-direction, the third integral is zero. The second
integral can be expressed as

 yz


zdz     z yz    yz  dz 
z
 z

h
z yz
2
   yz dz
|
h
2
0
since  yz  y ,  h 2   0

We will define
z
z
M  z y dz
=> positive moment is clockwise
Q   yz dz
Therefore, the second equilibrium becomes
M
Q  0
y
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  z  yz


 f z  dz  0
z
y

h 2
h
Equation (3)
2
 
2
 z
dz   z | , which is because we assumed  z  0
The first term simplifies to 
z
h 2
h
throughout the beam. Even if we allowed this stress, the integral is zero if we assume zero z-direction
tractions on the top and bottom. The second and third terms simplify as follows
¶s yz
ò ¶y
dz =
¶
¶Q
s yz dz =
ò
¶y
¶y
Where:
Define:
fz 
z
z
Q   yz dz  transverse shear stress resultant
f z dz
Therefore, the third equilibrium becomes
Summary
N
 fy  0
y
Q
 fz  0
y
M
Q  0
y
(valid since “h” is constant)
Q
 fz  0
y
z
z
z
z
z
N   y dz
M  z y dz
Q   yz dz
fy 
f y dz
fz 
f z dz
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Beam Constitutive Relations


Assume plane stress ( no x-direction stresses)
Assume  z  0

Assume
 yz  0
d
yz
z
i
0
y
Result: uniaxial - relation
 y  E  y   T  where  T  T
For a laminated beam, E varies piecewise
Now let’s get the “beam constitutive equations”
Axial stress resultant
N    y dz
  E  y   T  dz
  E y dz   E T dz
  E   yo  zk  dz   E T dz
N  A yo  Bk  N T
or
A   Edz
where
or
B   zEdz
N  N T  A yo  Bk
N T   E T dz
Moment
M   z y dz
  zE  y   T  dz
  zE y dz   zE T dz
  zE   yo  kz  dz   zE T dz
or
M  B yo  Dk  M T
where
or
D   z 2 Edz
M  M T  B yo  Dk
M T   zE T dz
The integral expressions for A, B, and D can be expressed as summations… do it!
The beam constitutive relations can be expressed in matrix form as
N OL
A BO
 O
L
L
N OL
M
P
M
P
M
P
M
P


M
P
M
P
M
P
M
NM QM
Nk P
Q
NM P
QNB DQ
T
yo
T
N T and M T  “Equivalent Mechanical Loads” to obtain deformation
Note: This is all per unit width
Uncoupling…ie B = 0
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Offset z-coordinate so that
z  z  zo
z
zEdz  0  ”B” “matrix”
z
z z
( z  zo ) E dz  0
zEdz  zo Edz  0
zo  constant
 zo 
z
z
zEdz
Edz
Assume piecewise variation (layers)
zo 
 Ei ezi21  zi2 j
 Ei bzi 1  zi g
1
2
Warning: not possible for plates!
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Shear Stress in a Beam
(no thermal effects, no body force in x-direction)
Warning: These notes are from another class that used some different labeling;
 The coordinate axes are different than used elsewhere for beams in mema613

The moment is defines to be
we use
M = - EI
M = EI
d 2v
. Elsewhere in MEMA 613 notes
dx 2
d 2v
.
dx 2
However, this file is self-consistent… so just be aware of the differences as you read.
y
x
h
b
z
x-equilibrium for 3D continuum
 xx  xy  xz


fx 0 :
x
y
z
If we assume no variation in the z-direction, the equation becomes
b gc h
b xy
b xx

0
x
y

b xy
y
b
 xx
x
Integrate from -h/2 to location where you want to know shear stress. (location = y )
b xy
z
y
h 2
y
Since
h

2
dy  
h 2
b xy
z
y
z
y
y
b xx
dy
x
bg
h ) 0 we obtain
dy b y  xy ( y ) b( 
h / 2)  xy ( h 2 ) and  xy ( 
2
z
y
b xx
b y  xy ( y ) 
dy
x
h

2
bg
bgz
1
=>  xy ( y ) 
b y
y
b xx
dy
x
h

2
This can be simplified for special cases (in both of these cases there is no coupling between flexure and
extension)
Homogeneous, symmetric beam
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F
d vI
d v M
yM
and
   
G
J
I
H dx K dx EI
1
 F byM I
1
FFb I b M IJdy
 ( y ) 

dy  

yG GJM 
G
J
z
z
H
K
Hx HI K I x K
bbg
y
x
I
bbg
y
M
F
bI

V and we will assume that
But
G
Jis very small. (We will neglect.)
x
x H
IK
2
 xx E 
y
2
2
xx
2
y
y
h

2
h

2
xy
bgz
bg z
y
y
ybV

V
dy 
bydy
I
b
y
I
h
h


2
2
1
 xy ( y )  
b y
For the special case of b = constant in y-direction
Note:

V y2
I 2
 xy ( y ) 
y

V 2 h2

y  4
2
I
h

2
|
h2
3
s xy (0) = V
=V
= 1.5 times the average shear stress
8I
2b h
Symmetric, laminated beam
(i.e. no extension-flexure coupling)
z
y
 xy
1

b ( y)
d id y
 b x x
x
h

2
 x  E y
d 2v
d x2
  x  E y
M
EI
F
H
z
bg G
y
 xy
d 2v M

d x 2 EI
1

b y

IJ
K

M
bE y
dy
x
EI
h
2
Simplifications & observations

The EI and

Assume
M
do not vary with y for any beam
x
F
IJ
G
HK
 bE
is very small (we will neglect)
x EI
e
j
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 xy
1 1 M

b y EI  x
bg
z
y
bE yd y

But
h
2
M
 V
x
z
y
 xy
V

b y EI
bg
Comments:
 The shear s
b E yd y

xy
h
2
is independent of the shear modulus of the lamina. (only equilibrium was
considered in the derivation)

The variation of s
xy
is piecewise quadratic if E is constant in each layer and dimensions do not
change in the y-direction.

The variation of s
xy
is tied to the assumed variation of
exx vs. y.
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Transverse Stresses in a Homogeneous Beam (check for typos)
 2 h2 
 z    derived elsewhere for b constant w.r.t. thickness direction
4

b zz b yz

 bf z  0
z
y
 yz 
V
2
Integrate from -
h
to z
2

b zz z  b zz

    
z
h
2
h
z
b yz

 bf z dz
y

( 2) = 0
1) Assume s zz - h
b
 0 for simplicity
y
¶s yz - ¶V 1 æ z h2 ö
=>
=
z - ÷
¶y
¶y 2I ç
4ø
è
2)
Recall:
V
 f z bh  0
y
Recall:


 zz z  
  f b h 2  z
2
 zz z   f z
 zz
2
z
h


z
1
2
 yz
y
h2
 f z bh

V
  f z bh
y
1  z h2 
z  
2 
4
4  bf  dz
 b2 h  2 h2 
 b  2  z  4  dz
h 
2
z
z
(assume fz=constant)

b2 h  z 3 h2 z  z
z   f z bz 
 
 |
2  3
4  h

2



 zz z   f z bz 

b 2 h 12  z 3 h 2 z   z
6b  z 3 h 2 z   z



f
bz

|




 
 |
z
2 bh3  3
4  h
h2  3
4  h

2
2

Note that for this beam theory, there is no transverse normal stress (per equilibrium
considerations) unless there is a distributed load in the transverse direction.
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Solution of Coupled Equations for General Laminated Beam
N
 fy  0
y
Q
 fz  0
y
M
Q  0
y
(1)
(2)
(3)
Combine (2) and (3):
M
Q
y
Constitutive Relations: N A
2 M
f z 0
y 2

F
IJ
G
H K
v
2 w
B  2
y
y
M B
(4)
F
IJ
G
H K
v
2 w
D  2
y
y
Write (1) and (4) in terms of v and w: (note: v= axial displacement where z=0)
F
IJ
G
H K
F  wIJ f
v
B
 DG
y
Hy K
A
2v
3w

B

 fy  0
y 2
y 3
3
4
3
4
z
(5)
0
(6)
______________________________________________________________________________
Rewrite (5) as
2v B  3w  f y
 
  . Substitute this into equation (6) to obtain
y 2 A  y 3  A
 4 w 
  B  3w  f y 
B   3     D   4   fz  0
y  A  y  A 
 y 
For simplicity, let’s assume B does not vary. We then obtain

B 2   4 w    B 
D


 
   f y   fz  0
A  y 4  y  A 

Now we have one fourth order equation in terms of just w(y). It should be noted that we could have derived
an equation in terms of just v(y).
If the distributed loads are not too complicated, this ODE can be integrated four times to obtain w(y) in
terms of four integration constants. For example, if the only distributed load was
f z  ay , we would
obtain
To obtain the complete general solution, the expression for w(y) is substituted into
equation (5), which is must be solved to obtain v(y). Integrating equation (5) twice will
introduce two more integration constants. At this point we have expressions for w(y) and
w=
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v(y) in terms of 6 integration constants. These integration constants are determined by
imposing the BC’s. For example, here is one possible set of BC’s.
bg
vbg
L v
v 0 v1
2
bg
wbg
L w
w 0 w1
2
w
0  1
y
w
L  2
y
bg
bg
There are six equations and six intergation constants.
We can solve for the integration constants.
The BC’s above are all kinematic BC’s, but in general would be some mix of kinematic and force type
BC’s.
D:\106744654.doc
Transformation of ABD Matrix Due to Shift in z-Coordinate System