Chapter 19: Micro-Optical-Electro-Mechanical Devices
Source: Hunsperger, sixth edition (2009), Problems 19.1–19.9, printed pages 419–420.
Units and boundary conditions
Use Table 19.2 values \(E=190\) GPa, \(G=73\) GPa, \(\nu=0.28\), and \(\rho=2300\,{\rm kg/m^3}\). The pressures printed in kg/m2 are interpreted as kilogram-force per square metre, so multiply by \(g=9.80665\,{\rm m/s^2}\). Plate results assume clamped boundaries, small deflection, no residual tension, and an isotropic approximation for crystalline silicon.
Problem 19.1 — Stress and strain of a loaded silicon rod
Working. A 3 g mass exerts \(F=0.003g=0.02941995\) N. For diameter 1 mm,
Answer. Axial stress magnitude is 37.46 kPa and extension strain magnitude is \(1.972\times10^{-7}\). “Longitudinal” in part (c) describes the same axial direction. If lateral strain was intended, its magnitude is \(\nu|\epsilon_a|=5.5202\times10^{-8}\), with opposite sign.
Check. Under the usual extension-positive convention the axial strain is positive and lateral strain negative. The book’s tension-negative stress convention changes signed labels, not magnitudes.
Problem 19.2 — Clamped square silicon plate
Working. Convert pressure to \(P=98066.5\) Pa. For side \(a=0.007\) m and thickness \(t=150\,\mu{\rm m}\), form flexural rigidity \(D_p=Et^3/[12(1-\nu^2)]\). The chapter’s plate relations give
Answer. Centre deflection 5.137 micrometres; maximum longitudinal stress 65.80 MPa; transverse stress 18.42 MPa; fundamental frequency approximately 47.94 kHz.
Check. Deflection is about 3.4 percent of thickness, consistent with the small-deflection plate approximation.
Problem 19.3 — Uniformly pressure-loaded cantilever
Working. The observation point is \(x=L-0.5\,{\rm mm}=1.0\,{\rm mm}\) from the clamp. For width \(b=0.5\) mm and thickness 0.25 mm, \(I_b=bt^3/12\). Convert surface pressure to line load: \(q=(100g)b=0.4903325\,{\rm N/m}\).
Answer. Deflection 1.404 nm; maximum bending stress 0.1059 MPa; unloaded-beam fundamental approximately 162.6 kHz.
Check. A pressure in N/m2 cannot be inserted directly where the beam formula requires N/m. The beam model also neglects fluid loading and shear corrections.
Problem 19.4 — A ten-gram point load at the free end
Working. Use \(F=0.01g=0.0980665\) N at \(x=L\):
At the same observation point, these give 0.46246 micrometres and maximum stress 28.243 MPa. If the 10 g load is an attached mass, the first-mode estimate is
Answer. Static deflection 0.4625 micrometres and stress 28.24 MPa; approximately 528 Hz for an attached 10 g mass.
Check. Reusing the unloaded 162.6 kHz frequency is appropriate only if the force is applied without added moving mass. A physical 10 g weight dominates this tiny beam’s effective mass.
Problem 19.5 — Formal torsion restoring torque
Working. Apply the two-beam relation with \(\ell=200\,\mu{\rm m},w=t=100\,\mu{\rm m}\):
For \(\theta=\pi/4\), the magnitude is \(8.1141\times10^{-3}\,{\rm N\,m}\).
Answer. The linear model predicts about 8.11 mN m opposing the rotation.
Check. This very large twist over a short, thick support implies large shear strain. The formal calculation is not evidence that crystalline silicon can sustain the specified elastic rotation without fracture; the linear torsion assumptions need checking.
Problem 19.6 — Circular-plate centre deflection
Working. The clamped isotropic plate solution follows from \(D_p\nabla^4w=P\) and \(w(a)=w'(a)=0\):
For diameter 8 mm, thickness 200 micrometres, and the stated pressure, this gives \(w(0)=2.85404\,\mu{\rm m}\).
Answer. Approximately 2.854 micrometres from the clamped-plate equation. The book prints 265 instead of 256 in Eq. (19.18); using that printed denominator gives 2.75712 micrometres.
Check. The denominator 256 follows directly by substituting radius \(a=d/2\) into the plate solution. The discrepancy is not an alternative pressure-unit conversion.
Problem 19.7 — Length relaxation after release
Working. Strain magnitude is \(|\epsilon|=20\,{\rm MPa}/280\,{\rm GPa} =7.14286\times10^{-5}\). For a 1000 micrometre beam, relaxation of that tensile strain changes length by \(|\Delta L|=L|\epsilon|=0.0714286\,\mu{\rm m}\).
Answer. About 71.43 nm shortening when the initial tensile stress is released. In the book’s tension-negative convention, the signed relaxation result is \(\Delta L=-71.43\) nm.
Check. A beam subjected to a newly applied tensile force would instead lengthen. Distinguish release of prestress from application of a load.
Problem 19.8 — Three MOEM mechanical categories
Working. The chapter groups devices by their compliant structure: membranes, cantilevers, and torsion plates. A membrane deflects under pressure, a cantilever bends about its clamp, and torsion supports rotate a plate about an axis.
Answer. Examples are an optically read pressure diaphragm, a cantilever-supported tunable cavity mirror, and a torsional scanning mirror or optical switch.
Check. “Sensor, actuator, and switch” classify device functions, not the three structural categories emphasized by this chapter.
Problem 19.9 — Circular-plate deflection away from the centre
Working. For diameter 12 mm, thickness 250 micrometres, and \(r=4\) mm,
Evaluate the centre deflection with the same pressure and elastic constants as 19.6, then multiply by this factor.
Answer. \(w(4\,{\rm mm})=2.28324\,\mu{\rm m}\) using denominator 256. The printed denominator 265 gives 2.20569 micrometres.
Check. Deflection approaches zero with zero slope at the clamped edge and is maximal at the centre.