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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. —9x 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 faiure 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 = proexponential [mars] Ga = activation energy [mecl] R = gas constant [3.314 JimolK]
T = absolute temperature [K] Nonsteady 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 0.950.90 0320903970
. — . 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 —. Dx_2..
z and erf(z) values are given 5 250! 422'
"I Table 61 °° _
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 JImoIK)[In (2.3x 1o—5 m’i'sln _)] T1300 K1027° C ...
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 Winter '12
 DaRocha
 Deformation, Tensile strength, loaYieId Strength

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