User:Eml4500c.f08.gravy.mmm/HW2

FEA Notes and Homework 2 

Truss Problem description


Data:

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


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


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

Solving a Truss Problem with Statics


First take the sum of the moments about global node 3 and solve for $$R_3$$ which is the reaction at global node 3 in the y-direction

$$ \sum {M_1}= P(4cos(30)) +R_3(4cos(30) + 2sin(45))$$

$$R_3= \frac{-P(4cos(30))} {4cos(30) + 2cos(45)} = -4.97071$$

Then use this information to take the sum of the forces in the y-direction and solve for $$R_1$$ which is the reaction at global node 1 in the y-direction.

$$\sum{F_y}= -4.97071 + P+R_1$$

$$\ R_1= -2.02929 \,$$

By observation it can be determined that there are no forces in the x-direction so $$R_1$$ and $$R_3$$ are the only reaction forces

Solving a Truss Problem with Finite Element Analysis
1)Global Picture



$$\begin{Bmatrix} f_1 \\\vdots & \\ f_6 \end{Bmatrix}_{1X6}=\begin{bmatrix} & &  \\  & k & \\ & &   \end{bmatrix}_{6X6}*    \begin{Bmatrix} d_1 \\\vdots & \\ d_6 \end{Bmatrix}_{1X6}$$

$$\begin{matrix} global& &global& &global \\ force & &stiffness& &displacement \\  column &   &matrix  & &column \\ matrix & & & &matrix \end{matrix}$$

2) Element Picture

3) Global FD at element level:

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

$$\ k^{(e)}\,$$ = the element stiffness matrix

$$\ d^{(e)}\,$$= the element displacement matrix

$$\ f^{(e)}\,$$= the element force matrix

$$k^{(e)}\begin{bmatrix} (l^{(e)})^2 & l^{(e)}m^{(e)} & -(l^{(e)})^2 & -(l^{(e)}m^{(e)})  \\  l^{(e)}m^{(e)} & (m^{(e)})^2 & -(l^{(e)}m^{(e)}) & -(m^{(e)})^2  \\ -(l^{(e)})^2 & -(l^{(e)}m^{(e)}) & (l^{(e)})^2 & l^{(e)}m^{(e)}\\ -(l^{(e)}m^{(e)}) & -(m^{(e)})^2 & l^{(e)}m^{(e)} & (m^{(e)})^2    \end{bmatrix}_{4X4} = \underline{k^{(e)}}$$

$$\ k^{(e)}= \,$$ $$\frac{E^{(e)}A^{(e)}}{L^{(e)}}$$ axial stiffness of bar element "e" where e=1,2

$$\ l^{(e)},m^{(e)} = \,$$ director cosines of x axis (goes from 1 to 2) with respect to global (x,y coordinates)



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

$$\ m^{(e)}= \vec i \cdot \vec j= cos\left (\frac{\pi}{2}- \theta^{(e)} \right )= sin\left (\theta^{(e)} \right )\,$$

$$\ \vec \tilde{i}=cos\left(\theta^{(e)} \right)\vec i + sin\left( \theta^{(e)}  \right )\vec j \,$$

$$\ \vec \tilde{i} \cdot \vec i = \vec \tilde{i} \left( cos\left(\theta^{(e)} \right)\vec i + sin\left( \theta^{(e)}  \right )\vec j \right) i= cos\left(\theta^{(e)}  \right) \vec i \vec i + sin\left( \theta^{(e)}  \right ) \vec i  \,$$

$$\ \vec \tilde{i} \cdot \vec j= HW \,$$

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

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

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

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

$$\ \underline{k^{(1)}}=\begin{bmatrix} k_{11}^{(1)} & k_{12}^{(1)} & k_{13}^{(1)} & k_{14}^{(1)}  \\  k_{21}^{(1)} & k_{22}^{(1)} & k_{23}^{(1)} & k_{24}^{(1)}  \\ k_{31}^{(1)} & k_{32}^{(1)} & k_{33}^{(1)} & k_{34}^{(1)}\\ k_{41}^{(1)} & k_{42}^{(1)} & k_{43}^{(1)} & k_{44}^{(1)}    \end{bmatrix}_{4X4}= \left [k_{ij}^1 \right ]_{4X4} \,$$

i=row number =1,2....4

j=column number =1,2....4

$$\ k_{11}^{(1)}= k^{(1)}\left( l^{(1)} \right)^2=\frac{3}{4}*\frac{\sqrt{3}}{2}= \frac{9}{16} \,$$

$$\ k_{12}^{(1)}= k^{(1)}\left( l^{(1)}m^{(1)} \right)=\frac{3}{4}*\frac{1}{2}= \frac{3\sqrt{3}}{16} \,$$

$$\ k_{42}^{(1)}=k_{24}^{(1)}= -k^{(1)}\left( m^{(1)} \right)^2=\frac{-3}{4}*\frac{1}{2}= \frac{-3}{16} \,$$

Observation:

1) Only need to compute 3 numbers. Other coefficients have same absolute value, just differ by (+ -)

2) Matrix $$ \underline{k^{(1)}}$$ is synmetrix, $$k_{ij}^{(1)}=k_{ji}^{(1)}$$

$$k_{13}^{(e)}=k_{31}^{(e)}$$ (just interchange the row and column index)

In general, $$k_{ij}^{(e)}=k_{ji}^{(e)} or k^{(e) T} =k^{(e)}$$

The transpose of $$ k^{(e)}$$ is equal to $$ k^{(e)}$$

$$\ \underline{k^{1}}=\begin{bmatrix} k_{11}^{(e)} & k_{12}^{(e)} & k_{13}^{(e)} & k_{14}^{(e)}  \\   & k_{22}^{(e)} & k_{23}^{(e)} & k_{24}^{(e)}  \\  &  & k_{33}^{(e)} & k_{34}^{(e)}\\ sym. & &  & k_{44}^{(e)}    \end{bmatrix}_{4X4}= \left [k_{ij}^1 \right ]_{4X4} \,$$

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

$$\ \theta^{(2)} = \frac{-\pi}{4}= -45^o \,$$

$$\ l^{(2)}=cos\theta^{(2)}= cos (\frac{-\pi}{4})= \frac{\sqrt{2}}{2} \,$$

$$\ m^{(2)}=sin\theta^{(2)}= sin (\frac{-\pi}{4})= \frac{-\sqrt{2}}{2} \,$$