Transcript Diapositiva 1 - Politecnico di Milano

Bars and Beams
FEM Linear Static Analysis
Stiffness matrix formulation: bar element
•Uniform prismatic elastic bar of length L, elastic modululus E and
area A (2 dof)
•Only axial direct displacement are allowed
F11  F21 
AE
u1
L
Forces that must be applied to the nodes in order to maintain the
displacement state
  u1
AE
F11  F21 
u1
L
Nodal forces
  u2
F12  F22 
AE
u2
L
F1  F11  F12
F2   F21  F22
Advanced Design for Mechanical System - Lec 2008/10/07
2
Direct Method
d  u1 u2 
T
Nodal displacement vector
 F11
F
 21
F12  1  F1 
  

F22  1  F2 
AE  1  1 u1   F1 
  


L  1 1  u2   F2 


 
K
d
r
A column of K is a vector of nodal loads that must to be applied to the
element to sustain a deformation state in which:
• the corresponding nodal d.o.f has unit value;
• all other nodal d.o.f. are zero.
•For example with u1=1 and u2=0
AE  1  1 1  F1 
  


L  1 1  0 F2 
 F1  AE  1 
 F11 
 
 u1  

F

F

1
L
 
 2
 21 
Advanced Design for Mechanical System - Lec 2008/10/07
3
Formal procedure
• The direct methods can produce a stiffness matrix only for simple
elements, where the relations between nodal displacements and
loads are known.
• For most general elements a general formula for k must be used
• It can be derived by stating that work is done by nodal loads that are
applied to create nodal displacements and this work is stored in the
element as elastic strain energy
• Stiffness matrix
k   ΒT EBdV
B strain displacement matrix
E elastic modulus
dV increment of the element volume
Advanced Design for Mechanical System - Lec 2008/10/07
4
Shape function matrix
• To obtain B we write the axial displacement u of an arbitrary point on
the bar as linear interpolation between its nodal value u1 and u2
 L  x x   u1 
u
 

 L L  u2 
u  N1 N 2  d

N
N shape function matrix
• Each shape function Ni describe how u varies with x when the
corresponding d.o.f. ui is unity while the other is zero.
Advanced Design for Mechanical System - Lec 2008/10/07
5
Stiffness matrix
•Axial strain is the gradient of the displacement
du  d 
x 
  Nd  Bd
dx  dx 
 1 1
dove B  
 L L 
•For the bar problem, matrix E is simply the elastic modulus E, a scalar
• dV is Adx
The equation for k becomes
 1 L  1
k  
E 
1L  L
0
L
1
AE  1  1
Adx 

L
L  1 1 
The definition of k guarantees that it will be a Symmetric matrix
Advanced Design for Mechanical System - Lec 2008/10/07
6
Limitation of the two nodes bar element
•It can represent only a constant state of strain
•In terms of generalized coordinates
u  1   2 x
du
x 
 2
dx
•If axial force are applied only at nodes, the element agree exactly
with a mathematical model that represents the bar as straight line
having constant A and E between locations where axial forces are
applied
•If axial force are distributed along all part of the length or if the bar is
tapered, then the element is only approximate
•Distributed load can still be applied, in the form of equivalent forces
applied to nodes
Advanced Design for Mechanical System - Lec 2008/10/07
7
Simple plane beam element
• Resist in plane bending and transverse shear forces (4 dof)
A area of the cross section
E elastic modulus
I centroidal moment of inertia
Nodal displacements consist of lateral translations v1 and v2 and
rotations qz1 and qz2 about the z axis
d  v1 q z1 v2 q z 2 
T
The beam centerline has lateral displacement is v=v(x)
v=v(x) is cubic in x for a uniform prismatic beam loaded only at its ends
We will ignore transverse shear deformation, although commercial
software account for it
Advanced Design for Mechanical System - Lec 2008/10/07
8
Shape functions
One d.o.f. has unit value and all other d.o.f. are zero
Advanced Design for Mechanical System - Lec 2008/10/07
9
Stiffness matrix
kij are nodal forces and moments that must be applied to sustain the
assigned deformation state, positive direction are upward for forces and
counterclokwise for moment
To solve for the first column of k, the column vector
k11 k21 k31 k41 
T
the following conditions are used

k11 L3 k21 L2
v1  1 at node1 1 


3EI
2 EI
standardbeam deflectionformulas
3
2
k L k L
q Z 1  0 at node1 0   11  21 
2 EI
EI 
( forces) y  0
0  k11  k31
(m om ents) node 2  0

staticsequations
0  k21  k41  k11 L 
Advanced Design for Mechanical System - Lec 2008/10/07
10
Stiffness matrix
•Each deformation state yields terms in one column of k
•The result of this process is the element stiffness matrix
 12EI L3

2
6
EI
L
k
 12EI L3

2
6
EI
L

6 EI L2
 12EI L3
4 EI L
 6 EI L2
 6 EI L2
12EI L3
2 EI L
 6 EI L2
which operates on the vector of nodal d.o.f.
Advanced Design for Mechanical System - Lec 2008/10/07
6 EI L2 

2 EI L 
 6 EI L2 

4 EI L 
d  v1 q z1 v2 q z 2 
T
11
Formal procedure for simple beam element
• The general expression of stiffness matrix for flexural deflection is:
L
k   BT EI Bdx
0
where B is a matrix that yields curvature d2n/dx2 of the beam
element from the product B d.
1 T
In each case the expression: d Bd represents the strain
2
energy in an element under nodal displacement d.
In bars, strain energy depends on axial strain; in beams, strain
energy depends on curvature.
Advanced Design for Mechanical System - Lec 2008/10/07
12
In terms of generalized coordinates the lateral displacement is a
cubic function in x:
v  1  2 x  3 x2  4 x3
The j terms can be stated in terms of nodal d.o.f. making the
substitution at beam extremities, like, for example:
x  0 v  v1 q z  q z1 q z 
dv
dx
Thus an alternative form for the lateral displacement uses shape
function Ni
 v1 
q 
 
v  N1 N 2 N 3 N 4  z1   Nd
 v2 
q z 2 
where each Ni states the deflected shape associated with a
particular end translation of rotation
Advanced Design for Mechanical System - Lec 2008/10/07
13
Curvature in simple beam element
The curvature of the beam element is expressed in matrix form using
the derivative of the shape function and the nodal displacements
d 2v  d 2 
  2 Nd  Bd
2
dx  dx 
Where strain-displacement matrix B is:
 6 12x
B   2  3
L
 L
4 6x
  2
L L
6 12x
 3
2
L
L
Advanced Design for Mechanical System - Lec 2008/10/07
2 6x 
  2
L L 
14
Stress in simple beam element
Bending moment and flexural stresses are computed from curvature,
which in turn depends on nodal d.o.f. d.
If y is the distance from the neutral axis, the following holds:
d 2v
M  EI 2  EIBd
dx
My

I
Flexural moment M caused by nodal displacement d varies linearly
with x in each element
Advanced Design for Mechanical System - Lec 2008/10/07
15
Limitation of the simple plane beam element
•Under the usual restriction, that the beam is initially straight,
linearly elastic, without taper,a beam subjected to forces and
moment only at nodes, has a deflected shape that is cubic in x, just
as described by the shape functions Ni
•A FE model built of beam elements provides an exact solution
when force and/or moment are applied to its nodes only.
•A uniformly distributed load produces a beam deflection that is
fourth degree in x.
•Accordingly, beam elements are inexact under distributed loads,
but exact results are approached as more and more elements are
used in the FE model.
Advanced Design for Mechanical System - Lec 2008/10/07
16
2D Beam Element
• It is a combination of a bar element and a simple beam element, and
it might also be called a plane frame element.
• It resists axial stretching, transverse shear force, and bending in one
plane.
• Combining bar and simple beam stiffness matrices, the stiffness
matrix of the 2D beam element is the following, indicating on the right
the d.o.f. on which k operates:
Advanced Design for Mechanical System - Lec 2008/10/07
17
3D beam element
Web of beam section
Local reference frame
global reference frame
A beam element in a general purpose FE program has three-dimensional
capability and may also be called a “space beam” element.
A global reference frame XYZ is introduced, while the longitudinal axis of the
beam lies along a local x axis, defined by the coordinates of nodes 1 and 2, of a
local reference frame xyz.
In order to orient the local frame xyz, a node 3 is introduced, whose coordinates
serve to orient the xy plane in XYZ space. In the example the xy plane is placed
along the web of the beam section. No d.o.f. are associated to node 3.
Advanced Design for Mechanical System - Lec 2008/10/07
18
d.o.f. of 3D beam element
6 dof for each node
12 dof for each element
At each node the element has six d.o.f. (three displacements and three
rotations)
The matrix k is formulated in the local reference frame, then k is
transformed so that global d.o.f. replace local d.o.f. at each node.
The element resists forces in any direction and moments about any axis.
Advanced Design for Mechanical System - Lec 2008/10/07
19
The following data are needed to define the 3D beam element:
•Nodal coordinates (nodes 1, 2 and 3), for beam and its section
orientation
•Parameters:
A
E, G
Iyy Izz
J
fy, fz
cross section area
Young modulus and shear modulus (or Poisson’s
coefficient n)
section inertia moment about local y and z axis
torsion constant
shear deformation factors
Advanced Design for Mechanical System - Lec 2008/10/07
20
3D beam element: some remarks
L
Additional data are needed for stress computation, such as the appropriate
distance y in the flexure formula.
If the section is noncircular, note that J is not the polar moment of inertia of the
cross sectional area A. J is a property of the cross section, such that the correct
relative rotation of nodes a and 2 under the torque T is given by TL/GJ.
Often (especially for open sections, like the one of the example), is much lower
than the polar moment of A
Advanced Design for Mechanical System - Lec 2008/10/07
21
Properties of k (element) and K (structure)
 Stiffness matrices k e K are symmetric. This is true for any
element or structure when there is a linear relationship between
applied loads and resulting deformations
 Each diagonal coefficient is positive. This corresponds the say that,
in a configuration where the corresponding d.o.f is the only one
existing, a load ri and its displacement di must be directed in the
same direction
ri  kiidi
 A negative diagonal coefficient kii would mean that a load and its
displacement are oppositely directed, which is unreasonable
Advanced Design for Mechanical System - Lec 2008/10/07
22
Avoiding singularities for K
 A structure that is unsupported or inadequately supported has a
singular stiffness matrix K, and the software will not be able to
solve the equation:
KD=R
 To prevent singularity, supports must be sufficient to prevent all
possible rigid-body motions.
 These are motions that produce no deformation of the structure.
 For example, in a one element structure made of a simple beam
element (4 d.o.f.):
- if unsupported, it can have two rigid-body motions in the xy plane
lateral translation and rotation about a point
- is adequately supported if at least two d.o.f. are prescribed,
expect that the two rotations are prescribed, in such case a rigid
translation is possible.
Advanced Design for Mechanical System - Lec 2008/10/07
23
 Adequate support make the structure statically determined
 More than adequate support make the support reaction statically
indeterminate, but this is perfectly acceptable and it does not
complicate the FE process in any way.
 A structure may have a singular k matrix because it contains a
mechanism.
 Imagine a straight beam, attached to a rigid support at each end,
and modeled by two beam element. The model is stable and has
not mechanism.
 If the two beam element are replaced by two bar elements The
structure is a mechanism because two collinear bar elements can
not resist a lateral force applied to the central node.
Advanced Design for Mechanical System - Lec 2008/10/07
24
Mechanical loads
•Load may be applied as:
- a concentrated force or a moment directly to a node
- as force or moment distributed along a line
- as a surface pressure.
•Body force loading, which acts at every material point in the body include
self weight (gravity) and acceleration (inertia forces)
•Thermal loading comes from temperature changes.
•Moment load can be applied to a node if at least one element connected to
that node has rotational d.o.f. in its stiffness formulation.
•Input data to software consists of the magnitude, direction, and node
associated with the force or moment.
Advanced Design for Mechanical System - Lec 2008/10/07
25
Line loads
• Axial force, weight or resistance to acceleration , (line load q [N/m])
uniformly distributed on a bar element, must be converted into equivalent
nodal loads
•Total load on an element is qL, Half of this load is applied to each node
• If two collinear elements of length La and Lb are connected, the node
they share will be loaded by a total force qLa/2+qLb/2.
•Most software are able to accomplish the conversion, the user need only
tell the software what the loading is and where it acts.
Advanced Design for Mechanical System - Lec 2008/10/07
26
Transversal loads
• Theory indicates that a transversal load (q [N/m]) uniformly distributed
on a beam element converted into equivalent nodal loads that consists
of forces and moments.
• These loads are the support reactions (direct opposite) for a uniform
beam fixed at both ends and uniformly loaded.
•If elements of equal length and equal distributed loads are assembled,
moment loads cancel at nodes shared by two elements.
Advanced Design for Mechanical System - Lec 2008/10/07
27
Stresses
•A FE software calculates the displacements first, and form these the stresses
are obtained.
• Nodal displacements are more accurate than stresses, because stresses
are proportional to strains, and strains are derivatives of the displacements.
•Stresses are well approximated at the centre of the element, as well as the
mean stress on the element
•Stresses at boundaries are usually not well approximated.
Maximum value not well approximated
Advanced Design for Mechanical System - Lec 2008/10/07
28