4. A uniform simply supported beam is subjected to a...

4. A uniform simply supported beam is subjected to a concentrated force $F_0$ at the midspan (at $x = \frac{l}{2}$). Assuming zero initial conditions, derive an expression for the dynamic response (displacement distribution) of the beam.

Transcript

00:01 So here in this question we are given a continuous beam whose length is given that is equals to l that is a cantilever and subjected to a vertical bending deflection which is w of x of t.

00:11 The beam is having mass to the length ratio which is m of x and it is being stiffness which is ei of x.

00:18 So here in the first part of the question we have to derive an analytical integral expression for the kinetic energy of the beam.

00:24 So we can say that for that delta of k is equals to w which is equals to f multiplied by the delta of s that is equals to ma multiplied by the delta of s.

00:33 Let's say this is the equation number 1.

00:35 We are considering about a kinematic equation.

00:39 So this kinematic equation from here will be v square is equals to v naught square plus 2a multiplied by the delta of s.

00:46 So a multiplied by the delta of s is equals to v square minus v naught square divided by 2.

00:51 So the equation 1 becomes delta of k is equals to m multiplied by the v raised to the power 2 minus v naught raised to the power 2 divided by 2.

00:59 So delta of k is this.

01:02 We can say that delta of k now becomes equals to 1 divided by 2 mv square minus 1 divided by 2 mv naught square where this term is 0 because kinetic energy has rest is 0.

01:13 As we know that normal kinetic energy is 1 divided by 2 mv square.

01:16 So delta of k from here is equals to w which is equals to integral of f over r multiplied by the dr which is equals to ma multiplied by dr.

01:26 So delta of k from here become equals to m integral over dv divided by the dt multiplied by the dr.

01:33 So delta of k become equals to m integral over dr divided by dt multiplied by the dv.

01:38 So delta of k become equals to m integral over v multiplied by the dv.

01:42 So delta of k become equals to 1 divided by 2 mv square minus 1 divided by 2 mv node square where this term is equals to 0 because kinetic energy at rest is 0.

01:52 So kinetic energy here in this case become equals to 1 divided by 2 mv square.

01:58 Hence this is the answer to the first part of this question...

Question

4. A uniform simply supported beam is subjected to a concentrated force $F_0$ at the midspan (at $x = \frac{l}{2}$). Assuming zero initial conditions, derive an expression for the dynamic response (displacement distribution) of the beam.

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