Fundamentals of Machine Component Design 5th Edition Juvinall Solutions Manual

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2-1
SOLUTION (2.1D)
Known: Definitions are needed for the terms: free-body diagram, equilibrium
analysis, internal loads, external loads, and three-force members.
Find: Write definitions of above terms.
Analysis:
1. A free-body diagram is a drawing or a sketch of a body (or part of a body)
that shows all the forces from the surroundings acting on that body. The
forces could be caused by gravitational attraction, centrifugal acceleration,
magnetic repulsion or attraction, or another body.
2. An equilibrium analysis is an analytical method that employs the basic
equations of equilibrium to determine unknown loads (forces and/or
moments).
3. An internal load is a load internal to the body and not from the surroundings.
4. An external load is a load from the surroundings that acts on a member.
5. A three-force member is a body that has three forces from the surroundings acting
on the body. If the body is in static equilibrium the forces (or projection of the
forces) all pass through a common point and the sum of the three vector forces is
zero.

2-2
SOLUTION (2.2)
Known: The Iron Arms forearm grips exercises the forearm by resisting the rotation of
the handle grips – see textbook FIGURE P2.2. The Iron Arms consists of a handle
which rotates inside of the housing that compresses a helical spring. The dimensions of
the Iron Arms forearm grip are known.
Find: Draw a free body diagram of each spring, each handle, the housing (rings) and the
Iron Arms assembly.
Assumptions:
1. No gravitational forces, frictional forces or handle torque act on the components.
2. The unit is at static equilibrium at the rotated position.
Schematic and Given Data: See FIGURE P2.2 in the textbook.
Analysis:

2-3
SOLUTION (2.3)
Known: The Iron Arms forearm grips exercises the forearm by resisting the rotation of
the handle grips – see textbook FIGURE P2.2. The Iron Arms consists of a handle
which rotates inside of the housing that compresses a helical spring. The dimensions of
the Iron Arms forearm grip is known.
Find: Draw a free body diagram of each spring, each handle, the housing (rings) and the
Iron Arms assembly when the handles have each been rotated 90 degrees inside the housing
from a grip torque.
Assumptions:
3. No gravitational forces or frictional forces are applied to the components.
4. The unit is at static equilibrium at the rotated position.
5. A pure torque is applied to each handle grip.
Schematic and Given Data: See FIGURE P2.2 in the textbook.
Analysis:

2-4
SOLUTION (2.4)
Known: A steel cable is tensioned using two methods.
Find: What is the tension in the steel cable if (a) one person on each end pulls with a force
of 75 lb and (b) one end of the cable is attached directly and permanently to a tree and a
person pulls on the other end with a force of 75 lb?
Analysis: The tension in the steel cable is 75 lb for case (a) and case (b).
SOLUTION (2.5D)
Known: Forces act on a person when walking.
Find: What forces act on a person when walking on a level roadway? What forces act on a
person walking on a level belt of a treadmill? Which activity takes more effort?
Assumptions:
1. The treadmill belt velocity is equal to the velocity of the person walking on level ground.
2. The roadway is firm, clear of debris and possesses a coefficient of friction similar to the
treadmill belt.
3. The person uses the same shoes for both activities.
Analysis:
1. The forces acting on a person walking on a level roadway are:
(a) Vertical force resisting gravity through the shoe/ground interface
(b) Drag force from the surrounding air on the walker. This could be an assisting
or retarding force relative to the path of travel.
(c) Traction force at the shoe/ground interface providing the means for forward
travel.
2. The forces acting on a person walking on a level treadmill belt are:
(a) Vertical force resisting gravity through the shoe/belt interface
(b) Traction force from the shoe to the tread belt interface.
3. The person walking outside generally requires more effort due to wind resistance
forces for calm atmospheric conditions. For this reason a treadmill should be raised
to 0.5° or 1.0° to simulate outdoors walking.
Comments: The net sum of tread belt forces results in three conditions. If the tractive
force applied by the user is less than the tractive force from the tread belts movement
the result is the user will move in the direction the tread belt rearward (and ultimately
off the treadmill!). If the two tractive forces are nearly equivalent the user remains
relatively stationary. If the user’s applied tractive force is larger than the tread belt
tractive force the user will move forwards relative to the treadmill.

2-5
SOLUTION (2.6Dnew)
Known: A motorcycle of weight W is shown in textbook Figure P2.6D. The two tires
carry the weight of the motorcycle and passenger(s) as well as the forces generated
during braking and steering. The motorcycle has a wheelbase of length L. The center of
gravity is a distance c forward of the rear axle and a distance of h above the ground.
The coefficient of friction between the pavement and the tires is µ.
Find: Draw a free-body diagram for the motorcycle for (a) rear wheel braking only,
(b) front wheel braking only, and (c) front and rear wheel braking. Also, determine
the magnitudes of the forces exerted by the roadway on the two tires during braking
for the above cases.
Schematic and Given Data:
CG
L
h
R F
c
B µ R
µ F
ma
W
Assumptions:
1. The friction force is constant during braking.
2. The friction coefficient is identical for the front and rear tire-road interface.
3. The vehicle deceleration is uniform.
Analysis: Part (c) front and rear wheel braking.
1. From summation of moments (positive clockwise) about point B,
-F(L) + (ma)h + cW = 0
2. From summation of forces in the vertical direction,
+ R - W + F = 0
3. From summation of forces in the horizontal direction,
+(ma) - µR - µF = 0
4. Solving by eliminating (ma) and R gives
- FL + µhW + cW = 0
5. Solving for F gives, F =
W( c + µh )
L
Comment:
1. With the motorcycle stationary, the static force on the front tire is F = Wc/L.
2. Note that the reverse effective deceleration force (F=ma) is equal to the friction
forces (µR and µF) that the pavement exerts on the front and rear tires.

2-6
SOLUTION (2.7new)
Known: A motorcycle of weigh, W = 1000 lb is shown in Figure P2.6D of the
textbook. The two tires carry the weight of the motorcycle and passenger(s) as well as
the forces generated during braking and steering. The motorcycle has a wheelbase of
length L = 70 in. The center of gravity is a distance c = 38 in. forward of the rear axle
and a distance of h = 24 in. above the ground. The coefficient of friction between the
pavement and the tires is µ = 0.7.
Find: Determine the forces on the rear wheel of the motorcycle for (a) rear wheel
braking only, and (b) front wheel braking only. (Also, draw a free-body diagram for
the motorcycle.)
A
CG
L
h
R F
c
B µ R
µ F
ma
W
Assumptions:
2. The friction coefficient is identical for the front and rear tire-road interface.
Analysis:
Part (a) -- rear wheel braking only.
1. From summation of moments (positive clockwise) about point A,
R(L) + (ma)h - W(L-c) = 0
+ R - W + F = 0
+(ma) - µR = 0
4. Combining the equations of part 1 and part 3 by eliminating (ma) gives
R(L) + (µR)h - (L-c)W = 0
5. Solving for R gives, R =
W (L - c)
L + µh
=
1000 lb (70 in. - 38 in.)
70 in. + 0.7 (24 in.)
= 368.7 lb
6. The friction force on the rear tire (µR) = 0.7 (368.7) = 258 lb.

2-7
Analysis: Part (b) -- front wheel braking only.
1. From summation of moments (positive clockwise) about point B,
-F(L) + (ma)h + cW = 0
+ R - W + F = 0
+(ma) - µF = 0
4. Combining the equations of part 1 and part 3 by eliminating (ma) gives
- FL + (µF)h + cW = 0
5. Solving for F gives, F = cW
(L - µh )
=
38 in.(1000 lb)
70 in. - 0.7 (24 in.)
= 714.3 lb
6. The rear wheel force is R = W - F = 1000 - 714.3 = 286.6 lb
Comment:
1. With the motorcycle stationary, the static force on the front tire is F = Wc/L.
2. Note that the reverse effective deceleration force (F = ma) is equal to the friction
forces (µR and/or µF) that the pavement exerts on the front and/or rear tires.
SOLUTION (2.8new)
Known: An automobile of weight W and wheelbase L slides while braking on
pavement having a given coefficient of friction. The location of the center of gravity is
specified.
Find: Draw a free-body diagram of the automobile.
CG
L
h
W
R F
c
A
µ R
µ F
Assumptions:
3. The motor exerts negligible torque on the wheels (the motor is disconnected).
Analysis:
1. From summation of moments at point R,

2-8
MR∑ = -F(L) - Ah + Wc = 0
Fy∑ = 0 = R - W + F = 0
Fx∑ = 0 = A + µR + µF = 0
4. Solving by eliminating A and R gives
-FL + µhW + Wc = 0
W( c + µh )
L
Comment: For a 4000 lb vehicle, with L =120 in., h = 26 in., c = 70 in., and µ = 0.7:
1. F = 4000 lb
70 in + (0.7) (26 in )
120
= 2940 lb
2. With the vehicle stationary, the static force on the two front tires is 2330 lb.
SOLUTION (2.9 new)
Known: An automobile of weight W = 4000 lb, wheel base L = 117 in., c =65 in. and h
= 17.5 in slides while braking on pavement with a coefficient of friction of 0.7. (The
location of the center of gravity is specified.)
Find: Determine the force on each of the rear tires. (Start by drawing a free-body
diagram of the automobile.)
CG
L
h
W
R F
c
A
µ R
µ F
Assumptions:
3. The motor exerts negligible torque on the wheels (the motor is disconnected).
4. The friction force is shared equally by each of the rear wheels.

2-9
Analysis:
1. From summation of moments at point R,
MR∑ = -F(L) - Ah + Wc = 0
Fy∑ = 0 = R - W + F = 0
Fx∑ = 0 = A + µR + µF = 0
4. Solving by eliminating A and F gives
RL - WL + µhW + Wc = 0
5. Solving for R gives, R =
W( L - c - µh )
L
6. For a 4000 lb vehicle, with L =117 in., h = 17.5 in., c = 65 in., and µ = 0.7:
7. R =
W( L - c - µh)
L
=
4000 lb( 117 in. - 65 in. - (0.7)(17.5 in.)
117 in.
= 1358 lb
8. The upward force on each rear tire is 1358/2 = 679 lb. The friction force on each
rear tire force is (0.7)(679 lb) = 475.3 lb. The total force then on each rear tire is
679
2
+475.3
2
= 827.6 lb.
SOLUTION (2.10new)
Known: An automobile of weight Wcar and wheelbase L slides while braking on
pavement with a given coefficient of friction. The location of the center of gravity is
specified (as in Problem 2.8). The automobile is towing a one-axle trailer of weight
Wtrailer.
Find: Determine the minimum stopping distance for the automobile and trailer
assuming (a) no braking on the trailer and (b) full braking on the trailer. What is the
minimum stopping distance for the automobile if it is not towing a trailer?
CG
W trailer
L
h
R F
c
µ R
carW
A
µ F
CG
T
µ T
B

2-10
Assumptions:
3. The center of gravity of the trailer and the car are at the same height from the
ground.
4. The coupler connecting the trailer to the car is horizontal and at the same height as
the center of gravity for the trailer.
5. The center of gravity for the trailer is directly over the trailer tires.
6. The coupler does not transmit a bending moment.
Analysis:
1. We first determine the stopping distance, S, for full braking on the car and the
trailer.
2. The kinetic energy of the car and trailer is given by KE = (1/2) mcarV2
car + (1/2)
mtrailer V2
trailer.
3. The work done in stopping the car is µWcarS where S is the minimum stopping
distance and µ is the maximum tire-pavement friction coefficient.
4. The work done in stopping the trailer is µWtrailerS where S is the minimum stopping
distance and µ is the maximum tire-pavement friction coefficient.
5. The kinetic energy of the car and trailer equals the work done to stop the car and
trailer or KE = µWcarS + µWtrailerS.
6. Solving for S, we have
S =
1
2
[mcar+mtrailer] V
2
µmcarg + µmtrailerg
7. With no braking on the trailer, the pavement will not exert a friction force on the
trailer tires and
S =
1
2
[mcar+mtrailer] V
2
µmcarg
8. If the car is not towing a trailer,
S =
1
2
[mcar] V
2
µmcarg

2-11
Comment:
1. With a 4000 lb car, a 2000 lb trailer, µ = 0.7, and a speed of 60 mph (88 feet/s), the
stopping distance for the car without the trailer is S = (1/2)V2
/µg = 171.8 feet.
This is also the stopping distance for the car with the trailer braking with the same
tire-pavement friction coefficient. The stopping distance for the car and trailer
without trailer braking is
[mcar+mtrailer]
mcar
= 1.5 times farther.
2. An interesting exercise is to calculate the forces acting on the car and trailer during
braking (while considering the necessity of the assumptions that were given).
SOLUTION (2.11new)
Known: An automobile of weight W and wheelbase L slides while braking on
pavement with a given coefficient of friction. The location of the center of gravity is
specified. The automobile is traveling downhill at a grade of 10:1.
Find: Draw a free-body diagram of the automobile as it is traveling downhill.
CG
L h
R
F
c
A
µR
µ F
!
ma
W
!
y
x
Assumptions:
Analysis:
1. From summation of moments (positive clockwise) about point B, where B is the
contact “point” of the rear tire with the roadway and where A = ma (the reverse
effective deceleration force),
-F(L) + Ah + c(Wcos θ) + h(Wsin θ) = 0
2. From summation of forces in the y-direction,
R - Wcosθ + F = 0

2-12
3. From summation of forces in the x-direction,
A + Wsinθ - µR - µF = 0
4. Solving by eliminating A and R gives
- FL + h[-Wsinθ + µ(Wcosθ - F) + µF] + cWcosθ + hWsinθ = 0
W( c + µh )
L
cos !
Comment: For a 4000 lb vehicle, with L =120 in., h = 26 in., c = 70 in.,
θ = tan-1
(1/10) = 5.71o
, and µ = 0.7, we have F = 2925.4 lb.
SOLUTION (2.12D)
Known: A vertical wall channel C holds solid cylindrical rods A and B of known
density. The width of channel C is not given.
Find: Select a metal for rods A and B. Draw free body diagrams for rod A, rod B, and
channel C. Determine the magnitude of the forces acting on A, B, and C.
B
A
d
d
!
w = 4 in.
d
2
d
2
d
2
d
2
+ dsin!
d = diameter = 2.5 in.
G H
w
C
g
rod length =2.0 in.
d cos !
Decisions:
1. Select steel which has a known density of ρ = 0.28 lbm/in.3.
2. Select w = 4.0 in. for analysis.

2-13
Assumptions:
1. The channel is open upward and supported on the bottom by two knife
edges at G and H.
2. The friction forces between the contacting bodies are negligible.
3. The rods A and B and channel C are in static equilibrium.
4. The force of gravity is the only body force.
5. The weight of the channel C is negligible.
Analysis:
1. From the free body diagram for A: D = W
sin !
, C = W
tan !
.
2. From the free body diagram for B: E = W
tan !
, F = 2W.
C
B
W
D
F
E
A
W
D
!
!
C = (W/tan !)
E = (W/ tan !)
F = 2W
d
2
d
2
HG
w
+ d sin !d
2
3. For w = 4 in., d = 2.5 in, and rod length L = 2.0 in.; the rod mass = ρV = (0.28
lbm/in.3)(2.0 in.)(π)(1.25)2/4 = 0.687 lbm.
4. The weight of each rod is W = F = ma = (0.687 lbm)(32.2 ft/s2)/gc = 0.687 lb.

2-14
5. θ = cos-1((4-d)/d) = 53.13o.
6. D = .858 lb, C = .515 lb, E = .515 lb, F = 1.374 lb.
7. From the free body diagram for the channel, the forces G and H can be obtained
from force equilibrium.
Comments:
1. An assumption had to be made about the forces that the surroundings exert on
channel C. In order to simplify the analysis, it was assumed that point forces from
the "ground" acted at G and H.
2. To apply the equations of force equilibrium, the channel C should be supported
such that it is in stable equilibrium.
3. To consider the container as a free body, all the forces from the surroundings
acting on the body must be shown.

2-15
SOLUTION (2.13Dnew)
Known: Two spheres A and B are in a container C.
Find: Draw the free body diagrams of A, B, and C. Also determine the forces acting
on these bodies.
L
B
125 N
A
1000 N
D
d
!
A
1000 N
C
D
B
125 N
D
F
E
C = (1000/tan !) N
E = (1000/ tan !) N
F = 1125 N
G H
d
2
d
2
(2/3)L
d
2
+ (D + d) sin !
2

2-16
Assumptions:
1. The container is cylindrical in shape and supported on two knife edges, each at a
distance of (1/3)L (where L is the diameter of the container cross section) from the
base circle center or (2/3)L apart.
2. The friction forces between the contacting bodies are negligible.
3. The sphere and containers are in static equilibrium.
4. The force of gravity is the only body force.
Analysis:
1. From the free body diagram for A, D = 1000
sin !
N, C = 1000
tan !
N.
2. From the free body diagram for B, E = 1000
tan !
N, F = 1125 N.
B
A
D
d
!
G H
d
2
d
2
(2/3)L
d
2
+ (D + d) sin !
2
3. From the above diagram and the free body diagram for C, the forces G and H can
be obtained.
Comments:
1. The geometry of the container must be clearly determined before proceeding to
solve this problem. For example, if the container is assumed to be rectangular in
shape, then the spheres would not be in stable equilibrium. Also, until the
container shape is defined it is impossible to draw the free body diagram.
2. The container should be supported such that it is in stable equilibrium. To
consider the container as a free body, all the forces from the surroundings acting
on the body must be shown.

2-17
SOLUTION (2.14new)
Known: The geometry and the loads acting on a pinned assembly are given.
Find: Draw a free-body diagram for the assembly and determine the magnitude of the
forces acting on each member of the assembly.
45˚
45˚
1000 mm
1500 N 1500 N
45˚
45˚
A D
B C
Link 5
Link 3
Link 1
Link
2
Link
4
Assumptions:
1. The links are rigid.
2. The pin joints are frictionless.
3. The weight of the links are negligible.
4. The links are two force members and are either in tension or compression.
Analysis:
1. We first draw a free-body diagram of the entire structure.
45˚
45˚
1500 N 1500 N
45˚
45˚
A
D
B C
1
1
1
2 2
Ay
Ax
Dy
2. Taking moments about point A and assuming clockwise moments to be positive,
MA = 0 = 1500(2) + 1500(1) - Dy(1)!
3. Solving for Dy gives Dy = 4500 N.
4. Summation of forces in the y-direction and assuming vertical forces positive,
Fy = 0 = Ay + Dy - 1500 - 1500 = Ay + Dy - 3000.!

2-18
5. Since, Dy = 4500 N, Ay = 3000 - Dy = -1500 N.
6. Fx = 0! gives, Ax = 0.
7. We now draw a free-body diagram for a section at C.
45˚
1500 N
DC
CB
8. Fx = 0 = - CB + DC sin 45˚ and!
Fy = 0 = DC sin 45˚ - 1500!
9. Solving simultaneous equations gives DC = 2121 N, CB = 1500 N.
10. We now draw a free-body diagram for a section at A.
45˚
1500 N
AB
DA
11. Fy = 0 = AB sin 45˚- 1500!
Fx = 0 = AB cos 45˚- DA!
12. Solving simultaneous equations gives AB = 2121 N, DA = 1500 N.
13. We now draw a free-body diagram for a section at D.
45˚
1500 N
2121 N
BD
D
B
C
A
4500 N

2-19
14. Fy = 0 = 4500 - BD - 2121(sin 45˚)!
Hence, BD = 3000 N.
15. The free-body diagrams for links DC, BC, AB, AD, and BD are:
B
A
1500 2 = 2121
2121
B C
1500 1500
D
B
3000
3000
DA
15001500
C
D
1500 2 = 2121
2121
16. We can now draw a free-body diagram of pin B:
B
2121 N
3000 N
1500 N
1500 N
45˚
17. Checking for static equilibrium at pin B gives:
Fx = 2121 cos 45˚ - 1500 = 0!
Fy = 1500 + 2121 sin 45˚ - 3000 = 0!
18. We can also draw a free-body diagram for pin C:
C
2121 N
1500 N
1500 N
45˚

2-20
19. Checking for static equilibrium at pin C gives:
Fx = 1500 - 2121 cos 45˚ = 0!
Fy = 1500 - 2121 sin 45˚ = 0!
Comment: From force flow visualization, we determine that links 1, 3, and 4 are in
compression and that links 2 and 5 are in tension.
SOLUTION (2.15new5e)
Known: The geometry and the loads acting on a pinned assembly are given.
Find: Draw a free-body diagram for the assembly and determine the magnitude of the
forces acting on each member of the assembly.
45˚
45˚
1000 mm
750 N 750 N
45˚
45˚
A D
B C
Link 5
Link 3
Link 1
Link
2
Link
4
Assumptions:
1. The links are rigid.
2. The pin joints are frictionless.
3. The weight of the links are negligible.
4. The links are two force members and are either in tension or compression.
Analysis:
1. We first draw a free-body diagram of the entire structure.

2-21
45˚
45˚
750 N 750 N
45˚
45˚
A
D
B C
1
1
1
2 2
Ay
Ax
Dy
2. Taking moments about point A and assuming clockwise moments to be positive,
!MA = 0 = 750(2) + 750(1) - Dy(1)
3. Solving for Dy gives Dy = 2250 N.
4. Summation of forces in the y-direction and assuming vertical forces positive,
!Fy = 0 = Ay + Dy - 750 - 750 = Ay + Dy - 1500.
5. Since, Dy = 2250 N, Ay = 1500 - Dy = -750 N.
6. Fx = 0! gives, Ax = 0.
7. We now draw a free-body diagram for a section at C.
45˚
750 N
DC
CB
8. Fx = 0 = - CB + DC sin 45˚ and!
!Fy = 0 = DC sin 45˚ - 750
9. Solving simultaneous equations gives DC = 1060.6 N, CB = 750 N.
10. We now draw a free-body diagram for a section at A.

2-22
45˚
750 N
AB
DA
11. !Fy = 0 = AB sin 45˚- 750
Fx = 0 = AB cos 45˚- DA!
12. Solving simultaneous equations gives AB = 1060.6 N, DA = 750 N.
13. We now draw a free-body diagram for a section at D.
45˚
750 N
1060.6 N
BD
D
B
C
A
2250 N
14. !Fy = 0 = 2250 - BD - 1060.6(sin 45˚)
Hence, BD = 1500 N.
15. The free-body diagrams for links DC, BC, AB, AD, and BD are:

2-23
B
A
B C
750 750
D
B
1500
DA
C
D
1060.6
1060.6
1060.6
1060.6
750 750
1500
16. We can now draw a free-body diagram of pin B:
B
1060.6 N
1500 N
750 N
750 N
45˚
17. Checking for static equilibrium at pin B gives:
!Fx = 1060.6 cos 45˚ - 750 = 0
!Fy = 750 + 1060.6 sin 45˚ - 1500 = 0

2-24
18. We can also draw a free-body diagram for pin C:
C
1060.6 N
750 N
750 N
45˚
19. Checking for static equilibrium at pin C gives:
!Fx = 750 - 1060.6 cos 45˚ = 0!
!Fy = 750 - 1060.6 sin 45˚ = 0
Comment: From force flow visualization, we determine that links 1, 3, and 4 are in
compression and that links 2 and 5 are in tension.
SOLUTION (2.16new)
Known: A 1800 rpm motor is rotating a blower at 6000 rpm through a gear box having
a known weight.
Find: Determine all loads acting on the gear box when the motor output is 1 hp, and
sketch the gear box as a free-body in equilibrium.
10 in.
Gear box
Blower
6000 rpm
Motor
1 hp
1800 rpm
20 lb
Assumption: The friction losses in the gear box are negligible.
Analysis:

2-25
Gear box
20 lb
10.50 lb in.
7.55 lb
12.45 lb
35.01 lb in.
A
B
Direction of
rotation
Direction of
rotation
1. From Eq. (1.3), T = 5252 • W
n =
5252(1)
1800
T = 2.92 lb•ft = 35.01 lb•in. (motor shaft)
2. To the blower, T = 35.01 ( )1800 rpm
6000 rpm = 10.50 lb•in. (to blower)
3. Mounting torque reaction = 35.01 - 10.50 = 24.51 lb•in.
4. Mounting forces = 24.51 lb•in./ 10 in. = 2.45 lb. The mounting force acts upward
at A and downward at B.
5. Add 10 lb acting upward at A and B to support the gravity load, giving 12.45 lb
upward at A and 7.55 lb upward at B.
SOLUTION (2.17new5e)
Known: A 1800 rpm motor is rotating a blower at 6000 rpm through a gear box having
a weight of 40 lb.
Find: Determine all loads acting on the gear box when the motor output is 1 hp, and
sketch the gear box as a free-body in equilibrium.
10 in.
Gear box
Blower
6000 rpm
Motor
1 hp
1800 rpm
40 lb

2-26
Analysis:
Gear box
40 lb
10.50 lb in.
7.55 lb + 10 lb = 17.55 lb
12.45 lb + 10 lb = 22.45 lb
35.01 lb in.
A
B
Direction of
rotation
Direction of
rotation
1. From Eq. (1.3), T = 5252 • W
n =
5252(1)
1800
T = 2.92 lb•ft = 35.01 lb•in. (motor shaft)
2. To the blower, T = 35.01 ( )1800 rpm
6000 rpm = 10.50 lb•in. (to blower)
3. Mounting torque reaction = 35.01 - 10.50 = 24.51 lb•in.
4. Mounting forces = 24.51 lb•in./ 10 in. = 2.45 lb. The mounting force acts upward
at A and downward at B.
5. Add 20 lb acting upward at A and B to support the gravity load, giving 22.45 lb
upward at A and 17.55 lb upward at B.
Comment: In textbook problem 2.16, the gear box weights 20 lb.
SOLUTION (2.18new)
Known: The motor operates at constant speed and develops a torque of 100 lb-in.
during normal operation. A 5:1 ratio gear reducer is attached to the motor shaft; i.e., the
reducer output shaft rotates in the same direction as the motor but at one-fifth the motor
speed. Rotation of the reducer housing is prevented by the “torque arm,” pin-connected
at each end as shown in Fig. P2.18. The reducer output shaft drives the load through a
flexible coupling. Gravity and friction can be neglected..
Find: Determine the loads applied to (a) the torque arm, (b) the motor output shaft, and
(c) the reducer output shaft.

2-27
500 lb in.
Torque
80 lb
80 lb
100 lb in.
Torque
640 in.lb
Bending
Analysis:
1. The force in the torque arm is 80 lb tension.
2. The loads on the reducer input shaft are 100 lb in. torque, plus 80 lb vertical load
and 640 in lb bending moment in the plane of the motor face.
3. The load on the reducer output shaft is 500 lb in. torque.

2-28
SOLUTION (2.19new)
Known: The motor operates at constant speed and develops a torque of 200 lb-in.
during normal operation. A 5:1 ratio gear reducer is attached to the motor shaft; i.e., the
reducer output shaft rotates in the same direction as the motor but at one-fifth the motor
speed. Rotation of the reducer housing is prevented by the “torque arm,” pin-connected
at each end as shown in Fig. P2.18. The reducer output shaft drives the load through a
flexible coupling. Gravity and friction can be neglected..
Find: Determine the loads applied to (a) the torque arm, (b) the motor output shaft, and
(c) the reducer output shaft.
1000 lb
in.
Torque
160 lb
160 lb
200 lb in.
Torque
1280 in.lb
Bending
Analysis:
1. The force in the torque arm is 160 lb tension.
2. The loads on the reducer input shaft are 200 lb in. torque, plus 160 lb vertical load
and 1280 in lb bending moment in the plane of the motor face.
3. The load on the reducer output shaft is 1000 lb in. torque.
Comment: In textbook problem 2.18, the motor develops a torque of 100 lb in. during
operation.

2-29
SOLUTION (2.20)
Known: The drawing in Fig. P2.20 shows the engine, transmission, and propeller shaft
of a prototype automobile. The transmission and engine are not bolted together but are
attached separately to the frame. The transmission weighs 100 lb, receives an engine
torque of 100 lb-ft at A through a flexible coupling, and attaches to the propeller shaft
at B through a universal joint. The transmission is bolted to the frame at C and D.
Assume that the transmission ratio is -3; i.e., reverse gear with propeller shaft speed = -
1/3 engine speed.
Find: Draw the transmission as a free body in equilibrium.
CG
6 in.6 in.
350 lb
450 lb
300 lb ft
Torque
100 lb ft
Torque
Weight
of 100 lb
Rotation
Rotation
Output
Input
Analysis: A free body diagram is given above.
SOLUTION (2.21new)
Known: The drawing in Fig. P2.20 shows the engine, transmission, and propeller shaft
of a prototype automobile. The transmission and engine are not bolted together but are
attached separately to the frame. The transmission weighs 50 lb, receives an engine
torque of 100 lb-ft at A through a flexible coupling, and attaches to the propeller shaft
at B through a universal joint. The transmission is bolted to the frame at C and D.
Assume that the transmission ratio is -3; i.e., reverse gear with propeller shaft speed = -
1/3 engine speed.
Find: Draw the transmission as a free body in equilibrium.

2-30
CG
6 in.6 in.
375 lb
425 lb
300 lb ft
Torque
100 lb ft
Torque
Weight
of 50 lb
Rotation
Rotation
Output
Input
D
C
Analysis:
1. The force on the transmission at C is 425 lb upward.
2. The force on the transmission at D is 375 lb downward.
3. A free body diagram is given above.
Comment: In textbook problem 2.20, the transmission weights only 50 lb, yet
transmits a torque of 100 lb-ft during operation.
SOLUTION (2.22)
Known: An electric fan motor supported by mountings at A and B delivers a known
torque to fan blades which, in turn, push air forward with a known force.
Find: Determine all loads acting on the fan and sketch it as a free-body in equilibrium.
50 mm50 mm
A B
Clockwise rotation
Force against
the wind = 20 N
T = 2 N•m
The motor delivers a torque,
of T = 2 Nm to the fan blades.
Assumption: The gravity forces can be ignored.

2-31
Analysis:
A B
Clockwise rotation
20 N
20 N
10 N
10 N
20 N
2 N•m
1. The torque exerted on the blades by the wind is 2 N•
m counterclockwise.
2. Mounting forces = (2 N•m)/(0.1 m) = 20 N. Thus, 20 N is exerted upward at A and
downward at B.
SOLUTION (2.23new)
Known: An electric fan motor supported by mountings at A and B delivers a torque of
4 N•m to fan blades which, in turn, push air forward with a force of 40 N.
Find: Determine all loads acting on the fan and sketch it as a free-body in equilibrium.
50 mm50 mm
A B
Clockwise rotation
Force against
the wind = 40 N
of T = 4 N m to the fan blades.
T = 4 N m
Assumption: The gravity forces can be ignored.
Analysis:

2-32
A B
Clockwise rotation
40 N
40 N
20 N
20 N
40 N
4 N•m
m counterclockwise.
downward at B.
3. The force on the motor mount is 40 N upward at A and 40 N downward at B.
SOLUTION (2.24)
Known: A pump is driven by a motor integrally attached to a gear reducer. Shaft C is
attached to Cʹ′, face A is attached to Aʹ′, and face B is attached to Bʹ′.
Find: Sketch the connecting tube and show all loads acting on it.
A
A'
B
B'
C'
C
Pump
450 rpm
Connecting
tube
Motor
1.5 kW
1800 rpm4:1 ratio
gear reducer
Direction
of rotation
Assumption: The components are in static equilibrium.
Analysis:
1. From Eq. (1.2), motor torque, T = 9549 • W
n =
9549(1.5)
1800
= 7.96 N•m
2. Reducer output torque = Pump input torque = 7.96(4) = 31.84 N•m

2-33
A
A'
B
B'
C'
C
Pump
450 rpm
Connecting
tube
Motor
1.5 kW
1800 rpm4:1 ratio
gear reducer
Motor and gear
reducer are attached
to a fixed support
Direction of
rotation
31.84 N•m
31.84 N•m
31.84 N•m
31.84 N•m
SOLUTION (2.25new)
Known: A motor integrally attached to a gear reducer drives a pump. Shaft C is
attached to Cʹ′, face A is attached to Aʹ′, and face B is attached to Bʹ′.
Find: Calculate the output torque if the 4:1 ratio gear reducer has an efficiency of 95%.
A
A'
B
B'
C'
C
Pump
450 rpm
Connecting
tube
Motor
1.5 kW
1800 rpm4:1 ratio
gear reducer
Direction
of rotation
Assumption: The components are in static and thermal equilibrium.
Analysis:
1. From Eq. (1.2), motor torque, T = 9549 • W
n =
9549(1.5)
1800
= 7.96 N•m

2-34
2. Reducer output torque = Pump input torque = 7.96(4) = 31.84 N•m for 100%
efficiency.
3. efficiency = hpout/hpin x 100% = 0.95 x 100% = 95%
4. hpout = Tout
• nout = hpin x efficiency = 1.5 kW x 0.95 = 1.425 kW
5. Gear reducer output torque = Pump input torque = (7.96)(4)(.95) = (.95)(31.84
N•m) = 30.25 N•m
6. A free-body diagram is show below for the motor, gear reducer, connecting tube,
and pump where the gear reducer efficiency is 95%.
A
A'
B
B'
C'
C
Pump
450 rpm
Connecting
tube
Motor
1.5 kW
1800 rpm4:1 ratio
gear reducer
Motor and gear
reducer are attached
to a fixed support
Direction of
rotation
N•m30.25
N•m30.25
N•m30.25
N•m30.25
N•m30.25
N•m30.25
Comment: In textbook problem 2.24, the gear reducer has an efficiency of 100%.
■

2-35
SOLUTION (2.26)
Known: An engine and propeller rotate clockwise viewed from the propeller end. A
reduction gear housing is bolted to the engine housing through the bolt holes shown.
The power and angular velocity of the engine are known.
Find:
(a) Determine the direction and magnitude of the torque applied to the engine housing
by the reduction gear housing.
(b) Determine the magnitude and direction of the torque reaction tending to rotate
(roll) the aircraft.
(c) Find an advantage of using opposite-rotating engines with twin-engine propeller-
driven aircraft.
Engine
150 hp
3600 rpm
Reduction gear
ratio = 1.5
Propeller
Assumption: The friction losses are negligible.
Analysis:
1. From Eq. (1.3), engine torque, T = 5252 • W
n =
5252(150)
3600
= 219 lb ft
2. Reduction gear torque, T = 219(1.5) = 328 lb ft

2-36
Engine
150 hp
3600 rpm
Reduction gear
ratio = 1.5
Propeller
A
328 lb ft
328 lb ft from aerodynamic forces
328 lb ft
109 lb ft
109 lb ft
219 lb ft
219 lb ft
328
A
lb
328
A
lb*
*
3. The attachment forces apply an equal and opposite torque of 328 lb•ft ccw
tending to "roll" the airplane--see (*) in the above figure.
4. Thus, the torque applied to the engine housing by the reduction gear housing is
109 lb ft counter-clockwise, and the torque reaction tending to rotate the aircraft is
328 lb ft counter-clockwise. ■
5. Torque reactions applied to the air frame by the two engines cancel. (This
produces bending in the connecting structure, but does not require a compensating
roll torque from the aerodynamic control surfaces.) ■

2-37
SOLUTION (2.27)
Known: A marine engine delivers a torque of 200 lb-ft to a gearbox that provides a
reverse rotation of -4:1.
Find: Determine the torque required to hold the gearbox in place.
Rotation
200 lb ft
Torque
Rotation
Output
800 lb ft
Torque
Mounting
Torque, Tm
Input
Analysis:
1. The summation of moments about the axis of the shaft must be zero. Therefore,
Toutput + Tinput + Tmounting = 0.
2. The input torque is in the same direction as the input rotation. The output torque is
in a direction opposite the output rotation, and it is known that the output rotation
is opposite the input rotation. Therefore the input and output torques act on the
reducer shafts in the same direction.
3. Also, Tinput = 200 lb ft, Toutput = (4)( 200 lb ft) = 800 lb ft.
4. Therefore we have, Tmounting = -(Toutput + Tinput) = -(200 + 800) = -1,000 lb ft.
Comment: The directions of the torques and the shaft rotations are shown in the above
diagram.
SOLUTION (2.28new)
Known: A marine engine delivers a torque of 400 lb-ft to a gearbox that provides a
reverse rotation of -4:1.
Find: Determine the torque required to hold the gearbox in place.

2-38
Rotation
400 lb ft
Torque
Rotation
Output
1600 lb ft
Torque
Mounting
Torque, Tm
Input
Analysis:
Toutput + Tinput + Tmounting = 0.
is opposite the input rotation. Therefore the input and output torques act on the
reducer shafts in the same direction.
3. Also, Tinput = 400 lb ft, Toutput = (4)(400 lb ft) = 1600 lb ft.
4. Therefore we have, Tmounting = -(Toutput + Tinput) = -(400 + 1600) = -2,000 lb ft.
Comment: The directions of the torques and the shaft rotations are shown in the above
diagram.
SOLUTION (2.29)
Known: A motor delivers 50 lb-ft torque at 2000 rpm to an attached gear reducer. The
reducer and motor housings are connected together by six bolts located on a 12-in.-dia.
circle, centered about the shaft. The reducer has a 4:1 ratio. Neglect friction and
weight.
Find: Estimate the average shearing force carried by each bolt.

2-39
Rotation
50 lb ft
Torque
Rotation
Output
200 lb ft
Torque
Typical force,
Fbolt Input
Assumption:
1. Neglect friction and weight.
2. The components are in static equilibrium.
Analysis:
Toutput + Tinput + Tbolts = 0.
is in the same direction as the input rotation. Therefore the input and output
torques act on the reducer shafts in opposite directions.
3. Also, Tinput = 50 lb ft, Toutput = (-4)(50 lb ft) = - 200 lb ft.
4. Therefore we have, Tbolts = -(Toutput + Tinput) = -(50 - 200) = + 150 lb ft.
5. The force on each bolt, Fbolts can be calculated from the following equation:
Tbolts = 6 Fbolts rbolt circle = 6 Fbolts (0.5 feet) = 150 lb ft.
Yielding Fbolts = 50 lb (per bolt)
Comment: The directions of the torques, the shaft rotations, and the typical bolt force
acting on the reducer are shown in the above diagram.

2-40
SOLUTION (2.30D)
Known: A single-cylinder reciprocating compressor has a crankshaft, connecting rod,
piston, and frame. The piston is 60o
before head-end dead center on the compression
stroke.
Find: Sketch the crankshaft, connecting rod, piston, frame and the entire compressor as
separate free bodies for the 60o
condition.
Assembly
T

2-41
Analysis:
Assembly
T
1 Frame
T
b
Fgas
F41
F21
Engine Mounting
Couple = (F41) b
F41
Rod3
F43
F23
4
Piston
Fgas
F14
F34
Engine Mounting Couple =
T = (F32) a.
So,(F32) a = (F41) b
T
Crank2
F12
F32
a
T = (F32) a
Force Diagrams
F12
F32 F43
F23
Fgas
F14
F34F41
F21
Fgas
4321

2-42
SOLUTION (2.31)
Known: A gear reduction unit and a propeller of an outboard boat operate with a
known motor torque and a known thrust.
Find: Show all external loads acting on the assembly.
500 mm
Forward
direction
of boat
thrust = 400 N
2:1 ratio bevel gear
150 mm
Y
X
Z
20 N•m
Rotation
Propeller
rotation
Assumption: The effects of gravity and friction are negligible.

2-43
Analysis:
500 mm
Forward
direction
of boat travel
2:1 ratio bevel gears
are inside this housing.
150 mm
Y
X
Z
MZ = 20 N•m
MZ = 20 N•m
MY = 200 N•m
MX = 40 N•m
MX = 40 N•m
400 N
Propeller
rotation
Mounting flange
400 N
Suggested notation
Y
Z
X
MX
MZ
MY

2-44
SOLUTION (2.32)
Known: A rider is applying full weight to one pedal of a bicycle.
Find: Draw as free-bodies in equilibrium:
(a) The pedal, crank, and pedal sprocket assembly.
(b) The rear wheel and sprocket assembly.
(c) The front wheel.
(d) The entire bicycle and rider assembly.
400 600
160
800 N
330 R
100 R
40 R
330 R
Assumptions:
1. The bicycle can be treated as a two-dimensional machine.
2. The bicycle weight is negligible.
Analysis:
1. For the pedal, crank, and pedal sprocket assembly, the chain force is F =
800(160/100) = 1280 N
100 R
160
1280 N
1280 N
800 N
800 N
2. For the rear wheel and sprocket assembly,
rear wheel gravity load = 800(440/1000) = 352 N

2-45
rear wheel friction force = 1280(40/330) = 155.2 N
horizontal bearing force = 1280 + 155.2 = 1435.2 N
352 N
352 N
155.2 N
1435.2 N
1280 N
3. For the front wheel, front wheel gravity load = 800(560/1000) = 448 N
448 N
448 N
4. For the entire bicycle and rider assembly, the drawing is given below.
800 N
352 N
155.2 N
448 N

2-46
Comments: The drawing does not show the rearward 155.2 N inertia force necessary
to establish ∑FH = 0. It would be located thru the center of gravity of the cycle-plus-
rider, the location of which is not given. Since this vector would be at some distance
"h" above the ground, the resulting counter clockwise couple, 155.2 h, would be
balanced by decreasing the vertical force on the front wheel and increasing the vertical
force on the rear wheel, both by (155.2 h/1000) N.
SOLUTION (2.33new)
Known: A small rider is applying full weight to one pedal of a bicycle.
Find: Draw as free-bodies in equilibrium:
(a) The pedal, crank, and pedal sprocket assembly.
(b) The rear wheel and sprocket assembly.
(c) The front wheel.
(d) The entire bicycle and rider assembly.
400 600
160
400 N
330 R
100 R
40 R
330 R
Assumptions:
1. The bicycle can be treated as a two-dimensional machine.
2. The bicycle weight is negligible.
Analysis:
1. For the pedal, crank, and pedal sprocket assembly, the chain force is F =
400(160/100) = 640 N

2-47
100 R
160
640 N
1280 N
400 N
400 N
2. For the rear wheel and sprocket assembly,
rear wheel gravity load = 400(440/1000) = 176 N
rear wheel friction force = 640(40/330) = 77.5 N
horizontal bearing force = 640 + 77.5 = 717.5 N
176 N
176 N
77.5 N
717.5 N
640 N
3. For the front wheel, front wheel gravity load = 400(560/1000) = 224 N
224 N
224 N
4. For the entire bicycle and rider assembly, the drawing is given below.

2-48
400 N
176 N
77.5 N
224 N
Comments: The drawing does not show the rearward 77.5 N inertia force necessary to
establish ∑FH = 0. It would be located thru the center of gravity of the cycle-plus-rider,
the location of which is not given. Since this vector would be at some distance "h"
above the ground, the resulting counter clockwise couple, 77.5 h, would be balanced by
decreasing the vertical force on the front wheel and increasing the vertical force on the
rear wheel, both by (77.5 h /1000) N.
SOLUTION (2.34)
Known: A solid continuous round bar is shown in Fig. P2.34 and can be viewed as
comprised of a straight segment and a curved segment – segments 1 and 2. We are to
neglect the weight of the member.
Find: Draw free body diagrams for segments 1 and 2. Also, calculate the force and
moments acting on the ends of both segments.

2-49
(1)
Top View
x
R!
P
Side View
(2)
(3)
(1) (2)
(3)
bending
axis
torque
axis
L
Assumption: The weight of the round bar is negligible.
Analysis:
1. For section 1 to 2, M = Px
2. For section 2 to 3, M = P(L cos θ + R sin θ) and T = P[L sin θ + R(1 - cos θ)].
SOLUTION (2.35)
Known: The spring clip shown in Fig. P2.35 has a force P acting on the free end. We
are to neglect the weight of the clip.
Find: Draw free-body diagrams for segments 1 and 2 – straight and curved portions of
the clip. Also, determine the force and moments acting on the ends of both segments.

2-50
!
M
P
R
L
P
Segment 1
Segment 2
Assumption: The weight of the clip is negligible.
Analysis:
!
M
R
P P
M
P
L
P
M
x
3
2
2
1
1. In the straight section 1 to 2, M = Px, and the shear force V = P. At section 2,
M = PL.
2. In the curved section 2 to 3, M = P(L + R sin θ). The shear force V = P and the
moment M = PL at section 2 and at section 3.
Comment: Note that at the top of the curved section the member is in axial
compression.
SOLUTION (2.36)

2-51
Known: A semicircular bar of rectangular cross section has one pinned end -- see Fig.
P2.36. The free end is loaded as shown.
Find: Draw free-body diagrams for the entire semicircular bar and for a left portion of
the bar. Discuss what influence the weight of the semicircular bar has on this problem.
P !
R
Assumptions:
1. Deflections are negligible.
2. The friction forces at the pinned end are negligible.
3. The semicircular bar is in static equilibrium.
4. The weight of the semicircular bar is negligible except where we address the effect
of weight, then the force of gravity is the only body force.
Analysis:
1. A free body diagram for the entire member is shown below (ignoring the weight of
the bar).
PP
R
2. A free body diagram for a left portion of the member is shown below.
M
P !
F
V
P
3. At any section, θ, the loads are M = PR sinθ, F = P sinθ, and V = P cosθ.

2-52
Comments: The weight for each small segment can be added in the free body diagram
at the segment center of mass. An application of the equations of force equilibrium will
establish the forces F, V and P, and the moment M at section θ.
SOLUTION (2.37)
Known: A bevel gear reducer with known input and output angular velocity is driven
by a motor delivering a known torque of 12 N•m. The reducer housing is held in place
by vertical forces applied at mountings A, B, C and D.
Find: Determine the forces applied to the reducer at each of the mountings:
(a) Assuming 100% reducer efficiency.
(b) Assuming 95% reducer efficiency.
100 mm
100 mm
A
B
C D
Bevel gear
reducer
Attached
to load
600 rpm
1800 rpm
Attached
to motor
12 N•m
Assumption: The bevel gear reducer is in static equilibrium.
Analysis:

2-53
100 mm
100 mm
A
B
C
D Bevel gear
reducer
Attached
to load
600 rpm
1800 rpm
Attached
to motor
T = 36 N•m
(34.2)
T = 12 N•m
120 N
120 N
360 N
(342 N)360 N
(342 N)
Note: 95% efficiency in parenthesis.
l

2-54
SOLUTION (2.38new)
by vertical forces applied at mountings A, B, C and D.
100 mm
100 mm
A
B
C D
Bevel gear
reducer
Attached
to load
600 rpm
1800 rpm
Attached
to motor
24 N•m
Analysis:

2-55
100 mm
100 mm
A
B
C
D Bevel gear
reducer
Attached
to load
600 rpm
1800 rpm
Attached
to motor
240 N
240 N
720 N
(684 N)
720 N
(684 N)
T = 72N•m
(68.4)
N•mT = 24

2-56
SOLUTION (2.39new)
by vertical forces applied at mountings A, B, C and D. Note that AB = CD = 50 mm.
A
B
C D
Bevel gear
reducer
Attached
to load
600 rpm
1800 rpm
Attached
to motor
12 N•m
50 mm
50 mm
Analysis:

2-57
A
B
C
D Bevel gear
reducer
Attached
to load
600 rpm
1800 rpm
Attached
to motor
240 N
240 N
720 N
(684 N)
720 N
(684 N)
T = 36N•m
(34.2)
T = 12 N•m
50 mm
50 mm

2-58
SOLUTION (2.40)
Known: A motor applies a known torque to the pinion shaft of a spur gear reducer.
Find: Sketch free-bodies in equilibrium for
(a) The pinion and shaft assembly.
(b) The gear and shaft assembly.
(c) The housing.
(d) The entire reducer assembly.
8 in.
Rear bearings
4 in.
Front bearings
6 in.
Output shaft
Mountings
200 lb ft
Reducer assy.
Motor input torque
200 lb ftFront bearings
2 in. dia. pinion
6 in. dia. gear
Output
Housing and
gear-shaft assy.
details
Assumptions:
1. The effect of gravity is negligible.
2. The forces between the gears act tangentially.

2-59
Analysis:
8 in.
4 in.
6 in.
1200
1200
1200
1200
1200
1200
1200
1200
600
600
600
600
600
600
600600
2400
2400
600 lb ft
200 lb ft
Output
(Output shaft)
600 lb ft
200 lb ft
(Input shaft)
6 in. dia. gear
2 in. dia. pinion
Reducer assy.
Housing and gear-shaft assy.
Front
bearings

2-60
SOLUTION (2.41new)
Given: A rim and hub are connected by spokes (springs) as shown in Figure P2.41.
The spokes are each tightened to a tension of 20 lb.
Find: Draw a free body diagram of (a) the hub, (b) the rim, (c) one spring, and (d) one-
half (180o
) of the rim.
Schematic and Given Data: See Figure P2.41.
F
Hub
Rim
Spring
Assumption:
1. The weight of each component can be ignored.
2. The rim and hub change from circular shapes to oval shapes when the spokes are
tightened.
3. The rim and hub are of homogeneous material that has the same modulus of
elasticity in tension and compression.
4. The cross section of the rim and hub are each uniform.
5. The maximum stress does not exceed the proportional limit.
Analysis:
1. The hub has two opposed radial 20 lb forces pulling it apart.
2. The ring had two opposed radial 20 lb forces pulling inward.
3. Each spring has a 20 lb force on each end placing the spring in tension.
4. A free body diagram of half (180o
) of the rim shows a 20 lb force pulling radially
inward and a compressive force of 10 lb acting on each “cut” end of the half (180o
)
ring. At each cut end we show unknown moment M and shear force V.

2-61
(a) the hub
20 lb
20 lb
(b) the rim
20 lb
20 lb
(c) one spring
(d) one-half of the rim
20 lb
20 lb
20 lb
10 lb 10 lb
M
M
V V
Comments:
1. The circular ring may be regarded as a statically indeterminate beam, and can be
analyzed by Castigliano’s method – see Section 5.8 and 5.9 of the textbook.
Section 5.9 discusses the case of redundant reactions -- a statically indeterminate
problem.
2. The compressive force in the ring is given by F = - ½ W sin θ, where W is the
force in the spring (the inward force exerted by the spring on the ring), and
where θ is defined in the figure below. The shear force in the ring is given by
V = - ½ W cos θ. See the figure below for terminology.

2-62
!
V
M
Top of ring
F
R
x
y
Directions positive
as shown
Bottom of
ring
3. Roark, Formulas for Stress and Strain, gives the following equations obtained
apparently by using Castigliano’s method:
M = WR( 0.3183 – ½sin θ
) = WR( 0.3183 – [1/(2sin θ
)]
δx = + 0.137 WR3
/EI (increase in diameter in the x-direction)
δy = - 0.149 WR3
/EI (decrease in diameter in the y-direction)
4. In the above equations, W= inward force of the spring on the ring, I =
moment of inertia of ring cross section, E = modulus of elasticity, M =
bending moment, F = circumferential tension, V = radial shear at an angular
distance θ from the bottom of the ring, δx = change in horizontal diameter,
and δy = change in vertical diameter.
5. Note that:
max (+M) = 0.3183WR at bottom and top (θ = 0, θ = 180)
max (-M) = - 0.1817WR at sides (θ = π/2, θ = - π/2)
F (at bottom and top) = 0
V (at bottom and top) = - ½ W
6. If the ring is connected to the hub by N spokes rather than with two spokes, Roark
points out that the formulas can be combined by superposition so as to cover
almost any condition of loading and support likely to occur.

2-63
SOLUTION SOLUTION (2.42)
Known: An engine rotates with a known angular velocity and delivers a known torque
to a transmission which drives a front and rear axle.
Find: Determine the forces applied to the free-body at A, B, C, and D.
X
Y Y
12 in.
24 in.
Engine
2400 rpm
100 lb ft torque
A
B
C D
Rear drive shaft-1200 rpm
Front drive shaft-1200 rpm
Rear axle
(not part of
free-body)
X
Transmission
2.0 ratio
Left-front wheel
axle shaft-400 rpm
Right-front wheel
axle shaft-400 rpm
Assumptions:
1. The friction and gravity forces are negligible.
2. The mountings exert only vertical forces.
3. All four wheels have full traction.
Analysis:

2-64
X
Y Y
12 in.
24 in.
Engine
2400 rpm
100 lb ft torque
A
B
C D
Rear drive shaft-1200 rpm
Front drive shaft-1200 rpm
Transmission
2.0 ratio
Left-front wheel
axle shaft-400 rpm
Right-front wheel
axle shaft-400 rpm
X
(T = 100 lb ft)
100 lb ft
150 lb ft
100 lb
100 lb
150 lb
150 lb
150 lb ft
Assume: All four wheels have full traction.
1. Drive shaft torque =
(Engine torque)(Transmission ratio)
Number of drive shafts
= (100)(2)/2 = 100 lb ft
2. Wheel torque =
(Drive shaft torque)(Axle ratio)
Number of wheels per drive shaft
= (100)(3)/2 = 150 lb ft
3. Therefore, A: 150 lb down ■
B: 150 lb up ■
C: 100 lb down ■
D: 100 lb up ■
SOLUTION (2.43D)
Known: A mixer is supported by symmetric mounts at A and B. The motor torque
should be 20 N.m to 50 N.m. The motor delivers a torque to mixing paddles which, in
turn, stir a fluid to be mixed.
Find: Determine the forces acting on the mixer. Sketch a free-body of the mixer.

2-65
A B
Motor
Direction
of rotation
200 mm
g
Mass of mixer
system = 50 kg
The motor delivers a torque of
20 Nm to 50 Nm to the mixer
blades.
x
y
z
Decision: A motor with a maximum torque output of 50 N•
m is selected for analysis.
Assumption: The fluid forces on the paddles create a torque to oppose the rotation of
the paddles. Other fluid forces can be neglected (e.g. the paddles are buoyed up by the
weight of the displaced fluid).
Analysis:
1. The torque exerted on the paddles by the fluid is 50 N•
m maximum.
2. Mounting forces to resist torque = (50 N•
m)/(0.2 m) = 250 N. Thus, a force of 250
N is exerted at A and at B.
3. Gravitational forces apply F = ma = (50 kg)(9.8 m/s2) = 490 N. Thus, there is a
245 N force upward at A and at B.
4. The forces are shown on the free body diagram for the maximum motor torque of
50 N•
m.
A B
Direction
of rotation
200 mm
g
Mass of mixer
system = 50 kg
245 N 245 N
250 N
250 N
490 N

2-66
Comment: For torques less than 50 N•m, the resultant mounting forces will be smaller.
SOLUTION (2.44new)
Known: A rear wheel driven vehicle travels at a steady speed. The forces opposing the
motion of the vehicle are (i) the drag force, Fd, imposed on a vehicle by the surrounding
air, (ii) the rolling resistance force on the tires, Fr , and (iii) the forces of the road acting
on the tires. The vehicle has a weight W.
Find: Draw a free body diagram of the rear wheel driven vehicle. Describe how the
free body diagram changes if the accelerator pedal is pushed and the vehicle starts
accelerating.
Assumptions: The vehicle is operating initially at steady state conditions.

2-67
Comment: The acceleration of the vehicle results from an increase in traction force,
and the additional force to acceleration the vehicle is directly related to its mass and its
acceleration; i.e., F = ma. ■
SOLUTION (2.45new)
Known:
A front wheel driven vehicle travels at a steady speed. The opposing forces are the (i)
drag force, Fd , imposed on a vehicle by the surrounding air and the (ii) rolling
resistance force on the tires, Fr , opposing the motion of the vehicle, and (iii) the forces
of the road acting on the tires. The vehicle has a weight W.
Find: Draw a free body diagram for a front wheel driven vehicle. Describe how the
free body diagram changes if the accelerator pedal is pushed and the vehicle starts
accelerating.
Assumptions: The vehicle is initially operating at steady state conditions.

2-68
Comment: Free body diagrams are shown above. Although the front tire force(s)
decreases with acceleration, this is only a traction problem if the front wheel(s) loses a
grip on the road. ■

2-69
SOLUTION (2.46new)
Known: The handles are approximately 2.5 inches long and the handle wire is 1/16
inch in diameter. The spring clip is shaped like a triangle when closed (end view) and
has two legs each 1-1/4 inches long and a third connecting side that is 1 inch long. The
spring clip is approximately 2 inches wide.
Find: Draw a free body diagram for a large size binder clip where the clip is opened
and being readied to fasten together a stack of loose 8.5 x 11 inch sheets of paper. Also,
draw free-body diagrams for the handles and a diagram for the spring steel clip.
F
Handle
F
Handle
Closed binder clip
(a) Side view of spring binder clip with applied force F but where F is less than a force required
to open the spring clip
Closed spring clip
Width = 2.0 in.
1.0 in.
(b) View of spring showing width

2-70
Assumption:
1. The weight of each component of the binder clip can be neglected.
2. The externally applied forces are equal and opposite to each other and collinear.
Analysis: A free body of the binder clip with external forces opening the clip appears
as shown below. Also shown are free body diagrams for the spring and the two
handles.
F
F
Handle
Handle
Spring
2.5 in.
1.25 in.
F
F
2F
F
F
2F
2F
F
F
Handle
2F
Comments:
1. The diagram of the binder clip is representative of the contact on the handle
and spring when collinear external forces are applied.
2. The diagram above does not show the bending of the handles and the bending
of the legs and back of the spring. ■

2-71
SOLUTION (2.47D)
Known: An electric squirrel cage blower motor supported by mountings at A and B
delivers a known torque to fan blades.
Find: Determine all loads acting on the motor and sketch it as a free-body in
equilibrium.
A B
Direction of
shaft rotation
T =1 N•m
of T = 1 N m to the blower.
75 mm to 150 mm
mass of blower
system = 15 kg
g
Assumption: The air flows only in the radial direction and exerts a resisting rotational
torque on the squirrel cage.
Decision: A width between A and B of 100 mm is selected for analysis.
Analysis:
A B
10 N
147 N
10 N 10 N
73.5 N 73.5 NShaft
rotation
1 N•mT =
m.
downward at B.

2-72
3. Gravitational forces apply F = ma = (15 kg)(9.8 m/s2) = 147 N downward at the
center of gravity of the blower unit. Thus, there is a 73.5 N force upward at A and
at B.

2-73
SOLUTION (2.48)
Known: The geometry and dimensions of the gear and shaft assembly are known.
Find: Draw a free-body diagram of the assembly. Also draw the free-body diagrams
for gear 1, gear 2 and the shaft.
FAV
FAH
FA
FCV
FCH
FC
45 mm
30 mm
20 mm
Gear 1
Gear 2
20o
20o
Bearing B
Bearing D
Assumption: Gravity forces are negligible.
Analysis:
FAV
FAH
FA
FCV
FCH
FC
RV1
RV2
RH2
RH1
45 mm
30 mm
20 mm
Gear 1
Gear 2
20o
20
o

2-74
FAV
FAH
FCV
FCH
RV1
RV2
RH2
RH1 T
T
Gear 1
FAV
FAH
T
Gear 2
FCV
FCH
T

2-75
SOLUTION (2.49)
Known: The geometry and dimensions of the gear and shaft assembly are shown in
Figure P2.48. The force FA applied to gear 1 is 550 N.
Find: Determine the magnitude of force FC and list the assumptions.
Assumptions:
1. Frictional losses in the bearings can be neglected.
2. The gears are rigidly connected to the shafts.
3. The shaft is in static equilibrium or operating in a steady state condition.
X
Y
Z
FAV
FAH
FA
FCV
FCH
FC
45 mm
30 mm
20 mm
Gear 1
Gear 2
20o
20o
Bearing B
Bearing D
Analysis:

2-76
X
Y
Z
FAV
FAH
FA
FCV
FCH
FC
45 mm
30 mm
20 mm
Gear 1
Gear 2
20o
20o
Bearing B
Bearing D
, dia = 24 mm
= 550 N
, dia = 50 mm
= 1145.8 N
1. Since
!
T= 0" about the axis of the shaft, we have
!
T=" FA cos 20
o
(25 mm) – FC •
cos 20
o
(12 mm) = 0.
2. Solving the above equation, gives FC = FA•(25/12) = 1145.8 N ■
Comment: Recall that gear 1 has a diameter of 50 mm and gear 2 has a diameter of 24
mm; i.e., gear 1 has a radius of 25 mm and gear 2 has a radius of 12 mm.

2-77
SOLUTION (2.50)
Known: The geometry and dimensions of a gear shaft are shown in Figure P2.48. The
force FA applied to gear 1 is 1000 N.
Find: Determine the forces at bearing D and list the assumptions.
Assumptions:
1. There is no thrust load on the shaft.
2. The free body diagram determined in problem 2.48 is accurate.
4. Gravity forces and shaft deflection are negligible.
5. The location of the bearing loads can be idealized as points.
X
Y
Z
FAV
FAH
FA
FCV
FCH
FC
45 mm
30 mm
20 mm
Gear 1
Gear 2
20o
20o
Bearing B
Bearing D
Analysis:

2-78
X
Y
Z
FAV
FAH
FA
FCV
FCH
FC
45 mm
30 mm
20 mm
Gear 1
Gear 2
20o
20o
Bearing B
Bearing D
, dia = 24 mm
= 1000 N
, dia = 50 mm
= 2083.3 N
1. Summing moments about the X axis gives FC = FA (25/12) = 1000(25/12) = 2083.3
N.
2. Using the free body diagram created, we can directly write the equations of
equilibrium for the moments about the bearing B.
3. The sum of the moments about the Z-axis at B is given by
ΣMZB = 0 = FDV(45+30) + (FA cos 20o
)(30) + (FC sin 20o
)(20)
Solving for FDV = -565.9 N ■
4. Similarly for the Y axis
ΣMYB = 0 = FDH(45+30) – (FA sin 20o
)(30) – (FC cos 20o
)(20)
Solving for FDH = 658.9 N ■
Comments: Summing moments about the Y axis, ΣMYD = 0 and then the Z axis, ΣMZD
= 0 at bearing D yields bearing forces at B of FBV = 338.7 N and FBH = -2275 N. We
can check the answers by verifying that the sum of the forces in the Y direction and
then the Z direction are both equal to zero.

2-79
SOLUTION (2.51)
Known: The geometry and dimensions of a gear shaft are shown in Figure P2.48. The
force FC applied to gear 2 is 750 N.
Find: Determine the forces at bearing B and list the assumptions.
Assumptions:
1. There is no thrust load on the shaft.
2. The free body diagram determined in problem 2.48 is accurate.
4. Gravity forces and shaft deflection are negligible.
5. The location of the bearing loads can be idealized as points.
X
Y
Z
FAV
FAH
FA
FCV
FCH
FC
45 mm
30 mm
20 mm
Gear 1
Gear 2
20o
20o
Bearing B
Bearing D
Assumption: Gravity forces are negligible.
Analysis:

2-80
X
Y
Z
FAV
FAH
FA
FCV
FCH
FC
45 mm
30 mm
20 mm
Gear 1
Gear 2
20o
20o
Bearing B
Bearing D
, dia = 24 mm
= 360 N
, dia = 50 mm
= 750 N
1. Summing moments about the X axis gives FA = FC (12/25) = 750(12/25) =360N
2. Using the free body diagram created, we can directly write the equations of
equilibrium for the moments about the bearing D.
3. The sum of the moments about the Z-axis at D is given by
ΣMZB = 0 = FBV(45+30) + FA cos 20o
(45) - FC sin 20o
(45 + 30 + 20)
Solving for FBV = 121.9 N ■
4. Similarly for the Y-axis
ΣMYB = 0 = -FBH(45+30) + FA sin 20o
(45) - FC cos 20o
(45 + 30 + 20)
Solving for FBH = -818.8 N ■
Comments: Summing moments about the Z axis and then the Y axis at bearing D
yields bearing forces at D of FDV = -203.7 N and FDH = 237.2 N. We can check the
answers by verifying that the sum of the forces in the Y direction and then the Z
direction are both equal to zero.

2-81
SOLUTION (2.52)
Known: The solid continuous member shown in textbook Figure P2.52 can be viewed
as comprised of several straight segments. The member is loaded as shown. We are to
neglect the weight of the member.
Find: Draw free-body diagrams for the straight segments 1, 2, and 3. Also, determine
the magnitudes (symbolically) of the force and moments acting on the straight
segments.
(1)
L/4
L/2
L
(2)
(3)
P
L/4
P
Analysis:
1. Because of symmetry, we need to look only at one half of the member.
(1)
x
(2)(3)
L/4
P
L/2
y
L/2
(4)
2. In Section 1 to 2, the axial compressive force, F = P.
3. In Section 2 to 3, the moment, M = Px, and the shear force,V = P.
4. In Section 3 to 4, the moment, M = PL/2, and the axial tensile force, F = P.

2-82
SOLUTION (2.53)
Known: A gear exerts the same known force on each of two geometrically different
steel shafts supported by self-aligning bearings at A and B.
Find: Draw shear and bending moment diagrams for each shaft.
200 300
140
300200
100 N 100 N
50 N
Analysis:
200 300
140
300200
100 N 100 N
50 N
V
M
V
M
24 N
74 N40 N60 N
60 N
-40 N
74 N
-26 N
-50 N
12 N•m 14.8 N•m
7 N•m
FB = 100 200
500
= 40 N FB = -100(200) + 50(640)
500
= 24 N
A BAB

2-83
SOLUTION (2.54)
Known: Six shaft loading configurations are shown in Fig. P2.54. For each
configuration, the 2-in. diameter steel shaft is supported by self-aligning ball bearings at
A and B; and a special 6-in. diameter gear mounted on the shaft caused the forces to be
applied as shown.
Find: Determine bearing reactions, and draw appropriate shear and bending moment
diagrams for each shaft and gear loading configuration.
8 in
A
B
A
B
12 in
12 in
16 in
3 in 3 in
(c) Radial and axial loads (f) Radial and axial loads
600 lb
1200 lb
600 lb
1200 lb
Analysis:

2-84
(c) Radial and axial loads (f) Radial and axial loads
8 in
A
B
12 in
3 in
600 lb
1200 lb
A
B
12 in
16 in
3 in
600 lb
1200 lb
250 lb
A
B
600 lb
1200 lb
600 lb
550 lb 650 lb
A
B
600 lb
1200 lb
600 lb
550 lb
1200 lb
V V
- 250 lb
-650 lb
1450 lb
M M
4400 in lb
2600 in lb
1800 in lb
-3000 in lb

2-85
SOLUTION (2.55)
Known: A pulley of known radius is attached at its center to a structural member. A
cable wrapped 90o around the pulley carries a known tension.
Find:
(a) Draw a free-body diagram of the structure supporting the pulley.
(b) Draw shear and bending moment diagrams for both the vertical and horizontal
portions of the structure.
12 in.48 in.
27 in.100 lb
100 lb12 in.
Cable
Cable
pulley rad.
Assumption: The weight of the pulley and the supporting structure is negligible.
Analysis:
1. Free-body diagram of the structural member:
M = 7500 lb in.
100 lb
100 lb
100 lb
100 lb
M = 100 (27+48) = 7500 lb in.

2-86
2. Vertical portion of member:
100 lb
100 lb
100 lb
100 lb
100 lb
- 2700 lb in.
00
M V
+
M = 2700 lb in
3. Horizontal portion of member:
7500 lb in.
100 lb
100 lb
100 lb
100 lb
2700 lb in.
100 lb
-7500 lb in.
-2700 lb in.
0
0
V
M

2-87
SOLUTION (2.56)
Known: A bevel gear is attached to a shaft supported by self-aligning bearings at A and
B, and is driven by a motor. The axial force, radial force, and tangential force are
known.
Find:
(a) Draw (to scale) axial load, shaft torque, shear force, and bending moment
diagrams for the shaft.
(b) Determine the values of axial load and torque along the shaft.
40 60
A B
Takes thrust Does not take thrust
Motor attaches
to this end
of shaft
50
Fa= 1000N
Fr= 600N
Ft = 2000N
Assumptions:
1. The weight of the gear and shaft is negligible.
2. The bearing at A takes all the thrust load.
Analysis:
(a) Since bearing B carries no axial thrust load, Bx = 0. A free body diagram of the
bevel gear and shaft is:
40 60
A B
50
1000N
600N
2000N
y
x
z
Tx
Bz
By
x
A
z
A
yA

2-88
From force equilibrium:
1. ∑MzA = 0 : By =
600(40) - 1000(50)
100
= -260 N
2. ∑Fy = 0 : Ay - 600 + By = 0; Ay = 860 N
3. ∑MyA = 0 : (2000)(40) - (Bz)(100) = 0; Bz = 800 N
4. ∑Mxx = 0 : Tx - (2000)(50) = 0; Tx = 100,000 N mm
5. ∑Fx = 0 : Ax - 1000 N = 0; Ax = 1000 N
6. ∑Fz = 0 : Az + Bz - 2000 = 0; Az = 1200 N
The answers are:
40 60
A B
50
600 N
1000 N
1000 N
860 N
260 N
34.4 N•m
-15.6 N•m
2000 N
N= 860Ay
= 260 NBy
1200 N
800 N
100 N m
Vx-y
Mx-y
y
x
z
Tx
Bz
x
A
zA
=
=
= =

2-89
48.0 N•m
1200 N
800 N
Vx-z
Mx-z
1000 N
100 N mTxx
Fx
(b) The compressive force between the gear and the bearing A is 1000 N. The torque
between the gear and the bearing B is 50 mm times the tangential gear force, Ft.
For Ft = 2000 N, this torque is (2 kN)(50 mm) = 100 N•m.

2-90
SOLUTION (2.57)
Known: The shaft with bevel gear is supported by self-aligning bearings A and B.
Gear loads are known.
Find: Draw axial load, shaft torque, shear force, and bending moment diagrams for the
shaft.
A B
Pump shaft is
coupled to this
end of shaft
200 N
100 N
50 150
100
1000 N
Assumptions:
1. The weight of the gear and shaft is negligible.
2. The bearing at A takes all the thrust load.
Analysis: A free body diagram of the shaft with bevel gear is:
A B
200 N
100 N
50 150
100
1000 N
Bz
By
x
A
zA yA
Tx
y
x
z

2-91
Force equilibrium requires:
1. ∑MzA = 0 : By =
100(100) - 200(50)
200
= 0 N
2. ∑Fy = 0 : Ay - 200 + By = 0; Ay = 200 N
3. ∑MyA = 0 : (1000)(50) - (Bz)(200) = 0; Bz = 250 N
4. ∑Mxx = 0 : Tx - (1000)(100) = 0; Tx = 100 N m
5. ∑Fx = 0 : Ax - 100 = 0; Ax = 100 N
6. ∑Fz = 0 : Az + Bz - 1000 = 0; Az = 750 N
The answers are:
A B
200 N
100 N
50 150
100
100 N
200 N
1000 N
750 N
250 N
100 N m
200 N
200(50) = 10,000 N mm
Vx-y
Mx-y
y
x
z

2-92
750 N
750(50) = 37,500 N mm
Mx-z
Vx-z
250 N
100 N m
100 N
Txx
Fx

2-93
SOLUTION (2.58)
Known: The shaft with a bevel gear and a spur gear is supported by self-aligning
bearings A and B. Neither end of the shaft is connected to another component. The
gear loads are known except for the transmitted force on the spur gear.
shaft.
A B
30
40
8020 20
200 N
400 N
200 N
1000 N
Pt
Assumptions:
1. The weight of the components is negligible.
2. The bearing at A carries all the thrust load.
Analysis: A free body diagram of the shaft with bevel gear is:
A B
30
40
8020 20
200 N
400 N
200 N
1000 N
Pt
y
x
z
xA
yB
yA
zA
Bz

2-94
Force equilibruim requires:
1. ∑Fx = 0 : Ax - 200 = 0; Ax = 200 N
2. ∑Mxx = 0 : - (30)Pt + (1000)(40) = 0; Pt = 1333.33 N
3. ∑MyA = 0 : (1000)(20) - (Bz)(100) - (120)Pt = 0; Bz = -1400 N
4. ∑Fz = 0 : Az - 1000 - 1400 + 1333.33 = 0; Az = 1066.7 N
5. ∑MzA = 0 : 400(20) - 200(40) - By(100) + 200(120) = 0; By = 240 N
6. ∑Fy = 0 : Ay - 400 + By - 200 = 0; Ay = 360 N
The answers are:
A B
30
40
8020 20
200 N
400 N
200 N
200 N
360 N 240 N
360 N
-40 N
200 N
7200 N•mm
-4000 N•mm
-800 N•mm
1000 N
Pt
1066.7 N 1400 N
y
x
z
Vx-y
Mx-y

2-95
-40 N
1066.7
! N
Vx-z
Mx-z
66.7
1333.33
N•m21.33
26.66 N•m
66.7 N
400 N•mTxx
200 N
Fx

2-96
SOLUTION (2.59)
Known: A shaft has two bevel gears, and neither end of the shaft is connected to
another component. The gear loads are known except for the transmitted force on one
bevel gear.
shaft.
A B
250 N
100 N
120140300200
150
500 N
150 N
1500 N
Pt
y
x
z
Assumptions:
1. The weight of the components is negligible.
2. The bearing at A carries all the thrust load.
Analysis: A free body diagram of the shaft with two bevel gears is:
A B
250 N
100 N
120140300200
150
500 N
150 N
1500 N
xA
yB
yA
zA
Bz
Pt

2-97
From force equilibrium:
1. ∑Mxx = 0 : - (120)Pt + (1500)(150) = 0; Pt = 1875 N
2. ∑Fx = 0 : Ax + 150 - 100 = 0; Ax = -50 N
3. ∑MyA = 0 : (1500)(200) - (Bz)(500) - (640)Pt = 0; Bz = -1800 N
4. ∑Fz = 0 : Az - 1500 - 1800 + 1875 = 0; Az = 1425 N
5. ∑MzA = 0 : 150(150) + 500(200) - By(500) - 250(640) +100(120) = 0;
By = -51 N
6. ∑Fy = 0 : Ay - 500 + By + 250 = 0; Ay = 301 N
The answers are:
A B
250 N
100 N
120140300200
150
500 N
150 N
50 N
301 N
51 N
1500 N
301 N
-199 N
-250 N
60.2 N•m
82.7 N•m
-12 N•m
Vx-y
Mx-y 23 N•m
y
x
z

2-98
1425 N
-75 N
-1875 N
Mx-z
Vx-z
N•m285
N•m262.5
100 N
N•m-225
Fx
Txx
-50 N
SOLUTION (2.60)
Known: A static force, F, is applied to the tooth of a gear that is keyed to a shaft.
Find: Identify the stresses in the key, and write an equation for each. State the
assumptions, and discuss briefly their effects.

2-99
r
t
b
R
t/2
A
A
F
L
Key
Section on A-A
Assumption: The compressive forces on each side of the key are uniform.
Analysis:
1. Compression on key sides
b
t
(σc • L •
t
2
)r ≈ FR ;
Hence, σc ≈ 2FR
(L) (t) (r)
or more precisely,
σcL • t
2
(r - t
4
) = FR
so, σc =
2FR
Lt (r - t/4)
2. Key shear
τ (bL)r = FR ; hence, τ = FR
bLr

2-100
Comment: The compressive forces on each side of the key will most probably not be
uniform because of key cocking.
SOLUTION(2.61)
Known: A screw with a square thread is transmitting axial force F through a nut with n
threads engaged.
Find: Identify the types of stresses in the threaded portion of the screw and write an
equation for each. State the assumptions made, and discuss briefly their effect.
F
t
D
d
Assumption: The assumptions will be stated in the analysis section.
Analysis:
1. Compression at interface. Assuming uniform stress distribution, we have
σ =
F
π(D - d)n
■
(The bending of the thread would tend to concentrate the stress toward the inside
diameter and also produce a tensile stress at the thread root. Geometric inaccuracy may
concentrate the load at one portion of a thread.)
2. Shear at the base of threads. Assuming uniform distribution of τ across the
cylindrical failure surface, we have
τ = F
πdnt
■
(The stress concentration would create a higher stress in the thread root. The effect of
thread helix angle is neglected.)

2-101
SOLUTION (2.62new)
Known: A force P is applied as shown at the end of a thin walled, metal, cylindrical
container that is open on the right end and closed on the left end. The container has a
diameter of 3 inches and is 6 inches long.
Find: Sketch the force flow lines. Use the force flow concept to locate the critical
sections and/or critical surfaces for the container.
P
Flat plate
Assumptions:
1. The force P will not permanently deform the cylinder.
2. The left end of the cylinder is stiff compared to the right end.
3. The flat plate is thick and will remain flat and will experience negligible elastic
deformation compared to the right end of the cylinder; e.g., the flat plate is
relatively rigid and will not deflect in bending.
4. The material is uniform in elasticity and strength.
Analysis: The right end will deform elastically under the load with a small amount of
force flowing toward the left end as the force P increases. The two critical locations are
at the periphery of the open end of the cylinder 90 degrees from the force P where the
bending moment (stress) is highest. A simple experiment would establish failure
points.
P

2-102
Comment: An experiment with a thin walled, cylindrical container, open at one end
and closed on the other end could also be conducted to obtain a better estimate of the
length of an equivalent ring (no closed end). With the deflection characteristics of an
equivalent ring, i.e., the force deflection curve (ring spring constant), a spring model
comprised of multiple rings could be developed and an analysis could be employed to
study the load sharing along the length of the container. Equations for the deflection of
various cylindrical rings are well known. For example, see Roark, “Formulas for Stress
and Strain”.
SOLUTION (2.63)
Known: A total gas force F is applied to the top of a piston.
Find:
(a) Copy the drawing and sketch the force paths through the piston, through the piston
pin, and into the connecting rod.
(b) Identify the stresses in the piston pin and write an equation for each. State the
assumptions made, and discuss briefly their effect.
a a2a
d
Total gas force = F
a
2a
d
Total gas force = F
a
1
1
2
Assumption: The assumptions are stated in the analysis section.
Analysis:
1. Compression with piston and with rod: σ = Force
Projected area
= F*
2ad
■
2. Transverse shear stress, τ =
2F
Δ
πd
2 (for a solid pin) ■
3. Bending of pin (stresses depend on fit and rigidity of the members.)

2-103
_______________
* Assumes uniform axial distribution of stress which would not be strictly true due
to pin bending.
Δ Assumes uniform stress distribution. Actual stresses may be greater at top and
bottom.

2-104
SOLUTION (2.64)
Known: A force P is applied to an engine crankshaft by a connecting rod. The shaft is
supported by main bearings A and B. Torque is transmitted to an attached member
through flange F.
Find:
(a) Draw the shaft, and show all loads necessary to place it in equilibrium as a free-
body.
(b) Starting with P and following the force paths through the shaft to the flange,
identify the locations of potentially critical stresses.
(c) Making appropriate simplifying assumptions, write an equation for each.
d D
A B
L
F
f tt
d D
LLR
P P
A
d D
L
f tt
d D
LLR
P P
A
PR PR
P
2
, P
2
2
3
5
4
P
2
on each
main journal
Analysis:
1. Where "P" is applied to the crankpin, the compressive stress (assuming uniform
stress distribution) is given by:
σ =
P
Projected Area
= P/DL

2-105
2. The shear stress at section 2 (assuming uniform stress distribution) is:
τ =
P
2π(D
2
- d
2
)/4
: τ =
2P
π(D
2
- d
2
)
3. The shear stress at section 3 (assuming a uniform distribution): τ = P/2tA
4. The torsional stress at section 4 (neglecting stress concentration):
τ = Tc
J
=
(PR)(D/2)
π
32
(D
4
- d
4
)
= 16PRD
π(D
4
- d
4
)
5. The shear stress at cylindrical section 5:
τ =
T
πDf(D/2)
=
2PR
πD
2
f
6. Bending stresses are also present, the magnitudes of which depend on rigidities of
the shaft and associated components, and on the fits between these components.
SOLUTION (2.65)
Known: In Figure P2.65, all the joints are pinned and all links have the same length L
and the same cross-sectional area A. The central joint (pin) is loaded with a force P.
Find: Determine the force in the bars.
L
45 o
F
L
L
P
(P - F)/! 2(P - F)/! 2
45 o
Assumptions:
1. The weight of the members are negligible.
2. Buckling will not occur in the lower links.

2-106
Analysis:
1. Let F be the tensile force in the upper link.
2. From Σ Fvertical = 0, the lower link forces are (P-F)/√2. ■
SOLUTION (2.66)
Known: We are to repeat Problem 2.65 except that the top link has a cross-sectional
area of A, and the two lower links have a cross-sectional area of A'. All the joints are
pinned and all links have the same length L. The central joint (pin) is loaded with a
force P.
Find: Determine (a) the force in the bars, and (b) the ratio A/A' that will make the
force in all the links numerically equal.
L
45 o
F
L
L
P
(P - F)/! 2(P - F)/! 2
45 o
Assumptions:
1. The weight of the members are negligible.
2. Buckling will not occur in the lower links.
Analysis:
1. Let F be the tensile force in the upper link. Then from Σ Fvertical = 0, the lower link
forces are (P-F)/√2. ■
2. The upper link has a cross sectional area A, and the lower links each have a cross
sectional area A'.
3. The force P causes the center pivot to deflect downward. We define the deflection
as δ -- see the figure below.

2-107
4. Since the upper member is subjected to the axial tension force, F, we can calculate
the deflection of the upper member as δupper = FL/AE.
5. Since the each lower member is subjected to the axial tension force, (P-F)/√2, we
can calculate the deflection of each lower member as δlower = [(P-F)/√2]L/(A'E).
L
45 o
F
L
L
P
(P - F)/! 2(P - F)/! 2
45 o
"
DD
# A
A'A'
6. From the above diagram, we have cos 45o
= Δ/δ = 1/√2.
7. If the links carry equal loads, then (P-F)/√2 = F, or F = P
2+1
8. Combining equations shows that A'/A= √2. ■
SOLUTION (2.67)
Known: A "T" bracket, attached to a fixed surface by four bolts, is loaded at point E.
Find:
(a) Copy the drawing and sketch paths of force flow going to each bolt.
(b) Determine the division of load among the four bolts.

2-108
A
B
E
C
D
A
B
E
C
D
Assumptions
1. The T-bracket deflection is negligible.
2. The stiffness between point E and the plate through bolts B and C is twice the
stiffness between point E and the plate through bolts A and D.
A
B
C
D
F
6
F
6
F
3
F
3
F
Analysis:
1. The force flow is shown in the above schematic.
2. If all "springs" deflect equally, bolts "B" and "C" each carry twice the load of
bolts "A" and "D".

2-109
A
B
C
D
k
k
2k
2kk = stiffness, N/mm
A
B
C
D
F
6
F
6
F
3
F
3
F
SOLUTION (2.68)
Known: A stiff horizontal bar, supported by four identical springs, is subjected to a
known center load.
Find: Determine the load applied to each spring.

2-110
k
k
k k
aa
100 N
100 N
40
40
40
40
40
40
40
40
20
20
20
20
20
20
Assumption: The k of the horizontal bar is much greater than the k of the springs.
Analysis:
1. The upper springs each deflect only half as much as the lower springs, hence
they carry only half as much load.
2. Let L = load carried by each lower spring. Then,
2L + L/2 = 100 N and L = 40 N
3. In summary, the lower springs carry 40 N, the upper springs 20 N. ■
SOLUTION (2.69)
Known: A horizontal nonrigid bar with spring constant k is supported by four identical
springs each with spring constant k. The bar is subjected to a center load of 100N.
Find: Determine the load applied to each spring.

2-111
k
k
k k
aa
k
k k
aa
100 N
! beam
B
k
! top spring
! bottom
spring
! top spring
Assumption: The k of the horizontal bar is equal to the k of the springs.
Analysis:
1. The deflection of the two upper springs added together minus the deflection of the
beam equals the deflection of each lower spring; i.e, (Δtop spring + Δtop spring) – Δbeam =
Δbottom spring.
2. From a free body diagram of the loaded beam, where F is the force in the upper
springs and P is the force in each lower springs, we have, F + 2P = 100 N.
3. Also, Δtop spring = F/k, Δbottom spring = P/k, and Δbeam = (100 – F)/k.
4. Combining the equations and solving yields F = 42.8 N and P = 28.6 N.
5. In summary, the lower springs carry 28.6 N, the upper springs 42.8 N. ■
SOLUTION (2.70)
Known: A "T" bracket, attached to a fixed surface by four bolts, is loaded at point E.
Find: Determine the maximum force F that can be applied to the bracket: (a) if the
bolts are brittle and each one fractures at a load of 6000 N, (b) if the bolts are ductile
and each bolt has a yield strength of 6000 N.

2-112
A
B
E
C
D
A
B
C
D
k
k
2k
2kk = stiffness, N/mm

2-113
A
B
C
D
F
6
F
6
F
3
F
3
F
Assumptions:
1. The T-bracket deflection is negligible.
2. The stiffness between point E and the plate through bolts B and C is twice the
stiffness between point E and the plate through bolts A and D.
Analysis:
(a) Bolts "B" and "C" each carry F/3. They fracture at F/3 = 6000 N; hence
maximum bracket force is 18,000 N. ■
(b) Bolts "B" and "C" begin to yield at F = 18,000 N, but permit sufficient
elongation to allow "F" to be increased with bolts "A" and "D" also yielding;
hence the maximum bracket force is 24,000 N. ■
SOLUTION (2.71--alternate problem with different yield strengths)
Known: Two plates are joined with straps and a single row of rivets (or bolts). Plates,
straps, and rivets are all made of ductile steel having yield strengths in tension,
compression, and shear of 300, 300, and 170 MPa respectively.
Find:
(a) Calculate the force F that can be transmitted across the joint per pitch P, of joint
width, based on the rivet shear strength.
(b) Determine minimum values of t, tʹ′ , and P that will permit the total joint to transmit
this same force (thus giving a balanced design).
(c) Determine the efficiency of the joint (ratio of joint strength to strength of a
continuous plate).

2-114
P
tF F
Rivet dia. = 10 mm
t !
t !
Syt= 300 MPa
Syc= 300 MPa
Sys = 170 MPa
Assumption: The frictional forces between the plates and straps are negligible.
Analysis:
(a) Each pitch involves transmitting force "F" through 1 rivet in double shear:
F = 2( )!d
2
4
• Sys = 2 (25π mm2)(170 MPa) = 26,700 N ■
(b) For plate and strap to have equal tensile strength and equal compressive strength
(at rivet interface), t = 2tʹ′.
The compressive load carrying capacity (at rivet interface) is
F = Projected area • Syc :
26,700 N = 10 t mm2 • 300 MPa. Hence, t = 8.90 mm; tʹ′ = 4.45 mm. ■
The tensile load carrying capacity (at rivet interface) is F = (P -10) t • Syt :
26,700 N = (P - 10)(8.90) mm2 • 300 MPa
P = 20 mm ■
(c) Efficiency =
Joint strength
Strength of a continuous plate
=
26,700
Syt(t)(P) =
26,700
(300 MPa)(8.90 mm)(20 mm)
= 0.50 = 50% ■

2-115
SOLUTION (2.71)
Known: Two plates are joined with straps and a single row of rivets (or bolts). Plates,
straps, and rivets are all made of ductile steel having yield strengths in tension,
compression, and shear of 284, 284, and 160 MPa respectively.
Find:
(a) Calculate the force F that can be transmitted across the joint per pitch P, of joint
width, based on the rivet shear strength.
(b) Determine minimum values of t, tʹ′, and P that will permit the total joint to transmit
this same force (thus giving a balanced design).
(c) Determine the efficiency of the joint (ratio of joint strength to strength of a
continuous plate).
P
tF F
Rivet dia. = 10 mm
t !
t !
Syt = 284 MPa
Syc= 284 MPa
Sys = 160 MPa
Assumption: The frictional forces between the plates and straps are negligible.
Analysis:
(a) Each pitch involves transmitting force "F" through 1 rivet in double shear:
F = 2( )!d
2
4
• Sys = 2 (25π mm2)(160 MPa) = 25,133 N ■
(b) For plate and strap to have equal tensile strength and equal compressive strength
(at rivet interface), t = 2tʹ′.

2-116
The compressive load carrying capacity (at rivet interface) is
F = Projected area • Syc :
25,133 N = 10 t mm2 • 284 MPa. Hence, t = 8.85 mm; tʹ′ = 4.425 mm. ■
The tensile load carrying capacity (at rivet interface) is F = (P -10) t • Syt :
25,133 N = (P - 10)(8.85) mm2 • 284 MPa.
Hence, P = 20 mm. ■
(c) Efficiency =
Joint strength
Strength of a continuous plate
=
25,133
Syt(t)(P) =
25,133
(284 MPa)(8.85 mm)(20 mm)
= 0.50 = 50% ■
SOLUTION (2.72)
Known: Plates of known thickness are butted together and spliced using straps and
rivets. A double-riveted joint is used. Tensile, compression, and shear strengths of all
materials are known.
Find: Determine the pitch, P, giving the greatest joint strength. Comment on how this
compares with the strength of a continuous plate.
d = 40 mm
t !
t !
t
Syt= 200 MPa
Syc= 200 MPa
Sys = 120 MPa
t = 20 mm
t ! = 10mm
Assumption: The friction forces between plates and straps are negligible.

2-117
Analysis:
3
21
!
8
75
4 6
7
1. Failure of both force paths together
Failure load at 1: F = (P - 40)(20)(200)
But 8 is more critical: F = (P - 80)(20)(200) ---(a)
2. Failure of each path individually
Outer row of rivets:
At 3: F = (P - 40)(10)(200)
At 4: F = π(20)2120 = 150,796 N (single shear)
At 5: F = (40)(10)(200) = 80,000 N
The inner row of rivets:
At 2: F = (P-80)(20)(200)
At 6: F = 2π(20)2120 = 301,592 N (double shear)
At 7: F = (40)(20)200 = 160,000 N (7ʹ′ is the same)
3. The outer row will have a strength of 80,000 N if,
2000 (P - 40) ≥ 80,000
or P ≥ 80 ■
4. The inner row will have a strength of 160,000 N if,
4000 (P - 80) ≥ 160,000
or, P ≥ 120
5. For 8 to have a strength of
80,000 + 160,000 = 240,000 N:
(P - 80)(20)(200) = 240,000, or P = 140 mm ■
6. Continuous plate strength = P(20)(200) = (140)(20)(200) = 560,000 N
7. Joint strength = 240
560
= 43% of continuous plate strength ■
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