Tài liệu shi20396 chương 5 ppt

Chapter 5 5-1 (a) k = F y ; y = F k 1 + F k 2 + F k 3 so k = 1 (1/k 1 ) + (1/k 2 ) + (1/k 3 ) Ans. (b) F = k 1 y + k 2 y + k 3 y k = F/y = k 1 + k 2 + k 3 Ans. (c) 1 k = 1 k 1 + 1 k 2 + k 3 k =  1 k 1 + 1 k 2 + k 3  −1 5-2 For a torsion bar, k T = T/θ = Fl/θ, and so θ = Fl/k T . For a cantilever, k C = F/δ, δ = F/k C . For the assembly, k = F/y, y = F/k = lθ +δ So y = F k = Fl 2 k T + F k C Or k = 1 (l 2 /k T ) + (1/k C ) Ans. 5-3 For a torsion bar, k = T/θ = GJ /l where J = πd 4 /32 . So k = πd 4 G/(32l) = Kd 4 /l . The springs, 1 and 2, are in parallel so k = k 1 + k 2 = K d 4 l 1 + K d 4 l 2 = Kd 4  1 x + 1 l − x  And θ = T k = T Kd 4  1 x + 1 l − x  Then T = kθ = Kd 4 x θ + Kd 4 θ l − x k 2 k 1 k 3 F k 2 k 1 k 3 y F k 1 k 2 k 3 y shi20396_ch05.qxd 8/18/03 10:59 AM Page 106 Chapter 5 107 Thus T 1 = Kd 4 x θ; T 2 = Kd 4 θ l − x If x = l/2 , then T 1 = T 2 . If x < l/2 , then T 1 > T 2 Using τ = 16T /πd 3 and θ = 32Tl/(Gπd 4 ) gives T = πd 3 τ 16 and so θ all = 32l Gπd 4 · πd 3 τ 16 = 2lτ all Gd Thus, if x < l/2 , the allowable twist is θ all = 2xτ all Gd Ans. Since k = Kd 4  1 x + 1 l − x  = π Gd 4 32  1 x + 1 l − x  Ans. Then the maximum torque is found to be T max = πd 3 xτ all 16  1 x + 1 l − x  Ans. 5-4 Both legs have the same twist angle. From Prob. 5-3, for equal shear, d is linear in x. Thus, d 1 = 0.2d 2 Ans. k = π G 32  (0.2d 2 ) 4 0.2l + d 4 2 0.8l  = π G 32l  1.258d 4 2  Ans. θ all = 2(0.8l)τ all Gd 2 Ans. T max = kθ all = 0.198d 3 2 τ all Ans. 5-5 A = πr 2 = π(r 1 + x tan α) 2 dδ = Fdx AE = Fdx Eπ(r 1 + x tan α) 2 δ = F π E  l 0 dx (r 1 + x tan α) 2 = F π E  − 1 tan α(r 1 + x tan α)  l 0 = F π E 1 r 1 (r 1 +l tan α) l x ␣ dx F F r 1 shi20396_ch05.qxd 8/18/03 10:59 AM Page 107 108 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Then k = F δ = π Er 1 (r 1 +l tan α) l = EA 1 l  1 + 2l d 1 tan α  Ans. 5-6  F = (T + dT) + w dx − T = 0 dT dx =−w Solution is T =−wx + c T | x=0 = P +wl = c T =−wx + P + wl T = P +w(l − x) The infinitesmal stretch of the free body of original length dx is dδ = Tdx AE = P +w(l − x) AE dx Integrating, δ =  l 0 [P +w(l − x)] dx AE δ = Pl AE + wl 2 2AE Ans. 5-7 M = wlx − wl 2 2 − wx 2 2 EI dy dx = wlx 2 2 − wl 2 2 x − wx 3 6 + C 1 , dy dx = 0 at x = 0 , І C 1 = 0 EIy = wlx 3 6 − wl 2 x 2 4 − wx 4 24 + C 2 , y = 0 at x = 0 , І C 2 = 0 y = wx 2 24EI (4lx − 6l 2 − x 2 ) Ans. l x dx P Enlarged free body of length dx w is cable’s weight per foot T ϩ dT w dx T shi20396_ch05.qxd 8/18/03 10:59 AM Page 108 Chapter 5 109 5-8 M = M 1 = M B EI dy dx = M B x +C 1 , dy dx = 0 at x = 0 , І C 1 = 0 EIy = M B x 2 2 + C 2 , y = 0 at x = 0 , І C 2 = 0 y = M B x 2 2EI Ans. 5-9 ds =  dx 2 + dy 2 = dx  1 +  dy dx  2 Expand right-hand term by Binomial theorem  1 +  dy dx  2  1/2 = 1 + 1 2  dy dx  2 + ··· Since dy/dx is small compared to 1, use only the first two terms, dλ = ds −dx = dx  1 + 1 2  dy dx  2  − dx = 1 2  dy dx  2 dx І λ = 1 2  l 0  dy dx  2 dx Ans. This contraction becomes important in a nonlinear, non-breaking extension spring. 5-10 y =− 4ax l 2 (l − x) =−  4ax l − 4a l 2 x 2  dy dx =−  4a l − 8ax l 2   dy dx  2 = 16a 2 l 2 − 64a 2 x l 3 + 64a 2 x 2 l 4 λ = 1 2  l 0  dy dx  2 dx = 8 3 a 2 l Ans. y ds dy dx ␭ shi20396_ch05.qxd 8/18/03 10:59 AM Page 109 110 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 5-11 y = a sin π x l dy dx = aπ l cos π x l  dy dx  2 = a 2 π 2 l 2 cos 2 π x l λ = 1 2  l 0  dy dx  2 dx λ = π 2 4 a 2 l = 2.467 a 2 l Ans. Compare result with that of Prob. 5-10. See Charles R. Mischke, Elements of Mechanical Analysis, Addison-Wesley, Reading, Mass., 1963, pp. 244–249, for application to a nonlinear extension spring. 5-12 I = 2(5.56) = 11.12 in 4 y max = y 1 + y 2 =− wl 4 8EI + Fa 2 6EI (a − 3l) Here w = 50/12 = 4.167 lbf/in, and a = 7(12) = 84 in, and l = 10(12) = 120 in. y 1 =− 4.167(120) 4 8(30)(10 6 )(11.12) =−0.324 in y 2 =− 600(84) 2 [3(120) − 84] 6(30)(10 6 )(11.12) =−0.584 in So y max =−0.324 −0.584 =−0.908 in Ans. M 0 =−Fa − (wl 2 /2) =−600(84) − [4.167(120) 2 /2] =−80 400 lbf ·in c = 4 − 1.18 = 2.82 in σ max = −My I =− (−80 400)(−2.82) 11.12 (10 −3 ) =−20.4 kpsi Ans. σ max is at the bottom of the section. shi20396_ch05.qxd 8/18/03 10:59 AM Page 110 Chapter 5 111 5-13 R O = 7 10 (800) + 5 10 (600) = 860 lbf R C = 3 10 (800) + 5 10 (600) = 540 lbf M 1 = 860(3)(12) = 30.96(10 3 ) lbf · in M 2 = 30.96(10 3 ) + 60(2)(12) = 32.40(10 3 ) lbf · in σ max = M max Z ⇒ 6 = 32.40 Z Z = 5.4in 3 y| x=5ft = F 1 a[l − (l/2)] 6EIl  l 2  2 + a 2 − 2l l 2  − F 2 l 3 48EI − 1 16 = 800(36)(60) 6(30)(10 6 )I (120) [60 2 + 36 2 − 120 2 ] − 600(120 3 ) 48(30)(10 6 )I I = 23.69 in 4 ⇒ I /2 = 11.84 in 4 Select two 6 in-8.2 lbf/ft channels; from Table A-7, I = 2(13.1) = 26.2in 4 , Z =2(4.38) in 3 y max = 23.69 26.2  − 1 16  =−0.0565 in σ max = 32.40 2(4.38) = 3.70 kpsi 5-14 I = π 64 (1.5 4 ) = 0.2485 in 4 Superpose beams A-9-6 and A-9-7, y A = 300(24)(16) 6(30)(10 6 )(0.2485)(40) (16 2 + 24 2 − 40 2 ) + 12(16) 24(30)(10 6 )(0.2485) [2(40)(16 2 ) − 16 3 − 40 3 ] y A =−0.1006 in Ans. y| x=20 = 300(16)(20) 6(30)(10 6 )(0.2485)(40) [20 2 + 16 2 − 2(40)(20)] − 5(12)(40 4 ) 384(30)(10 6 )(0.2485) =−0.1043 in Ans. % difference = 0.1043 −0.1006 0.1006 (100) = 3.79% Ans. R C M 1 M 2 R O A O B C V (lbf) M (lbf •in) 800 lbf 600 lbf 3 ft 860 60 O Ϫ540 2 ft 5 ft shi20396_ch05.qxd 8/18/03 10:59 AM Page 111 112 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 5-15 I = 1 12  3 8  (1.5 3 ) = 0.105 47 in 4 From Table A-9-10 y C =− Fa 2 3EI (l +a) dy AB dx = Fa 6EIl (l 2 − 3x 2 ) Thus, θ A = Fal 2 6EIl = Fal 6EI y D =−θ A a =− Fa 2 l 6EI With both loads, y D =− Fa 2 l 6EI − Fa 2 3EI (l +a) =− Fa 2 6EI (3l + 2a) =− 120(10 2 ) 6(30)(10 6 )(0.105 47) [3(20) + 2(10)] =−0.050 57 in Ans. y E = 2Fa(l/2) 6EIl  l 2 −  l 2  2  = 3 24 Fal 2 EI = 3 24 120(10)(20 2 ) (30)(10 6 )(0.105 47) = 0.018 96 in Ans. 5-16 a = 36 in, l = 72 in, I = 13 in 4 , E = 30 Mpsi y = F 1 a 2 6EI (a − 3l) − F 2 l 3 3EI = 400(36) 2 (36 − 216) 6(30)(10 6 )(13) − 400(72) 3 3(30)(10 6 )(13) =−0.1675 in Ans. 5-17 I = 2(1.85) = 3.7in 4 Adding the weight of the channels, 2(5)/12 = 0.833 lbf/in, y A =− wl 4 8EI − Fl 3 3EI =− 10.833(48 4 ) 8(30)(10 6 )(3.7) − 220(48 3 ) 3(30)(10 6 )(3.7) =−0.1378 in Ans. ␪ A a D C F B a E A shi20396_ch05.qxd 8/18/03 10:59 AM Page 112 Chapter 5 113 5-18 I = πd 4 /64 = π(2) 4 /64 = 0.7854 in 4 Tables A-9-5 and A-9-9 y =− F 2 l 3 48EI + F 1 a 24EI (4a 2 − 3l 2 ) =− 120(40) 3 48(30)(10 6 )(0.7854) + 80(10)(400 − 4800) 24(30)(10 6 )(0.7854) =−0.0130 in Ans. 5-19 (a) Useful relations k = F y = 48EI l 3 I = kl 3 48E = 2400(48) 3 48(30)10 6 = 0.1843 in 4 From I = bh 3 /12 h = 3  12(0.1843) b Form a table. First, Table A-17 gives likely available fractional sizes for b: 8 1 2 , 9, 9 1 2 , 10 in For h: 1 2 , 9 16 , 5 8 , 11 16 , 3 4 For available b what is necessary h for required I? (b) I = 9(0.625) 3 /12 = 0.1831 in 4 k = 48EI l 3 = 48(30)(10 6 )(0.1831) 48 3 = 2384 lbf/in F = 4σ I cl = 4(90 000)(0.1831) (0.625/2)(48) = 4394 lbf y = F k = 4394 2384 = 1.84 in Ans. choose 9" × 5 8 " Ans. b 3  12(0.1843) b 8.5 0.638 9.0 0.626 ← 9.5 0.615 10.0 0.605 shi20396_ch05.qxd 8/18/03 10:59 AM Page 113 114 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design 5-20 Torque = (600 −80)(9/2) = 2340 lbf · in (T 2 − T 1 ) 12 2 = T 2 (1 − 0.125)(6) = 2340 T 2 = 2340 6(0.875) = 446 lbf, T 1 = 0.125(446) = 56 lbf  M 0 = 12(680) − 33(502) + 48R 2 = 0 R 2 = 33(502) − 12(680) 48 = 175 lbf R 1 = 680 − 502 + 175 = 353 lbf We will treat this as two separate problems and then sum the results. First, consider the 680 lbf load as acting alone. z OA =− Fbx 6EIl (x 2 + b 2 −l 2 ); here b = 36 ", x = 12 ", l = 48 ", F = 680 lbf Also, I = πd 4 64 = π(1.5) 4 64 = 0.2485 in 4 z A =− 680(36)(12)(144 + 1296 − 2304) 6(30)(10 6 )(0.2485)(48) =+0.1182 in z AC =− Fa(l − x) 6EIl (x 2 + a 2 − 2lx) where a = 12 " and x = 21 + 12 = 33 " z B =− 680(12)(15)(1089 + 144 − 3168) 6(30)(10 6 )(0.2485)(48) =+0.1103 in Next, consider the 502 lbf load as acting alone. 680 lbf A C B O R 1 ϭ 510 lbf R 2 ϭ 170 lbf 12" 21" 15" z x R 2 ϭ 175 lbf680 lbf AC BO R 1 ϭ 353 lbf 502 lbf 12" 21" 15" z x shi20396_ch05.qxd 8/18/03 10:59 AM Page 114 Chapter 5 115 z OB = Fbx 6EIl (x 2 + b 2 −l 2 ), where b = 15 " , x = 12 ", l = 48 ", I = 0.2485 in 4 Then, z A = 502(15)(12)(144 + 225 − 2304) 6(30)(10 6 )(0.2485)(48) =−0.081 44 in For z B use x = 33 " z B = 502(15)(33)(1089 +225 −2304) 6(30)(10 6 )(0.2485)(48) =−0.1146 in Therefore, by superposition z A =+0.1182 −0.0814 =+0.0368 in Ans. z B =+0.1103 −0.1146 =−0.0043 in Ans. 5-21 (a) Calculate torques and moment of inertia T = (400 −50)(16/2) = 2800 lbf ·in (8T 2 − T 2 )(10/2) = 2800 ⇒ T 2 = 80 lbf, T 1 = 8(80) = 640 lbf I = π 64 (1.25 4 ) = 0.1198 in 4 Due to 720 lbf, flip beam A-9-6 such that y AB → b = 9, x = 0, l = 20, F =−720 lbf θ B = dy dx     x=0 =− Fb 6EIl (3x 2 + b 2 −l 2 ) =− −720(9) 6(30)(10 6 )(0.1198)(20) (0 + 81 −400) =−4.793(10 −3 ) rad y C =−12θ B =−0.057 52 in Due to 450 lbf, use beam A-9-10, y C =− Fa 2 3EI (l +a) =− 450(144)(32) 3(30)(10 6 )(0.1198) =−0.1923 in 450 lbf720 lbf 9" 11" 12" O y A B C R O R B A CB O R 1 R 2 12" 502 lbf 21" 15" z x shi20396_ch05.qxd 8/18/03 10:59 AM Page 115 [...]... 200(43 ) 1 .56 6(1 05 ) 2 − − = E I1 I2 2 =− 30(106 ) 1.28(104 ) 1 .56 6(1 05 ) + 0.24 85 0.7 854 = −0.016 73 in Ans −x ∂M = ∂Q 2 [1200x − 100(x − 4)2 ] − x dx 2 shi20396_ ch 05. qxd 8/18/03 10 :59 AM Page 129 129 Chapter 5 5-49 O x 3 Fa 5 l A O 4 Fa 5 l 4 F 5 A A 3 F 5 a a 3 F 5 B B 3 F 5 x 4 F 5 4 F 5 AB ∂M =x ∂F M = Fx N= 3 F 5 ∂N 3 = ∂F 5 T = 4 Fa 5 ∂T 4 = a ∂F 5 M1 = 4 Fx ¯ 5 ∂ M1 4 = x ¯ ∂F 5 M2 = 3 Fa 5 ∂ M2... 6 6 3 B: x = 0 .5 m, y B = −6.70(10−4 ) m = −0.670 mm Ans C: x = 1 m, yC = 38 85 3 1 15 830 − (1 ) + (1 − 0 .5) 3 − 62.0 45( 1) 3) 167.2(10 6 6 = −2.27(10−3 ) m = −2.27 mm Ans D: x = 1 .5, 38 85 1 15 830 − (1 .53 ) + (1 .5 − 0 .5) 3 3) 167.2(10 6 6 10 − (103 )(1 .5 − 1) 3 − 62.0 45( 1 .5) 3 = −3.39(10−4 ) m = −0.339 mm Ans yD = shi20396_ ch 05. qxd 8/18/03 10 :59 AM Page 1 35 1 35 Chapter 5 5-60 y FBE 50 0 lbf 3" A 3"... ∂F 5 OA δB = 1 ∂u = ∂F EI + = I = δB = = l 0 Fl AE J = 2I, 64Fa 3 9 + 4 3Eπd 25 (3 /5) F(3 /5) l (4 /5) Fa(4a /5) l + AE JG F x(x) dx + 0 1 EI 9 Fa 3 + 3E I 25 π 4 d , 64 a 4 Fx ¯ 5 + 16 25 A= 4Fl πd 2 E 4 1 x dx + ¯ ¯ 5 EI Fa 2l JG + l 0 3 3 Fa a dx ¯ 5 5 Fl 3 EI 16 75 + 9 25 Fa 2l EI π 2 d 4 + 16 25 32Fa 2l πd 4 G + 16 75 64Fl 3 Eπd 4 E 4F 400a 3 + 27ld 2 + 384a 2l + 256 l 3 + 432a 2l 4 75 Ed G + 9 25 Ans... 3(π/4)(0.0801) 2 σs = 65. 36 = 21 300 psi = 21.3 kpsi Ans (π/4)(0.06 252 ) 5- 51 σb = 0.9( 85) = 76 .5 kpsi Ans (a) Bolt stress Bolt force Cylinder stress σc = − π 4 Fb = 6(76 .5) (0.3 752 ) = 50 .69 kips Fb 50 .69 = − 15. 19 kpsi Ans =− Ac (π/4)(4 .52 − 42 ) (b) Force from pressure P= π D2 π(42 ) p= (600) = 754 0 lbf = 7 .54 kip 4 4 50 .69 Ϫ Pc 50 .69 ϩ Pb 6 bolts Fx = 0 x Pb + Pc = 7 .54 (1) P ϭ 7 .54 kip Pb L Pc L =... 0.0044 = −0.0 45 87 in Ans shi20396_ ch 05. qxd 8/18/03 10 :59 AM Page 121 121 Chapter 5 5-31 (0.1F)(1 .5) TL = = 9.292(10−4 ) F JG (π/32)(0.0124 )(79.3)(109 ) θ= Due to twist δ B 1 = 0.1(θ) = 9.292(10 5 ) F Due to bending F(0.13 ) F L3 = = 1 .58 2(10−6 ) F 3E I 3(207)(109 )(π/64)(0.0124 ) δB2 = δ B = 1 .58 2(10−6 ) F + 9.292(10 5 ) F = 9. 450 (10 5 ) F 1 = 10 .58 (103 ) N/m = 10 .58 kN/m Ans 9. 450 (10 5 ) k= 5- 32 F A... )(0.1667)(−1.4668)(10−3 ) = −26.67(63 ) + C1 (6) C1 = 55 2 .58 lbf · in2 yB = 1 [−26.67(183 ) + 40(18 − 6) 3 + 55 2 .58 (18)] 6 )(0.1667) 10(10 = −0.0 45 87 in 5- 41 I1 = R1 = Ans π (1 .54 ) = 0.24 85 in4 64 200 (12) = 1200 lbf 2 For 0 ≤ x ≤ 16 in, M = 1200x − I2 = π 4 (2 ) = 0.7 854 in4 64 MրI 200 x −4 2 2 x shi20396_ ch 05. qxd 8/18/03 10 :59 AM Page 1 25 1 25 Chapter 5 1200x M = − 4800 I I1 1 1 − I1 I2 x − 4 0 − 1200... Mechanical Engineering Design σB E = 10 45. 2 = 13 627 psi = 13.6 kpsi Ans (π/4) (5/ 16) 2 σD F = − 45. 2 = 58 9 psi Ans (π/4) (5/ 16) 2 yA = 1 (−11 52 2) = −0.007 68 in Ans 1 .5( 106 ) yB = 250 3 1 − (3 ) + 4136.4(3) − 11 52 2 = −0.000 909 in Ans 1 .5( 106 ) 3 yD = 250 3 1 10 45. 2 59 0.4 − (9 ) + (9 − 3) 3 + (9 − 6) 3 + 4136.4(9) − 11 52 2 6) 1 .5( 10 3 6 6 = −4.93(10 5 ) in Ans 5- 61 ␪ F Q (dummy load) ∂M = R(1 − cos... /∂ F)L dx + ∂F JG −F x(−x) d x + ¯ ¯ ¯ 0 0.1F(0.1)(1 .5) JG 0.015F F (0.13 ) + 3E I JG Where π (0.012) 4 = 1.0179(10−9 ) m4 64 J = 2I = 2.0 358 (10−9 ) m4 I = δB = F k= 5- 48 0.001 3(207)(109 )(1.0179)(10−9 ) + 0.0 15 = 9. 45( 10 5 ) F −9 )(79.3)(109 ) 2.0 358 (10 1 = 10 .58 (103 ) N/m = 10 .58 kN/m Ans 9. 45( 10 5 ) From Prob 5- 41, I1 = 0.24 85 in4 , I2 = 0.7 854 in4 For a dummy load ↑ Q at the center 200 Q x − 4... τ1 = 3 πd π(7/8)3 τ2 = 16(969.4) = 252 8 psi Ans π(1. 25) 3 5- 56 10 kip 5 kip FA RO FB x RC (1) Arbitrarily, choose RC as redundant reaction Fx = 0, (2) 10(103 ) − 5( 103 ) − R O − RC = 0 R O + RC = 5( 103 ) lbf (3) δC = RC ( 15) [10(103 ) − 5( 103 ) − RC ]20 [5( 103 ) + RC ] − (10) − =0 AE AE AE −45RC + 5( 104 ) = 0 ⇒ RC = 1111 lbf Ans R O = 50 00 − 1111 = 3889 lbf Ans 5- 57 w A MC B a C RC RB l x (1) Choose... dθ = = 5- 67 π P R3 π2 2E I 2π 2 (3π 2 − 8π − 4) P R 3 8π EI π 4 + π2 +4 π 4 − 2π 2 − 4π + 2π Ans Must use Eq (5- 34) A = 80(60) − 40(60) = 2400 mm2 R= ( 25 + 40)(80)(60) − ( 25 + 20 + 30)(40)(60) = 55 mm 2400 Section is equivalent to the “T” section of Table 4 -5 rn = 60(20) + 20(60) = 45. 9 654 mm 60 ln[( 25 + 20)/ 25] + 20 ln[(80 + 25) /( 25 + 20)] e = R − rn = 9.0 35 mm Straight section y z Iz = 30 mm 50 mm . 15& quot; z x R 2 ϭ 1 75 lbf680 lbf AC BO R 1 ϭ 353 lbf 50 2 lbf 12" 21" 15& quot; z x shi20396_ ch 05. qxd 8/18/03 10 :59 AM Page 114 Chapter 5 1 15 z OB = Fbx 6EIl (x 2 +. C 1 (6) C 1 = 55 2 .58 lbf ·in 2 y B = 1 10(10 6 )(0.1667) [−26.67(18 3 ) + 40(18 − 6) 3 + 55 2 .58 (18)] =−0.0 45 87 in Ans. 5- 41 I 1 = π 64 (1 .5 4 ) = 0.24 85 in 4 I 2 = π 64 (2 4 )