Learning topic
Inertia Forces
Use D’Alembert’s principle to calculate inertia forces and dynamic stress in accelerating bars and rotating components, including distributed inertia loading.
When a structural member moves with acceleration, its mass produces an inertia effect that must be considered when determining internal forces and stresses. D'Alembert's principle provides a convenient way to formulate such problems.
D'Alembert's principle formally reduces a motion problem to equilibrium equations by adding inertia forces to the active forces:
$$\mathbf F_i=-m\mathbf a.$$
For translational motion of a particle:
$$\sum \mathbf F+\mathbf F_i=0.$$
For rotation of a rigid body about a fixed axis, the corresponding inertia moment is:
$$M_i=-J\varepsilon.$$
- $\mathbf F_i$ — inertia force;
- $m$ — mass;
- $\mathbf a$ — acceleration;
- $M_i$ — inertia moment;
- $J$ — mass moment of inertia about the rotation axis;
- $\varepsilon$ — angular acceleration.
Inertia forces are calculation quantities directed opposite to the corresponding accelerations. After introducing them, internal forces and stresses can be determined by methods analogous to static analysis.
Translational motion
For a body of mass $m$ with translational acceleration $a$, the magnitude of the inertia force is $ma$ and its direction in the D'Alembert calculation scheme is opposite to the acceleration. If mass is distributed along a member, the inertia loading may also be distributed.
For a bar with constant mass per unit length $m_l$ and uniform acceleration $a$ along its length, the magnitude of the distributed inertia load is:
$$q_i=m_l a.$$
Stress in an accelerating bar
If a straight uniform bar of length $L$ and area $A$ accelerates along its own axis, the axial force at a section must accelerate the portion of mass on one side of that section. Therefore $N$ varies along the length and normal stress is $\sigma=N/A$.
For example, if a bar of density $\rho$ is pulled at one end and the entire bar has axial acceleration $a$, then for a section at distance $x$ from the free end:
$$N(x)=\rho Aax,\qquad \sigma(x)=\rho ax.$$
The largest stress occurs near the end through which the accelerating force is transmitted.
Rotational motion
For angular acceleration $\varepsilon$, an inertia moment $J\varepsilon$ is introduced. During steady rotation with angular velocity $\omega$, material points have centripetal acceleration $a_n=\omega^2r$, so rotating components develop stresses associated with distributed mass forces.
Example: thin ring
For an idealized thin ring of material density $\rho$, radius $r$, and angular velocity $\omega$, the circumferential stress in a simple membrane model is:
$$\sigma_\theta=\rho\omega^2r^2.$$
Thick disks, complex rims, and nonuniform rotating components require more detailed stress models.
Calculation procedure
- Determine the law of motion and accelerations of the masses.
- Determine concentrated or distributed inertia forces.
- Add them to the calculation scheme according to D'Alembert's principle.
- Determine internal force resultants.
- Calculate stresses and displacements using the usual strength-of-materials relations within the adopted model.
Limitation
If acceleration changes rapidly, vibration develops, or deformation waves propagate through the member, a quasi-static inertia-force representation may be insufficient. A full dynamic analysis is then required.
About this topic
Analyzing machine components moving with acceleration (hoist cables, flywheel rims, rotating shafts) requires incorporating inertia forces. This section applies D'Alembert's principle to convert dynamic equilibrium into static equivalent systems for stress analysis.