SOLUTIONS MANUAL
, Chapter 1
Problems 1-1 through 1-4 are for student research.
1-5
(a) Point vehicles
v
x
cars v 42.1v − v2
Q= = =
hour x 0.324
Seek stationary point maximum
dQ 42.1 − 2v
=0= ∴ v* = 21.05 mph
dv 0.324
42.1(21.05) − 21.052
Q* = = 1368 cars/h Ans.
0.324
(b) v
l x l
2 2
µ ¶−1
v 0.324 l
Q= = +
x +l v(42.1) − v2 v
Maximize Q with l = 10/5280 mi
v Q
22.18 1221.431
22.19 1221.433
22.20 1221.435 ←
22.21 1221.435
22.22 1221.434
1368 − 1221
% loss of throughput = = 12% Ans.
1221
22.2 − 21.05
(c) % increase in speed = 5.5%
21.05
Modest change in optimal speed Ans.
,2 Solutions Manual • Instructor’s Solution Manual 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 implementa-
tion and answers are not known, minimizing instead of maximizing is the largest error
one can make.
Σ
FV = F1 sin θ − W = 0
Σ
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)
F1 W l1
A1 = = , l2 =
S S sin θ cos θ
¯ F2 ¯ W cos θ
A2 = =
¯S¯ S sin θ
µ ¶
l2 W l2 W cos θ
fom = −¢γ +
cos θ S sin θ S sin θ
µ ¶
−¢γ Wl 2 1 + cos2 θ
=
S cos θ sin θ
Set leading constant to unity
θ◦ fom
θ* = 54.736◦ Ans.
0 −∞ fom* = −2.828
20 − 5 .86
30 −4.04 Alternative:
−3.22 µ ¶
40 d 1 + cos2 θ
45 = 0
−3.00 dθ cos θ sin θ
50 −2.87
54.736 And solve resulting tran-
−2.828
60 scendental for θ*.
—2.886
Check second derivative to see if a maximum, minimum, or point of inflection has been
found. Or, evaluate fom on either side of θ *.
, Chapter 1 3
1-7
(a) x1 + x2 = X1 + e1 + X2 + e2
error = e = (x1 + x2) − ( X1 + X2)
= e1 + e2 Ans.
(b) x1 − x2 = X1 + e1 − ( X2 + e2)
e = (x1 − x2) − ( X1 − X2) = e1 − e2 Ans.
(c) x1x2 = ( X1 + e1)( X2 + e2)
e = x1x2 − X 1 X 2 = X 1e2 +µ X2e1 + e1¶e2
e e
= X1e2 + X2e1 = X1 X2 + Ans.
. 1 2
X1 X2
µ ¶
x1 X1 + e1 X1 1 + e1/ X1
(d) = =
x2 X2 + e2 X2 1 + e2/ X2
µ ¶−1 µ ¶µ ¶
e2 . e2 e1 e2 . e1 e2
1+ =1− and 1+ 1− =1+ −
X2 X2 X1 X2 X 1 X2
µ ¶
x1 X1 . X1 e1 e2
e= − − Ans.
=
x2 X2 X2 X1 X2
1-8 √
(a) x1 = 5 = 2.236 067 977 5
X1 = 2.23 3-correct digits
√
x2 = 6 = 2.449 487 742 78
X2 = 2.44 3-correct digits
√ √
x1 + x2 = 5 + 6 = 4.685 557 720 28
√
e1 = x1 − X1 = 5 − 2.23 = 0.006 067 977 5
√
e2 = x2 − X2 = 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 = X1 + X2 + e
= 2.23 + 2.44 + 0.015 557 720 28
= 4.685 557 720 28 (Checks) Ans.
(b) X1 = 2.24, X2 = 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 = X1 + X2 + e
= 2.24 + 2.45 + (−0.004 442 279 72)
= 4.685 557 720 28 Ans.
, Chapter 1
Problems 1-1 through 1-4 are for student research.
1-5
(a) Point vehicles
v
x
cars v 42.1v − v2
Q= = =
hour x 0.324
Seek stationary point maximum
dQ 42.1 − 2v
=0= ∴ v* = 21.05 mph
dv 0.324
42.1(21.05) − 21.052
Q* = = 1368 cars/h Ans.
0.324
(b) v
l x l
2 2
µ ¶−1
v 0.324 l
Q= = +
x +l v(42.1) − v2 v
Maximize Q with l = 10/5280 mi
v Q
22.18 1221.431
22.19 1221.433
22.20 1221.435 ←
22.21 1221.435
22.22 1221.434
1368 − 1221
% loss of throughput = = 12% Ans.
1221
22.2 − 21.05
(c) % increase in speed = 5.5%
21.05
Modest change in optimal speed Ans.
,2 Solutions Manual • Instructor’s Solution Manual 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 implementa-
tion and answers are not known, minimizing instead of maximizing is the largest error
one can make.
Σ
FV = F1 sin θ − W = 0
Σ
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)
F1 W l1
A1 = = , l2 =
S S sin θ cos θ
¯ F2 ¯ W cos θ
A2 = =
¯S¯ S sin θ
µ ¶
l2 W l2 W cos θ
fom = −¢γ +
cos θ S sin θ S sin θ
µ ¶
−¢γ Wl 2 1 + cos2 θ
=
S cos θ sin θ
Set leading constant to unity
θ◦ fom
θ* = 54.736◦ Ans.
0 −∞ fom* = −2.828
20 − 5 .86
30 −4.04 Alternative:
−3.22 µ ¶
40 d 1 + cos2 θ
45 = 0
−3.00 dθ cos θ sin θ
50 −2.87
54.736 And solve resulting tran-
−2.828
60 scendental for θ*.
—2.886
Check second derivative to see if a maximum, minimum, or point of inflection has been
found. Or, evaluate fom on either side of θ *.
, Chapter 1 3
1-7
(a) x1 + x2 = X1 + e1 + X2 + e2
error = e = (x1 + x2) − ( X1 + X2)
= e1 + e2 Ans.
(b) x1 − x2 = X1 + e1 − ( X2 + e2)
e = (x1 − x2) − ( X1 − X2) = e1 − e2 Ans.
(c) x1x2 = ( X1 + e1)( X2 + e2)
e = x1x2 − X 1 X 2 = X 1e2 +µ X2e1 + e1¶e2
e e
= X1e2 + X2e1 = X1 X2 + Ans.
. 1 2
X1 X2
µ ¶
x1 X1 + e1 X1 1 + e1/ X1
(d) = =
x2 X2 + e2 X2 1 + e2/ X2
µ ¶−1 µ ¶µ ¶
e2 . e2 e1 e2 . e1 e2
1+ =1− and 1+ 1− =1+ −
X2 X2 X1 X2 X 1 X2
µ ¶
x1 X1 . X1 e1 e2
e= − − Ans.
=
x2 X2 X2 X1 X2
1-8 √
(a) x1 = 5 = 2.236 067 977 5
X1 = 2.23 3-correct digits
√
x2 = 6 = 2.449 487 742 78
X2 = 2.44 3-correct digits
√ √
x1 + x2 = 5 + 6 = 4.685 557 720 28
√
e1 = x1 − X1 = 5 − 2.23 = 0.006 067 977 5
√
e2 = x2 − X2 = 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 = X1 + X2 + e
= 2.23 + 2.44 + 0.015 557 720 28
= 4.685 557 720 28 (Checks) Ans.
(b) X1 = 2.24, X2 = 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 = X1 + X2 + e
= 2.24 + 2.45 + (−0.004 442 279 72)
= 4.685 557 720 28 Ans.
