University of Florida/Eml4507/s13.team3.Guzy

Problem 2.5: Finite Analysis Of A Spring System




The free body diagram is defined by the matrix EDU>> Edof=[1 1 2;2 2 3;3 2 3];

Next, create the load force matrix EDU>> f=[0 100 0];

Next each element stiffness matrix is created EDU>> K= [0 0 0;0 0 0;0 0 0]; EDU>> k=1500; EDU>> ep1=k; EDU>> ep2=2*k; EDU>> Ke1=springle(ep1); EDU>> Ke2=springle(ep2);

Now we create the global stiffness matrix EDU>> K=assem(Edof(1,:),K,Ke2); EDU>> K=assem(Edof(2,:),K,Ke1); EDU>> K=assem(Edof(3,:),K,Ke2);

Boundary conditions are now applied EDU>> bc=[1 0; 3 0]; EDU>> [a,r]=solveq(K,f,bc);

Displacements are found: EDU>> ed1=extract(Edof(1,:),a) ed1 = 0     0.0133  EDU>> ed2=extract(Edof(2,:),a) ed2 = 0.0133     0   EDU>> ed3=extract(Edof(3,:),a) ed3 = 0.0133     0

Spring forces are evaluated EDU>> es1=spring1s(ep2,ed1) es1 = 40 EDU>> es2=spring1s(ep1,ed2) es2 = -20 EDU>> es3=spring1s(ep2,ed3) es3 = -40 By hand:


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$$\displaystyle f=kd$$
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$$\displaystyle f = force$$
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$$\displaystyle k = stiffness$$
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$$\displaystyle d = displacement$$
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In matrix form:

$$ \begin{bmatrix} f_{1x} \\ f_{2x} \\ \end{bmatrix} = \begin{bmatrix} k & -k \\ -k & k \\ \end{bmatrix} \begin{bmatrix} d_{1x} \\ d_{2x} \\ \end{bmatrix} $$

First we create a matrix to show the systems Degree of Freedoms.

The first column represents each spring(#1,#2,#3). The second and third column represent the connectivity of each node.

$$ DOF = \begin{bmatrix} 1 & 1 & 2\\ 2 & 2 & 3\\ 3 & 2 & 3\\ \end{bmatrix} $$

The following is a matrix of the Force vector

$$ F = \begin{bmatrix} F_{1x}\\ F_{2x}\\ F_{3x}\\ \end{bmatrix} $$

Next we need each element stiffness matrix for every spring:

Element stiffness matrix for Spring #1:

$$ k_{1} = \begin{bmatrix} 3000 & -3000 \\ -3000 & 3000 \\ \end{bmatrix}= \begin{bmatrix} 3000 & -3000 & 0\\ -3000 & 3000 & 0\\ 0 & 0 & 0\\ \end{bmatrix} $$

Element stiffness matrix for Spring #2:

$$ k_{2} = \begin{bmatrix} 1500 & -1500 \\ -1500 & 1500 \\ \end{bmatrix}= \begin{bmatrix} 0 & 0 & 0\\ 0 & 1500 & -1500\\ 0 & -1500 & 1500\\ \end{bmatrix} $$

Element stiffness matrix for Spring #3:

$$ k_{3} = \begin{bmatrix} 3000 & -3000 \\ -3000 & 3000 \\ \end{bmatrix}= \begin{bmatrix} 0 & 0 & 0\\ 0 & 3000 & -3000\\ 0 & -3000 & 3000\\ \end{bmatrix} $$

To get the global stiffness matrix, we add up each element stiffness matrix:

$$ K = \begin{bmatrix} 3000 & -3000 & 0\\ -3000 & 3000 & 0\\ 0 & 0 & 0\\ \end{bmatrix}+ \begin{bmatrix} 0 & 0 & 0\\ 0 & 1500 & -1500\\ 0 & -1500 & 1500\\ \end{bmatrix}+ \begin{bmatrix} 0 & 0 & 0\\ 0 & 3000 & -3000\\ 0 & -3000 & 3000\\ \end{bmatrix}= \begin{bmatrix} 3000 & -3000 & 0\\ -3000 & 7500 & -4500\\ 0 & -4500 & 4500\\ \end{bmatrix} $$

Plugging each matrix into Hooke's Law we obtain:

$$ \begin{bmatrix} 3000 & -3000 & 0\\ -3000 & 7500 & -4500\\ 0 & -4500 & 4500\\ \end{bmatrix} \begin{bmatrix} d_{1x} \\ d_{2x} \\ d_{3x} \\ \end{bmatrix}= \begin{bmatrix} F_{1x}\\ F_{2x}\\ F_{3x}\\ \end{bmatrix} $$

Because node one and three are attached to the wall, their displacements are zero:


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$$\displaystyle d_{1x}=0$$
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$$\displaystyle d_{3x}=0$$
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Now we solve the follwing equation:

$$ \begin{bmatrix} 3000 & -3000 & 0\\ -3000 & 7500 & -4500\\ 0 & -4500 & 4500\\ \end{bmatrix} \begin{bmatrix} 0 \\ d_{2x} \\ 0 \\ \end{bmatrix}= \begin{bmatrix} F_{1x}\\ F_{2x}\\ F_{3x}\\ \end{bmatrix} $$

Solving the System we obtain the following equations:


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$$\displaystyle 3000*0 -3000*d_{2x}+0*d_{3x} = F_{1x}$$
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$$\displaystyle -3000*0+7500*d_{2x}-4500*d_{3x} = F_{2x}$$
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$$\displaystyle 0*0-4500*d_{2x}+4500*d_{3x} = F_{3x}$$
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Simplifying the above equations we obtain:
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$$\displaystyle -3000*d_{2x} = F_{1x}$$
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$$\displaystyle 7500*d_{2x}-4500*d_{3x} = F_{2x}$$
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$$\displaystyle -4500*d_{2x}+4500*d_{3x} = F_{3x}$$
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Solving we obtain:


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$$\displaystyle F_{1x}= 40 $$
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Plugging this value in


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$$\displaystyle {\frac{40}{-3000}}=d_{2x}=-0.01333 $$
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Using substitution for the remaining values, we obtain magnitudes of:

$$ d_{1x}= 0 $$

$$ d_{2x}= 0.01333 $$

$$ d_{3x}= 0 $$

$$ F_{1x}= 40 $$

$$ F_{2x}= 20 $$

$$ F_{3x}= 40 $$


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