User:Eml4500.f08.ateam.didomenico/hw 2

A-Team Teampage

Solving for element forces of Element 1
$$k^{(1)}d^{(1)} = f^{(1)}$$

$$\begin{Bmatrix} f^{(1)}_1 \\ f^{(1)}_2\\ f^{(1)}_3\\ f^{(1)}_4 \end{Bmatrix}$$ = $$\begin{bmatrix} 0.5625 & 0.32776 & -0.5625 & -0.32476 \\ 0.32476 & 0.1875 & -0.32476 & -0.1875 \\ -0.5625 & -0.32476 & 0.5625 & 0.32476 \\ -0.32476 & -0.1875 & 0.32476 & 0.1875 \end{bmatrix} $$ $$\begin{Bmatrix} 0 \\ 0 \\ 4.352 \\ 6.1271 \end{Bmatrix}$$

This relation can be reduced to the following by the observation that the first two terms in the element displacement matrix are zero. This is due to matrix algebra causing the first two columns of the element stiffness matrix will completely drop out.

 $$\begin{Bmatrix} f^{(1)}_1 \\ f^{(1)}_2\\ f^{(1)}_3\\ f^{(1)}_4 \end{Bmatrix}$$ = $$\begin{bmatrix} -0.5625 & -0.32476 \\ -0.32476 & -0.1875 \\ 0.5625 & 0.32476 \\ 0.32476 & 0.1875 \end{bmatrix} $$ $$\begin{Bmatrix} 4.352 \\ 6.1271 \end{Bmatrix}$$

Carrying out this relation reveals the element force matrix.

 $$\begin{Bmatrix} f^{(1)}_1 \\ f^{(1)}_2\\ f^{(1)}_3\\ f^{(1)}_4 \end{Bmatrix}$$ = $$\begin{Bmatrix} -4.4378 \\ -2.5622 \\ 4.4378 \\ 2.5622 \\ \end{Bmatrix}$$

In which, $$f^{(1)}_1$$ and $$f^{(1)}_2$$ are reaction forces while $$f^{(1)}_3$$ and $$f^{(1)}_4$$ are internal forces.

Element 1 is in equilibrium because the reaction forces and internal forces offset each other and the sum of the moments about any point on Element 1 is zero.



$$\sum F_x = f^{(1)}_1 + f^{(1)}_3 = 0$$

$$\sum F_x = f^{(1)}_2 + f^{(1)}_4 = 0$$

$$\sum M_{any} = 0$$

Solving for element forces of Element 2


$$P^{(1)}_1 = \sqrt{f^{(1)}_1 + f^{(1)}_2}$$

Accounting for the applied force
In order to proceed, the applied force P must be accounted for in the free-body diagram and the equilibrium of all nodes must be verified. The equilibrium of global nodes 1 and 3 were previously verified.



Statics method
There is a second method to solve for the reactions and internal forces of the 2-bar truss depicted here. This method uses the principles of statics, but also must employ the Euler Cut Principle. The Euler Cut Principle must be utilized because the truss is considered statically indeterminate.



Method 1



Method 2

There are multiple ways of applying the Euler Cut Principle to the truss. Each way is determined simply by where the cut is made in the truss system. Depending on this choice, there are two or three resulting free-body diagrams from which the forces are derived.