User:Eml4500.f08.RAMROD.F/HW2

=Friday September 19 2008=

Element 1:

K(1) · d(1) = F(1)

$$= \begin{bmatrix} 0.5625 & 0.032476 & -0.5625 & -0.032476 \\ 0.032476 &

0.1875 & -0.032476 & -0.1875 \\ -0.5625 & -0.032476 & 0.5625 & 0.032476 \\ -0.032476 & -0.1875 & 0.032476 & 0.1875 \end{bmatrix} \begin{Bmatrix} 0 \\ 0 \\ 4.352 \\ 6.1271 \end{Bmatrix}$$

You may simplify the matrix because of the boundary conditions e.g. the first two numbers in the displacement matrix are zero.

$$= \begin{bmatrix} -0.5625 & -0.032476 \\ -0.032476 & -0.1875 \\ 0.5625 & 0.032476 \\ 0.032476 & 0.1875 \end{bmatrix} \begin{Bmatrix} 4.352 \\ 6.1271 \end{Bmatrix}$$

$$=\begin{Bmatrix} -4.4378 \\ -2.5622 \\ +4.4378 \\ +2.5622 \end{Bmatrix} = \begin{Bmatrix} F_1^1 \\ F_2^1 \\ F_3^1 \\ F_4^1 \end{Bmatrix}$$

Reactions forces are actually $$F_1^1$$ & $$F_2^1$$

Internal forces are actually $$F_3^1$$ & $$F_4^1$$

Element 1:

$$P_1^1 = \sqrt((F_1^1)^2 + (F_2^1)^2)$$

$$P_1^1 = \sqrt((-4.4378)^2 + (-2.5622)^2$$

$$ P_1^1 = 5.69064                     $$



Element 2:

$$P_1^2 = \sqrt((F_1^2)^2 + (F_2^2)^2)$$

$$P_1^2 = \sqrt((-4.4378)^2 + (-2.5622)^2$$

$$P_1^2 = 5.69064  $$



Method Two: Statics
Solving the two bar truss system using Euler Cut Principle.



Back to FE Solution: How to bring P back into the big picture?

Ans: Using the Euler cut method to isolate node two, equilibrium can be established at global node 2. Statics can be applied to the system by using summation of the forces in the x and y directions. This provides two equations with two unknowns that can be solved by using a system of equations.



$$\sum{F_x}= P_1cos30 - P_2cos45=0$$

$$\sum{F_y}=7- P_1cos30 -P_2cos45=0$$

$$\sum{F_x}= P_1 \frac{\sqrt{3}}{2} - P_2\frac{\sqrt{2}}{2}=0$$

$$\sum{F_y}=7- \frac{P_1}{2}-P_2\frac{\sqrt{2}}{2}=0$$

$$P_1=5.122436\;$$

$$P_2 = 6.27603\;$$