(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.

*Examples*

Impact, drop test, etc.

**PROBLEM**

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.

**Solution:**

Again apply the three steps outlined previously.

*Step 1: Find the Element Stiffness Equations*

*Element 1:*

*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

**PROBLEM **

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.

**Solution:**

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,

Deflections:

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

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