User:Eml4500.f08.group/Homework2/Homework Problems

=SOLVE USING FEA=

For the first part of our homework, the 2-truss system shown in the figure below had to be analyzed and solved using the F.E.A. procedure. Where P is an applied downward force of 7, R1x, R1y, R3x, and R3y are the reactions in the x and y directions at nodes 1 and 3 respectively.

The element 1 is inclined at an angle of 30 degrees relative to the x axis while element 2 is declined at an angle of 45 degrees.



L1 = 4 L2 = 2 A1 = 1

A2 = 2 E1 = 3

E2 = 5

The next step is to draw the Free Body Diagram for the 1st element and determine its stiffness matrix. This Free Body Diagram is shown in the following figure and the stiffness matrix is defined as follows.



where k(e) = k(1) = E(1)*A(1)/L(e), l(e) = cos(θ(e)), m(e) = sin(θ(e))



In the case of element 1,

k(1) = 3*1/4 = 3/4 = 0.75

l(1) = cos(30) = 0.866

m(1) = sin(30) = 0.500

The resulting stiffness matrix for element 1 is, k(1) $$ = \begin{bmatrix} 0.5625&0.32476&-0.5625&-0.32476 \\ 0.32476&0.1875&-0.32476&-0.1875\\ -0.5625&-0.32476 & 0.5625&0.32476\\ -0.32476&-0.1875&0.32476&0.1875 \end{bmatrix} $$

Next it is necessary to calculate the stiffness matrix for the second element.

k(2) = 5*2/2 = 5



l(2) = cos(-45) = 0.500

m(2) = sin(-45) = -0.500

The resulting stiffness matrix for element 2 is, k(2) $$ = \begin{bmatrix} 2.5&-2.5&-2.5&2.5 \\ -2.5&2.5&2.5&-2.5\\ -2.5&2.5&2.5&-2.5\\ 2.5&-2.5&-2.5&2.5 \end{bmatrix} $$

Given that each element stiffness matrix is a 4x4 (because there are 4 unknown forces), it is next necessary to assemble a global (6x6) force-displacement matrix. This is done by recognizing that

F(1)1 = R(1)x

F(1)2 = R(1)y

F(2)3 = R(3)x

F(2)4 = R(3)y

F(1)3 = F(2)1 = Px

F(1)4 = F(2)2 = Py

Applying the above relationships to the global (6x6) matrix yields the following

K = $$ \begin{bmatrix} 0.5625&0.32476&-0.5625&-0.32476&0&0 \\ 0.32476&0.1875&-0.32476&-0.1875&0&0 \\ -0.5625&-0.32476&3.0625&-2.1752&-2.5&2.5 \\ -0.32476&-0.1875&-2.1752&2.6875&2.5&-2.5 \\ 0&0&-2.5&2.5&2.5&-2.5 \\ 0&0&2.5&-2.5&-2.5&2.5 \end{bmatrix} $$

Next, we observe that since node 1 and node 3 are pinned and fixed. This means that no displacement can occur at these points and thus d11, d12, d23, and d24 = 0. We have also been given that Py = 7 and is oriented vertically Px = 0. Applying these observations into the equation F = KD yields the following system.

$$ \begin{bmatrix} 0.5625&0.32476&-0.5625&-0.32476&0&0 \\ 0.32476&0.1875&-0.32476&-0.1875&0&0 \\ -0.5625&-0.32476&3.0625&-2.1752&-2.5&2.5 \\ -0.32476&-0.1875&-2.1752&2.6875&2.5&-2.5 \\ 0&0&-2.5&2.5&2.5&-2.5 \\ 0&0&2.5&-2.5&-2.5&2.5 \end{bmatrix} $$ * $$\begin{bmatrix} 0\\ 0\\ D_3\\ D_4\\ 0\\ 0 \end{bmatrix} $$ = $$ \begin{bmatrix} R_{x1}\\ R_{y1}\\ 0\\ 7\\ R_{x3}\\ R_{y3} \end{bmatrix}$$

These equations can be used to solve for D3 and D4 (the x and y displacement of node 2).

$$D_3 = 4.353$$

$$D_4 = 6.1271$$

We can now plug these values back into the matrix equation and solve for the unknown reactions at nodes 1 and 2. When we do this, we find:

$$R_{x1} = -4.4378$$

$$R_{y1} = -2.5622$$

$$R_{x3} = 4.4378$$

$$R_{y3} = -4.4378$$

=SOLVE USING STATICS=

To verify these results, it is possible to solve the same problem using statics. Thus can been done by realize that in a truss, any reaction is directed along the axis of the truss. When the reactions at nodes 1 and 3 are combined into a single reaction forces (R1 and R3)we can solve the problem as follows.

ΣFy = R1sin(30) + R2sin(45) = 7

ΣFx = -R1cos(30) + R2cos(45) =

$$\begin{bmatrix} sin(30)&sin(45) \\ -cos(30)&cos(45) \end{bmatrix}$$ * $$\begin{bmatrix} R1\\ R3 \end{bmatrix}$$ = $$\begin{bmatrix} 7\\ 0 \end{bmatrix}$$

R1 = 5.12436

R3 = 6.27603

Plugging these values into the above equations, we find that

Rx1 = -5.12436*cos(30) = -4.4378

Ry1 = -5.12436*sin(30) = -2.562

Rx3 = 6.27603*cos(45) = 4.4378

Ry3 = -6.27603*sin(45) = -4.4378

These values agree perfectly with the ones attained using the Finite Element method thus confirming the results.



=PROVE THE EXAMPLE IS A 2-FORCE BODY=

The above statics solutions assumes a 2-force body and that the solutions attained via this assumption match those of the solutions attained using Finite Element Analysis. Given that the two solutions sets match, this proves that the trusses in the example problem are indeed 2-force bodies.

=NODE 2 IN EQUILIBRIUM=

In order for node 2 to be in equilibrium, the sum of the forces in the y direction have to equal to 0 as to the sum of the forces in the x direction. That is

R3sin(45) + R1sin(30) = 7

R3cos(45) - R1 cos(30) = 0

(6.27603)(0.707107) + (5.12436)(0.5) = 7

(6.27603)(0.707107) - (5.12436)(0.866025) = 0

Thus the sum of the forces in both the x and y directions equal 0, i.e. the node does not move and is in equilibrium.