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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