Shaft Deflection⚠ unverified
Mechanical / Shafts · Compute the simple beam deflection of a shaft under an applied moment
Parameters
| Input | Symbol | Unit | Default | Description |
|---|---|---|---|---|
| M | M | N*m | 1.0 | Applied bending moment |
| E | E | Pa | 1.0 | Young's modulus of the shaft material |
| I | I | m^4 | 1.0 | Second moment of area of the cross section |
| L | L | m | 1.0 | Length |
| Output | Symbol | Unit | Description |
|---|---|---|---|
| result | y | m | Deflection, in metres (m). Returns ``0.0`` when ``E * I`` is not positive |
The science & history
Understanding the Parameters
- Bending moment $M$ — the applied end moment; deflection is directly proportional to it.
-
Bending stiffness $EI$ — the product of material stiffness $E$ and section stiffness $I$ (Rectangular Moment of Inertia gives $I = \pi d^4/64$ for a round shaft). Deflection is inversely proportional to $EI$, so — because $I \propto d^4$ — a modest diameter increase stiffens the shaft dramatically. This is the main lever for controlling deflection.
-
Length $L$ — enters as $L^2$ for this end-moment case (and as $L^3$ for transverse point/distributed loads). Either way, deflection grows steeply with span, which is why extra bearings or shorter overhangs are so effective.
Derivation (Approaching a Proof)
The deflection follows from integrating the Euler–Bernoulli moment–curvature relation $EI\,y'' = M(x)$. For a cantilever fixed at $x = 0$ and free at $x = L$, loaded by a constant end moment $M$, the internal moment is uniform, $M(x) = M$:
$$EI\,\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} = M.$$
Integrate once for the slope, applying zero slope at the fixed end ($y'(0) = 0$):
$$EI\,y'(x) = M x.$$
Integrate again for the deflection, applying zero deflection at the fixed end ($y(0) = 0$):
$$EI\,y(x) = \frac{M x^2}{2} \;\Longrightarrow\; y(x) = \frac{M x^2}{2 E I}.$$
Evaluating at the free end $x = L$ gives the tip deflection
$$y = \frac{M L^2}{2 E I}.$$
The two constants of integration were fixed by the fixed-end (cantilever) boundary conditions; different supports or load types change those conditions and hence the coefficient and the power of $L$.
Dimensional check. $[y] = \dfrac{\text{N}\cdot\text{m}\cdot\text{m}^2}{(\text{N/m}^2)\,\text{m}^4} = \dfrac{\text{N}\cdot\text{m}^3}{\text{N}\cdot\text{m}^2} = \text{m}$. ✓
History and Development
Beam-deflection analysis is classical Euler–Bernoulli theory (18th–19th century, completed by Navier). For shafts, deflection and slope limits at bearings and gears are codified design constraints (Shigley gives allowable slopes for bearing types and deflection limits for gear meshes). Lateral stiffness also sets the shaft's first critical (whirl) speed (Critical Speed Shaft), tying deflection directly to rotordynamics.
Related Concepts: Beam Bending Stress, Rectangular Moment of Inertia, Critical Speed Shaft, Shaft Diameter Bending, Section Modulus
Notes: This is the cantilever end-moment case ($ML^2/2EI$); use the appropriate beam-deflection formula for your actual load and supports. Deflection/slope limits at gears and bearings often govern before stress does. $I = \pi d^4/64$ for a solid round shaft.