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Unformatted text preview: 2.13 Rod ED is made of steel (E = 29 x 106 psi) and is used to brace the axially
compressed member ABC. The maximum force that can be developed in member
ED is 0.02P. If the stress must not exceed 18 ksi and the maximum change in
length of BD must not exceed 0.001 times the length of ABC, determine the
l P = 13" kiPS smallest—diameter rod that can be used for member BD. Problem 2.13 F'BD = 0.02P = (0.09.)(Jso\= 2.61695 = 2.6XI0315. ConsiJem‘na s'l'ressi 6’: 13 ksi = [8*1’0: PM
Re _
6‘7:  A=E€= 72% = 0.194% M COnEEJEJ‘Cnz a$rmm+¢\om: $=(0.00H(}W§3= O."+‘I iv“
5 Fol—m. _ Featan (2.6ylo3)(54) _ 0.033g2in2. .. _————————————————_, Larger area Saverus. A = 0!‘4WN l‘n,‘ =3: 1 3 _ [L75 _ I HXOJQWHS _ 
A Lfd ' d ’ Tr  o!  OJ'ilal \Yj,‘ Problem 2. 1 4 2.14 The 4‘mmdiameter cable BC is made ofa steel with E = 200 GPa. Knowing
that the maximum stress in the cable must not exceed 190 MPa and that the
elongation of the cable must not exceed 6 mm, ﬁnd the maximum load P that can be
applied as shown. Lac = 1/ 91+ 4‘ = 7.2m m Use loaf A3 as apree loaaly. P
 FB‘
QEMA = o 3.5? — (Glﬁjim Pack o
1: = 0.?509 FBC Ax
Cans/Jena? aijouaéﬂ: S‘h‘ess: 6: IqOXIOG Pa A7
A = %al1' = Egacm'r = [2.566 #10“ m‘
6* z %‘ .1 FM: 5A = 090 x;0‘\(22.5;5xzo“) = 2.388MO‘ N
Conga/aria? ajﬁordwlaje epomjn+:‘an= a = g x/odm
—6 q I
g 7 FngL'I‘G 1‘: a 025%”: ilioo “to ————x 6 xloﬁ) : 2.041. X103 U
Swapper Vaduz. gave/As. lie—— 2.07/Xl03 N P . 0.9593: F“ = (031509“;ow:x;o‘)=l.qzzwo’u Prmss IN a Problem 2 16 2.16 The specimen shown is made from a lin.diameter cylindrical steel rod with
' two 1.5in.outerdiameter sleeves bonded to the rod as shown. Knowing that E =
29 x 106 psi, determine (a) the load P so that the total deformation is 0.002 in., (b)
1%in. diameter the corresponding deformation of the central portion BC. P ' A /
lin. diameter B Sum P =(RQX'O‘XO0013C6ﬂas)” = magnum: P: 91.53 Mn «:3 (H = "‘ T:13 if: ’ ————Z:fiﬁ*g‘3’<a.zm) S=l.25"l><lo’5{m_sn Problem 2.24 2.24 For the structure in Prob. 2.23, determine (a) the distance h so that th: deformations in members AB, BC, CD and AD are equal to 0.04 in., (b) tht
corresponding tension in member AC. 2.23 Members AB and CD are 1;~in.diameter steel rods, and members BC and AD
7 are Findiameter steel rods. When the turnbuckle is tightened, the diagonal
member AC is put in tension. Knowing that E = 29 x 106 psi and h = 4 ﬂ, determine the largest allowable tension in AC so that the deformations in members
AB and CD do not exceed 0.04 in. (Q3 SinHes? Use Join'l‘ B 0.5 m‘Free body. 8 F3:
FV/DM simupar ‘l’n‘anjpesj
Lo
RB_1_F® F» F‘" F» F” h '5
'y, ‘ T ' E2
F3; 5
' Ear“ i 3 N'An 9e Geome‘l'Ii For ea, um? daton 4+l‘0145) 1 b A
S“ = 58‘ A FELAIZB : ‘— CA): ‘0' FAB = 7; ' A: Fer. Equajihnj expressions ‘Fov‘ FAB, problem 2 26 2.26 Each ofthe links AB and CD is made of steel (E = 29 x 106 psi) and has a
uniform rectangular cross section of % x l in. Determine the largest load which
can be suspended ﬁom point E if the deﬂection of E is not to exceed 0.01 in. Staci{cs ‘ FN‘ E, = O
Awe“. 6F IQIwPI
Aaﬂm = max = O
LequH:
" L = 8 in.
. _ FEEL. __ (1.5?)(m _ _ .6
DeFov‘wmi'ncms. SAB EA W  LBDSQVID P
_ F L _ 2.5?)(23 : 6
SC” ?F ' (muu‘MOJS) Ziggwlo P
DC'FIe—e'ifons. Pofh‘i‘ B. 53': 5,45 Sg=.6552>‘10“P Q
POM C. 5; = Sm 5: = 2.753910"? L
Geomeiréﬁ plumage.
55 C E
B
‘55
F—IDIni— [gin—>1
— C 6
e: 53+3c: Lessnxlo P+ 7.758exlo P = oﬂulsguodp LB; ‘0
SE = 3., + LcEe = Q.7s.Séx[D'°P+(Is‘)(o.'I‘H38>‘IO"P) = 737ﬁ3x/6‘P\ Limf'i‘ihj VAIUC a? 5". SE 1' 0.0] in. Lmﬂma VaJue a??? 7.3'H3xzo"P = 0.0!
P: [055 J]: P: LOéé hrs :1 ...
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This note was uploaded on 12/01/2009 for the course CVEN 233 taught by Professor Gardoni during the Spring '09 term at Texas A&M.
 Spring '09
 Gardoni

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