User:Eml4500.f08.team.guan/hw2

7.3-7.5

Element Stiffness Matrices
In order to analyze the two different elements of the two-bar truss system, their individual stiffness matrices must be calculated. The notation for the elemental stiffness matrix can be denotes with an underlined, lower case "k" with a superscript "e" where "e" equals either 1 or 2 corresponding to its respective element. It can also be represented by an actual four by four matrix or an underlined k with a superscript "e" and a subscript "ij" uncased in brackets as seen below. "e" again denotes the element number, "i" denotes the row number, and "j" denotes the column number. Here in this example, "i" and "j" range from 1 to 4. $$\displaystyle k^{(e)}_{matrix} $$ = $$\begin{bmatrix} k_{11} & k_{12} &k_{13} &k_{14} \\ k_{21} & k_{22} &k_{23} &k_{24} \\ k_{31} & k_{32} &k_{33} &k_{34} \\ k_{41} & k_{42} &k_{43} &k_{44} \\ \end{bmatrix} $$ =$$\begin{bmatrix} k^{(e)}_{ij} \\ \end{bmatrix} $$ However, an important aspect of the stiffness matrix is noticed using a more specific representation of this four b four matrix in this problem. From previous notes, the stiffness matrix is seen below. $$\displaystyle k^{(e)}_{matrix} $$ = $$k^{(e)}\begin{bmatrix} (l^{(e)})^2 & l^{(e)}m^{(e)} & -(l^{(e)})^2 & -l^{(e)}m^{(e)} \\ l^{(e)}m^{(e)} & (m^{(e)})^2 &  -l^{(e)}m^{(e)} & -(m^{(e)})^2 \\ -(l^{(e)})^2 & -l^{(e)}m^{(e)} &  (l^{(e)})^2 & l^{(e)}m^{(e)} \\ -l^{(e)}m^{(e)} & -(m^{(e)})^2 & l^{(e)}m^{(e)} & (m^{(e)})^2 \\ \end{bmatrix} $$ As mentioned before, l and m and their element number superscripts are the director cosines of the inclined local axis (x_tilda) from local node 1 to local node 2 with respect to the global (x,y) coordinate system. However, more importantly is that this element stiffness matrix representation shows that the matrix is symmetrical about the diagonal ($$k^{(1e)}_{ij}=k^{(1e)}_{ji}$$). This also means that its transpose is equal to itself. This characteristic of the stiffness matrix allows for you to check your work upon calculation. If the matrix does not turn out to be symmetrical, there may be an error in your calculations. Now it is necessary to calculate the values of the stiffness matrix. Let us examine the element stiffness matrix of element one. Recall that element one has an elastic modulus (E) of 3, length (L) of 4, a cross sectional area (A) of 1, and an inclination angle (theta) of 30 degrees. A picture of element one is below in the collapsible box to better recall the characteristics.  $$l^{(1)}=cos(\theta^{(1)})=cos(30), m^{(1)}=sin(\theta^{(1)})=sin(30), k^{(1)}=\frac{E^{(1)}A^{(1)}}{L^{(1)}}$$ $$k^{(1)}_{11}=k^{(1)}(l^{(1)})^2=\frac{3}{4}(\frac{\sqrt{3}}{2})^2=\frac{9}{16} $$ $$k^{(1)}_{12}=k^{(1)}l^{(1)}m^{(1)}=\frac{3}{4}(\frac{3}{2})\frac{1}{2}=\frac{3\sqrt{3}}{17}$$ $$k^{(1)}_{42}=-k^{(1)}(m^{(1)})^2=\frac{3}{4}(\frac{1}{2})^2=\frac{-3}{16}$$ One can observe that only three values of the matrix need to be calculate because all of the values in the matrix have one of these three values' magnitudes. Noticing this can save an ample amount of computation time. Similarly, element stiffness matrix of element two can be calculated. Recalling the properties of element two, it has an elastic modulus (E) of 5, length (L) of 2, a cross sectional area (A) of 2, and an inclination angle (theta) of -45 degrees. A picture of element two is below in the collapsible box to better recall the characteristics. These images are not drawn to scale.

 $$l^{(2)}=cos(\theta^{(2)})=cos(-45), m^{(2)}=sin(\theta^{(2)})=sin(-45), k^{(2)}=\frac{E^{(2)}A^{(2)}}{L^{(2)}}$$ $$k^{(2)}_{11}=k^{(2)}(l^{(2)})^2=(5)\frac{2}{4}=\frac{5}{2} $$ Now, one can observe for this element stiffness matrix that the absolute values of all coefficients $$k^{(2)}_{ij}$$ are the same. Therefore, you can compute one coefficient to fill in the entire matrix. This is due to the director cosines of this element being equal with an inclination angle of -45 degrees. These element stiffness matrices all only have four displacement degrees of freedom thus they are represented by a four by four matrix. However, the global stiffness matrix has a total of six displacement degrees of freedom.