Mechanics of a Growing Bar

We consider a rod of width $w$ and time-varying height $h(t)$ that grows symmetrically on its apical and basal sides. Material is deposited with a constant velocity and is added in a completely stress-free state.

1. Kinematics and Equilibrium

The stress-strain relation for the rod is given by: \(\sigma = E(\epsilon - \epsilon_p)\) where $E$ is the Young’s modulus, $\epsilon$ is the total strain, and $\epsilon_p$ is the pre-strain (or growth strain).

Normal Force

The normal force balance is: \(N = \int_{-h(t)/2}^{h(t)/2} E(\epsilon(x, t) - \epsilon_p(x, y)) \, dy\)

Defining the average pre-strain over the cross-section as $\epsilon_0(x,t) = \frac{1}{h(t)} \int_{-h(t)/2}^{h(t)/2} \epsilon_p(x,y) \, dy$, the normal force can be rewritten as: \(N = A(t)E(\epsilon(x,t) - \epsilon_0(x,t))\) where $A(t) = w h(t)$ is the cross-sectional area.

Bending Moment

The bending moment is defined as: \(M = -\int_{-h(t)/2}^{h(t)/2} y E(\epsilon(x, t) + y\kappa(x, t) - \epsilon_p(x, y)) \, dy\)

We implicitly define the natural curvature, $\kappa_0$, through the first moment of the pre-strain field: \(\kappa_0(x, t)\, I(t) = \int_{-h/2}^{h/2} y\, \epsilon_p(x, y)\, dy\) where $I(t)$ is the second moment of area.

Interpretation of Natural Curvature: This definition makes $\kappa_0$ the pre-strain-weighted average curvature — it captures how the pre-strain is distributed asymmetrically through the thickness. Intuitively, if more pre-strain is accumulated on the apical side ($y > 0$) than the basal side, the bar tends to curve, and $\kappa_0$ quantifies that tendency.

Using this definition, the bending moment simplifies to: \(M = -EI(\kappa - \kappa_0)\) Thus, the bar is stress-free in bending when the actual curvature $\kappa$ equals the natural curvature $\kappa_0$. This is entirely analogous to the natural length of a spring — $\kappa_0$ is the curvature the bar would adopt if unconstrained.

2. Growth Conditions

The growth function $h(t)$ is a monotonically increasing function of time $t$.

For any material point at vertical coordinate $y > h(0)/2$, there exists a unique birth time $t_y>0$ such that the boundary of the rod reaches its position, i.e., $h(t_y)/2 =

y

$.

This relation can be inverted to find the time of deposition:

\[t_y = h^{-1}(2|y|)\]

Because material is added in a stress-free state, the stress must vanish at the boundary at the moment of deposition. This yields the stress-free state condition at birth:

\[\epsilon(x, t_y) + y \kappa(x, t_y) = \epsilon_p(x, y)\]

Equivalently, we can write

\[\epsilon(x, t) \pm \frac {h(t)} 2 \kappa(x, t) = \epsilon_p(x, \pm h(t)/2)\]

3. Evolution Equations

We recall that the two natural strain measures are defined as:

\[\epsilon_0(x,t) = \frac{1}{A(t)}\int_{-h/2}^{h/2} \epsilon_p(x,y)\,dy, \qquad \kappa_0(x,t) = \frac{1}{I(t)}\int_{-h/2}^{h/2} y\,\epsilon_p(x,y)\,dy\]

with $A(t) = h(t)$ and $I(t) = h^3(t)/12$/

Evolution of $\epsilon_0$

Differentiate $\epsilon_0 A = \int_{-h/2}^{h/2} \epsilon_p\,dy$ with respect to time using the Leibniz rule:

\[\dot{\epsilon}_0 A + \epsilon_0 \dot{A} = \frac{\dot{h}}{2}\left[\epsilon_p\!\left(x,+\tfrac{h}{2}\right) + \epsilon_p\!\left(x,-\tfrac{h}{2}\right)\right]\]

Substituting the boundary pre-strain $\epsilon_p(x, \pm h/2) = \pm\frac{h}{2}\kappa + \epsilon$:

\[\frac{\dot{h}}{2}\left[\left(\tfrac{h}{2}\kappa + \epsilon\right) + \left(-\tfrac{h}{2}\kappa + \epsilon\right)\right] = \dot{h}\,\epsilon\]

Since $\dot{A} = \dot{h}$,

\[\boxed{\dot{\epsilon}_0 = \frac{\dot{A}}{A}(\epsilon - \epsilon_0)=\frac{\dot A}{A}\frac{N}{EA}.}\]

The natural axial strain is pulled toward the current axial strain $\epsilon$ at a rate proportional to the fractional growth rate.

Evolution of $\kappa_0$

Differentiate $\kappa_0 I = \int_{-h/2}^{h/2} y\,\epsilon_p\,dy$ with Leibniz:

\[\dot{\kappa}_0 I + \kappa_0 \dot{I} = \frac{\dot{h}}{2}\left[\frac{h}{2}\,\epsilon_p\!\left(x,+\tfrac{h}{2}\right) + \left(-\frac{h}{2}\right)\epsilon_p\!\left(x,-\tfrac{h}{2}\right)\right]\]

Substituting the boundary pre-strains:

\[\frac{\dot{h}}{2}\left[\frac{h}{2}\left(\tfrac{h}{2}\kappa+\epsilon\right) - \left(-\frac{h}{2}\right)\left(-\tfrac{h}{2}\kappa+\epsilon\right)\right]\] \[= \frac{\dot{h}}{2}\left[\frac{h}{2}\!\left(\tfrac{h}{2}\kappa+\epsilon\right) + \frac{h}{2}\!\left(\tfrac{h}{2}\kappa-\epsilon\right)\right] = \frac{\dot{h}\,h^2}{4}\kappa = \dot{I}\,\kappa\]

Dividing through by $I$:

\[\dot{\kappa}_0 = \frac{\dot{I}}{I}(\kappa - \kappa_0)=\frac{\dot I}{I}\frac{M}{EI}\]