Mechanical engineering question

Question

There is a plant growing at a constant mass rate (kg/s). It develops into a large tree that has radius r(t) that grows thicker over time and maintains constant vertical deflection at one end. How much does the tree thicken as a function of time for the constant deflection to be maintained? The tree has a density of p. Calculate an expression for r(t) in terms of the initial radius r_0, m, g, p, L, and time (t).

Answer 1

Here is my attempt at this problem:

I am taking the tree to be a cantilever beam, deflecting due to its self-weight. As the tree grows in length and radius, the deflection at the tip remains the same.

deflection at the tip of a cantilever beam with a distributed load can be expressed as

Δ = wL⁴/(8EI) -----> (equation 1)

Note that since the length and radius are changing in time, I, w, and L are actually I(t), w(t), and L(t), but I will omit this notation for the sake of readability.

w = pgπr² (density • g • area)

I = πr⁴/4 (moment of inertia for a circle)

So we can plug these into equation 1 and simplify:

Δ = pgL⁴/(2Er²)

rearranging this to solve for L:

L = [2EΔ/(pg)]^1/4 • √r

Let's set [2EΔ/(pg)]^1/4 to c_Δ for readability, so

L = c_Δ • √r -----> (equation 2)

and recall L and r are the only variables that change with time.

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Now let's look at the mass:

p = m/V (density is mass/volume)

Note that m and V are actually m(t) and V(t), but I will omit this notation for the sake of readability.

V = πr²L so

p=m/(πr²L)

solving for m,

m = pπr²L -----> (equation 3)

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Since the plant is growing at a constant mass rate, let's differentiate equation 3, and note that m' is a constant, so we will replace it with c_m:

m' = c_m = pπ(2r•r'•L + r²•L') -----> (equation 4)

Realizing we now have L and L', let's differentiate equation 2 to obtain L':

L' = c_Δ•r' / (2√r) -----> (equation 5)

Now we have L and L', and can substitute both into equation 4:

c_m = pπ{2r•r'•c_Δ•√r + r²•c_Δ•r' / (2√r)}

Now we have a differential equation in terms of r. When we simplify, the terms in the {curly brackets} add nicely:

c_m/(pπ) = 5/2•r^3/2•r'•c_Δ

so we get

r^3/2•r' = 2c_m/(5pπc_Δ)

If we set 2c_m/(5pπc_Δ) to c_r,

r^3/2•r' = c_r -----> (equation 6)

I used wolfram alpha to solve this differential equation:

r(t) = (5/2)^2/5•(c_r•t + c₁)^2/5 -----> (equation 7)

Note that we defined c_r, but c₁ comes from the constant in solving a differential equation, so we can use boundary conditions to solve for c₁.

We will use r(0) = r_o:

r(0) = r_o = (5/2)^2/5•(c₁)^2/5

solving for c₁, we get

c₁ = (2/5)•r_o^5/2

So we now have an equation for r(t):

r(t) = (5/2)^2/5•(c_r•t + (2/5)•r_o^5/2)^2/5

where c_r = 2c_m/(5pπc_Δ)

and c_Δ = [2EΔ/(pg)]^1/4

and c_m is the constant mass rate

and Δ is the constant deflection at the tip

and E is the elastic modulus of the tree

p, g, and r_o were given.

Please let me know if you see any mistakes in my work, since there are so many steps