Class 9 - Indeterminate Axial Members

Structural Analysis (6th Edition)

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Unformatted text preview: Offiféfifivég (’33 Eégzéw gfigmfirié’ffimé @ fi§g§ $32 aafiffig {$3 “$2 fig“ ggfiéfé @Qywxyagmgm *9 i? A: mamfi :55; Lzméfl W ifigaélm {fifiifié : fig; g“; {figw ‘ Ckw gfifig‘i mama? :aw $24M @fi?€3w%§%§ gfim %&5fi&é 5M»; b23136; é; $i§§é “11% %%i%%fim¢ {71‘ fl LE3: A L i g? 2 Q5! ? E; W é‘éfigéf’fi“? Maggi" @Efiksm Swag “7&6 QME MW i vmgngg : If, @Ufifi‘iam : {663: ,MWWVfl/w” «*«QNWM’ figéé ggfia 1:: $3 Z? “can: ,. thaw §T$+»?‘3Q gm mm} 617? {25%;th Sfiw éfiwg j? big $3532 fiQE; @Qfi?§a§mi 3 $ng 33$ C {lgEfifififé gamma; $3» mm; 5 MAM fig? WQQ mg fm figmgwél a”; 9 g Lg‘g g Q o 91%” E E if wggbg ggmg; %§%} 9ng iéfigkgggfii fixgmggfgig; 33g “#%@6 Hgfimgg Mam 5M?w~z$ gafm?’§gmm, gamw 3%wa g??? L; gig? 4%: Wflj 6mm fifiigfi Problem 2.440 A rigid bar of weight W = 800 N hangs from three equally spaced vertical wires, two of steel and one of aluminum (see figme). The wires also support a load P aeting at the midpoint of the bar. The diameter of the steel wires is 2 mm, and the diameter of the aluminum wire is 4 mm. What load Pallow can be supported if the allowable stress in the steel wires is 220 MPa and in the aluminum wire is 80 MPa? (Assume 7E5 = 210 GPa and Ea = 70 GPa.) Rigid bar of weight W Problem 2.440 A rigid bar of weight W = 800 N hangs from three equally spaced vertical wires, two of steel and one of aluminum (see figure). The wires also support a load P acting at the midpoint of the bar. The diameter of the steel wires is 2 mm, and the diameter of the aluminum wire is4 mm. What load Pallow E5 = 210 GPa and ElI = 70 GPa.) can be supported if the allowable stress in the steel wires is 220 MP3 and in the aluminum Wire is 80 MP3? (Assume SECTION 2.4 Statically Indeterminate Structures 97 STEEL WIRES 5=2mm (TS ALUMINUM WIRES A=4mm EA : 70 GPa = 220 MPa (7A = 80 MPa E5 = 210 GPa FREE-BODY DIAGRAM 0F RIGID BAR F5 is Fs lP+W EQUATION 0F EQUILIBRIUM EFVBI’I : 0 2F5+FA~P~W=0 EQUATION 0F COMPATIBILITY FORCE DISPLACEMENT RELATIONS 8 S 5 25A _ FSL EA 5 A _ FAL EAAA (Eq- 1) (Eq- 2) (Eqs. 3, 4) Rigid bar of weight W SOLUTION 0F EQUATIONS Substitute (3) and (4) into Eq. (2): F51. _ FAL (Eq 5) ESAS EAAA ' Solve simultaneously Eqs. (1) and (5): AA F = P + --——’1—~—~—) . 6 A ( W)(EAAA + 255A: (Eq ) F ~ (P + W)( :A’ ) (Eq 7) ‘ EAAA + 2ESAS ' STRESSES IN THE WIRES F A (P + W)EA = - = -—————-————- E . 8 “A AA EAAA + 253A: ( q ) F, _ (P + W)E, ‘7‘ '" A: ' EAAA + 2155As (Eq' 9) ALLOWABLE LOADS (FROM figs. (8) AND (9)) 0A PA = jig-(EAAA + 2E,A5) — W (Eq. 10) A 0.5 P: : E(EAAA + ZEsAs) - W (Eq. ll) SUBSTITUTE NUMERICAL VALUES INTO EQS. (10) AND (11): As = E:- (2 mm)2 = 3.1416 mm2 AA = 3T— (4 mm)2 -—= 12.5664 mm2 4 PA = 1713N PS = 1504N Steel governs. Palm 2 1500 N 4—— ...
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