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02-01ChapGere.0012

02-01ChapGere.0012 - b 5 200 mm k 5 3.2 kN/m a 5 45° and P...

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74 CHAPTER 2 Axially Loaded Members D ETERMINE THE ANGLE S elongation of spring S 2 S 1 b (cos cos ) For the spring: F k ( S ) F bk (cos cos ) Substitute F into Eq. (1): P cos bk (cos cos )(sin ) (Eq. 2) This equation must be solved numerically for the angle . D ETERMINE THE DISTANCE L 2 L 1 2 b cos 2 b cos 2 b (cos cos ) or P bk cot u cos u cos 0 From Eq. (2): Therefore, (Eq. 3) N UMERICAL RESULTS b 8.0 in. k 16 lb/in. 45 P 10 lb Substitute into Eq. (2): 0.078125 cot cos 0.707107 0 (Eq. 4) Solve Eq. (4) numerically: Substitute into Eq. (3): 1.78 in. u 35.1 2 P b cot u 2 b ¢ cos u cos u P cot u bk cos cos u P cot u bk Problem 2.2-14 Solve the preceding problem for the following data:
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Unformatted text preview: b 5 200 mm, k 5 3.2 kN/m, a 5 45°, and P 5 50 N. Solution 2.2-14 Framework with rigid bars and a spring See the solution to the preceding problem. Eq. (2): Eq. (3): d 5 2 P k cot u P bk cot u 2 cos u 1 cos a 5 N UMERICAL RESULTS b 5 200 mm k 5 3.2 kN/m a 5 45 8 P 5 50 N Substitute into Eq. (2): 0.078125 cot u 2 cos u 1 0.707107 5 (Eq. 4) Solve Eq. (4) numerically: Substitute into Eq. (3): d 5 44.5 mm u 5 35.1 8 A-PDF Split DEMO : Purchase from www.A-PDF.com to remove the watermark...
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