# Changes

,  17:16, 13 April 2010
Line 31: Line 31:
Several key engineering aspects of the platform have to be reviewed. First, we have to look at the structural rigidity of the "floor". Making the assumptions that the beams used at 2"x4" wooden beams we can say that the [http://en.wikipedia.org/wiki/Young's_modulus Youngs Modulus] of the wood is 12GPa. Next we use Equation 1 to determine the beams moment of inertia.

Several key engineering aspects of the platform have to be reviewed. First, we have to look at the structural rigidity of the "floor". Making the assumptions that the beams used at 2"x4" wooden beams we can say that the [http://en.wikipedia.org/wiki/Young's_modulus Youngs Modulus] of the wood is 12GPa. Next we use Equation 1 to determine the beams moment of inertia.
−
$I=\frac{1}{12}bh^3$&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(1)
+
$I=\frac{1}{12}bh^3$&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(1)<ref name="Mechanical of Materials">["'Mechanicals of Materials': Engineering Textbook". R. C. Hibbeler. Seventh Edition. 2008.  Copyright Pearson Education. 2008]</ref>

From this we determine the moment of inertia to be  $1.11x10^{-6} m^4$. These are the material constant elected to be using in the platform. Now we set the maximum defection of the beams to be 1cm. with this information a short iterative process can be applied to find the optimum beam length to load. To do this the floor is models as a series of beam with fixed ends and a uniform applied load. From this model Equation 2 is derived to describe the deflection

From this we determine the moment of inertia to be  $1.11x10^{-6} m^4$. These are the material constant elected to be using in the platform. Now we set the maximum defection of the beams to be 1cm. with this information a short iterative process can be applied to find the optimum beam length to load. To do this the floor is models as a series of beam with fixed ends and a uniform applied load. From this model Equation 2 is derived to describe the deflection
−
$EIv=\frac{\omega L}{12}x^3-\frac{\omega}{24}x^4-\frac{M'}{24}x^2$&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(2)
+
$EIv=\frac{\omega L}{12}x^3-\frac{\omega}{24}x^4-\frac{M'}{24}x^2$&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(2)<ref name="Mechanical of Materials"></ref>

Because the beam is redundant the moment M' must be solved for from the boundary conditions. Once done the moment is found to be

Because the beam is redundant the moment M' must be solved for from the boundary conditions. Once done the moment is found to be
−
$M'=\frac{\omega L^2}{12}$&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(3)
+
$M'=\frac{\omega L^2}{12}$&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(3)<ref name="Mechanical of Materials"></ref

Substituting Equation 3 back into Equation 2 and solving for the length L we are left with

Substituting Equation 3 back into Equation 2 and solving for the length L we are left with
−
$L=\sqrt[4]{\frac{-384EIv}{\omega}}$&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(4)
+
$L=\sqrt[4]{\frac{-384EIv}{\omega}}$&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(4)<ref name="Mechanical of Materials"></ref

We determine the ultimate strength of wood to be:

We determine the ultimate strength of wood to be:
−
$\sigma_{ut}=3.0MPa$ for tension
+
$\sigma_{ut}=3.0MPa$ for tension<ref name="Mechanical of Materials"></ref
−
$\sigma_{uc}=30.0MPa$ for compression
+
$\sigma_{uc}=30.0MPa$ for compression<ref name="Mechanical of Materials"></ref
−
$\sigma_{us}=6.3MPa$ for shear
+
$\sigma_{us}=6.3MPa$ for shear<ref name="Mechanical of Materials"></ref

to check that the load does not exceed any of these material limits Equations 5 and 6 are used to calculate the shear stress and normal stress respectively

to check that the load does not exceed any of these material limits Equations 5 and 6 are used to calculate the shear stress and normal stress respectively
−
$\sigma_s=\frac{V}{A}$&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(5)
+
$\sigma_s=\frac{V}{A}$&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(5)<ref name="Mechanical of Materials"></ref

For this double fixed beam the shear is:

For this double fixed beam the shear is:
−
$V=\frac{\omega L}{2}$&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(5.1)
+
$V=\frac{\omega L}{2}$&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(5.1)<ref name="Mechanical of Materials"></ref

With the area of the beam

With the area of the beam
−
$A=bh$&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(5.2)
+
$A=bh$&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(5.2)<ref name="Mechanical of Materials"></ref
−
$\sigma_n=\frac{M' y}{I}$&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(6)
+
$\sigma_n=\frac{M' y}{I}$&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(6)<ref name="Mechanical of Materials"></ref

By substituting Equation 1 and 3 in to the above formula the normal stress at any point 'y' can be found.

By substituting Equation 1 and 3 in to the above formula the normal stress at any point 'y' can be found.
123

edits