Study Guide with Lab Manual for Jeffus' Welding: Principles and Applications, 8th
8th Edition
ISBN: 9781305494701
Author: Larry Jeffus
Publisher: Cengage Learning
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Chapter 24, Problem 5R
As stricter standards are applied to the welding process, what happens to the cost of the weldment?
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A flanged bolt coupling consists of two concentric rows of bolts. The inner row has 6 nos. of 16mm diameterbolts spaced evenly in a circle of 250mm in diameter. The outer row of has 10 nos. of 25 mm diameter bolts spaced evenly in a circle of 500mm in diameter. If the allowable shear stress on one bolt is 60 MPa, determine the torque capacity of the coupling. The Poisson’s ratio of the inner row of bolts is 0.2 while that of the outer row is 0.25 and the bolts are steel, E =200 GPa.
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10: The constant wall thickness of a steel tube with the cross sectionshown is 2 mm. If a 600-N-m torque is applied to the tube. Use G = 80 GPa forsteel.1. Find the shear stress (MPa) in the wall of the tube.2. Find the angle of twist, in degrees per meter of length.
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A torque wrench is used to tighten the pipe shown.Dimensions: S1 = 400 mm; S2 = 250 mm; S3 = 100 mmModulus of Rigidity G = 78 GPa1. The diameter of the solid pipe is 20 mm. How much is themaximum force P (N) that can be applied based on theallowable shear stress of 60 MPa?2. For a hollow pipe with 50 mm outside diameter and is 6 mmthick, compute for the maximum force P (kN) that can beapplied such that the angle of twist at A does not exceed 5degrees.3. The torque applied to tighten the hollow pipe is 200 N-m.Given: Pipe outside diameter = 50 mm Pipe thickness = 6 mmSolve for the resulting maximum shear stress (MPa) in the pipe.
Chapter 24 Solutions
Study Guide with Lab Manual for Jeffus' Welding: Principles and Applications, 8th
Ch. 24 - Why is it important to select the correct welding...Ch. 24 - What are codes and standards?Ch. 24 - What is the difference between welding codes or...Ch. 24 - What might influence the selection of a particular...Ch. 24 - As stricter standards are applied to the welding...Ch. 24 - What can happen if welding standards are lax?Ch. 24 - What are three commonly used codes?Ch. 24 - What is a WPS?Ch. 24 - Prob. 9RCh. 24 - What is the form called that records the...
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- Correct answer and complete fbd only. I will upvote. 6: The shaft carries a total torque T0 that is uniformly distributedover its length L. Determine the angle of twist (degrees) of the shaft in termsif T0 = 1.2 kN-m, L = 2 m, G = 80 GPa, and diameter = 120 mm.arrow_forward2. Calculate the force in all members of the trusses shown using the method of joints. A 5525 lb C 3500 lb BY 14'-0" D 10'- 0" 6250 lb 10'- 0" Earrow_forwardCorrect answer and complete fbd only. I will upvote. 8: The steel rod fits loosely inside the aluminum sleeve. Both components are attached to a rigid wall at A andjoined together by a pin at B. Because of a slight misalignmentof the pre-drilled holes, the torque T0 = 750 N-m was appliedto the steel rod before the pin could be inserted into theholes. Determine the torque (N-m) in each component afterT0 was removed. Use G = 80 GPa for steel and G = 28 GPa foraluminumarrow_forward
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