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stress and strain(tut_soln)

# stress and strain(tut_soln) - (a M OnAn = P sin(I V P V...

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Unformatted text preview: (a) M: OnAn = P sin (I: V P V: InAn : P COS {h o 1 5 N - -— o o - tan 1 +3 : tan 5 = 59.0 n , 2 (b) A : 50(100) : 5000 mm2 = 0.005 m A : .A 0 = ﬂ; 0.005833 m2 “ S1n sin 59° Cr A 6 3 P - “‘3" “ = —-————5(10 )(0'005833) : 34.0(10 ) N : 34.0 kN max H sin (t) sin 590 Aiternatively, P : rmaxA“ : ——-—3(1°6)(°‘005833) = 34.0(103) N : 34.0 kN m“ COS ¢ cos 59° 2. Let T be the tension in the rope, moment about A gives: 500><Tm50O + 400><Tsin500 = (400+700) x15 :> T = 26.28 kN Safety factor for rope = 100/ 26.28 = 3.81 Let Fx and Fy be the horizontal & vertical forces that the pin exerts on ABC. Consideration of horizontal and vertical force equilibrium give: Fx = -T cos 500 = -16.89 1<N,jFy : 15 -T sin 500: -5.12 kN Noting that the pin is in double shear, the shear stress is: 17.65x1000 =112.4 moor/4 MPa Safety factor for pin = 350/1124 : 3.11 (a) A t : (30 — 12)(10) : 180 mm2 = 180(10'6) m2 PWt : 2Ata : 2(130)(1o‘5)(350)(105) = 125(103) N Amid : (25 — 12)(25) : 325 mm2 : 325(10“6) m2 Pmid : Ame = 325(10‘6)(350)(106) : 113.8(103) N (b) Ab : 2dt : 2(12)(10) : 240 mm2 = 240(10'5) m2 —6 Pb = Abo : 240(10 )(650)(106) : 156C103) N _n: 2 It 2 _ (0) AS — Z d 2 1(12) : 113.10 mm 2: 113.10(10 6) m2 PS : 2A3t : 2(113.10)(10‘6)(240)(106) = 54.3(103) N (d) AS : 2Lt : 2(18)(10) : 360 mm2 : 360(10'6) m2 P5 = 2A5: : 2(360)(1o'5)(300)(106) : 216(103) N PAB : 370 RM (t) PBC = 370 — 2(210) : —50 kN : 50 kN (C) PCD : 37o — 2(210) + 2(125) : 200 RM (T) A : 50(50) = 2500 mm2 : 2500(10‘6) m2 Pmax 370(103) 6 2 (a) «max = A : ‘______?E' : 148.0(10 ) N/m : 148.0 MPa (T) 2500(10 ) - 3 3 (b) 5 I E E% = 370(10 )(3) + —50(10 )(2) 2500(10'5)(210)(109) 2500(10‘6)(210)(109) 200(103)(1.5) 2500(10‘6)(210)(109) 3 + : 2.495(10_ ) m E 2.50 mm 5. Let F be the tensile force acting on the bolt, then the comgressive force acting on the collar must be F. FL Elongation of the stressed portion of the bolt: 55 U S 3 FL Compressed length of collar: 5 A: 2 AA! E AI 8 +5 :1; “ ’“ 81000 F-50x10‘3 + F-50x10-3 3 600x10‘5 -200 900x10—6 -70 ‘ 8000 F = 3.098 x 10'4 GN F _ 3.098x10‘4 G = — — ‘— 2 2 S A 600x10’6 +.5164GPa +516.4MPa F - 3.098x10'4 6 = _ — K: — — _ Ag A“ 900x10‘6 0.3443GPa —- 344.3MPa 6. (a) Total strain = mechanical strain + thermal strain BIL = OLE + or?" 8=£+OLLT AE (b) The problem is statically indeterminate. Vertical force equilibrium gives: TA + T3 + T C = P About B, moment equilibrium gives: 21‘}: + P = 211; Compatibility condition: 253 = 6;; + 50 Using the result of part (a), the last condition becomes: 2TBL _ TAIL + (TCL — + (1LT) AE AE AE i) 2T3=TA+Tc+m4ET Solving (i), (ii) and (iii), we obtain TA = (7P — 20LAET)/12, T3 = (P + ocAET)/3, TC = (P — 2aAET)/12 TAL L 5A = = — (7P — 2004M) AE 12AE T L 8C = C— + arr = —L—(P — 2aAET)+ aLT AE 12AE Rotation = (5c — Big/L : a T _ P in the clockwise direction 2 A E (i) (ii) _ _ -5 55 - usLs AT _ 11.9(10 )(300)(—45) : —0.16065 mm _ _ -6 6a - aan AT _ 22.5(10 )(300)(~45) = —0.30375 mm 5p = 10(oa — 59) - 58 = 10(0.30375 — 0.16065) — 0.16065 _ 21.27035 mm a 1.270 mm A : E(150)2 : 17,671 mm2 A = E(45)2 = 1590-4 mm D 4 b 4 .. -3 _ - 5p + 5b - 0.15110 ) Pp ' Pb ' P PLP PLb _3 A E + A E : 0.15(10 ) p p b b P(200)(10 3) + ,__E£3992£19;il_______ : 0 15(10‘3) 17,571(10'6)(2.10)(109) 1590.4(10‘6)(100)(109) P : 18.976(103) N a 18.98 kN (a) g z a. : ——l§L§l§—:Ew : 1.0738(106) N/m2 2 1.074 MPa (T) ' P p 17,671(10 ) : 11.932(106) N/m2 2 11.93 MPa (T) G :P _ 18,976 b 1590.4(10'5) (b) 0 : P P ApEp 11,671(10'6)(2.10)(109) a = 200 + 5P = 200 + 0.1023 2 200.1023 mm P L -3 P ——1§4§1§33991319——l———— = 0.10221(10‘3) m a 0.1023 mm ...
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stress and strain(tut_soln) - (a M OnAn = P sin(I V P V...

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