User:Eml4507.s13.team4ever.pickett

Problem 1.3
Description:	For the model 2-D truss with 2 inclined elements, provide the numerical values for the element stiff matrices and find the coordinates of the global nodes 1 and 3, with the global node 2 at the origin of the global coordinate system.



Img1.3.1: Obtained from the Eml4500.f08.1.djvu p.13



Img1.3.2: Obtained from the Eml4500.f08.1.djvu p.17



Img1.3.3: Obtained from the Eml4500.f08.1.djvu p.16

$$

\begin{array}{lcr} \Theta_{1} = 30^{0} \\

\Theta_{2} = \frac{\pi}{4}

\end{array} $$

>> l1 = cos(30*pi/180);

>> m1 = sin(30*pi/180);

>> k1 = zeros(4,4)

>> k1(1,1) = l1^2;

>> k1(2,2) = m1^2;

>> k1(3,3) = l1^2;

>> k1(4,4) = m1^2;

>> k1(1,2) = l1*m1;

>> k1(1,3) = -(l1^2);

>> k1(1,4) = -(l1*m1);

>> k1(2,3) = -(l1*m1);

>> k1(2,4) = -(m1^2);

>> k1(3,4) = l1*m1;

>> for i = 1:4

for j = 1:4

if k1(i,j)==0

k1(i,j)=k1(j,i)

end

end

end

>> k1

k1 =

0.7500   0.4330   -0.7500   -0.4330    0.4330    0.2500   -0.4330   -0.2500   -0.7500   -0.4330    0.7500    0.4330   -0.4330   -0.2500    0.4330    0.2500

$$

\begin{array}{lcr} k^{(1)}= \frac{E^{(1)}*A^{(1)}}{L^{(1)}}=\frac{(3)*(1)}{(4)}=0.75 \end{array} $$

>> k1 = k1*0.75

k1 =

0.5625   0.3248   -0.5625   -0.3248    0.3248    0.1875   -0.3248   -0.1875   -0.5625   -0.3248    0.5625    0.3248   -0.3248   -0.1875    0.3248    0.1875

>> l2 = cos(pi/4);

>> m2 = sin(pi/4);

>> k2(1,1) = l2^2;

>> k2(2,2) = m2^2;

>> k2(3,3) = l2^2;

>> k2(4,4) = m2^2;

>> k2(1,2) = l2*m2;

>> k2(1,3) = -(l2^2);

>> k2(1,4) = -(l2*m2);

>> k2(2,3) = -(l2*m2);

>> k2(2,4) = -(m2^2);

>> k2(3,4) = l2*m2;

>> for i = 1:4

for j = 1:4

if k2(i,j)==0

k2(i,j)=k2(j,i)

end

end

end

k2 =

0.5000   0.5000   -0.5000   -0.5000    0.5000    0.5000   -0.5000   -0.5000   -0.5000   -0.5000    0.5000    0.5000   -0.5000   -0.5000    0.5000    0.5000

$$

\begin{array}{lcr} k^{(2)}= \frac{E^{(2)}*A^{(2)}}{L^{(2)}}=\frac{(5)*(2)}{(2)}=5 \end{array} $$

>> k2 = k2*5

k2 =

2.5000   2.5000   -2.5000   -2.5000    2.5000    2.5000   -2.5000   -2.5000   -2.5000   -2.5000    2.5000    2.5000   -2.5000   -2.5000    2.5000    2.5000

The coordinated of global nodes one and three can be represented as seen below with the global node two being the origin.

$$

\begin{array}{lcr} node_{2} = (0 \hat{\underline{i}}, 0 \hat{\underline{j}}) \\

node_{3} = 4*(cos(45^{^{0}}) \hat{\underline{i}}, -sin(45^{^{0}}) \hat{\underline{j}}) = (2.8284 \hat{\underline{i}}, -2.8284 \hat{\underline{j}}) \\

node_{1} = 2*(-cos(30^{^{0}}) \hat{\underline{i}}, -sin(30^{^{0}}) \hat{\underline{j}}) = (-1.7320 \hat{\underline{i}}, -1 \hat{\underline{j}})

\end{array} $$

Problem 1.4
“Description:	Consider a plane truss consisting of three bars with the properties E = 200GPa, A1 = 6.0 x 10-4 m2, A2 = 3.0 x 10-4 m2, A3 = 10.0 x 10-4 m2, and loaded by a single force P = 80 kN. The corresponding finite element model consists of three elements and eight degrees of freedom.” (calfem p.234)



Img1.4: Obtained from the calfem manual (calfem.pdf) p.234

The first step is to define a topology matrix: Edof

>> Edof = [1,1,2,5,6;2,5,6,7,8;3,3,4,5,6]

Edof =

1    1     2     5     6     2     5     6     7     8     3     3     4     5     6

K is the stiffness matrix and f is the force matrix

>> K = zeros(8);

>> f = zeros(8,1);

>> f(6)=-80e3

f =

0          0           0           0           0      -80000           0           0

E is the modulus of elasticity. A1, A2, and A3 are the corresponding areas of each truss. Combined, we make the property matrices for each truss in the system, defined as ep1, ep2, and ep3.

>> E=2.0e11;

>> A1=6.0e-4;

>> A2=3.0e-4;

>> A3=10.0e-4;

>> ep1=[E A1]

ep1 =

1.0e+011 *

2.0000   0.0000

>> ep2=[E A2];

>> ep3=[E A3];

The coordinates of each connection of the truss system are defined below.

>> ex1=[0,1.6];

>> ex2=[1.6,1.6];

>> ex3=[0,1.6];

>> ey1=[0,0];

>> ey2=[0,1.2];

>> ey3=[1.2,0];

Using the calfem34 designed function bar2e, we are able to construct the element stiffness matrices as seen below.

>> Ke1=bar2e(ex1,ey1,ep1)

Ke1 =

1.0e+007 *

7.5000        0   -7.5000         0         0         0         0         0   -7.5000         0    7.5000         0         0         0         0         0

>> Ke2=bar2e(ex2,ey2,ep2)

Ke2 =

1.0e+007 *

0        0         0         0         0    5.0000         0   -5.0000         0         0         0         0         0   -5.0000         0    5.0000

>> Ke3=bar2e(ex3,ey3,ep3)

Ke3 =

64000000  -48000000   -64000000   48000000   -48000000    36000000    48000000  -36000000   -64000000    48000000    64000000  -48000000    48000000   -36000000   -48000000   36000000

The next step is to create the global stiffness matrix by using the topology and assem function.

>> K=assem(Edof(1,:),K,Ke1);

>> K=assem(Edof(2,:),K,Ke2);

>> K=assem(Edof(3,:),K,Ke3);

>> K

K =

1.0e+008 *

0.7500        0         0         0   -0.7500         0         0         0         0         0         0         0         0         0         0         0         0         0    0.6400   -0.4800   -0.6400    0.4800         0         0         0         0   -0.4800    0.3600    0.4800   -0.3600         0         0   -0.7500         0   -0.6400    0.4800    1.3900   -0.4800         0         0         0         0    0.4800   -0.3600   -0.4800    0.8600         0   -0.5000         0         0         0         0         0         0         0         0         0         0         0         0         0   -0.5000         0    0.5000

Using the physical displacements for this problem, the solveq function is used to solve the system of equations for the displacement vector a and the support force vector r.

>> bc= [1,0;2,0;3,0;4,0;7,0;8,0];

>> [a,r]=solveq(K,f,bc)

a =

0        0         0         0   -0.0004   -0.0012         0         0

r =

1.0e+004 *

2.9845        0   -2.9845    2.2383    0.0000    0.0000         0    5.7617

The normal forces for section 1, 2, and 3 are calculated using the bar2s function that came with the calfem34 toolbox.

>> ed1=extract(Edof(1,:),a);

>> N1=bar2s(ex1,ey1,ep1,ed1)

N1 =

-2.9845e+004

>> ed2=extract(Edof(2,:),a);

>> N2=bar2s(ex2,ey2,ep2,ed2)

N2 =

5.7617e+004

>> ed3=extract(Edof(3,:),a);

>> N3=bar2s(ex3,ey3,ep3,ed3)

N3 =

3.7306e+004