(Dynamic Response/Time-History Analysis) Structure response to arbitrary, time-dependent loading.
TRANSIENT RESPONSE ANALYSIS
(Dynamic Response/Time-History Analysis)
Structure response to arbitrary, time-dependent loading.
Compute responses by integrating through time:
B. Modal Method
First, do the transformation of the dynamic equations using the modal matrix before the time marching:
Then, solve the uncoupled equations using an integration method. Can use, e.g., 10%, of the total modes (m= n/10).
Uncoupled system, Fewer equations,
No inverse of matrices,
More efficient for large problems.
1Cautions in Dynamic Analysis
Symmetry: It should not be used in the dynamic analysis (normal modes, etc.) because symmetric structures can have antisymmetric modes.
Mechanism, rigid body motion means = 0. Can use this to check FEA models to see if they are properly connected and/or supported.
Input for FEA: loading F(t) or F( ) can be very complicated in real applications and often needs to be filtered first before used as input for FEA.
Impact, drop test, etc.
In the spring structure shown k1 = 10 lb./in., k2 = 15 lb./in., k3 = 20 lb./in., P= 5 lb. Determine the deflection at nodes 2 and 3.
Again apply the three steps outlined previously.
Step 1: Find the Element Stiffness Equations
Step 2: Find the Global stiffness matrix
Now the global structural equation can be written as above.
Step 3: Solve for Deflections
The known boundary conditions are: u1 = u4 = 0, F3 = P = 3lb. Thus, rows and columns 1 and 4 will drop out, resulting in t following matrix equation,
Solving, we get u2 = 0.0692 & u3 = 0.1154
In the spring structure shown, k1 = 10 N/mm, k2 = 15 N/mm, k3 = 20 N/mm, k4 = 25 N/mm, k5 = 30 N/mm, k6 = 35 N/mm. F2 = 100 N. Find the deflections in all springs.
Here again, we follow the three-step approach described earlier, without specifically mentioning at each step.
Now, apply the boundary conditions, u1 = u4 = 0, F2 = 100 N. This is carried out by deleting the rows 1 and 4, columns 1 and 4, and replacing F2 by 100N. The final matrix equation is,
Spring 1: u4 – u1 = 0
Spring 2: u2 – u1 = 1.54590
Spring 3: u3 – u2 = -0.6763
Spring 4: u3 – u2 = -0.6763
Spring 5: u4 – u2 = -1.5459
Spring 6: u4 – u3 = -0.8696