Tài liệu shi20396 chương 4 pdf

Chapter 4 4-1 1 R C R A R B R D C A B W D 1 2 3 R B R A W R B R C R A 2 1 W R A R Bx R Bx R By R By R B 2 1 1 Scale of corner magnified W A B (e) (f) (d) W A R A R B B 1 2 W A R A R B B 11 2 (a) (b) (c) shi20396_ch04.qxd 8/18/03 10:35 AM Page 50 Chapter 4 51 4-2 (a) R A = 2 sin 60 = 1.732 kN Ans. R B = 2 sin 30 = 1 kN Ans. (b) S = 0.6 m α = tan −1 0.6 0.4 + 0.6 = 30.96 ◦ R A sin 135 = 800 sin 30.96 ⇒ R A = 1100 N Ans. R O sin 14.04 = 800 sin 30.96 ⇒ R O = 377 N Ans. (c) R O = 1.2 tan 30 = 2.078 kN Ans. R A = 1.2 sin 30 = 2.4 kN Ans. (d) Step 1: Find R A and R E h = 4.5 tan 30 = 7.794 m ۗ +  M A = 0 9R E − 7.794(400 cos 30) − 4.5(400 sin 30) = 0 R E = 400 N Ans.  F x = 0 R Ax + 400 cos 30 = 0 ⇒ R Ax =−346.4N  F y = 0 R Ay + 400 − 400 sin 30 = 0 ⇒ R Ay =−200 N R A =  346.4 2 + 200 2 = 400 N Ans. D C h B y E x A 4.5 m 9 m 400 N 3 4 2 30° 60° R Ay R A R Ax R E 1.2 kN 60° R A R O 60°90° 30° 1.2 kN R A R O 45Њ Ϫ 30.96Њ ϭ 14.04Њ 135° 30.96° 30.96° 800 N R A R O O 0.4 m 45° 800 N ␣ 0.6 m A s R A R O B 60° 90° 30° 2 kN R A R B 2 1 2 kN 60° 30° R A R B shi20396_ch04.qxd 8/18/03 10:35 AM Page 51 52 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design Step 2: Find components of R C on link 4 and R D ۗ +  M C = 0 400(4.5) − (7.794 −1.9)R D = 0 ⇒ R D = 305.4N Ans.  F x = 0 ⇒ (R Cx ) 4 = 305.4N  F y = 0 ⇒ (R Cy ) 4 =−400 N Step 3: Find components of R C on link 2  F x = 0 ( R Cx ) 2 + 305.4 − 346.4 = 0 ⇒ (R Cx ) 2 = 41 N  F y = 0 ( R Cy ) 2 = 200 N 4-3 (a) ۗ +  M 0 = 0 −18(60) + 14R 2 + 8(30) − 4(40) = 0 R 2 = 71.43 lbf  F y = 0: R 1 − 40 + 30 + 71.43 − 60 = 0 R 1 =−1.43 lbf M 1 =−1.43(4) =−5.72 lbf · in M 2 =−5.72 − 41.43(4) =−171.44 lbf · in M 3 =−171.44 − 11.43(6) =−240 lbf · in M 4 =−240 + 60(4) = 0 checks! 4" 4" 6" 4" Ϫ1.43 Ϫ41.43 Ϫ11.43 60 40 lbf 60 lbf 30 lbf x x x O AB CD y R 1 R 2 M 1 M 2 M 3 M 4 O V (lbf) M (lbf• in) O C C DB A B D E 305.4 N 346.4 N 305.4 N 41 N 400 N 200 N 400 N 200 N 400 N Pin C 30° 305.4 N 400 N 400 N 200 N 41 N 305.4 N 200 N 346.4 N 305.4 N (R Cx ) 2 (R Cy ) 2 C B A 2 400 N 4 R D (R Cx ) 4 (R Cy ) 4 D C E Ans. shi20396_ch04.qxd 8/18/03 10:35 AM Page 52 Chapter 4 53 (b)  F y = 0 R 0 = 2 + 4(0.150) = 2.6 kN  M 0 = 0 M 0 = 2000(0.2) + 4000(0.150)(0.425) = 655 N · m M 1 =−655 + 2600(0.2) =−135 N · m M 2 =−135 + 600(0.150) =−45 N · m M 3 =−45 + 1 2 600(0.150) = 0 checks! (c)  M 0 = 0: 10R 2 − 6(1000) = 0 ⇒ R 2 = 600 lbf  F y = 0: R 1 − 1000 + 600 = 0 ⇒ R 1 = 400 lbf M 1 = 400(6) = 2400 lbf · ft M 2 = 2400 − 600(4) = 0 checks! (d) ۗ +  M C = 0 −10R 1 + 2(2000) + 8(1000) = 0 R 1 = 1200 lbf  F y = 0: 1200 − 1000 − 2000 + R 2 = 0 R 2 = 1800 lbf M 1 = 1200(2) = 2400 lbf · ft M 2 = 2400 + 200(6) = 3600 lbf · ft M 3 = 3600 − 1800(2) = 0 checks! 2000 lbf 1000 lbf R 1 O O M 1 M 2 M 3 R 2 6 ft 2 ft2 ft AB C y M 1200 Ϫ1800 200 x x x 6 ft 4 ft A O O O B Ϫ600 M 1 M 2 V (lbf) 1000 lbf y R 1 R 2 400 M (lbf •ft) x x x V (kN) 150 mm200 mm 150 mm 2.6 Ϫ655 M (N• m) 0.6 M 1 M 2 M 3 2 kN 4 kN/m y A O O O O BC R O M O x x x shi20396_ch04.qxd 8/18/03 10:35 AM Page 53 54 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design (e) ۗ +  M B = 0 −7R 1 + 3(400) − 3(800) = 0 R 1 =−171.4 lbf  F y = 0: −171.4 − 400 + R 2 − 800 = 0 R 2 = 1371.4 lbf M 1 =−171.4(4) =−685.7 lbf · ft M 2 =−685.7 − 571.4(3) =−2400 lbf · ft M 3 =−2400 + 800(3) = 0 checks! (f) Break at A R 1 = V A = 1 2 40(8) = 160 lbf ۗ +  M D = 0 12(160) − 10R 2 + 320(5) = 0 R 2 = 352 lbf  F y = 0 −160 + 352 − 320 + R 3 = 0 R 3 = 128 lbf M 1 = 1 2 160(4) = 320 lbf · in M 2 = 320 − 1 2 160(4) = 0 checks! (hinge) M 3 = 0 − 160(2) =−320 lbf · in M 4 =−320 + 192(5) = 640 lbf · in M 5 = 640 − 128(5) = 0 checks! 40 lbf/in V (lbf) O O 160 Ϫ160 Ϫ128 192 M 320 lbf 160 lbf 352 lbf 128 lbf M 1 M 2 M 3 M 4 M 5 x x x 8" 5" 2" 5" 40 lbf/in 160 lbf O A y BD C A 320 lbf R 2 R 3 R 1 V A A O O O C M V (lbf ) 800 Ϫ171.4 Ϫ571.4 3 ft 3 ft4 ft 800 lbf400 lbf B y M 1 M 2 M 3 R 1 R 2 x x x shi20396_ch04.qxd 8/18/03 10:35 AM Page 54 Chapter 4 55 4-4 (a) q = R 1 x −1 − 40x − 4 −1 + 30x − 8 −1 + R 2 x − 14 −1 − 60x − 18 −1 V = R 1 − 40x − 4 0 + 30x − 8 0 + R 2 x − 14 0 − 60x − 18 0 (1) M = R 1 x − 40x − 4 1 + 30x − 8 1 + R 2 x − 14 1 − 60x − 18 1 (2) for x = 18 + V = 0 and M = 0 Eqs. (1) and (2) give 0 = R 1 − 40 + 30 + R 2 − 60 ⇒ R 1 + R 2 = 70 (3) 0 = R 1 (18) − 40(14) + 30(10) + 4R 2 ⇒ 9R 1 + 2R 2 = 130 (4) Solve (3) and (4) simultaneously to get R 1 =−1.43 lbf, R 2 = 71.43 lbf. Ans. From Eqs. (1) and (2), at x = 0 + , V = R 1 =−1.43 lbf, M = 0 x = 4 + : V =−1.43 − 40 =−41.43, M =−1.43x x = 8 + : V =−1.43 −40 + 30 =−11.43 M =−1.43(8) − 40(8 − 4) 1 =−171.44 x = 14 + : V =−1.43 −40 + 30 + 71.43 = 60 M =−1.43(14) − 40(14 − 4) + 30(14 − 8) =−240 . x = 18 + : V = 0, M = 0 See curves of V and M in Prob. 4-3 solution. (b) q = R 0 x −1 − M 0 x −2 − 2000x − 0.2 −1 − 4000x − 0.35 0 + 4000x − 0.5 0 V = R 0 − M 0 x −1 − 2000x − 0.2 0 − 4000x − 0.35 1 + 4000x − 0.5 1 (1) M = R 0 x − M 0 − 2000x − 0.2 1 − 2000x − 0.35 2 + 2000x − 0.5 2 (2) at x = 0.5 + m, V = M = 0 , Eqs. (1) and (2) give R 0 − 2000 − 4000(0.5 − 0.35) = 0 ⇒ R 1 = 2600 N = 2.6 kN Ans. R 0 (0.5) − M 0 − 2000(0.5 − 0.2) − 2000(0.5 − 0.35) 2 = 0 with R 0 = 2600 N, M 0 = 655 N · m Ans. With R 0 and M 0 , Eqs. (1) and (2) give the same V and M curves as Prob. 4-3 (note for V, M 0 x −1 has no physical meaning). (c) q = R 1 x −1 − 1000x − 6 −1 + R 2 x − 10 −1 V = R 1 − 1000x − 6 0 + R 2 x − 10 0 (1) M = R 1 x − 1000x − 6 1 + R 2 x − 10 1 (2) at x = 10 + ft, V = M = 0 , Eqs. (1) and (2) give R 1 − 1000 + R 2 = 0 ⇒ R 1 + R 2 = 1000 10R 1 − 1000(10 − 6) = 0 ⇒ R 1 = 400 lbf , R 2 = 1000 − 400 = 600 lbf 0 ≤ x ≤ 6: V = 400 lbf, M = 400x 6 ≤ x ≤ 10: V = 400 −1000(x − 6) 0 = 600 lbf M = 400x − 1000(x − 6) = 6000 − 600x See curves of Prob. 4-3 solution. shi20396_ch04.qxd 8/18/03 10:35 AM Page 55 56 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design (d) q = R 1 x −1 − 1000x − 2 −1 − 2000x − 8 −1 + R 2 x − 10 −1 V = R 1 − 1000x − 2 0 − 2000x − 8 0 + R 2 x − 10 0 (1) M = R 1 x − 1000x − 2 1 − 2000x − 8 1 + R 2 x − 10 1 (2) At x = 10 + , V = M = 0 from Eqs. (1) and (2) R 1 − 1000 − 2000 + R 2 = 0 ⇒ R 1 + R 2 = 3000 10R 1 − 1000(10 − 2) − 2000(10 − 8) = 0 ⇒ R 1 = 1200 lbf , R 2 = 3000 − 1200 = 1800 lbf 0 ≤ x ≤ 2: V = 1200 lbf, M = 1200x lbf ·ft 2 ≤ x ≤ 8: V = 1200 −1000 = 200 lbf M = 1200x − 1000(x − 2) = 200x + 2000 lbf · ft 8 ≤ x ≤ 10: V = 1200 − 1000 − 2000 =−1800 lbf M = 1200x − 1000(x − 2) − 2000(x −8) =−1800x + 18 000 lbf · ft Plots are the same as in Prob. 4-3. (e) q = R 1 x −1 − 400x − 4 −1 + R 2 x − 7 −1 − 800x − 10 −1 V = R 1 − 400x − 4 0 + R 2 x − 7 0 − 800x − 10 0 (1) M = R 1 x − 400x − 4 1 + R 2 x − 7 1 − 800x − 10 1 (2) at x = 10 + , V = M = 0 R 1 − 400 + R 2 − 800 = 0 ⇒ R 1 + R 2 = 1200 (3) 10R 1 − 400(6) + R 2 (3) = 0 ⇒ 10R 1 + 3R 2 = 2400 (4) Solve Eqs. (3) and (4) simultaneously: R 1 =−171.4 lbf, R 2 = 1371.4 lbf 0 ≤ x ≤ 4: V =−171.4 lbf, M =−171.4x lbf ·ft 4 ≤ x ≤ 7: V =−171.4 −400 =−571.4 lbf M =−171.4x − 400(x − 4) lbf · ft =−571.4x + 1600 7 ≤ x ≤ 10: V =−171.4 − 400 + 1371.4 = 800 lbf M =−171.4x − 400(x − 4) + 1371.4(x − 7) = 800x − 8000 lbf · ft Plots are the same as in Prob. 4-3. (f) q = R 1 x −1 − 40x 0 + 40x − 8 0 + R 2 x − 10 −1 − 320x − 15 −1 + R 3 x − 20 V = R 1 − 40x + 40x − 8 1 + R 2 x − 10 0 − 320x − 15 0 + R 3 x − 20 0 (1) M = R 1 x − 20x 2 + 20x − 8 2 + R 2 x − 10 1 − 320x − 15 1 + R 3 x − 20 1 (2) M = 0 at x = 8 in ∴ 8R 1 − 20(8) 2 = 0 ⇒ R 1 = 160 lbf at x = 20 + , V and M = 0 160 − 40(20) + 40(12) + R 2 − 320 + R 3 = 0 ⇒ R 2 + R 3 = 480 160(20) − 20(20) 2 + 20(12) 2 + 10R 2 − 320(5) = 0 ⇒ R 2 = 352 lbf R 3 = 480 − 352 = 128 lbf 0 ≤ x ≤ 8: V = 160 − 40x lbf, M = 160x −20x 2 lbf · in 8 ≤ x ≤ 10: V = 160 −40x + 40(x − 8) =−160 lbf , M = 160x − 20x 2 + 20(x −8) 2 = 1280 − 160x lbf · in shi20396_ch04.qxd 8/18/03 10:35 AM Page 56 Chapter 4 57 10 ≤ x ≤ 15: V = 160 −40x + 40(x −8) + 352 = 192 lbf M = 160x − 20x 2 + 20(x −8) + 352(x − 10) = 192x − 2240 15 ≤ x ≤ 20: V = 160 − 40x + 40(x −8) + 352 − 320 =−128 lbf M = 160x − 20x 2 − 20(x −8) + 352(x − 10) − 320(x − 15) =−128x + 2560 Plots of V and M are the same as in Prob. 4-3. 4-5 Solution depends upon the beam selected. 4-6 (a) Moment at center, x c = (l − 2a)/2 M c = w 2  l 2 (l − 2a) −  l 2  2  = wl 2  l 4 − a  At reaction, |M r |=wa 2 /2 a = 2.25, l = 10 in, w = 100 lbf/in M c = 100(10) 2  10 4 − 2.25  = 125 lbf · in M r = 100(2.25 2 ) 2 = 253.1 lbf · in Ans. (b) Minimum occurs when M c =|M r | wl 2  l 4 − a  = wa 2 2 ⇒ a 2 + al − 0.25l 2 = 0 Taking the positive root a = 1 2  −l +  l 2 + 4(0.25l 2 )  = l 2  √ 2 − 1  = 0.2071l Ans. for l = 10 in and w = 100 lbf, M min = (100/2)[(0.2071)(10)] 2 = 214.5 lbf · in 4-7 For the ith wire from bottom, from summing forces vertically (a) T i = (i + 1)W From summing moments about point a,  M a = W(l − x i ) − iWx i = 0 Giving, x i = l i + 1 WiW T i x i a shi20396_ch04.qxd 8/18/03 10:35 AM Page 57 58 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design So W = l 1 + 1 = l 2 x = l 2 + 1 = l 3 y = l 3 + 1 = l 4 z = l 4 + 1 = l 5 (b) With straight rigid wires, the mobile is not stable. Any perturbation can lead to all wires becoming collinear. Consider a wire of length l bent at its string support:  M a = 0  M a = iWl i + 1 cos α − ilW i + 1 cos β = 0 iWl i + 1 (cos α −cos β) = 0 Moment vanishes when α = β for any wire. Consider a ccw rotation angle β , which makes α → α +β and β → α − β M a = iWl i + 1 [cos(α +β) − cos(α − β)] = 2iWl i + 1 sin α sin β . = 2iWlβ i + 1 sin α There exists a correcting moment of opposite sense to arbitrary rotation β . An equation for an upward bend can be found by changing the sign of W . The moment will no longer be correcting. A curved, convex-upward bend of wire will produce stable equilibrium too, but the equation would change somewhat. 4-8 (a) C = 12 + 6 2 = 9 CD = 12 − 6 2 = 3 R =  3 2 + 4 2 = 5 σ 1 = 5 + 9 = 14 σ 2 = 9 − 5 = 4 2␾ s (12, 4 cw ) C R D ␶ 2 ␶ 1 ␴ 1 ␴ ␴ 2 2␾ p (6, 4 ccw ) y x ␶ cw ␶ ccw W i W il i ϩ 1 T i ␤ ␣ l i ϩ 1 shi20396_ch04.qxd 8/18/03 10:35 AM Page 58 Chapter 4 59 φ p = 1 2 tan −1  4 3  = 26.6 ◦ cw τ 1 = R = 5, φ s = 45 ◦ − 26.6 ◦ = 18.4 ◦ ccw (b) C = 9 + 16 2 = 12.5 CD = 16 − 9 2 = 3.5 R =  5 2 + 3.5 2 = 6.10 σ 1 = 6.1 + 12.5 = 18.6 φ p = 1 2 tan −1 5 3.5 = 27.5 ◦ ccw σ 2 = 12.5 − 6.1 = 6.4 τ 1 = R = 6.10 , φ s = 45 ◦ − 27.5 ◦ = 17.5 ◦ cw (c) C = 24 + 10 2 = 17 CD = 24 − 10 2 = 7 R =  7 2 + 6 2 = 9.22 σ 1 = 17 + 9.22 = 26.22 σ 2 = 17 − 9.22 = 7.78 2␾ s (24, 6 cw ) C R D ␶ 2 ␶ 1 ␴ 1 ␴ 2 2␾ p (10, 6 ccw ) y x ␴ ␶ cw ␶ ccw x 12.5 12.5 6.10 17.5Њ x 6.4 18.6 27.5Њ 2␾ s (16, 5 ccw ) C R D ␶ 2 ␶ 1 ␴ 1 ␴ 2 2␾ p (9, 5 cw ) y x ␴ ␶ cw ␶ ccw 3 5 3 3 3 18.4Њ x x 4 14 26.6Њ shi20396_ch04.qxd 8/18/03 10:35 AM Page 59 [...]... θ(deg) 0 0.125 0.225 0.325 0 .42 5 0 .47 5 0.902 5 0.889 087 0.859 043 0.811 831 0. 747 45 0 0.708 822 3.8 3.585 398 3 .41 3 717 3. 242 035 3.070 3 54 2.9 84 513 0 0.10 0.20 0.30 0 .40 0 .45 1037.9 1022.5 987.9 933.6 859.6 815.1 0 0.10 0.20 0.30 0 .40 0 .45 4. 825 4. 621 4. 553 4. 576 4. 707 4. 825 1200 1000 800 T (lbf • in) shi20396_ ch 04. qxd 600 40 0 200 0 0 0.1 0.2 0.3 ri (in) 0 .4 0.5 shi20396_ ch 04. qxd 82 8/18/03 10:36 AM... 4. 57 y φp = 4. 57 1 5 90 + tan−1 = 61.0◦ cw 2 8 x 61Њ 23 .43 τ1 = R = 9 .43 4, φs = 61◦ − 45 ◦ = 16◦ cw 14 x 16Њ 14 9 .43 4 9 + 19 = 14 2 shi20396_ ch 04. qxd 8/18/03 10:35 AM Page 61 61 Chapter 4 4-9 (a) ␶1 ␶ cw C= y (12, 7cw) R ␴2 D ␴1 ␴ 2␾p σ1 = 4 + 10.63 = 14. 63 σ2 = 4 − 10.63 = −6.63 2␾s x ␶2 ␶ ccw 12 + 4 =8 2 R = 82 + 72 = 10.63 CD = C ( 4, 7ccw) 12 − 4 =4 2 φp = 14. 63 69 .4 1 8 90 + tan−1 = 69 .4 ccw 2 7... = 45 ◦ − 13.28◦ = 31.72◦ cw 10 22.36 x 31.72Њ 10 (c) ␶1 ␶ cw 2␾s (Ϫ10, 9cw) ␴2 2␾p D ␴ ␴1 C R (18, 9ccw) y ␶ ccw −10 + 18 =4 2 10 + 18 CD = = 14 2 R = 142 + 92 = 16. 64 C= x σ1 = 4 + 16. 64 = 20. 64 σ2 = 4 − 16. 64 = −12. 64 ␶2 φp = 12. 64 1 14 90 + tan−1 = 73.63◦ cw 2 9 x 73.63Њ 20. 64 τ1 = R = 16. 64, φs = 73.63◦ − 45 ◦ = 28.63◦ cw 4 x 28.63Њ 4 16. 64 shi20396_ ch 04. qxd 8/18/03 10:35 AM Page 65 65 Chapter 4. .. 8/18/03 10:36 AM Page 85 85 Chapter 4 (τmax ) rd = Round: 16 T 16T 3. 545 T = = 3 3/2 π d π(4A/π) ( A) 3/2 (τmax ) sq 4. 8 = 1.3 54 = (τmax ) rd 3. 545 Square stress is 1.3 54 times the round stress 4- 45 s= √ A, d= Ans 4A/π Square: Eq (4- 44) with b = c, β = 0. 141 θsq = Tl Tl = 4G 0. 141 c 0. 141 ( A) 4/ 2 G Round: θrd = Tl Tl 6.2832T l = = JG ( A) 4/ 2 G (π/32) (4A/π) 4/ 2 G θsq 1/0. 141 = 1.129 = θrd 6.2832 Square has... ji = 0 WT = 1 04. 4, W1 = W2 = W3 = W4 = 1 04. 4 = 26.1 kip 4 (1200 − 238) 2 M1 = (1 04. 4) = 20 128 kip · in 4( 1200) 47 6 = 238 in, Wheel 1: d1 = 2 Wheel 2: i−1 d2 = 238 − 84 = 1 54 in (1200 − 1 54) 2 M2 = (1 04. 4) − 26.1( 84) = 21 605 kip · in = Mmax 4( 1200) Check if all of the wheels are on the rail 84" 77" 84" 315" xmax 600" 600" (b) xmax = 600 − 77 = 523 in (c) See above sketch (d) inner axles 4- 23 (a) D c1... psi 2 (2)(1) Ans shi20396_ ch 04. qxd 8/18/03 10:36 AM Page 77 77 Chapter 4 4-26 Mmax = wl 2 8 ⇒ σmax = wl 2 c 8I ⇒ w= 8σ I cl 2 (a) l = 12(12) = 144 in, I = (1/12)(1.5)(9.5) 3 = 107.2 in4 w= 8(1200)(107.2) = 10 .4 lbf/in 4. 75( 144 2 ) Ans (b) l = 48 in, I = (π/ 64) ( 24 − 1.2 54 ) = 0.6656 in4 w= 8(12)(103 )(0.6656) = 27.7 lbf/in 1 (48 ) 2 Ans 8(12)(103 )(2.051) = 57.0 lbf/in 1.5 (48 ) 2 Ans l = 48 in, I = (1/12)(2)(33... A-17, select 1 3 /4 in τstart = 16(2)(12 605) = 23.96(103 ) psi = 23.96 kpsi 3) π(1.75 (b) design activity 4- 43 ω = 2πn/60 = 2π(8)/60 = 0.8378 rad/s T = dC = 1000 H = = 11 94 N · m ω 0.8378 16T πτ 1/3 = From Table A-17, select 45 mm 4- 44 s= √ A, d = 16(11 94) π(75)(106 ) 1/3 = 4. 328(10−2 ) m = 43 .3 mm Ans 4A/π Square: Eq (4- 43) with b = c 4. 8T c3 4. 8T (τmax ) sq = ( A) 3/2 τmax = shi20396_ ch 04. qxd 8/18/03... ⇒ 150(106 ) = 4. 023(1 04 ) 150(106 ) 1/3 4. 023(1 04 ) 540 0(d/2) = (π/32)[d 4 − (0.75d) 4 ] d3 = 6 .45 (10−2 ) m = 64. 5 mm From Table A-17, the next preferred size is d = 80 mm; ID = 60 mm π (0.0 84 − 0.0 64 ) = 2. 749 (10−6 ) mm4 32 540 0(0.030) = 58.9(106 ) Pa = 58.9 MPa τi = 2. 749 (10−6 ) Ans J= (b) Ans 4- 42 63 025(1) 63 025H = = 12 605 lbf · in n 5 16T 16T 1/3 16(12 605) τ= ⇒ dC = = 3 πτ π( 14 000) πdC T =... MPa 30 − 10 = 10 MPa τ1/2 = 2 Ans 10 + 20 = 15 MPa τ2/3 = 2 30 + 20 = 25 MPa τmax = τ1/3 = 2 4- 19 ␶ (MPa) ␶1/3 ␶2/3 ␶1/2 Ϫ20 10 Ans From Eq (4- 15) σ 3 − (1 + 4 + 4) σ 2 + [1 (4) + 1 (4) + 4( 4) − 22 − ( 4) 2 − (−2) 2 ]σ −[1 (4) (4) + 2(2)( 4) (−2) − 1( 4) 2 − 4( −2) 2 − 4( 2) 2 ] = 0 σ 3 − 9σ 2 = 0 30 ␴ (MPa) shi20396_ ch 04. qxd 70 8/18/03 10:36 AM Page 70 Solutions Manual • Instructor’s Solution Manual to Accompany... 2 Am = (1 − 0.05) 2 − 4 rm − rm = 0.952 − (4 − π)rm 4 L m = 4( 1 − 0.05 − 2rm + 2πrm /4) = 4[ 0.95 − (2 − π/2)rm ] T = 2Am tτ = 2(0.05)(11 500) Am = 1150Am Eq (4- 45): Eq (4- 46): θ(deg) = θ1 l T L m l 180 T L m (40 ) 180 180 = = π 4G A2 t π 4( 11.5)(106 ) A2 (0.05) π m m = 9.9 645 (10 4 ) T Lm A2 m Equations can then be put into a spreadsheet resulting in: ri 0 0.10 0.20 0.30 0 .40 0 .45 rm Am Lm ri T(lbf · . R 1 =−1 .43 lbf, M = 0 x = 4 + : V =−1 .43 − 40 = 41 .43 , M =−1 .43 x x = 8 + : V =−1 .43 40 + 30 =−11 .43 M =−1 .43 (8) − 40 (8 − 4) 1 =−171 .44 x = 14 + : V =−1 .43 40 . in) O C C DB A B D E 305 .4 N 346 .4 N 305 .4 N 41 N 40 0 N 200 N 40 0 N 200 N 40 0 N Pin C 30° 305 .4 N 40 0 N 40 0 N 200 N 41 N 305 .4 N 200 N 346 .4 N 305 .4 N (R Cx ) 2 (R Cy ) 2 C B A 2 40 0