EBK 3I-EBK: WELDING PRINCIPLES & APPLIC
8th Edition
ISBN: 9780176919764
Author: Jeffus
Publisher: VST
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Chapter 7, Problem 33R
Sketch the proper end shape of a soapstone that is to be used for marking metal.
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Chapter 7 Solutions
EBK 3I-EBK: WELDING PRINCIPLES & APPLIC
Ch. 7 - Using Table 7-1, list the six different fuel gases...Ch. 7 - What metals can be cut with the oxyfuel gas...Ch. 7 - What other term is used to refer to the OFC...Ch. 7 - What is a combination welding and cutting torch?Ch. 7 - State one advantage of owning a combination...Ch. 7 - State one advantage of owning a dedicated cutting...Ch. 7 - What is a mixing chamber? Where is it located?Ch. 7 - Define the term equal-pressure torch. How does it...Ch. 7 - How does an injector-type mixing chamber work?Ch. 7 - State the advantages of having two oxygen...
Ch. 7 - Why are some copper alloy cutting tips...Ch. 7 - Using Table 7-4, answer the following: a . Oxygen...Ch. 7 - What determines the amount of preheat flame...Ch. 7 - What can happen if acetylene is used on a tip...Ch. 7 - Why are some propane and natural gas tips made...Ch. 7 - What types of tip seals are used with cutting...Ch. 7 - If a cutting tip sticks in the cutting head, how...Ch. 7 - How can cutting torch tip seals be repaired?Ch. 7 - What is used to reduce the high cylinder or system...Ch. 7 - What do the two pressure gauges on a regulator...Ch. 7 - Why must the gas pressure be released and the...Ch. 7 - What should be done if the torch flashes back?Ch. 7 - What is the purpose of a reverse flow valve?Ch. 7 - Why must the reverse flow valve and the flashback...Ch. 7 - How can a hose be checked for leaks?Ch. 7 - Why is the oxygen valve turned on before starting...Ch. 7 - Why does the preheat flame become slightly...Ch. 7 - What causes the tiny ripples in a hand cut?Ch. 7 - Why is a slight forward torch angle helpful for...Ch. 7 - Why should cans, drums, tanks, or other sealed...Ch. 7 - Why is the torch tip raised as the cutting lever...Ch. 7 - Why are the preheat holes not aligned in the kerf...Ch. 7 - Sketch the proper end shape of a soapstone that is...Ch. 7 - What are two methods you can use to determine what...Ch. 7 - What is the best way to set the oxygen pressure...Ch. 7 - Why is it important to have extra ventilation...Ch. 7 - What factors regarding a cut can be read from the...Ch. 7 - What is hard slag?Ch. 7 - Why is it important to make good-quality cuts?Ch. 7 - When using an ordinary welding table, what can be...Ch. 7 - Describe the methods of controlling distortion...Ch. 7 - List three things that can become a problem when...Ch. 7 - How does cutting small-diameter pipe differ from...
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- Hi, can you please help me .Identify and justify suitable analytical techniques of the scenario below, bearing in mind the kinds of information being handled to reach a conclusion (methodology). A child swing set was discovered to have failed at the fixing at the top of the chains connecting the seat to the top of the swing set. A 12 mm threaded steel bolt, connecting the shackle to the top beam, failed at the start of the threaded region on the linkage closest to the outside side of the swing set . The linkage and bolts were made of electro galvanised mild steel . The rigid bar chain alternatives and fixings were of the same material and appeared to be fitted in accordance with guidelines. The yield strength of the steel used is 260 MPa and the UTS is 380 MPa. The bolt that failed was threaded using a standard thread with a pitch (distance between threads) of 1.75 mm and a depth of approximately 1.1 mm. The swing set in question had been assigned to ‘toddlers’ with the application of…arrow_forwardHi, can you please define and calculate the failure mode of the linkage that failed on the swing (images added) : A child swing set was discovered to have failed at the fixing at the top of the chains connecting the seat to the top of the swing set. A 12 mm threaded steel bolt, connecting the shackle to the top beam, failed at the start of the threaded region on the linkage closest to the outside side of the swing set . The linkage and bolts were made of electro galvanised mild steel . The rigid bar chain alternatives and fixings were of the same material and appeared to be fitted in accordance with guidelines. The yield strength of the steel used is 260 MPa and the UTS is 380 MPa. The bolt that failed was threaded using a standard thread with a pitch (distance between threads) of 1.75 mm and a depth of approximately 1.1 mm. The swing set in question had been assigned to ‘toddlers’ with the application of a caged-type seat. However, the location was within the play area not…arrow_forwardPage 11-68. The rectangular plate shown is subjected to a uniaxial stress of 2000 psi. Compute the shear stress and the tensile developed on a plane forming an angle of 30° with the longitud axis of the member. (Hint: Assume a cross-sectional area of unity) 2000 psi 2000 psi hparrow_forward
- 11-70. A shear stress (pure shear) of 5000 psi exists on an element. (a) Determine the maximum tensile and compressive stresses caused in the element due to this shear. (b) Sketch the element showing the planes on which the maximum tensile and compressive stresses act.arrow_forward11-20. An aluminum specimen of circular cross section, 0.50 in. in diameter, ruptured under a tensile load of 12,000 lb. The plane of failure was found to be at 48° with a plane perpendicular to the longitudinal axis of the specimen. (a) Compute the shear stress on the failure plane. (b) Compute the maximum tensile stress. (c) Compute the tensile stress on the failure plane. hparrow_forwardA long flat steel bar 13 mm thick and 120 mm wide has semicircular grooves as shown and carries a tensile load of 50 kN Determine the maximum stress if plate r= 8mm r=21mm r=38mmarrow_forward
- Problem 13: F₁ = A =250 N 30% Determine the moment of each of the three forces about point B. F₂ = 300 N 60° 2 m -3 m B 4 m F3=500 Narrow_forward3 kN 3 kN 1.8 kN/m 80 mm B 300 mm D an 1.5 m-1.5 m--1.5 m- PROBLEM 5.47 Using the method of Sec. 5.2, solve Prob. 5.16 PROBLEM 5.16 For the beam and loading shown, determine the maximum normal stress due to bending on a transverse section at C.arrow_forward300 mm 3 kN 3 kN 450 N-m D E 200 mm 300 mm PROBLEM 5.12 Draw the shear and bending-moment diagrams for the beam and loading shown, and determine the maximum absolute value (a) of the shear, (b) of the bending moment.arrow_forward
- CORRECT AND DETAILED SOLUTION WITH FBD ONLY. I WILL UPVOTE THANK YOU. CORRECT ANSWER IS ALREADY PROVIDED. I REALLY NEED FBD. The cantilevered spandrel beam shown whose depth tapers from d1 to d2, has a constant width of 120mm. It carries a triangularly distributed end reaction.Given: d1 = 600 mm, d2 = 120 mm, L = 1 m, w = 100 kN/m1. Calculate the maximum flexural stress at the support, in kN-m.2. Determine the distance (m), from the free end, of the section with maximum flexural stress.3. Determine the maximum flexural stress in the beam, in MPa.ANSWERS: (1) 4.630 MPa; (2) 905.8688 m; (3) 4.65 MPaarrow_forwardCORRECT AND DETAILED SOLUTION WITH FBD ONLY. I WILL UPVOTE THANK YOU. CORRECT ANSWER IS ALREADY PROVIDED. I REALLY NEED FBD A concrete wall retains water as shown. Assume that the wall is fixed at the base. Given: H = 3 m, t = 0.5m, Concrete unit weight = 23 kN/m3Unit weight of water = 9.81 kN/m3(Hint: The pressure of water is linearly increasing from the surface to the bottom with intensity 9.81d.)1. Find the maximum compressive stress (MPa) at the base of the wall if the water reaches the top.2. If the maximum compressive stress at the base of the wall is not to exceed 0.40 MPa, what is the maximum allowable depth(m) of the water?3. If the tensile stress at the base is zero, what is the maximum allowable depth (m) of the water?ANSWERS: (1) 1.13 MPa, (2) 2.0 m, (3) 1.20 marrow_forwardCORRECT AND DETAILED SOLUTION WITH FBD ONLY. I WILL UPVOTE THANK YOU. CORRECT ANSWER IS ALREADY PROVIDED. I NEED FBD A short plate is attached to the center of the shaft as shown. The bottom of the shaft is fixed to the ground.Given: a = 75 mm, h = 125 mm, D = 38 mmP1 = 24 kN, P2 = 28 kN1. Calculate the maximum torsional stress in the shaft, in MPa.2. Calculate the maximum flexural stress in the shaft, in MPa.3. Calculate the maximum horizontal shear stress in the shaft, in MPa.ANSWERS: (1) 167.07 MPa; (2) 679.77 MPa; (3) 28.22 MPaarrow_forward
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