example2 - Midterm Exam ME572 (Prof. Pan) Name 3:40PM -...

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Unformatted text preview: Midterm Exam ME572 (Prof. Pan) Name 3:40PM - 5:10 PM, October 23, 2006 A formula sheet is allowed. Box your answers and include units. 1. We consider a tensile bar under uniaxial cyclic loading conditions. Consider a material element on the bar surface. The coordinate system of the material element is shown. The material element is subjected to constant amplitude loading conditions from the minimum normal stress of O to the maximum normal stress 0m in the x1 direction as shown. Consider a damage plane for the material element with the normal making an angle of g1) with respect to the x1 axis. A damage parameter D for this plane can be defined as a linear combination of the shear stress range Arm to the plane and the maximum normal stress (0' rm )max to the plane in a cycle as D = ATM + k(0',m)max where k is determined as 0.4 to fit the experimental fatigue life data. The shear stress range ATM is defined as the difference between the maximum shear stress and the minimum shear stress in a cycle. (a) Determine the value of (I) for the maximum value of D based on the stress transformation rule (or Mohr’s circle). This value of ¢ should define the damage plane orientation under constant amplitude cyclic loading conditions. (b) In experiments, however, small shallow surface cracks are usually observed near the direction where the maximum shear stress range occurs. Determine the value of ¢ where the shear stress range ATM is maximum using the stress transformation rule (or Mohr’s circle). X2. S‘l' Yes 5 2. Consider a fcc single crystal material element with the coordinate axes parallel to the axes of the unit cell. The material element is subject to combined normal and shear stresses as shown. The tensile normal stress has a magnitude of 100 MPa and the shear stress has a magnitude of 100 MPa as shown. We consider the (1 1 0) slip plane and two possible slip directions. The slip direction sA is in the direction of [-1 l -l] and the slip direction SE is in the direction of [-1 1 l]. (a) Determine the traction on the (l 1 0) plane. (b) Determine the magnitude of the normal stress on the (1 l 0) plane. (0) Determine the shear stress on the (1 1 0) plane as a vector. (d) Determine the magnitude of- the resolved shear stress in the sA or [—1 1 -1] direction on the (l l 0) plane as shown. (e) Determine the magnitude of the resolved shear stress in the 5B or [-1 1 1] direction on the (l 1 0) plane as shown. (f) Which direction of the slip will be first activated if the combined normal and shear stresses increases proportionally? l 3. A thin-walled cylindrical pressure vessel has the mean radius r of 100 mm and the wall thickness t of 10 mm. The vessel is under the internal pressure p and a torque T .1 Two strain gages A and B are located on the outer wall surface in the middle part of the vessel as shown. For the material element near the strain gages, the local - coordinate parallel to the longitudinal direction is denoted as x , the coordinate in the circumferential direction is denoted as y , and the coordinate in the radial or out-of- plane direction is denoted as z. The strain gages are used to measure the normal strains in the x direction and the direction making 45° with respect to the x direction as shown, respectively. The measured strains are used to infer the pressure p and the torque T . The strain gage A gives a tensile strain of 100 y and the strain gage B gives a tensile strain of 400 y. Note that Young’s modulus E is 210 GPa and Poisson’s ratio v is 0.3 for the vessel material. Use the formulae based on the thin— walled pressure vessel assumption. (a) Determine the pressure p inside the vessel, (b) Determine the torque T , (c) Determine the principal stresses. Make a sketch of a material element and indicate the principal directions, (d) Determine the maximum shear stress. Make a sketch of a material element and indicate the planes of the maximum shear stress. ‘ “P MP0... / / / Lag Pfindpafl Strefies carnal {*5 CITF‘E‘Qfir—on- z—q /~ =(/——U;)3+I+I / ~3€#@>=o- MI+IVIJ~+M5 =- O f/IL" Mb 2'0 (I/l[+(I/IL :0 M; + M: 1 0 ' (fin/61 (tr-7W4] wage», ex7s~tg ml ‘ : for ml+m,,-—o ML ‘ mum/1», =0 lm1+ML 10~ PHILIP PARK CIVIL & ENVIRONMENTAL ENGINEERING UNIVERSITY OF MICHIGAN PHILIP PARK CIVIL 8: ENVIRONMENTAL ENGINEERING UNIVERSITY OF MICHIGAN fi @, Midterm Exam 3:40PM — 5:10 PM,- October 23, 2006 ME572 (Prof. Pan) Name , A formula sheet is allowed. Box your answers and include units. 1. We consider a tensile bar under uniaxial cyclic loading conditions. Consider a material element on the bar surface. The coordinate system of the material element is shown. The material element is subjected to constant amPlitude loading conditions from the minimum normal stress of 0 to the maximum normal stress am in the x1 direction as shown. Consider a damage plane for the material element with the normal making an angle of ¢ with respect to the x1 axis. A damage parameter D for this plane can be defined as a linear combination of the shear stress range AT,u to the plane and the maximum normal stress (0,”, )1m to the plane in a cycle as D_ = ATM + k(o‘,m )m _ r where k is determined» to fit the experimental fatigue life data. The shear stress range A1," is defined as the difference between the maximum shear stress and the minimum shear stress in a cycle. (a) Determine the value of q) for the maximum value of D based on the stress transformation rule (or Mohr’s circle). This value of (I) should define the damage ' plane orientation under constant amplitude cyclic loading conditions. ' (b) In experiments, however, small shallow surface cracks are usually observed near the direction Where the maximum shear stress range occurs. Determine the value of ¢ where the shear stress range A1,“ is maximum using the stress transformation rule (or Mohr’s circle). ' x2. 2. Consider a fee single crystal material element with the coordinate axes parallel to the ' axes of the unit cell. The material element is subject to combined normal and shear stresses as shown. The tensile normal stress has a magnitude of 100 MPa and the shear stress has a magnitude of 100 MPa as shown. We consider the (1 1 O) slip plane and two possible slip directions. The slip direction sA is in the direction of [-1 1 -1] and the slip direction sB is in the direction of [-1 1 1]. (a) Determine the traction on the (1 1 0) plane. (b) Determine the magnitude of the normal stress on the (1 1 0) plane. (c) Determine the shear stress on the (1 1. 0) plane as .a vector. (d) Determine the magnitude of- the resolved shear stress in the sA or [-1 1 -1] direction on the (l 1 0) plane as shown. (e) Determine the magnitude of the resolved shear stress in the 5B or [-1 1 1] direction on the (1 l 0) plane as shown. (t) Which direction of the slip will be first activated if the combined normal and shear messes increases proportionally? )) 3..A thin—walled cylindrical pressure vessel has the (mean radius r of 100 mm and the wall thickness t of 10 mm. The vessel is under the internal pressure p and a torque T .' Two strain gages A and B are located on the outer wall surface in the middle part of the vessel as shown. For the material element near the strain gages, the local - coordinate parallel to the longitudinal direction is denoted as x , the coordinate in the circumferential direction is denoted as y ,’ and the coordinate in the radial or out-of— plane direction is denoted as 2:. The strain gages are used to measure the normal strains in the x direction and the direction making 45° with respect to the x direction as shown, respectively. The measured strains are used to infer the pressure p and the torque T. The strain gage A gives a tensile strain of 100 p and the strain gage B gives a tensile strain of 400 ,u. Note that Young’s modulus E is 210 GPa and Poisson’s ratio V is 0.3 for the vessel material. Use’the formulae based on the thin- walled pressure vessel assumption. (a) Determine the pressure p inside the vessel, (b) Determine the torque T , . (c) Determine the principal stresses. Make a sketch of a material element and indicate the principal directions, ' ' ' ((1) Determine the maximum shear stress. Make a sketch of a material element and indicate the planes of the maximum shear stress. _ ' at) = 0~ on“ ’19 =A’ZZS+ k A2745 = 221M ’_ 224/74. (UPI-HDMM: (WMVX. morMa—é, A: =0. 54 La) In; €15 {or (W 19 (IQITMI CI mm SAM!" Z ~9~ gob/em. , (I } Q3744 ,Mofir’: (We/e. I; ~ . w: 3L<U~7+$M3W~ mass-=49 +97— %” ' PHILIP PARK CIVIL & ENVIRONMENTAL ENGINEERING UNIVERSITY OF MICHIGAN AB UM)“ —.—— 35 : 7 (Liaisiék ~&,c957a.2z9) 1 0- coggs : o,¢97n¢’ = f0” fl "(WW rt- ' ../ 1 . (0;; )mm :» - " Y'W‘ (MSJMM 7* m way - 2 C2) My? Wmaffm #me 2 st¢ ¢7n¢ I ~sfn§5 ms¢ 0‘ [973: o cb"lI[C—s]=b‘l~ ~53 OJ 8 C . PHILIP PARK CIVIL & ENVIRONMENTAL ENGINEERING UNIVERSITY OF MICHIGAN C3) “(771% CWC/w/s firmu/a" N I—T 1:0“ng ADM/“e fl = m¢g+§n¢§ — U" Cosgfi? I I I, 0 O PHILIP PARK CIVIL & ENVIRONMENTAL ENGINEERING UNIVERSITY OF MICHIGAN A22: = U‘Cos¢ 97/19/ (Dam 2 (305255 C5”. PHILIP PARK CIVIL & ENVIRONMENTAL ENGINEERING UNIVERSITY OF MICHIGAN PHILIP PARK CIVIL & ENVIRONMENTAL ENGINEERING UNIVERSITY OF MICHIGAN fif-T} i1=$€i°€§a~+~>> rm 1%: EOX‘Ji: mp 115 = $3 2 +557 ,3 + .7: w ,__. .é “ 7-: ("$3 f§+>§fl (OI) (fflzv I 935»: f3: (‘Q-Iéfy-Q) Iriifzz: This!” 50 M. M WSA=~J6=<I+|—2) =0, 0"] (BA, : 0 5 Ce) [IfssIz O “85 - fat—(“Q +21%) fill '2 "'71 o : _SQ —~ 3'00 4? [~33 A; (513 WCI+I+2>— [6‘- 0L 21:4ch affine 313% WMEMWE "‘MMWWW" _ f ( x/ik 4546‘ (07$ 5% 07// AC actrfuafeof, PHILIP PARK CIVIL & ENVIRONMENTAL ENGINEERING UNIVERSITY OF MICHIGAN h - 7 -' (4i) ft/Egfi Wmsfirrmaflon . m I [93: 0 0 ° %::(%+s}) o (00/00 0/00 0 [023: PHILIP PARK CIVIL & ENVIRONMENTAL ENGINEERING UNIVERSITY OF MICHIGAN PHILIP PARK CIVIL & ENVIRONMENTAL ENGINEERING UNIVERSITY OF MICHIGAN (a) In? PHILIP PARK CIVIL & ENVIRONMENTAL ENGINEERING UNIVERSITY OF MICHIGAN J53 _ I I I En ’ E 03:- “ "g ‘37; — E (IOF—OIBXLT) TE 3%? “fix ’O'flfl’m _ '- QIOX/og MP0» a / Eu :- 400 X/O—é : «‘7'; (M0 ~54. ' 9 z: . A; "—7- .7 ' “,3. 120 5/» > I1: lat/L ‘ EL mengP m; : G“ (A811,) G: .lC/H/fl : ,1X/,;~ q ’ UT; : T max/0'3 .— ,,_‘ F6 “lax/£5" g lX/Kg X ' T: lax/05‘ x1/0X/03 MW”.-. ~meM.....~.....~..~E.,__¢.. X $6 I” X /0 ” fix 1 X /, 5 ~ 75K2/0Xi7f .L_ I g ‘ /.5 *X/0 =.~/3,,7¢é>w MM, WNW”--. IT: my MAXI vv—~m..u—__._.WM.._._... - u. (5% : (of :— 10: MP0, 0: T ~ @256 w” Vim/4’“ " gm}??? “ 2%1/ MM PHILIP PARK CIVIL & ENVIRONMENTAL ENGINEERING UNIVERSITY OF MICHIGAN C8I+SZIL CSL'I'SELL 0"“ C -S -3 a“ swap. "Sal-I138”, o S C O 0 E95 0 o c "a ,, +1032, L + 915; ~sc 2,, "I-Cci—s’fiéufi C 9 in O PHILIP PARK CIVIL & ENVIRONMENTAL ENGINEERING “655. #8495). +cs a; 5’2” 10” aL—I 01a; I ‘ // 0 'v UNIVERSITY OF MICHIGAN (9:46” ~> Co$&=:7n61= 093w = 921% z Cams?» 61: [£3 = g’ (Emladé‘n) %=%%=é®flfi%fl Poo/M 2 (00+4-LS‘ «Plain, .15}; '= 7:7? 3/; -~ _ __ E‘ 3.10 X IO3 -5 IL— é} \ .121; — 10+”) 3.2,; 2X“ 3 X175‘X/0 r‘*- ~———«—__._._E I 22‘34 MPGLW [$3 2 51‘s- 22.>_I 0 .129.) ID? 0 O O 0 7‘6 gIt/e Effie/“#41492. , L ‘0} Cfg‘f— Clog—~03) +(22‘1I) Q} = o‘ *%<®V¢m9%+23£;4%%)=a f————-~—~~.-..E?,._Iff. 1-1, Mr 2: We"; 4 x9027,’17‘“' U} = 9L -~~-~ ~ -: 70971515437 ®= WI / v2=4436 ,0; :o, I. ~éoI65L 12m 0 WI 2 o 2.2mm —§,sz o L o o ~H3.H,l (m5 PHILIP PARK CIVIL & ENVIRONMENTAL ENGINEERING UNIVERSITY OF MICHIGAN —50.6HLM,+12‘.2-J. (m =0 12M"); —<?\MLM.\=O $mx= 0. 33% &=J7,oug2” M PHILIP PARK CIVIL & ENVIRONMENTAL ENGINEERING UNIVERSITY OF MICHIGAN -\V PHILIP. PARK CIVIL & ENVIRONMENTAL ENGINEERING Maw Q) UNIVERSITY OF MICHIGAN ’3 53%.g‘ hd‘w ~ ‘ E C§3+€} \ 0. />rlrp Armcefom g (av/~4— [0'39 (pm VFW/571* a: / > EMA6f :— , E 0;:- Dgé * 2 (9/; :f- 4/423 ‘ MM; 1’3 7/76: 0/775/1 I} car/r7ord/Zg iD/ch/L. PHILIP PARK CIVIL & ENVIRONMENTAL ENGINEERING UNIVERSITY OF MICHIGAN ...
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This note was uploaded on 10/13/2009 for the course ME 572 taught by Professor Pan during the Spring '07 term at University of Michigan-Dearborn.

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example2 - Midterm Exam ME572 (Prof. Pan) Name 3:40PM -...

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