Natural Vibration of Beam - Algebraic Query

  • Thread starter bugatti79
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715
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/?i=determinant+of++(2Z/l^3{{6,3l},{3l,2l^2}}-((omega^2+p+A+l)/420)*{{156,22l},{22l,4l^2}})+=0 (Z=EI)

I dont see how he arrives at this quadratic expression or how he determined ##\lambda## ...thanks
 
715
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/?i=determinant+of++(2Z/l^3{{6,3l},{3l,2l^2}}-((omega^2+p+A+l)/420)*{{156,22l},{22l,4l^2}})+=0 (Z=EI)

I dont 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.
 
715
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/?i=determinant+of++(2Z/l^3{{6,3l},{3l,2l^2}}-((omega^2+p+A+l)/420)*{{156,22l},{22l,4l^2}})+=0 (Z=EI)

I dont 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

Homework Helper
1,381
10
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).
 
715
1
##\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 isnt that kg.m^3/m=kg/m^4?...
 
715
1
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

Science Advisor
Homework Helper
6,953
291
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

Homework Helper
1,381
10
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 isnt 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.
 
715
1
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...
 

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