SHIGLEY'S MECH.ENGINEERING DESIGN-EBK>I
10th Edition
ISBN: 9781259489563
Author: BUDYNAS
Publisher: INTER MCG
expand_more
expand_more
format_list_bulleted
Concept explainers
Videos
Textbook Question
Chapter 5, Problem 2P
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 plane stress states:
5-2 Repeat Prob. 5–1 with the following principal stresses obtained from Eq. (3–13):
- (a) σA = 100 MPa, σB = 100 MPa
- (b) σA = 100 MPa, σB = –100 MPa
- (c) σA = 100 MPa, σB = 50 MPa
- (d) σA = 100 MPa, σB = –50 MPa
- (e) σA = 100 MPa, σB = –100 MPa
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
The stresses on the surface of a hard bronze component are shown in the figure below. The yield strength of the bronze is σY = 345 MPa.a) What is the factor of safety predicted by the maximum-shear-stress theory of failure for the stress state shown? Does the component fail according to this theory?b) What is the value of the Mises equivalent stress for the given state of plane stress?c) What is the factor of safety predicted by the failure criterion of the maximum-distortion energy theory of failure? Does the component fail according to this theory?
A flange bolt coupling consist of 8-12.7mmꬾ steel bolts on a circle diameter of 304.80mm & 6-12.7mmꬾ steel bolts on a circle diameter of 228.60 mm. is subjected to a 950KN-mm torque. The shear stress developed in the outer circle of the flange bolt coupling is ___ MPa.
Forces of F1=8 kN in the z direction from the B point, F2=5 kN in the z direction from the C point, and F3=10 kN in the y direction act on the arm in the figure. The lengths of the arm are also given as L1=0.15 m and L2=0.16 m. The radius r in the a-a section taken over the arm is r=0.2 m and it is desired to determine the stress state at point A. slip of sleeve material module is G=75 Gpa. Accordingly,
a) Find the shear stress due to the shear force at point A.
b) Find the shear stress due to the torsional moment at point A.
c) Find the total shear stress at point A.
d) Find the normal stress due to the normal force at point A.
e) Find the normal stress due to the bending moment at point A.
f) Find the total normal stress at point A.
Chapter 5 Solutions
SHIGLEY'S MECH.ENGINEERING DESIGN-EBK>I
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
- A propeller shaft for a small yacht is made of a solid steel bar 104 mm in diameter. The allowable stress in shear is 48 MPa, and the allowable rate of twist is 2.0° in 3.5 meters. (a) Assuming that the shear modulus of elasticity is G = 80 GPa, determine the maximum torque that can be applied to the shaft. (b) Repeat part (a) if the shaft is now hollow with an inner diameter of 5d18. Compare values to corresponding values from part (a).arrow_forwardTwo sections of steel drill pipe, joined by bolted flange plates at Ä are being tested to assess the adequacy of both the pipes. In the test, the pipe structure is fixed at A, a concentrated torque of 500 kN - m is applied at x = 0.5 m, and uniformly distributed torque intensity t1= 250 kN m/m is applied on pipe BC. Both pipes have the same inner diameter = 200 mm. Pipe AB has thickness tAB=15 mm, while pipe BC has thickness TBC= 12 mm. Find the maximum shear stress and maximum twist of the pipe and their locations along the pipe. Assume G = 75 GPa.arrow_forwardThe stepped shaft shown in the figure is required to transmit 600 kW of power at 400 rpm. The shaft has a full quarter-circular fillet, and the smaller diameter D1= 100 mm. If the allowable shear stress at the stress concentration is 100 MPa, at what diameter will this stress be reached? Is this diameter an upper or a lower limit on the value of D2?arrow_forward
- A circular steel tube with an outer diameter of 75 mm and inner diameter of 65 mm is subjected to torques 7"at its ends. Calculate the maximum permissible torque 7jliajl if the allowable normal strain is ea= 5 X 10"4 Assume that G = 15 GPa.arrow_forwardA Steel shaft is subjected to an end thrust producing a stress of 90 MPa and the minimum shearing stress on the surface arising from torsion is 60 MPa .The yield point stress of the material in simple tension was found to be 300 MPa. calculate the factor of safety of the shaft according to the following theory 1)maximum shear stress theoryarrow_forwardThe factors of safety at point H predicted by the maximum-distortion-energy theory (von Mises criterion) can be calculated as The von Mises equivalent stresses at point H can be calculated as MPaarrow_forward
- The shaft AB made of steel has an outside diameter of 165 mm and a wall thickness of 9.5 mm. The shaft is subjected to an axial compression load of P = 156 kN and a torque T = 12 kN.m, which act in the directions shown in the Figure. The yield strength of the steel is Y = 248 MPa and a minimum factor of safety = 2.0 is required by specification. Consider the point K and determine whether the shaft satisfies the specifications according to the maximum-distortion-energy theory.arrow_forwardQ#8: The 0.06R (mm)-diameter bar is made of a brittle material with the ultimate strengths of 0.15R MPa in tension and 0.25R MPa in compression. The bar carries a bending moment and a torque, both of magnitude M. (a) Use the maximum normal stress theory to find the largest value of M that does not cause rupture. (b) Is the value of M found in Part (a) safe according to Mohr’s theory of failure? Note : Use R=980 Please See attached Picture For Details. Please Use Graphs For Mohr's Circlearrow_forwardThe figure (attached) shows a belt pulley mechanism which is loaded statically. The shaft is made of AISI 1030 steel with the yield strength of 480 MPa. Using distortion energy theory (DET), determine the diameter of the shaft with a factor of safety of 2.arrow_forward
- The load on a bolt consists of an axial thrust of 10 kN, with transverse shear force of 6 kN.Calculate the diameter of the bolt according to (a) maximum principal stress theory,(b) maximum shear stress theory, and (c) strain energy theory. Take factor of safety to be 3,Which is a safe theory? yield strength= 285 , v=0.3arrow_forwardanswer this ; The factors of safety predicted at point H by the maximum shear stress theory of failure (Tresca criterion)arrow_forwardThe compound axial member shown consists of a 20-mm-diameter solid aluminum [E=70 GPa] segment (1), a 24 mm-diameter solid alurminum segment (2), and a 16-mm-diameter solid steel [E=200 GPa) segment (3) Determine the displacements of member 1, member 2 and member 3. Answer in positive value if the deformation is elongation consequently. answer in negative value if the deformation is shortening.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
Pressure Vessels Introduction; Author: Engineering and Design Solutions;https://www.youtube.com/watch?v=Z1J97IpFc2k;License: Standard youtube license