03-02ChapGere.0027 - B 2 1 2.4 in. 3.0 in. 4 R 5 30.12 in....

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Solution 3.8-9 Bar with a hole 230 CHAPTER 3 Torsion B A x d 2 T 0 L /2 L /2 T B T A L 5 50 in. L /2 5 25 in. d 2 5 outer diameter 5 3.0 in. d 1 5 diameter of hole 5 2.4 in. T 0 5 Torque applied at distance x Find x so that T A 5 T B E QUILIBRIUM (1) R EMOVE THE SUPPORT AT END B f B 5 Angle of twist at B I PA 5 Polar moment of inertia at left-hand end I PB 5 Polar moment of inertia at right-hand end (2) 2 T 0 ( L / 2) GI PA f B 5 T B ( L / 2) GI PB 1 T B ( L / 2) GI PA 2 T 0 ( x 2 L / 2) GI PB 5 T 0 2 T A 5 T B T A 1 T B 5 T 0 Substitute Eq. (1) into Eq. (2) and simplify: C OMPATIBILITY f B 5 0 S OLVE FOR x : S UBSTITUTE NUMERICAL VALUES : x 5 50 in. 4
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Unformatted text preview: B 2 1 2.4 in. 3.0 in. 4 R 5 30.12 in. x 5 L 4 B 2 1 d 1 d 2 4 R I PB I PA 5 d 2 4 2 d 1 4 d 2 4 5 1 2 d 1 d 2 4 x 5 L 4 3 2 I PB I PA x I PB 5 3 L 4 I PB 2 L 4 I PA f B 5 T G B L 4 I PB 1 L 4 I PA 2 x I PB 1 L 2 I PB 2 L 2 I PA R x T L /2 I PB T B I PA A-PDF Split DEMO : Purchase from www.A-PDF.com to remove the watermark...
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This note was uploaded on 09/20/2009 for the course COE 3001 taught by Professor Armanios during the Spring '08 term at Georgia Institute of Technology.

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