
Precision Machining Technology (MindTap Course List)
2nd Edition
ISBN: 9781285444543
Author: Peter J. Hoffman, Eric S. Hopewell, Brian Janes
Publisher: Cengage Learning
expand_more
expand_more
format_list_bulleted
Textbook Question
Chapter 5.2, Problem 5RQ
Name two material shapes that can be properly held in a four-jaw chuck.
Expert Solution & Answer

Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
Stress, ksi
160
72
150-
140
80
70
༄ ྃ ༈ ཎྜ རྦ ༅ ཎྜ ྣཧྨ ➢
130
120
110
100
90
2.0
2.8
3.6
4.4
5
Wire diameter, mm
6.0
6.8
2
7.6
8.4
Compression and extension springs.
ASTM A227 Class II
Light service
Average service
0.020
0.060
0.100
0.140
0.180
0.220
0.260
0.300
0.340
0.380
0.420
0.460
0.500
Wire diameter, in
Torsional stress due to initial tension, ksi
10
४
20
Preferred
range
100
Stress, MPa
9.2
10.0
10.8
11.6
12.4
1100
1035
965
895
825
760
Severe service
690
620
550
50
150
3456789 10 11 12 13 14 15 16
Spring index, C = DJD
FIGURE 18-21 Recommended torsional shear stress in an extension spring due to initial tension (Data from Associated
Spring, Barnes Group, Inc.)
50
200
485
Stress, MPa
Bolted Joint Design
Bolted Frames
Total Force due to door weight: P =
240
lb
Number of Bolts: N =
Distance to Bolt C/L: a =
4
N/A
Bolt Material -
Allowable shear stress of bolt material: T₂ =
x Distance from Bolt centroid to bolt: x =
y Distance from Bolt centroid to bolt: y =
Degrees per Radian-
Results
y-Load on each bolt: F, =
Moment resisted by bolt pattern: M =
Radial distance from Bolt centroid to bolt: r =
Sum squares of all radial distances: Σr²
Force on each bolt to resist moment: F, -
Angle for force composition: e=
X-Force on each bolt to resist moment: F-
y-Force on each bolt to resist moment: Fly
Total y-Force on each bolt: Fy =
Resultant force on bolt 1: R₁ =
Required shear stress area for a bolt: A₂ =
ASTM Grade
A307 Steel
10,000
0
psi
from Table 20-1
3.0
57.296
in
degrees
lb per bolt
lb-in
Formula
FS-P/N
M-Px XB
r = (x² + y²)0.5
in²
Σ
4r²
Mr
F₁ =
Στ
lb
degrees
lb
lb
lb
Minimum Bolt Diameter: Din =
Rounded up Bolt Diameter: D =
55
P.
1.5 in
2 in (3x)
1 in
This bracket…
University of Babylon
Collage of Engineering/
Al-Musayab
Department of Automobiles
Final Examination/ Stage: 3rd
Notes:
Answer 4 questions only
2023-2202
Subject: Theory of vehicles
Date: 2023\06\10-Saturday
Time: Three Hours
Course 2nd Attempt 1st
Q1: A Hooke's coupling connects two shafts whose axes are inclined at 30°. The
of the driven shaft? Find the maximum value of retardation or acceleration and
driving shaft rotates uniformly at 600 rpm. What are the extreme angular velocities
state the angle where both will occur.
(12.5 Marks)
Q2: Four masses, A, B, C, and D), revolve at equal radii and are equally spaced
along a shaft. The mass B is 7 kg, and the radius of C and D make angles of 90°
and 240°, respectively, with the radius of B. Find the magnitude of the masses A,
C, and D and the angular position of A so that the system may be completely
balanced.
(12.5 Marks)
Q3: A cam has straight worked faces that are tangential to a base circle of diameter
90 mm. The follower is a roller…
Chapter 5 Solutions
Precision Machining Technology (MindTap Course List)
Ch. 5.1 - List the four main parts of the engine lathe.Ch. 5.1 - What are the two main purposes of the lathe...Ch. 5.1 - What part of the lathe is used to set the feed...Ch. 5.1 - Prob. 4RQCh. 5.1 - Prob. 5RQCh. 5.1 - What is the purpose of the leads crew of a lathe?Ch. 5.1 - What two functions can the lathe tailstock...Ch. 5.1 - The standard taper in most lathe tailstocks is the...Ch. 5.1 - Define the swing and the bed length of a lathe.Ch. 5.2 - What is the special name for the type of jaw-type...
Ch. 5.2 - The most common variation of the above chuck has...Ch. 5.2 - Name two material shapes that can be properly held...Ch. 5.2 - List two advantages of using a self-centering...Ch. 5.2 - Name two material shapes that can be properly held...Ch. 5.2 - List three benefits of holding a workpiece between...Ch. 5.2 - List three potential advantages of using an...Ch. 5.2 - List three characteristics of a workpiece that...Ch. 5.2 - What type of mandrel would be ideal for gripping a...Ch. 5.2 - Name the type of tailstock center that raid be...Ch. 5.2 - What two auxiliary devices can be used to...Ch. 5.2 - Explain the differences between the two auxiliary...Ch. 5.2 - Name the device that is used to transmit the...Ch. 5.2 - Which two tool posts are the most efficient if...Ch. 5.2 - Which device may be used for either toolholding or...Ch. 5.3 - If a 0.050" depth of cut is taken on the diameter...Ch. 5.3 - A lathe cross slide uses a diameter-reading...Ch. 5.3 - In what units are feed rates measured for lathe...Ch. 5.3 - Are deeper cuts used for roughing or finishing...Ch. 5.3 - Calculate spindle RPM and machining time for...Ch. 5.3 - List three safety precautions related to clothing...Ch. 5.3 - What two materials are most commonly used for...Ch. 5.3 - What feature of a lathe cutting tool has a direct...Ch. 5.3 - Is a left-hand or right-hand tool normally used...Ch. 5.3 - What part of the lathe is used to feed the tool...Ch. 5.3 - When facing, why should the tool not be fed past...Ch. 5.3 - Should a left-hand or right-hand tool be used when...Ch. 5.3 - When and how should chips he removed from the work...Ch. 5.3 - What are two reasons for center drilling on the...Ch. 5.3 - When drilling and reaming on the lathe, how are...Ch. 5.3 - How can hole depth be controlled during drilling...Ch. 5.3 - What are two reasons boring may be selected to...Ch. 5.3 - Why must extra care be taken when performing...Ch. 5.3 - How can a tap be aligned when threading a hole on...Ch. 5.3 - Briefly define form cutting.Ch. 5.3 - How do grooving and cutoff speeds compare to...Ch. 5.3 - How can tool binding be overcome when cutting deep...Ch. 5.3 - List the two basic knurl patterns.Ch. 5.3 - How is knurling different from other lathe...Ch. 5.3 - List and briefly describe the two different types...Ch. 5.4 - The distance of actual contact of two mating...Ch. 5.4 - What feature of mating threads determines the...Ch. 5.4 - How many classes of fit are there in the Unified...Ch. 5.4 - Determine the major diameter limits for the...Ch. 5.4 - Determine the minor diameter limits for the...Ch. 5.4 - Determine the pitch diameter limits for the...Ch. 5.4 - Determine the approximate compound-rest in-feed...Ch. 5.4 - What is the name for the rotating device that...Ch. 5.4 - When threading, what is the reason for feeding the...Ch. 5.4 - Prob. 10RQCh. 5.4 - Why should the depth of cut be reduced for each...Ch. 5.4 - What dimension of the thread is measured by using...Ch. 5.4 - What measuring tool is used to visually inspect...Ch. 5.4 - List two applications of Acme threads.Ch. 5.4 - What type of thread is machined on a tapered...Ch. 5.5 - Briefly define a taper.Ch. 5.5 - What is the difference between an included angle...Ch. 5.5 - What does TPI stand for in relation to tapers?Ch. 5.5 - What are the TPI and TPF of a part with end...Ch. 5.5 - What are the corresponding centerline and included...Ch. 5.5 - What is the corresponding centerline angle of a...Ch. 5.5 - What is the limitation of the tool bit taper...Ch. 5.5 - What must be known to use the compound-rest taper...Ch. 5.5 - What taper turning methods allow use of the lathes...Ch. 5.5 - What two steps can be taken to eliminate backlash...Ch. 5.5 - The TPI specified on a print is 0.030". If...Ch. 5.5 - If TPF is 0.42", how much movement should register...Ch. 5.5 - What is the benefit of using the offset tailstock...Ch. 5.5 - What are two ways to reduce uneven pressure on...Ch. 5.5 - Calculate tailstock setover for a 13.5" part with...
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.Similar questions
- Problem 18-26 Added Extension Springs Spring Material ASTM A227 Modulus of Elasticity of the Material in Shear: G 1.150E+07 psi Average Service Max Operating Load: F₁ = 100 lb Max Length between attachment points: L₁ = 60.00 in 20.00 lb 26.00 1.400 Min Operating Load: F₁ = Min Length between attachment points: L₁ = Maximum Outside Diameter = in in Results Note: you select a wire diameter from the "US steel wire gage" column in table 18-2 Formula k = AF/AL k = (F0-F1)/(Lo - L₁) Spring Rate: k = lb/in Assumed Trial Outside Diameter: OD = Assumed Trial Mean: D ma Assumed Design Stress in Spring: Tda in 1.070 in 102,000 psi Assumed Wahl Factor: K = 1.2 Calculated Wire Diameter: Dwa Actual Wire Diameter: Dw Actual outer diameter: OD = Actual inner diameter: ID= Spring Index: C = See Figure 18-8 Dw= [8KF Dm πTd 1/3 in 5' 5' 5' 5' This corresponds to US Steel 9 wire gage ID = Dm - Dw C = Dm/Dw 4C - 1 0.615 K = + 4C - с Wahl Factor: K = 8KFDm 8KFC T = TD πD Stress in Spring at F = Fo: To psi…arrow_forwardCHAIN DRIVE DESIGN Initial Input Data: Application: Garage Door Opener Drive type: AC Motor Driven machine Chain and Sprocket to pull the door up Degrees per Radian: 57.2958 degrees Sprocket Diameter: D = 1.690 in Number of strands: Chain number: 1 40 Service factor: 1.3 Table 7-10 No. of teeth Computed Data: Actual Motor Power Input: 0.000 hp Sprocket Speed (for sprocket attached to gear shaft) Design power: 0.00 rpm 0 hp 11 12 0.06 0.15 0.29 0.56 0.99 1.09 1.61 2.64 TABLE 7-7 Horsepower Ratings-Single Strand Roller Chain No. 40 0.500 inch pitch 10 25 50 100 180 200 300 500 700 900 1000 1: 0.06 0.14 0.27 0.52 0.91 1.00 1.48 2.42 3.34 4.25 4.70 ! 3.64 4.64 5.13 13 0.07 0.16 0.31 0.61 1.07 1.19 1.75 2.86 3.95 5.02 5.56 Design Decisions-Chain Type and Teeth Numbers: 14 Chain number: Use Table 7-7 Chain pitch: p = in 15 Number of Teeth: N = Per Table 7-7 16 0.08 0.20 0.39 0.75 1.32 1.46 2.15 3.52 0.07 0.17 0.34 0.66 1.15 1.28 1.88 3.08 0.08 0.19 0.36 0.70 1.24 1.37 2.02 3.30 4.55 5.80…arrow_forwardInput Data: Torque needed to overcome rolling friction in rollers, slides and other moving parts, except for Motor and Worm Gear the worm gear T₁ = Length of travel of door: Time for door to open or close: LD = 50 lb-in. 90 in t= 12.5 seconds Pitch diameter for chain sprocket: DPC 1.690 in Weight of Door: P = No. of worm threads: Nw = Worm Pitch diameter: Dw Diametral pitch: Pd Normal pressure angle: Degrees per Radian: Number of gear teeth: Calculated Data: Linear velocity of door and chain (in/sec): Linear velocity of door and chain (ft/min): Output Speed of Gear and Sprocket: Upward Force due to Weight of Door: Фо = = NG= 240 lb 2 1.250 in 12 14.5 degrees 57.2958 degrees 28 Vα= in/sec VC= ft/min NG = rpm FD lb Net Upward Force on Door: Fou lb Torque on gear ignoring rolling friction: TG = lb-in. Formula = FDU FD-2 x Fo (note: Fo is the Max Operating load of the extension springs). This is also the initial tension in the chain. TG = FDU X DPC/2 This is the also the torque on the…arrow_forward
- Q5/A: A car with a track of 1.5 m and a wheelbase of 2.9 m has a steering gear mechanism of the Ackermann type. The distance between the front stub axle pivots is 1.3 m. The length of each track arm is 150 mm, and the length of the track rod is 1.2 m. Find the angle turned through by the outer wheel if the angle turned through by the inner wheel is 30°. (6 Marks) Q5/B: Write True on the correct sentences and False on the wrong sentences listed below:- 1- In automobiles, the power is transmitted from the gearbox to the differential through bevel gears. 2- The minimum radius circle drawn to the cam profile is called the base circle. 3- The Proell governor, compared to the Porter governor, has less lift at the same speed. 4- The balancing of rotating and reciprocating parts of an engine is necessary when it runs at a slow speed. (6.5 Marks) ***Best of Luck *** جامعة بابل UNIVERSITY OF BABYLON Examiner: Mohanad R. Hameed Head of Department: Dr. Dhyai H. Jawadarrow_forwardUniversity of Babylon Collage of Engineering/ Al-Musayab Department of Automobiles Mid Examination/ Stage: 3rd Subject: Theory of Vehicles Date: 14 \ 4 \2025 Time: 1.5 Hours 2025-2024 Q1: The arms of a Porter governor are 250 mm long. The upper arms are pivoted on the axis of revolution, but the lower arms are attached to a sleeve at a distance of 50 mm from the axis of rotation. The weight on the sleeve is 600 N and the weight of each ball is 80 N. Determine the equilibrium speed when the radius of rotation of the balls is 150 mm. If the friction is equivalent to a load of 25 N at the sleeve, determine the range of speed for this position. Q2: In a loaded Proell governor shown in Figure below each ball weighs 3 kg and the central sleeve weighs 25 kg. The arms are of 200 mm length and pivoted about axis displaced from the central axis of rotation by 38.5 mm, y=238 mm, x=303.5 mm, CE 85 mm, MD 142.5 mm. Determine the equilibrium speed. Fe mg E M 2 Q3: In a spring loaded Hartnell type…arrow_forwardusing the theorem of three moments, find all the reactions and supports, I need the calculations onlyarrow_forward
- Q.5: (10 Marks) Select the correct answer (choose 10 only) 1. The forward whirling speed is ......... the static structure tilting speed. (a) Less than (b) Higher than (c) equal to 2. The divergence between the forward and backward whirling speeds increases as: (a) The rotating speed increase (b) the polar moment of inertia increases (c) Both (a) and (b) (d) do not change 3. Increasing the system natural frequency can be done by: (a) add masses (b) adding braces and supports (c) increase damping 4. The amplitude of vibration due to external force can be reduced by: (a) Increasing damping (b) Decreasing damping (c) Increasing mass 5. Tuned absorbers are used to: (a) Shift the natural frequency (b) increase damping (c) Increase stiffness 6. Accelerometers sensors contains: г (a) Piezoelectric materials (b) Magnet and coil (c) coil only 7. Increasing the stiffness of the system causes: (a) Less transmitted force (b) more transmitted force (c) Transmitted force does not change 8. The…arrow_forwardQ.1: (15 Marks) Find the first three natural frequencies and mode shapes of the axial and torsional vibration for a steel shaft free at both ends, having a length of 3 m. Find the subsequent axil motion if the shaft is subjected to the following initial conditions, given that E = 210 GPa, G=80 GPa, p = 7800 kg/m³: f(x)=0 v(x) = {1 2.8arrow_forwardQ.4: (15 Marks) A uniform rotor of mass 500 kg and diametral moment of inertia of 20 kg.m², is supported by identical short bearings of stiffness 1 MN/m in the horizontal and vertical directions. If the distance between the bearings is 0.6 m: (a) What is the corresponding polar moment of inertia if the backward whirling speed is 80% of the static structure tilting natural frequency? (b) Determine the forward whirling speed. 45.27arrow_forwardUniversity of Babylon Collage of Engineering/ Al-Musayab Department of Automobiles Mid Examination/ Stage: 3rd Subject: Theory of Vehicles Date: 14 \ 4 \2025 Time: 1.5 Hours 2025-2024 Q1: The arms of a Porter governor are 250 mm long. The upper arms are pivoted on the axis of revolution, but the lower arms are attached to a sleeve at a distance of 50 mm from the axis of rotation. The weight on the sleeve is 600 N and the weight of each ball is 80 N. Determine the equilibrium speed when the radius of rotation of the balls is 150 mm. If the friction is equivalent to a load of 25 N at the sleeve, determine the range of speed for this position. Q2: In a loaded Proell governor shown in Figure below each ball weighs 3 kg and the central sleeve weighs 25 kg. The arms are of 200 mm length and pivoted about axis displaced from the central axis of rotation by 38.5 mm, y=238 mm, x=303.5 mm, CE 85 mm, MD 142.5 mm. Determine the equilibrium speed. Fe mg E M 2 Q3: In a spring loaded Hartnell type…arrow_forwardQ.2: (15 Marks) = 1400 For the following system, determine the first natural frequency using Dunkerley's equation, Given that the disk has moment of inertia J = 2 kg.m², the shaft has G = 20 GPa, p kg/m³, polar moment of cross-sectional area of the shaft Ip = 8×108 m². 500 mm 220 mm k=200 N/m FOF m=1 kg 14.14 56.56. W слarrow_forwardQ.2: (15 Marks) = 1400 For the following system, determine the first natural frequency using Dunkerley's equation, Given that the disk has moment of inertia J = 2 kg.m², the shaft has G = 20 GPa, p kg/m³, polar moment of cross-sectional area of the shaft Ip = 8×108 m². 500 mm 220 mm k=200 N/m FOF m=1 kg 14.14 56.56. W слarrow_forwardarrow_back_iosSEE MORE QUESTIONSarrow_forward_ios
Recommended textbooks for you
- Precision Machining Technology (MindTap Course Li...Mechanical EngineeringISBN:9781285444543Author:Peter J. Hoffman, Eric S. Hopewell, Brian JanesPublisher:Cengage Learning

Precision Machining Technology (MindTap Course Li...
Mechanical Engineering
ISBN:9781285444543
Author:Peter J. Hoffman, Eric S. Hopewell, Brian Janes
Publisher:Cengage Learning
HOME SHOP JIGS & FIXTURES PART 1, TYPES OF JIGS & ACCESSORIES AND THE THEORIE BEHIND THE TOOLS; Author: THATLAZYMACHINIST;https://www.youtube.com/watch?v=EXYqi42JimI;License: Standard Youtube License