User:Eml4500.f08.gravy.mmm/HW3

12.1
Derivation of element Force Displacement with respect to global coordinate system.

$$\ k^{(e)}_{4x4} d^{(e)}_{4x1}=f^{(e)}_{4x1} \,$$ equation 1



12.2
The force displacement can be written as

$$ \ K^{(e)} * \begin{bmatrix} 1 & -1 \\ -1 &1 \end{bmatrix}_{\hat{K}_{2x2}^{(e)}}* \begin{Bmatrix} q_1^{(e)} \\ q_2^{(e)} \end{Bmatrix}_{q_{2x1}^{(e)}} = \begin{Bmatrix} p_1^{(e)} \\ p_2^{(e)} \end{Bmatrix}_{p_{2x1}^{(e)}} \,$$ equation 2

$$\ q_i^{(e)} = \,$$ axial disp.of element at local node i

$$\ p_i^{(e)} = \,$$ axial disp.of element at local node i

Goal: Derive Equation 1 from Equation 2

Want the relationship between $$\underline {q^{(e)}}$$ and $$\underline {d^{(e)}}$$ also $$\underline {p^{(e)}}$$ and $$\underline {f^{(e)}}$$

12.3
These relationships can be expressed in the form:

$$\ \underline {q^{(e)}_{2x1}}= \underline { T^{(e)}_{2x4}} * \underline {d^{(e)}_{4x1}} \,$$

Consider the displacement vector of local node 1 denoted by

$$ \vec{d^{(e)}_{1}} $$



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

$$\ q_i^{(e)} = \,$$ axial displacement of node 1 is the orthogonal projection of the displacement.

12.4
Vector $$ \vec{d^{(e)}_{1}} $$ of node 1 on the axis $$\tilde{x}$$ of element $$e^{(1)}$$

$$\ \vec{d^{(e)}_{1}}* \vec{\tilde{i}} \,$$

$$\ = ( d_1^{(e)}*\vec{i} + d_2^{(e)}*\vec{j}) \cdot \vec{\tilde{i}} \,$$

$$\ = d_1^{(e)}* (\vec{i} \cdot \vec{\tilde{i}}) + d_2^{(e)}*(\vec{j} \cdot \vec{\tilde{i}} ) \,$$

$$\ (\vec{i} \cdot \vec{\tilde{i}} )= cos \theta^{(e)} = l^{(e)} \,$$

$$\ (\vec{j} \cdot \vec{\tilde{i}} )=sin \theta^{(e)} = m^{(e)} \,$$

$$\ q_1^{(e)}=l^{(e)}*d_1^{(e)} + m^{(e)}* d_2^{(e)} \,$$

$$\ = \begin{bmatrix} l^{(e)} & m^{(e)} \end{bmatrix}_{1x2} * \begin{Bmatrix} d_1^{(e)} \\ d_2^{(e)} \end{Bmatrix}_{2x1} \,$$

12.5
Similarly for node 2 : HW.

$$\ q_2^{(e)}=l^{(e)}*d_1^{(e)} + m^{(e)}* d_2^{(e)}= \begin{bmatrix} l^{(e)} & m^{(e)} \end{bmatrix}_{1x2} * \begin{Bmatrix} d_3^{(e)} \\ d_4^{(e)} \end{Bmatrix} \,$$

$$ \begin{Bmatrix}q_1^{(e)} \\ q_2^{(e)} \end{Bmatrix}_{\underline{q_{2x1}^{(e)}}} = \begin{bmatrix} l^{(e)} & m^{(e)} & 0& 0\\0 & 0 & l^{(e)} & m^{(e)} \end{bmatrix}_{\underline{T_{2x4}^{(e)}}}* \begin{Bmatrix} d_1^{(e)} \\ d_2^{(e)} \\d_3^{(e)} \\ d_4^{(e)} \end{Bmatrix}_{\underline{d_{4x1}^{(e)}}}$$

$$\ \underline{ q_{2x1}^{(e)} }= \underline{ T_{2x4}^{(e)} }* \underline{ d_{4x1}^{(e)} } \,$$

14.1
Similarly (same argument) :

$$\ \begin{Bmatrix} P_1^{(e)}\\P_2^{(e)} \end{Bmatrix}_{\underline{ P_{2x1}^{(e)} } } = \underline{ T_{2x4}^{(e)} } * \begin{Bmatrix} f_1^{(e)} \\ f_2^{(e)} \\ f_3^{(e)} \\ f_4^{(e)} \end{Bmatrix} _{\underline{ f_{4x1}^{(e)} } } \,$$

$$\underline{ P^{(e)} } = \underline{ T^{(e)}* f^{(e)} }$$

Recall element axial Force Displacement relation:

$$\ \underline{\hat{k}_{2x2}^{(e)}} * \underline{q_{2x1}^{(e)}}= \underline{P_{2x1}^{(e)}} \,$$

14.2
$$\underline{\hat{k}^{(e)}} *( \underline{ T^{(e)} } * \underline{ d^{(e)} })_{\underline{q^{(e)}}} = ( \underline{ T^{(e)} } * \underline{ f^{(e)} })_{\underline{P^{(e)}}}$$

Goal: Want to have $$\ \underline{k^{(e)}} * \underline{d^{(e)}}= \underline{f^{(e)}} \,$$

So "move " $$\underline{ T^{(e)}}$$ from the right hand side to the left hand side by multiplying the equation by $$ \underline{ T^{(e) -1}}$$ (inverse of $$\underline{ T^{(e)}} $$)

Unfortunately, $$\underline{ T^{(e)}}$$ is a rectangular matrix (2x4) so you can not invert it.

Ans: $$[\underline{ T_{4x2}^{(e)T} } * \underline{\hat{k}_{2x2}^{(e)}} * \underline{ T_{2x4}^{(e)}} ]_{is the matrix we are looking for} * d_{4x1}^{(e)}= f_{4x1}^{(e)}$$

14.3
$$\ \underline{k^{(e)} } * \underline{d^{(e)} }= \underline{f^{(e)} } \,$$

1) $$\ \underline{k^{(e)} } = \underline{T^{(e)T}} * \underline{\hat{k}^{(e)} }* \underline{T^{(e)} } \,$$

HW: verify above.

Justification for 1) : PVW (later see 10-1 for fist approach of PVW Reduction of Global Force Displacement relation)

$$ \underline{K_{6x6}}* \underline{d_{6x1}}= \underline{F_{6x1}} which reduces to \underline{K_{2x2}}* \underline{d_{2x2}}= \underline{F_{2x2}}$$

Remark: Why not solve as follows

$$ \underline{d} = \underline{K^{-1}}* \underline{F} $$ ?

The reason is that K can't be unversed because it is singular

14.4
Singular matrix has a determinate of 0 so it is not invertible

Recall: need to compute $$\frac{1}{det(K)}$$ to find K-1

Why? For an unconstrained structure system there are three possible rigid body motion in 2-D (2 translation, and 1 rotation)

HW: compute eigenvalues values of K6x6 and make observation about the number of zero eigenvalues

Dyn. eval Problem : $$\underline{K}_{stiff. mat} *v= \lambda_{related to vibration frequency}*M_{mass mat. }*v$$

14.5
Zeros corresponds to zero stored elastic energy or rigid body modes. modes=mode shapes=eigenvectors

//(end Remark)

17.1


$$ \sum{F_x} =0$$

$$ \sum{F_y} =0$$

$$ \sum{M_A} =0$$ trivial

Question: How about $$\sum{M_B}?$$ It still equals 0



$$ \sum{M_B} = \vec{BA'}\times \vec{F}$$

for A' on line of action of $$\vec{F}$$

$$\vec{BA'}= \vec{BA} + \vec{AA'}$$

17.2
$$ \sum{M_B}=(\vec{BA}+ \vec{AA'})\times \vec{F}$$ $$ = \vec{BA}\times F + \vec{AA'}\times \vec{F}= 0 $$

Back to 3-bar truss

Node A is in equilibrium : $$\sum_{i=0}{\vec{F_i}} =\vec{0}$$

$$ \sum_i{M_{B,i}}=\sum_i{\vec{BA'_i}}\times \vec{F_i}$$

A'i= any point on line of action of $$\vec{F_i}$$

$$ \sum_i{M_{B,i}}=\sum_i{\vec{BA}}\times \vec{F_i} \vec{BA} \times \sum_i{\vec{F_i}} = 0 $$



Where $$K_{33}= k^{(1)}_{33}+ k^{(2)}_{11}+k^{(3)}_{11} $$ and

$$K_{34}= k^{(1)}_{34}+ k^{(2)}_{13}+k^{(3)}_{12} $$

HW: fill out the rest

hw
From previous lecture

Element Length

$$\ L^{(1)}=4 \,$$

$$\ L^{(2)}=2 \,$$

Young's Modules

$$\ E^{(1)}=3 \,$$

$$\ E^{(2)}=5 \,$$

Cross-sectional Area

$$\ A^{(1)}=1 \,$$

$$\ A^{(2)}=2 \,$$

Inclination angle


 * $$\ \theta^{(1)} = 30^o \,$$


 * $$\ \theta^{(2)} = -45^o \,$$


 * $$\ \theta^{(1)} = 30^o \,$$

Since $$ \ P_{(e)} = T_{(e)} * f_{(e)}\,$$

And the T matrix is

$$\ T^{(e)} = \begin{bmatrix} l^{(e)} & m^{(e)}& 0 & 0 \\ 0 & 0 & l^{(e)} & m^{(e)}\end{bmatrix} \,$$

we can solve for P

The f matrix was solved in a previous lecture and is:

$$\ F= \begin{bmatrix} -4.4378 \\ -2.5622 \\ +4.4378 \\ +2.5622 \end{bmatrix}_{4X1}\,$$

l m and k were also solved in a previous lecture and are:

$$\ l^{(1)}=cos\theta^{(1)}= cos 30^o= \frac{\sqrt{3}}{2} \,$$

$$\ m^{(1)}=sin\theta^{(1)}= sin 30^o= \frac{1}{2} \,$$

$$\ k^{(1)}= \frac{E^{(1)}A^1}{L^1} \frac{\left(3\right ) \left(1\right )}{4}= \frac {3}{4}\,$$

$$\ l^{(2)}=cos\theta^{(2)}= cos -45^o= .707 \,$$

$$\ m^{(2)}=sin\theta^{(2)}= sin -45^o= -.707 \,$$

$$\ k^{(1)}= \frac{E^{(5)}A^1}{L^1} \frac{\left(2\right ) \left(1\right )}{2}= 5\,$$

$$T^{(1)}$$ and $$T^{(2)}$$ are solved as the following:

$$\ T^{(1)} = \begin{bmatrix} \frac{1}{2} & \frac{1}{2} & 0 & 0 \\ 0 & 0 & \frac{1}{2} & \frac{1}{2}\end{bmatrix} \,$$

$$\ T^{(2)} = \begin{bmatrix} .707 & -.707 & 0 & 0 \\ 0 & 0 & .707 & -.707\end{bmatrix} \,$$

With T and f the P s were found as being,

$$P_{2}^{(1)}$$= $$P_{1}^{(2)}$$=5.1243

so