advanced32 min studyLesson 164 in Physics

Build local momentum balance for a solid

Connect surface tractions, stress divergence, body forces, inertia, and boundary data in a continuum equation of motion.

Before you begin

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The question for today

What you’ll understand

Derive the local equation of motion from a material-volume force balance.
Keep body force per mass distinct from force per volume.
Identify boundary and initial data needed for a solid mechanics problem.

A nonzero stress does not automatically accelerate a solid. Opposite faces may transmit large forces that cancel. Acceleration instead responds to the imbalance between neighboring tractions together with distributed body forces. The stress divergence captures that local imbalance. Deriving it from Newton's law shows why stress gradients, force-density units, and boundary data must be considered together before adding a material's stress–strain law.

Apply Newton's law to the same moving material

Take a material volume that follows the solid rather than a fixed box through which matter can flow. Its rate of change of momentum equals the total surface traction plus the body force acting throughout its volume. The surface traction on its outward normal is supplied by the Cauchy stress tensor. A divergence theorem converts this surface integral into a volume integral. Because the material region can be chosen arbitrarily within a smooth part of the body, the integrands give a local balance.

Use $b_i$ for body force per unit mass, such as gravitational acceleration, with units $\mathrm{m\,s^{-2}}$ . Then $\rho b_i$ is force per current volume. With the force-index-first convention, the ith stress-divergence component is $\partial\sigma_{ij}/\partial x_j$ . The acceleration is material acceleration, following a particular piece. A finite spatial description uses $D\mathbf v/Dt=\partial_t\mathbf v+(\mathbf v\cdot\nabla)\mathbf v$ , so a steady velocity field does not necessarily imply zero material acceleration.

\rho\frac{D v_i}{Dt}=\frac{\partial\sigma_{ij}}{\partial x_j}+\rho b_i

Every term is a force density in N/m³: density times acceleration equals stress gradient plus body force per volume. Spatial derivatives and Cauchy stress refer to the current configuration.↔ Scroll wide equations horizontally.

Equilibrium is a balance, not an absence of stress

In static equilibrium the acceleration vanishes, leaving a differential constraint on the stress field. For a vertical bar, gravity creates a stress gradient because an upper cross-section must support more material below it than a lower cross-section. Uniform tension without a body force instead has zero stress divergence in the interior. It can still transmit a nonzero force between its ends, just as a stationary rope can carry tension without accelerating.

Check this equation first by units and then by a small-element free-body diagram. Stress measured in pascals must be differentiated with respect to length before it can balance a volumetric force. If a symbol instead denotes a body force already measured in newtons per cubic metre, multiplying it by density again would be incorrect. The divergence equation also does not determine all six stress components by itself; kinematics and constitutive equations supply the additional structure.

\frac{d\sigma_{zz}}{dz}-\rho g=0,\qquad \sigma_{zz}(0)=0

For a uniform vertical bar with z measured upward from a free lower end, gravity has bz = −g. Static equilibrium gives dσzz/dz = ρg. The free-end condition sets the integration constant.↔ Scroll wide equations horizontally.

A differential equation needs physically compatible data

On a boundary one typically prescribes either displacement components or the corresponding traction components. A clamped end constrains motion and develops reaction forces; those reactions are part of the solution, not independent extra data to specify arbitrarily. A free surface has zero applied traction, which does not require every component of the interior stress tensor to vanish. Specifying all displacements and all tractions on the same boundary generally overconstrains a problem unless the two sets happen to be compatible.

For dynamics, initial displacement and initial velocity are needed as well. In a small-deformation formulation about a fixed reference state, material acceleration is approximated by the second time derivative of displacement and reference/current coordinates are identified to the retained order. That approximation must not silently be substituted into a finite-deformation problem. Pure traction problems also admit rigid translations and rotations unless constraints or conventions remove them; force and moment balance are necessary for a static solution.

\mathbf u=\overline{\mathbf u}\ \text{on }\Gamma_u,\qquad \boldsymbol\sigma\mathbf n=\overline{\mathbf t}\ \text{on }\Gamma_t,\qquad \Gamma_u\cap\Gamma_t=\varnothing

This schematic split prescribes either the full displacement or full traction on separate boundary pieces; mixed component conditions are also possible. Dynamic solutions additionally require consistent initial displacement and velocity.↔ Scroll wide equations horizontally.

Work it through

The stress in a hanging bar

A uniform vertical bar of density $2000\,\mathrm{kg\,m^{-3}}$ , length $3\,\mathrm m$ , and area $0.01\,\mathrm{m^2}$ hangs from its upper end. Use $g=10\,\mathrm{m\,s^{-2}}$ and neglect deformation when evaluating its weight. Find stress at height z above the free lower end and the support force.

Choose z upward. Static balance is $d\sigma_{zz}/dz=\rho g=20000\,\mathrm{N\,m^{-3}}$ . The lower end has zero traction, so $\sigma_{zz}(0)=0$ .
Integrate: $\sigma_{zz}(z)=20000z\,\mathrm{Pa}$ when z is measured in metres. At the upper end, stress is $60000\,\mathrm{Pa}=60\,\mathrm{kPa}$ in tension.
Multiply top stress by area: $F=60000\times0.01=600\,\mathrm N$ . Independently the mass is $\rho AL=60\,\mathrm{kg}$ , whose weight is also 600 N.

The result

The traction gradient supports the distributed weight. A material law would be needed to calculate how far this bar elongates, but not to establish this statically determined axial force.

The living notebook

Explore the model

Interactive lab

Interactive lab · fields

A field can vary from point to point

Set rotation Ω to zero and predict how reversing source strength q reverses the displayed vectors. Inspect the softened field q r/(r² + ε), then add Ω to see a rotational contribution. This illustrates spatially varying vector fields only; it does not show a stress tensor, its divergence, or a solid satisfying momentum balance.

Local field directionSource strength

Fx at (1, 0): 0.833
Fy at (1, 0): 0.5
Field magnitude there: 0.972

Source strength q1Rotation Ω0.5

t = 0 s

The softened illustration is q·r/(r² + 0.2) + Ω(−y, x). Moving markers indicate local direction, not integrated particle paths or a Maxwell solution. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA local field value and its spatial variation carry different information.

Pause & reason

Check your understanding

Interactive checks need JavaScript. You can still read the explanation.

Read the checkpoint explanation

Stress has units N/m² and differentiation with respect to position adds a factor of inverse length. The resulting N/m³ balances density times acceleration, not a total force.

Your turn

Try it yourself

A one-dimensional material has constant density $5\,\mathrm{kg\,m^{-3}}$ , no body force, and $\sigma_{xx}=12x\,\mathrm{Pa}$ when x is in metres. Find its instantaneous material acceleration. Explain why uniform stress would give a different answer.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Differentiate stress before dividing by density. The coefficient 12 has units Pa/m.

Compare with the solution

The stress gradient is $12\,\mathrm{N\,m^{-3}}$ , hence $a_x=12/5=2.4\,\mathrm{m\,s^{-2}}$ . A spatially uniform stress has zero gradient, so with no body force it would produce zero local acceleration. Boundary conditions and a constitutive model would still be needed to construct a complete motion realizing either field.

Keep the idea

Take it with you

Stress divergence measures the local imbalance of transmitted surface forces.
Body force per mass must be multiplied by density to enter a volumetric momentum balance.
Boundary conditions, constitutive laws, and dynamic initial data turn a balance identity into a predictive problem.

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