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Unformatted text preview: MASSACHUSETTS INSTITUTE OF TECHNOLOGY DEPARTMENT OF MECHANICAL ENGINEERING CAMBRIDGE, MASSACHUSETTS 02139 2.002 MECHANICS AND MATERIALS II SOLUTION for HOMEWORK NO. 1 Distributed : Wednesday, September 10, 2003 Due : Wednesday, September 17, 2003 Problem 1 30 mm 5 mm y x Steel: E=208 [GPa] P=? 200 mm x1=? 1 m Figure 1: schematic drawing of Problem 1 This beam is under bending and shear. Strain gauges are located at the top surface of the beam. On the top surface, there is zero transverse shear. Thus the beam is only in a stress state due to bending at the location of the strain gauges. We must know how to compute the stress due to a bending moment and relate that stress to the strain in order to solve this problem. First, the axial stress due to bending is π ( x, y ) = − M ( x ) y (1) I where I and y on the top of surface of the beam are the following bh 3 I = 12 (2) h y = 2 (3) 1 M can be found via taking a section cut in the beam and balancing the moments. Be sure to use proper sign convention M ( x ) = − P ( L − x ) (4) Rewrite (1) using (2), (3), (4) π ( x ) = − ( − P ( L − x )) h = 6 P ( L − x ) (5) 2 bh 3 bh 2 12 This is in the LinearElastic regime so, π ( x ) = Eσ ( x ) (6) Rewrite (5) with (6) to get σ ( x ) = 6 P ( L − x ) (7) Ebh 2 We have two values of strains σ 1 and σ 2 at positions x 1 and x 2 so we now have two equations (7), (8), with three unknowns x 1 , x 2 , and P σ 1 = 6 P ( L − x 1 ) (8) Ebh 2 σ 2 = 6 P ( L − x 2 ) (9) Ebh 2 The third equation needed comes from the geometry condition given in the problem statement x 2 − x 1 = 200 × 10 − 3 = d (10) Now we just have 3 unknowns and 3 equations so it can be solved anyway you like. One was is shown below To solve for P , combine equations (8), (9), (10) 6 P 6 P d σ 1 − σ 2 = Ebh 2 ( x 2 − x 1 ) = (11) Ebh 2 Now solve (11) for P , and plug in values. We know 2 σ 1 σ 2 = = 1200 × 10 − 6 900 × 10 − 6 then, P = ( σ 1 − σ 2 ) Ebh 2 6 d P = 300 × 10 − 6 · 30 × 10 − 3 [ m ] · (5 × 10 − 3 ) 2 [ m 2 ] · 208 × 10 9 [ N/m 2 ] 6 · 200 × 10 − 3 [ m ] = 39[ N ] (12) Knowing P we can get the positions of the strain gauges. Solve (8) for x 1 . Ebh 2 σ 1 x 1 = L − 6 P 2 = 1[ m ] 208 × 10 9 [ N/m 2 ] · 30 × 10 − 3 [ m ] · (5 × 10 − 3 ) 2 [ m ] 1200 × 10 − 6 · − 6 · 39[ N ] = 1 − . 8 = . 2[ m ] (13) Now get x 2 x 2 = x 1 + d (14) x 2 = . 2 m + . 2 m = . 4 m (15) Problem 2 The natural frequency of a simple harmonic oscillator depends on both the stiffness of the restoring (elastic) member in the system and the mass which is being accelerated/decelerated.accelerated/decelerated....
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This note was uploaded on 02/23/2012 for the course MECHANICAL 2.002 taught by Professor Davidparks during the Spring '04 term at MIT.
 Spring '04
 DavidParks
 Mechanical Engineering

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