hw9_2.73_2.76 - Ex‘l‘VEJ’ 1" E M09599” m LY u...

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Unformatted text preview: Ex‘l‘VEJ’ 1" E M09599”: m LY u W1 Mum? +om c7=E 6; o.» s, =—€—;a:2+ v81“ Problem 2.75 2.73 In many situations it is known that the normal stress in a given direction is ~ zero for example 0', = 0 in the case of the thin plate shown. For this casevwgnch is known as plane stress, show that if the strains 8x and 6,, have been dete experimentally, we‘can express 0;, cry and a, as follows: 3, + vay 8 +V£ V‘ __ 0' =E——y x x. 1_V2 y 1— V2 89+ V=—E—:. Problem 2.76 or 6; t 2.74 In many situations physical constraints preven; strain from occurring in a: given direction, for eXarnple £2 = 0 in the case shown, where longitudm movement of the long prism is prevented at every pomt. Plane sections perpendicular to the longitudinal axis remain plane and the same dlstance apart. Show that for this situation, which is known as plane strain, we can express 0,, 8x and 3,, as follows: 0'2 % V(0‘x + 0y) 5y = %[(l— V2)O‘y — V(l+ V)0‘x] 2.75 The plastic block shown is bonded to a fixed base and to a- horizontal rigid - plate to which a force P is applied. Knowing that for the plastic' used G = 55 ksi, determine the deflection of the plate when P = 9 kips. Consider the 'ppas‘l'ic bfioak. The Slfieawm‘nfi I ‘Pbl‘e E Callie} is ‘P = 7x21331912. The area is A: (3.5)(5.5)= H.125 ital 3 p q. * T = Z:- = 2:21;, = 0.0085006 OJ) '1 k 7‘ = (2.2 Kaoogfooé) = o. 0137 in, a 2.76 'A vibration isolation unit consists of two blocks of hard rubber bonded to plate AB and to rigid supports as shown. For the type and grade of rubber used ran = 220 psi and G = 1800 psi. Knowing that a centric vertical force of magnitude P = 3.2 kips must cause a 0.1 in. vertical deflection of the plate AB, determine the smallest allowable dimensions a and b of the block. can 57wa +l4€ “liaise" \byock W ‘i’dm’fl V173 Lit. I‘i’: Cawwies a shearer-vi $011.42 e 04.7 ‘lb JP— ‘3 “I . 2 The aheqn'ng stress. is ’2: '-' J32: ~ - -m- . 2‘ ow P£1U1V?L’i A ~ 2,: - (23(2203 \ 7.2727 In But A ‘~‘ L30? 5 Hence in = EA—O ‘= 2.42 in. low”: 2.42M. "M Use to:- 2J-sz :9; am! 225-.- 220 PM gbfimh‘nfi strain, Y ‘—‘ :5: 7— iii: = O. ”1222. 0'2 — 1/(0'x + 0y) 1 "”x = EKI‘ V501 - Va+ no.1 84’ I ég‘fixt‘sy- v62): % :hEL[CI'”1363 - VCl+v36xj [-Vgx +63, e 2J2 (5&6; )] q '5 " 'E 71:2}; ’ ”67-52 F5" 7” = 7’;- = :éiifi, 0.0085006 S -: h 7‘ = (2.2 Kabogfooé) : 0.0137 in. ‘ an “Papas‘l’ic Bfiock. The sbear-‘nfl ‘POWC? Caf"!‘¢?.ol) is ’P = 7x103 21:. The area is A = (3.5)(5.5)= 151.25%" 2.76 A vibration isolation unit consists of two blocks of hard rubber bonded to plate AB and to rigid supports as shown. For the type and grade of rubber used 1311 = 220 psi and G = 1800 psi. Knowing that a centric vertical force of magnitude P = 3.2 kips must cause a 0.1 in. vertical deflection of the plate AB, determine the smallest allowable dimensions a and b of the block. ‘ Consider +Le Wobbler 52ml: w “Hue We} hf. I'l', Counties a. SheamM-a“ {ox-4:; equal +9 ‘5']?- TWWE 1:3; 3 _ 3.2“503 _ . 2 22,-- (23(2203 ~ 7.2727 In w Pequfeerl A '5 But A = (3.0? la » b = ‘A’ = 2.42 in; Hence is,“ = 2.42. m. M {it’lé‘j (3:)? 220 Pfil' O ‘= O. IQ’ZZZ ...
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