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# HW5%20solution - EEL5225 Principles of MEMS Transducers HW5...

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Principles of MEMS Transducers Page 1 of 7 Prepared by D. Arnold October 27, 2010 EEL5225 Principles of MEMS Transducers HW5 Fall 2010 Semester Assigned: Wednesday, 10/13 Due: Monday, 10/25 1. The deflection curve for a clamped-clamped beam is given by, w ( x ) = pl 4 2 Eh 3 x 2 l 2 2 x 3 l 3 + x 4 l 4 Λ Ν Μ Ξ Π Ο . Using the center deflection, w ( l /2), as the reference point, derive the lumped mass and compliance for the beam. Solution: The deflection curve is given by: w ( x ) = pl 4 2 Eh 3 x 2 l 2 2 x 3 l 3 + x 4 l 4 Λ Ν Μ Ξ Π Ο Center deflection: w ( l /2) = w c = pl 4 2 Eh 3 1 4 2 8 1 16 Λ Ν Μ Ξ Π Ο = pl 4 32 Eh 3 The deflection curve can be written as: w ( x ) = 16 w c x 2 l 2 2 x 3 l 3 + x 4 l 4 Λ Ν Μ Ξ Π Ο Equivalent compliance: We need to calculate the total potential energy in the system. Consider infinitesimal strip of the beam: Recall: W PE e dq = Fdw And in the linear case: W PE = 1 2 e q = 1 2 Fw For the infinitesimal element dW PE ( x ) = 1 2 F ( x ) w ( x ) = 1 2 p b dx Force w ( x ) b l h Pressure, p (Distributed Force) Clamped Ends x

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Principles of MEMS Transducers Page 2 of 7 Prepared by D. Arnold October 27, 2010 The total energy is now obtained by integration: W PE = dW PE ( x ) 0 l = 1 2 bpw ( x ) dx 0 l = 1 2 bp w ( x ) dx 0 l = = 1 2 bp 16 w c x 2 l 2 2 x 3 l 3 + x 4 l 4 Λ Ν Μ Ξ Π Ο dx 0 l = 8 bpw c x 3 3 l 2 2 x 4 4 l 3 + x 5 5 l 4 Ρ Σ ΢ Τ Φ Υ 0 l l / 30 = 4 bpw c l 15 Noting that w c = pl 4 32 Eh 3 p = 32 Eh 3 w c l 4 and substituting for p, W PE = 4 bw c l 15 32 Eh 3 w c l 4 = 128 Ebh 3 w c 2 15 l 3 Now, equate this to the energy to of a spring with compliance C m = 1 / k m assuming the center displacement w c . 1
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## This note was uploaded on 01/17/2011 for the course EEL 5225 taught by Professor Arnold during the Fall '08 term at University of Florida.

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HW5%20solution - EEL5225 Principles of MEMS Transducers HW5...

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