Natural Vibration of Beam - Algebraic Query

In summary, in the given conversation, the author is solving a matrix equation involving a quadratic characteristic equation in ##\omega^2##. The author determines ##\lambda## by setting the determinant of the matrix to 0, which results in the quadratic equation ##15120-1224 \lambda + \lambda^2=0##. However, it is unclear how the author arrives at this expression or how he determines the units of ##\lambda##. It is also noted that the coefficient ##\rho## in the equation is confusing, as it is stated as "mass density per unit length" which does not seem to match the units of kg/m3 given the other terms in the equation.
  • #1
bugatti79
794
1
Folks,

Just wondering what the author is doing here, given

##\displaystyle \left (\frac{2EI}{L^3} \begin {bmatrix} 6&3L\\3L&2L^2 \end {bmatrix} -\omega^2 \frac{\rho A L}{420} \begin {bmatrix} 156&22L\\22L&4L^2 \end{bmatrix} \begin {bmatrix} W\\ \theta \end {bmatrix} \right )=\begin {bmatrix} 0\\ 0 \end {bmatrix}##

He proceeds to set the determinant of the matrix above to 0 to obtain a quadratic characteristic equation in ##\omega^2##

The equation being

##15120-1224 \lambda + \lambda^2=0##, where ##\displaystyle \lambda=\frac{\rho A L^4}{EI} \omega^2##

Using wolfram http://www.wolframalpha.com/input/?...mega^2+p+A+l)/420)*{{156,22l},{22l,4l^2}})+=0 (Z=EI)

I don't see how he arrives at this quadratic expression or how he determined ##\lambda## ...thanks
 
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  • #2
bugatti79 said:
Folks,

Just wondering what the author is doing here, given

##\displaystyle \left (\frac{2EI}{L^3} \begin {bmatrix} 6&3L\\3L&2L^2 \end {bmatrix} -\omega^2 \frac{\rho A L}{420} \begin {bmatrix} 156&22L\\22L&4L^2 \end{bmatrix} \begin {bmatrix} W\\ \theta \end {bmatrix} \right )=\begin {bmatrix} 0\\ 0 \end {bmatrix}##

He proceeds to set the determinant of the matrix above to 0 to obtain a quadratic characteristic equation in ##\omega^2##

The equation being

##15120-1224 \lambda + \lambda^2=0##, where ##\displaystyle \lambda=\frac{\rho A L^4}{EI} \omega^2##

Using wolfram http://www.wolframalpha.com/input/?...mega^2+p+A+l)/420)*{{156,22l},{22l,4l^2}})+=0 (Z=EI)

I don't see how he arrives at this quadratic expression or how he determined ##\lambda## ...thanks

Ah, I have it. The first equation in the alternative form in wolfram gives the clue.
 
  • #3
bugatti79 said:
Folks,

Just wondering what the author is doing here, given

##\displaystyle \left (\frac{2EI}{L^3} \begin {bmatrix} 6&3L\\3L&2L^2 \end {bmatrix} -\omega^2 \frac{\rho A L}{420} \begin {bmatrix} 156&22L\\22L&4L^2 \end{bmatrix} \begin {bmatrix} W\\ \theta \end {bmatrix} \right )=\begin {bmatrix} 0\\ 0 \end {bmatrix}##

He proceeds to set the determinant of the matrix above to 0 to obtain a quadratic characteristic equation in ##\omega^2##

The equation being

##15120-1224 \lambda + \lambda^2=0##, where ##\displaystyle \lambda=\frac{\rho A L^4}{EI} \omega^2##

Using wolfram http://www.wolframalpha.com/input/?...mega^2+p+A+l)/420)*{{156,22l},{22l,4l^2}})+=0 (Z=EI)

I don't see how he arrives at this quadratic expression or how he determined ##\lambda## ...thanks

Why do we have ##\displaystyle \lambda=\frac{\rho A L^4}{EI} \omega^2##? Why isn't it some other combination of units? In other words, what should the units of lambda be?

If we do a units balance on this to which I calculate

##\displaystyle\frac{\frac{kg}{m^4}m^2m^4 }{\frac{\frac{kgm}{s^2}}{m^2}m^4}\frac{rad^2}{s^2}=\frac{rad^2}{m}## where ##\rho## is the mass density per metre...thanks
 
  • #4
bugatti79 said:
Why do we have ##\displaystyle \lambda=\frac{\rho A L^4}{EI} \omega^2##? Why isn't it some other combination of units? In other words, what should the units of lambda be?

If we do a units balance on this to which I calculate

##\displaystyle\frac{\frac{kg}{m^4}m^2m^4 }{\frac{\frac{kgm}{s^2}}{m^2}m^4}\frac{rad^2}{s^2}=\frac{rad^2}{m}## where ##\rho## is the mass density per metre...thanks

##\lambda## should be dimensionless. You can tell because the quadratic equation involves different powers of ##\lambda##, so it can't have dimensions.

Are you sure that ##\rho## is a linear mass density? It looks like it should be kg/m3.

Also, keep in mind that "radians" are dimensionless units (they have dimensions of length/length).
 
  • #5
Mute said:
##\lambda## should be dimensionless. You can tell because the quadratic equation involves different powers of ##\lambda##, so it can't have dimensions.

Not sure I understand this. For example on a slight tangent of topic if we consider a 1 dimensional bar element whose interpolation function is a quadratic in x which is the distance along the bar in metres...

##u_h^e(x)=c_1+c_2x+c_3x^2## This is not dimensionless...


Mute said:
Are you sure that ##\rho## is a linear mass density? It looks like it should be kg/m3.

Also, keep in mind that "radians" are dimensionless units (they have dimensions of length/length).

Well this problem stems from the Euler-Bernoulli Beam Theory and the book states that ##\rho## is "mass density per unit length" so isn't that kg.m^3/m=kg/m^4?...
 
  • #6
bugatti79 said:
Not sure I understand this. For example on a slight tangent of topic if we consider a 1 dimensional bar element whose interpolation function is a quadratic in x which is the distance along the bar in metres...

##u_h^e(x)=c_1+c_2x+c_3x^2## This is not dimensionless...

Ah, I think I know what you mean. The LHS of above is not 0 so dimensions are allowed.

Its when an equation is set to 0 with different powers of ##\lambda## like you say, that they must be dimensionless, right...?

thanks
 
  • #7
Mute said:
Are you sure that ##\rho## is a linear mass density? It looks like it should be kg/m3.
Agreed, since the mass of the beam appears to be ##\rho A L## in your matrix equation.

Also, keep in mind that "radians" are dimensionless units (they have dimensions of length/length).

##\omega## is the angular frequency in radians/secons, not an angle in radians.
 
  • #8
bugatti79 said:
Not sure I understand this. For example on a slight tangent of topic if we consider a 1 dimensional bar element whose interpolation function is a quadratic in x which is the distance along the bar in metres...

##u_h^e(x)=c_1+c_2x+c_3x^2## This is not dimensionless...

The difference with this example is that your coefficients ##c_i## each have different units to compensate for the units of the different powers of x.

In the quadratic for ##\lambda##, you assigned no numbers to the numerical coefficients, and since you were following a book that appears to have left all dimensionful quantities as variables, I assumed those coefficients were dimensionless. Are the coefficients supposed to have units?

Well this problem stems from the Euler-Bernoulli Beam Theory and the book states that ##\rho## is "mass density per unit length" so isn't that kg.m^3/m=kg/m^4?...

The book says that word-for-word? That is a very confusing statement. One usually says "mass per unit length" or "mass per unit volume", but I have not heard "mass density per unit length". I think it must be a typo or something. Even if we were to take that at face value, is the density a linear density or a volumetric density?


AlephZero said:
##\omega## is the angular frequency in radians/secons, not an angle in radians.

Yes, I was not referring to the angular frequency as a whole, only the factor of radians that remained in bugatti79's calculation, so I mentioned that radians are actually dimensionless.
 
  • #9
Mute said:
The difference with this example is that your coefficients ##c_i## each have different units to compensate for the units of the different powers of x.

In the quadratic for ##\lambda##, you assigned no numbers to the numerical coefficients, and since you were following a book that appears to have left all dimensionful quantities as variables, I assumed those coefficients were dimensionless. Are the coefficients supposed to have units?
Not sure I understand what you are saying about the numerical coefficients? The coefficients of ##\lambda## and ##\lambda^2## are -1224 and 1 respectively, right?


Mute said:
The book says that word-for-word? That is a very confusing statement. One usually says "mass per unit length" or "mass per unit volume", but I have not heard "mass density per unit length". I think it must be a typo or something. Even if we were to take that at face value, is the density a linear density or a volumetric density?

Yes. That is all the book says as above. So perhaps it is a typo...

I have never heard of a volumetric density...
 

1. What is the concept of natural vibration of a beam?

Natural vibration of a beam refers to the inherent tendency of a beam to vibrate at specific frequencies without any external force or disturbance. It is a result of the elastic properties of the beam and its boundary conditions.

2. How is the natural vibration of a beam calculated?

The natural vibration of a beam is calculated using the Euler-Bernoulli beam theory, which takes into account the beam's material properties, geometry, and boundary conditions. This theory results in a differential equation, the solution of which gives the natural frequencies and mode shapes of the beam.

3. What factors affect the natural vibration of a beam?

The natural vibration of a beam is affected by several factors, including its material properties, geometry, boundary conditions, and applied loads. Changes in any of these factors can alter the natural frequencies and mode shapes of the beam.

4. How does the natural vibration of a beam impact its structural integrity?

The natural vibration of a beam can have a significant impact on its structural integrity. If the natural frequencies of a beam coincide with the frequencies of external forces, resonance can occur, leading to excessive vibrations and potential failure of the beam. Therefore, it is essential to consider the natural vibration of a beam during the design and analysis process.

5. Can the natural vibration of a beam be controlled?

Yes, the natural vibration of a beam can be controlled by altering its material properties, geometry, or boundary conditions. Additionally, adding dampers or using tuned mass dampers can also help reduce the effects of natural vibration on the beam's structural integrity.

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