881_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

881_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

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Problem 11.5-7 A wide-flange member (W 10 ± 30) is compressed by axial loads that have a resultant P ² 20 k acting at the point shown in the figure. The material is steel with modulus of elasticity E ² 29,000 ksi. Assuming pinned-end conditions, determine the maximum permissible length L max if the deflection is not to exceed 1/400th of the length. SECTION 11.5 Columns with Eccentric Axial Loads 875 Solution 11.5-7 Column with eccentric axial load Wide-flange member: Pinned-end conditions. Bending occurs about the weak axis (axis 2-2). Maximum allowable deflection From Table E-1: k ² A P EI ² 0.006426 in. ³ 1 e ² 5.810 in. 2 ² 2.905 in. I ² 16.7 in. 4 ² L 400 ( ² d ) P ² 20 k E ² 29,000 ksi L ² length (inches) W 10 * 30 D EFLECTION AT MIDPOINT (E Q . 11-51) Rearrange terms and simplify: ( NOTE: angles are in radians) Solve the equation numerically for the length
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Unformatted text preview: L : L 150.5 in. M AXIMUM ALLOWABLE LENGTH L max 150.5 in. 12.5 ft ; sec (0.003213 L ) 1 L 1162 in. L 400 (2.905 in.) [sec (0.003213 L) 1] d e a sec kL 2 1 b P = 20 k W 10 30 Problem 11.5-8 Solve the preceding problem (W 250 44.8) if the resultant force P equals 110 kN and E 200 GPa. Solution 11.5-8 Bending occur about the weak axis (axis 2-2) k 0.000281 mm 1 k A P EI e 74 mm e b 2 b 148 mm I 6.95 * 10 6 mm 4 P 110 kN d L 400 W 250 * 44.8 E 200 GPa Deflection at midpoint Solve for the length L ; L max 3.14 m L 400 e a sec a kL 2 b 1 b d e a sec a kL 2 b 1 b 11Ch11.qxd 9/27/08 3:39 PM Page 875...
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