In nature, the failure of structural members usually result from large torque because of twisting or bending rather than due to tensile or compressive strains. This process of structural breakdown is called buckling and in cases of tall cylindrical structures like trees, the torque is caused by its own weight bending the structure. Thus the vertical through the centre of gravity does not fall within the base. The elastic torque caused because of this bending about the central axis of the tree is given by $\dfrac{Y \pi r^{4}}{4R}$. $Y$ is the Young's modulus, $r$ is the radius of the trunk and $R$ is the radius of curvature of the bent surface along the height of the tree containing the centre of gravity (the neutral surface). Estimate the critical height of a tree for a given radius of the trunk.

In nature, the failure of structural members usually result from large torque because of twisting or bending r

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\dfrac{Y \pi r^{4}}{4R}

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Concept understanding — Youngs Modulus

Young’s Modulus: The Stretchiness of a Solid

When you pull on a rubber band, it stretches easily. When you pull on a steel rod of the same size, it barely moves. Both are elastic — they return to their original shape when you let go — but they resist stretching very differently. Young’s modulus is the number that tells you exactly how much a material resists being stretched or compressed lengthwise.

The Intuition: Stiffness per Unit Size

Think of a spring. A stiff spring requires a large force to stretch it a little. A soft spring stretches a lot with a small force. Young’s modulus is like the “stiffness” of a material, but it’s cleverly designed to be independent of the object’s shape and size.

If you take a thick steel rod and a thin steel wire of the same length, the rod is harder to stretch. That’s because you’re pulling on more material. Young’s modulus removes this size effect — it tells you the stiffness of the material itself, not the particular piece you’re holding.

The Precise Definition

Young’s modulus ( $E$ or $Y$ ) is defined as the ratio of tensile stress to tensile strain, as long as the material obeys Hooke’s law (the deformation is reversible and proportional to the force).

Y = \frac{\text{Tensile Stress}}{\text{Tensile Strain}}

Let’s break down the two parts.

Tensile Stress ( $\sigma$ ) is the force per unit area. If you pull with a force $F$ on a rod of cross-sectional area $A$ , the stress is:

\sigma = \frac{F}{A}

Stress has units of pressure — pascals (Pa) or $\text{N/m}^2$ . It tells you how “intense” the pulling is, regardless of the rod’s thickness.

Tensile Strain ( $\varepsilon$ ) is the fractional change in length. If the original length is $L_0$ and it stretches by $\Delta L$ , the strain is:

\varepsilon = \frac{\Delta L}{L_0}

Strain is a pure number — it has no units. A strain of 0.01 means the rod stretched by 1% of its original length.

Putting it together:

Y = \frac{F/A}{\Delta L / L_0} = \frac{F L_0}{A \Delta L}

What the Number Tells You

A high Young’s modulus means the material is very stiff — it takes a huge stress to produce even a tiny strain. Steel has $Y \approx 200 \times 10^9$ Pa. A low Young’s modulus means the material is easily stretched. Rubber has $Y \approx 0.01 \times 10^9$ Pa — about 20,000 times smaller than steel.

Note

Young’s modulus is only valid in the elastic region — where the material returns to its original shape after the force is removed. If you stretch too far (past the elastic limit), the material deforms permanently or breaks, and Young’s modulus no longer applies.

A Worked Example

A steel wire of length 2.0 m and cross-sectional area $1.0 \times 10^{-6} \text{ m}^2$ is pulled by a force of 100 N. How much does it stretch? (Young’s modulus of steel = $2.0 \times 10^{11}$ Pa)

From $Y = \frac{F L_0}{A \Delta L}$ , rearrange:

$\Delta L = \frac{F L_0}{A Y} = \frac{100 \times 2.0}{(1.0 \times 10^{-6}) \times (2.0 \times 10^{11})} = \frac{200}{2.0 \times 10^5} = 1.0 \times 10^{-3} \text{ m} = 1.0 \text{ mm}$ …

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A tree buckles under its own weight when the gravitational torque exceeds the elastic restoring torque. Balancing these at the critical height gives $h_{\text{crit}} \propto r^{2/3}$ , specifically $h_{\text{crit}} = \left(\frac{4Y}{\rho g}\right)^{1/3} r^{2/3}$ .

Why trees have a height limit

A tall tree is a cantilever beam loaded by its own weight. As it grows taller, gravity tries to bend it over. The tree resists through elastic forces in the trunk—compression on the concave side, tension on the convex side—that produce a restoring torque. Buckling occurs when the gravitational torque from the tree's weight acting through its center of mass exceeds the maximum elastic torque the trunk can supply.

The problem gives us the elastic restoring torque for a bent cylindrical trunk:

\tau_{\text{elastic}} = \frac{Y \pi r^4}{4R}

where $R$ is the radius of curvature of the neutral (unbent central) surface. At the critical height, the tree is on the verge of collapse: the gravitational torque equals the elastic torque, and $R$ becomes comparable to the tree's height itself.

Finding the critical height

1. Gravitational torque about the base

Consider a tree of height $h$ , uniform density $\rho$ , and trunk radius $r$ . Its mass is

m = \rho \cdot \pi r^2 h

and its weight $W = \rho \pi r^2 h g$ acts at the center of gravity, roughly at height $h/2$ . When the tree bends slightly, the vertical line through the center of mass is displaced horizontally from the base. For small bending with radius of curvature $R$ , the horizontal displacement of the top is approximately $\delta \sim \frac{h^2}{2R}$ (from the geometry of a circular arc). The center of mass displaces by roughly half that, so the lever arm is of order $\frac{h^2}{4R}$ .

The gravitational torque trying to topple the tree is then

\tau_{\text{grav}} \sim \rho \pi r^2 h g \cdot \frac{h^2}{4R} = \frac{\rho \pi r^2 g h^3}{4R}.

2. Elastic restoring torque

The trunk resists bending with the given elastic torque:

\tau_{\text{elastic}} = \frac{Y \pi r^4}{4R}.

Notice both torques are inversely proportional to $R$ —the tighter the bend (smaller $R$ ), the larger both torques become.

3. Critical condition

At the critical height $h_{\text{crit}}$ , the tree is marginally stable:

\tau_{\text{grav}} = \tau_{\text{elastic}}.

Substituting:

\frac{\rho \pi r^2 g h_{\text{crit}}^3}{4R} = \frac{Y \pi r^4}{4R}.

The factor $\frac{\pi}{4R}$ cancels: …

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Why trees have a height limit

Finding the critical height

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