Mecanica

Philadelphia University Mechanical Engineering Design 8th eng.ahmad jabali

Chapter 1
Problems 1-1 through 1-4 are for student research. 1-5 (a) Point vehicles
v x

Q= Seek stationary point maximum

cars v 42.1v − v 2 = = hour x 0.324

dQ 42.1 − 2v =0= ∴ v* = 21.05 mph dv 0.324 Q* = (b) 42.1(21.05) − 21.052 = 1368 cars/h Ans. 0.324
v l 2 x l 2

000001 0.324 v l = Q= + 2 x +l v(42.1)− v v Maximize Q with l = 10/5280 mi v 22.18 22.19 22.20 22.21 22.22 Q 1221.431 1221.433 1221.435 ← 1221.435 1221.434

−1

% loss of throughput = (c) % increase in speed

1368 − 1221 = 12% 1221

Ans.

22.2 − 21.05 = 5.5% 21.05 Modest change in optimal speed Ans.

Philadelphia University Mechanical Engineering Design 8th eng.ahmad jabali
2
Solutions Manual • Instructor’s SolutionManual to Accompany Mechanical Engineering Design

1-6 This and the following problem may be the student’s first experience with a figure of merit. • Formulate fom to reflect larger figure of merit for larger merit. • Use a maximization optimization algorithm. When one gets into computer implementation and answers are not known, minimizing instead of maximizing is the largest error one can make. FV = F1sin θ − W = 0 FH = −F1 cos θ − F2 = 0 From which F1 = W/sin θ F2 = −W cos θ/sin θ fom = −$ = −¢γ (volume) . = −¢γ(l1 A1 + l2 A2 ) W F1 l1 = , l2 = A1 = S S sin θ cos θ A2 = W cos θ F2 = S S sin θ l2 W cos θ l2 W + cos θ S sin θ S sin θ 1 + cos2 θ cos θ sin θ

fom = −¢γ = Set leading constant to unity θ◦ 0 20 30 40 45 50 54.736 60

−¢γ W l2 S

fom −∞ −5.86 −4.04 −3.22 −3.00 −2.87 −2.828−2.886

θ* = 54.736◦ fom* = −2.828 Alternative:

Ans.

d 1 + cos2 θ =0 dθ cos θ sin θ And solve resulting transcendental for θ*.

Check second derivative to see if a maximum, minimum, or point of inflection has been found. Or, evaluate fom on either side of θ*.

Philadelphia University Mechanical Engineering Design 8th eng.ahmad jabali
Chapter 1

3

1-7 (a) x1 + x2 = X 1 + e1 + X 2 + e2error = e = (x1 + x2 ) − ( X 1 + X 2 ) = e1 + e2 Ans. (b) x1 − x2 = X 1 + e1 − ( X 2 + e2 ) e = (x1 − x2 ) − ( X 1 − X 2 ) = e1 − e2 Ans. (c) x1 x2 = ( X 1 + e1 )( X 2 + e2 ) e = x1 x2 − X 1 X 2 = X 1 e2 + X 2 e1 + e1 e2 e2 e1 . Ans. + = X 1 e2 + X 2 e1 = X 1 X 2 X1 X2 (d) x1 X 1 + e1 X1 = = x2 X 2 + e2 X2 1+ e= 1-8 (a) e2 X2
−1

1 + e1 / X 1 1 + e2 / X 2 and 1+ e1 X1 1− e2 X2 e1 e2 . =1+ − X1 X2e2 . =1− X2

x1 X1 . X1 − = x2 X2 X2 x1 =

e2 e1 − X1 X2

Ans.

√ 5 = 2.236 067 977 5

X 1 = 2.23 3-correct digits √ x2 = 6 = 2.449 487 742 78 X 2 = 2.44 3-correct digits √ √ x1 + x2 = 5 + 6 = 4.685 557 720 28 √ e1 = x1 − X 1 = 5 − 2.23 = 0.006 067 977 5 √ e2 = x2 − X 2 = 6 − 2.44 = 0.009 489 742 78 √ √ e = e1 + e2 = 5 − 2.23 + 6 − 2.44 = 0.015 557 720 28 Sum = x1 + x2 = X 1 + X 2 +e = 2.23 + 2.44 + 0.015 557 720 28 = 4.685 557 720 28 (Checks) Ans. (b) X 1 = 2.24, X 2 = 2.45 √ e1 = 5 − 2.24 = −0.003 932 022 50 √ e2 = 6 − 2.45 = −0.000 510 257 22 e = e1 + e2 = −0.004 442 279 72 Sum = X 1 + X 2 + e = 2.24 + 2.45 + (−0.004 442 279 72) = 4.685 557 720 28 Ans.

Philadelphia University Mechanical Engineering Design 8th eng.ahmad jabali
4
Solutions Manual • Instructor’sSolution Manual to Accompany Mechanical Engineering Design

1-9 (a) (b) (c) (d) (e) (f) (g) (h) (i) 1-10 (a) (b) (c) (d) (e) (f) (g) (h) (i) 1-11 (a) σ = 200 = 13.1 MPa 15.3 42(103 ) = 70(106 ) N/m2 = 70 MPa (b) σ = 6(10−2 ) 2 1200(800) 3 (10−3 ) 3 = 1.546(10−2 ) m = 15.5 mm 3(207)109 (64)103 (10−3 ) 4 1100(250)(10−3 ) = 9.043(10−2 ) rad = 5.18◦ 9 )(π/32)(25) 4 (10−3 ) 4 79.3(10 l = 1.5/0.305 = 4.918ft = 59.02 in σ = 600/6.89 = 86.96 kpsi p = 160/6.89 = 23.22 psi Z = 1.84(105 )/(25.4) 3 = 11.23 in3 w = 38.1/175 = 0.218 lbf/in δ = 0.05/25.4 = 0.00197 in v = 6.12/0.0051 = 1200 ft/min = 0.0021 in/in V = 30/(0.254) 3 = 1831 in3 σ = 20(6.89) = 137.8 MPa F = 350(4.45) = 1558 N = 1.558 kN M = 1200 lbf · in (0.113) = 135.6 N · m A = 2.4(645) = 1548 mm2 I = 17.4 in4 (2.54) 4 = 724.2 cm4 A =...