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PS4-11-solution - Homework Solution#4 MEMS MECE E4212...

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1 Homework Solution #4 MEMS – MECE E4212 Problem #1 (a) k F = δ m N E E E E L EI k / 2 ) 3 1 ( ) 6 10 )( 6 50 ( 12 1 * 9 160 * 3 3 3 3 3 = = = m m N N E µ δ 5 . 0 / 2 6 1 = = (b) Neglecting Point Load: L E n n 2 ) 1 2 ( ρ π ω = E ~ 160E9Pa 3 3 3 3 2300 1 100 1 10 * 3 . 2 m Kg m cm gr cm g = = ρ ( ) ) 6 1000 ( 2 ) / ( 2300 ) / ( 9 160 1 2 3 2 m E m Kg m N E n n = π ω kHz kHz n n 1 . 13 ) 1 2 ( 1 . 13 = = ω for n = 1 ( n is the mode of vibration ) Or alternatively solve for natural frequency the following way: m k n = ω kg x m m kg V m 9 18 3 10 15 . 1 ) 10 )( 10 )( 50 )( 1000 ( 2300 = = = ρ ; but for a distributed mass, an equivalent mass of 33 * m / 140 can be used in the mass-spring relation: sec 10 9 . 85 ) 5 . 0 )( 2300 )( 33 )( 12 ( ) 3 1 ( 140 ) 6 10 )( 6 50 )( 9 160 ( 3 ) ( ) 33 )( 12 ( 140 3 ) 33 ( ) 140 ( 3 3 3 3 3 3 3 rad x E E E E Ltb L Ebt m L EI n = = = = ρ ω
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2 kHz rad x n 67 . 13 2 1 sec 10 9 . 85 3 = = π ω Or a more accurate answer can be found from Vibrations textbooks, where for a distributed mass, the bending vibration frequency is: 4 2 L m EI L n n β ω = where L m is the distribute mass per unit length, and n β a correction coefficient. For the fundamental mode ( n =1), 1 ` β = 1.875 (from reference text). For higher modes, 2 ` β = 4.694 and 3 ` β = 7.855. This gives an estimate of ω 1 = 84 kHz compared to the earlier answers. (c)(d) I My x = σ where: m t y µ 5 2 / = = beam of root at ) 1000 )( 1 ( m N M µ µ = 3 12 1 bt I = MPa x 2 . 1 = σ Tensile on convex side and compressive on concave side. 6 5 . 7 160 2 . 1 = = = E GPa GPa E x σ ε ( ) 2 3 6 12 ) 2 ( bt FL bt t FL = = σ 2 / 6 Ebt FL = ε (e) Max Strain : is proportional to L. So as L ↑⇒ ↑⇒ ε σ M , but inversely proportional to b and t 2 , so strain will decrease with increasing b , t .
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