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# Sol HW3 - AI Problem 250 2.50 A 4-ﬁ concrete post is...

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Unformatted text preview: AI Problem 250 2.50 A 4-ﬁ concrete post is reinforced by four steel bars, each of Knowing that E, z 29 x 106 psi, .2; = 5.5 x 104/01; and Ec= 3.5 x 106 >< 10'6/°F, determine the normal stresses induced in the steel and in 3. temperature rise of 80°F . As 1 Li Eda : (413%); = 1.76715“ in‘ ACE A-—A\$ = 81~1Jé7i§ -— c2.233 :n‘ Le’i‘ P; be ‘4‘)1: +¢h¢3mfﬂe in “Hue conévse-l-e_ Fad" Equieoﬂgyﬁuwx b.1041: Zero ‘beeaJ .‘Hue Compressive 'Fosl‘ce {A *Hne \$01)!» 5419) won-33 3‘3 “Pa. P5 — _ __ P —-) EsAsq-dstlxr) _ ﬂ+ 015M! 5:1 5:4- otJAT) Ma‘l’o‘zﬁn3: sczas E; + 04441") = 15:15 + Gem-'3 (51A; + {Es—‘59P: = (04V seam-r3 ; 1 - l ((3.6xzo“)_cg2.2333 + (Qcifo‘XIJé-WSQ P” 4 7 03\$ 10‘ — 5.5xlo" )(80) Pc ? Buzz-wara M03 )5 13:, T — 3.33% ><1og 19L A \ 'n ' '3_ . . S'i'wass I‘m isles/9 65? ~135— 1' ‘W “ *— L888 vlos Psi A5 1.7mm - 3 = ‘ 2.53 Two steel bars (13. '= 200 GPa and a; = 11.7 x 10%;) are used to reinforce a PrOb'em 2-58 brass bar (Eb = 105 are, ab 220.9 x 106/°C)which is subjected to aload P = 25 kN. 'When the steel bars were fabricated, the distance between the centers of the holes i‘\ which were to ﬁt on the pins was made 0.5 mm smaller than the 2 m needed[ The steel bars were then placed in an oven to increase their length so that they would just 15 mm ﬁt on the pins. Following fabrication, the temperature in the steel bars dropped back to room temperature. Determine (a) the increase in temperature that was required to ﬁt the steel bars on the pins, (1)) the stress in the brass bar aﬂer the load is applied to J m if 5 mm Brass (370 Required +eynPev~ae+W€ change “pov- ‘Feeio Na eel-i on - _ S1- ? 0.5 Hm I 0-5YlO—3m eJ'TemPewce-i‘UWE change. require—El +5 éxPamei est-e6? beer in.) “Hr-ii: amoUﬁ+ Steel 40mm 5T: L015 AT} 0.5>406:(2.OO)(II.7>!/O“)(AT)3 A7“ = 0.53065 = (1)0132 lo“)(ATl AT: 21.363 “(1 ' 21.4w; .4 (5) Once QSSSM-Ezpeji a I-lensfl’e ¥oree Par aieueﬂups in “He s'l'eeep «Moi o. compWESsive elk-wee P” deueﬂops in ‘Hie brass) in omelet“ +4 gym/\de 'Huz stew? any? can-+wa ‘Hte brass. Epdntjdd'iom oi: S+E€Dp; A5 = 400 Mn:- 3 ‘f‘CbOMIQ-G M1 I [:VL _:.!Pi‘(2_oo -? :- (3935 ‘ ASE!) =(L’00x'0‘ﬁcgocxlcef5r = 25 HO P COM‘l‘r‘eoi'Eo-e Oi: hV‘é-s‘b : A}. t (40 )(15 3 ‘- 600 MM?- = 600 *IO" "'1'- :Ir '5‘ b } ' PL " WM"— = 31.7% my"? (3'35: A585 H (600 xlo'sxltﬁwo'l) 89+ (Spls --l~ I (5,»); riser-36,an 4v 'Hne. -I'ni'|"l.aj amoun'l' JFMI‘S'PF'i' J (Sf)s+(5,,)b= 0.91073 55.7% We?“? 9"; 05,40“ P" = asnwo‘ N Si‘resses aide +0 ‘Fasion‘cxiion 7f 3. Steel): 6;": g— = w: 22.03x/o‘Pe. = 22.03 we. 5 3 ems: 61‘": 7%“ = “3.5% = —/‘l.(8rio‘PQ =-i‘-l.6‘3 M'PQ. Te “Hiese stresses mus—l be “Wee-l Hue s‘tmsses aloe +9 44¢ 25'“! food. Con i’l'n veal P'robﬁé’m 2.5.8 Con'l'l'nued ‘ For ‘Hme enable; 190mg?) +‘IE Mall-“ILEDMQJ o’e‘FOWmaJJ'1'on {5 +Ae Séme ‘Por “HIE Shae) Qua, 1‘4”: LMS'S . §l be ‘Hle Mon-[4044;] apfsrapacme/J". AJSOJ 161‘ 135 ammo! Pb he +lne aapaluh‘avxnj "Powces o‘eUeJopwl in ‘er s+ee¢9 aw? EMS; resped‘tveﬁy. 5' = t - ASES AbEI, . PB _ ‘AgEs 5' = (wooxlo‘ﬁmmwoﬂ 5. : Howe; 8- L 2.0:: R: = 5': W 5‘ .-. 3:.5wo‘3' TaJmJ P= “P5 4- P], = REX/03 N Home: S + 3/.5x10‘ 8’ = 25*!03’ 5': 3%-Gswo" m :_ Hey/053(3Q7.c\$xlo") = 9.6;“ do? N PL: (3;.s'xro‘K3c-amrwo" 1 = “we xxos' u x 3 ,_ P _ -H u 3 AJJ s‘l’r‘ess Joe. +9 JFan‘aJHm 6; = 3%.??310‘ + 22.03x/0‘ = 57.0xl06’Pq :- 520 Mm 3-€3 MPQ A II 61, = 13.36 wto‘ - Ream" = acaxio“ Pa Problem 2.73 2.73 In many situations physical constraints prevent strain ﬁom occurring in a given direction, for example «'5‘Z = 0 in the case shown, Where longitudinal movement of the long prism is prevented at every point. Plane sections perpendicular to the longitudinal axis remain plane and the same distance apart. Show that for the this situation, which is known as plane strain, we can express 0;, 5;: and 6; as follows: oz = 1/(0'x + 0y) £1: = [0* V5035 - V(1+ ﬂay] _£y = Ellﬂ— V2)0'y~_ V(1+ Max] A E -é—(S‘x-I v63,—wﬁ):é—[5’w-y63~vl(6;+63ﬂ %[(l'vzjgx " A .. ‘ ‘F‘wfﬂwu Problem 291 *2.91 A composite cube with 40-mm side:; and the properties sho glass polymer ﬁbers aligned in the x direction. The cube is consuam deformations in the y and 2 directions and is subjected to a tensile 1061661” 65- ‘ the x direction. Determine (a) the change in th: length of the cube in ﬁll: at diIeEtio' A E1 = 50 GPa vt2 = 0.254 (b) the stresses Ox, 0,, and (72. ' ' Ey = 152. GPa 1121‘ = 0.254. 2 = 15.2 GPa v = 0.428 SAITBSS- +0 — S'h‘am‘w equa+fons cure 0'7 VﬂﬂTy V cr V V: Ex : __.-‘__ ﬂ A4 (p, ﬂ : J. £1, E), E2 ' Ex 4}, C V 0' 0' V 0’ V V a H k zy x + _y_ zy 2 2 i: _’L .— y E .y c 1 E) as) 3’ yxz‘g-x Vﬂo—J’ U 52 z i — ‘ +14 = VJ: ') Ex 15y E, Ex (6 The ms+wain+ ﬁohisk'ama we 5:], 1 0 amp? Ez : 0_ 051313- (2) and C3) wiij +1“: consa'l'v‘m‘n'i‘ ans’Hw 393m: ,1. 6,3, — 2352»: 5.11 6M (7} 335:! E; Ex .? f— _ 1) <2) ,5-‘3 6'3, - ﬁg: 5'2 =17 \$3; or 6:, --r-_g_>.#zsd;‘ = momma 6‘; ﬁg:sz 4* .El‘iﬁ; 3 033,5“ 6} or [email protected] + 6; =0.ov7a;¢ 6; Sch/""3 sl'wvgﬁnewsf; 63/ = 6": a 0434*?93 8; 7‘ I' V . \ L - —l-‘ L. y 21m. 05m? (‘0 M (W £x_. Ex 3‘, _ ?6“Z Ex = 3’; l — 7C9. 254 )(Oszﬁ) - (ca. 2900:. l34§%)] 6,. T 0.931'42 6k ' E: A (LI-OKLJo} = 1600 NHL : 1600 xio-G w," P 65x 03 6", .AT— -_- W = umgzgxlog 'PcL (033142) Luisa? “"333 = 756,73 xIO—e E)! = 59 gig? (CC! 5% .2 Lx 5;: = um r75! .78YIO-‘) 1" 'o_ 0303 hm .4: 6;! 7‘ 40- QKSYIOBPQ = 90-.Q MPO. 4 5‘), = 6‘1 = (0.134363)(40-62§W10‘)= 5.43 #0qu = 5. 5+2 MFA. ‘1 ...
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