User:Eml4500.f08.Ateam.Slevinski/Homework7

Lecture Notes - 11/17/08
Element 1: $$E_1^{(1)} = 2$$  $$ E_2^{(1)} = 4$$ $$A_1^{(1)} = 0.5$$ $$ A_2^{(1)} = 1.5$$

Element 2: $$E_1^{(1)} = 2$$  $$ E_2^{(1)} = 4$$ $$A_1^{(1)} = 0.5$$ $$ A_2^{(1)} = 1.5$$

Model Frame With Two Elements:

In the prevoius picture, the frame has a rigid connection between the two elements. This means that when the frame is deformed, there is no change in the angle between the elements. This leads to bending moments at either end of each element. We will now consider the free body diagrams of each element of the frame:





It must be noted that in the diagrams, $$d_i^{(e)}$$ and $$f_i^{(e)}$$ are generalized forces:    $$d_3^{(e)}$$ and $$d_6^{(e)}$$ are the rotational degrees of freedom and $$f_3^{(e)}$$ and $$f_6^{(e)}$$ are the bending moments.

Now we can consider the global degrees of freedom for the frame.



With the two $$6x6$$ element stiffness matrices, we can develop the $$9x9$$ global stiffness matrix $$\underline{K}_{9x9} = A\underline{k}_{6x6}^{(e)}$$



Lecture Notes - 11/21/08
$$[\varepsilon ] = \frac{[du]}{[dx]} = \frac{L}{L} = 1$$

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

$$[A] = L^2$$

$$[I] = L^4$$

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

$$[\tilde{k}_{11}\tilde{d}_1] = [\tilde{k}_{11}][\tilde{d}_1] = F$$

$$[\tilde{k}_{23}\tilde{d}_3] = \frac{[6][E][I]}{L^2} = \frac{(\frac{F}{L^2})L^4}{L^2} = F$$



Using the principle of virtual work and focusing only on the bending effect, the expression for the matrix $$\tilde{k}$$ becomes:

$$\frac{\partial^2 }{\partial x^2}[(EI)\frac{\partial^2v }{\partial x^2}] - f_t(x) = m(x)\ddot{v}$$