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midterm1 - I p ’/’ i Exam#2 ME572(Prof Pan Name I 4...

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Unformatted text preview: I. p; ’/’ i / Exam #2 ME572 (Prof. Pan) Name I 4' / a”; K. 10:00AM ‘ 12300 PM, December 14, 2007 or 4:00PM - 6:00 PM, December 17, 2007 Open book, open note, and no homework solutions. Box your answers and include units. 1. Consider a slab of sheet metal subject to equal biaxial resultant tensile force L in the x1 and x2 directions as shown. The dimensions in the x1 and x2 directions are denoted as w and the thickness of the sheet is denoted as t. The original dimensions of the sheet are denoted as we and to. The Mises yield criterion is used to describe the plastic behavior of the sheet. Since the plastic deformation is large, rigid plastic behangL is assumed here. The relation of the equivalent tensile stress 6 and the equivalent tensile strain E is given as 6 = K?” . Determine the maximum load Lmx in terms of K, n, Wu and to. 2. We consider a polymeric material under uniaxial loading conditions as shown. The uniaxial tensile yield stress 0', is 60 MPa but the uniaxial compressive yield stress 0‘ is 75 MPa. We can model the yielding of the polymer by the Drucker-Prager C yield function as f =0. +f3lwm -0ge =0 where 0e is the effective stress, ,u is the pressure sensitivity, 0' is the mean stress, I" and age is the generalized effective stress. Here, the effective stress 0e is defined as 3 0‘2 : (—2— 01710;)” 2 where 0;. are the deviatoric stresses. The mean stress am is . - o . . u . . . defined as 0',” = For Simpliclty, we consrder this material as a rigid perfectly plastic material. Therefore, age is a constant. (a) Determine the value of the pressure sensitivity ,u . (b) Now we conduct a pure shear test where we apply a pure shear deformation to a material element, determine the shear yield stress based on the yield function. (0) Consider a material element under uniaxial tensile loading conditions in the x] direction. Determine the plastic Poisson ratio Vp (= —d82”2 / d611 = —d£3”3/d£f1) with the assumption of the associated (normality) plastic flow rule. (d) Consider a material element under uniaxial compressive loading conditions in the x1 direction. Determine the plastic Poisson ratio V p (2 —a’€2"§/d£{’l =-d83’g/d€{’,) With the assumption of the associated (normality) ' plastic flow rule. ’( In transformation—toughened ceramics, the phase transformation is mainly controlled by the hydrostatic stress. For simplicity, when the mean stress 0'," equals to a critical value 0",, phase transformation takes place. The phase transformation criterion is defined as ' 0m = O-kk = 0-! Assume that the linear elastic asymptotic crack-tip field is applicable to determine the phase transformation size and shape. Consider a crack under plane strain mode I loading conditions. The in-plane asymptotic elastic crack-tip stresses including the T stress are K’ cos§(1—sin§sin£)+T U”=,/2m 2 2 2 K’ cosg (1+ sin —6—sin E) a” : 1/2727” 2 2 2 K, a. 6’ 36 cos—sm—cos— 0X" 1/ 2727 2 2 2 Derive the analytical expression for the transformation radius r, as a function of 6 for a given set of K I , 0', and T near a crack tip under plane strain mode I loading conditions. Give the expression for the transformation radius r, for T = 0 also. U.) 4. Consider an axisymmetric problem of two centrally loaded circular diaphragms. The cross sections along the symmetry plane of the connected circular diaphrang are shown. The two circular diaphragms have a radius a and the thickness h as shown. The two diaphragms are connected along the circumference. A load P is applied along the center line of the upper diaphragm and a load P is applied along the center line of the lower diaphragm as shown. The displacement w at the load application point for the upper diaphragm and the lower diaphragm can be derived from the plate bending theory as _ 3Pa2(l—v2) W 47th3 where E is the Young’s modulus and V is the Poisson’s ratio. (a) Derive the expression for the energy release rate in terms of the geometric parameter, the material parameters, and P. (b) Specify the fracture mode (mode I, II or mixed mode) for the crack in this diaphragm specimen and give an argument for your answer. Derive the stress intensity factor solution for the crack. <4? I C Pn’r/f Pafl? . H A/ B: : KE’” 3 o 192E€rm¢rVLQ me r77 ferns 470 K, 04,41; gig, Is .» WM; rimfressx’é7/f’fg O = Z 6 3 L 64 M [MD ah: Mt . .: O—Moé Xéjofo £29: 42f Wu 0 W: 4! I ~ (Ta/oi; K2; a 7M mag, 0&5?— "79 8%”); __ a2 (a a) — out: >6 8%; :3 9P<59=A7Z , 192, , g ‘9 \ 4) Ham ‘6) 8 ‘Sfress £122,619. ($.41: O— E“: = 330‘ LIZ] U 0 0 EEC/J: écr 0 O o 6‘ o o (fa-c7 @ o o 0 0 o —§U I/on R7383 ~a OILS}: g/ Wffi7°”fi/Q 00 5X1". HajaI — flasflc '5 if: s: S) OIEII=OQ512 : "é'dxgay ' 22(22222"3\:533. L615 5“ £ng = I2 7’ 0 Li]: 8 ° 0 0 dz 9 a 3 ° 0 0 #146 O O 9‘18 PHILIP PARK CIVIL & ENVIRONMENTAL ENGINEERING UNIVERSITY OF MICHIGAN —I r MK (AEDM K 2, = 2m KCAEYM - 1m I< (22>M" \ }<()_5)”‘ I<(15)W [Dot—~Aal : O. (A: 3 =5. fimm. LMOI OCCM r: PHILIP PARK CIVIL & ENVIRONMENTAL ENGINEERING UNIVERSITY OF MICHIGAN gbeeI/f . g—OLJ“ ‘ I‘S La“) ’bé‘f’c'Y‘mJM/L, @ WWW +emian. ,/ (pm/a ESE]: 0? O o"? CQ/J': IF EC? 0 o ow; I 3 I ’io‘l ,L C7 7. Z CI IL L -L N L33 2*I1’3‘7WI I‘TII (CI—*1:- +Wslom U2=I7l Cow/1F fizz-CE 65 W = “TEli = 35:“ O 3. ”fwfmn tart : m “Iii/g _ 1: O‘ I 3 07/“ 0% D7 (Hg = Ufie U? VIE me‘bresjoé fag—Jr ,0. :I: “U? "FEDTFOO‘LQ = c). 97(é‘4) : $9. W: “127%,er :07 (VI/Dog- jg. _ PHILIP PARK CIVIL & ENVIRONMENTAL ENGINEERING UNIVERSITY OF MICHIGAN W1<l+f1§2 3. age lilac-re. Cf“: {acMDak 60 50+ 73- : 66:8 Lam? ’ UT ‘0 '~ 9% LOW QTz’Wp" 7r~fg : 0-64, 60+ (-63 ~ 741 25 5/4“ ’ 6/“ 504—7? : ¢f_éo a; ___ Ii W60) _ V‘Kf; M {Mar “173: = WM $160+ gxomc; my WA, : “a gfiod‘figf 7 66196 r4?” .L—flk "l‘ PHILIP PARK CIVIL & ENVIRONMENTAL ENGINEERING UNIVERSITY OF MICHIGAN do) Pure 9W fies? Cons7ol.2r 5L tomEm fest 0 [Bibi—£3. PHILIP PARK CIVIL & ENVIRONMENTAL ENGINEERING UNIVERSITY OF MICHIGAN Cc) Pmch Po?sson‘s mtfo'. VJ’"’0I2f (I WAVA/Xiafl. “694437814. PHILIP PARK CIVIL & ENVIRONMENTAL ENGINEERING UNIVERSITY OF MICHIGAN .L—C —2__ l X‘ S; Y; ._ A \ VIE f ” A 7‘ ,__.” _ E I + I; 5+” PHILIP PARK CIVIL & ENVIRONMENTAL ENGINEERING UNIVERSITY OF MICHIGAN GI) VF mmfrcsfran +c51‘ n? = 0M. xyi JE F- .110) A8“) I JongvFfi’ +3/Ax& WA? and mmfaresion PHILIP PARK CIVIL & ENVIRONMENTAL ENGINEERING UNIVERSITY OF MICHIGAN %’DI>I€IM 3 phase "firmsqgrmk orffefion 5;“ = CE I o flame sfffifm. rmooiz I, > 532 — Maggy £2%:0~ EICE~WUHUgflfo K\ "’4' _ V/Cz .32“ F K1603: : PI- K & ’ K; g :g .3 I J—Dié XlC‘oS: +7” 4 Vim Yléos: +T? : Ci.” E K1 xLCOngvI-T? : @‘t r) Phage {Ta/“5% zon Tjfl ' Hi Wmflmfl I A -, I L , Kg: . _—_- u ‘1 ~ 2 It m; I ((—3) a; m GIMME) W) I I PHILIP PARK CIVIL & ENVIRONMENTAL ENGINEERING UNIVERSITY OF MICHIGAN 47KB If C41) kviSérIl/e Hue (daft-Se mfg f P AWITeoI "9 jeaoI é! £42 maféfr'ej F: 9&5f7f. cmflme 4* tag WU. PHILIP PARK CIVIL & ENVIRONMENTAL ENGINEERING UNIVERSITY OF MICHIGAN Dial—Ills FfIaftITD/I‘ Ipr (yr?on I °FIW m5 Griz -E/—(/<ZS’) e I ' - (H71) 2— sttm/n f1 _ ? KI) #Ixrkz/mmetr/‘c 5“. p/Mze, shat/32. 9: £453 (5). ’9 57/3) 9:1. PHILIP PARK CIVIL & ENVIRONMENTAL ENGINEERING UNIVERSITY OF MICHIGAN PHILIP PARK CIVIL & ENVIRONMENTAL ENGINEERING UNIVERSITY OF MICHIGAN ...
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