912_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

912_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1

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Unformatted text preview: 11Ch11.qxd 9/27/08 906 2:27 PM Page 906 CHAPTER 11 Columns Assume L/r is less than 55: Assume L/r is greater than 55: Eq. (11-84a): sallow 30.7 0.23 (L/r) ksi or 20.20 30.7 0.23 (L/r) L r Solve for L/r: (45.65)r L max L 6 55 r 45.65 17.98 in. ‹ ok or ; d pd 2 4 A r 23.62 in. L r d 4 23.62 in. d/4 39.34 k 50.09 pd 2/4 P A sallow 0.6 m d2 2 (94.48/d)2 8.280 in.4 1.696 in. dmin (b) FIND dmin IF L (L/r)2 54,000 50.09 d4 457 mm 54,000 ksi Eq. (11-84b): sallow L/r 94.48/d 94.48 in. d ; 43.1 mm 94.48/1.696 55.7 7 55 ‹ ok (ksi) Problem 11.9-27 A solid round bar of aluminum having diameter d (see figure) is compressed by an axial force P 10 k. The bar has pinned supports and is made of alloy 6061-T6. (a) If the diameter d 1.0 in., what is the maximum allowable length Lmax of the bar? (b) If the length L 20 in., what is the minimum required diameter dmin? Solution 11.9-27 Aluminum bar Alloy 6061-T6 Pinned supports (K (a) FIND Lmax IF d A r pd 2 4 AA sallow (b) FIND dmin IF L P 1). A 1.0 in. 0.7854 in.2 I d 4 I P A 20 in. 10 k pd 4 64 pd 2 4 P A sallow d 4 r 10 k 12.73 pd 2/4 d2 Eq. (11-85b): sallow 10 k 2 0.7854 in. 12.73 ksi or Assume L/r is less than 66: 12.73 d 2 0.126 (L/r) ksi d4 1.597 in.4 or 20.2 0.126 (L/r) L/r 80/d L max (59.29)r 59.29 14.8 in. (ksi) L 6 66 r ; ‹ ok 51,000 ksi (L/r)2 (80/d)2 20.2 Solve For L/r: 80 in. d 51,000 Eq.(11-85a): sallow L r 20 in. d/4 Assume L/r is greater than 66: 0.2500 in. 12.73 L r dmin 80/1.12 1.12 in. ; 71 7 66 ‹ ok ...
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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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