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midterm2 - Midterm Exam ME572(Prof Pan Name IDA/GE Pﬂfk...

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Unformatted text preview: Midterm Exam. ME572 (Prof Pan) Name IDA/GE Pﬂfk . 3:40PM — 5:00 PM, October 24, 2007 A formula sheet is allowed. Box your answers and include units. 1. For a given Cartesian coordinate system, consider the stress tensor 11.1 6=1111VJPa 111 (a) Determine the principal stresses and the corresponding principal directions. Comment on the stress state and the principal directions. (b) Determine the maximum shear stress for the stress tensor.‘ Make a sketch 'of a material element with respect to the principal directions and indicate the plane where the maximum shear stress is acting on. " 2. Consider a hexagonal closed packed (hcp) single crystal material element with the / \' coordinate axes as shown. The ratio of the length c to the length a is about 1.57 to ‘ 1.89. We consider the slips on the basal plane as shown. We use the usual vector ' notation to denote the slip directions (we do not use the Miller-Bravais indices here I for simplicity). The material element is subject to the stress tensor with respect to the 1 . coordinate system as shown. The stress tensor is 100 100 100 0' = 100 100 100 MP3 100 100 100 The slip plane has the normal in the direction: of [0 0 1]. The slip direction sA is in the direction of [2 -1 0], the slip direction .53 is in the direction of [0 1 0], and the slip direction sC is in the direction of [-2 -I 0]. (a) Determine the traction on the slip plane with the normal in the direction of [0 0 1]. (b) Determine the magnitude of the normal stress on the slip plane. (0) Determinethe shear stress on the slip plane as a vector. (d) Determine the ma nitude of stress ' e sA direction on the slip plane. ' (e) Determine the magnitude of the resolvedshear stress in the sB direction on the slip plane. . \T (f) Determine the magnitude of the resolved shear stress in the sC direction on the slip plane. (g) Which direction of the slip will be ﬁrst activated if the stresses increase proportionally? - x3 \_ 3. A thin-walled rubber tube has the inner radius of Ri of 25 mm and the outer radius R, of 30 mm. The length of the rubber tube is long and denoted as L. The rubber tube is used to shrink (or expansion) ﬁt to a steel bar as shown. "The length of the rubber does not change after the fit. The steel bar has the radius of R, of 26 mm. The rubber tube has Young’s modulus E of 3.5 MPa and Poisson’s ratio v of 0.5. The steel has a much larger Young’s modulus than the rubber. We can treat the steel bar as rigid for this ﬁt. Based on the cylindric ordinate iysttanmasshownym (a) Determine the'outer radius r0 of the rubber tube after the ﬁt. (b) Determine the strain tensor of the rubber material on the outside surface of the tube. Write the strain tensor with 6 components in the matrix form. (c) Determine the stress tensor of the rubber material on the outside surface of the tube. Write the stress tensor with 6 components in the matrix form. ((1) Determine the maximum shear stress. Make a sketch of a material element and indicate the planes of the maximum shear stress. - L4) Prfmcffaa Wges Z: alwectran. ‘ D‘éﬁlpgiimﬁf +13 —— o Iu= 0E1: = 3 MP“ _ IL=IZ(V7432'DJE* _C5;3—v3-) zgzL(3x3~—?}*=O, IB=0IetIEIII== I+I+l-l-I-I =0. 7?) Di = U177 :0. =9 An oIfrectTon noITmIar +0 6L Per?” - W ml! I32 ml or 441 M —'=9 Cf/l’m/rl'mj State 071) stress, In)L E: Cﬁzo © wnTowTd stress PHILIP PARK CIVIL & ENVIRONMENTAL ENGINEERING UNIVERSITY OF MICHIGAN PHILIP PARK CIVIL & ENVIRONMENTAL ENGINEERING UNIVERSITY OF MICHIGAN C4) waﬁ-m. a7? (’41 3/17; F/M PHILIP PARK CIVIL & ENVIRONMENTAL ENGINEERING UNIVERSITY OF MICHIGAN MAI-M10; (01> :Eaﬁgot 4;,» C3). wIﬁch oITnectTan MII Iae acéTva'éeJ‘ PHILIP PARK CIVIL & ENVIRONMENTAL ENGINEERING UNIVERSITY OF MICHIGAN 3, 7Z7n ova/”EX ”4556/“ fuée. PIE/183'" 3 P) '= 3—th Radar :790 2, 30mm . R3624 : E3 : 3—6MM . (a) Oder mills Y‘o V11 2&6 MM: 1/— : W. .E‘ecéérn © ﬁrm 071) mAéer‘ (Ema—t cémfpl. 24R t <730l4790~1)7T/«’ = Q3123") 7'5 ~ 293;: Roi'R‘kvL 17:3— 2 aom—JS‘L +344; 2 7w ._. \ , L ”'9__/Z__':___________ WW "7: é”: ~44»; PHILIP PARK (3er & ENVIRONMENTAL ENGIE 'ING UNIV'SITY OF‘MiCHl-G en t ' €99 Tm WAS . 5 21,1 (L/Lo) - . 24M \ 59“ ”Z” 20 1 ﬁlm/76 7-?0‘048. SH": [01 ﬁiﬁ- = ~40»? % ~405; , V 6‘ —z . L: E j = ——0.032_ O 0 O yrDJ—f 0 K d o @‘T see mx'é fax. 7 \ M0 VO(WLL 461W 5 1' ; 5 J l i I i f i I i ! ‘ ( l 3 x I I I 3 l 1 3 C0) Stress fwsar‘ a} 0L4%e,r‘ Swwg’mce) 4-6. 31/1:ng O?" * O l C~Pyea S‘M“7[L\<‘€> Ear; ——y (ma—ICE?) = ~a.o”§;.¢><3.ﬁ' PHILIP PARK CIVIL & ENVIRONMENTAL ENGINEERING UNIVERSITY OF MICHIGAN /’\ (a? ) 7 74¢ mmfwum 3/1141"- (TAM 3 @640); '/(/o s/Lear mmﬁo/szs r71 [51‘] => ”mails/Lear occur at 945°,“ PHILIP PARK CIVIL & ENVIRONMENTAL ENGINEERING UNIVERSITY OF MICHIGAN ...
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