Solved: Blowing in the Wind In September 1989, Hurricane Hugo hammered the coast of

Chapter 3, Problem 9

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Blowing in the Wind  In September 1989, Hurricane Hugo hammered the coast of South Carolina with winds estimated at times to be as high as 60.4 m/s(135 mi/h). Of the billions of dollars in damage, approximately $420 million of this was due to the market value of loblolly pine (Pinus taeda) lumber in the Francis Marion National Forest. One image from that storm remains hauntingly bizarre: all through the forest and surrounding region, thousands upon thousands of pine trees lay pointing exactly in the same direction, and all the trees were broken 5 -8 meters from their base. In September 1996, Hurricane Fran destroyed over 8.2 million acres of timber forest in eastern North Carolina. As happened seven years earlier, the planted loblolly trees all broke at approximately the same height. This seems to be a reproducible phenomenon, brought on by the fact that the trees in these planted forests are approximately the same age and size.

                                           

In this problem, we are going to examine a mathematical model for the bending of loblolly pines in strong winds, and then use the model to predict the height at which a tree will break in hurricane-force winds.* Wind hitting the branches of a tree transmits a force to the trunk of the tree. The trunk is approximately a big cylindrical beam of length L, and so we will model the deflection y(x) of the beam of length L, and so we will model the deflection y(x) of the tree with the static beam equation \(E I y^{(4)}=w(x)\) (equation (4) in this section), where x is distance measured in meters from ground level. Since the tree is rooted into the ground, the accompanying boundary conditions are those of a cantilevered beam: y(0)=0, \(y^{\prime}(0)=0\) at the rooted end, and \(y^{\prime \prime}(L)=0\), \(y^{\prime \prime \prime}(L)=0\) at the free end, which is the top of the tree.

(a) Loblolly pines in the forest have the majority of their crown (that is, branches and needles) in the upper 50% of their length, so let's ignore the force of the wind on the lower portion of the tree. Furthermore, let's assume that the wind hitting the tree's crown results in a uniform load per unit length \(w_{0}\). In other words, the load on the tree is modeled by

\(w(x)= \begin{cases}0, & 0 \leq x<L / 2 \\ w_{0}, & L / 2 \leq x \leq L\end{cases}\)

We can determine y(x) by integrating both sides of \(E I y^{(4)}=w(x)\). Integrate w(x) on [0, L/2] and then on [L/2, L] to find an expression for EIy"' (x) on each of these intervals. Let \(c_{1}\) be the constant of integration on [0, L/2] and \(c_{2}\) be the constant of integration on [L/2, L]. Apply the boundary condition \(y^{\prime \prime \prime}(L)=0\) and solve for \(c_{2}\). Then find the value of \(c_{1}\) that ensures continuity of the third derivative \(y^{\prime \prime \prime}\) at the point x=L/2.

(b) Following the same procedure as in part (a) show that

\(E I y^{\prime}(x)= \begin{cases}\frac{w_{0}}{8}\left(-2 L x^{2}+3 L^{2} x\right), & 0 \leq x \leq L / 2 \\ \frac{w_{0}}{48}\left(8 x^{3}-24 L x^{2}+24 L^{2} x-L^{3}\right), & L / 2 \leq x \leq L .\end{cases}\)

Integrate \(E I y^{\prime}\) to obtain the deflection y(x).

(c) Note that in our model y(L) describes the maximum amount by which the loblolly will bend. Compute this quantity in terms of the problem's parameters.