883_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

883_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

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Unformatted text preview: 11Ch11.qxd 9/27/08 3:39 PM Page 877 SECTION 11.5 Columns with Eccentric Axial Loads Solution 11.5-10 Fixed-free column d deflection at the top Use Eq. (11-51) with L/2 replaced by L: d e (sec kL 1) (This same equation is obtained in Prob. 11.5-9.) NUMERICAL DATA E 73 GPa b P 50 kN d I (1) 14 [b 12 100 mm t 8 mm b 2 30 mm e SOLVE FOR L FROM EQ. (1) d 1+ e e e+d sec kL cos kL L L AP e+d e kL 1 e arccos k e+d EI arccos P A EI arccos k e e+d e e+d (b 2t)4] 50 mm 4.1844 * 106 mm4 MAXIMUM ALLOWABLE LENGTH AP Substitute numerical data into Eq.(2). EI (2) arccos L max e e+d 2.4717 m e e+d 0.625 0.89566 radians (2.4717 m)(0.89566) 2.21 m ; Problem 11.5-11 Solve the preceding problem for an aluminum column with b 6.0 in., t 0.5 in., P 30 k, and E 10.6 * 103 ksi. The deflection at the top is limited to 2.0 in. Solution 11.5-11 Fixed-free column d deflection at the top Use Eq. (11-51) with L/2 replaced by L: d e (sec kL 1) (This same equation is obtained in Prob. 11.5-9.) sec kL cos kL e e+d L L e+d e kL k 10.6 * 103 ksi b P 30 k 14 [b 12 d 6.0 in. t 0.5 in. b 2.0 in. e 3.0 in. 2 (b 2t)4] 55.917 in.4. MAXIMUM ALLOWABLE LENGTH P A EI arccos EI e arccos AP e+d 1 e arccos k e+d (1) E I SOLVE FOR L FROM EQ. (1) d 1+ e NUMERICAL DATA e e+d AP Substitute numerical data into Eq. (2). EI e 0.60 140.56 in. e+d e arccos 0.92730 radians e+d L max (140.56 in.)(0.92730) ; 130.3 in. 10.9 ft 877 ...
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