User:EML4500.f08.A-team.VandenBerg/HW2

 Comment: This is HW2 located in the user namespace of Lukas Vandenberg EML4500.f08.A-team.VandenBerg 20:15, 26 September 2008 (UTC)

Axial Stiffness
k(e)d(e)=f(e)

where:


 * k(e) is a 4x4 column matrix, which is the element stiffness matrix for element "e", where e=1,2.


 * d(e) is a 4x1 column matrix, which is the element displacement matrix for element "e".


 * f(e) is a 4x1 column matrix, which is the element force matrix for element "e".

In the book on page 225, the second equation from the bottom states:

$$ \begin{array}{cccc} & \begin{matrix} \quad~ d^{(e)}_1 \quad~ & \quad~ d^{(e)}_2 \quad~ & \quad~ d^{(e)}_3 \quad~ & \quad~ d^{(e)}_4 \quad~ \end{matrix} \\ k^{(e)} & \begin{bmatrix} (\ell^{(e)})^2 & \ell^{(e)}m^{(e)} & -(\ell^{(e)})^2 & -\ell^{(e)}m^{(e)} \\ \ell^{(e)}m^{(e)} & (m^{(e)})^2 & -\ell^{(e)}m^{(e)} & -(m^{(e)})^2 \\ -(\ell^{(e)})^2 & -\ell^{(e)}m^{(e)} & (\ell^{(e)})^2 & \ell^{(e)}m^{(e)} \\ -\ell^{(e)}m^{(e)} & -(m^{(e)})^2 & \ell^{(e)}m^{(e)} & (m^{(e)})^2 \end{bmatrix} & = & k^{(e)} \end{array} $$

Also: k(e) = E(e)A(e)/L(e), which is the actual stiffness of the bar element "e". It is also the equivalent of the spring constant "k".

Director Cosines
l(e) and m(e) are the director cosines of the $$\tilde{x}$$ axis which goes from local node number one to local node number 2 with respect to (w.r.t) the global (x,y) coordinate system.

6-2

Meeting 7; Wednesday, 10 September 2008 7-1

2-Bar Truss System Model
continued....

Element 1 Stiffness:
Θ(1) = 30°

l(1) = cosΘ(1) = cos30° = √3/2

k(1) = E(1)A(1)/L(1) = (3)(1)/(4) = 3/4

$$ k^{(1)} = \begin{bmatrix} k^{(1)}_{11} & k^{(1)}_{12} & k^{(1)}_{13} & k^{(1)}_{14} \\ k^{(1)}_{21} & k^{(1)}_{22} & k^{(1)}_{23} & k^{(1)}_{24} \\ k^{(1)}_{31} & k^{(1)}_{32} & k^{(1)}_{33} & k^{(1)}_{34} \\ k^{(1)}_{41} & k^{(1)}_{42} & k^{(1)}_{43} & k^{(1)}_{44} \end{bmatrix} = \begin{bmatrix} k^{(1)}_{ij} \end{bmatrix} $$

i: row index (=1,2,3,4)

j: column index (=1,2,3,4)

(e): element

4x4 matrix

Observations:

1)Only three numbers need to be computed; other coefficients have the same absolute values just differing by a "+" or "-".

2)Matrix k(1) is symmetric, i.e., kij(1) = kji(1)

k13(1) = k31(1)

just interchanging the row and column indices

k11(1) = k(1) x (l1)2

= (3/4)(√3/2)2 = 9/16

k12(1) = k1(l(1)m(1))

= 3√3/16

k42(1) = -k1(m(1))2

= -3/16

In general kij(1) = kji(1) or k(e)T = k(e) 7-3


 * T = Transpose

The transpose of k(e) is equal to k(e)



When filling in a matrix, the upper triangular portion is the only part that needs to be filled in because the lower triangular portion is symmetrical.

Element 2 Stiffness:
7-4

k(2) = E(2)A(2)/L(2) = (5)(2)/(2) = 5

Θ(2) = -π/4

l(2) = cos(-π/4) = √2/2

m(2) = sin(-π/4) = -√2/2

k(2) = [kij(2)]4x4

k11(2) = k(2)(l(2))2 = (5)(√2/2)2 = 2.5

Observations:

1) The absolute values of all the coefficients kij(e), e=2, (i,j) = 1, ..., 4 are the same.  Comparing one coefficient to another by adding a "+" or a "-".

2) k(2)T = k(2), i.e., k(2) sym.

Element Force Displacement(FD) Relation
7-5

k(e)d(e) = f(e)

with:

k(e) being a 4x4 matrix,

d(e) being a 4x1 matrix,

f(e) being a 4x1 matrix.

and e = 1,2

$$ d^{(e))} = \begin{bmatrix} d^{(e)}_1 \\ d^{(e)}_2 \\ d^{(e)}_3 \\ d^{(e)}_4 \end{bmatrix}

f^{(e))} = \begin{bmatrix} f^{(e)}_1 \\ f^{(e)}_2 \\ f^{(e)}_3 \\ f^{(e)}_4 \end{bmatrix} $$

Global FD Relation ("free-free structure)
Kd = F

where:

K is an nxn matrix,

d is an nx1 matrix, and

F is an nx1 matrix.

there, n = 6 (page 5-5)

Meeting 8, Friday 12, September 2008 8-1

Control from page 7-5:

In compact notation, the equation yields:

[Kij]{dj} = {Fi}

with:

[Kij] being a 6x6 matrix,

{dj} being a 6x1 matrix, and

{Fi} being a 6x1 matrix.

(or more generally noted as nxn)

$$\sum^6_{j = 1} k_{ij} d_j = F_i ~ where ~ i = 1,...,6$$

K nxn = {Kij}nxn = This is the global stiffness matrix

d nx1 = {dj}nx1 = This is the global displacement matrix

F nx1 = {Fi}nx1 = This is the global force matrix 8-2 Recall from page 7-2 the element FD rel....

k4x4(e)d4x1(e) = f4x1(e)

k4x4(e) = [kij(e)] = element stiffness matrix

d4x1(e) = {di(e)} = element displacement matrix

f4x4(e) = {fi(e)} = element force matrix

Element Matrices to Global Matrices
The How to... of going from element to global matrices (i.e. stiffness, displacement, and force)

An assembly process is how it's done!

First things first, the correspondence between the element displacement d.o.f.'s and global d.o.f.'s must be identified. 8-3 Global level: page 5-5

{d1, dSubscript text2 ,....., d6}

Element level: page 4-3, and page 4-2


 * Element 1: {d1(1), d2(1), d3(1), d4(1)}


 * Element 2: {d1(2), d2(2), d3(2), d4(2)}

Global-level DOF Identifying
(Beginning of Assembly Process)

Node 1:

d1 = d1(1)


 * (1) = element 1


 * 1 = first DOF


 * d1 = first Global DOF (LHS)

d2 = d2(1)

Node 2:

d3 = d3(1) = d1(2)

d4 = d4(1) = d2(2)


 * Global Node 2 has 2 displacement components regardless of notation.

Node 3:

d5 = d3(2)

d6 = d4(2)

Conceptual Step of Assembly:
(Topology of K)

The overlap is a result of global node 2 identifications.

Meeting 9, September 15, 2008



Kundefinedk(1)11 = 9/16

Kundefinedk(1)12 = .3247595

K33 = k(1)33 + k(2)11 = 9/16 + 5/2 = 3.0625

K34 = K34 = k(1)34 + k(2)12 = .3247595 + 5/2

The rest of the stiffness matrix can be computed similarly

Elimination of known dofs:

reduce global FD rel.

d1 = d2 = d5 = d6 = 0

EML4500.f08.A-team.kirley 20:20, 26 September 2008 (UTC)



Applying fixed B.C.'s => deleting corresponding collumns i nthe global stiffness matrix K.

By principle of Virtual Work (PVW) => deleting also corresponding rows (here, rows 1,2,3,5,6). Corresponding rows of F are also deleted.

Resulting FD relation





What is K-1 (without calculator) d1 K = (K33 K44) - (K43 K34)



K*K-1 = K-1*K = I



$$K^{-1} \neq \frac 1 {det K} ~ K^{-1}$$

Element 1: k (1) d (1) = f (1)

| 0.5625  0.32476  ... ...   |          |  0  |    |                             |          |  0  |    |                             |          |  0  |    |                             |4x4       |  0  |4x1

| -4.4348  |       |  f1(1)  | =  |  -2.5622  |  =    |  f2(1)  | | +4.4378  |       |  f3(1)  | | +2.5522  |       |  f4(1)  | f1(1) & f2(1) = reactions f3(1) & f4(1) = internal forces

Observation: Element 1 is in equilibrium summation Fx = f1(1) + f3(1) = 0 summation Fy = f2(1) + f4(1) = 0

p1(1) = [(f1(1))2 + f2(1))2]1/2

Two methods for solving 2-bar truss with Euler cut principle

Method 1:

Method 2:

Back to FE solution: to bring p back into the big picture --> equilibrium of global node n(circled)

Homework Problems for HW 2
9/8/08 P. 1

Write out k(1) and k(2).

First begin with:

Now find the axial stiffness, k:

Next multiply through the stiffness matrix by k to create the k(1) matrix:

The same goes for the k(2) matrix:

and finally:

9/8/08 P. 2

Prove that dotting ĩ with j gives the director cosine equal to sine of the angle theta:

P. 3

Compare the results of statically determining the reactions to the finite element method:

Since this is a statically determinate system, statics can be used to calculate the reaction forces on each of the element. Start by summing the forces on node 2.

Since P is known (P = 7), there are 2 equations with 2 unknowns making this easily solvable. After rearranging the equations and substituting, it can be found that F12 = 5.124 and that F23 = 6.276. The reaction forces can be split up as follows:

These reaction forces are exactly the same as those computed by the MatLab code. Using statics involved less complex mathematics (no matrices), however, rearranging and substituting equations, especially when trig functions are involved can become confusing and tedious for more complex systems. The finite element method reduces everything down to a few matrix multiplications and can easily be put into a computer for computation and saves a great deal of time and headache. The stiffness matrices are the same for either hand or computer calculations, which is not unreasonable to believe since it is merely a scalar being multiplied through a matrix. Displacement matrices were also nearly the same but it can be imagined that if a more complicated system were analyzed that a computer would be a better choice since it will be less apt to make mistakes and can carry out numbers to more decimal places.

Contributions for HW2
Contributors:

Lukas Vandenberg EML4500.f08.A-team.VandenBerg 20:15, 26 September 2008 (UTC)

Brian Morford --EML4500.f08.A-team.morford 20:31, 26 September 2008 (UTC)

Joseph Rieth --EML4500.f08.A-team.rieth 20:13, 19 September 2008 (UTC)

Joseph Melvin --EML4500.f08.A-team.melvin 20:23, 19 September 2008 (UTC)

Brian Kirley --EML4500.f08.A-team.kirley 20:32, 19 September 2008 (UTC)

Daniel Robinson --Eml4500.f08.a-team.robinson 20:32, 19 September 2008 (UTC)