User:EML4507.s13.team4ever.Bonner/Bonnerreport3p5

= Problem R3.5 =

On my honor, I have neither given nor recieved unauthorized aid in doing this assignment.

Input Constants
EDU>> E = 70E9;

EDU>> InL=0.009;

EDU>> OutL=0.012;

EDU>> F = 1000;

Begin Computation of variables
EDU>> InA=InL^2;

EDU>> OutA=OutL^2;

EDU>> A = OutA-InA;

EDU>> L1=sqrt(5);

EDU>> L2=2;

EDU>> K1=E*A/L1;

EDU>> K2=E*A/L2;

EDU>> l=cos(-26.56);

EDU>> m=sin(-26.56);

EDU>> K1a=K1*[l^2 l*m -l^2 -l*m 0 0;

l*m m^2 -l*m -m^2 0 0;

-l^2 -l*m l^2 l*m 0 0;

-l*m -m^2 l*m m^2 0 0;

0 0 0 0 0 0;

0 0 0 0 0 0];

EDU>> K2a = K2*[0 0 0 0 0 0;

0 0 0 0 0 0;

0 0 1 0 -1 0;

0 0 0 0 0 0;

0 0 -1 0 1 0;

0 0 0 0 0 0];

Cut down matrices to do calculations
EDU>> A = [K(3,:);K(4,:)]

A =

1.0e+06 *

Columns 1 through 4

-0.0404   0.2792    2.2454   -0.2792    0.2792   -1.9319   -0.2792    1.9319

Columns 5 through 6

-2.2050        0         0         0

EDU>> B = [A(:,3) A(:,4)]

B =

1.0e+06 *

2.2454  -0.2792   -0.2792    1.9319

EDU>> Fa = [0;-1000];

Final Results from Matlab
EDU>> d = B\Fa

d =

1.0e-03 *

-0.0655  -0.5271

EDU>> d1=[0;0;d(1);d(2);0;0]

d1 =

1.0e-03 *

0        0   -0.0655   -0.5271         0         0

Reaction Forces
EDU>> F = K*d1

F =

1.0e+03 *

-0.1445   1.0000         0   -1.0000    0.1445         0

To test the member in tension
$$\sigma=\frac{F}{A}$$ $$\sigma=\frac{1010N}{6.3*10^{-5}m^{2}}=16.03 MPa$$

The stress is well under the yield strength so this value is fine.

====To test the member in compression $$P_{cr}=\frac{\pi^{2}EI}{L}$$ $$P_{cr}=\frac{\pi^{2}(70*10^{9})*(1.18*10^{-9})}{2}=408.4 N$$

We can compare this to the actually force of 144.5 N.

Optimize
To optimize for the lowest weight while still being safe, we can turn the above matlab program into an .m file and adjust the dimensions of the cross section. When optimized, the outer member will be 12mm and inner hole will be 10.5mm.

EDU>> [u v]=Report3(0.012,0.0105)

u =

1.0e+03 *

-0.1445   1.0000   -0.0000   -1.0000    0.1445         0

v =

247.0138

For the above calculation, u is the stress in each member and v is P critical for bending. This is well within the range to not buckle with a factor of safety of 1.2 and the tensile stress is also well within the factor of safety of 2.

Calfem Verification
EDU>> A = 0.012^2-0.009^2;

EDU>> E = 70E9;

EDU>> Edof = [1 1 2 3 4;

2 3 4 5 6];

EDU>> K = zeros(6);

EDU>> f = zeros(6,1);

EDU>> f (4) = -1000;

EDU>> ep = [E A];

EDU>> Ex = [0 2;

2 0];

EDU>> Ey = [1 0;

0 0];

EDU>> for i = 1:2

Ke = bar2e(Ex(i,:),Ey(i,:),ep);

K = assem(Edof(i,:),K,Ke);

end;

EDU>> bc = [1 0;2 0;5 0;6 0];

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

a =

0        0   -0.0009   -0.0043         0         0

r =

1.0e+03 *

-2.0000   1.0000    0.0000    0.0000    2.0000         0

When verified with calfem, the reactions didn't match the numbers from matlab or hand calculations. The displacement seems to be closer to what would be expected in real life applications however the reaction forces seem strange. We were not able to find the cause for this error.