FEM 7 Beams and Plates.ppt

Beams
Beams are slender members that are used for supporting
transverse loading.
Beams with cross sections that are symmetric with respect to
plane of loading are considered here. A general horizontal beam is
shown in Fig.8.1. Figure 8.2 shows the cross section and the
bending stress distribution.
1

Figure 8.1 (a) Beam loading and (b) deformation of the neutral axis
2

Figure 8.2 Beam section and stress distribution
3

M
y
I
  
E

 
2
2
d M
dx EI


(8.1)
(8.2)
(8.3)
where σ is the normal stress, ε is the normal strain, M is the bending
moment at the section, υ is the deflection of the centroidal axis at x,
and I is the moment of inertia of the section about the neutral axis
(z-axis passing through the centroid).
4

Potential – Energy Approach
The strain energy in an element of length dx is
2
2
2
1
2
1
2
A
A
dU dAdx
M
y dA dx
EI

 
 
  
 


5

Noting that
2
A
y dA
 is the moment of inertia I, e have
2
2
1
2
M
dU dx
EI
 (8.4)
where Eq.8.3 is used, the total strain energy in the beam is given
by
2
2
2
0
1
2
L
d
U EI dx
dx

 
  
 
 (8.5)
6

The potential energy of the beam is then given by
2
2
'
2
0 0
1
2
L L
m m k k
m k
d
EI dx p dx P M
dx

  
 
   
 
 
 
  (8.6)
where p is the distributed load per unit length, Pm is the point
load at point m, Mk is the moment of the couple applied at
point k, is the deflection at point m, and '
k

is the slope at point k.
m

7

The beam is divided into four elements, as shown in Fig.8.4. Each
node has two degrees of freedom, transverse displacement and
slope or rotation. The vector
Q = [Q1, Q2,…….Q10]T (8.13)
FINITE ELEMENT FORMULATION
8

e 1 2
1 1 2
2 2 3
3 3 4
4 4 5
Local
Global
Fig.8.4 Finite element discretization
9

Figure. 8.5 Hermite shape functions
10

represents the global displacement vector. For a single element,
the local degrees of freedom are represented by
q = [q1,q2,q3,q4]T (8.14)
The shape functions for beam element are
2 3
, 1,2,3,4
i i i i i
H a b c d i
  
     (8.15)
11

The conditions given in the following table must be satisfied:
H1 H’1 H2 H’2 H3 H’3 H4 H’4
= -1
= 1
1
0
0
0
0
0
1
0
0
1
0
0
0
0
0
1
The coefficients ai, bi, ci and di can be easily obtained by imposing
these conditions.
12

2 3
1
2 2 3
2
2 3
3
2 2 3
4
1 1
(1 ) (2 ) (2 3 )
4 4
1 1
(1 ) ( 1) (1 )
4 4
1 1
(1 ) (2 ) (2 3 )
4 4
1 1
(1 ) ( 1) ( 1 )
4 4
H or
H or
H or
H or
   
    
   
    
    
     
    
      
(8.16)
The Hermite shape functions can be used to write υ in the form
  1 1 2 3 2 4
1 2
d d
H H H H
d d
 
   
 
   
   
   
   
(8.17)
13

1 2
1 2 2 1
1 1
2 2
2 2
x x x
x x x x
 

 
 
 
 
(8.18)
Since ℓe = x2 – x1 is the length of the element, we have
2
e
dx d
 (8.19)
The coordinates transform by the relationship,
14

2
e
d d
d dx
 

 (8.20)
Noting that rule dυ/dξ evaluated at nodes 1 and 2 is q2 and q4,
respectively, we have
1 1 2 2 3 3 4 4
( )
2 2
e e
H q H q H q H q
      (8.21)
which may be denoted as
Hq
  (8.22)
The chain rule   
/ / /
dv d dv d dx d
  
 gives us
15

where
1 2 3
, , , 4
2 2
e e
H H H H H
 
  
 
(8.23)
In the total potential energy of the system, we consider the
integrals as summations over the integrals over the elements.
The element strain energy is given by
2
2
2
1
2
e
e
d
U EI dx
dx

 
  
 
 (8.24)
16

From Eq. (8.20),
2 2
2 2 2
2 4
e e
d d d d
and
dx d dx d
   
 
 
Then, substituting = Hq, we obtain

2
2 2 2
4 2 2
16
T
T
e
d d H d H
q q
dx d d

 
     

     
     
(8.25)
17

2
2
3 1 3 3 1 3
, , , ,
2 2 2 2 2 2
e e
d H
d
 
 

    
 
 
   
 
 
(8.26)
On substituting  
/ 2
e
dx d
 and Eq. 8.25 and 8.26 in Eq. 8.24,
   
 
 
2 2
2 2
2 2
1
3
1
2
2
2
9 3 9 3
1 3 1 3
4 8 4 8
1 3 1 9
3
1 3
4 8 16
1 8
2
9 3
1 3
4 8
1 3
4
e e
e e e
T
e
e
e
e
EI
U q d q
Symmetric
     
 
 

  



 
   
 
 
 
 
   
 
 
  
 
 
 
  
 
 
 
 
 
 

 
 
 
 
 
 (8.27)
18
we get

Each term in the matrix needs to be integrated. Note that
1 1 1
2
1 1 1
2
0 2
3
d d d
    
  
  
  
  
This results in the element strain energy given by
1
2
T e
e
U q k q
 (8.28)
19

where the element stiffness matrix is
2 2
3
2 2
12 6 12 6
6 4 6 2
12 6 12 6
6 2 6 4
e e
e e e e
e
e e
e
e e e e
EI
k

 
 

 

 
  
 

 
 
(8.29)
which is symmetric.
20

The load contributions from the uniformly distributed load p in the
element is
Load Vector
1
1
2
e
e
p
p dx H d q
 

 
 
 
  (8.32)
On substituting for H from Eqs. 8.16 and 8.23 and integrating, we
obtain
e
eT
p dx f q
 
 (8.33)
21

where
2 2
, , ,
2 12 2 12
T
e e e e e
p p p p
f
 
 
 
 
(8.34)
22

e
p dx


1
2
T T
Q KQ Q F
 
0
T T
KQ F
 
 

This equivalent load on an element is shown in Fig.8.6. The same
result is obtained by considering the term in
Eq.8.12 for the Galerkin formulations. The point loads Pm and Mk
are readily taken care of by introducing nodes at the points of
application. On introducing the local-global correspondence, from
the potential-energy approach, we get
and from Galerkin’s approach, we get
where = arbitrary admissible global virtual displacement vector
(8.35)
(8.36)
23

Boundary Considerations
 
2
1
2
r
C Q a to
 
When the generalized displacement value is specified as a for
the degree of freedom (dof) r, we follow the penalty approach
and add
KQ = F (8.37)
a=known generalized displacement
Figure 8.7 Boundary conditions for a beam 24

Using the bending moment and shear force equations
SHEAR FORCE AND BENDING MOMENT
2
2
d dM
M EI V and Hq
dx dx


  
we get the element bending moment and shear force:
   
1 2 3 4
2
6 3 1 6 3 1
e e
e
EI
M q q q q
   
 
     
  (8.38)
25

 
1 2 3 4
3
6
2 2
e e
e
EI
V q q q q
    (8.39)
These bending moment and shear force values are for the
loading as modeled using equivalent point loads. Denoting
element and equilibrium loads as R1, R2, R3, and R4, we note
that
26

1 1
2
2 2
2 2
3
3 3
2 2
4 4
2
2
12 6 12 6
6 4 6 2 12
12 6 12 6
2
6 2 6 4
12
e
e e
e
e e e e
e
e e e
e e e e
e
p
R q
p
R q
EI
R q p
R q
p

 
 
 

 
     
 
     
  
     
 
   
  
 
     
 
     
 
     
 
   
     
 
     
 
     
 

   
   
 
 
 
(8.40)
27

It is easily seen that the first term on the right key. Also note that
the second term needs to be added only on elements with
distributed load. In books on matrix structural analysis, the
previous equations are written directly from element equilibrium.
Also, the last vector on the right side of the equation consists of
terms that are called fixed-end reactions. The shear forces at the
two ends of the element are V1 = R1 and V2 = -R3. The end bending
moments are M1 = -R2 and M2 = R4.
28

Example 8.1
For the beam and loading shown in Fig.E8.1, determine (1) the
slopes at 2 and 3 and (2) the vertical deflection at the midpoint of
the distributed load.
29

Solution: We consider the two elements formed by the three
nodes. Displacements Q1,Q2,Q3, and Q5 are constrained to be zero,
and Q4 and Q6 need to be found. Since the lengths and sections are
equal, the element matrices are calculated form Eq.8.29 as follows:
  
9 6
5
3 3
200 10 4 10
8 10 /
1
EI
N m

 
  
1 2 5
1 2 3 4
3 4 5 6
12 6 12 6
6 4 6 2
8 10
12 6 12 6
6 2 6 4
1
2
k k
e Q Q Q Q
e Q Q Q Q

 
 

 
  
 
  
 

 


30

We note that global applied loads are F4 = -1000 N.m and F6 =
+1000 N.m obtained from as seen in Fig.8.6. We
2
/12,
p
use here the elimination approach presented. Using the connectivity,
we obtain the global stiffness after elimination.
     
   
1 2 2
44 22 24
2 2
42 44
5 8 2
8 10
2 4
k k k
K
k k
 

 
 
 
 
   
 
31

The set of equations is given by
4
5
6
8 2 1000
8 10
2 4 1000
Q
Q

 
   
 
   
  
   
 
The solution is
4
4
4
6
2.679 10
4.464 10
Q
Q


 
 
   

   

 
   
For element 2, q1 = 0, q2 =0, q3 = Q4, and q4 = q6. To get vertical
deflection at the midpoint of the element, use 0:
Hq at
 
 
32

   
2 4 4 6
4 4
5
0 0
2 2
1 1 1 1
2.679 10 4.464 10
2 4 2 4
8.93 10
0.0893
e e
H Q H Q
m
mm

 

   
     
     
     
     
 

33

Prof .N. Siva Prasad, Indian Institute of Technology Madras
In many engineering applications, beams are supported
on elastic members. Shafts are supported on ball, roller,
or journal beatings. Large beams are supported on elastic
walls.
BEAMS ON ELASTIC SUPPORTS

Based on the nature of stress state, plates are classified
as
1.Thick plate: The ratio of thickness to least dimension on
plan exceeds 1/10 may be taken as belonging to this class.
It is defined by a complete set of differential equations of
three dimensional theory of elasticity.
PLATE BENDING

2. Thin plates: The ratio of thickness to span does not
exceed 1/10 and the maximum deflection w is less
than h/10 to h/5. Small deflection in which the
membrane stresses are very small compared to
flexural stresses under deformation due to transverse
loading.
3. Thin plates with large deflection: Flexural stresses
are accompanied by relatively large tensile or
compressive stresses in the middle plane. These
membrane stresses significantly affect the bending
moment.

Classical thin plate theory (referred to as Kirchhoff’s theory)
is based upon the assumptions.
1. A lineal element of the plate normal to the midsurface, on
application of the load(a) undergoes at most a translation
and rotation and (b) remains to the deformed mid-surface.
2. The stresses normal to the plate can be neglected.
Thin Plate Theory

From assumptions 1(a) a lineal element through the
thickness does not elongate or contract. 1(b) the shear
strains γxz and γyz becomes zero, i.e., γxz = γyz = 0. (2)
implies that σz = 0. Thus the three-dimensional problem has
been reduced to a two-dimensional problem.

(a) Detailed illustration

(b) Vectorial notation
Figure 1 External and internal forces on the element of
the middle surface

The strain distribution corresponding to Eq. 2 is given by
2
2
2
2
x x
y y
u w
z z k
x x
w
z z k
y y

 
   
 
 
   
 
Basic Relationships
/2 /2
/2 /2
h h
x x y y
h h
M z dz M z dz
 
 
 
 

and
/2
/2
h
xy yx xy
h
M M z dz


   1(a)
Similarly, the stress resultants Qx and Qy are given by
/2 /2
/2 /2
h h
x xz y yz
h h
Q dz Q dz
 
 
 
  1(b)
Let u, v and w be the displacements at any point (x,y,z) and
the variation of the displacement u and v across the thickness
can be expressed in terms of the displacement w as

w w
u z v z
x y
 
  
 
(2)
2
2
xy xy
u v w
z z K
y x x y

  
   
   
(3)
where
2 2 2
2 2
2
,
x y xy
w w w
k k and k
x y x y
  
  
   
(4)

and the shear strains γxz = γyz =0. Thus the problem is
reduced to a plane stress problem. The general
constitutive law for plane stress is given by
11 12 13
21 22 23
31 32 33
x x
y y
xy xy
C C C
C C C
C C C


 
   

 
   
 
 
   
 
   
 
 
   
(5)
In the case of plates, it is convenient to regard the stress
resultants {M}T = {Mx, My,Mxy} instead of {σ}T = { σx σy τxy}.
Thus substituting the Eqs.(5 and 3) in Eq(1a) we get

11 12 13
3
21 22 23
31 32 33
12
x x
y y
xy xy
M C C C k
h
M C C C k
C C C
M k
   
 
   
 

   
 
   
 
 
   
i.e. {M} = [Cf] {ke}
where {ke}T = [kx ky kxy] (6)
This equation is exactly similar to {σ} = [C] {ε}. In case of
isotropic plates, the constitutive matrix is given by

2
(1 )
xy p
w
M D
x y


 
 
(8)
where
3
2
12(1 )
p
Eh
D



In general, there are three classes of displacement functions
which can be chosen for plate element and they are as
follows.
DISPLACEMENT FUNCTIONS

1. Class, C2: The assumed function w(x,y) has
continuous second derivatives (and hence curvatures)
at element corners and inside the element.
2. Class, C1: The assumed function for w(x,y) has
continuous first derivatives but may have
discontinuous corner curvatures.
3. Class, C0: The assumed function only is continuous
and independent functions are assumed to represent
the variation of w and slopes.

PLATE BENDING ELEMENTS
Rectangular Plate Element with 12 Degrees of Freedom
Fig. 2 Rectangular element with 12 degrees of freedom

The displacement filed is given by
2 2 3 2 2
1 2 3 4 5 6 7 8 9
3 3 3
10 11 12
w x y x xy y x x y xy
y x y xy
        
  
        
  
(9)
The three variables at each corner node are
b
i i
w w
w and
x y
 
 
 
 
 
 
   
i.e.  
T
i i
i i
w w
d w
x y
 
 
 
 
 
 
 
 
 
 
 
 
(10)

We would like to study the compatibility of displacement
and slopes along the interface edges. Consider the edge
1-2 which is assumed to be the interelement boundary
between two elements A and B joining along this edge
(Fig 3). The displacement w along this edge is given by

(a) Displacement Variation

(b) Transverse Slope Variation
Fig. 3 Compatibility conditions along the 1 -2 edge

Mindlin’s Theory to Include Shear Deformation in Plates
The three assumptions made in Mindlin’s Theory of Plates
are given below:
i. The deflections of the plate, w, are small.
ii. Normals to the plate midsurface before deformation
remain straight but are not necessarily normal to it after
deformation.
iii. Stresses normal to the midsurface are negligible.
Rectangular Plate Element with 16 Degree
of Freedom

The second assumption to account for shear deformation
is different from Kirchhoff’s theory while the first and third
assumptions are the same for both the theories.
Referring to Fig.5 (b), in which x

transverse shear strain for a section x = constant, the total
denotes an average
rotation θy can be expressed as,
y x
w
x
 

 

(11)

and similarly for section y = constant (Fig 5(a) ).
x y
w
x
 

  

(12)
Hence, the average shear deformations, and are given by
x
 y

x y
w
x
 

 

y x
w
y
 

 

(13)

(a) (b)

Fig. 4 Rotation of the normals about x and y axes
considering average shear deformation
(c)

z

y x
w w
and
x y
 
 
 
 
Equations 13 are based on the assumptions that the total
rotations θx and θy are small and the transverse strain
is negligible. In the limit tening to thin plate range the shear
strains tend to become zero and hence
The expression for strain energy to include the contribution
due to shear can be written as
   
2
2
1
2
s x y
U xGA dxdy
 
 
 
 
 
 (14)

Substituting for x
 and y
 from Eqs 13, we get, for isotropic
plate,
 
  2
2 2
3
1
2
3
2
24 1
24 1
s y x
Eh w w
U GA dxdy
Eh x y

  

   
  
 
 
 
    
 
 
 
 
  
   
 
 
 
 (15)
Since
2(1 )
E
G



, Eq. 15 can be written as,
 
 
2
2
3
2
2
6 1
24 1
s y x
x
Eh w w
U dxdy
h x y

 

 
  
 
 
    
 
 
 
 
  
 
 
 
 (16)

Therefore, the total strain energy is given by,
b s
U U U
  (18)
Displacement Model
The geometry of the element is given by
FOUR NODED ISOPARAMETRIC ELEMENT
(PLATE4)

4
1
4
1
i i
i
i i
i
x N x
y N y






(1)
The variation of displacements within an element is
expressed in terms of the nodal values as
4
1
i i
i
w N w



4
1
4
1
x i xi
i
y i yi
i
N
N
 
 






(2)

Fig. 10.6 Details of bilinear plate element.

Figure 1 shows the stress resultants (forces per unit
length) acting on a shell element.
THIN SHELL THEORY
Figure 1 Stress resultants

Flat Plate Elements
The earliest attempts at constructing suitable shell elements
were based on combining membrane elements with P late
bending elements
The difficulties and the shortcomings of the flat plate element
used for the analysis of shells are
1. The behavior of the shells as represented by the
differential equations is not approached in the limit of
refinement of flat plate approximation.
REVIEW OF SHELL ELEMENTS

2. The discontinuities of slope between adjacent
plate elements may produce bending moments
in the regions of shells where they do not exist.
3. The coupling of membrane and bending effects
due to curvature of the shell is absent in the
interior of the individual elements.

Four general difficulties encountered in the development
of curved elements are
1. The choice of an appropriate shell theory as there
are quite a number of theories being available.
2. The description of the geometry of the elements
using the given element data.
Curved Shell Elements

3. The satisfaction of rigid body modes of behavior is
acute in curved shell analysis.
4. The inter element compatibility condition is difficult to
achieve as pointed out in the case of plate elements.

BILINEAR DEGENERATED SHELL ELEMENT
The following two assumptions are made:
1.Normals to the midsurface remain straight after
deformation. Thus, the formulation include transverse
shear deformation and Kirchhoff- Love hypothesis is not
assumed.
2.Stresses normal to the midsurface are zero.

The four noded element is evolved from an eight noded
solid element. The midsurface enclosed by four straight
sides forms a hyperbolic paraboloid. The shell element is
shown in Fig. The shape function to describe the
midsurface in terms of natural coordinates is the same as
given by Eq. for two dimensional isoparametric element.
Thus,
Shape Functions for Geometry and Displacement

  
1
4 1 1 1,......,4
i i i
N rr s s i
    (1)
where ri and si are the natural coordinates of node i
The thickness of the shell element in the direction normal to
the midsurface at each node is required and is specified as
input.
3
4
3
1
3
1
2
i i
i i i i
i
i i
x l
x
y N y th m
z z n

 
   
 
 
     
 
     
     
     
 
 (2)

where xi, yi, zi are the global coordinates of the midsurface
node i.
hi is the thickness at node i and
l3i, m3i and n3i are the normal unit vector at node i

Fig. Four noded shell element

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