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
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Textbook Question
Chapter 5, Problem 55P
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.
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Required information
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.
0x²
=104 MPa, 0y = 19 MPa, and Txy = -19 MPa
I
The factor of safety from the maximum-shear-stress theory is 3.8
3.45
and the factor of safety from the distortion-energy theory is
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:
1. A ductile hot-rolled steel bar has a 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 state:
75 MPa
50 MPa
50 MPa
2. 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: 0A = 30 kpsi, OB = 15 kpsi.
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...
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- Required information 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. 0x = -46 MPa, Oy = -80.5 MPa, and Txy = -46 MPa The factor of safety from the maximum-shear-stress theory is 3.56, and the factor of safety from the distortion-energy theory is 3.3arrow_forwardA rod specimen of ductile cast iron was tested in a torsion-testing machine. The rod diameter was 14 mm, and the rod length was 440 mm. When the applied torque reached 225.3 N-m, a shear strain of 2105 microradians was measured in the specimen. What was the angle of twist in the specimen? Calculate the shear stress in the specimen. Answer: T = MPaarrow_forwardThe state of stress at a point in member is given by: 50 60 100 60 40 80 MPa 100 80 20 A torsion test performed on a specimen made of the same material shows that yielding occurs at a shearing stress of 140 MPa. Examine using maximum distortion energy theory if yielding will be occur under the state fo stress.arrow_forward
- A rod specimen of ductile cast iron was tested in a torsion-testing machine. The rod diameter was 16 mm, and the rod length was 360 mm. When the applied torque reached 254.8 N-m, a shear strain of 1780 microradians was measured in the specimen. What was the angle of twist in the specimen? Part 1 Calculate the shear stress in the specimen. Answer: T= i MPaarrow_forwardA rod specimen of ductile cast iron was tested in a torsion-testing machine. The rod diameter was 14 mm, and the rod length was 440 mm. When the applied torque reached 225.3 N-m, a shear strain of 2105 microradians was measured in the specimen. What was the angle of twist in the specimen? Calculate the shear stress in the specimen. Answer: T= 418.16 Use Hooke's Law to calculate the shear modulus of the specimen. answer: MPa G = i MPaarrow_forward1. An AISI 1018 steel has a yield strength, Sy = 295 MPa. Using the distortion energy theory for the plane stress state ox = -30 MPs, Oy = -65 MPa, and Txy = 40 MPa. 1) Determine the factor of safety. 2) Plot the safety envelope of the material. 3) Plot the load line. 4) Estimate the factor of safety by graphical method (refer to 4-1 Ductile Failure I, P10).arrow_forward
- 2. 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_forwardA 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_forward1. A ductile hot-rolled steel bar has a 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 state: 75 MPa 50 MPa 50 MPaarrow_forward
- In the picture there is a sketch of a socket wrench. Assume the wrench is held at a fixed point “A”. The yield stress of the material is known to be 400 MPa. Answer the questions below Describe the stresses at point “A” and their causes and calculate the stresses. Determine the factor of safety against yield assuming the Tresca yield criteria. Determine the factor of safety against yield assuming the von Mises yield criteria using both principal stresses and “Cartesian” stresses. Do your values match or not, and is this expected? Explain. Do the calculated values make sense with the respect to the Tresca value? Explain, why or why not?arrow_forwardProblem 1: The stress state at a critical point on a steel part (Sy = 250 MPa) is shown below. Determine whether yielding occurs according to the Maximum Distortional Energy (Von Mises) failure theory. 60 MPa 40 MPa 70 MPaarrow_forwardA rod specimen of ductile cast iron was tested in a torsion-testing machine. The rod diameter was 24 mm, and the rod length was 280 mm. When the applied torque reached 255.0 N-m, a shear strain of 1620 microradians was measured in the specimen. What was the angle of twist in the specimen? Answer: Ф= iarrow_forward
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