# I.E. Irodov Problems in General Physics, 1.304

• Luke Tan
In summary: Therefore, the deflection at a given radius is not just the contribution of the fluid particles between that radius and the axis, but also the contribution from higher radii.In summary, the conversation discusses finding the bending moment on an element a distance ##x## away from the axis of rotation using an integral equation. There is a discrepancy between the given integral expression and the solution in the textbook. The conversation also explores other approaches to solving the problem, such as considering the bending moment about a parallel axis or using the cantilever beam method. It is determined that the deflection is not just due to the mass of the steel plate but also the inertial force in the rotating frame.
Luke Tan
Homework Statement
A steel plate of thickness h has the shape of a square whose side equals l, with h << l. The plate is rigidy fixed to a vertical axle OO which is rotated with a constant angular acceleration ##\beta##. Find the deflection ##\lambda##, assuming the sagging to be small
Relevant Equations
$$EI\frac{d^2y}{dx^2}=N(x)$$
My struggle here comes from finding the bending moment ##N(x)##. My working is as follows.

We want to find the bending moment on an element a distance ##x## away from the axis of rotation. To do so, let us consider the bending moment due to the force on an element ##\xi>x## away from the axis of rotation. The perpendicular force on this element is ##\xi\beta dm##. Hence, the total bending moment on this element would be
$$N=\int_x^l(\xi-x)(\xi\beta \mu d\xi)$$
Where ##\mu=\rho l h## is the mass per unit length. However, the solutions manual has this expression

(Solutions to I.E. Irodov's Problems in General Physics, Volume 1, Second Edition, Singh)

Where the integrand is ##\xi^2##, rather than ##\xi(\xi-x)##. I initially thought that the solutions manual was wrong, but it seems like sources all over the internet have ##\xi^2## under the integral sign. Can someone explain to me what I did wrong?

Thanks!

etotheipi
A diagram would help. Is the plate in a horizontal or vertical plane? Where is the axle in relation to the plate?

haruspex said:
A diagram would help. Is the plate in a horizontal or vertical plane? Where is the axle in relation to the plate?
I forgot to attach the given diagram, here it is (the axle is vertical, thus the plate is vertical)

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Then I agree with your integral.

Yes I also agree with yourself and @haruspex. The bending moment ##N(x)## is applied to the element about a vertical axis passing through an arbitrary point on the cross section connecting the element and the rest of the square. The solution they give computes it about the axis OO, which doesn't make sense!

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Lnewqban
Just in case anyone's interested, there's another way you could look at the solution that makes it slightly less wrong (but still wrong).

The internal forces acting on the outer element are comprised of a shear force and a bending moment. The easiest way to solve it would be to, as you did, balance moments about the point ##x##, since then the shear force has no moment. However, we could also balance the moments on this element about the axis OO.

The bending moment is, as always, produced by a couple so its moment about the axis OO is exactly the same as that about a parallel axis through the point ##x## - which is what you get if you evaluate your integral. Except now we also need to account for the contribution of the shear force, at a distance ##x## from OO, which we can compute by balancing forces in the axial direction: $$F_s = \int_x^l \xi \beta dm = \int_x^l \xi \beta \rho h l d\xi = \frac{1}{2}\beta \rho h l (l^2 - x^2)$$ $$\tau_s = \frac{1}{2}x\beta \rho h l (l^2 - x^2)$$ You will see that if you do the integral you wrote down, this term ##\tau_s## will be the difference between it and the book answer.

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Lnewqban
etotheipi
Lnewqban said:
What does create the deflection, only the mass of the steel plate or some additional drag force from the fluid surrounding it?
Could you solve the problem as a cantilever beam under uniformly distributed load?

https://www.engineeringtoolbox.com/cantilever-beams-d_1848.html

In the rotating frame, there's an inertial force proportional to the tangential acceleration ##\beta## that results in a non-uniform loading on the structure. So the cantilever approach is definitely a good one.

You'll see that I have you to thank for post #6

Lnewqban
etotheipi said:
In the rotating frame, there's an inertial force proportional to the tangential acceleration ##\beta## that results in a non-uniform loading on the structure. So the cantilever approach is definitely a good one.

You'll see that I have you to thank for post #6
Thank you very much for the clarification, etotheipi.
In that case, the inertial load is not uniformly distributed, but dependent on the radius.

etotheipi

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