User:Eml4500.f08.group/Homework3/Lecture Notes

=Force-Distance Relationship on the Global System=

Derivation of element fd with respect to the global coordinate system:

eq (1) k (e) d (e) = f (e)



eq (2) $$k^{(e)}\begin{bmatrix} 1 &-1 \\ -1& 1 \end{bmatrix} \begin{Bmatrix} q^{(e)}_{1}\\ q^{(e)}_{2} \end{Bmatrix} = \begin{Bmatrix} P^{(e)}_{1}\\ P^{(e)}_{2} \end{Bmatrix}$$

q(e)i = axial displacement of element e at the local node i. P(e)i = axial force of element e at local node i.

We can derive eq(1) from eq(2) and we can find a relationship between q(e)i and d(e)i as well as the relationship between P(e)i and f(e)i. These relationships can be expressed in the form:

$$q^{(e)}=T^{(e)}d^{(e)}$$

Consider the displacement vector of the local node 1, denoted by d(e)1

$$\vec{d}_{1}^{(e)}=d_{1}^{(e)}\vec{i}+d_{2}^{(e)}\vec{j}$$

$$q_{1}^{(e)}$$ is equal to the axial displacement of node 1, which is the orthogonal projection of the displacement vector $$\vec{d}_{1}^{(e)}$$ of node 1 on the x tilda axis of element e. Therefore:

$$q_{1}^{(e)}=\vec{d}_{1}^{(e)}\cdot \vec{\tilde{i}}$$ $$=(d_{1}^{(e)}\vec{i}+d_{2}^{(e)}\vec{j})\cdot \vec{\tilde{i}}$$ $$=d_{1}^{(e)}(\vec{i}\cdot \vec{\tilde{i}})+d_{2}^{(e)}(\vec{j}\cdot \vec{\tilde{i}})$$ From the definition of l(e) equal to cosθ(e) and m(e) equal to sinθ(e), the above equation reduces to: $$q_{1}^{(e)}=l^{e}d_{1}^{e}+m^{e}d_{2}^{e}$$ In matrix form reduces to: $$q_{1}^{(e)}=\begin{bmatrix} l^{(e)} & m^{(e)} \end{bmatrix} \begin{Bmatrix} d_{1}^{(e)} \\ d_{2}^{(e)} \end{Bmatrix}$$

HW: Similarly for node 2:

$$\vec{d}_{3}^{(e)}=d_{3}^{(e)}\vec{i}+d_{4}^{(e)}\vec{j}$$

$$q_{2}^{(e)}$$ is equal to the axial displacement of node 2, which is the orthogonal projection of the displacement vector $$\vec{d}_{3}^{(e)}$$ of node 2 on the x tilda axis of element e. Therefore:

$$q_{2}^{(e)}=\vec{d}_{3}^{(e)}\cdot \vec{\tilde{i}}$$ $$=(d_{3}^{(e)}\vec{i}+d_{4}^{(e)}\vec{j})\cdot \vec{\tilde{i}}$$ $$=d_{3}^{(e)}(\vec{i}\cdot \vec{\tilde{i}})+d_{4}^{(e)}(\vec{j}\cdot \vec{\tilde{i}})$$ From the definition of l(e) equal to cosθ(e) and m(e) equal to sinθ(e), the above equation reduces to: $$q_{2}^{(e)}=l^{e}d_{3}^{e}+m^{e}d_{4}^{e}$$ In matrix form reduces to: $$q_{2}^{(e)}=\begin{bmatrix} l^{(e)} & m^{(e)} \end{bmatrix} \begin{Bmatrix} d_{3}^{(e)} \\ d_{4}^{(e)} \end{Bmatrix}$$

Therefore the relationship between q(e) and d(e) is expressed through the form $$q^{(e)}=T^{(e)}d^{(e)}$$ where $$T=\begin{bmatrix} l^{(e)} & m^{(e)}&0&0\\ 0&0&l^{(e)} & m^{(e)} \end{bmatrix}$$ Leaving the resultant relationship as: $$\begin{Bmatrix} q_{1}^{(e)}\\ q_{2}^{(e)} \end{Bmatrix}=\begin{bmatrix} l^{(e)} & m^{(e)}&0&0\\ 0&0&l^{(e)} & m^{(e)} \end{bmatrix} \begin{Bmatrix} d_{1}^{(e)}\\ d_{2}^{(e)} \\ d_{3}^{(e)} \\ d_{4}^{(e)}

\end{Bmatrix} $$

Finding the Stiffness Matrix
If you use the same argument, $$\begin{Bmatrix} p^{(e)}_{1}\\ p^{(e)}_{2} \end{Bmatrix}_{2x1} = \bar{T}^{(e)}_{2x4} \begin{Bmatrix} f^{(e)}_{1}\\ f^{(e)}_{2}\\ f^{(e)}_{3}\\ f^{(e)}_{4} \end{Bmatrix}_{4x1}$$

This yields the symbolic simplification:

P (e) = T (e) f (e)

For each node, the relationship can be expressed as:

q (e)2x1 = T (e)2x4 d (e)4x1

Substituting variables into the element axial Force-Distance Relationship:

$$\hat{k}^{(e)}_{2x2} q^{(e)}_{2x1}= p^{(e)}_{2x1}$$ → $$\hat{k}^{(e)} (T^{(e)}d^{(e)})= (T^{(e)}f^{(e)})$$

The goal is to have k (e) d (e) = f (e). Therefore, we want to move the T (e) from the right hand side to the left hand side by multiplying both sides by T (e)-1 (or the inverse of T). But because T is not a square matrix, it cannot be inverted.

k (e) d (e)= f (e) → $$[T^{(e)T}_{4x2} (transpose) \hat{k}^{(e)}_{2x2}T^{(e)}_{2x4}]_{(4x4)}d^{(e)}_{4x1} = f^{(e)}_{4x1}$$

HW: Verify that $$k^{(e)} = T^{(e)T}\hat{k}^{(e)}T^{(e)}$$ $$\bar{T}^{(e)T}k^{(e)}\begin{bmatrix} 1 & -1\\ -1 & 1 \end{bmatrix}\begin{bmatrix} l^{(e)} & m^{(e)} & 0 & 0\\ 0 & 0 & l^{(e)} & m^{(e)} \end{bmatrix} = \begin{bmatrix} l^{(e)} & 0\\ m^{(e)} & 0\\ 0 & l^{(e)}\\ 0 & m^{(e)} \end{bmatrix}k^{(e)}\begin{bmatrix} l^{(e)} & m^{(e)} & -l^{(e)} & -m^{(e)}\\ -l^{(e)} & -m^{(e)} & l^{(e)} & m^{(e)} \end{bmatrix} = k^{(e)}\begin{bmatrix} (l^{(e)})^{2} & l^{(e)}m^{(e)} & -(l^{(e)})^{2} & -(l^{(e)}m^{(e)})\\ l^{(e)}m^{(e)} & (m^{(2)})^2 & -(l^{(e)}m^{(e)}) & -(m^{(2)})^2\\ -(l^{(e)})^{2} & -(l^{(e)}m^{(e)}) & (l^{(e)})^{2} & l^{(e)}m^{(e)}\\ -(l^{(e)}m^{(e)}) & -(m^{(2)})^2 & l^{(e)}m^{(e)} & (m^{(2)})^2 \end{bmatrix} = \bar{k}^{(e)}$$

Principle of Virtual Work: K 6x6 d 6x1 = F 6x1 → K' 2x2 d' 2x1 = F' 2x1

We cannot solve for d as d = K -1 F because of the singularity of K -1 where the det[ K ]=0 and therefore we cannot invert K.

In 2D motion, there are three possible rigid body motions: two in translation and one rotation

HW: To find the eigenvalues: $$det\begin{bmatrix} k_{11}^{(1)} -\lambda & k_{12}^{(1)} & k_{13}^{(1)} & k_{14}^{(1)} & 0 & 0\\ k_{21}^{(1)} & k_{22}^{(1)} - \lambda & k_{23}^{(1)} & k_{24}^{(1)} & 0 & 0\\ k_{31}^{(1)} & k_{32}^{(1)} & (k_{33}^{(1)}  k_{11}^{(2)}) - \lambda & (k_{34}^{(1)}   k_{12}^{(2)}) & k_{13}^{(2)} & k_{14}^{(2)}\\ k_{41}^{(1)} & k_{42}^{(1)} & (k_{43}^{(1)}  k_{21}^{(2)}) & (k_{44}^{(1)}   k_{22}^{(2)}) - \lambda  & k_{23}^{(2)} & k_{24}^{(2)}\\ 0 & 0 & k_{31}^{(2)} & k_{32}^{(2)} & k_{33}^{(2)} - \lambda & k_{34}^{(2)}\\ 0 & 0 & k_{41}^{(2)} & k_{42}^{(2)} & k_{43}^{(2)} & k_{44}^{(2)} - \lambda \end{bmatrix} = 0$$ This can be reduced to (because of the Principle of Virtual Work) $$det\begin{bmatrix} (k_{33}^{(1)} + k_{11}^{(2)}) - \lambda & (k_{34}^{(1)} + k_{12}^{(2)})\\ (k_{43}^{(1)} + k_{21}^{(2)}) & (k_{44}^{(1)} + k_{22}^{(2)}) - \lambda \end{bmatrix} = 0$$ Solving becomes $$[(k_{33}^{(1)} + k_{11}^{(2)}) - \lambda]*[(k_{44}^{(1)} + k_{22}^{(2)}) - \lambda] - [(k_{34}^{(1)} + k_{12}^{(2)})(k_{43}^{(1)} + k_{21}^{(2)})] = 0$$ But because K43=K34, this can be written as $$\lambda ^{2}- \lambda(k_{33}^{(1)} + k_{11}^{(2)} + k_{44}^{(1)} + k_{22}^{(2)}) + [(k_{33}^{(1)} + k_{11}^{(2)})(k_{44}^{(1)} + k_{22}^{(2)}) - (k_{34}^{(1)} + k_{12}^{(2)})^{2}] = 0 $$ Therefore, the eigenvalues are λ = 0,$$(k_{33}^{(1)} + k_{11}^{(2)})(k_{44}^{(1)} + k_{22}^{(2)}) - (k_{34}^{(1)} + k_{12}^{(2)})^{2}$$, and $$k_{33}^{(1)} + k_{11}^{(2)} + k_{44}^{(1)} + k_{22}^{(2)}$$ There is a zero eigenvalue because there is no rotational component of the system. This means that there is no elastic energy stored in the structure in the rotational mode. → Dynamic eigenvalue as realted to vibration frequency: K v =λ M  v, where K is the stiffness matrix, λ is the eigenvalue, and M is the mass matrix.

Meeting 15 (9/29/08)
Closing the Loop Between The FEM and statistics: virtual displacement



Two Bar Truss System:

From p. 11-2: Free body diagram (FBD) of the 2 force bodies of elements 1 and 2

=> By statics the reactions are unknown and therefore the member forces, $$ p_1^{(1)}, p_2^{(2)}$$, we can compute the axial displacement degrees of freedom or the amount of extension of the bars. This is on p.12-2: $$\displaystyle q_2^{(1)} = \frac{p_1^{(1)}}{k^{(1)}} = \frac{p_2^{(1)}}{k^{(1)}}$$, as shown in figure on p.11-2 $$\displaystyle q_2^{(1)} = A C$$ $$\displaystyle q_1^{(1)} = 0$$ since it is fixed on node 1 \displaystyle q_1^{(2)} = - \frac{p_2^{(2)}}{k^{(2)}} $$\displaystyle q_2^{(2)} = 0$$ since it is fixed on node 3

From the above results how can we obtain the displacement degrees of freedom of node 2?





Note:

Use the following method to solve:

$$\displaystyle u_y = R sin\alpha = R \alpha$$ ($$\alpha$$ small)

$$\displaystyle u_x =$$ axial displacement $$\displaystyle = R (1 - cos\alpha = 0$$  (first order)

Meeting 16 (10/1/08)
Closing the loop cont'd:

Look at an infinitesimal displacement (similar to the virtual displacement assosciated with the Principal of Virtual Work):



$$AC = \frac {|P_2^{(1)}|}{k^{(1)}} = \frac {|5.1243|}{0.75} = 6.8324$$

$$AB = \frac {|P_1^{(2)}|}{k^{(2)}}$$

Next, find the $$(x,y)$$ coordinates of points B and C:

$$(x_B, y_B) $$ $$(x_C, y_C) $$

Now there are 2 unknowns left, the x and y coordinates of point D, $$(x_D, y_D)$$. So, we need to find equations for line AB and line BC.

A approach to finding the desired information is given as follows:

$$\vec{PQ} = |\vec{PQ}| \hat{\tilde{i}} = PQ[cos\theta \hat{\tilde{i}} + sin\theta \hat{\tilde{j}}] $$

$$ = (x-x_p) \hat{\tilde{i}} + (y-y_p) \hat{\tilde{j}}$$

$$ => x-x_p = |PQ|cos\theta $$

and

$$ => y-y_p = |PQ|sin\theta $$

$$=> \frac{y-y_p}{x-x_p} = \tan\theta = >y-y_p = \tan\theta(x-x_p)$$

Thus, the equation for a line perpendicular to the above line and passing through point p would be given by:

$$y-y_p = \tan(\theta + \frac{\pi}{2})(x-x_p)$$

$$(x_D, y_D) = $$

Turning now to vector AD:

$$\vec{AD} = (x_D - x_A)\hat{i} + (y_D - y_A)\hat{j}$$

Since point A was chosen to be at the origin, we are left with:

$$\vec{AD} = (x_D)\hat{i} + (y_D)\hat{j}$$

By definition, we have:

$$\vec{AD} = d_3\hat{i} + d_4\hat{j}$$, where $$\vec{AD}$$ is the displacement vector of point A, and $$y_D = d_4$$ and $$x_D = d_3$$ from the finite element method (FEM).

Example: 3-Bar Truss System



$$\theta^{(1)} = 30 E^{(1)} = 2 A^{(1)} = 3 L^{(1)} = 5 $$

$$\theta^{(2)} = -30 E^{(2)} = 4 A^{(2)} = 1 L^{(2)} = 5 $$

$$\theta^{(3)} = 45? E^{(3)} = 3 A^{(3)} = 2 L^{(3)} = 10 $$

$$P = 30$$

Local node numbering should be carried out in such a manner that will make assembling the global stiffness matrix, K, convenient. A convenient local node numbering is demonstrated in the FBD diagram below:



17-1


The problem shown in the figure is a statically indeterminate structure. Though there are 3 equations (ΣFx, ΣFy, ΣMA) and 3 unknowns (R1, R2, and R3) this structure is still statically indeterminate because summing the moments about point A ends up becoming a useless equation. Since each member is a two-force body, it can bee seen through examination that each reaction force acts on a line that runs through point A. Thus there is no moment arm and no moment. The end result is an equation stating 0 = 0. This point can be proved statically as well.

HW: justiy using statics.

ΣMa=0

$$R1sin(\theta 1)*L1cos(\theta 1)+R2sin(\theta 2)*L2cos(\theta 2)-R3sin(\theta 3)*L3cos(\theta 3) +R3cos(\theta 3)*L3sin(\theta 3)-R2cos((\theta 2)*L2sin(\theta 2)-R1cos(\theta 1)*L1sin(\theta 1)=0$$

Note that every tern cancels and the end result is 0 = 0.

17-2


Question: What about summing the moments about any other point? Answer: The result it the same (yields 0 = 0). Proof:

ΣMB=(B*A x F) = (B*A' x F)

B*A' = B*A + A*A'

MB=(B*A+A*A') x (F) = (B*A x F) + (A*A' x F)

(A*A' x F) = 0 because it is along the same line of action

Thus MB=(B*A x F)

17-3
Global Stiffness Matrix $$\begin{bmatrix} K11 & K12 & K13 & K14 & K15 & K16 & K17 & K18\\ K21 & K22 & K23 & K24 & K25 & K26 & K27 & K28\\ K31 & K32 & K33 & K34 & K35 & K36 & K37 & K38\\ K41 & K42 & K43 & K44 & K45 & K46 & K47 & K48\\ K51 & K52 & K53 & K54 & K55 & K56 & K57 & K58\\ K61 & K62 & K63 & K64 & K65 & K66 & K67 & K68\\ K71 & K72 & K73 & K74 & K75 & K76 & K77 & K78\\ K81 & K82 & K83 & K84 & K85 & K86 & K87 & K88 \end{bmatrix}$$

Element Stiffness matrix (same for elements 1, 2, and 3) $$\begin{bmatrix} k11 & k12 & k13 & k14\\ k21 & k22 & k23 & k24\\ k31 & k32 & k33 & k34\\ k41 & k42 & k43 & k44 \end{bmatrix}$$

HW: Define Global Stiffness Matrix in terms of Element Stiffnesses.

K11=k(1)11 K12=k(1)12 K13=k(1)13 K14=k(1)14 K15=0 K16=0 K17=0 K21=k(1)21 K22=k(1)22 K23=k(1)23 K24=k(1)24 K25=0 K26=0 K27=0 K28=0 K31=k(1)31 K32=k(1)32 K33=k(1)33+k(2)11+k(3)11 K34=k(1)34+k(3)12+k(3)12 K35=k(2)13 K36=k(2)14 K37=k(3)13 K38=k(3)14 K41=k(1)41 K42=k(1)42 K43=k(1)43+k(2)21+k(3)21 K44=k(1)44+k(2)22+k(3)22 K45=k(2)23 K46=k(2)24 K47=k(3)23 K48=k(3)24 K51=0 K52=0 K53=k(2)31 K54=k(2)32 K55=k(2)33 K56=k(2)34 K57=0 K58=0 K61=0 K62=0 K63=k(2)41 K64=k(2)42 K65=k(2)43 K66=k(2)44 K67=0 K68=0 K71=0 K72=0 K73=k(3)31 K74=k(3)32 K75=0 K76=0 K77=k(3)33 K78=k(3)34 K81=0 K82=0 K83=k(3)41 K84=k(3)42 K85=0 K86=0 K87=k(3)43 K88=k(3)44