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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 8in. 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.6032 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—mmdiameter 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 120mm gage length, determine the modulus of elasticity,
the modulus of rigidity, and Poisson’s ratio of the material. SOLUTION V*15mmdiameter A : £16.2 : 1:05)]. : 7G.7[5'mmt __ 7é.715¥I06 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
_ _ Oé2 _ _ ‘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 = % = lGWZX’O’  _  _ 3 r  3
20%;!) 3;;  225,,  (osstesqzwo) 0557%(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 2.00in. gage
length, (b) in the width of portion AB of the test coupon, (c) in the thickness of portion SOLUTION AB, (d) in the crosssectional area of portion AB. A =(‘i_)(i—) : 003125 WI 50011: 6mm  ————. [9,2x103 Psi _ gs I‘LIZMD3
Ex  E ‘ ———— 24 ,, .0, = camMO“ 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 “ ('aEXHaxzvlo“) = 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;] = WICoﬁ‘tXlzono‘3 + leoxm‘] (an saw/max = (100m)(754.02x16‘) = 0.0754 mm 2.71 A fabric used in airinﬂated 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: loamn 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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 Fall '09
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