MECHANICS  CASE STUDY SOLUTION

Observation Platform 

A 6 ft by 7 ft observation platform is supported by a single 6x8 wood post. The
maximum weight of the observer is 300 lb. It is assumed that the observer can stand
within 6 inches of the railing in any corner. If the observer stands in one of
the four corners, the distance from the post center will be 3 ft and 2.5 ft.
The observer weight will cause a bending moment about both the y and z axis
that are in the plane of the platform, as shown in the diagram. This results in
unsymmetric bending stress. The observer weight also causes a direct axial
compression force on the post that is in addition to the bending moments. 



Observer Location and Post Dimensions 

Bending Moment


The unsymmetric bending moment is best determined by calculating the moment
about the y and z axis separately. These moments are
M_{y} = (2.5 ft)(12 in/1ft)(300 lb) = 9,000 inlb
M_{z} = (3 ft)(12 in/1 ft)(300 lb) = 10,800 inlb
Both moments are positive since they generate a moment vector that points along
both positive axes. The
righthand rule is helpful in determining the moment
vector direction where the fingers rotate around the axes in the rotation direction
of the moment and the thumb points in the direction of the moment vector. 



Moment of Inertia



The moment of inertia needs to be calculated for the member cross section. The
exact dimensions of the 6x8 wood member is 5.5 in by 7.5 in (See Structure
Shapes appendix).
I_{y} = (5.5 in)^{3} (7.5 in)
/ 12 = 104.0 in^{4}
I_{z} = (7.5 in)^{3} (5.5 in)
/ 12 = 193.4 in^{4} 





Neutral Axis (NA)

Neutral Axis Location 

To help locate the point where the bending stress is maximum, it is useful to
plot the NA. This is done by setting the unsymmetric bending stress equation
equal to zero.
0 = M_{y} z / I_{y}  M_{z} y
/ I_{z}
0 = 9,000 z / 104.0  10,800 y / 193.4
y/z =86.54/55.84 = 1.550
This line is plotted in the diagram at the left. Notice, the furthest perpendicular
point from the NA is either point A or B. Point A will give the highest compression
stress where point B will give the highest tension stress. 





Bending Stress



The unsymmetric bending stress can be calculated from the equation,
Point A (diagram above) is one of two points that are the furthest away from the neutral axis which will give the largest bending stress. The location of point A is y = 3.75 in, and z = 2.75 in, which gives
σ_{b} = 9,000 (2.75) / 104.0  10,800 (3.75) / 193.4
=  238.0  209.4 psi
=  447.4 psi
The negative sign indicates a compression stress. 





Normal Axial Compression Stress

Bending and Axial Stress Distribution 

In addition to the bending stress, there is an axial stress caused
by the vertical weight of the observer. This stress is
σ_{a} = P / A = 300 lb / (5.5 in)(7.5 in) = 7.273
psi
This stress is significantly lower than the bending stress and could be ignored,
but it is always good to check.
The total compression stress is
σ = σ_{b} + σ_{a} =
 447.4  7.3 =  454.7 psi 


