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'1 L J' l J (300 During a tension test of a large plate with a | ' ' ' central crack the crack opening 5(0) was i recorded as a function of the load....

So, we use the second equation and reduce it to 4K/(Esqrt(πa)).. and where do we use this? I know there's Irwin's model, but I'm unsure of how to apply this or if that's even correct. Any help is appreciated to get on the right track.

‘1 L J'— l J (300 During a tension test of a large plate with a
| — ' ' ' central crack the crack opening 5(0) was
i recorded as a function of the load. The recorded ‘1
8(0) 3’ i
' x i curve deviated from a straight line. The
' l deviation from linearity may be explained partly
! l by the appearance of a plastic zone at the crack
| l .. tip and partly by crack growth timing the
l l l l 1600 loading. (a) How large deviation from linearity will be expected due to the plastic zone
at the crack tip? The remote stress 0,, is cry/2. (b) The deviation from linearity of the recorded curve appeared to be 30 per
symmetry is maintained. tie. the two crack tips move the same distance).
The material is linearly elastic. ideally plastic with yield strength 6y. Plane
stress is at hand so that the Dngdale model may be used. The crack opening
displacement (COD) is 6(0) _ 807 1n] + sin(ao_l20y) 7—3 costtt (IJZGY) and if 0,, &lt;&lt; 0., one obtains ﬂ: 4K1 {1+ﬁ K + }
a BM 24 WE HINTS: Use the second formula above for pure elastic case by ignoring the second term
as the yield stress is very high compared om. The remote stress for part (b) is the same as remote stress in part (a).
Express Part (b) answer in terms of half crack length “a”. Ideally Plastic means elastic perfectly plastic.

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