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Unformatted text preview: budynas_SM_ch05.qxd 11/29/2006 15:00 Page 122 FIRST PAGES 122 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 5-8 See Prob. 5-7 for plot. (a) For all methods: (b) BCM: All other methods: (c) For all methods: (d) MNS: BCM: MM: n= n= n= n= n= n= n= 1.55 OB = = 1.5 OA 1.03 1.4 OD = = 1.75 OC 0.8 OE 1.55 = = 1.9 OC 0.8 5.2 OL = = 7.6 OK 0.68 5.12 OJ = = 6.2 OF 0.82 2.85 OG = = 3.5 OF 0.82 3.3 OH = = 4.0 OF 0.82 5-9 Given: S y = 42 kpsi, Sut = 66.2 kpsi, ε f = 0.90. Since ε f > 0.05, the material is ductile and thus we may follow convention by setting S yc = S yt . Use DE theory for analytical solution. For σ , use Eq. (5-13) or (5-15) for plane stress and Eq. (5-12) or (5-14) for general 3-D. (a) σ = [92 − 9( −5) + ( −5) 2 ]1/2 = 12.29 kpsi n= 42 = 3.42 Ans. 12.29 42 = 3.21 Ans. 13.08 42 = 3.60 Ans. 11.66 42 = 4.29 Ans. 9.798 (b) σ = [122 + 3(32 )]1/2 = 13.08 kpsi n= (c) σ = [( −4) 2 − ( −4)( −9) + ( −9) 2 + 3(52 )]1/2 = 11.66 kpsi n= (d) σ = [112 − (11)(4) + 42 + 3(12 )]1/2 = 9.798 n= budynas_SM_ch05.qxd 11/29/2006 15:00 Page 123 FIRST PAGES Chapter 5 123 B (d) 1 cm 10 kpsi G O C A E D (b) A H B (a) F (c) For graphical solution, plot load lines on DE envelope as shown. (a) σ A = 9, σ B = −5 kpsi OB 3.5 n= = = 3.5 Ans. OA 1 12 2 2 12 ± (b) σ A , σ B = 2 n= (c) σ A , σ B = n= + 32 = 12.7, −0.708 kpsi 4.2 OD = = 3.23 OC 1.3 −4 − 9 ± 2 4−9 2 2 + 52 = −0.910, −12.09 kpsi 4.5 OF = = 3.6 Ans. OE 1.25 11 − 4 2 2 11 + 4 ± (d) σ A , σ B = 2 n= 5-10 + 12 = 11.14, 3.86 kpsi OH 5.0 = = 4.35 Ans. OG 1.15 This heat-treated steel exhibits S yt = 235 kpsi, S yc = 275 kpsi and ε f = 0.06. The steel is ductile ( ε f > 0.05) but of unequal yield strengths. The Ductile Coulomb-Mohr hypothesis (DCM) of Fig. 5-19 applies — confine its use to first and fourth quadrants. budynas_SM_ch05.qxd 11/29/2006 15:00 Page 124 FIRST PAGES 124 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design (a) σx = 90 kpsi, σ y = −50 kpsi, σz = 0 σ A = 90 kpsi and σ B = −50 kpsi. For the fourth quadrant, from Eq. (5-31b) 1 1 n= = = 1.77 Ans. ( σ A / S yt ) − ( σ B / Suc ) (90/235) − ( −50/275) (b) σx = 120 kpsi, τx y = −30 kpsi ccw. σ A , σ B = 127.1, −7.08 kpsi. For the fourth quadrant 1 n= = 1.76 Ans. (127.1/235) − ( −7.08/275) (c) σx = −40 kpsi, σ y = −90 kpsi, τx y = 50 kpsi . σ A , σ B = −9.10, −120.9 kpsi. Although no solution exists for the third quadrant, use S yc 275 n=− =− = 2.27 Ans. σy −120.9 (d) σx = 110 kpsi, σ y = 40 kpsi, τx y = 10 kpsi cw. σ A , σ B = 111.4, 38.6 kpsi. For the first quadrant S yt 235 n= = = 2.11 Ans. σA 111.4 Graphical Solution: OB 1.82 = = 1.78 (a) n = OA 1.02 (b) n = (c) n = (d) n = OD 2.24 = = 1.75 OC 1.28 2.75 OF = = 2.22 OE 1.24 2.46 OH = = 2.08 OG 1.18 1 in 100 kpsi G H (d ) B O C D (b) A A B (a) E F (c) budynas_SM_ch05.qxd 11/29/2006 15:00 Page 126 FIRST PAGES 126 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 5-12 . Since ε f < 0.05, the material is brittle. Thus, Sut = Suc and we may use MM which is basically the same as MNS. (a) σ A , σ B = 9, −5 kpsi 35 = 3.89 Ans. 9 (b) σ A , σ B = 12.7, −0.708 kpsi n= 35 = 2.76 Ans. 12.7 (c) σ A , σ B = −0.910, −12.09 kpsi (3rd quadrant) n= n= 36 = 2.98 Ans. 12.09 35 = 3.14 Ans. 11.14 B (d) σ A , σ B = 11.14, 3.86 kpsi n= 1 cm 10 kpsi H G O A E (b) A Graphical Solution: 4 OB = = 4.0 Ans. (a) n = OA 1 (b) n = (c) n = 3.45 OD = = 2.70 Ans. OC 1.28 3.7 OF = = 2.85 Ans. (3rd quadrant) OE 1.3 F (c) (d ) C D B (a) OH 3.6 = = 3.13 Ans. (d) n = OG 1.15 5-13 Sut = 30 kpsi, Suc = 109 kpsi Use MM: (a) σ A , σ B = 20, 20 kpsi Eq. (5-32a): n= 30 = 1.5 Ans. 20 30 = 2 Ans. 15 (b) σ A , σ B = ± ( 15) 2 = 15, −15 kpsi Eq. (5-32a) (c) σ A , σ B = −80, −80 kpsi For the 3rd quadrant, there is no solution but use Eq. (5-32c). Eq. (5-32c): n=− 109 = 1.36 Ans. −80 n= ...
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This note was uploaded on 02/22/2010 for the course MECHANICAL ME 342 taught by Professor Namıkçıblak during the Spring '08 term at Yeditepe Üniversitesi.

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