All 20 Chapters Covered
SOLUTION MANUAL
, Chapter 1
Problems 1-1 through 1-4 are for student research.
1-5 Impending motion to left
E
1 1
f f
A B
G
Fcr F
D C cr
Facc
Consider force F at G, reactions at B and D. Extend lines of action for fully-developed fric-
tion D E and B E to find the point of concurrency at E for impending motion to the left. The
critical angle is θcr . Resolve force F into components Facc and Fcr . Facc is related to mass and
acceleration. Pin accelerates to left for any angle 0 < θ < θcr . When θ > θcr , no magnitude
of F will move the pin.
Impending motion to right
E⬘ ⭈E
1 1
f f
A B
G
d
⬘
Fcr F⬘ ⬘
D C ⬘
cr
F⬘acc
Consider force F ′ at G, reactions at A and C. Extend lines of action for fully-developed fric-
tion AE ′ and C E ′ to find the point of concurrency at E ′ for impending motion to the left. The
critical angle is θcr′ . Resolve force F ′ into components Facc
′ and F ′ . F ′ is related to mass
cr acc
and acceleration. Pin accelerates to right for any angle 0 < θ ′ < θcr′ . When θ ′ > θcr′ , no mag-
nitude of F ′ will move the pin.
The intent of the question is to get the student to draw and understand the free body in
order to recognize what it teaches. The graphic approach accomplishes this quickly. It is im-
portant to point out that this understanding enables a mathematical model to be constructed,
and that there are two of them.
This is the simplest problem in mechanical engineering. Using it is a good way to begin a
course.
What is the role of pin diameter d?
Yes, changing the sense of F changes the response.
,2 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
1-6
(a) y Fy = −F − f N cos θ + N sin θ = 0 (1)
F
T
Fx = f N sin θ + N cos θ − =0
r
T
r x F = N (sin θ − f cos θ) Ans.
T = Nr( f sin θ + cos θ)
N
fN
Combining
1 + f tan θ
T = Fr = KFr Ans. (2)
tan θ − f
(b) If T → ∞ detent self-locking tan θ − f = 0 ∴ θcr = tan−1 f Ans.
(Friction is fully developed.)
Check: If F = 10 lbf, f = 0.20, θ = 45◦ , r = 2 in
10
N= = 17.68 lbf
−0.20 cos 45◦ + sin 45◦
T
= 17.28(0.20 sin 45◦ + cos 45◦ ) = 15 lbf
r
f N = 0.20(17.28) = 3.54 lbf
θcr = tan−1 f = tan−1 (0.20) = 11.31◦
11.31° < θ < 90°
1-7
(a) F = F0 + k(0) = F0
T1 = F0r Ans.
(b) When teeth are about to clear
F = F0 + kx2
From Prob. 1-6
f tan θ + 1
T2 = Fr
tan θ − f
( F0 + kx2 )( f tan θ + 1)
T2 = r Ans.
tan θ − f
1-8
Given, F = 10 + 2.5x lbf, r = 2 in, h = 0.2 in, θ = 60◦ , f = 0.25, xi = 0, x f = 0.2
Fi = 10 lbf; Ff = 10 + 2.5(0.2) = 10.5 lbf Ans.
, Chapter 1 3
From Eq. (1) of Prob. 1-6
F
N=
− f cos θ + sin θ
10
Ni = = 13.49 lbf Ans.
−0.25 cos 60◦ + sin 60◦
10.5
Nf = 13.49 = 14.17 lbf Ans.
10
From Eq. (2) of Prob. 1-6
1 + f tan θ 1 + 0.25 tan 60◦
K = = = 0.967 Ans.
tan θ − f tan 60◦ − 0.25
Ti = 0.967(10)(2) = 19.33 lbf · in
Tf = 0.967(10.5)(2) = 20.31 lbf · in
1-9
(a) Point vehicles
v
x
cars v 42.1v − v 2
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* = = 1367.6 cars/h Ans.
0.324
(b) v
l x l
2 2
−1
v 0.324 l
Q= = +
x +l v(42.1) − v 2 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
,4 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
22.2 − 21.05
(c) % increase in speed = 5.5%
21.05
Modest change in optimal speed Ans.
1-10 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.
FV = F1 sin θ − W = 0
FH = −F1 cos θ − F2 = 0
From which
F1 = W/sin θ
F2 = −W cos θ/sin θ
fom = −S = −¢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
l2 W l2 W cos θ
fom = −¢γ +
cos θ S sin θ S sin θ
−¢γ W l2 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:
d 1 + cos2 θ
40 −3.22
45 −3.00 =0
dθ cos θ sin θ
50 −2.87
54.736 −2.828 And solve resulting tran-
60 −2.886 scendental 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*.
, Chapter 1 5
1-11
(a) x1 + x2 = X 1 + e1 + X 2 + e2
error = 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
. e1 e2
= X 1 e2 + X 2 e1 = X 1 X 2 + Ans.
X1 X2
x1 X 1 + e1 X 1 1 + e1 / X 1
(d) = =
x2 X 2 + e2 X 2 1 + e2 / X 2
−1
e2 . e2 e1 e2 . e1 e2
1+ =1− and 1+ 1− =1+ −
X2 X2 X1 X2 X1 X2
x1 X 1 . X 1 e1 e2
e= − = − Ans.
x2 X2 X2 X1 X2
1-12 √
(a) x1 = 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.
,6 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
1-13
(a) σ = 20(6.89) = 137.8 MPa
(b) F = 350(4.45) = 1558 N = 1.558 kN
(c) M = 1200 lbf · in (0.113) = 135.6 N · m
(d) A = 2.4(645) = 1548 mm2
(e) I = 17.4 in4 (2.54) 4 = 724.2 cm4
(f) A = 3.6(1.610) 2 = 9.332 km2
(g) E = 21(1000)(6.89) = 144.69(103 ) MPa = 144.7 GPa
(h) v = 45 mi/h (1.61) = 72.45 km/h
(i) V = 60 in3 (2.54) 3 = 983.2 cm3 = 0.983 liter
1-14
(a) l = 1.5/0.305 = 4.918 ft = 59.02 in
(b) σ = 600/6.89 = 86.96 kpsi
(c) p = 160/6.89 = 23.22 psi
(d) Z = 1.84(105 )/(25.4) 3 = 11.23 in3
(e) w = 38.1/175 = 0.218 lbf/in
(f) δ = 0.05/25.4 = 0.00197 in
(g) v = 6.12/0.0051 = 1200 ft/min
(h) ǫ = 0.0021 in/in
(i) V = 30/(0.254) 3 = 1831 in3
1-15
200
(a) σ = = 13.1 MPa
15.3
42(103 )
(b) σ = = 70(106 ) N/m2 = 70 MPa
6(10−2 ) 2
1200(800) 3 (10−3 ) 3
(c) y = = 1.546(10−2 ) m = 15.5 mm
3(207)(6.4)(109 )(10−2 ) 4
1100(250)(10−3 )
(d) θ = 4 9 −3 4
= 9.043(10−2 ) rad = 5.18◦
79.3(π/32)(25) (10 )(10 )
1-16
600
(a) σ = = 5 MPa
20(6)
1
(b) I = 8(24) 3 = 9216 mm4
12
π
(c) I = 324 (10−1 ) 4 = 5.147 cm4
64
16(16)
(d) τ = = 5.215(106 ) N/m2 = 5.215 MPa
π(253 )(10−3 ) 3
, Chapter 1 7
1-17
120(103 )
(a) τ = = 382 MPa
(π/4)(202 )
32(800)(800)(10−3 )
(b) σ = = 198.9(106 ) N/m2 = 198.9 MPa
π(32) 3 (10−3 ) 3
π
(c) Z = (364 − 264 ) = 3334 mm3
32(36)
(1.6) 4 (79.3)(10−3 ) 4 (109 )
(d) k = = 286.8 N/m
8(19.2) 3 (32)(10−3 ) 3
, Chapter 2
2-1
(a) 12
10
8
6
4
2
0
60 70 80 90 100 110 120 130 140 150 160 170 180 190 200 210
(b) f/(Nx) = f/(69 · 10) = f/690
x f fx f x2 f/(Nx)
60 2 120 7200 0.0029
70 1 70 4900 0.0015
80 3 240 19 200 0.0043
90 5 450 40 500 0.0072
100 8 800 80 000 0.0116
110 12 1320 145 200 0.0174
120 6 720 86 400 0.0087
130 10 1300 169 000 0.0145
140 8 1120 156 800 0.0116
150 5 750 112 500 0.0174
160 2 320 51 200 0.0029
170 3 510 86 700 0.0043
180 2 360 64 800 0.0029
190 1 190 36 100 0.0015
200 0 0 0 0
210 1 210 44 100 0.0015
69 8480 1 104 600
8480
Eq. (2-9) x̄ = = 122.9 kcycles
69
1/2
1 104 600 − 84802 /69
Eq. (2-10) sx =
69 − 1
= 30.3 kcycles Ans.
SOLUTION MANUAL
, Chapter 1
Problems 1-1 through 1-4 are for student research.
1-5 Impending motion to left
E
1 1
f f
A B
G
Fcr F
D C cr
Facc
Consider force F at G, reactions at B and D. Extend lines of action for fully-developed fric-
tion D E and B E to find the point of concurrency at E for impending motion to the left. The
critical angle is θcr . Resolve force F into components Facc and Fcr . Facc is related to mass and
acceleration. Pin accelerates to left for any angle 0 < θ < θcr . When θ > θcr , no magnitude
of F will move the pin.
Impending motion to right
E⬘ ⭈E
1 1
f f
A B
G
d
⬘
Fcr F⬘ ⬘
D C ⬘
cr
F⬘acc
Consider force F ′ at G, reactions at A and C. Extend lines of action for fully-developed fric-
tion AE ′ and C E ′ to find the point of concurrency at E ′ for impending motion to the left. The
critical angle is θcr′ . Resolve force F ′ into components Facc
′ and F ′ . F ′ is related to mass
cr acc
and acceleration. Pin accelerates to right for any angle 0 < θ ′ < θcr′ . When θ ′ > θcr′ , no mag-
nitude of F ′ will move the pin.
The intent of the question is to get the student to draw and understand the free body in
order to recognize what it teaches. The graphic approach accomplishes this quickly. It is im-
portant to point out that this understanding enables a mathematical model to be constructed,
and that there are two of them.
This is the simplest problem in mechanical engineering. Using it is a good way to begin a
course.
What is the role of pin diameter d?
Yes, changing the sense of F changes the response.
,2 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
1-6
(a) y Fy = −F − f N cos θ + N sin θ = 0 (1)
F
T
Fx = f N sin θ + N cos θ − =0
r
T
r x F = N (sin θ − f cos θ) Ans.
T = Nr( f sin θ + cos θ)
N
fN
Combining
1 + f tan θ
T = Fr = KFr Ans. (2)
tan θ − f
(b) If T → ∞ detent self-locking tan θ − f = 0 ∴ θcr = tan−1 f Ans.
(Friction is fully developed.)
Check: If F = 10 lbf, f = 0.20, θ = 45◦ , r = 2 in
10
N= = 17.68 lbf
−0.20 cos 45◦ + sin 45◦
T
= 17.28(0.20 sin 45◦ + cos 45◦ ) = 15 lbf
r
f N = 0.20(17.28) = 3.54 lbf
θcr = tan−1 f = tan−1 (0.20) = 11.31◦
11.31° < θ < 90°
1-7
(a) F = F0 + k(0) = F0
T1 = F0r Ans.
(b) When teeth are about to clear
F = F0 + kx2
From Prob. 1-6
f tan θ + 1
T2 = Fr
tan θ − f
( F0 + kx2 )( f tan θ + 1)
T2 = r Ans.
tan θ − f
1-8
Given, F = 10 + 2.5x lbf, r = 2 in, h = 0.2 in, θ = 60◦ , f = 0.25, xi = 0, x f = 0.2
Fi = 10 lbf; Ff = 10 + 2.5(0.2) = 10.5 lbf Ans.
, Chapter 1 3
From Eq. (1) of Prob. 1-6
F
N=
− f cos θ + sin θ
10
Ni = = 13.49 lbf Ans.
−0.25 cos 60◦ + sin 60◦
10.5
Nf = 13.49 = 14.17 lbf Ans.
10
From Eq. (2) of Prob. 1-6
1 + f tan θ 1 + 0.25 tan 60◦
K = = = 0.967 Ans.
tan θ − f tan 60◦ − 0.25
Ti = 0.967(10)(2) = 19.33 lbf · in
Tf = 0.967(10.5)(2) = 20.31 lbf · in
1-9
(a) Point vehicles
v
x
cars v 42.1v − v 2
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* = = 1367.6 cars/h Ans.
0.324
(b) v
l x l
2 2
−1
v 0.324 l
Q= = +
x +l v(42.1) − v 2 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
,4 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
22.2 − 21.05
(c) % increase in speed = 5.5%
21.05
Modest change in optimal speed Ans.
1-10 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.
FV = F1 sin θ − W = 0
FH = −F1 cos θ − F2 = 0
From which
F1 = W/sin θ
F2 = −W cos θ/sin θ
fom = −S = −¢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
l2 W l2 W cos θ
fom = −¢γ +
cos θ S sin θ S sin θ
−¢γ W l2 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:
d 1 + cos2 θ
40 −3.22
45 −3.00 =0
dθ cos θ sin θ
50 −2.87
54.736 −2.828 And solve resulting tran-
60 −2.886 scendental 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*.
, Chapter 1 5
1-11
(a) x1 + x2 = X 1 + e1 + X 2 + e2
error = 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
. e1 e2
= X 1 e2 + X 2 e1 = X 1 X 2 + Ans.
X1 X2
x1 X 1 + e1 X 1 1 + e1 / X 1
(d) = =
x2 X 2 + e2 X 2 1 + e2 / X 2
−1
e2 . e2 e1 e2 . e1 e2
1+ =1− and 1+ 1− =1+ −
X2 X2 X1 X2 X1 X2
x1 X 1 . X 1 e1 e2
e= − = − Ans.
x2 X2 X2 X1 X2
1-12 √
(a) x1 = 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.
,6 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
1-13
(a) σ = 20(6.89) = 137.8 MPa
(b) F = 350(4.45) = 1558 N = 1.558 kN
(c) M = 1200 lbf · in (0.113) = 135.6 N · m
(d) A = 2.4(645) = 1548 mm2
(e) I = 17.4 in4 (2.54) 4 = 724.2 cm4
(f) A = 3.6(1.610) 2 = 9.332 km2
(g) E = 21(1000)(6.89) = 144.69(103 ) MPa = 144.7 GPa
(h) v = 45 mi/h (1.61) = 72.45 km/h
(i) V = 60 in3 (2.54) 3 = 983.2 cm3 = 0.983 liter
1-14
(a) l = 1.5/0.305 = 4.918 ft = 59.02 in
(b) σ = 600/6.89 = 86.96 kpsi
(c) p = 160/6.89 = 23.22 psi
(d) Z = 1.84(105 )/(25.4) 3 = 11.23 in3
(e) w = 38.1/175 = 0.218 lbf/in
(f) δ = 0.05/25.4 = 0.00197 in
(g) v = 6.12/0.0051 = 1200 ft/min
(h) ǫ = 0.0021 in/in
(i) V = 30/(0.254) 3 = 1831 in3
1-15
200
(a) σ = = 13.1 MPa
15.3
42(103 )
(b) σ = = 70(106 ) N/m2 = 70 MPa
6(10−2 ) 2
1200(800) 3 (10−3 ) 3
(c) y = = 1.546(10−2 ) m = 15.5 mm
3(207)(6.4)(109 )(10−2 ) 4
1100(250)(10−3 )
(d) θ = 4 9 −3 4
= 9.043(10−2 ) rad = 5.18◦
79.3(π/32)(25) (10 )(10 )
1-16
600
(a) σ = = 5 MPa
20(6)
1
(b) I = 8(24) 3 = 9216 mm4
12
π
(c) I = 324 (10−1 ) 4 = 5.147 cm4
64
16(16)
(d) τ = = 5.215(106 ) N/m2 = 5.215 MPa
π(253 )(10−3 ) 3
, Chapter 1 7
1-17
120(103 )
(a) τ = = 382 MPa
(π/4)(202 )
32(800)(800)(10−3 )
(b) σ = = 198.9(106 ) N/m2 = 198.9 MPa
π(32) 3 (10−3 ) 3
π
(c) Z = (364 − 264 ) = 3334 mm3
32(36)
(1.6) 4 (79.3)(10−3 ) 4 (109 )
(d) k = = 286.8 N/m
8(19.2) 3 (32)(10−3 ) 3
, Chapter 2
2-1
(a) 12
10
8
6
4
2
0
60 70 80 90 100 110 120 130 140 150 160 170 180 190 200 210
(b) f/(Nx) = f/(69 · 10) = f/690
x f fx f x2 f/(Nx)
60 2 120 7200 0.0029
70 1 70 4900 0.0015
80 3 240 19 200 0.0043
90 5 450 40 500 0.0072
100 8 800 80 000 0.0116
110 12 1320 145 200 0.0174
120 6 720 86 400 0.0087
130 10 1300 169 000 0.0145
140 8 1120 156 800 0.0116
150 5 750 112 500 0.0174
160 2 320 51 200 0.0029
170 3 510 86 700 0.0043
180 2 360 64 800 0.0029
190 1 190 36 100 0.0015
200 0 0 0 0
210 1 210 44 100 0.0015
69 8480 1 104 600
8480
Eq. (2-9) x̄ = = 122.9 kcycles
69
1/2
1 104 600 − 84802 /69
Eq. (2-10) sx =
69 − 1
= 30.3 kcycles Ans.
