Question

Could someone please tell me why max stress goes to zero in this answer.

YIELDING CRITERIA FOR DUCTILE MATERIAL - EXAMPLE 1.1 0=+ *29) +=54.500 + |(40240) + z = . =-00+78 100 MPa 10x - 0 Ob = Ox toy 2 * 2) + Tžy - On = -00-00 2 17-00+ 0 / 2 -> 00= -00 – Txy - 2 ] + Try / 0o = 200 MPa = -100 MPa Oa = -00 + Txy Oq = -200 MPa + 100 MPa Op = -0. – Txy - Op = -200 MPa – 100 MPa -> Op = -300 MPa | No yielding Omax = 0 Omin = -300 MPa 325 MPa F. 300 MPa 2Tmax = Omax - Omin = 0– (-300 MPa) = 300 MPa < 325 MPa 83 0o = 240 MPa 0a = -00 + Txy Oq = -240 MPa + 100 MPa - Oq = -140 MPa Op = -340 MPa Yielding occurs The state of plane stress shown occurs in a machine component made of a steel with Oy=325 MPa. Using the maximum-shearing-stress criterion, determine whether yield will occur when (a) 0,=200 MPa, (b) 0.=240 MPa, (c) 0,=280 MPa. If yield does not occur, determine the corresponding factor of safety. Op = -00 - Txy Omax = 0 Ob = -240 MPa - 100 MPa Omin = -340 MPa F.S.= 325 MPa 340 MPa = 0 9 2Tmax = Omax - Omin = 0) – (-340 MPa) = 340 MPa > 325 MPa 0o = 280 MPa da = -00+ Txy Oa = -180 MPa Oq = -280 MPa + 100 MPa Ox = -00 Op = -00 - Txy Ob = -280 MPa – 100 MPa Op = -380 MPa Yielding occurs Oy = 325 MPa Oy = -0. Omax = 0 Omin = -380 MPa 325 MPa F.S.= Txy = 100 MPa 380 MPa = 0.855 2Tmax = Omax - Omin = 0) – (-380 MPa) = 380 MPa > 325 MPa

Answer 1

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se com os p/15/N] see Tmax = max of 11 J 102-03 1 luz let's take first case & where vo=200MP oi= -300mia , 02=-10 and mia, 02=-100ml Os=0 d it 2-o'plane condition is given = 1 take 183=0 Tmaxmax of /7-3004100), 5100-01,135020) Zmax = max of / 100, 50,150) Tmax= 150 MB) Similarly other parts.