A steel plate of thickness h has the shape of a square whose side equals l, with `h lt lt l`. The plate is rigidly fixed to a vertical axle OO which is rotated with a constant angular acceleration `beta` (figure). Find the deflection `lambda`, assuming the sagging to be small.

A steel plate of thickness h has the shape of a square whose side equa

1.	A steel plate of thickness h has the shape of a square whose side equals l, with `h lt lt l`. The plate is rigidly fixed to a vertical axle OO which is rotated with a constant angular acceleration `beta` (figure). Find the deflection `lambda`, assuming the sagging to be small.
Answer» The deflection of the plate can be noticed by going to a co-rotating frame. In this frame each element of the plate experiences a pseudo force proportional to its mass. These forces have a moment which constitutes the bending moment of the problem. To calculate this moment we note that the acceleration of an element at a distance `xi` from the axis is `a=xibeta` and the moment of the force exerted by the section between x and l is `N=rholhbetaunderset(x)overset(l)intxi^2dxi=1/3rholhbeta(l^3-x^3)`. From the fundamental equation `EI(d^2y)/(dx^2)=1/3rholhbeta(l^3-x^3)`. The moment of inertia `I=underset(-h//2)overset(+h//2)intz^2ldz=(lh^3)/(12)`. Note that the natural surface (i.e. the surface which contains lines which are neither stretched nor compressed) is a vertical plane here and z is perpendicular to it. `(d^2y)/(dx^2)=(4rhobeta)/(Eh^2)(l^3-x^3)`. Integrating `(dy)/(dx)=(4rhobeta)/(Eh^2)(l^3x-x^4/4)+c_1` Since `(dy)/(dx)=0`, for `x=0`, `c_1=0`. Integrating again, `y=(4rhobeta)/(Eh^2)((l^3x^2)/(2)-(x^5)/(20))+c_2` `c_2=0` because `y=0` for `x=0` Thus `lambda=y(x=l)=(9rhobetal^5)/(5Eh^2)`

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A steel plate of thickness h has the shape of a square whose side equals l, with `h lt lt l`. The plate is rigidly fixed to a vertical axle OO which is rotated with a constant angular acceleration `beta` (figure). Find the deflection `lambda`, assuming the sagging to be small.

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