LECTURE+36+COE-3001-A

Supports a uniform load intensity q acng

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Unformatted text preview: at distance 2a/3 from support A and uniform load of intensity (q) acts between B and C. Determine: 1.  Deflec=on δB at the internal hinge 2.  angle of rota=on ΘA at support A SOLUTION: Consider the superposi=on of two bema problems 1.  Simple beam AB of length (a) 2.  Can=lever beam BC of length (b) 3.  Two beams linked together by a pin connec=on at B 4/9/13 M. Mello/Georgia Tech Aerospace 25 EXAMPLE 8- 8 2a 3 X MA = 0 2a + F (a) = 0 3 2P F= 3 P Equal and opposite force acts downward at the end of the can=lever ☞ We have tabulated solu=ons to this problem in Appendix G ☞ superpose distributed load solu=on solu=on (case 1 Appendix G- 1) with Point load solu=on (Case 4 Appendix G- 1) 4/9/13 M. Mello/Georgia Tech Aerospace 26 EXAMPLE 8- 8 ☞ replace L with b and P with F=2P/3 B 4/9/13 qb4 F b3 = + 8EI 3EI B qb4 2 P b3 = + 8EI 9EI M. Mello/Georgia Tech Aerospace 27 EXAMPLE 8- 8 •  Solve for angle of rota=on at support A: ☞ Solu=on consists of two parts 1.  Angle BAB’ in Figure (d) produced by rigid body mo=on associated with downward displacement of the hinge. 2.  Rota=on produced by bending of beam AB( (beam AB’) as a simple beam) qb4 2 P b3 B ( ✓A ) 1 = + ☞ Angle BAB’ : (✓A )1 = a 8aEI 9aEI ☞ angle of rota=on at the end of a simply supported beam with a concentrated load ☞ Refer to case 5 Appendix G- 2 4/9/13 M. Mello/Georgia Tech Aerospace 28 EXAMPLE 8- 8 ☞ replace a by 2a/3 , b by a/3, and L by a in our problem (✓A )2 = 4/9/13 M. Mello/Georgia Tech Aerospace P 2a 3 a 3 a+ 6aEI a 3 29 EXAMPLE 8- 8 ( ✓A ) 2 = P 2a 3 a 3 a+ 6aEI a 3 4P a2 ( ✓A ) 2 = 81EI ☞ combine solu=ons to obtain rota=on at support A: ✓ = ( ✓A ) 1 + ( ✓A ) 2 qb4 2P b3 4P a2 ✓= + + 8aEI 9aEI 81EI 4/9/13 M. Mello/Georgia Tech Aerospace 30...
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