User:Eml4500.f08.delta 6.castrillon/3-bar space truss

3-Dimensional Space Truss (MATLAB)
A three-bar space truss is discussed in the course textbook by Asghar Bhatti on page 230. This truss system contains elements with these properties:

a1 = 200; a2 = 600; e = 200000; P = 20000;
 * E$$^{(1)}$$=200,000
 * E$$^{(2)}$$=200,000
 * E$$^{(3)}$$=200,000


 * A$$^{(1)}$$=200 mm
 * A$$^{(2)}$$=200 mm
 * A$$^{(3)}$$=600 mm

The truss system has four nodes, three coplanar and supported by a wall, and belong the the three elements, which meet at the free-moving fourth node. Also, a 20-kN force acts upon the fourth global node in the negative y-direction.

The following MATLAB code shows the colaculation of the system reactions and nodal dofs. It is easy to note that only the fourth global node will showacse any displacement due to the force P.

First defining some functions:

These functions serve to set up each element's stiffness matrix and display the results. The bulk of the code is below, which constructs the stiffness matrix, determines the dof values, and the reactions. The results are also displayed below.

The deformation of this space truss due to the force P and its code is shown below:

This is the truss output from the above code.

x-axis perspective:



y-axis perspective:



z-axis perspective:



(-2,-2,3) perspective:



(1.5,2,2) perspective:



Statics Approach
The three-bar space truss explored above proves to be statically determinate because its reactions were readily able to be determined. Determining these results was possible from the fact that there were enough constraints on the truss, i.e. the fixed nodes on the wall. Since this truss system is statically determinate, then a statics approach can be developed.



This truss system is loaded by -P$$\hat{j}$$.

Defining the vectors from global node 4 to each of the other three global nodes:


 * 41=0.96$$\hat{i}$$+1.92$$\hat{j}$$-2$$\hat{k}$$


 * 42=-.144$$\hat{i}$$+1.44$$\hat{j}$$-2$$\hat{k}$$


 * 43=-2$$\hat{k}$$

And their magnitudes:


 * $$|41|=\sqrt{0.96^2+1.92^2+(-2)^2}=2.9339$$


 * $$|42|=\sqrt{(-1.44)^2+1.44^2+(-2)^2}=2.8543$$


 * $$|43|=\sqrt{(-2)^2}=2$$

Defining the forces along each element (negative for compression) in vector notation:


 * T$$_{41}$$ = $$T_{41}$$ λ$$_{41}$$ =$$T_{41}\frac{\mathbf{T_{41}}}{|41|}$$=$$(0.3272\mathbf{\hat{i}}+0.6544\mathbf{\hat{j}}-0.6817\mathbf{\hat{k}})T_{41}$$


 * T$$_{42}$$ = $$T_{42}$$ λ$$_{42}$$ =$$T_{42}\frac{\mathbf{T_{42}}}{|42|}$$=$$(-0.5045\mathbf{\hat{i}}+0.5045\mathbf{\hat{j}}-0.7007\mathbf{\hat{k}})T_{42}$$


 * T$$_{43}$$ = $$T_{43}$$ λ$$_{43}$$ =$$T_{43}\frac{\mathbf{T_{43}}}{|43|}$$=$$(-\mathbf{\hat{k}})T_{43}$$

Applying the knowledge that the forces will all balance out in all directions, we can solve for T$$_{41}$$, T$$_{42}$$, and T$$_{43}$$:

ΣF$$_x$$:


 * 0.3272$$T_{41}$$-0.5045$$T_{42}$$=0

ΣF$$_y$$:


 * 0.6544$$T_{42}$$+0.5045$$T_{42}$$-P=0

ΣF$$_z$$:


 * -$$T_{43}$$-0.6817$$T_{41}$$-0.7007$$T_{42}$$=0

Three unknowns and three equations allows us to find the reactions along the axis of the elements:


 * $$T_{41}$$=-2,314.533 N (compression)


 * $$T_{42}$$=20,374.691 N (tension)


 * $$T_{43}$$=13,214.269 N (tension)

Notice that these exact reaction values are equal to the values attained with the FEA method above.