202hw17_soln - P robiem 2 _56 2.56 Knowing that a 0.02-in....

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Unformatted text preview: P robiem 2 _56 2.56 Knowing that a 0.02-in. gap exists when the temperaurre is 75 c’F. determine (a) I the temperature at which the normal stress in the aluminum bar will be equal [0 -ll - ksi, (b) the corresponding exact length of the aluminum bar. | 5;: ~11 kg: : ~szozps.‘ P = — 61,45 (11y103l(cz-s)= 30.8 1/03 its. Bronze Aluminum A = 2,4 in.2 A 2 2.8 in,2 N ; 13 E =15 ><106 si E 210,6Xltlfigsi 5% la” "3 “be 11° (1 = 12 x 10“ /°F a 2 12,9 x 10‘ /°F p LL PL : -————- J, ———“"~ SP EbAu EA. Ad. P __ 30.3v103) H) + (30.81103 Ia‘) ,,E::j::::].f (Brice-XX. “ll ((0.6xto£l(z.8) : 30. G57 «10‘3 m. AVa-‘pql712 eflornja/l'fovt 'lzur +lIe-rwxap mp4nsi'on 2' ST T 0.01 + 30.657Hd3 =- £0.657xlo" .‘V._ 80+ 57 : Lbdbwfl’) + LQdQLATl .— 04 lClelo"XA'l-l + G‘glOQ-WIO“)(AT) = (400-0104) AT Equdlina (%O.QKJO‘C)AT T 50.é§7x(o_3 AT: 12;.60': to) Tm =14” AT - 75+ (26.6 r 201,6”F A _ _ PL Lt) 5h» Ladifllfl 53%: _ -c _ $30.8xtoglél8) _ O -3. _ (18)(,l2.‘7xlol(12é.€l uaéxlocxzm - l-"HZHO m Lam = l8 + (0.7I1YlO-3 r 13,0107 in_ A Problem 2.62 A : 0.01073: I51. 2 mml 1‘ H. ZXIO‘L m1 P: 2.79103 N 6: Le 2-75xzo3 x A 4 l‘LZXIQ“ : Msamuo‘ Pa _ _§_x_ _ waiz‘ryzo" 8* ' E ' locum" 5:: t 81 r weft : (a) L.- 0.050 m 5x : 00) W: 0.0|2\m m 7: : 0.00% m ‘3 (At A: AA: A—Aur : (XXQOMXQOOIQ X» ZHELHIUC ‘) : ~8.2.5 V/u—c' n1 r LE, : (0-0503(7ie.15x/o“l r w°(\+zj\7':,,(l+ at r won (H EJJHEL 4 5,93 Wpto 4631 4 he7iirj,lié‘ +emm» 7: 2.62 A 2.75 kN tensile load is applied to a test coupon made from 1.6—mm flat steel plate (E = 200 GPa, V: 0.30). Determine the resulting change (a) in the 50—min gage length, (b) in the width of portion AB of the test coupon, (c) in the thickness of portion AB, (d) in the cross—sectional area of portion AB. : 7l6. 15x10" ~CO.3O)(7/é. stlo“) : — 214.9%:0" gag/We“ m 0-0358 mm A 6 gj : w 53 r (o.onx\(—2IH.84 Ho") :~rz.57wto' w) “0. 00158 mm «a .4; :51 : (0.0olé\(—ZH.8H x1043 :—3L}37xlo"‘ m ~ 0.0003‘f'37 mm «I. A0 : W0 ta 1 'L —- 0. 00325 mm 1‘ Problem 2.63 2.63 A standard tension test is used to determine the properties of an experimental plastic. The test specimen is a %-in.—diameter rod and it is subjected to a 800 lb p tensile force. Knowing that an elongation of 0.45 in. and a decrease in diameter of 0.025 in. are observed in a 5-in. gage length. determine the modulus ofelasticity, the modulus of rigidity, and Poisson’s ratio of the material. A = $91 = JE(=§)Z= 0-302.796 m‘ i in. diameter 5.0m. b P ___ 300 AM, L 6:Y = A£= 3—273-32672— = Laney/0‘ Pa...- IM 8] = ii:= iii: 0:50.090 5, :. j” = jig—25— : —o.o4o E = = W3 =’ 2?.‘773X10393C E=2¢1O XIOS vsi “‘5 7) : — g— : _ ‘gfggj = 0-444% 7/: 0.4% «an E 22:773 )((03 : _——. L- m =1 S a" ___ v 3 . G 2(l+zl) (211+0jiqqu13 lo'ozqylo ’7 G‘ l0.03 (0 P‘H Problem 2.79 2.79 An elastomeric bearing (G = 0.9 MPa) is used to support a bridge girder as shown to provide flexibility during earthquakes. The beam must not displace more than 10 mm when a 22 kN lateral load is applied as shown. Knowing that the maximum allowable shearing stress is 420 kPa, determine (a) the smallest allowable dimension b, (b) the smallest required thickness a. sh eam'nfi "Fl—'1ch P 7: 22"“): N Shem/4n») shes; f ’ 420 VIO'S Pa ~— 3. - i=3: axio‘ _ X -3 m ,L- A .. A ,E m -52.32/ /o m : 52.38! x103 mi” A = (200 mm b) A 52.38: x103 brg—m=’“§;‘gmf2€2mm <1 -L_ 420x103 _ x -3 7A- G _ Oqfloc. — 4é6.é7 lo =i: ’0” _ =2/.4 m 4 Problem 2 97 2.97 For P = 8.5 kips, determine the minimum plate thickness 1* required ifthe ' allowable stress is 18 ksi. 131 ‘ l 1 From Fug 2.6? a. K = 2. 22 _ 1:2 _ KP _ KP Gw‘ A"? ‘ dint - t d; 6;." : 2.12')(8.5) _ ~ in t (1.200%) ‘ 0'3.” ' . . .1 M He iQ-JfleJr ‘D= 2-1m) rag = LG m :3): - $2— - 1.375 Y‘s: % = 0.375 in a: : #7:? = 0.2344 . _ _ LE _ KB FPOK Fifi E K “ 6;“: ' AMI“ ' d8: 1‘; KP = ("7"‘(8'9 0.502m. ‘ 6.36m (LéW = The page» Var/Fae .is .‘H‘LE requfwed vu‘vxl mum Ppafe {elite/161615 t: Dog-Win. “ 2.126 The uniform wire ABC, of unsu-etched length 21, is attached to the supports shown and a vertical load P is applied at the midpoint B. Denoting by A the cross- sectional area of the wire and by E the modulus of elasticity, Show thal. for 6 << I, the deflection at the midpoint B is P 5 2 13" AE Problem 2.126 2 3 n9 _ P 9 PP EflonSq'l'lon $53 — 2AES DEF/Peal“)?! 49 Foam :Hie walrl‘ “hf‘fatvijle. (19+ 8,33; = 1’2 + $1 a S eh 3‘: 42+ 2A0 5” + 3.3 u?“ .= 215,5 (0 + lie) 2’ 2p ‘5” 21? “BE; ABS 3N 13,93. N 3_E__ S n AE “gflj‘ AB .4 ...
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202hw17_soln - P robiem 2 _56 2.56 Knowing that a 0.02-in....

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