chapters solutions
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advance mechanicsTRANSCRIPT
Chapter 1 Analysis of Stress
1.1. P = 4.27 kN
1.4. P = 32.5 kN, θ = 26.56°
1.6. Pall = 38.3 kN
1.8. σx′ = 156 MPa, τx′y′ = –131 MPa
1.10. σx′ = –13.15 MPa, τx′y′ = –15.67 MPa
1.11. F = 67.32 kN/m3
1.14. Fx = Fy = Fz = 0
1.21.
a. σ = 153.3 MPa,
b. σ = 23.1 MPa
1.26.
a. σ1 = 121 MPa, σ = –71 MPa, τmax = 96 MPa
b. σ1 = 200 MPa, σ2 = –50 MPa, τmax = 125 MPa
1.31.
a. σx = 38.68 MPa, σy = 12.12 MPa, τxy = 7 MPa
b. σ1 = 40.41 MPa, σ2 = 10.39 MPa,
1.34. P = 3πpr2
1.37. τmax = 48 MPa, θp = 29°
1.38. σw = 61.5 MPa, τw = 26.3 MPa
1.42.
a. σ = –0.237τo, τ = 0.347τo
b. σ1 = 3.732τo, σ2 = –0.268τo,
1.44. σ1 = –σ2 = 51.96 MPa,
1.46. σ1 = 46.17 MPa, σ2 = 13.83 MPa,
1.49.
a. σx = 100 MPa, τxy = –30 MPa
b. σ1 = 110 MPa,
1.53. P = 1069 kN, p = 467 kPa
1.55.
a. σx = 186 MPa,
b. σ1 = 188 MPa, τmax = 101 MPa,
1.58. p = 494 kPa
1.62. σ1 = 66.06 MPa, l1 = 0.956, m1 = 0.169, n1 = 0.242
1.64. σ1 = 24.747 MPa, σ2 = 8.48 MPa, σ3 = 2.773 MPa, l1 = 0.647, m1 = 0.396, n1 = 0.652
1.69.
a. σ1 = 12.049 MPa, σ2 = –1.521 MPa, σ3 = –4.528 MPa, l1 = 0.618, m1 =
0.533, n1 = 0.577
1.73. σ = 52.25 MPa, τ = 36.56 MPa
1.75.
a. τ13 = 8.288 MPa, τ12 = 6.785 MPa, τ23 = 1.503 MPa
1.79. σoct = 12 MPa, τoct = 9.31 MPa
1.82.
a. σ1 = 108.3 MPa, σ2 = 51.7 MPa, σ3 = –50 MPa,
1.85. σ = –12.39 MPa, τ = 26.2 MPa, px = 16.81 MPa, py = –3.88 MPa, pz = –23.30 MPa
Chapter 2 Strain and Material Properties
2.4. c0 = 2a0, c1 = 2(a1 + b0)
2.9.
a. εx = 667 µ, εy = 750 µ, γxy = –1250 µ
b. ε1 = 1335 µ, ε2 = 82 µ,
2.11. ε1 = 758 µ, ε2 = –808 µ,
2.13. ∆QB = 0.008 mm, ∆AC = 0.016 mm
2.15.
a. γmax = 200 µ, θs = 45°
b. εx = 350 µ, εy = 250 µ, γxy = –173 µ
2.17. ε1 = –59 µ, ε2 = –1141 µ,
2.21.
a. J1 = –3 × 10–4, J2 = –44 × 10
–8, J3 = 58 × 10–12
b. εx′ = 385 µ
c. ε1 = 598 µ, ε2 = –126 µ, ε3 = –772 µ
d. γmax = 1370 µ
2.28.
a. ∆L = 5.98 mm,
b. ∆d = –0.014 mm
2.33.
a. ε1 = 1222 µ, ε2 = 58 µ
b. γmax = 1164 µ,
c. (γmax)t = 1222 µ
2.36. ∆BD = 0.283/E m
2.39. εx = 522 µ, εy = 678 µ, γxy = –1873 µ
2.41. σx = 72 MPa, σy = 88 MPa, σz = 40 MPa, τxy = 16 MPa, τyz = 64 MPa, τxz = 0
2.46. (b) ∆V = –2250 mm3
2.48. σ3: σ2: σ1 = 1 : 1.086:1.171, σ1 = 139.947 MPa, σ2 = 129.757 MPa, σ3 = 119.513
MPa
2.52. εx = γ(a – x)/E, εy = –vεx, σx = γ(a – x), γxy = τxy = σy = 0. Yes.
2.55. U1 = P2L/2EA, U2 = 5U1/8, U3 = 5U1/12
2.61.
a. U = 60.981T2a/πd4G,
b. U = 2.831 kN · m
2.67. Uov = 3.258 kPa, Uod = 38.01 kPa
2.69.
Chapter 3 Problems in Elasticity
3.1.
(b)
3.2.
3.14. All conditions, except on edge x = L, are satisfied.
3.16.
3.19. Yes, ε1 = 32.8 µ,
3.20. σx = σy = EαT1/(v – 1), εz = 2vαT1/(1 – v) + αT1
3.24. Px = –161.3 kN
3.31.
a. (σx)elast. = P/0.512L, (σx)elem. = P/0.536L
b. (σx)elast. = P/1.48L, (σx)elem. = P/3.464L
3.33.
a. (σ)elast. = 19.43F/L, (σx)elem. = 20.89F/L
(τxy)elast. = 5.21F/L, (τxy)elem. = 2.8F/L
3.38. r = 6.67 mm, d = 26.7 mm
3.42. T = 153.5 N · m
3.47.
a. σ1 = 36.5 MPa,
b. τmax = 16.96 MPa,
c. σoct = 11.3 MPa, τoct = 17.85 MPa
3.49.
a. σc = 1233 MPa,
b. σc = 1959 MPa
3.51. σc = 418 MPa, b = 0.038 mm
3.53. a = 2.994 mm, b = 1.605 mm, σc = 505.2 MPa
3.58. σc = 1014.7 MPa
Chapter 4 Failure Criteria
4.1. Pall = 707 kN
4.3.
a. Yes,
b. No
4.6.
a. σyp = 152.6 MPa,
b. σyp = 134.9 MPa
4.8. t = 8.45 mm
4.9.
a. d = 27.95 mm,
b. d = 36.8 mm
4.11.
a. T = 31.74 kN · m,
b. T = 26.05 kN · m
4.14.
a. R = 932 N,
b. R = 959 N
4.16.
a. p = 6.466 MPa,
b. p = 5.6 MPa
4.22. (b) σ1 = 75 MPa, σ2 = –300 MPa
4.25.
a. τ = σu/2,
b.
4.30. t = 9.27 mm
4.36. p = 11.11 MPa
4.39. t = 0.973 mm
4.43. Pmax = 18.7 kN
4.46. τmax = 274.3 MPa, φmax = 4.76°
Chapter 5 Bending of Beams
5.4. Mo = 266.8 N · m
5.6.
a. P = 3.6 kN,
b. P = 3.36 kN
5.9.
a. φ = –12.17°,
b. σA = 136.5 MPa
5.10. (b) σx = px3/Lth2,
(c) (σx)elast. = 0.998(σx)elem.
5.13. p = 3.88 kN/m
5.14. P = 9320 N
5.16. (a) σA = 35.3 MPa,
(b) σB = 23.5 MPa
(b) rZ = –205.48 m
5.19.
a. τmax/σmax = h/L,
b. pall = 10.34 kN/m
5.20. P = 15.6 kN
5.24. e = 4R/π
5.28. R = –13pL/32
5.30. v = Mox2(x – L)/4EIL, RB = 3Mo/2L
5.34. P = 1.614 kN
5.35.
a. σmax = –177 MPa,
b. σmax = –176.2 MPa
5.45.
a. σθ = –182.3P,
b. δp = 215.65P/E m
Chapter 6 Torsion of Prismatic Bars
6.2. D = 60.9 mm
6.6. (τmax)B = 130.4 MPa
6.13. T = 1.571 kN · m
6.14.
a. τe > τc;
b. Te > Tc
6.16. T = 256.5 kN · m
6.18. k = Gθ/2a2(b –1)
6.21. θA = aT/2r4G, θB = 2θA
6.23.
6.26. τmax = 76.8 MPa, θ = 0.192 rad/m
6.32.
a. C = 2.1 × 10–7G, τmax = 112,860T
6.33. θ = 0.1617 rad/m
6.35. θ = 2T/9Ga3t
6.37. τmax = 5.279 MPa, θ = 0.0131 rad/m
6.39. τ2 = τ4 = τmax = 50.88 MPa, θ = 0.01914 rad/m
Chapter 7 Numerical Methods
7.3. τB = 0.0107Gθ
7.6. v(L) = 7PL3/32EI
7.10. vmax = 0.01852pL3/EI
7.12. vB = –Pa3/16EI, θA = –Pa
2/16EI
7.14. vmax = 1.68182ph4/EI, θmax = –0.02131pL
3/EI
7.15. RA = RB = P/2, M = PL/8
7.18. (b) u2 = 2PL/11AE, u3 = –3PL/11AE
(c) R1 = 2P/11, R4 = 9P/11
7.19. (b) u2 = PL/9AE, u3 = PL/18AE
(c) R1 = 2P/3, R4 = P/3
7.24. (c) u2 = 0.9 mm, v2 = –3.02 mm
(d) R1x = 4.5 kN, R1y = 6 kN, R3y = –4.5 kN
(e) F12 = –7.5 kN, F23 = 36 kN
7.27. v2 = –5pL4/384EI
7.29. v2 = –5pL3/48EI, θ2 = PL
2/8EI
7.32. (c) v3 = –7PL3/12EI, θ3 = –3PL
2/4EI, θ2 = –PL2/4EI
(d) R1 = –3P/2, R2 = 5P/2, M1 = PL/2, M2 = –PL
7.35. (c) u2 = 1.0 mm, v2 = –3.9 mm
(d) R1x = 7.63 kN, R1y = 10.18 kN, R3x = –7.5 kN, R3y = 0
7.42. {σx, σy, τxy}a = {66.46, 6.65, –92.12} MPa
Chapter 8 Axisymmetrically Loaded Members
8.2. pi = 51.2 MPa
8.3.
a.
b. rx = 27.12 mm
8.5.
a. pi = 1.6po,
b. pi = 1.16po
8.7. t = 0.59 m
8.11.
a. t = 0.825di,
b. ∆d = 0.0074 mm
8.13.
a.
b.
8.16. T = 5.073 kN · m
8.21. ∆ds = 0.23δ0 m
8.23. σθ,max = 1.95EbEs(T2 – T1)/(Es + 3Eb)105
8.26.
a. pi = 155.6 MPa,
b. pi = 179.3 MPa
8.30. Shaft: σθ = σr = –49.4 MPa
Cylinder: σθ,max = 63.8 MPa, σr,max = 49.4 MPa
8.32. T = 101.79 N · m
8.36. ω = 36,726 rpm
8.40.
a. σθ,max = 108.68 MPa,
b. ω = 4300 rpm
8.44.
a. σθ,max = 554.6 MPa
Chapter 9 Beams on Elastic Foundations
9.1. P = 51.18 kN
9.3.
9.5. vmax = 3.375 mm, σmax = 103.02 MPa
9.6. P = 41.7 kN
9.8. σmax = 196.3 kN
9.11.
9.14. v = –MLf2(βx)/2β2EI
9.16.
a. v = 10 mm;
b. vL = 15 mm, vR = 5 mm
9.17. vC = 0.186 mm
9.18. vC = 2.81 × 10–8 m, θE = 5 × 10–7 rad
Chapter 10 Applications of Energy Methods
10.3. vp = (11Pc1a4/12E) + (7Pc2a
3/3E) + (Pa3/3EI2)
10.6. δA = 3PL3/16EI, θA = 5PL
2/16EI
10.9. δD = Fab2/8EI
10.11.
10.14. δD = 8.657WL/AE
10.17. vc = pL4/96EI
10.18. NBD = 0.826P, NAB = –0.131P, NBC = 0.22P
10.23. RA = –3Mo/2L, RB = 3Mo/2L, MA = Mo/2
10.27. RA = 4poL/10, MA = poL2/15, RB = poL/10
10.29.
10.32. δE = 41.5(P/AK)3
10.37. NA = P/2π, MA = PR/4
10.43. v = Px2(3L – x)/6EI
10.44. v = Pc2(L – c)2/4EIL
Chapter 11 Stability of Columns
11.1. F = 357.5 kN
11.3. Fall = 12.93 kN
11.5. σcr = 183.2 MPa
11.9. L = 4.625 m
11.12.
a. 6.25%,
b. Pall = 17.99 kN
11.15.
a. σcr = 34 MPa,
b. σall = 35.25 MPa
11.17. Le = L
11.19.
a. ∆T = δ/2αL,
b. ∆T = (δ/2αL) + (π2I/4L2Aα)
11.21. Bar BC fails as a column; Pcr = 1297 N
11.23. σcr = 59.46 MPa
11.31.
a. σmax = 93.71 MPa, vmax = 150.6 mm
11.33. P = 0.89π2EI/L2
11.35. Pcr = 12EI/L2
11.37. Pcr = 9EI1/4L
2
11.38.
11.43. Pcr = 16EI1/L2
Chapter 12 Plastic Behavior of Materials
12.2. α = 42.68°
12.5. σmax = 3Mh/4I
12.7.
a. Pmax = 471 kN,
b. (δAB)p = 0, (δBC)p = 7.66 mm
12.9. P = 189.8 kN
12.13. M = 11ah2σyp/54
12.15.
a. Myp = 15.36 kN · m, M = 21.12 kN · m
12.18. f = 1.7
12.19. f = 1.4
12.22. P = 46.18 kN
12.25. Pu = 9Mu/2L
12.28.
a. Mu = 2Myp,
b. Mu = 16b(b3 – a3)Myp/3π(b
4 – a4)
12.36. to = 6.3 mm
12.41.
a. pu = 25.53 MPa,
b. pu = 29.48 MPa
Chapter 13 Plates and Shells
13.2. εmax = 1250 µ, σmax = 274.7 MPa
13.4. (b) w = Ma(x2 – y2)/2D(1 – v)
13.6. (b) po = 12.05 kPa
13.8. w = Mo(a2 – r2)/2D(1 + v), σr,max = σθ,max = 6Mo/t
2
13.10. n = 3.33
13.11.
13.12. d = 3.85 mm, Mmax = 320 N
13.16.
a. t = 18.26 mm,
b. wmax = 0.336 mm
13.21. p = 1.2 MPa
13.23.
a. t = 11.11 mm,
b. t = 15.6 mm,
c. t = 17.1 mm
13.29. σmax = 2.48 MPa
13.30.
13.31. t = 0.791 mm
13.34.
