All 20 Chapters Covered
SOLUTION MANUAL
, Chapter 1
Problems 1-1 through 1-4 are for student research.
Impending motion to left
1-5
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 DE and BE 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
d G
F F
D cr
C
cr
Facc
Consider force F ′ at G, reactions at A and C. Extend lines of action for fully-developed fric-
tion AE ′ and CE ′ to find the point of concurrency at E ′ for impending motion to the left.
The critical angle is θc′r . Resolve force F ′ into components Fa′ cc and Fc′r . Fa′ cc is related to
mass and acceleration. Pin accelerates to right for any angle 0 < θ ′ < θc′r . When θ ′ > θc′r ,
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)
T
F
Fx = f N sin θ + N cos θ −
=0
T
x F = N (sin θ − f cos θ) r
Ans.
r
T = Nr ( f sin θ + cos θ)
N Combining
fN
1 + f tan θ
T = Fr tan θ −
= KFr Ans. (2)
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 + kx 2
From Prob. 1-6
f tan θ + 1
T2 = Fr tan θ − f
( F0 + kx 2 )( f tan θ + 1)
T2 = r tan θ − Ans.
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
N= F
— f cos θ + sin θ
10 = 13.49 lbf Ans.
Ni =
−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 = = tan
tan θ − f − 0.25 = 0.967 Ans.
60◦
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 − v2
Q= = =
hour x 0.324
Seek stationary point maximum
dQ 42.1 − 2v
∴ v* 21.05 mph
=
= 0 = 0.324
dv 2
Q* = 42.1(21.05) − 21.05
= 1367.6 cars/h Ans.
0.324v
(b)
l x l
2 2
Q= v 0.324 −1
x = l
+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 = 12% Ans.
% loss of 1221
throughput
,4 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
22.2 − 21.05 = 5.5%
(c) % increase in speed 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 F1 = W/sin θ
which F2 = −W cos θ/sin θ
fom = −S = −¢γ (volume)
.
= −¢γ(l1 A1 + l2 A2)
F1 W l1
A1 = , l2 =
= S sin θ cos θ
S
F2 W cos θ
S sin θ
A2 = =
S l2 W l2 W cos θ
fom = −¢γ cos S sin θ +
S sin θ
θ
−¢γ Wl2 1 + cos2 θ
= cos θ sin θ
S
Set leading constant to unity
θ◦ fom θ* = 54.736◦ Ans.
0 −∞ fom* = −2.828
20 −5.86
Alternative:
30 −4.04
40 −3.22 d 1 + cos2 θ
=0
45 −3.00 dθ cos θ sin θ
50 −2.87 And solve resulting
54.736 −2.828 tran- scendental for θ*.
60 −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 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 = X1e2 + X2e 1 + e1e2
. e1 e2 Ans.
= X 1 e 2 + X 2 e1 = X 1 X 2
X1 +
x1 X 1 + e1 X2
X 1 1 + e 1/ X 1
= =
(d)
x2 X 2 + e2 X2 1 + e 2/ X 2
−1
e2 . e2 e1 e2 . e1 e2
1+ =1— and 1+ 1− =1
X2 X2 X1 X2 + −
x1 e2 X1 X2
X 1 . X 1 e1 Ans.
e= − = −
x2 X2 X 2 X1 X 2
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 + x 2 =5 + 6 = 4.685 557 720 28
√
e1 = x 1 − X 1 = 5 − 2.23 = 0.006 067 977 5
√
e2 = x 2 − X 2 = − 2.44 = 0.009 489 742 78
6 √
√
e = e1 + e2 = 5 − 2.23 + 6 − 2.44 = 0.015 557 720 28
Sum = x 1 + x 2 = 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
√ − 2.45 = −0.000 510 257 22
e2 = 6
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.
,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) 6 2
= 70 MPa
(b) σ = = 70(10 ) N/m
6(10−2)2
1200(800) 3(10−3)3
−2
(c) y = = 1.546(10 ) m = 15.5 mm
3(207)(6.4)(109)(10−2)4
1100(250)(10−3 )
−2 ◦
(d) θ = = 9.043(10 ) rad = 5.18
79.3(π/32)(25) 4(109)(10−3)4
1-16
600
(a) σ = = 5 MPa
20(6)
1 3 = 9216 mm 4
(b) I = 8(24)
12 4 −1 4
π 4
) = 5.147 cm
(c) I = 32 (10
64 16(16)
6 2
= 5.215 MPa
(d) τ = = 5.215(10 ) N/m
π(25 3)(10−3)3
, Chapter 1 7
1-17
120(103)
(a) τ = 2)
= 382 MPa
(π/4)(20
32(800)(800)(10 −3) 6 2
(b) σ = = 198.9 MPa
−3) 3 = 198.9(10 ) N/m
ππ(32)3(10
4 4 3
− 26 ) = 3334 mm
(c) Z = (36
32(36)4
(d) k = (1.6) (79.3)(10−3)4(109)
= 286.8 N/m
8(19.2)3(32)(10− 3)3
, 8 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
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TO DOWNLOAD THE FULL PDF
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SOLUTION MANUAL
, Chapter 1
Problems 1-1 through 1-4 are for student research.
Impending motion to left
1-5
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 DE and BE 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
d G
F F
D cr
C
cr
Facc
Consider force F ′ at G, reactions at A and C. Extend lines of action for fully-developed fric-
tion AE ′ and CE ′ to find the point of concurrency at E ′ for impending motion to the left.
The critical angle is θc′r . Resolve force F ′ into components Fa′ cc and Fc′r . Fa′ cc is related to
mass and acceleration. Pin accelerates to right for any angle 0 < θ ′ < θc′r . When θ ′ > θc′r ,
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)
T
F
Fx = f N sin θ + N cos θ −
=0
T
x F = N (sin θ − f cos θ) r
Ans.
r
T = Nr ( f sin θ + cos θ)
N Combining
fN
1 + f tan θ
T = Fr tan θ −
= KFr Ans. (2)
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 + kx 2
From Prob. 1-6
f tan θ + 1
T2 = Fr tan θ − f
( F0 + kx 2 )( f tan θ + 1)
T2 = r tan θ − Ans.
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
N= F
— f cos θ + sin θ
10 = 13.49 lbf Ans.
Ni =
−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 = = tan
tan θ − f − 0.25 = 0.967 Ans.
60◦
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 − v2
Q= = =
hour x 0.324
Seek stationary point maximum
dQ 42.1 − 2v
∴ v* 21.05 mph
=
= 0 = 0.324
dv 2
Q* = 42.1(21.05) − 21.05
= 1367.6 cars/h Ans.
0.324v
(b)
l x l
2 2
Q= v 0.324 −1
x = l
+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 = 12% Ans.
% loss of 1221
throughput
,4 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
22.2 − 21.05 = 5.5%
(c) % increase in speed 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 F1 = W/sin θ
which F2 = −W cos θ/sin θ
fom = −S = −¢γ (volume)
.
= −¢γ(l1 A1 + l2 A2)
F1 W l1
A1 = , l2 =
= S sin θ cos θ
S
F2 W cos θ
S sin θ
A2 = =
S l2 W l2 W cos θ
fom = −¢γ cos S sin θ +
S sin θ
θ
−¢γ Wl2 1 + cos2 θ
= cos θ sin θ
S
Set leading constant to unity
θ◦ fom θ* = 54.736◦ Ans.
0 −∞ fom* = −2.828
20 −5.86
Alternative:
30 −4.04
40 −3.22 d 1 + cos2 θ
=0
45 −3.00 dθ cos θ sin θ
50 −2.87 And solve resulting
54.736 −2.828 tran- scendental for θ*.
60 −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 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 = X1e2 + X2e 1 + e1e2
. e1 e2 Ans.
= X 1 e 2 + X 2 e1 = X 1 X 2
X1 +
x1 X 1 + e1 X2
X 1 1 + e 1/ X 1
= =
(d)
x2 X 2 + e2 X2 1 + e 2/ X 2
−1
e2 . e2 e1 e2 . e1 e2
1+ =1— and 1+ 1− =1
X2 X2 X1 X2 + −
x1 e2 X1 X2
X 1 . X 1 e1 Ans.
e= − = −
x2 X2 X 2 X1 X 2
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 + x 2 =5 + 6 = 4.685 557 720 28
√
e1 = x 1 − X 1 = 5 − 2.23 = 0.006 067 977 5
√
e2 = x 2 − X 2 = − 2.44 = 0.009 489 742 78
6 √
√
e = e1 + e2 = 5 − 2.23 + 6 − 2.44 = 0.015 557 720 28
Sum = x 1 + x 2 = 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
√ − 2.45 = −0.000 510 257 22
e2 = 6
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.
,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) 6 2
= 70 MPa
(b) σ = = 70(10 ) N/m
6(10−2)2
1200(800) 3(10−3)3
−2
(c) y = = 1.546(10 ) m = 15.5 mm
3(207)(6.4)(109)(10−2)4
1100(250)(10−3 )
−2 ◦
(d) θ = = 9.043(10 ) rad = 5.18
79.3(π/32)(25) 4(109)(10−3)4
1-16
600
(a) σ = = 5 MPa
20(6)
1 3 = 9216 mm 4
(b) I = 8(24)
12 4 −1 4
π 4
) = 5.147 cm
(c) I = 32 (10
64 16(16)
6 2
= 5.215 MPa
(d) τ = = 5.215(10 ) N/m
π(25 3)(10−3)3
, Chapter 1 7
1-17
120(103)
(a) τ = 2)
= 382 MPa
(π/4)(20
32(800)(800)(10 −3) 6 2
(b) σ = = 198.9 MPa
−3) 3 = 198.9(10 ) N/m
ππ(32)3(10
4 4 3
− 26 ) = 3334 mm
(c) Z = (36
32(36)4
(d) k = (1.6) (79.3)(10−3)4(109)
= 286.8 N/m
8(19.2)3(32)(10− 3)3
, 8 Solutions Manual • Instructor’s Solution Manual to Accompany Mechanical Engineering Design
THOSE WERE PREVIEW PAGES
TO DOWNLOAD THE FULL PDF
CLICK ON THE L.I.N.K
ON THE NEXT PAGE
