Natural Vibration of Beam - Algebraic Query

[bugatti79]

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

 

[bugatti79]

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.

 

[bugatti79]

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

 

[Mute]

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).

 

[bugatti79]

$\lambda$ 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...

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?...

 

[bugatti79]

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

 

[AlephZero]

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.

 

[Mute]

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?

$\omega$ 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.

 

[bugatti79]

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?

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...