% $Id: 21m735-tn3.tex,v 1.6 2004/04/02 09:13:04 dan Exp $
\documentclass[11pt]{article}
\usepackage{setspace}
\include{21m735-preamble}

\begin{document}
\technote{3}{Dan Ports}{2004/02/29}{drkp@mit.edu}
\tntitle{A Unified Case Formula For Platform Loading}
\onehalfspacing

A natural concern for the technical designer faced with a platform
design problem is how much load the platform will be able to support.
This note presents a unified case formula for identifying the maximum
permissible load for a simple platform under normal loading
conditions. It considers the most common sources of platform failure.
Though the final result is a large and intimidating formula, it is
nonetheless expressed in a form suitable for rapid numerical
evaluation through a spreadsheet, programmable calculator, or other
mathematical tool.

A few notes:
\begin{itemize}
\item For maximum flexibility, no design factor is applied. The
  designer will want to make an appropriate choice based on the
  situation and reduce the rating accordingly.
\item An evenly distributed load is assumed. If the load is especially
  uneven, approaching a point load, this result may not apply.
\item These calculations do not take into account off-axis or other
  complex loading conditions. In particular, it does not take into
  account any cross-bracing 
\item The surface of the platform is not taken into account.
\item The weight of beams and surface materials should be subtracted
  from the allowable load.
\item A standard $\frac{l}{240}$ deflection criteria is used.
\item Standard indoor theatrical conditions are used. When necessary,
  the most conservative assumptions (e.g. load duration of 10 years)
  are employed. For less conservative conditions, the load rating can
  be improved by applying the appropriate adjustment factors to the
  allowable stress constants.
  
\end{itemize}

\newtheorem*{defs}{Definitions}
\begin{defs}
  Let a platform $\mathscr{P} = <l,w,h>$ be a simple
  platform\footnote{This rather odd bit of notation expresses the
    fact that we are defining a simple platform in terms of its
    length, width, and height.  A pure mathematician might point out
    that by doing so we have defined a \emph{platform space} that is a
    subset of $\R^{+^3}$. This is an excellent reason to keep pure
    mathematicians far away from the theater.} with length $l$ and
  width $w$. It is supported by four columns of height $h$ at the four
  corners.

  We require the following data about the materials
  used to construct the platform:
  \begin{itemize}
  \item For the beam in the $x$ direction, the modulus of elasticity
  $E_x$, moment of inertia $I_x$, allowable bending stress $F_{b,x}$,
  and section modulus $S_x$.
  \item For the beam in the $y$ direction, the analogous quantities
  $E_y$, $I_y$, $F_{b,y}$, and $S_y$.
  \item For the columns (in the $z$ direction), the cross-sectional
  column area $A$, the modulus of elasticity $E_z$, the allowable
  compressive stress $F_c^*$, and the minimum dimension $d$ of the column.
  \end{itemize}
  
\end{defs}

\newtheorem*{foo}{Result}
\begin{foo}
  For a platform $\mathscr{P}$ defined as above, we want to find the
  maximum allowable load per square foot, which we call
  $P(\mathscr{P})$.  We can compute this by taking the smallest of
  five values:
  \[P(\mathscr{P}) = \min
  \begin{cases}
    0.64 \frac{E_x I_x}{1200 l^3 w} \\
    0.64 \frac{E_y I_y}{1200 w^3 l} \\
    \frac{8 F_{b,x} S}{w l^2}\\
    \frac{8 F_{b,y} S}{l w^2} \\
    \frac{4}{l w} A F_c^* \left( 0.625 - 0.1875 \frac{E_z d^2}{h^2 F_c^*} - \sqrt {0.03516
      \frac{E_z^2 d^4}{h^4 F_c^{*^2}} - 0.1406 \frac{E_z d^2}{h^2 F_c^*} +
      0.3906} \right) 
\end{cases}
  \]
\end{foo}
\begin{proof}[Derivation]
  We first consider the beams running along the long axis of the
  platform. These beams can fail due to bending or shear, or they can
  provide an unsatisfactory level of deflection (the $\frac{l}{240}$
  criteria previously mentioned). In practice, under theatrical
  loading conditions, beam failure due to shear is very rare, and
  minimizing deflection generally is the limiting factor.

  The deflection $\Delta$ is given by
  \[\Delta = \frac{5 p l^4}{384 E I}\]
  where $p$ is the weight per linear foot of beam length. Since we are
  interested in maximum weight per square foot, we adjust it by
  half the width, the area supported by each beam: the maximum
  allowable weight in square feet, $P = \frac{2p}{w}$. Applying the
  $\frac{l}{240}$ deflection criteria,
  \begin{align*}
    \frac{l}{240} &\ge \frac{5 P l^4 w}{768 E I} \\
    P &\le \frac{768 E I}{1200 l^3 w} = 0.64 \frac{E I}{1200 l^3 w}
  \end{align*}
  This gives us the maximum allowable weight due to deflection in the
  length axis. By interchanging length and width, we come to the
  corresponding limit due to deflection in the other beam:
  \[P \le 0.64 \frac{E I}{1200 w^3 l}\]

  Failure due to bending is also a concern. This imposes the
  following constraint:
  \begin{align*}
    M_{\max} &\le F_b S \\
    \frac{pl^2}{8} &\le F_b S \\
    \frac{P w l^2}{8} &\le F_b S \\
    P &\le \frac{8 F_b S}{w l^2}
  \end{align*}
  and in the other beam,
  \[ P \le \frac{8 F_b S}{l w^2} \]

  Next, we consider the columns supporting the platform. We assume the
  load is distributed evenly among the four columns, i.e. that each
  column supports a weight of $\frac{P l w}{4}$. We first identify the
  critical buckling design value
  \[F_{cE} = 0.3\frac{E d^2}{h^2}\]
  and then compute the column stability factor
  \begin{align*}
    C_p &= \frac{1+\frac{F_{cE}}{F_c^*}}{1.6} - \sqrt{
    \left[\frac{1+\frac{F_{cE}}{F_c^*}}{1.6}\right]^2 -
    \frac{\frac{F_{cE}}{F_c^*}}{0.8}}\\
   &= 0.625 - 0.625 \frac{F_{cE}}{F_c^*} - \sqrt{0.3906
    \frac{F_{cE}^2}{F_c^{*^2}} - 0.469 \frac{F_{cE}}{F_c^*} + 0.3906}\\
   &= 0.625 - 0.1875 \frac{E d^2}{h^2 F_c^*} - \sqrt {0.03516
     \frac{E^2 d^4}{h^4 F_c^{*^2}} - 0.1406 \frac{E d^2}{h^2 F_c^*} + 0.3906}
  \end{align*}
  which leads to the maximum allowable compressive load
  \[ P \le \frac{4}{l w} A F_c^* \left( 0.625 - 0.1875 \frac{E d^2}{h^2 F_c^*} - \sqrt {0.03516
      \frac{E^2 d^4}{h^4 F_c^{*^2}} - 0.1406 \frac{E d^2}{h^2 F_c^*} +
      0.3906} \right) \]


We take the minimum of these maximum loads, i.e. the most restrictive,
and it is our final result for the maximum allowable load.
\end{proof}

\section*{References}
\begin{itemize}
\item Holden and Sammler, \emph{Structural Design for the Stage},
  Focal Press 1999
\item Lots and lots of math
\end{itemize}


\end{document}
