User:Eml4500.f08.ateam.shah/Week 7

FEA Nov 21, 2008
$$ [\epsilon]=\frac{[du]}{[dx]}=\frac{[L]}{[L]}=1 $$

$$ [\sigma]=[\epsilon]=\frac{F}{L^2} $$

$$ [\frac{EA}{L}]=[k_{11}]=\frac{(\frac{F}{L^2}{L^2})}{L}=\frac{F}{L} $$

$$ [k_{11}d_1]=[k_{11}][d_1]=F $$

$$ [k_{23}d_3]=[k_{23}][d_3]=[\frac{[6][E][I]}{L^2]}]=\frac{((I)\frac{F}{L^2}(L^4))}{L^2}=F $$

FD relationship is derived in global coordinate system from the element force displacement relationship which is in the local coordinate system. That is, obtain

k (e)6x6 d (e)6x1= f (e)6x1 k (e)6x6= T (e)T6x6 k (e)6x6 T (e)6x6

from k (e)6x6 d (e)6x1= f (e)6x1



\begin{bmatrix} d_1\\ d_2\\ d_3\\ d_4\\ d_5\\ d_6\\ \end{bmatrix} =    \begin{bmatrix} R & R & 0 & 0 & 0 & 0\\ R & R & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0\\     0 & 0 & 0 & R & R & 0\\ 0 & 0 & 0 & R & R & 0\\ 0 & 0 & 0 & 0 & 0 & 1\\    \end{bmatrix} \begin{bmatrix} d_1\\ d_2\\ d_3\\ d_4\\ d_5\\ d_6\\ \end{bmatrix} $$

Where d is a 6x1 matrix and, the transformation matrix T is a 6x6 matrix.

Plot the deformed shape of the figure above. Derivation of k, using the Principal of Virtual Work (PVW), foucing only on bending effects leads to

$$ \frac{d^2}{dx^2}(EI\frac{d^2v}{dx^2})-f_t (x)=m(x)\ddot{v} $$

Dec 5, 2008
Note: Equation 2 from the previous lecture P 39-2 was used for dimensional analysis.

[u]=L
 * P 31-4 [N1]=[N4]=1

Using Equation 2 determine that [N1][d1]=[N4][d4]

[v]=L
 * [N2][d2]=L

Thus


 * [N3][d3]=L

Derivation of the beam shaped function N2, N3, N5, and N6 have been previously determined in Lecture 37-4, without the use of ft (distributed transmitted load) and the inertia force.


 * $$ m \ddot{v} (static) =\frac{d^2}{dx^2}(EI\frac{d^2x}{dx^2})$$

Further consider constant values for E & I, the equation simplifies down to $$\frac{d^4V}{dx^4}=0$$

When the above equation is integrated four times, four constant values are determined to be C0, C1, C2, and C4.

$$V(x)=C_0+C_1x^1+C_2x^2+C_3x^3$$

To obtain N2(x), use $$ V(0)=1 $$ $$ V(L)=0 $$

$$ \frac{dV(0)}{dx}=0 $$

$$ \frac{dV(L)}{dx}=0 $$

Use the boundary conditions listed above to solve for C0 and C3.

$$ V(0)=1=C_0 $$

$$ V(L)=1+C_1L+C_2L^2+C_3L^3 $$

$$ \frac{dV(x)}{dx}=C_1+2C_2x+3C_3x^2 $$

$$ \frac{dV(0)}{dx}=C_1=0 $$

$$ \frac{dV(L)}{dx}=2C_2L+3C_3L^2=0 $$

$$ C_3=\frac{-2C_2}{3L} $$

$$ 0=1+C_2L^2+\frac{-2C_2L^2}{3} $$

$$ C_2=\frac{-3}{L^2} $$

$$ C_3=\frac{2}{L^3} $$

For N3: Boundary Conditions are

$$V(0)=V(L)=0$$ $$\frac{dV(0)}{dx}=0$$ $$\frac{dV(L)}{dx}=0$$

For N5: Boundary Conditions are

$$V(0)=0$$ $$V(L)=1$$ $$\frac{dV(0)}{dx}=0$$ $$\frac{dV(L)}{dx}=0$$

For N6: Boundary Conditions are

$$V(0)=0$$ $$V(L)=0$$ $$\frac{dV(0)}{dx}=0$$ $$\frac{dV(L)}{dx}=L$$

See Lecture 39-1 for the plots of N5 and N6

Derivation of the coefficients of the K (element stiffness matrix) coefficients with EA: Done coefficients with EI: To be Done $$k_{22}=\frac{12EI}{L^2}=\int_{0}^{L}\frac{d^2N_2}{dx^2}(EI)\frac{d^2N_2}{dx^2}dx$$