Diaphragm pressure sensor mechanics – dependence on diameter For a thin, circular diaphragm used as a pressure-sensing element (with material, thickness, corrugation design, and clamping held constant), the central deflection under uniform pressure varies with the diaphragm diameter d according to which proportionality?

Introduction / Context:
Elastic elements (diaphragms, Bourdon tubes, bellows) convert pressure into displacement. For thin, flat or corrugated circular diaphragms, small-deflection plate theory provides simple power-law relations linking tip/center deflection to geometry. Understanding how deflection scales with diameter helps in selecting diaphragm size for desired sensitivity at a given pressure range.

Given Data / Assumptions:

• Uniform pressure loading over the diaphragm surface.
• Clamped circular edge (common construction).
• Material (E, v), thickness t, and corrugation details fixed.
• Linear small-deflection regime.

Concept / Approach:
For a clamped, circular, thin plate in the small-deflection regime, central deflection w is proportional to p * a^4 / (E * t^3) times a function of Poisson’s ratio, where a is the plate radius (a = d/2). Thus w ∝ a^4 for constant p, t, and material, which implies w ∝ d^4. Corrugations effectively lower the bending stiffness, increasing compliance, but the fundamental diameter dependence in the linear regime remains quartic when other factors are held constant.

Step-by-Step Solution:

Verification / Alternative check:
Design charts and sensor vendor notes show rapid sensitivity increase with diameter (fourth-power), motivating the use of larger diaphragms for low-pressure ranges.

Why Other Options Are Wrong:

• d, d^2, d^3: Underestimate the strong geometric leverage of plate bending; experimental and theoretical results show quartic dependence.

Common Pitfalls:
Confusing membrane (tension-dominated) behaviour with plate bending; in pure membrane action, scaling differs, but standard instrument diaphragms are designed for elastic plate behaviour within calibrated spans.

Final Answer:
d^4