Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering)
10th Edition
ISBN: 9780073398204
Author: Richard G Budynas, Keith J Nisbett
Publisher: McGraw-Hill Education
expand_more
expand_more
format_list_bulleted
Textbook Question
Chapter 5, Problem 53P
5-39* to 5-55* For the problem specified in the table, build upon the results of the original problem to determine the minimum factor of safety for yielding. Use both the maximum-shear-stress theory and the distortion-energy theory, and compare the results. The material is 1018 CD steel.
Problem Number Original Problem, Page Number
Problem Number | Original Problem Page Number |
5–39* 3–68 | 151 |
5–40* 3–69 | 151 |
5–41* 3–70 | 151 |
5–42* 3–71 | 151 |
5–43* 3–72 | 152 |
5–44* 3–73 | 152 |
5–45* 3–74 | 152 |
5–46* 3–76 | 153 |
5–47* 3–77 | 153 |
5–48* 3–79 | 153 |
5–49* 3–80 | 153 |
5–50* 3–81 | 154 |
5–51* 3–82 | 154 |
5–52* 3–83 | 154 |
5–53* 3–84 | 154 |
5–54* 3–85 | 155 |
5–55* 3–86 | 155 |
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solutionStudents have asked these similar questions
5-63
The figure shows a shaft mounted in bearings at A and D and having pulleys at B and C. The
forces shown acting on the pulley surfaces represent the belt tensions. The shaft is to be made of
AISI 1035 CD steel. Using a conservative failure theory with a design factor of 2, determine the
minimum shaft diameter to avoid yielding.
f-in R
300 Ibf
50 lbf
59 Ibf
392 lbf D
Problem 5-63
C 6 in
8-in D.
8 in
B
8 in
please i want a clear solution
The plate of the figure is subjected to a bending moment with irregular cycles, which are
repeated. In the graphic one of this cycles is represented in terms of stress which appears in
each section whose height is h. The piece is made of ductile steel. Determine the number
of repetitions of the sequence which the piece can resist before the failure takes place due
to fatigue considering a reliability of 95 %.
Data: Sult = 1.000 MPa
Syp 3D800 Mра
thickness e = 4 mm
H = 10 cm
h = 5 cm
r=1 cm
ka = 0,72
kp = 0,95
S(MPa)
M
400
h
300
200
100
W
-100
-200
Chapter 5 Solutions
Shigley's Mechanical Engineering Design (McGraw-Hill Series in Mechanical Engineering)
Ch. 5 - A ductile hot-rolled steel bar has a minimum yield...Ch. 5 - A ductile hot-rolled steel bar has a minimum yield...Ch. 5 - A ductile hot-rolled steel bar has a minimum yield...Ch. 5 - A ductile hot-rolled steel bar has a minimum yield...Ch. 5 - A ductile hot-rolled steel bar has a minimum yield...Ch. 5 - Prob. 6PCh. 5 - 5-7 to 5-11 An AISI 1018 steel has a yield...Ch. 5 - 5-7 to 5-11 An AISI 1018 steel has a yield...Ch. 5 - 5-7 to 5-11 An AISI 1018 steel has a yield...Ch. 5 - 5-7 to 5-11 An AISI 1018 steel has a yield...
Ch. 5 - 5-7 to 5-11 An AISI 1018 steel has a yield...Ch. 5 - A ductile material has the properties Syt = 60...Ch. 5 - Prob. 13PCh. 5 - Prob. 14PCh. 5 - Prob. 15PCh. 5 - 5-14 to 5-18 An AISI 4142 steel QT at 800F...Ch. 5 - 5-14 to 5-18 An AISI 4142 steel QT at 800F...Ch. 5 - 5-14 to 5-18 An AISI 4142 steel QT at 800F...Ch. 5 - A brittle material has the properties Sut = 30...Ch. 5 - Repeat Prob. 519 by first plotting the failure...Ch. 5 - For an ASTM 30 cast iron, (a) find the factors of...Ch. 5 - For an ASTM 30 cast iron, (a) find the factors of...Ch. 5 - Prob. 23PCh. 5 - For an ASTM 30 cast iron, (a) find the factors of...Ch. 5 - 5-21 to 5-25 For an ASTM 30 cast iron, (a) find...Ch. 5 - 5-26 to 5-30 A cast aluminum 195-T6 exhibits Sut =...Ch. 5 - 5-26 to 5-30 A cast aluminum 195-T6 exhibits Sut =...Ch. 5 - 5-26 to 5-30 A cast aluminum 195-T6 exhibits Sut =...Ch. 5 - 5-26 to 5-30 A cast aluminum 195-T6 exhibits Sut =...Ch. 5 - 5-26 to 5-30 A cast aluminum 195-T6 exhibits Sut =...Ch. 5 - 5-31 to 5-35 Repeat Probs. 526 to 530 using the...Ch. 5 - 5-31 to 5-35 Repeat Probs. 526 to 530 using the...Ch. 5 - Repeat Probs. 526 to 530 using the modified-Mohr...Ch. 5 - Repeat Probs. 526 to 530 using the modified-Mohr...Ch. 5 - Repeat Probs. 526 to 530 using the modified-Mohr...Ch. 5 - This problem illustrates that the factor of safety...Ch. 5 - For the beam in Prob. 344, p. 147, determine the...Ch. 5 - A 1020 CD steel shaft is to transmit 20 hp while...Ch. 5 - For the problem specified in the table, build upon...Ch. 5 - For the problem specified in the table, build upon...Ch. 5 - 5-39 to 5-55 For the problem specified in the...Ch. 5 - Prob. 42PCh. 5 - For the problem specified in the table, build upon...Ch. 5 - For the problem specified in the table, build upon...Ch. 5 - Prob. 45PCh. 5 - 5-39 to 5-55 For the problem specified in the...Ch. 5 - Prob. 47PCh. 5 - For the problem specified in the table, build upon...Ch. 5 - For the problem specified in the table, build upon...Ch. 5 - For the problem specified in the table, build upon...Ch. 5 - For the problem specified in the table, build upon...Ch. 5 - 5-39 to 5-55 For the problem specified in the...Ch. 5 - 5-39 to 5-55 For the problem specified in the...Ch. 5 - For the problem specified in the table, build upon...Ch. 5 - For the problem specified in the table, build upon...Ch. 5 - Build upon the results of Probs. 384 and 387 to...Ch. 5 - Using F = 416 lbf, design the lever arm CD of Fig....Ch. 5 - A spherical pressure vessel is formed of 16-gauge...Ch. 5 - This problem illustrates that the strength of a...Ch. 5 - Prob. 60PCh. 5 - A cold-drawn AISI 1015 steel tube is 300 mm OD by...Ch. 5 - Prob. 62PCh. 5 - The figure shows a shaft mounted in bearings at A...Ch. 5 - By modern standards, the shaft design of Prob. 563...Ch. 5 - Build upon the results of Prob. 340, p. 146, to...Ch. 5 - For the clevis pin of Prob. 340, p. 146, redesign...Ch. 5 - A split-ring clamp-type shaft collar is shown in...Ch. 5 - Prob. 68PCh. 5 - Prob. 69PCh. 5 - Prob. 70PCh. 5 - Two steel tubes have the specifications: Inner...Ch. 5 - Repeal Prob. 5-71 for maximum shrink-fit...Ch. 5 - Prob. 73PCh. 5 - Two steel lubes are shrink-filled together where...Ch. 5 - Prob. 75PCh. 5 - Prob. 76PCh. 5 - Prob. 77PCh. 5 - Prob. 78PCh. 5 - Prob. 79PCh. 5 - Prob. 80PCh. 5 - Prob. 81PCh. 5 - For Eqs. (5-36) show that the principal stresses...Ch. 5 - Prob. 83PCh. 5 - A plate 100 mm wide, 200 mm long, and 12 mm thick...Ch. 5 - A cylinder subjected to internal pressure pi has...
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
- Could the torsion be either positive or negative and result the same factor of safety?arrow_forwardPlease answer asap I will give u ggod rating if u do not copy. thank u and please solve by yourself.arrow_forward2. Consider a bar of AISI 1015 cold-drawn steel. Using the distortion-energy and maximum-shear-stress theories to determine the factors of safety for a stress state with the following plane principal stresses: 04 = 30 kpsi, OB = 15 kpsi.arrow_forward
- A ductile hot-rolled steel bar has a minimum yield strength in tension and compression of 350 MPa. Using the distortion-energy and maximum-shear-stress theories determine the factors of safety for the following principal stresses:arrow_forwardPlease don't provide handwritten solution .....arrow_forwardSy=370 MPaarrow_forward
- A bar of steel has the minimum properties Se = 40 kpsi, Sy = 60 kpsi, and Sut = 80 kpsi. The bar is subjected to a steady torsional stress of 6 kpsi and an alternating bending stress of 23 kpsi. Find the factor of safety guarding against a static failure and either the factor of safety guarding against a fatigue failure or the expected life of the part. For the fatigue analysis, use 1. Modified Goodman criterion 2. Gerber criterion 3. Morrow criterion Take τa = 0 kpsi and σm = 0 kpsi. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. What are the values of the alternating and mean von Mises stresses? The alternating von Mises stress is kpsi. The mean von Mises stress is kpsi.arrow_forwardThis problem illustrates that the factor of safety for a machine element depends on the particular point selected for analysis. Here you are to compute factors of safety, based upon the distortion-energy theory, for stress elements at A and B of the member shown in the figure. This bar is made of AISI 1006 cold-drawn steel and is loaded by the forces F = 0.55 kN, P = 8.0 kN, and T = 30 N m = 280 MPa 100 mm A B d=20 mmarrow_forwardProblem 2 A sector of a steel plate with endurance limit Se = 400 MPa, and ultimate strength Sut = 1000 MPa, has a stress state defining with the following cycle parameters calculated with von Mises criterion: = 130 MPa ση σα max-Mises max-Mises = = 400 MPa. Calculate the useful life in cycles according to Goodman criterion considering an equivalent cycle of completely inverted stress. Take n = 1 and work it in in sector of high cycle and finite life.arrow_forward
- Can you please answer these questions? (3 and 4) thank youarrow_forwardRequired information This problem illustrates that the factor of safety for a machine element depends on the particular point selected for analysis. Here you are to compute factors of safety, based upon the distortion-energy theory, for stress elements at A and B of the member shown in the figure. This bar is made of AISI 1006 cold-drawn steel and is loaded by the forces F= 0.55 kN, P= 4 kN, and T=25 N-m. Given: Sy= 280 MPa. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. B 15-mm D. -100 mm What is the value of the axial stress at point A? The value of the axial stress at point A is MPa.arrow_forward6-10 A rotating shaft of 25-mm diameter is simply supported by bearing reaction forces R, and R2. The shaft is loaded with a transverse load of 13 kN as shown in the figure. The shaft is made from AISI 1045 hot-rolled steel. The surface has been machined. Determine (a) the minimum static factor of safety based on yielding. (b) the endurance limit, adjusted as necessary with Marin factors. (c) the minimum fatigue factor of safety based on achieving infinite life. (d) If the fatigue factor of safety is less than 1 (hint: it should be for this problem), then estimate the life of the part in number of rotations.arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
- Mechanics of Materials (MindTap Course List)Mechanical EngineeringISBN:9781337093347Author:Barry J. Goodno, James M. GerePublisher:Cengage Learning
Mechanics of Materials (MindTap Course List)
Mechanical Engineering
ISBN:9781337093347
Author:Barry J. Goodno, James M. Gere
Publisher:Cengage Learning
Understanding Failure Theories (Tresca, von Mises etc...); Author: The Efficient Engineer;https://www.youtube.com/watch?v=xkbQnBAOFEg;License: Standard youtube license