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:
\[\sigma = E(e- \epsilon_p)\]where $E$ is the Young’s modulus, and
\[e(x,y,t)=\epsilon(x,t)-y \kappa(x,t)\]is the total strain, and $\epsilon_p$ is the pre-strain (or growth strain).
Normal Force
The normal force balance is:
\[N = \int_{-h(t)/2}^{h(t)/2} E(\epsilon(x, t) - y\kappa(x, t) - \epsilon_p(x, y)) \, dy\]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:
\[N(x,t) = A(t)E(\epsilon(x,t) - \epsilon_0(x,t))\]where $A(t) = h(t)$ is the cross-sectional area.
Bending Moment
The bending moment is defined as:
\[M = \int_{-h(t)/2}^{h(t)/2} E(y^2\kappa(x, t) + y \epsilon_p(x, y)) \, dy\]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:
\[M(x,t) = EI(t)(\kappa(x,t) - \kappa_0(x,t))\]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 given monotonically increasing function of time $t$.
| For any material point at vertical coordinate $y$ satisfying $ | 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 at that particular $y$:
\[t_y = h^{-1}(2|y|).\]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:
\[\begin{aligned} 0&=\sigma\left(x,\pm \frac{h(t)}2,t\right)=E(e(x,\pm h(t)/2,t)-\epsilon_p(x,\pm h(t)/2,t)) \\ &=E\left(\epsilon(x, t) \mp \frac {h(t)} 2 \kappa(x, t) -\epsilon_p(x,\pm h(t)/2,t)\right) \end{aligned}\]which implies
\[\epsilon_p\left(x, \pm \frac{h(t)}2\right) = \epsilon(x, t) \mp \frac {h(t)} 2 \kappa(x, t).\]Equivalently, we can write
\[\epsilon(x, t_y) + y \kappa(x, t_y) = \epsilon_p(x, y).\]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 $\epsilon_p(x, \pm h/2) = \mp\frac{h}{2}\kappa + \epsilon$:
\[\frac{\dot{h}}{2}\left[\left(\tfrac{h}{2}\kappa + \epsilon\right) + \left(-\tfrac{h}{2}\kappa + \epsilon\right)\right] = \dot{h}\,\epsilon\]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 and substitute the boundary pre-strains:
\[\begin{aligned} \dot{\kappa}_0 I + \kappa_0 \dot{I} &= - \frac{\dot{h}}{2}\left[\frac{h}{2}\,\epsilon_p\!\left(x,+\tfrac{h}{2}\right) + \left(-\frac{h}{2}\right)\epsilon_p\!\left(x,-\tfrac{h}{2}\right)\right] \\ &= - \frac{\dot{h}}{2}\left[\frac{h}{2}\left(-\tfrac{h}{2}\kappa + \epsilon\right) - \frac{h}{2}\left(\tfrac{h}{2}\kappa + \epsilon\right)\right] = - \frac{\dot{h}\,h^2}{4}(-\kappa) = \dot{I}\,\kappa \end{aligned}\]Dividing through by $I$:
\[\dot{\kappa}_0 = \frac{\dot{I}}{I}(\kappa - \kappa_0)=\frac{\dot I}{I}\frac{M}{EI}\]4. Rate of Recoverable Energy and Dissipation
Taking the time derivative of the recoverable bending energy
\[\mathcal E = \int_0^L \frac{1}{2} EI (\kappa - \kappa_0)^2 \, dx\]gives:
\[\frac{d \mathcal E}{dt} = \int_0^L \left[ EI(\kappa - \kappa_0)(\dot{\kappa} - \dot{\kappa}_0) + \frac{1}{2} E\dot{I}(\kappa - \kappa_0)^2 \right] dx\]Using $M = EI(\kappa - \kappa_0)$ and the evolution equation $\dot{\kappa}_0 = \frac{\dot{I}}{I}(\kappa - \kappa_0)$, we obtain:
\[\int_0^L M\dot{\kappa} \, dx - \int_0^L \frac{1}{2} E\dot{I}(\kappa - \kappa_0)^2 \, dx = \frac{d \mathcal E}{dt}\]This shows that the rate of change of recoverable energy equals the mechanical work of the bending moments minus the dissipation $\mathcal{D}$, where:
\[\mathcal{D} = \int_0^L \frac{1}{2} E\dot{I}(\kappa - \kappa_0)^2 \, dx\]5. Energy Decomposition and Latent Energy
Assuming pure bending ($\epsilon = 0$), the total strain energy of the rod is:
\[U = \int_0^L \int_{-h/2}^{h/2} \frac{1}{2} E (-y\kappa - \epsilon_p)^2 \, dy \, dx\]We decompose the integrand by adding and subtracting $y\kappa_0$: $(-y\kappa - \epsilon_p) = -y(\kappa - \kappa_0) - (y\kappa_0 + \epsilon_p)$. Expanding the square yields three terms:
- Recoverable Energy:
- Latent Energy ($B$):
- Cross Term:
By the definition of natural curvature ($\kappa_0 I = -\int_{-h/2}^{h/2} y\epsilon_p \, dy$), the inner integral of the cross term evaluates exactly to zero. Thus, the total energy is simply the sum of the recoverable bending energy and the latent energy $B$.
6. Rate of Latent Energy
Taking the time derivative of the latent energy density $B$ using the Leibniz rule gives:
\[\dot{B} = \frac{\dot{h}}{4} E \left( \frac{h}{2}\kappa_0 + \epsilon_p(h/2) \right)^2 + \frac{\dot{h}}{4} E \left( -\frac{h}{2}\kappa_0 + \epsilon_p(-h/2) \right)^2 + \int_{-h/2}^{h/2} E(y\kappa_0 + \epsilon_p)y\dot{\kappa}_0 \, dy\]The integral term vanishes because $\int (y^2\kappa_0 + y\epsilon_p) \, dy = 0$. Substituting the boundary pre-strains $\epsilon_p(\pm h/2) = \mp \frac{h}{2}\kappa$ into the boundary terms yields:
\[\dot{B} = \frac{\dot{h}}{4} E \left[ \left( \frac{h}{2}(\kappa_0 - \kappa) \right)^2 + \left( -\frac{h}{2}(\kappa_0 - \kappa) \right)^2 \right] = \frac{E\dot{h}h^2}{8} (\kappa_0 - \kappa)^2\]Since the rate of change of the second moment of area is $\dot{I} = \frac{d}{dt}(h^3/12) = \frac{h^2\dot{h}}{4}$, we obtain the result:
\[\dot{B} = \frac{1}{2} E\dot{I} (\kappa - \kappa_0)^2\]This confirms that the dissipated external work is perfectly accounted for as the rate of accumulation of latent energy in the newly deposited material layers.