Question

A 30cm Stainless steel ruler is setup as a simply cantilever beam, represented in the figure below.

Dimensions of beam: 300mm (L) x 0.8 mm (T) x 26mm (H)

Density, p = 7700 kg/m³

Young's Modulus, E = 200 GPa

a) Determine the first natural frequency

b) Determine the first natural frequency, when a weight of 200 grams is added at the end of the beam (point 2).

c) List any assumptions/boundary conditions that were made.

Answer 1

Technical data, Length of steel ruler; 300mm = 0.3m=L Height : 26mm :-0'026m ih Thickness :-0.8mm. =0.008m =t Density :.7700 kg/m3 =8 E = Youngs modulus: 200 Gpa.= 2x10' N/m2=E Son: al Determine the first natural Frequency * To determine finest natus al frequency we know that wd = 1.8752/E (663/2) | 3C 550L upon simplifying 1.875? r Eh? 12514 wa= 1.8752 - 1-8752 2x1011 +0.0262 12x7700X 0410.34 wd=1494.21 rad/sec Natural frequency = wd fn = 1494.21 2XIT First natural frequency fn = 237 81 Hz. wa= 2T

by Determine natural frequency when mass of 200 grams is added at free end. In order to find is it we need to have Effective mass of system. which is given by Meff= BEI ام بي 3 12 where I = moment of Inertia of rectangular Section I= ht3 = 0.026 +0.0083 12 Is 1.10X109 m4 MEff= 3EI - 342x10lt 1.10x109 ² was 0.33 y 1494.219 Meff=0.009 kg Yotal mass M = Mefft mass at end of rules M = 0,009 to.2 Mato - M=0.210kg. weft = BEI 3+2x10x110x10 9 L34 0.3 340.210 weft = 341.17 rad's First Natural the quency with 200gms. at end of Steel suler is fo= were 21T In- 341.17 = 54.30 Hz 2T

cy the assumptions involved are * system is considered as continuos system & mass is considered to be distributed along with stiffness of shaft. Boundry conditions are at x=0, y )=0, dy (2) - dy at x=L d²4 (X) - - 0 diz - x -
Hope I have helped you