User:Eml4500.f08.team.figuerrez/hw2

=Model 2-bar truss system=

 Element 1:
Inclination angle $$\theta^{(1)} = 30 \deg$$

Element Length $$L^{(1)} = 4$$

Cross-sectional area $$A^{(1)} = 1$$

therefore the computed value for l(1) and m(1) are as follows:

$$l^{(1)} = cos \theta^{(1)}$$ $$l^{(1)} = cos {(30)}$$ $$l^{(1)} = 0.866$$

$$m^{(1)} = sin \theta^{(1)}$$ $$m^{(1)} = sin {(30)}$$ $$m^{(1)} = 0.5$$

Once the values for l(1) and m(1) are computed, the stiffness matrix for element 1 can be determined:

$$K^{(1)}_{11} = [K^{(1)}(l^{(1)})^{2}] = [\frac {3}{4})(\frac{\sqrt{3}}{2})^{2}] = (\frac {9}{16}) = 0.5625$$

$$K^{(1)}_{12} = [K^{(1)}(l^{(1)})*m^{(1)}] = [\frac {3\sqrt{3}}{16} = 0.325$$

$$K^{(1)}_{42} = -[K^{(1)}(m^{(1)})^{2}] = \frac {-3}{16} = -0.1875$$

Obs:


 * 1) You only need to compute 3 numbers. The other coefficients have the same absolute value; they just differ by +/-.
 * 2) Matrix K(1) is symmetric, i.e.,

$$K^{(1)}_{ij} = K^{(1)}_{ji}$$ $$K^{(1)}_{13} = K^{(1)}_{31}$$

$$ K^{(e)} = \begin{bmatrix} K^{(e)}_{11} & K^{(e)}_{12} & K^{(e)}_{13} & K^{(e)}_{14} \\

& K^{(e)}_{22} & K^{(e)}_{23} & K^{(e)}_{24} \\

& & K^{(e)}_{33} & K^{(e)}_{34} \\

& & & K^{(e)}_{44} \\ \end{bmatrix} $$

Here, the upper right "triangle" portion of the matrix has to be computed. The remaining values are symmetric and differ by +/-.

Element 2:
$$K^{(2)} = (\frac{E^{(2)}*A^{(2)}}{L^{(2)}}) = (\frac{5*2}{2}) = 5$$

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

$$K^{(2)} = [K^{(2)}_{ij}]_{(4x4)}$$

$$K^{(2)}_{11} = K^{(2)}(l^{(2)})^{2} = 5 * (\frac {\sqrt{2}}{2})^{2} = 2.5$$

Obj:

$$K^{(2)}_{ij}, e=2, (i,j) = 1,....,4$$ <p style="text-align:center;">are all the same: comp. 1 coefficient e <p style="text-align:center;">For other coefficients, add + or -
 * Absolute value of all coefficients:


 * K11(2)T = K(2), i.e. (K(2) is symmetric)

<p style="text-align:center;">$$K^{(e)}d^{(e)} = f$$ <p style="text-align:center;">e = 1,2
 * Element FD relationship:

<p style="text-align:center;">$$ d^{(e)} = \begin{bmatrix} d^{(e)}_{1} \\

d^{(e)}_{2} \\

d^{(e)}_{3} \\

d^{(e)}_{4} \\ \end{bmatrix}_{(4x1)} $$

<p style="text-align:center;">and

<p style="text-align:center;">$$ f^{(e)} = \begin{bmatrix} f^{(e)}_{1} \\

f^{(e)}_{2} \\

f^{(e)}_{3} \\

f^{(e)}_{4} \\ \end{bmatrix}_{(4x1)} $$

<p style="text-align:center;">$$K_{(nxn)} * d_{(nx1)} = f_{(nx1)}$$ <p style="text-align:center;">Here n = 6
 * Global FD relationship: