In September 1989, Hurricane Hugo hammered the coast of South Carolina with winds estimated at times to be as high as $60.4 \mathrm{~m} / \mathrm{s}(135 \mathrm{mi} / \mathrm{h})$. 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 runk is approximately a big cylindrical beam of length $L$, and so we will model the deflection $y(x)$ of the tree with the static beam equation $E I^{(4)}=w(x)$ (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 rootedend, 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 foresthave the majority of theircrown (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
$$
w(x)=\left\{\begin{array}{ll}
0, & 0 \leq x<L / 2 \\
w_{0}, & L / 2 \leq x \leq L
\end{array}\right.
$$
We can determine $y(x)$ by integrating both sides of $E I^{(4)}=w(x)$. Integrate $w(x)$ on $[0, L / 2]$ and then on $[L / 2, L]$ to find an expression for $E I y^{\prime \prime \prime}(x)$ 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
$$
E I y^{\prime}(x)=\left\{\begin{array}{ll}
\frac{w_{0}}{8}\left(-2 L x^{2}+3 L^{2} x\right), & 0 \leq x \leq L / 2 \\
\frac{w_{0}}{48}\left(8 x^{3}-24 L x^{2}+24 L^{2} x-L^{3}\right), & L / 2 \leq x \leq L
\end{array}\right.
$$
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.