HW8 - 2.63 In a standard tensile test a steel rod of-in...

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Unformatted text preview: 2.63 In a standard tensile test a steel rod of % -in. diameter is subjected to a tension force of 17 kips. Knowing that V: 0.3 and E = 29 X 106 psi, determine (a) the %-in. diameter 1” M elongation of the rod in an 8-in. gage length, (b) the change in diameter of the rod. I ’ PS PROBLEM 2.63 l7 kips L—Bin'—’l SOLUTION P = m up = mm? .95. Ag 32; dz: Mg}? = O.60|32 in _ .. |7~¢ID3 . 103 - 6' - ‘AE — m2 = 23.17XI03ps.‘ 2.5% = % = 7743on ‘ 3x = L?,. = (3.0301743: we“) = 7.80xlo‘3m = 0.00720 m. « 57 = ‘ 72 82: = -(O.3)(?79.‘13(I0")= ; 292.5 XIo" S, = olej = (%)(-2‘12.5>Ilo") =—2$é no“ {u = 43.000156 m- d PROBLEM 2 6 4 2.64 A standard tension test is used to determine the properties of an experimental ' plastic. The test specimen is a 15—mm-diameter rod and it is subjected to a 3.5 kN tensile force. Knowing that an elongation of 11 mm and a decrease in diameter of 0.62 mm are observed in a 120-mm gage length, determine the modulus of elasticity, the modulus of rigidity, and Poisson’s ratio of the material. SOLUTION V*15-mmdiameter A : £16.2- : 1:05)]. : |7G.7[5'mmt __ |7é.715¥I0-6 MI P = 3.5 ”[05 N 5: a? = .353???“ = ”-809” P“ én': -E!: (£1110 : "“1LGG7XIO.S 6 E = g = $205333 = Zlcxlo‘ Pa. - Zlé MP6. A : — O-$2 MM _ _ O-é2 _ _ ‘3 £7 _ :5": '5 _ 41.333 XIO ‘ 5 41.333XID“ . _ -_J. t ' 2 2. 7’ a “H.667Xl0—‘3 0'45“ 4 F E ‘ G _ _ 2’6“"? : 74.511106 ’Pc..= 76*.5 MPct “ = 20+») ' 204414507) 2.67 An aluminum plate (E = 74 GPa, v = 0.33) plate is subjected to a centric axial PROBLEM 2'67 load which causes a normal stress 0. Knowing that before loading, a line of slope 2:1 is scribed on the plate, determine the slope of the line when a = 125 MPa. SOLUTION The 5490,92 a:c+e\l‘ c)e£0rwm.+tou is +4” 9 :- Mr). 6" xo‘ - 5*: TEL = % = l-GWZX’O’ - _ - _ -3 r - -3 20%;!) 3;; - 225,, - (osstesqzwo) 0-557%(10 7. I — o . 00055743 = , A has = < = 1.7fi55I 4 E ‘+5 \+ 0.00\€841 I ~ PROBLEM 2.68 2.68 A 600 lb tensile load is applied to a test coupon made from k in. flat steel plate g (E = 29 X 10" psi, V: 0.30). Determine the resulting change (a) in the 2.00-in. gage length, (b) in the width of portion AB of the test coupon, (c) in the thickness of portion SOLUTION AB, (d) in the cross-sectional area of portion AB. A =(‘i_)(i—) : 0-03125 WI 50011: 6mm - ———-—-.- [9,2x103 Psi _ gs I‘LIZMD3 Ex - E ‘ ————- 24 ,, .0, = cam-MO“ i (a) 5x ‘ L48, = (2.0 )(ccz.o1x)o“) = 1.32“] >40" ta. ‘ 27:2,}- as, = -(o.3o\(cc2.ow(o“)= — 138.62xlo“ (lo) 5m». = may “ ('aEX-Haxzvlo“) = Ans wd“ .‘n 4 v (C) Snag“; Z5082 = (fi)(-I‘78.62>«Io“ = 42.41210“ in. I . (a) A: wt = w°(:+s_yit,(l+sz) {g = wot. (1+ 21 + s; + 81813 AA = A — A0 2 Wot°(£y' +51. “FE-’22.) f = CEBU; )(- 193.62 Mb" - 191.42 «10" + “31:53:“. +u~\\ = - 12.4: NJ6 in‘ 4 C PROBLEM 2.71 £2 = _EL(_v5’ _ 116:, + 5;] = WI-Cofi‘tXlzono‘3 + leoxm‘] (an saw/max = (100m)(754.02x16‘) = 0.0754 mm 2.71 A fabric used in air-inflated structures is subjected to a biaxial loading that results in normal stresses ox = 120 MPa and al= 160 MPa. Knowing that the properties of the fabric can be approximated as E = 87 GPa and V: 0.34, determine the change in length of (a) side AB, (b) side BC, (c) diagonal AC SOLUTION 5x = (Zoxto‘ Po. , (>3: 0) 6, = Icovlo‘ ‘Pa. Ex= ELCS} — v63 — vG‘Z) I 87Xlo‘7 = 759. 02 x10" [[20 Ho" — (0.3?)Cléofil0‘fl l z = (75Mm)(r.37w I0“)= 0.1023 n» A Lake) stale: OF‘V‘CgH hi'anjpe. ABC as a.) L) anal c 7. c. = 03413 owed» al.‘¥l5ewon+a‘ais L», caijjus 2C alc = 2an + 25 db dc = % ala. + ‘E’ AB Bu+ 0.: loo mMJ b: loam-n mm =12: m». 0.0-: gm SAC = 91c 0‘0759 ham ab = 33‘ = O.|37O WM [Do 75 -- O.l028 : . '25. 25, ( ) OJZQG mm ‘ (0.0754) + ...
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