Calculating Transverse Wave Propagation in a Semi-Infinite Beam

Click For Summary

Homework Help Overview

The discussion revolves around calculating the transverse wave propagation in a semi-infinite beam subjected to a transient force. Participants are exploring the relationships between displacement, velocity, acceleration, and strain at various positions along the beam, as well as the implications of the wave equation in this context.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to relate wave properties to the force acting on the beam and expresses uncertainty about the next steps. Some participants provide equations relevant to the wave motion and discuss the implications of boundary conditions. Others inquire about MATLAB implementation and share their own challenges with similar assignments.

Discussion Status

Participants are actively engaging with the problem, sharing equations and discussing the dynamics of the beam. There is an ongoing exploration of the mathematical relationships involved, and while some guidance has been offered, there is no explicit consensus on the approach to take next.

Contextual Notes

Participants are working under the constraints of homework assignments, which may impose specific requirements for calculations and the use of software tools like MATLAB. There is also a focus on deriving expressions based on boundary conditions and the nature of the wave propagation in a semi-infinite medium.

Trevorman
Messages
20
Reaction score
2
1. A transversely directed transient force F(t) acts at the free end of a semi-infinite beam.
a) Show how displacement, velocity, acceleration and strain at an arbitrary position along the beam can be determined.
b) Calculate (MATLAB) the transversal acceleration (or an other quantity) at an arbitrary position. Assume suitable parameters and a force history.
c) Make the calculation for several positions thus illustrating the propagation of the wave.

Homework Equations


##\hat{v} = Ae^{i\beta x} + Be^{-i\beta x} +Ce^{-\beta x} + De^{\beta x}##
##\hat{T} = -E I \hat{v}^{\prime \prime \prime} = -E I \frac{\partial^3 \hat{v}}{\partial x^3}##
##\hat{T} = -\frac{1}{2}\hat{F}##

## v = \sum_n \hat{v} e^{-i \omega t} ##

Where ##v## is the displacement
##T## transverse force in the beam (given from free body diagram)
##E## Youngs modulus
A,B,C,D is just constants

The Attempt at a Solution


What i know is that I can calculate the axial velocity and acceleration
Velocity
##\dot{v} = \sum_n - \hat{v} \omega e^{-i \omega t}##
Acceleration
##\dot{v} = \sum_n \hat{v} \omega^2 e^{-i \omega t}##

Also, since it is a semi-infinite beam, there will only be a wave going in one direction.
therefore the
##\hat{v} = A e^{-\beta x} + B e^{- i \beta x}##

I need to relate this to the force acting on the beam and do not know how to proceed...
 
Last edited by a moderator:
Physics news on Phys.org
The wave equation which is the simplest dynamic model for transverse motion of a beam is displayed below if it helps.

## v^{\prime \prime \prime \prime} + \frac{\rho A}{E I} \ddot{v} =0 ##
 
For those who have interest in this, I can give you the answer right away. Derivate the displacement with the boundary conditions, solve out all constants and you will get a very pretty expression that looks like this

##
\hat{v}(x,\omega) = \frac{\hat{F}(\omega)}{4(EI)^{\frac{1}{4}}i \omega^{\frac{3}{2}}(\rho A)^{\frac{3}{4}}}\left[ e^{-i\left[\omega^2 \frac{\rho A}{EI} \right]^{\frac{1}{4}}x} -ie^{-\left[\omega^2 \frac{\rho A}{EI} \right]^{\frac{1}{4}}x} \right]##
Which could be expressed
##\hat{v}(x,\omega) = \hat{F}(\omega) \cdot \hat{H}(x,\omega)##
 
Hej Trevor, did you ever manage with the MATLAB script?
Im doing a similar asssignment, and I am not getting the script to behave properly.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
19
Views
3K
Replies
4
Views
2K
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K