User:Eml4500.f08.team.foskey.ckf/HW2

Statically Indeterminant?
Proving that the truss problem is not Statically Indeterminate

At first glance, the truss problem appears to be statically indeterminate. The truss has four unknown reaction forces, and there are only three equations to use to solve for those values. However, the truss can be broken up into its components. Each of the components is a two force member. To further simplify things, a new coordinate system is chosen such that the x’-axis is along each of the elements.

Solving the Statics equations
In element 1, if we take the sum of the moments around node 2, we find that force f (1)2 must equal zero. Taking the sum of the moments about node 1, we find that force f (1)4 is related to the known force P, the known angle force P acts upon, and the known length of the element.

$$ \sum M_{2}^{(1)} = 0 = f_{2}^{(1)}*L^{1} $$ therefore $$ f_{2}^{(1)} = 0$$

$$ \sum M_{1}^{(1)} = 0 = f_{4}^{(1)}*L^{1}+ Pcos \theta *L^{1} $$ therefore $$ f_{4}^{(1)} = -Pcos \theta $$

In element 2, taking the sum of node1 and node 2 results in f (2)2 and f (2)4 both being zero.

$$ \sum M_{1}^{(2)} = 0 = f_{4}^{(2)}*L^{2} $$ therefore $$ f_{4}^{(2)} = 0$$

$$ \sum M_{2}^{(2)} = 0 = -f_{2}^{(2)}*L^{2} $$ therefore $$ f_{2}^{(2)} = 0$$

Combining the elements again will show that force f (1)3 corresponds to force f (2)2, therefore force f (1)3 is zero as well. Similarly, force f (1)4 corresponds to force f (2)1, and the two forces are equal.

$$ f_{3}^{(1)}=f_{2}^{2} = 0 $$ $$ f_{1}^{(2)}=f_{4}^{1} = -Pcos \theta $$

Returning to element 2, and summing the forces in the x’ direction and setting them equal to zero gives force f (2)3 being equal to –f (2)1. Returning to element 1, and setting the sum of the forces in the x’ direction equal to zero allows us to obtain the value of force f (1)1.

$$ f_{3}^{(2)}= -f_{1}^{2} = Pcos \theta $$

$$ \sum F_{x'}^{(1)} = 0 = f_{3}^{(1)} + Pcos \theta + f_{1}^{(1)} $$

$$ = -(f_{3}^{(1)} + Psin \theta ) = -Psin \theta $$

Thus all of the forces acting on the truss can be found by using statics equations, and taking into account the two force members. The Finite Element Method is not necessary to solve this problem.

Verification of two force member forces
Since the element bars that make up the truss are two force members, the axial forces on each end of the bars should be equal and opposite. This can be shown by calculating the resultant forces for each end of each bar.

Element 1
As shown earlier, by changing the local coordinate system, the axial forces are found. These were:

$$ f^{(1)}_{1} = -Psin \theta $$ $$f^{(1)}_{3} = 0 $$

where \ theta is the angle between force P and the axis of the bar, and $$f^{(1)}_{1}$$ and $$f^{(1)}_{3}$$ are the resultant axial forces for each end of the element. Knowing that in the global figure, element 1 was at a 30 degree angle from the horizontal, and that force P acted vertically, the \theta in the element 1 figure is 60 degrees. For the elemental node 2 of element 1, since the force $$f^{(1)}_{3}$$ is zero, the axial force component is composed of only the axial component of force P, which is $$Psin \theta$$. Also knowing that P = 7, we can calculate that the axial forces at each end of the element are 5.124. Each of these forces act outward from the element.

Element 2
As previously shown, the axial forces in element 2 were:

$$f^{(2)}_{1} = -Pcos \theta$$ $$f^{(2)}_{3} = Pcos \theta$$ Again, the axial forces at each end should be equal and opposite, as the element is a two force member. Using the original problem data with P = 7, the axial forces in element 2 are found to be 6.276. These forces too act outward from the center of the element.

Director Cosines Proof
Proving that    m (e)  =  sinθ (e)

Since $$ m^{(e)} = \hat \tilde i \bullet \hat j $$

and $$ \hat \tilde i = cos \theta ^{(e)} \hat i + sin \theta^{(e)} \hat j $$ expanding $$ \hat\tilde i \bullet \hat j= (cos \theta ^{(e)} \hat i + sin \theta^{(e)} \hat j ) \bullet \hat j $$

$$ = cos \theta ^{(e)} \hat i \bullet \hat j + sin \theta^{(e)} \hat j \bullet \hat j $$

Since $$ \hat i \bullet \hat j = 0 $$ and $$ \hat j \bullet \hat j= 1 $$ we obtain $$m^{(e)} = \hat \tilde i \bullet \hat j = cos \theta ^{(e)} * 0 + sin \theta ^{(e)}*1 $$ proving that $$ m^{(e)}=sin\theta ^{(e)}$$

The Stiffness Matrices
Solving the k1 and k2 stiffness matrices

Using the data for the truss problem and the equation for calculating the elemental stiffness matrix, we can simply plug in the known values into the equations. The end results for the stiffness matrices are:

<p style="text-align:center;">$$ k^{(1)} = \frac{3*1}{4}$$ $$ \begin{bmatrix} (\frac{\sqrt{3}}{2})^{2} & (\frac{\sqrt{3}}{2} * \frac{1}{2}) & -(\frac{\sqrt{3}}{2})^{2} & -(\frac{\sqrt{3}}{2} * \frac{1}{2}) \\

( \frac{\sqrt{3}}{2}* \frac{1}{2}) & (\frac{1}{2})^{2} & - ( \frac{\sqrt{3}}{2} * \frac{1}{2}) & - ( \frac{1}{2})^{2} \\

-(\frac{\sqrt{3}}{2})^{2} & -(\frac{\sqrt{3}}{2} * \frac{1}{2})& (\frac{\sqrt{3}}{2})^{2} & (\frac{\sqrt{3}}{2} * \frac{1}{2}) \\

-(\frac{\sqrt{3}}{2} * \frac{1}{2})& -(\frac{1}{2})^{2} & (\frac{\sqrt{3}}{2} * \frac{1}{2}) &(\frac{1}{2})^{2} \\ \end{bmatrix} $$

<p style="text-align:center;">$$ = \begin{bmatrix} \frac{9}{16} & \frac{3\sqrt{3}}{16} & -\frac{9}{16} & -\frac{3\sqrt{3}}{16} \\ \frac{3\sqrt{3}}{16} & \frac{3}{16} & -\frac{3\sqrt{3}}{16} & -\frac{3}{16} \\ -\frac{9}{16} & -\frac{3\sqrt{3}}{16} & \frac{9}{16} & \frac{3\sqrt{3}}{16} \\ -\frac{3\sqrt{3}}{16} & -\frac{3}{16} & \frac{3\sqrt{3}}{16} & \frac{3}{16} \\ \end{bmatrix} $$

<p style="text-align:center;">and

<p style="text-align:center;">$$ k^{(2)} = \frac{5*2}{2} \begin{bmatrix} (\frac{\sqrt{2}}{2})^{2} & (\frac{\sqrt{2}}{2} * -\frac{\sqrt{2}}{2}) & -(\frac{\sqrt{2}}{2})^{2} & -(\frac{\sqrt{2}}{2} * -\frac{\sqrt{2}}{2}) \\ (\frac{\sqrt{2}}{2} * -\frac{\sqrt{2}}{2}) & -(\frac{\sqrt{2}}{2})^{2} & -(\frac{\sqrt{2}}{2} * -\frac{\sqrt{2}}{2}) & -(\frac{\sqrt{2}}{2})^{2} \\ -(\frac{\sqrt{2}}{2})^{2} & -(\frac{\sqrt{2}}{2} * -\frac{\sqrt{2}}{2}) & (\frac{\sqrt{2}}{2})^{2} & (\frac{\sqrt{2}}{2} * -\frac{\sqrt{2}}{2}) \\ -(\frac{\sqrt{2}}{2} * -\frac{\sqrt{2}}{2}) & -(\frac{\sqrt{2}}{2} * -\frac{\sqrt{2}}{2}) & (\frac{\sqrt{2}}{2} * -\frac{\sqrt{2}}{2}) \\ \end{bmatrix} $$

<p style="text-align:center;">$$ = \begin{bmatrix} \frac{10}{4} & -\frac{5}{2} & -\frac{10}{4} & \frac{5}{2} \\ -\frac{5}{2} & \frac{10}{4} & \frac{5}{2} & -\frac{10}{4} \\ -\frac{10}{4} & \frac{5}{2} & \frac{10}{4} & -\frac{5}{2} \\ \frac{5}{2} & -\frac{10}{4} & -\frac{5}{2} & \frac{10}{4} \\ \end{bmatrix} $$

Equilibrium of the Node
In the Finite Element Method of solving the two bar truss problem, bring in the force P back into the problem requires looking at the equilibrium of global node 2. Using the Euler Cut Principle to isolate global node 2, we can see that in order for the node to be in equilibrium, the axial forces in each element must cancel out the force P. This can easily be shown by resolving the axial forces into their horizontal and vertical components, and then setting the sum of the forces in each direction equal to zero.

<p style="text-align:center;">$$\sum F_{horizontal} = f^{(1)}_{1}*cos(30) + f^{(2)}_{3}*cos(-45) = -5.124cos(30) + 6.276cos(-45)=0$$

$$\sum F_{vertical} = f^{(1)}_{1}*sin(30) + f^{(2)}_{3}*cos(-45) + P =-5.124sin(30) + -6.276sin(-45) + 7 =0$$