How Does Bending Affect the Shape of a Clamped Rod with a Mass?

[Logistics]

Hey guys this is my question

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A long thin rod is clamped vertically at its lower end and a mass M is attached to its upper end. The coordinates (x, y) of any point on it satisfy the equation

$$EI\frac{d^2x}{dy^2} = Mg(a-x)$$

where E, I and a are constants. Given that x = 0 when y = 0 and x = a when y = L show that

$$x = a [1- \frac{sin\omega(L-y)}{sin\omega L}]$$

where $$\omega^2 = \frac{Mg}{EI}$$. (Hint: Let z = a - x).

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Now I have got this here:

$$z = a - x \Rightarrow \frac{dz}{dy} = -\frac{dx}{dy} \Rightarrow \frac{d^2z}{dy^2} = -\frac{d^2x}{dy^2}$$

Thus:

$$EI\left(-\frac{d^2z}{dy^2}\right) = Mgz$$

$$\frac{d^2z}{dy^2} + \frac{Mg}{EI}z = 0$$


I'm thinking that it's correct so far, How would I solve that?. ie. Continue on from here


Could someone help me out please


Thanks

 

[OlderDan]

Hey guys this is my question

----------------------------------------------------------------------

A long thin rod is clamped vertically at its lower end and a mass M is attached to its upper end. The coordinates (x, y) of any point on it satisfy the equation

$$EI\frac{d^2x}{dy^2} = Mg(a-x)$$

where E, I and a are constants. Given that x = 0 when y = 0 and x = a when y = L show that

$$x = a [1- \frac{sin\omega(L-y)}{sin\omega L}]$$

where $$\omega^2 = \frac{Mg}{EI}$$. (Hint: Let z = a - x).

----------------------------------------------------------------------


Now I have got this here:

$$z = a - x \Rightarrow \frac{dz}{dy} = -\frac{dx}{dy} \Rightarrow \frac{d^2z}{dy^2} = -\frac{d^2x}{dy^2}$$

Thus:

$$EI\left(-\frac{d^2z}{dy^2}\right) = Mgz$$

$$\frac{d^2z}{dy^2} + \frac{Mg}{EI}z = 0$$


I'm thinking that it's correct so far, and by differentiating the equation above I should get my answer. But I have no idea how to differentiate that thing above :(


Could someone help me out please


Thanks

Assuming what you have done is correct, the equation you have generated for z is a well known equation. If y were time, your equation would be saying acceleration is proportional to minus the displacement. This is the equation of a harmonic oscillator. Assume z as a function of y is harmonic (sine or cosine of ky) and take two derivatives. It will all fall into place.

 

[Logistics]

So I get this ?

z = a - x = a*sinw(L-y) / sinwL

dz/dy = -w*acosw(L-y) / sinwL

d^2z/dy^2 = -w^2*asinw(L-y) / sinwL

= -w^2 * 2


If this is what you were referring to, what happens now ?

 

[da_willem]

$$\frac{d^2z}{dy^2} + Cz = 0$$ $$C=\frac{Mg}{EI}$$
$$\frac{d^2z}{dy^2} =-Cz$$
$$z(y)=?$$

 

[Logistics]

I'm not following :(((

 

[HallsofIvy]

It's hard to give any help without knowing what background you have, what kind of course, this is, etc. In particular, it seems very peculiar that you would be doing a problem like this without know how to solve a linear second order differential equation with constant coefficients.

 

[Logistics]

It's hard to give any help without knowing what background you have, what kind of course, this is, etc. In particular, it seems very peculiar that you would be doing a problem like this without know how to solve a linear second order differential equation with constant coefficients.

We have studied the things you just mentioned. linear, 2nd order, homogenous, inhomogenous, auxillary equations etc.


It's just we never did anything like that, we were always dive, simpler questions. The lecturer just isn't on the same level as we are :(


Edit: If someone did the working out, I'm sure I could figure it out and next time be able to do it myself :)

 

[whozum]

OlderDan was trying to point out that the resulting equation was similar to that of a harmonic oscillator. da_willem then showed you the final differential equation you needed to solve to find z(y). The final result should be very similar to the equations of SHM.

I haven't taken Diff Eq so I can't be more specific, sorry.

 

[OlderDan]

So I get this ?

z = a - x = a*sinw(L-y) / sinwL

dz/dy = -w*acosw(L-y) / sinwL

d^2z/dy^2 = -w^2*asinw(L-y) / sinwL

= -w^2 * 2


If this is what you were referring to, what happens now ?

Where did the *2 come from? You mean *z. So

d^2z/dy^2 + w^2*z = 0

Compare this to your DE and identify omega. Put your expression for z into the defining equation for z and solve it for x.