Trusses and Transformations  Basic Ideas - ``123  Springs and Assemblies of

Application of Loads and Constraints

Before we can obtain a solution to equations (1.6) we have to modify the system above to account for

• applied loads
•  constraints

Let us account for the applied loads first. From Fig. 1.4 the applied load of F at node 3 implies -f23 = F. Since there are no applied loads at node 2, -f21=-f22=0. Modifying (1.6) we get

  

Now applying the constraint (node 1 is constrained  )

  

Moving the first column to the right hand side

  

The first equation reduces now to -k1 u2 = -f11, a redundant equation in u2. Dropping it we are left with

  

We are left with 2 equations for the 2 unknowns u2 and u3. Thus, the net effect of applying the constraint was that of deleting the row and column corresponding to the displacement that was constrained. This shortcut will work as long as the constraint is homogeneous i.e. ui=0. If a non-zero constraint is to be applied then the procedure outlined above must be conducted. Note that we did not have to deal with the unknown reaction at the wall f11!

• Step 4: Obtain the Solution

We can solve (1.10) to get values for u2,u3. Let k1=2000,k2=1000 and F=10 then





u3=.015 and u2=.005.

Exercise

Use the approach above to obtain the displacements for the three rod structure shown below.

   
Figure: 3 rod structure
 

Mon Mar 15 10:37:42 EST 1999