16_ch 12 Mechanical Design budynas_SM_ch12

16_ch 12 Mechanical Design budynas_SM_ch12 -...

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Unformatted text preview: budynas_SM_ch12.qxd 12/04/2006 15:24 FIRST PAGES Page 319 319 Chapter 12 µ = 10.8 MPa · s Fig. 12-13: 10.8 13 S = 0.055 fr = 2.23, c T= = 0.0457 = 0.874, ho = 0.13 c 2.23(0.0457)(102 ) 978(106 ) = 59.5°C 1 + 1.5(0.8742 ) 200(254 ) Ta v = 55°C + 59.5°C = 84.7°C 2 O.K. From Eq. (12-22) Q s = (1 + 1.5 2 ) π ps rc3 3µl = [1 + 1.5(0.8742 )] π (200)(0.0423 )(25) 3(10)(10−6 )(25) = 3334 mm3 /s h o = 0.13(0.042) = 0.005 46 mm or 0.000 215 in Trumpler: h o = 0.0002 + 0.000 04(50/25.4) = 0.000 279 in Tmax = Ts + Not O.K. T = 55°C + 63.7°C = 118.7°C or 245.7°F O.K. Pst = 4000 kPa or 581 psi n = 1, as done Not O.K. Not O.K. There is no point in proceeding further. 12-18 So far, we’ve performed elements of the design task. Now let’s do it more completely. First, remember our viewpoint. The values of the unilateral tolerances, tb and td , reflect the routine capabilities of the bushing vendor and the in-house capabilities. While the designer has to live with these, his approach should not depend on them. They can be incorporated later. First we shall find the minimum size of the journal which satisfies Trumpler’s constraint of Pst ≤ 300 psi. W Pst = ≤ 300 2dl W 2d 2 l /d ≤ 300 dmin = ⇒ d≥ W 600(l /d ) 900 = 1.73 in 2(300)(0.5) ...
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