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 3P
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:
5-3 Repeat Prob. 5–1 for a bar of AISI 1030 hot-rolled steel and:
- (a) σx = 25 kpsi, σy = 15 kpsi
- (b) σx = 15 kpsi, σy = –15 kpsi
- (c) σx = 20 kpsi, τxy = –10 kpsi
- (d) σx = –12 kpsi, σy = 15 kpsi, τxy = –9 kpsi
- (e) σx = –24 kpsi, σy = –24 kpsi, τxy = –15 kpsi
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
The 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.
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?
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.
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
- Two 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_forwardBar of steel, (yield strenght Sy = 462 MPa) is subjected to the following stresses; σx = 171 MPa , σy = -144 MPa , τxy = 175 MPa Using the Distortion-Energy Theory determine the factor of safety and check is the bar will fail or not.arrow_forwardA bar AISI 1035 steel forged and heat treated with minimum yield strength of 570MPa is subjected to bending moment of 290Nm and a torsional moment of 310Nm. Assuming a factor of safety of 2, determine the diameter of rod using the following theories of failure (i) Maximum Shear stress theory and (ii) Distortion Energy theory (iii) Maximum stress theory, also specify which diameter is considered as safearrow_forward
- A load P is supported by two springs arranged in series. The upper spring has 20 turns of 19 mm diameter wire on a mean diameter of 150 mm. The lower spring consists of 15 turns of 12 mm diameter wire on a mead diameter of 125 mm. Determine the spring index of the lower spring in MPa if the total deformation is 76 mm and G of 83 GPa.arrow_forwardOn the figure shown, a ring which is supported by the spring, is subjected to a concurrent forces P1 = 2000 N, and P2 = 2500 N. The ring is being kept in an equilibrium state by the spring attached. Determine the shearing stress of the spring (using the first equation) if the mean radius is 100 mm, G = 30 GPa, number of turns is 10, and diameter of the spring is 10 mm. Also, determine the deformation of the spring.Show complete solution with FBDarrow_forwardanswer this ; The factors of safety predicted at point H by the maximum shear stress theory of failure (Tresca criterion)arrow_forward
- The figure shows a section of a shaft made of annealed steel with yield strength of 600 MPa and ultimate tensile strength of 700 MPa. Diameters of the shaft are D=30 mm and d=20 mm. Fillet radius of the shoulder section is 2.5 mm. The shaft is rotating and subjected to steady bending moment of M=51 Nm, steady torsion of T=58 Nm and steady axial load of F=14 kN. The operating temperature of the shaft is 20 ᵒC and reliability is 99%. The shaft has ground surface.Calculate the mean component of the shear stress (in MPa)arrow_forwardThe figure shows a section of a shaft made of annealed steel with yield strength of 600 MPa and ultimate tensile strength of 700 MPa. Diameters of the shaft are D=30 mm and d=20 mm. Fillet radius of the shoulder section is 2.5 mm. The shaft is rotating and subjected to steady bending moment of M=51 Nm, steady torsion of T=58 Nm and steady axial load of F=14 kN. The operating temperature of the shaft is 20 ᵒC and reliability is 99%. The shaft has ground surface. Calculate the alternating component of the normal stress (in MPa). (If there is no alternating component of the normal stress, enter "0" in the answer box)arrow_forwardA bolt is subjected to a tensile load of 18kN and a shear load of 12kN. The material has an yield stress of 328.6Mpa. Taking factor of safety as 2.5, Determine the core diameter of bolt according to the following theories of failure. Take possion’s ratio as 0.298 (i) Maximum Shear stress theory and (ii) Distortion Energy theory (iii) Maximum stress theory (iv) maximum strain energy theory (v) Maximum total strain energy theory.arrow_forward
- An aluminum alloy is to be used for a solid drive shaft such that it transmits 30 hp at 1200 rev>min. Using a factor of safety of 2.5 with respect to yielding, determine the smallest-diameter shaft that can be selected based on the maximum shear stress theory. sY = 10 ksi.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_forwardHelical coil springs are subjected to direct shear and torsional stresses. What is the factor for direct shear for a spring having a spring index of 8.2?0.0671.0611.0670.061arrow_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
Everything About COMBINED LOADING in 10 Minutes! Mechanics of Materials; Author: Less Boring Lectures;https://www.youtube.com/watch?v=N-PlI900hSg;License: Standard youtube license