User:Eml4500.f08.group/Homework2/Lecture Notes

= Solving 2 Element 2D Truss=

Global Picture


As a standard, mark local load numbers in a manner in which they increase along the x axis. As shown above, the global system can be separated into two separate local systems, where P can only be applied only to one arbitrary local system. On the bottom right local node system, taking the moment about either node shows that Fx(1) = Fy(2) = 0.

Solution:

ΣFx = F1cosθ1 + Pcosθ2 + F2cosθ3

ΣFy = F1sinθ1 + Psinθ2 + F2sinθ3

ΣM2 = 0 (not useful)

ΣM1 = -Pcosθ2L1sinθ1 + Psinθ2L1cosθ1 + F2cosθ3(L1sinθ1 - L2sinθ3) + F2sinθ3(L1cosθ1 + L2cosθ3)

Here are three variables and three equations which can be solved by arithmetic.


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!Contribution Statically Indeterminate Systems, from team Carbon
 * A statically indeterminate system cannot be solved by the three classical equations of statics:
 * A statically indeterminate system cannot be solved by the three classical equations of statics:

ΣFx=0

ΣFy=0

ΣM=0 (take the sum of moments about any arbitrary node)

If the number of unknown forces exceed the number of useful static equilibrium equations, then the system is statically indeterminate. In these cases, mechanics equations such as that of deformation may be used to solve the system. Some examples are below:



Contribution by group.w
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Global Force Displacement at element level


k is the element stiffness matrix for element e (4x4 in this case)

d is the the element displacement matrix (4x1 in this case)

f is the element force matrix (4x1 in this case)

Matrix p 225 book, 2nd equation form the bottom





axial stiffness of bar elem "e". e = 1,2

l(e), m(e) = director cosines of ~x axis (goes from [1] to [2]) with respect to global (x,y) coords

→ l(e), m(e) = director cosines of x~ axis

l(e) = i~ • i = cosθ(e)

m(e) = i~ • j = cos(π/2 - θ(e)) = sinθ(e)

Note: Physically, the director cosines represent the components of $$\hat{\tilde{i}}$$ with respect to the basis $$(\hat{i},\hat{j})$$.




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!Contribution Director cosine derivation, from team Carbon
 * i~ = cosθ(e)i + sinθ(e)j
 * i~ = cosθ(e)i + sinθ(e)j

→ l(e) = i~ • i = (cosθ(e)i + sinθ(e)j) • i = cosθ(e)(i • i) + sinθ(e)(j • i), where i • i = 1 and j • i = 0, = cosθ(e)

→ m(e) = i~ • j = (cosθ(e)i + sinθ(e)j) • j = cosθ(e)(i • j) + sinθ(e)(j • j), where i • j = 0 and j • j = 1, = sinθ(e)

Contribution by group.w
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Note in the first matrix that there are only three numbers to compute in each matrix below, then change the respective sign. The first matrix is also symmetric, Kij(e) = Kji(e).

Note in the second matrix that the absolute value of all numbers are 5/2, and the transpose of the matrix is equal to the matrix (the matrix is symmetric).



The Force-Distance relationship: K(e)d(e) = F(e)



In compact notation:

[Kij]6x6{dj}6x1={Fi}6x1

Σkijdj=Fi, i=1,...6

Knxn=[Kij]nxn = global stiffness matrix

dnx1={dj}nx1 = global displacement matrix

Fnx1={Fi}nx1 = global force matrix

Recall the element FD relationship:

[kij]4x4{dj}4x1={fi}4x1

knxn=[kij]nxn = element stiffness matrix

dnx1={dj}nx1 = element displacement matrix

fnx1={fi}nx1 = element force matrix

Lower case designates the element matrix while upper case designates the global matrix.

How do you go from element matrices to global matrices? Through an assembly process:

Identify the correspondence between element displacement dof(s) (element1: {d1(1), d2(1), d3(1), d4(1)} element2: {d1(2), d2(2), d3(2), d4(2)})and global displacement dof(s)({d1, d2, d3, d4})

Global - Local dof identification:

 * Node 1:
 * d1 = d1(1)
 * d2 = d2(1)
 * Node 2:
 * d3 = d3(1) = d1(2)
 * d4 = d4(1) = d2(2)
 * Node 3:
 * d5 = d3(2)
 * d6 = d4(2)

Global conceptual step of assembly (topology of K): $$\begin{bmatrix} & &  |&  &k^{(2)}  &|\\ & &  |-&  -&  -& -  | \end{bmatrix} \begin{Bmatrix} d_{1}\\ d_{2}\\ d_{3}\\ d_{4}\\ d_{5}\\ d_{6} \end{Bmatrix} = \begin{Bmatrix} F_{1}\\ F_{2}\\ F_{3}\\ F_{4}\\ F_{5}\\ F_{6}
 * & &k^{(1)}  &|  &  & \\
 * & &k^{(1)}  &|  &  & \\

\end{Bmatrix}$$

Global stiffness matrix K:

$$\begin{bmatrix} k_{11}^{(1)}&k_{12}^{(1)}&k_{13}^{(1)}&k_{14}^{(1)} &0  &0 \\ k_{21}^{(1)}&k_{22}^{(1)} &k_{23}^{(1)}  &k_{24}^{(1)}  &0  &0 \\ k_{31}^{(1)}&k_{32}^{(1)} &k_{33}^{(1)}+k_{11}^{(2)}  &k_{34}^{(1)}+k_{12}^{(2)}  &k_{13}^{(2)}  &k_{14}^{(2)} \\ k_{41}^{(1)}&k_{42}^{(1)} &k_{43}^{(1)}+k_{21}^{(2)}  &k_{44}^{(1)}+k_{22}^{(2)}  &k_{23}^{(2)}  &k_{24}^{(2)} \\ 0& 0&  k_{31}^{(2)}&k_{32}^{(2)}  &k_{33}^{(2)}  &k_{34}^{(2)} \\ 0& 0&  k_{41}^{(2)}&k_{42}^{(2)}  &k_{43}^{(2)}  &k_{44}^{(2)} \end{bmatrix}$$

K11=k11(1)=$$\frac{9}{16}$$

K12=k12(1)=$$\frac{3\sqrt{3}}{16}$$

K33=k33(1)+k11(2)=$$\frac{9}{16} + \frac{5}{2}$$

K34=$$\frac{3\sqrt{3}}{16} + \frac{-5}{2}$$ = -2.1752

K44=$$\frac{3}{16} + \frac{5}{2}$$ = 2.6875

...etc

A general rule because of symmetry: K43=K34, K33 is not equal to K44

Elimination of known dofs by reducing the global FD relation matrices
Applying the fixed boundary equations, d1=d2=d5=d6=0, therefore d3 and d4 are the only displacement unknowns. Then by principle of virtual work (PVW) the rows(1,2,5,6) corresponding to the F supplied by the reactions are also deleted. Resulting in the FD relationship:

$$\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}$$

Applying P=7, Then F3=0, F4=7 Then solving for the displacement:

$$\begin{Bmatrix} d_{3}\\ d_{4}

\end{Bmatrix} \begin{bmatrix} K \end{bmatrix}^{-1} = \begin{Bmatrix} 0\\ p

\end{Bmatrix}$$

d3 = 4.354

d4 = 6.1271

Inverting Stiffness Matrix
What is $$\displaystyle \overline{\underline K}^{-1}$$, the inverse of $$\displaystyle \overline{\underline K}$$? How do you find it by hand (without calculator)?

We begin with the determinant of $$\displaystyle \overline{\underline K}$$

$$\displaystyle det(\overline{\underline K})=K_{33}K_{44}-K_{34}K_{43}$$

$$\overline{\underline K}^{-1} = \frac{1}{det(\overline{\underline K})} \begin{bmatrix} K_{44} &-K_{34} \\ -K_{43}  & K_{33} \end{bmatrix}$$

$$\overline{\underline K}^{-1} \overline{\underline K} =\overline{\underline I}= \begin{bmatrix} 1 & 0\\   0 & 1 \end{bmatrix}$$

$$=\frac{1}{det(\overline{\underline K})} \begin{bmatrix} det(\overline{\underline K}) & 0                           \\ 0                           & det(\overline{\underline K}) \end{bmatrix}$$

NOTE:
Do Not confuse the above with the transpose of K

$$\overline{\underline K}^{T} = \begin{bmatrix} K_{33} &-K_{43} \\ -K_{34}  & K_{44} \end{bmatrix}$$

$$\overline{\underline K}^{-1} \neq \frac{1}{det(\overline{\underline K})}\overline{\underline K}^{T}$$

Solving 2 Bar Truss Reactions
Back to the 2-bar truss system:

$$\begin{Bmatrix} d_3 \\d_4 \end{Bmatrix} = k\begin{Bmatrix} 0 \\P\end{Bmatrix} = \begin{Bmatrix} 4.352 \\6.1271\end{Bmatrix}$$

Compute Reactions using two methods, FEA and statics

What is Known
Global or element FD relations can be used (2 methods)

$$\displaystyle k^{(e)}d^{(e)} = f^{(e)} e=1,2 $$

Use the element Force-Displacement (FD) relations:

Element 1 (e=1)

$$\displaystyle k^{(1)}d^{(1)} = f^{(1)}$$

where: $$d^{(1)} = \begin{Bmatrix} 0 \\ 0 \\ 4.352 \\ 6.1271 \end{Bmatrix} = \begin{Bmatrix} d_1^{(1)} \\ d_2^{(1)} \\ d_3^{(1)} \\ d_4^{(1)} \end{Bmatrix}$$ $$f^{(1)} = \begin{Bmatrix} f_1^{(1)} \\ f_2^{(1)} \\ f_3^{(1)} \\ f_4^{(1)} \end{Bmatrix}$$
 * k is known
 * d is known
 * f is the unknown reaction

Element 2 (e=2)

$$\displaystyle k^{(2)}d^{(2)} = f^{(2)}$$

where: $$d^{(2)} = \begin{Bmatrix} 4.352 \\ 6.1271 \\ 0 \\ 0 \end{Bmatrix} = \begin{Bmatrix} d_1^{(2)} \\ d_2^{(2)} \\ d_3^{(2)} \\ d_4^{(2)} \end{Bmatrix}$$ $$f^{(2)} = \begin{Bmatrix} f_1^{(2)} \\ f_2^{(2)} \\ f_3^{(2)} \\ f_4^{(2)} \end{Bmatrix}$$
 * k is known
 * d is known
 * f is the unknown reaction

Using this information we can now solve for the reaction forces acting on each element

Element 1 Reaction Forces
From:

$$\underline{k} ^{(1)} \underline{d} ^{(1)} = \underline{f} ^{(1)}$$

we find that the reaction forces are equal to the solution of:

$$\underline{f} ^{(1)} = \begin{bmatrix} 0.5625 & 0.32476 &-0.5625 & -0.32476    \\  0.32476 & 0.1875  &-0.32476 & -0.1875       \\  -0.5625 & -0.32476  &0.5625 & 0.32476       \\  -0.32476 & -0.1875  &0.32476 & 0.1875       \\ \end{bmatrix} \begin{Bmatrix} 0 \\ 0 \\  4.352 \\  6.1271 \\ \end{Bmatrix}$$

The zeros in the 1 x 4 matrix eliminate the first two columns in the 4x4 matrix resulting in:

$$ \underline{f} ^{(1)} = \begin{bmatrix} &-0.5625 & -0.32476    \\  & -0.32476 & -0.1875       \\  &0.5625 & 0.32476       \\  & 0.32476 & 0.1875       \\ \end{bmatrix} \begin{Bmatrix} 4.352 \\ 6.1271 \\ \end{Bmatrix} = \begin{Bmatrix} -4.4378 \\ -2.5622 \\  4.4378 \\  2.5622 \\ \end{Bmatrix} = \begin{Bmatrix} \underline{f}_1 ^{(1)} \\ \underline{f}_2 ^{(1)} \\ \underline{f}_3 ^{(1)} \\ \underline{f}_4 ^{(1)} \\ \end{Bmatrix} $$

$$\underline{f}_1 ^{(1)}$$ and $$\underline{f}_2 ^{(1)}$$ are Reaction forces where as $$\underline{f}_3 ^{(1)}$$ and $$\underline{f}_4 ^{(1)}$$ are internal forces within the element. This element is a two force member as Verified Here. And so the following summations should be true, and are.

$$\sum_{} F_x = \underline{f}_1 ^{(1)} + \underline{f}_3 ^{(1)} = 0$$

$$\sum_{} F_y = \underline{f}_2 ^{(1)} + \underline{f}_4 ^{(1)} = 0$$

$$\displaystyle \sum_{} M_{any point} = 0$$

Element 2 Reaction Forces
$$P_1^{(1)} = [(f_1^{(1)})^2 + (f_2^{(1)})^2]^{1/2}$$

Statics Method
There are two methods that can be used to break up a truss, both are based on the Euler Cut Principle. Method 2 breaks up the truss segments further such that is is made up of nodes and 2-Force members


 * [[Image:EulerCutPrinciple.JPG]]

Back to the Finite Element Solution: How to bring $$\underline P$$ back into the big picture

The Equilibrium of node (2) is Verified Here