Exam 2 Cheat Sheet.docx

# Exam 2 Cheat Sheet.docx - Engineering Strain Engineering...

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Unformatted text preview: Engineering Strain Engineering Stress - Tensile strain (6,): - Lateral strain (5,): - Tensile stress, 0". ~ Shear stress, I: F £2 A! a _- Ad 0 =79 ”To Area, .40 "' Area A Q h E, - Shear strain (3/): A iF r = E E2] 9 i: A0 T'H/ —> Y=Axly=tan A0 Um {mm . y Both tensile and shear strain original cross— sectional ““3"“? Pl“ W‘W 9" . I .' ..: i __ __. m ”monks area before loaYieId Strength - Transition from elastic to iastic defamation is raduai . Yield strength: stress a which noticeable plas c deformatic Ductility plea - n-* «we non has occurred I - psctilijy: -amcunt of plastic deformation at failure: ' Elastic defamation igwlilgn 9,, - 0.002 ' El Specification 0f duwllﬁnrmanant and norllracovarable ‘UistrESSI +- Stre—s: Percent elongation: %EL -‘f_0. —9-x 100 or, = yield strength at a F _ -- Pemsntjsdgction' In area: 96R); :- A! A’ x 100 l .1 AI. Note: tor 5 cm sample s = 0.002 = A212 ‘0‘" Guam? ' AZ: 0.01 cm at le r (strain) f3=i1002 mmﬁlajuﬂ] use: _____ “I" ‘Tﬁursm; IeIasuc 5' [GPQDuesign/Safety Factors (cont) 1 GE? 9e- Design/Safety Factors 3" Example Problem: A cylindriqai rod, to be constructed from ' Equ'i’a Steel that has a yield strength of 310 MPa. is to withstand . Because of design uncertainties allowances must a load of 220,000 N without yielding Assuming a value at 4 be made to protect against unanticipated fai|ure hi9!” N SPBGliV a suitable bar dIameter. ““"“ - For structural applications. to protect against possibility \N d of failure—use working stress, a... and a Total number of 7.0,; _ -_ factor of safety, N i c /N\Steel rod. lattice. ago 000 N "y" _ 310 MP8 0/ yield strength I, 2’ 4 “w =Wy 2 F: 220300 N ' \ Depending on application Solving for the rod diameter dyields N is between L2 and 4 , Influence of Temperature on DiﬁusiOn - Diffusion coefﬁcient increases with increasing T D= D Dexp Fi‘l' °-"] D = diffusion coefficient [mars] Da = pro-exponential [mars] Ga = activation energy [mecl] R = gas constant [3.314 Jimol-K] T = absolute temperature [K] Non-steady State Diffusion (cont) Inf Example Problem (cont): Der We must now determine Irorn Twle 5.1 the value of zlor which me two error function is 0.8125. An interpolation is necessary as follows on : z_erlw 43.90 _ 03125—01970 m 0.95-0.90 03209-03970 . — . 00:] = Conc. at point rm: (.3 0‘ 3.95 3.3:}: timel f, Sut _ erf(z) a error function g 5 CU.” NOWSONBWID Z- X —. D--x_2.. z and erf(z) values are given 5 250! 422' "I Table 6-1 °° _ Tait 2 Distance from interface. 1 . I- i— _ (4 x IDFaI'I'I)a 1 h _ Example Problem (cont): ” 422; (4 . To solve tor the temperature at .. , which Dhastheabovevalue, T: H " we use a rearranged form 01 R(InD — Ia) Equation 5.9a ° From Table 5.2, for diffusion of C in FCC Fe DD = 2.3 x 10‘5 mzis _'__'-:f__:.: T - (3.314 JImoI-K)[In (2.3x 1o—5 m’i'sln _)] T-1300 K-1027° C ...
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• Winter '12
• DaRocha
• Deformation, Tensile strength, loaYieId Strength

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