11 10-jan-16 last course interpretations and properties of the stiffness matrix (cont’d) the dsm...
TRANSCRIPT
![Page 1: 11 10-Jan-16 Last course Interpretations and properties of the stiffness matrix (cont’d) The DSM for plane and space trusses](https://reader034.vdocuments.net/reader034/viewer/2022051401/5697c0221a28abf838cd31bf/html5/thumbnails/1.jpg)
11Apr 21, 2023
Last course Interpretations and properties of the
stiffness matrix (cont’d) The DSM for plane and space trusses
![Page 2: 11 10-Jan-16 Last course Interpretations and properties of the stiffness matrix (cont’d) The DSM for plane and space trusses](https://reader034.vdocuments.net/reader034/viewer/2022051401/5697c0221a28abf838cd31bf/html5/thumbnails/2.jpg)
2
Today
One-dimensional bar element
![Page 3: 11 10-Jan-16 Last course Interpretations and properties of the stiffness matrix (cont’d) The DSM for plane and space trusses](https://reader034.vdocuments.net/reader034/viewer/2022051401/5697c0221a28abf838cd31bf/html5/thumbnails/3.jpg)
3
Example
See Example 3.8 from D.L. Logan (2007).
E=1.2 x 106 psi for all elements.
See the Matlab code, Lect5_Truss2.m for the solution.
Answer: σ1=-945 psi; σ2=1440 psi; σ3=-2850 psi
![Page 4: 11 10-Jan-16 Last course Interpretations and properties of the stiffness matrix (cont’d) The DSM for plane and space trusses](https://reader034.vdocuments.net/reader034/viewer/2022051401/5697c0221a28abf838cd31bf/html5/thumbnails/4.jpg)
4
Homework 4 Given a space truss structure as
shown here. A=1.53e-3 m2; E= 75 GPa Determine the nodal
displacements, support reactions, and the element internal forces (use Matlab).
Use SAP2000 and compare your results.
Instead of use truss model in SAP2000, use rigid frame model and compare the results.Source: http://www.andrew.cmu.edu/course/24-ansys/htm/s4_truss.htm
![Page 5: 11 10-Jan-16 Last course Interpretations and properties of the stiffness matrix (cont’d) The DSM for plane and space trusses](https://reader034.vdocuments.net/reader034/viewer/2022051401/5697c0221a28abf838cd31bf/html5/thumbnails/5.jpg)
5
Truss Structures in SAP2000
In SAP2000, there is no special truss element To do analysis of a truss structure, use Frame
element If the truss is an ideal truss (see the truss
structure assumptions), you may either: Set the geometric Section properties j, i33, and i22 all
to zero, or Release both bending rotations, R2 and R3, at both
ends and release the torsional rotation, R1, at either end.
![Page 6: 11 10-Jan-16 Last course Interpretations and properties of the stiffness matrix (cont’d) The DSM for plane and space trusses](https://reader034.vdocuments.net/reader034/viewer/2022051401/5697c0221a28abf838cd31bf/html5/thumbnails/6.jpg)
6
One-dimensional Bar Element– Principal-of-Virtual Work Approach
![Page 7: 11 10-Jan-16 Last course Interpretations and properties of the stiffness matrix (cont’d) The DSM for plane and space trusses](https://reader034.vdocuments.net/reader034/viewer/2022051401/5697c0221a28abf838cd31bf/html5/thumbnails/7.jpg)
7
From direct to virtual work approach So far we have derived the spring element and the bar
element by directly applied the physical laws (direct approach), i.e. Hooke’s law and force equilibrium law.
More complex finite elements, however, may not be able to be formulated using the direct approach.
So we turn to alternative approach for deriving the element stiffness equation.
There are several methods for obtaining the stiffness equation, but in this course we are using the principle of virtual work.
![Page 8: 11 10-Jan-16 Last course Interpretations and properties of the stiffness matrix (cont’d) The DSM for plane and space trusses](https://reader034.vdocuments.net/reader034/viewer/2022051401/5697c0221a28abf838cd31bf/html5/thumbnails/8.jpg)
8
Basic concepts from the theory of elasticity Consider a 1D bar as shown here.
Assume that the bar has: Length L Cross-sectional area A Modulus of elasticity E Coefficient of thermal expansion α
x, u
b=b(x)
L
X=0 X=L
P
![Page 9: 11 10-Jan-16 Last course Interpretations and properties of the stiffness matrix (cont’d) The DSM for plane and space trusses](https://reader034.vdocuments.net/reader034/viewer/2022051401/5697c0221a28abf838cd31bf/html5/thumbnails/9.jpg)
9
A, E and α may vary along the bar, i.e. A=A(x), E=E(x), and α= α(x)
The bar is subjected to: Distributed axial loal b=b(x), in the unit of [force]/[length],
Note that this load is in the category of body force because it acts at every point in the body
Concentrated force P at the right end, and Temperature change T 0C (may also vary along the bar).
The loads and temperature change causes the bar deforms, which can be described completely by an axial displacement field,
x, u
b=b(x)
L
X=0 X=L
P
u=u(x)
![Page 10: 11 10-Jan-16 Last course Interpretations and properties of the stiffness matrix (cont’d) The DSM for plane and space trusses](https://reader034.vdocuments.net/reader034/viewer/2022051401/5697c0221a28abf838cd31bf/html5/thumbnails/10.jpg)
10
In the theory of elasticity, there are three basic set of equations, i.e. Strain-displacement equations Stress-strain equations Equations of equilibrium
To obtain strain-displacement equations, consider the free body diagram of the bar segment between sections at distances x and x+Δx:
u(x): displacement of the section at distance x u(x+Δx): displacement of the section at distance x+Δx
x x+Δx
u(x) u(x+Δx)
Δx
![Page 11: 11 10-Jan-16 Last course Interpretations and properties of the stiffness matrix (cont’d) The DSM for plane and space trusses](https://reader034.vdocuments.net/reader034/viewer/2022051401/5697c0221a28abf838cd31bf/html5/thumbnails/11.jpg)
11
The average strain in the bar segment is
If Δx approaches zero, then the average strain becomes the strain at any point in the section located at distance x, i.e.
( ) ( )u x x u x
x
u(x) u(x+Δx)
Δx
Δ 0
( ) ( )( ) lim ( ) ,x
x
u x x u x dx u x u
x d x
![Page 12: 11 10-Jan-16 Last course Interpretations and properties of the stiffness matrix (cont’d) The DSM for plane and space trusses](https://reader034.vdocuments.net/reader034/viewer/2022051401/5697c0221a28abf838cd31bf/html5/thumbnails/12.jpg)
12
Thus, the strain-displacement equation for the bar is
The strain, ε, at each point of the bar consists of two parts, i.e. The strain due to thermal expansion, The strain due to external loads,
If the bar is not restrained, then the thermal strain will not induce any thermal stress.
The strain that induces stress is
( )d u
xd x
0 T P
0 T
…(1)
![Page 13: 11 10-Jan-16 Last course Interpretations and properties of the stiffness matrix (cont’d) The DSM for plane and space trusses](https://reader034.vdocuments.net/reader034/viewer/2022051401/5697c0221a28abf838cd31bf/html5/thumbnails/13.jpg)
13
The stress-strain equation is given by Hooke’s law, viz.
Now, consider the force equilibrium of the bar segment:
: average axial force along the bar segment
Force equilibrium:
( )E T
b
σ(x) σ(x+Δx)
Δx
b
( ) ( ) ( ) ( ) 0x x A x x x A x b x
…(2)
![Page 14: 11 10-Jan-16 Last course Interpretations and properties of the stiffness matrix (cont’d) The DSM for plane and space trusses](https://reader034.vdocuments.net/reader034/viewer/2022051401/5697c0221a28abf838cd31bf/html5/thumbnails/14.jpg)
14
N(x): internal normal (axial) force in the bar at distance x If Δx approaches zero, then
( ) ( ) ( ) ( ) 0x x A x x x A x b x
( ) ( ) 0N x x N x b x
( ) ( )0
N x x N xb
x
0
( ) ( )lim ( ) 0x
N x x N xb x
x
σ(x) σ(x+Δx)
Δx
b
![Page 15: 11 10-Jan-16 Last course Interpretations and properties of the stiffness matrix (cont’d) The DSM for plane and space trusses](https://reader034.vdocuments.net/reader034/viewer/2022051401/5697c0221a28abf838cd31bf/html5/thumbnails/15.jpg)
15
Thus, the equation of equilibrium for the bar is
Or shortly,
We can obtain the governing equation for the 1D bar using Eqs. (1), (2) and (3). Multiplying Eq. (2) by A results in
( )( ) 0
dN xb x
dx
, 0xN b …(3)
( ) ( ( ) )N x EA x T …(4)
![Page 16: 11 10-Jan-16 Last course Interpretations and properties of the stiffness matrix (cont’d) The DSM for plane and space trusses](https://reader034.vdocuments.net/reader034/viewer/2022051401/5697c0221a28abf838cd31bf/html5/thumbnails/16.jpg)
16
Substituting Eq. (1) into Eq. (4), we obtain
Now substituting this equation into Eq. (3)
If E, A, α and T constant along the bar, then the equation simply becomes
( ) ( ( ) )N x EA x T …(4)
( )( ) ( )
du xN x EA T
dx …(5)
( )( ( )) ( ) 0
d du xEA T b x
dx dx
2
2
( )( ) 0
d u xEA b x
dx
![Page 17: 11 10-Jan-16 Last course Interpretations and properties of the stiffness matrix (cont’d) The DSM for plane and space trusses](https://reader034.vdocuments.net/reader034/viewer/2022051401/5697c0221a28abf838cd31bf/html5/thumbnails/17.jpg)
17
In conclusion, the governing equation for the bar is
Given the external loads and displacement boundary condition, utilizing this 2nd order differential equation we can obtain the displacement field, the strain field and the stress field.
x, u
b=b(x)
L
X=0 X=L
P
( )( ( )) ( ) 0
d du xEA T b x
dx dx
![Page 18: 11 10-Jan-16 Last course Interpretations and properties of the stiffness matrix (cont’d) The DSM for plane and space trusses](https://reader034.vdocuments.net/reader034/viewer/2022051401/5697c0221a28abf838cd31bf/html5/thumbnails/18.jpg)
1818Apr 21, 2023
Recall the example of bar structure
Find the exact solutions for the displacement, strain, and stress along the bar and compare to the finite element solutions.
0x Tx L
03A A 0A A
![Page 19: 11 10-Jan-16 Last course Interpretations and properties of the stiffness matrix (cont’d) The DSM for plane and space trusses](https://reader034.vdocuments.net/reader034/viewer/2022051401/5697c0221a28abf838cd31bf/html5/thumbnails/19.jpg)
1919Apr 21, 2023
For this problem, p(x)=0 and T=0. Thus, the governing equation can be written as
The area of a cross section at a distance x:
Boundary conditions:
x L
duEA F
dx
0T
2( ) ( 3)A x A x
L
T0 x L
00
xu
Displacement/geometrical/essential boundary condition
x ; uFH ( )( ( ) ) 0
d du xEA x
dx dx
Force/mechanical/natural boundary condition
![Page 20: 11 10-Jan-16 Last course Interpretations and properties of the stiffness matrix (cont’d) The DSM for plane and space trusses](https://reader034.vdocuments.net/reader034/viewer/2022051401/5697c0221a28abf838cd31bf/html5/thumbnails/20.jpg)
2020Apr 21, 2023
The solution of the governing equation is
The strain field is
The stress field is
T T
0 T
3( ) ln
2 3 2
FL Lu x
EA L x
T
0 T
( )( )
3 2
Ldu x Fx
dx EA L x
T
0 T
( ) ( )3 2
LFx E x
A L x
Displacement field
![Page 21: 11 10-Jan-16 Last course Interpretations and properties of the stiffness matrix (cont’d) The DSM for plane and space trusses](https://reader034.vdocuments.net/reader034/viewer/2022051401/5697c0221a28abf838cd31bf/html5/thumbnails/21.jpg)
21
Recall the finite element solutions
2 T
3 0
3/15
8 /15
D FLD EA
1 11 1 2 1
T 0
2
0.5 5x x
d d FE E
L A
2 22 2 2 1
T 0
2
0.5 3x x
d d FE E
L A
![Page 22: 11 10-Jan-16 Last course Interpretations and properties of the stiffness matrix (cont’d) The DSM for plane and space trusses](https://reader034.vdocuments.net/reader034/viewer/2022051401/5697c0221a28abf838cd31bf/html5/thumbnails/22.jpg)
2222Apr 21, 2023
Medan Perpindahan
0
0.1
0.2
0.3
0.4
0.5
0.6
0 0.2 0.4 0.6 0.8 1
X (L T)
u (
FL
T/E
A0)
Pendekatan
Eksak
![Page 23: 11 10-Jan-16 Last course Interpretations and properties of the stiffness matrix (cont’d) The DSM for plane and space trusses](https://reader034.vdocuments.net/reader034/viewer/2022051401/5697c0221a28abf838cd31bf/html5/thumbnails/23.jpg)
2323Apr 21, 2023
Medan Tegangan
0.00
0.20
0.40
0.60
0.80
1.00
1.20
0 0.2 0.4 0.6 0.8 1
X (L T)
( F
/A0 ) Pendekatan
Eksak
![Page 24: 11 10-Jan-16 Last course Interpretations and properties of the stiffness matrix (cont’d) The DSM for plane and space trusses](https://reader034.vdocuments.net/reader034/viewer/2022051401/5697c0221a28abf838cd31bf/html5/thumbnails/24.jpg)
24
Strain energy
Energy The capacity or ability to do work.
Work A product of a force and a distance in the direction
where the force moves. Elastic strain energy (or elastic deformation
energy) Energy stored in a solid body in the deformed state. The capacity of internal forces (or stresses) to do
work through deformation (strains) in the solid body.
![Page 25: 11 10-Jan-16 Last course Interpretations and properties of the stiffness matrix (cont’d) The DSM for plane and space trusses](https://reader034.vdocuments.net/reader034/viewer/2022051401/5697c0221a28abf838cd31bf/html5/thumbnails/25.jpg)
25
Strain energy for a bar
A bar has only one stress component along its axis (uniaxial stress condition)
Consider an infinitesimal element subjected to normal stress σx only
x xN dy dz
![Page 26: 11 10-Jan-16 Last course Interpretations and properties of the stiffness matrix (cont’d) The DSM for plane and space trusses](https://reader034.vdocuments.net/reader034/viewer/2022051401/5697c0221a28abf838cd31bf/html5/thumbnails/26.jpg)
26
Due to Nx, the element extends Δx
Assume that the material is elastic-linear
![Page 27: 11 10-Jan-16 Last course Interpretations and properties of the stiffness matrix (cont’d) The DSM for plane and space trusses](https://reader034.vdocuments.net/reader034/viewer/2022051401/5697c0221a28abf838cd31bf/html5/thumbnails/27.jpg)
27
The work done by Nx
Thus, the strain energy of the infinitesimal element is
![Page 28: 11 10-Jan-16 Last course Interpretations and properties of the stiffness matrix (cont’d) The DSM for plane and space trusses](https://reader034.vdocuments.net/reader034/viewer/2022051401/5697c0221a28abf838cd31bf/html5/thumbnails/28.jpg)
28
The strain energy per unit volume is then
The strain energy per unit volume is called strain energy density, U0
U0 is equal to the area under the stress-strain curve and above the strain axis.
The strain energy for the whole solid body:12 x x
V
U dV
![Page 29: 11 10-Jan-16 Last course Interpretations and properties of the stiffness matrix (cont’d) The DSM for plane and space trusses](https://reader034.vdocuments.net/reader034/viewer/2022051401/5697c0221a28abf838cd31bf/html5/thumbnails/29.jpg)
29
The strain energy for the bar:
x, u
b=b(x)
L
X=0 X=L
P
120
( ) ( ) L
U x x Adx
![Page 30: 11 10-Jan-16 Last course Interpretations and properties of the stiffness matrix (cont’d) The DSM for plane and space trusses](https://reader034.vdocuments.net/reader034/viewer/2022051401/5697c0221a28abf838cd31bf/html5/thumbnails/30.jpg)
30
Problem example Consider the following two bars
Suppose that the bars absorb the same amount of energy corresponding to the axial load
Neglecting stress concentration, compare the axial loads acting on bar 1, P1, and bar 2, P2
1 2
![Page 31: 11 10-Jan-16 Last course Interpretations and properties of the stiffness matrix (cont’d) The DSM for plane and space trusses](https://reader034.vdocuments.net/reader034/viewer/2022051401/5697c0221a28abf838cd31bf/html5/thumbnails/31.jpg)
31
Solution Bar 1
Bar 2
![Page 32: 11 10-Jan-16 Last course Interpretations and properties of the stiffness matrix (cont’d) The DSM for plane and space trusses](https://reader034.vdocuments.net/reader034/viewer/2022051401/5697c0221a28abf838cd31bf/html5/thumbnails/32.jpg)
32
Solution (cont’d)
![Page 33: 11 10-Jan-16 Last course Interpretations and properties of the stiffness matrix (cont’d) The DSM for plane and space trusses](https://reader034.vdocuments.net/reader034/viewer/2022051401/5697c0221a28abf838cd31bf/html5/thumbnails/33.jpg)
33
Homework 5 (due date: FEM Mid-sem Exam Time)1. Show how to obtain that the solutions for the
displacement and strain fields of the tapered bar.
T T
0 T
3( ) ln
2 3 2
FL Lu x
EA L x
T
0 T
( )( )
3 2
Ldu x Fx
dx EA L x
x ; uFH
![Page 34: 11 10-Jan-16 Last course Interpretations and properties of the stiffness matrix (cont’d) The DSM for plane and space trusses](https://reader034.vdocuments.net/reader034/viewer/2022051401/5697c0221a28abf838cd31bf/html5/thumbnails/34.jpg)
34
2. Compute the displacement and stress at the right end of the bar, i.e. u(LT) and σ(LT) by modeling the bar using finite elements. The area of each element is taken to be equal to the section area at the midpoint of the element.
a. Use one element, with L1=LT
b. Use two elements, with L1=L2=LT/2
c. Use three elements, with L1=L2=L3=LT/3
d. Use four elements, with the length of each element LT/4
![Page 35: 11 10-Jan-16 Last course Interpretations and properties of the stiffness matrix (cont’d) The DSM for plane and space trusses](https://reader034.vdocuments.net/reader034/viewer/2022051401/5697c0221a28abf838cd31bf/html5/thumbnails/35.jpg)
35
3. Examine the convergence of the finite element solutions in Problem #2 by:
a. Computing the relative error (%) for each solution, both for the displacement and the stress.
b. Draw convergence graphs for the displacement and the stress, with the abscissa is the number of element and the ordinate is the solutions, including the exact solution. Give your comment regarding the finite element solutions.
4. Prove that the exact strain energy in the bar is
T
0
ln 3
4
FLU
EA
![Page 36: 11 10-Jan-16 Last course Interpretations and properties of the stiffness matrix (cont’d) The DSM for plane and space trusses](https://reader034.vdocuments.net/reader034/viewer/2022051401/5697c0221a28abf838cd31bf/html5/thumbnails/36.jpg)
36
The principle of virtual work A virtual (imaginary) displacement is an imaginary and
very small change in the configuration of a system The principle of virtual work (or more precisely the
principle of virtual displacement) is stated as follows:
If a body in equilibrium is subjected to arbitrary virtual displacements associated with a compatible deformation of the body, the virtual work of external forces on the body is equal to the virtual strain energy of the internal stresses
Compatible (admissible) displacements are those that satisfy the boundary conditions and ensure that no discontinuities, such as voids or overlaps
![Page 37: 11 10-Jan-16 Last course Interpretations and properties of the stiffness matrix (cont’d) The DSM for plane and space trusses](https://reader034.vdocuments.net/reader034/viewer/2022051401/5697c0221a28abf838cd31bf/html5/thumbnails/37.jpg)
37
The principle of the virtual work
δU: the virtual strain energy of the internal stresses
δW: the virtual work of external forces on the body
X, u
Z, w
w
δw
δw is the virtual displacement
![Page 38: 11 10-Jan-16 Last course Interpretations and properties of the stiffness matrix (cont’d) The DSM for plane and space trusses](https://reader034.vdocuments.net/reader034/viewer/2022051401/5697c0221a28abf838cd31bf/html5/thumbnails/38.jpg)
38
The virtual strain energy for a bar
δε : virtual strain, which is the strain associated with the virtual displacement, i.e.
The external virtual work:
0( ) ( )
LU x x Adx
( ) ( )d
x u xdx
x, u
b=b(x)
L
X=0 X=L
P
0( ) ( ) ( )
LW u x b x dx P u L
![Page 39: 11 10-Jan-16 Last course Interpretations and properties of the stiffness matrix (cont’d) The DSM for plane and space trusses](https://reader034.vdocuments.net/reader034/viewer/2022051401/5697c0221a28abf838cd31bf/html5/thumbnails/39.jpg)
39
Notice that
Why is it not equal to the definition of actual strain energy?
120
( ) ( ) L
U x x dV
Discussion
![Page 40: 11 10-Jan-16 Last course Interpretations and properties of the stiffness matrix (cont’d) The DSM for plane and space trusses](https://reader034.vdocuments.net/reader034/viewer/2022051401/5697c0221a28abf838cd31bf/html5/thumbnails/40.jpg)
40
Formulation of bar element Consider the two-node bar element as follows
b(x) : body force along the bar element in the unit of force per unit length (e.g. kN/m)
u=u(x) : displacement field within the element, i.e. displacement as a function of point x within the element
0 x L
![Page 41: 11 10-Jan-16 Last course Interpretations and properties of the stiffness matrix (cont’d) The DSM for plane and space trusses](https://reader034.vdocuments.net/reader034/viewer/2022051401/5697c0221a28abf838cd31bf/html5/thumbnails/41.jpg)
41
Displacements at nodes 1 and 2:
Forces at nodes 1 and 2:
The first and fundamental step in the finite element formulation is to assume the displacement field within the element in terms of its nodal displacements
![Page 42: 11 10-Jan-16 Last course Interpretations and properties of the stiffness matrix (cont’d) The DSM for plane and space trusses](https://reader034.vdocuments.net/reader034/viewer/2022051401/5697c0221a28abf838cd31bf/html5/thumbnails/42.jpg)
42
Here it is assumed that the displacement field is a linear function
Let us now express the assumed function in terms of nodal displacements d1 and d2
![Page 43: 11 10-Jan-16 Last course Interpretations and properties of the stiffness matrix (cont’d) The DSM for plane and space trusses](https://reader034.vdocuments.net/reader034/viewer/2022051401/5697c0221a28abf838cd31bf/html5/thumbnails/43.jpg)
43
As a result,
![Page 44: 11 10-Jan-16 Last course Interpretations and properties of the stiffness matrix (cont’d) The DSM for plane and space trusses](https://reader034.vdocuments.net/reader034/viewer/2022051401/5697c0221a28abf838cd31bf/html5/thumbnails/44.jpg)
44
Written in matrix form
Thus we can write
![Page 45: 11 10-Jan-16 Last course Interpretations and properties of the stiffness matrix (cont’d) The DSM for plane and space trusses](https://reader034.vdocuments.net/reader034/viewer/2022051401/5697c0221a28abf838cd31bf/html5/thumbnails/45.jpg)
45
Where
N is also called the matrix of interpolation functions, because it interpolates the displacement field u=u(x) from the nodal displacements
L x x
L L
N
1
2
d
d
d
Matrix of shape functions
Vector of nodal displacements