User:Eml 4500.f08.delta 6.specht/New HW2

 The following is a revision of the Homework 2 assignment that was submitted for group Delta_6. The following link includes the differences between the two pages that were submitted.

http://en.wikiversity.org/w/index.php?title=User:Eml_4500.f08.delta_6.specht/New_HW2&diff=next&oldid=343668

--Eml 4500.f08.delta 6.specht 15:53, 9 December 2008 (UTC)

Unit Vector Multiplication
The value for m(e) can be calculated in the manner shown below. It is equal to the final solution that is found. It should be noted that the dot product of the i and j unit vectors equals zero, and the dot product of the two j unit vectors is equal to one.

Element Stiffness Matrices
The values that are used to calculate the scalar value k are shown below. The direction cosine matrix is multiplied by the scalar k value, which is the spring constant of each element. The value that is obtained from this operation is the stiffness matrix for each element. The values for the direction cosines of each element are as follows: l(1) = cos θ(1) = cos30° = $$\frac{\sqrt{3}}{2}$$ m(1) = sin θ(1) = sin30° = $$\frac{1}{2}$$ l(2) = cos θ(2) = $$cos(-\frac{\Pi }{4}) = \frac{\sqrt{2}}{2}$$ m(2) = sin θ(2) = $$\sin (-\frac{\Pi }{4}) = -\frac{\sqrt{2}}{2}$$ These values can then be substituted into the matrix of direction cosines in order to calculate the element stiffness matrices.

Verification of Equilibrium of element 1:
Note: The reactions for node 1 were calculated to be

F1(1) = -4.4278

F2(1) = -2.5622

F3(1) = 4.4278

F4(1) = 2.5622



Note: Negative values signify that the direction of the forces in element 1 were assumed to be in the wrong direction.

Redrawing the free body diagram with all positive forces shows more clearly how the resultant forces P1 and P2 cancel each other out.

Now,

F1(1) = 4.4278

F2(1) = 2.5622

F3(1) = 4.4278

F4(1) = 2.5622



Determining Resultants:

P­­­1= [(f1(1))2 + (f2(1))2]1/2= [(4.4278)2+ (2.5622)2]1/2= 5.12

P­­­2= [ (f3(1))2 + (f4(1))2]1/2= [(4.4278)2+ (2.5622)2]1/2= 5.12

Note: P1 = P2, thus element 1 is in equilibrium.

Statics Solution Method


P = 7 L(1) = 4 L(2) = 2 θ(1) = 30° θ(2) = 45°



Observation:

By using trigonometry and the angles given the unknown lengths can be determined. These lengths will be needed when analyzing the summation of the moments of each node.

$$\sin {30^{\circ} } = \frac{L_c}{4} \Rightarrow L_c = 4 \sin {30^{\circ} } = 2$$

$$\cos {30^{\circ} } = \frac{L_x}{4} \Rightarrow L_x = 4 \cos {30^{\circ} } = 3.46$$

$$\sin {45^{\circ} } = \frac{L_b}{2} \Rightarrow L_b = 2 \sin {45^{\circ} } = 1.4$$

$$\cos {45^{\circ} } = \frac{L_y}{2} \Rightarrow L_y = 2 \cos {45^{\circ} } = 1.4$$

Lz = Lx + Ly = 3.46 + 1.4 = 4.86

La = Lc - Lb = 2 - 1.4 = 0.6

Moment equations can now be analyzed since all lengths are known.

(1) M1: -(P)(Lx) + (F4)(Lz) + (F3)(La) = -(7)(3.46) + (F4)(4.86) + (F3)(0.6)

(2) M2: (P)(Ly) + (F1)(La) - (F2)(Lz) = (7)(1.4) + (F1)(0.6) - (F2)(4.86)

(3) M3: (F1)(Lc) - (F2)(Lx) - (F3)(Lb) + (F4)(Ly) = (F1)(2) - (F2)(3.46) - (F3)(1.4) + (F4)(1.4)

Note: There are 4 unknowns with only 3 equations. However, by inspection it is known that the summation of the forces in the x direction are equal to zero, giving us 4 equations with 4 unknowns. (4) $$\sum{F_x} = 0: F_1 - F_3 = 0 $$

All forces can now be determined by using the system of equations. Rearrange the equations solving for a common variable in order to calculate the first unknown force. Next, using the rearrange equations all unknown forces can be calculated.

(1) -24.22 + 4.86 F4 + 0.6 F3 = 0

(2) 9.8 + 0.6 F1 - 4.86 F2 = 0

(3) 2 F1 - 3.46 F2 - 1.4 F3 + 1.4 F4 = 0

(4) F1 = F3

Since F3 is already in terms of F1, F2 and F4 will also solved in terms of F1

(1) $$F_4 = \frac{24.22 - 0.6 F_3}{4.86} = \frac{24.22 - 0.6 F_1}{4.86}$$ (2) $$F_2 = \frac{9.8 + 0.6 F_1}{4.86}$$ (4) F3 = F1

Plugging in these relationships into equation (3), F1 and the rest of the forces can be solved for as follows:

(3) $$2 F_1 - 3.46 (\frac{9.8 + 0.6 F_1}{4.86}) - 1.4 F_1 + 1.4 (\frac{24.22 - 0.6 F_1}{4.86}) = 0 \Rightarrow $$

Combining like terms,

(3) $$2 F_1 - 6.976 - 0.427 F_1 - 1.4 F_1 + 6.976 - 0.173 F_1 = 0 \Rightarrow 0 = 0$$

Thus, F1 = F3 = 0

(2) $$F_2 = \frac{9.8 + 0.6(0)}{4.86} = 2.01$$ (1) $$F_4 = \frac{24.22 + 0.6(0)}{4.86} = 4.98$$

Results

F1 = 0 F2 = 2.01 F3 = 0 F4 = 4.98

Two Bar Truss Example (MATLAB)
It is important to understand how to solve a bar truss system employing a finite element MATLAB code in order to compare the resulting matrices with those obtained from hand calculations.

Take this truss example with the following: Data

MATLAB Code Note: this MATLAB code was obtained from Dr. L. Vu-Quoc website http://clesm.mae.ufl.edu/~vql/courses/fead/2008.fall/codes/twoBarTrussEx.txt

In order to run this MATLAB code the following functions are needed:

Note: this MATLAB code was obtained from the MATLAB.zip Chapter 9 folder given by the Student Companion Site Bhatti: Advanced Topics in Finite Element Analysis of Structures: With Mathematica and MATLAB Computations website http://bcs.wiley.com/he-bcs/Books?action=resource&bcsId=3105&itemId=0471648078&resourceId=7629

Note: this MATLAB code was obtained from the MATLAB.zip Common folder given by the Student Companion Site Bhatti: Advanced Topics in Finite Element Analysis of Structures: With Mathematica and MATLAB Computations website http://bcs.wiley.com/he-bcs/Books?action=resource&bcsId=3105&itemId=0471648078&resourceId=7629

Note: this MATLAB code was obtained from the MATLAB.zip Chapter 9 folder given by the Student Companion Site Bhatti: Advanced Topics in Finite Element Analysis of Structures: With Mathematica and MATLAB Computations website http://bcs.wiley.com/he-bcs/Books?action=resource&bcsId=3105&itemId=0471648078&resourceId=7629

Results

Testing the Steps
It is important to see how the steps to solve truss systems work.

Given


Take this truss example with the following data:

Step 1: Global Picture
Releasing the truss system from its supports and showing the reactions, we have:



There are 6 unknowns for which we need to solve for, therefore we need to analyze each element. The matrix equation is as follows.


 * $$\begin{pmatrix}

f_1\\ ...\\ f_6\end{pmatrix}$$ = $$\begin{bmatrix} & &  \\  & k &  \\ & &  \end{bmatrix}$$ $$\begin{pmatrix} d_1\\ ...\\ d_6\end{pmatrix}$$

which corresponds to


 * $$\mathbf{F}$$=$$\mathbf{K}$$ $$\mathbf{d}$$.

To solve for the reactions, we have to now examine the whole system by breaking it down into elements.

Step 2: Element Picture
Analyzing each element with its own force and dofs yields:





It is important to notice the chosen coordinate axis since they will allow for ease of analyzing each element. The tilda on the coordinate direction names designate a non-horizontal or non-vertical direction.

We now want to creat a matrix equation, just like for the global picture, for the element picture. The matrix equation consists on the basic force-distance relationship, however, this principle is applied at the element level. The notation is altered in that each term is designated to belong to the particular element with an upper script number in parenthesis so that we get, F(e)=k(e) d(e), named as the element force matrix, the element stiffness matrix, and the element distance matrix.

Step 3: Force-Displacement Relationship
To build the element stiffness matrix, we have to follow a "recipe" for calculating each term. This stiffness matrix will end up measuring 4x4.

The recipe is as follows, and in the wise words of Prof. Vu-Quoc, "The notation is beautiful here. It tells you no more, no less."


 * k$$^{(e)}$$ $$\begin{bmatrix}

l^{(e)2} & l^{(e)}m^{(e)} & -l^{(e)2} & -l^{(e)}m^{(e)}\\ l^{(e)}m^{(e)} & m^{(e)2} & -l^{(e)}m^{(e)} & -m^{(e)2}\\ -l^{(e)2} & -l^{(e)}m^{(e)} & l^{(e)2} & l^{(e)}m^{(e)}\\ -l^{(e)}m^{(e)} & -m^{(e)2} & l^{(e)}m^{(e)} & m^{(e)2} \end{bmatrix}$$ = k$$^{(e)}$$

where the axial stiffness k$$^{(e)}$$ of the bar element "e" is


 * $$k^{(e)}=\frac{E^{(e)} A^{(e)}}{L^{(e)}}$$,

where $$E^{(e)}$$ is the element's material Young's Modulus, $$A^{(e)}$$ is the element cross-sectional area, and $$L^{(e)}$$ the element length. Also, $$l^{(e)}$$ and $$m^{(e)}$$ are te director cosines of the rotated, and convenient, coordinate axes with respect to the positive, and horizontal, x-direction (the global axis).

It is very important to make the following observations for the efficiency to construct the stiffness matrix:
 * 1. It is only needed to compute 3 numbers, both of the squares and the product of eachother, and the determination of a negeative or a positive sign.
 * 2. The matrix k$$^{(e)}$$ is symmetric, which means that there is no need to calculate the bottom triangular part once the top has been determined, including the diagonal.

Director Cosines: The director cosines will yield the projections of vectors oriented with an inclination w.r.t. the positive x-axis on the horizontal or vertical directions. These horizontal and vertical directions are more precisely for this course the global x-y coordinates. From this diagram, we can derive our much needed $$l^{(e)}$$ and $$m^{(e)}$$ values. $$l^{(e)}$$ = $$\vec{\tilde{i}}$$ • $$\vec{\hat{i}}$$ = cosθ$$^{(e)}$$ $$m^{(e)}$$ = $$\vec{\tilde{i}}$$ • $$\vec{\hat{j}}$$ = sinθ$$^{(e)}$$ = cos(π/2-θ$$^{(e)})$$ Also, $$\vec{\tilde{i}}$$=cosθ$$^{(e)}$$•$$\vec{\hat{i}}$$+sinθ$$^{(e)}$$•$$\vec{\hat{j}}$$

Global stiffness matrix K.


 * {| style="text-align:center; border-collapse:collapse;" cellpadding="4"


 * rowspan="6" | K  =
 * rowspan="6" | $$\left.\begin{align} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \end{align} \right($$
 * title="stiffness matrix from element one" style="border-left: 1px solid black; border-top: 1px solid black;" | $$k_{11}^{(1)}$$
 * title="stiffness matrix from element one" style="border-top: 1px solid black;" | $$k_{12}^{(1)}$$
 * title="stiffness matrix from element one" style="border-top: 1px solid black;" | $$k_{13}^{(1)}$$
 * title="stiffness matrix from element one" style="border-top: 1px solid black; border-right: 1px solid black;" | $$k_{11}^{(1)}$$
 * 0
 * 0
 * rowspan="6" | $$\left)\begin{align} \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \end{align} \right.$$
 * title="stiffness matrix from element one" style="border-left: 1px solid black;" | $$k_{21}^{(1)}$$
 * title="stiffness matrix from element one" | $$k_{22}^{(1)}$$
 * title="stiffness matrix from element one" | $$k_{23}^{(1)}$$
 * title="stiffness matrix from element one" style="border-right: 1px solid black;" | $$k_{24}^{(1)}$$
 * 0
 * 0
 * title="stiffness matrix from element one" style="border-left: 1px solid black;" | $$k_{31}^{(1)}$$
 * title="stiffness matrix from element one" style="border-right: 1px solid black;" | $$k_{32}^{(1)}$$
 * title="stiffness matrix from element one and element two" style="border-top: 1px solid black;" | $$k_{33}^{(1)} + k_{11}^{(2)}$$
 * title="stiffness matrix from element one and element two" style="border-right: 1px solid black; border-top: 1px solid black;" | $$k_{34}^{(1)} + k_{12}^{(2)}$$
 * title="stiffness matrix from element two" style="border-top: 1px solid black;" | $$k_{13}^{(2)}$$
 * title="stiffness matrix from element two" style="border-right: 1px solid black; border-top: 1px solid black;" | $$k_{14}^{(2)}$$
 * title="stiffness matrix from element one" style="border-left: 1px solid black; border-bottom: 1px solid black;" | $$k_{41}^{(1)}$$
 * title="stiffness matrix from element one" style="border-right: 1px solid black; border-bottom: 1px solid black;" | $$k_{41}^{(1)}$$
 * title="stiffness matrix from element one and element two" style="border-bottom: 1px solid black;" | $$k_{43}^{(1)} + k_{21}^{(2)}$$
 * title="stiffness matrix from element one and element two" style="border-right: 1px solid black; border-bottom: 1px solid black;" | $$k_{44}^{(1)} + k_{22}^{(2)}$$
 * title="stiffness matrix from element two" | $$k_{23}^{(2)}$$
 * title="stiffness matrix from element two" style="border-right: 1px solid black;" | $$k_{24}^{(2)}$$
 * 0
 * style="border-right: 1px solid black;" | 0
 * title="stiffness matrix from element two" | $$k_{31}^{(2)}$$
 * title="stiffness matrix from element two" | $$k_{32}^{(2)}$$
 * title="stiffness matrix from element two" | $$k_{33}^{(2)}$$
 * title="stiffness matrix from element two" style="border-right: 1px solid black;" | $$k_{34}^{(2)}$$
 * 0
 * style="border-right: 1px solid black;" | 0
 * title="stiffness matrix from element two" style="border-bottom: 1px solid black;" | $$k_{41}^{(2)}$$
 * title="stiffness matrix from element two" style="border-bottom: 1px solid black;" | $$k_{42}^{(2)}$$
 * title="stiffness matrix from element two" style="border-bottom: 1px solid black;" | $$k_{43}^{(2)}$$
 * title="stiffness matrix from element two" style="border-right: 1px solid black; border-bottom: 1px solid black;" | $$k_{44}^{(2)}$$
 * }
 * 0
 * style="border-right: 1px solid black;" | 0
 * title="stiffness matrix from element two" | $$k_{31}^{(2)}$$
 * title="stiffness matrix from element two" | $$k_{32}^{(2)}$$
 * title="stiffness matrix from element two" | $$k_{33}^{(2)}$$
 * title="stiffness matrix from element two" style="border-right: 1px solid black;" | $$k_{34}^{(2)}$$
 * 0
 * style="border-right: 1px solid black;" | 0
 * title="stiffness matrix from element two" style="border-bottom: 1px solid black;" | $$k_{41}^{(2)}$$
 * title="stiffness matrix from element two" style="border-bottom: 1px solid black;" | $$k_{42}^{(2)}$$
 * title="stiffness matrix from element two" style="border-bottom: 1px solid black;" | $$k_{43}^{(2)}$$
 * title="stiffness matrix from element two" style="border-right: 1px solid black; border-bottom: 1px solid black;" | $$k_{44}^{(2)}$$
 * }
 * title="stiffness matrix from element two" style="border-bottom: 1px solid black;" | $$k_{43}^{(2)}$$
 * title="stiffness matrix from element two" style="border-right: 1px solid black; border-bottom: 1px solid black;" | $$k_{44}^{(2)}$$
 * }
 * }

The box, within the matrix on the top left, is the stiffness matrix from element one. The box to the lower right is the stiffness matrix from element two. Combined the two become the global stiffness matrix.

Where capital k (K) is the global stiffness coefficient and lowercase k (k) is the element stiffness coefficient. Below is the relationship between the global stiffness coefficients and the element stiffness coefficients for the global stiffness matrix above.

K11 = k11(1) ; K12 = k12(1) ; K13 = k13(1) ; K14 = k14(1) ; K15 = 0 ; K16 = 0

K21 = k21(1) ; K22 = k22(1) ; K23 = k23(1) ; K24 = k24(1) ; K25 = k25(1) ; K26 = k26(1)

K31 = k31(1) ; K32 = k32(1) ; K33 = k33(1) + k11(2) ; K34 = k34(1) + k12(2); K13 = k13(2) ; K14 = k14(2)

K41 = k41(1) ; K42 = k42(1) ; K43 = k43(1) + k21(2) ; K44 = k44(1) + k22(2); K23 = k23(2) ; K24 = k24(2)

K51 = 0 ; K52 = 0 ; K53 = k31(2) ; K54 = k32(2) ; K55 = k33(2) ; K56 = k34(2)

K61 = 0 ; K62 = 0 ; K63 = k41(2) ; K64 = k42(2) ; K65 = k43(2) ; K66 = k44(2) The calculation of the values in the k matrices for element 1 and element 2 are as follows: θ(1) = 30° l(1) = cos θ(1) = cos30° = $$\frac{\sqrt{3}}{2}$$ m(1) = sin θ(1) = sin30° = $$\frac{1}{2}$$ k(1) = E(1)·A(1)/ L(1) = $$\frac{3}{4}$$

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

k11(1) = k(1) · ( l(1) )2 = $$\frac{3}{4}$$ · $$(\frac{\sqrt{3}}{2})^{2}$$ = $$\frac{9}{16}$$

k12(1) = k(1) · l(1) · m(1) = $$\frac{3}{4}$$ · $$\frac{\sqrt{3}}{2}$$· $$\frac{1}{2}$$ = $$\frac{\sqrt{3}}{16}$$

k42(1) = - k(1) · ( m(1) )2 = - $$\frac{3}{4}$$ · $$(\frac{2})^{2}$$ = - $$\frac{3}{16}$$

Observation:

1) Only three numbers need to be computed. Other coefficients have the same absolute value just differ by sign (+ or -) 2) Matrix k(1) is symmetrical i.e., kij(1) = kji(1); just interchange the row and column indices

For example: k13(1) = k31(1)

In general,

kij(e) = kji(e) or kij(1)Т = k(1)

where Т represents the transpose of the matrix.

$$k^{(e)} = \begin{bmatrix} k_{11}^{(1)} & k_{12}^{(1)} & k_{13}^{(1)} & k_{14}^{(1)}\\ & k_{22}^{(1)} & k_{23}^{(1)} & k_{24}^{(1)}\\ & & k_{33}^{(1)} & k_{34}^{(1)}\\ & &  & k_{44}^{(1)} \end{bmatrix}$$

Since the matrix is symmetric about the diagonal, only the upper right triangle part needs to be calculated.

$$k^{(1)} = \begin{bmatrix} \frac{9}{16} & \frac{\sqrt{3}}{16} & -\frac{9}{16} & -\frac{\sqrt{3}}{16}\\ \frac{\sqrt{3}}{16} & \frac{3}{16} & -\frac{\sqrt{3}}{16} & -\frac{3}{16}\\ -\frac{9}{16} & -\frac{\sqrt{3}}{16} & \frac{9}{16} & \frac{\sqrt{3}}{16}\\ -\frac{\sqrt{3}}{16} & -\frac{3}{16} & \frac{\sqrt{3}}{16} & \frac{3}{16} \end{bmatrix}$$

Element 2:

k(2) = E(2)·A(2)/ L(2) = 5 θ(2) = $$-\frac{\Pi }{4}$$ l(2) = cos θ(2) = $$cos(-\frac{\Pi }{4}) = \frac{\sqrt{2}}{2}$$ m(2) = sin θ(2) = $$\sin (-\frac{\Pi }{4}) = -\frac{\sqrt{2}}{2}$$

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

Observation:

1) The absolute value of all coefficients kji(e), e=2 (i,j) = 1,...,4 are the same (|l| = |m|) → Only four coefficients need to be computed; all other coefficients add + or -. 2)kij(2)Т = k(2)

$$k^{(2)} = \begin{bmatrix} \frac{5}{2} & -\frac{5}{2} & -\frac{5}{2} & \frac{5}{2}\\ -\frac{5}{2} & \frac{5}{2} & \frac{5}{2} & -\frac{5}{2}\\ -\frac{5}{2} & \frac{5}{2} & \frac{5}{2} & -\frac{5}{2}\\ \frac{5}{2} & -\frac{5}{2} & -\frac{5}{2} & \frac{5}{2} \end{bmatrix}$$

Compact Notations between Global Diagrams and Element Diagrams
It is necessary to understand the relationships between the global free diagrams and the element free diagrams in order to solve the truss system of matrices going from the element matrices to the global matrices.

Global Free Diagram Relationship
The global free diagram relationship is given by the equation: $$\begin{bmatrix}K_{ij}^{}\end{bmatrix}_{6x6} \left\{d_{ij}\right\}_{6x1} = \left\{F_{i}\right\}_{6x1}$$ where the d and F  elements are composed by 6 x 6 matrices as following:

$$\sum_{j=1}^{6} K_{ij}d_{j}= F_{i}$$ where i goes from 1 through 6 since there are six equations
 * Global Stiffness Matrix:

$$K_{nxn} = \begin{bmatrix}K_{ij}^{}\end{bmatrix}_{nxn}$$
 * Global Displacement Matrix:

$$d_{nx1} = \left\{d_{j}^{}\right\}_{nx1}$$
 * Global Force Matrix:

$$F_{nx1} = \left\{F_{i}^{}\right\}_{nx1}$$

Element Free Diagram Relationship
The element free diagrams relationship is given by the equation: $$k_{4x4}^{(e)} d_{4x4}^{(e)} = f_{4x1}^{(e)}$$ where e indicates the element number. The d and f elements are composed by 4 x 1 matrices as following:

$$d_{4x1}^{(e)} = \left\{d_{j}^{(e)}\right\}_{4x1}$$ $$d_{4x1}^{(e)}=\begin{Bmatrix} d_{1}^{(e)} \\ d_{2}^{(e)} \\ d_{3}^{(e)} \\ d_{4}^{(e)} \end{Bmatrix}$$ $$f_{4x1}^{(e)} = \left\{f_{i}^{(e)}\right\}_{4x1}$$ $$f_{4x1}^{(e)}=\begin{Bmatrix} f_{1}^{(e)} \\ f_{2}^{(e)} \\ f_{3}^{(e)} \\ f_{4}^{(e)} \end{Bmatrix}$$ $$k_{4x4}^{(e)} = \begin{bmatrix}k_{ij}^{(e)}\end{bmatrix}_{4x4}$$ $$k_{4x4}^{(e)}=\mathbf \begin{bmatrix} K_{11}^{(e)} & K_{12}^{(e)} & K_{13}^{(e)} & K_{14}^{(e)}\\K_{21}^{(e)} & K_{22}^{(e)} & K_{23}^{(e)} & K_{24}^{(e)}\\K_{31}^{(e)} & K_{32}^{(e)} & K_{33}^{(e)} & K_{34}^{(e)}\\K_{41}^{(e)} & K_{42}^{(e)} & K_{43}^{(e)} & K_{44}^{(e)}\end{bmatrix}$$
 * Element Displacement Matrix:
 * Element Force Matrix:
 * Element Stiffness Matrix:

Since the matrix k is symmetric in general the upper triangular part of the matrix should transpose to the lower triangular part, i.e. $$k_{ij}^{(e)} = k_{ji}^{(e)}$$

Assembly Process
From the free body general relationships it is possible to generate new associations in order to solve the stiffness, displacement and forces matrices:


 * Global level: $$\mathbf\begin{Bmatrix} d_{1} & d_{2} & d_{3} & d_{4} & d_{5} & d_{6}\end{Bmatrix}$$


 * Element level:
 * Element 1: $$\mathbf\begin{Bmatrix} d_{1}^{(1)} & d_{2}^{(1)} & d_{3}^{(1)} & d_{4}^{(1)} & d_{5}^{(1)} & d_{6}^{(1)}\end{Bmatrix}$$
 * Element 2: $$\mathbf\begin{Bmatrix} d_{1}^{(2)} & d_{2}^{(2)} & d_{3}^{(2)} & d_{4}^{(2)} & d_{5}^{(2)} & d_{6}^{(2)}\end{Bmatrix}$$


 * Nodes:
 * Node 1: $$d_{2} = d_{1}^{(1)}$$
 * Node 2: $$d_{3} = d_{2}^{(1)}$$
 * $$d_{3} = d_{3}^{(1)} = d_{1}^{(2)}$$
 * $$d_{4} = d_{4}^{(1)} = d_{2}^{(2)}$$

Note: These points are from the same node therefore must move in the same direction =>> (same global number = same direction)
 * Node 3: $$d_{5} = d_{3}^{(2)}$$
 * $$d_{6} = d_{4}^{(2)}$$

Step 4: Elimination of known degrees of freedom (dofs)
This Process reduces the global force displacement relationship by eliminating the first and last two columns of the global stiffness matrix.

d1 = d2 = d5 = d6 = 0

The displacements above are zero. Therefore, columns 1, 2, 5 and 6 of the global stiffness matrix go to zero. This reduces the matrix to a 6x2.

The following displays the F matrix calculated two different ways.

$$ \textbf{K}\begin{bmatrix} d_{1}=0\\ d_{2}=0\\ d_{3}\\ d_{4}\\ d_{5}=0\\ d_{6}=0 \end{bmatrix} =\begin{bmatrix} K_{13} & K_{14}\\ K_{23} & K_{24}\\ K_{33} & K_{34}\\ K_{43} & K_{44}\\ K_{53} & K_{54}\\ K_{63} & K_{64} \end{bmatrix} \begin{bmatrix} d_{3}\\ d_{4} \end{bmatrix} = \textbf{F} $$

Here, we delete the corresponding columns in the global stiffness matix, K. Due to the Principle of Virtual Work (PVW), we must also delete the corresponding rows (rows 1, 2, 5, 6).

Below, we have the Resulting Force Displacement relation:

$$ \begin{bmatrix} K_{33} &K_{34} \\ K_{43} & K_{44} \end{bmatrix} \begin{bmatrix} d_{3}\\ d_{4} \end{bmatrix} = \begin{bmatrix} F_{3}\\ F_{4} \end{bmatrix} $$ where the F matrix is found below.

$$ \textbf{F} = \begin{bmatrix} F_{3}\\ F_{4} \end{bmatrix} = \begin{bmatrix} 0\\ P \end{bmatrix} $$ Before we continue, we will find the determinant and inverse of the K matrix. The determinant is simply equal to:

$$ det\; \textbf{K} = K_{33}K_{44}-K_{34}K_{43} $$

Now that we have the determinant, we can calculate the inverse using the formula below.

$$ \textbf{K}^{-1} = \frac{1}{det\; \textbf{K}} \begin{bmatrix} K_{44} & -K_{34}\\ -K_{43} & K_{33} \end{bmatrix} $$

The inverse of the K matrix cannot be found by using the transpose of the matrix, illustrated below.
 * NOTE*

$$ \textbf{K}^{T} = \begin{bmatrix} K_{33} & K_{34}\\ K_{43} & K_{44} \end{bmatrix} , where\; \textbf{K}^{-1}\neq \frac{1}{det\; \textbf{K}} \textbf{K}^{T} $$

With the inverse of the K matrix known, we can now find the unknown forces from the d matrix.

$$ \begin{bmatrix} d_{3}\\ d_{4} \end{bmatrix} = \textbf{K}^{-1}\begin{bmatrix} 0\\ P \end{bmatrix} = \begin{bmatrix} 4.352\\ 6.127 \end{bmatrix} $$

Step 5: Compute Reactions
There are two main methods for solving this 2-bar truss system. Method 1 For the first method, we will use the element force displacement relationship: $$ \textbf{k}^{e}\textbf{d}^{e}=\textbf{f}^{e}, where\; e=1,2 $$

We can now separate the truss into each individual bar, and compute the calculations for each bar alone.

Element 1:

$$where\; \textbf{d}^{(1)}= \begin{bmatrix} 0\\ 0\\ 4.352\\ 6.127 \end{bmatrix} = \begin{bmatrix} d_{1}^{(1)}\\ d_{2}^{(1)}\\ d_{3}^{(1)}\\ d_{4}^{(1)} \end{bmatrix} $$

Element 2:

$$ where\; \textbf{d}^{(2)}= \begin{bmatrix} 4.352\\ 6.127\\ 0\\ 0 \end{bmatrix}= \begin{bmatrix} d_{1}^{(2)}\\ d_{2}^{(2)}\\ d_{3}^{(2)}\\ d_{4}^{(2)} \end{bmatrix} since\; d_{3}=d_{1}^{(2)}, d_{4}=d_{2}^{(2)} $$

Step 6: Computation of Reaction Forces
The displacements for elements 1 and 2 are now known, and can be used in order to solve for the reaction forces at the pins. The force-displacement relationship that is shown below will be used to find the forces at the ends of each element.

The element stiffness matrix and the displacement matrix of element 1 can be reduced based on the values of d(1)1=d(2)2=0. These reduced matrices are shown below.

$$k^{(1)}*d^{(1)}=f^{(1)}= \left[ \begin{matrix} -0.5625&-0.32476\\  -0.32476&-0.1875\\  0.5625&0.32476\\   0.32476&0.1875 \end{matrix}\right]* \left\{ \begin{matrix} 4.352\\  6.1271 \end{matrix} \right\}$$

The forces that are induced in element 1 by the applied force P at node 2 are equivalent to:


 * $$f^{(1)}=\begin{Bmatrix} -4.4378 \\ -2.5622 \\ 4.4378 \\ 2.5622 \end{Bmatrix}=\begin{Bmatrix} f_1^{(1)} \\ f_2^{(1)} \\ f_3^{(1)} \\ f_4^{(1)} \end{Bmatrix}$$

The equations for static equilibrium are listed below. It is evident by the values of the forces in the element 1 force matrix that the first two equations are satisfied. The third equation is satisfied by taking the moment about either one of the nodes of element 1.

$$ \begin{array}{lr} \sum F_x=f_1^{(1)}+f_3^{(1)}=0&(1)\\ \sum F_y=f_2^{(1)}+f_4^{(1)}=0&(2)\\ \sum M_{Any\, Node}=0&(3)\\ \end{array} $$

The two elements of the 2-D Plane Truss problem are shown below. The forces that act along the axis of the element can be found by using the Pythagorean Theorem. It can be noted that the force that acts on one end of each element is equal and opposite to the force acting on the other end of the element. The equation to accomplish this is listed after the diagrams of the elements.





$$P_1^{\left(1\right)}=\left[\left(f_1^{\left(1\right)}\right)^2+\left(f_2^{\left(1\right)}\right)^2\right]^{\frac 12}$$

The forces that are found using the method described can be verified by applying the equations of static equilibrium across each element. The global picture can also be divided in two ways, by using the Euler Cut Principle, in order to perform this operation. This analysis is provided in the homework section at the beginning of this article. The methods that this can be cut are shown below. Each cut is denoted by a dashed circle. The first method includes the global node 2 in the main section of element 1, while the second method has node 2 being analyzed separately.



In order to verify the static equilibrium of the whole system at once, the equilibrium of node 2 can be established. The diagram for this is shown below, while the calculation is performed in the homework section at the beginning of the article. The forces shown are P (applied force), P(1)1 (axial force in element 1), P(2)2 (axial force in element 2).



Team Contributors
Iain Specht--Eml 4500.f08.delta 6.specht 14:24, 22 September 2008 (UTC) Micheal Trent Krueger--Eml4500.f08.delta 6.krueger 01:39, 26 September 2008 (UTC) Nicolas Castrillon --Eml4500.f08.delta 6.castrillon 01:40, 26 September 2008 (UTC) Diana Guzman--Eml4500.f08.delta 6.guzman 05:02, 26 September 2008 (UTC) Sean Lopez--Eml4500.f08.delta 6.lopez 05:25, 26 September 2008 (UTC) Eric Ramirez--Eml4500.f08.delta 6.ramirez 12:42, 26 September 2008 (UTC)