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 51P
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.
3–82* Repeat Prob. 3–80 with Fx = 75 lbf, Fy = −200 lbf, and Fz = 100 lbf.
3–80* The cantilevered bar in the figure is made from a ductile material and is statically loaded with Fy = 200 lbf and Fx = Fz = 0. Analyze the stress situation in rod AB by obtaining the following information.
(a) Determine the precise location of the critical stress element.
(b) Sketch the critical stress element and determine magnitudes and directions for all stresses acting on it. (Transverse shear may only be neglected if you can justify this decision.)
(c) For the critical stress element, determine the principal stresses and the maximum shear stress.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
Bar 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.
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 three principal stresses at a given point are σ1 =60 MPa, σ2 =−100 MPa, σ3 =−10 MPa. If the material has a yield stress of 250 MPa, estimate the factor of safety against yielding using (i) the maximum shear stress theory and (ii) von Mises’ 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
- 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.arrow_forwardThe 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_forwarda)Normal stress due to bending caused by V. Enter your answer in MPa b) Shear stress caused by T. Enter your answer in MPa c) Apply the Maximum-Distortion-Energy Theory and calculate Mises equivalent stress. Enter your answer in MPa d) Determine the factor of safety n.arrow_forward
- in the design of shafts made of ductile materials subjected to twisting moment and bending moment, the recommended theory of failure is (a) maximum principal stress theory (b) maximum principal strain theory (c) maximum shear stress theory (4) maximum strain-energy theoryarrow_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_forwardThe 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_forward
- Q#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 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_forwardB).shear stress caused by T C).APply miaximum-Distortion-energy Theory and calculate Mises equivalent stress. Enter your answer in MPa to 2 decimal places. D). determine factor of safety n (upto 2 decimal places)arrow_forward
- A 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_forwardDetermine the yield strength of a material required such that the component would not fail when subject to the following stresses sigma1 = 2 MPa and sigma 2= -15 MPa sigma 3 = 10 MPa). Use a yield criterion that assumes that yield failure will occur when the maximum shear stress in the complex system becomes equal to the limiting shear strength in a simple tensile test.arrow_forwardA single L8x6x1/2 (A36 steel) is used as a tension brace in a multi-story building. The 8 inch leg of the angle is attached to a gusset plate with two lines of four ¾ inch Ø bolts at a spacing of 3 inches. Determine the shear lag factor (consider Case 2 & 8) and the effective net area. Considering yielding and net section rupture, determine the design strength by LRFD and the allowable strength by ASD. (PLEASE SOLVE THIS ONE WITH THE BOLD TEXT)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