795_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

# 795_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

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Unformatted text preview: 09Ch09.qxd 9/27/08 1:37 PM Page 789 789 SECTION 9.10 Deflections Produced by Impact Solution 9.10-6 Overhanging beam DATA: W E 750 N bd 12 1.2 m. sallow 12 GPa 3 I LAB LBC in which a 2.4 m 45 MPa dst 1 (500 mm)(40 mm)3 12 2.6667 * 10 S bd 6 2 LAB: + L BC) (3) 3EI EQUATION (9-94): 2.6667 * 106 mm4 6 LBC and L W(L 2 )(L AB BC dmax 4 m dst (d2 st 2hdst)1/2 or 1 (500 mm)(40 mm)2 6 2h 1/2 b dst 1 + a1 + dmax dst (4) 133.33 * 103 mm3 133.33 * 10 6 MAXIMUM BENDING STRESS m3 For a linearly elastic beam, the bending stress s is proportional to the deflection d. DEFLECTION dC AT THE END OF THE OVERHANG 2h 1/2 b dst WL BC S M S s st 1 + a1 + dmax dst smax sst ‹ (5) (6) MAXIMUM HEIGHT h P load at end C Solve Eq. (5) for h: L length of span AB a length of overhang BC smax sst a From the answer to Prob. 9.8-5 or Prob. 9.9-3: dC Pa 2(L + a) 3EI Stiffness of the beam: k smax 2 b sst h P dC 3EI (1) a 2(L + a) a1 + 1 2a 2h 1/2 b dst smax b+1 sst dst smax smax ba a 2 sst sst 1+ 2h dst 2b (7) Substitute dst from Eq. (3), sst from Eq. (6), and sallow for smax: W(L 2 )(L AB + L BC) s allowS s allowS BC ba a 6EI WL BC WL BC MAXIMUM DEFLECTION (EQ. 9-94) h Equation (9-94) may be used for any linearly elastic structure by substituting dst W/k, where k is the stiffness of the particular structure being considered. For instance: SUBSTITUTE NUMERICAL VALUES INTO EQ. (8): Simple beam with load at midpoint: k Cantilever beam with load at free end: k W(L 2 )(L AB + L BC) BC 6 EI 48EI L3 3EI L3 s allowS WL BC Etc. For the overhanging beam in this problem (see Eq. 1): dst W k h 2 Wa (L + a) 3EI (2) or h 10 3 0.08100 m 3.3333 (0.08100 m) a 10 10 ba 3 3 360 mm ; 2b 0.36 m 2 b (8) ...
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## This note was uploaded on 12/22/2011 for the course MEEG 310 taught by Professor Staff during the Fall '11 term at University of Delaware.

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