Practical question for mechanical/Civil engineers

[Claudio L.]

(05-19-2015 01:45 PM)Claudio L. Wrote:  

(05-18-2015 05:37 AM)cyrille de brébisson Wrote:  Basically, I am trying to calculate the point of maximum deflection on a board set vertically and used to hold concrete in place while it is being poured in a vertical shaft.

In case you still want to know the point of maximum deflection, here's a formula:

\[ x=h * \left( 1 - \sqrt{1-\sqrt{ 8 \over 15 } } \right) \approx 0.48 * h\]

This is the height measured from the bottom at which maximum deflection would happen if you only had supports top and bottom.

And full formulae derivation, for those who care about the math:

\[ q=qmax-k*x \] (with x=height from the floor)

\[ qmax = \gamma * h * b \] ( \( \gamma \) = density of concrete, h = total height, b = width of the new pour)
and of course \( k= { qmax \over h } \) to make the pressure go to zero for x=h.

Now to get deflections we need to integrate a number of times:

\[ v = { \int_{}^{} p \cdot dx } = qmax \cdot x - k \cdot { x^2 \over 2 } + C1\]
\[ m = { \int_{}^{} v \cdot dx } = qmax \cdot { x^2 \over 2 } - k \cdot { x^3 \over 6 } + C1 \cdot x + C2 \]
\[ \phi = - { \int_{}^{} { m \over EI } \cdot dx } = - { 1 \over EI } \cdot \left[ qmax \cdot { x^3 \over 6 } - k \cdot { x^4 \over 24 } + C1 \cdot { x^2 \over 2 } + C2 \cdot x + C3 \right] \]
and finally:
\[ y = { \int_{}^{} \phi \cdot dx } = - {1 \over EI } \cdot \left[ qmax \cdot { x^4 \over 24 } - k \cdot { x^5 \over 120 } + C1 \cdot { x^3 \over 6 } + C2 \cdot { x^2 \over 2 } + C3 \cdot x + C4 \right] \]

The only thing left to do is to apply boundary conditions:

\[ \left. m=0 \right|_{x=0} \]
\[ \left. m=0 \right|_{x=h} \]

which allows to solve for C1 and C2 (C2 is trivially zero).

\[ 0 = qmax \cdot { h^2 \over 2 } - k \cdot { h^3 \over 6 } + C1 \cdot h \]
\[ C1 = k \cdot { h^2 \over 6 } - qmax \cdot { h \over 2 } \]

and at the supports:
\[ \left. y=0 \right|_{x=0} \]
\[ \left. y=0 \right|_{x=h} \]

to solve for C3 and C4 (C4=0 as well).

\[ 0 = - {1 \over EI } \cdot \left[ qmax \cdot { h^4 \over 24 } - k \cdot { h^5 \over 120 } + \left( k \cdot { h^2 \over 6 } - qmax \cdot { h \over 2 } \right) \cdot { h^3 \over 6 } + C3 \cdot h \right] \]

\[ C3 = k \cdot { h^4 \over 120 } - qmax \cdot { h^3 \over 24 } - \left( k \cdot { h \over 6 } - { qmax \over 2 } \right) \cdot { h^3 \over 6 } \]

and finally, the equation for deflection:

\[ y = - {1 \over EI } \cdot \left[ qmax \cdot { x^4 \over 24 } - k \cdot { x^5 \over 120 } + \left( k \cdot { h^2 \over 6 } - qmax \cdot { h \over 2 } \right) \cdot { x^3 \over 6 } + \left( k \cdot { h^4 \over 120 } - qmax \cdot { h^3 \over 24 } - \left( k \cdot { h \over 6 } - { qmax \over 2 } \right) \cdot { h^3 \over 6 } \right) \cdot x \right] \]

EI is the product of the elastic modulus of the material and the moment of inertia of the cross section of the board. I'll leave it for the reader to verify that \( y'=0 \) will result in a value of X per the formula given above (I didn't check).