User:Eml4500.f08.Ateam.Slevinski/Homework2

Class Notes - 9/10

Element 1: From the FBD of element 1, we see that the angle from the x-axis is 30°.

θ(1) = 30ο l(1) = cosθ(1) = cos30ο = $$\frac{\sqrt3}{2}$$ m(1) = sinθ(1) = sin30° = $$\frac{1}{2}$$

For element 1, the stiffness is given as: $$K^{(1)} = \frac{E^{(1)}A^{(1)}}{L^{(1)}}$$ = $$\frac{3}{4}$$

And the matrix of k(1) is given as:

$$ \underline K^{(1)}= K*\begin{bmatrix} k^{(1)}(l^{(1)})^2 & k^{(1)}l^{(1)}m^{(1)} & -k^{(1)}(l^{(1)})^2 & -k^{(1)}l^{(1)}m^{(1)} \\ k^{(1)}l^{(1)}m^{(1)} & k^{(1)}(m^{(1)})^2 & -k^{(1)}l^{(1)}m^{(1)} & -k^{(1)}(m^{(1)})^2 \\ -k^{(1)}(l^{(1)})^2 & -k^{(1)}l^{(1)}m^{(1)} & k^{(1)}(l^{(1)})^2 & k^{(1)}l^{(1)}m^{(1)} \\ -k^{(1)}l^{(1)}m^{(1)} & -k^{(1)}(m^{(1)})^2 & k^{(1)}l^{(1)}m^{(1)} & k^{(1)}(m^{(1)})^2 \end{bmatrix} $$

In this matrix, only three values need to be calulated. The rest only differ by being positive or negative. As you can see this matrix is symetric. Therefore,

 kij(1) = kji(1)

And in general:

 kij(e) = kji(e) and  k(e)T = k(e)

Element 2:

From the FBD of element 2, we see that the angle from the x-axis is -45°.

θ(2) = -45°

l(2) = cosθ(2) = cos(-45°) = $$\frac{\sqrt2}{2}$$

m(2) = sinθ(2) = sin(-45°) = $$\frac{-\sqrt2}{2}$$

For element 2, the stiffness is given as: $$K^{(2)} = \frac{E^{(2)}A^{(2)}}{L^{(2)}}$$ = 2.5

And the matrix of k(2) is given as:

$$ \underline K^{(2)}= K*\begin{bmatrix} k^{(2)}(l^{(2)})^2 & k^{(2)}l^{(2)}m^{(2)} & -k^{(2)}(l^{(2)})^2 & -k^{(2)}l^{(2)}m^{(2)} \\ k^{(2)}l^{(2)}m^{(2)} & k^{(2)}(m^{(2)})^2 & -k^{(2)}l^{(2)}m^{(2)} & -k^{(2)}(m^{(2)})^2 \\ -k^{(2)}(l^{(2)})^2 & -k^{(2)}l^{(2)}m^{(2)} & k^{(2)}(l^{(2)})^2 & k^{(2)}l^{(2)}m^{(2)} \\ -k^{(2)}l^{(2)}m^{(2)} & -k^{(2)}(m^{(2)})^2 & k^{(2)}l^{(2)}m^{(2)} & k^{(2)}(m^{(2)})^2 \end{bmatrix} $$