Connect 1-semester Access Card For Shigley's Mechanical Engineering Design
10th Edition
ISBN: 9780077591632
Author: Richard G Budynas; Keith J Nisbett
Publisher: McGraw-Hill Education
expand_more
expand_more
format_list_bulleted
Concept explainers
Videos
Textbook Question
Chapter 5, Problem 48P
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.
3–79* Repeat Prob. 3–77 with T = 900 lbf · in, a = 6 in, b = 5 in, c = 10 in, d = 1.375 in, e = 4 in, f = 10 in, and g = 6 in.
3–77* A torque T = 100 N · m is applied to the shaft EFG, which is running at constant speed and contains gear F. Gear F transmits torque to shaft ABCD through gear C, which drives the chain sprocket at B, transmitting a force P as shown. Sprocket B, gear C, and gear F have pitch diameters of a = 150, b = 250, and c = 125 mm, respectively. The contact force between the gears is transmitted through the pressure angle ϕ = 20°. Assuming no frictional losses and considering the bearings at A, D, E, and G to be simple supports, locate the point on shaft ABCD that contains the maximum tensile bending and maximum torsional shear stresses. Combine these stresses and determine the maximum principal normal and shear stresses in the shaft.
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?
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 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
Connect 1-semester Access Card For Shigley's Mechanical Engineering Design
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 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_forwardin 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_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_forwardFor a following loading that has generated the strain state below in the limit of isotropic elasticity, ?(epsilon) = 0.001 0.001 0 0.001 −0.002 0.0005 0 0.0005 0Find out how the stress state relates to yielding using a) Tresca criteriab) Von-Mises criteriaIf an experiment has shown that the yielding stress is 25 MPa. E=200 GPa and G=90 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_forward
- A loading condition is shown below. The wires at point A and D both made from a steel with the yield strength (Sy) of 190 MPa with 2 mm diameter. Using either maximum shear stress theory or distortion energy theory, determine the maximum load P that can be applied before yielding happens in the wires.arrow_forwardA mild steel shaft of 60 mm diameter is subjected to a bending moment of 2200 Nm and a torque T. If the yield point of the steel in tension is 200 MPa, find the maximum value of this torque without causing yielding of the shaft according to:1. The maximum principal stress theory;2. The maximum shear stress theory; and3. The maximum distortion strain energy theory of yieldingarrow_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
- 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_forwardSolve the preceding problem if the thickness of the steel plate is. t = 12 mm. the gage readings are x = 530 × 10-6 (elongation) and y = -210 -× l0-6 (shortening), the modulus is E = 200 GPa, and Poisson’s ratio is v = 0.30.arrow_forwardThe strength-to-weight ratio of a structural material is defined as its load-carrying capacity divided by its weight. For materials in tension, use a characteristic tensile stress obtained from a stress-strain curve as a measure of strength. For instance, either the yield stress or the ultimate stress could be used, depending upon the particular application. Thus, the strength-to-weight ratio RS/Wfor a material in tension is defined as Rs/w= in which a is the characteristic stress and 7 is the weight density. Note that the ratio has units of length. Using the ultimate stress Uas the strength parameter, calculate the strength-to-weight ratio (in units of meters) for each of the following materials: aluminum alloy 606I-T6, Douglas fir (in bending}, nylon. structural steel ASTM-A57.2, and a titanium alloy. Obtain the material properties from Tables [-1 and 1-3 of Appendix I. When a range of values is given in a table, use the average value.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