889_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

889_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

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SECTION 11.6 The Secant Formula 883 Problem 11.6-6 A circular aluminum tube with pinned ends supports a load kN acting at distance from the center (see figure). The length of the tube is m and its modulus of elasticity is GPa. If the maximum permissible stress in the tube is 20 MPa,what is the required outer diameter if the ratio of diameter is to be d 1 / d 2 ± 0.9? d 2 73 3.5 e ± 50 mm P ± 18 Solution 11.6-6 Aluminum tube Pinned ends. D ATA kN ± 20 MPa ± 0.9 S ECANT FORMULA (E Q . 11-59) () r ± 0.33634 d 2 ( r ± mm) r 2 ± I A ± 0.11313 d 2 2 ( d 2 ± mm; r 2 ± mm 2 ) d 2 ± mm; I ± mm 4 I ± p 64 ( d 2 4 ² d 1 4 ) ± p 64 [ d 2 4 ² (0.9 d 2 ) 4 ] ± 0.016881 d 2 4 P A ± 18,000 N 0.14923 d 2 2 ± 120,620 d 2 2 a P A ± MPa b d 2 ± mm; A ± mm 2 A ± p 4 ( d 2 2 ² d 1 2 ) ± p 4 [ d 2 2 ² (0.9 d 2 ) 2 ] ± 0.14923 d 2 2 s max ± P A c 1 + ec r 2 sec a L 2 r A P EA bd d 1 / d 2 s max L ± 3.5 m E ± 73 GPa e ± 50 mm P ± 18 S UBSTITUTE THE ABOVE EXPRESSIONS INTO E Q . (1): (2) S OLVE E Q . (2) NUMERICALLY : d 2 ± 131 mm ; ³ c 1
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