User:EML4500.f08.Ateam.Miller/homework4

Justification of Assembly of Elemental Stiffness Matrices into the Global Stiffness Matrix
The free body diagram at Node 2 yields the following force balance equations.

$$f_3^{(1)}+f_1^{(2)}=0\rightarrow (1)$$

$$f_4^{(1)}+f_2^{(2)}=\mathbf{P}\rightarrow (2)$$

Next, the elemental Force-Displacement relation,

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

can be used to rewrite Equation (1) as follows.

$$(1)\rightarrow [k_{31}^{(1)}d_1^{(1)}+k_{32}^{(1)}d_2^{(1)}+k_{33}^{(1)}d_3^{(1)}+k_{34}^{(1)}d_4^{(1)}]+ [k_{11}^{(2)}d_1^{(2)}+k_{12}^{(2)}d_2^{(2)}+k_{13}^{(1)}d_3^{(2)}+k_{14}^{(1)}d_4^{(2)}]$$

Similarly, Equation (2) can be written as follows.

$$(2)\rightarrow [k_{41}^{(1)}d_1^{(1)}+k_{42}^{(1)}d_2^{(1)}+k_{43}^{(1)}d_3^{(1)}+k_{44}^{(1)}d_4^{(1)}]+ [k_{21}^{(2)}d_1^{(2)}+k_{22}^{(2)}d_2^{(2)}+k_{23}^{(1)}d_3^{(2)}+k_{24}^{(1)}d_4^{(2)}]$$

Now, Equations (1) and (2) can be transformed into the global coordinate system as follows.

$$(1)\rightarrow [k_{31}^{(1)}d_1+k_{32}^{(1)}d_2+k_{33}^{(1)}d_3+k_{34}^{(1)}d_4]+ [k_{11}^{(2)}d_3+k_{12}^{(2)}d_4+k_{13}^{(1)}d_5+k_{14}^{(1)}d_6]$$

$$(2)\rightarrow [k_{41}^{(1)}d_1+k_{42}^{(1)}d_2+k_{43}^{(1)}d_3+k_{44}^{(1)}d_4]+ [k_{21}^{(2)}d_3+k_{22}^{(2)}d_4+k_{23}^{(1)}d_5+k_{24}^{(1)}d_6]$$

Now, the elemental stiffness matrices, k(e), can be assembled into the global stiffness matrix, K, through

$$\mathbf{K}=\mathbf{A}\mathbf{k}$$

where K is an n x n matrix, A is nel x 1 matrix and k is an ned x ned matrix. Here, n is the total number of degrees of freedom, nel is the total number of elements, ned is the total number of elemental degrees of freedom, and A is the assembly operator.

As we continue to justify the assembly of the elemental stiffness matrices into the global stiffness matrix, we must use the Principal of Virtual Work, as described previously, to obtain $$\mathbf{\overline{K}}$$ by eliminating the corresponding boundary conditions.

Once the Principal of Vitual Work is applied, the resulting elemental axial force equation and the elemental elemental stiffness equation can be written as follows.

$$\mathbf{q^{(e)}}=\mathbf{T^{(e)}}\mathbf{d^{(e)}}$$

$$\mathbf{k^{(e)}}=\mathbf{T^{(e)T}}\mathbf{\hat{k}^{(e)}}\mathbf{T^{(e)}}$$

Now, it is important to derive the Finite Element Method for Partial Differential Equations. Consider the Force-Displacement relation for a bar or spring with

$$kd=F$$

This implies that

$$kd-F=0\rightarrow (3)$$

or equivalently

$$W(kd-F)=0\rightarrow (4)$$

for all W.

Upon investigation, $$(3)\Rightarrow(4)$$ is trivial, but $$(4)\Rightarrow(3)$$ is not trivial.

Since (4) is valid for all W, select W=1. Now, (4) becomes $$1(kd-F)=0\Rightarrow(3)$$.