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hw6_soln - 1 Bar ABC has a lectangular cross section of 300...

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Unformatted text preview: 1. Bar ABC has a lectangular cross section of 300 mm 100 mm Attached 10d DB has a diamete: of 20 mm Moment Function Mix): As shown on figumu). «a! Mamet-l Functions 1119(1): As shown on figurefin). _ «a! Work Equation: For the slope a1 poim A. combincEq19~19 51:19-15 zo m 1 1. 1 37: {19de “N— IE1: ”LN-mg. = EL. {i—D.3333x}(20.fix)dz (11:20:01 + (—0.411561115001151 111: {a} B _ 30.0kN-m3 __104.167kN 03555101 941951104 A __ E! .45 {'H". I 30.011000) 104. 1611 1000) .—. 2001 10915101110331] _{;(010221]{200(109)1 = 41.991 ( 10") rad a 0.991( 10") m1 Ans The top of the beam 15 subjected to a temperatme = 200°C While the temperature of its bottom is W = 30°C. If at = 12( 10 My C determine the vertical 11L. 4“ ment of its end C due to the temperature Eu 1“ » 3:13 M1 C‘L—"JM $50 mm I I .4...“ In! 2 «m 1-x1112>uo*1.(—ss.m a c o 1‘15 dx =93‘3 mm Jr All 9-3. The hem rod has an E = 200 GM. G = 75 Gl’a. .5: :1 radius of 30 mm. Use the method of virtual work i: icicrminc the vertical deflection at C. Includc the pints of bending. shear, and torsional strain energy. y 2m Law! :11 dx K eix+ Wis 15! + [a {“2" :3 Whiz _.._. I {*xxfix) dz EJCx- LSKZI: mat: a E! E! 1Wm: 551N812” ”15%)“?de(«ax—mm} + a 0 6A 23 GA G! 2.4V! (:63; + Sistine]: + 3003} 190i WWGMOED‘ 75(10’1(fl{9.93}2 ?SE 20’ HIE—HOBBY 7 =2 0.022424; + QWQZ + 0.0314380 - .Efikh = 0.051493 =54.‘frnm .L Am {‘dv 2 “W {Q My Determine the reactions at the supports A and B. _mnstant. r: Reactiom‘: FBDfia) . 3E1“: =0: A, =0 Ans +T2r;=3; yang—15:0 [u L L 6 am a 0; B, (L)+M‘ {Ii—XE]: o [2] d of Superposition: Using 1hr. table in appendix C. the displacements m: m 7wL‘ .L u _ 9L3 _ 3,!) “5 “384E: ’ ' BE! ‘ 3.?! pnfibility condition mqu‘u-cs pl) 0:u3'+u5” 0: 7m!) + __B,L’ 3845’ 351 B _ 7wl. A , “ Tia" m 'ng 3_ inm Eqs.{l] and [2] yiakis, STWL 91:91:.2 A = —— MA = m Ans ...
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