Concept explainers
Why is it important to select the correct welding code or standard?

The reason for selecting the correct welding code or standard.
Answer to Problem 1R
The reason for selecting the correct welding code or standard is to judge the quality of weld.
Explanation of Solution
A welding code or standard is a list of rules and principles that may that may be applied to certain products or items. A welding code specifies certain specifications and requirements for a particular type of weld.
These are the detailed outlines established by various agencies and organizations that explains how to lay out a particular weld, how it should be performed and tested. These detailed outlines are known as codes, standards and specifications.
Codes, standards and specifications are a list of principles, rules and regulations that are needed to ensure that skilled welders across the world can produce the same type of weld with the same level of strength, safety and reliability.
Products which are fabricated as per a welding code or standard are usually inspected and checked in order to ensure the concurrence
There are some commonly used codes for welding. These are as follows.
- API standard 1104
- ASME Section IX
- AWS D1.1
- AWS SENSE
Selection of a code or standard is done by the designers or the engineers.
A correct welding code or standard is very necessary in order to judge a weld and compare the quality of the produced welded workpiece with that of ideal sample.
Conclusion:
The reason for selecting the correct welding code or standard is to judge the quality of weld.
Want to see more full solutions like this?
Chapter 24 Solutions
EBK 3I-EBK: WELDING PRINCIPLES & APPLIC
- For the first part: d = 4 cm, and Q = 15 Liters/s. For the second part Q = 6 Liters/s, and D = 80 cmarrow_forwardQ.2: (15 Marks) = 1400 For the following system, determine the first natural frequency using Dunkerley's equation, Given that the disk has moment of inertia J = 2 kg.m², the shaft has G = 20 GPa, p kg/m³, polar moment of cross-sectional area of the shaft Ip = 8×108 m². 500 mm 220 mm k=200 N/m FOF m=1 kg 14.14 56.56. W слarrow_forward1 Revolute four-bar mechanism, AB=60mm, BC=130mm, CD=140mm, AD=200mm,arrow_forwardCORRECT AND DETAILED HANDWRITTEN SOLUTION WITH FBD ONLY. I WILL UPVOTE THANK YOU. CORRECT ANSWER IS ALREADY PROVIDED. The roof truss shown carries roof loads, where P = 10 kN. The truss is consisting of circular arcs top andbottom chords with radii R + h and R, respectively.Given: h = 1.2 m, R = 10 m, s = 2 m.Allowable member stresses:Tension = 250 MPaCompression = 180 MPa1. If member KL has square section, determine the minimum dimension (mm).2. If member KL has circular section, determine the minimum diameter (mm).3. If member GH has circular section, determine the minimum diameter (mm).ANSWERS: (1) 31.73 mm; (2) 35.81 mm; (3) 18.49 mmarrow_forwardCORRECT AND DETAILED HANDWRITTEN SOLUTION WITH FBD ONLY. I WILL UPVOTE THANK YOU. CORRECT ANSWER IS ALREADY PROVIDED. 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 HANDWRITTEN SOLUTION WITH FBD ONLY. I WILL UPVOTE THANK YOU. CORRECT ANSWER IS ALREADY PROVIDED. 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 HANDWRITTEN SOLUTION WITH FBD ONLY. I WILL UPVOTE THANK YOU. CORRECT ANSWER IS ALREADY PROVIDED. 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_forwardA counter flow double pipe heat exchanger is being used to cool hot oil from 320°F to 285°F using cold water. The water, which flows through the inner tube, enters the heat exchanger at 70°F and leaves at 175°F. The inner tube is ¾-std type L copper. The overall heat transfer coefficient based on the outside diameter of the inner tube is 140 Btu/hr-ft2-°F. Design conditions call for a total heat transfer duty (heat transfer rate between the two fluids) of 20,000 Btu/hr. Determine the required length of this heat exchanger (ft).arrow_forward! Required information A one-shell-pass and eight-tube-passes heat exchanger is used to heat glycerin (cp=0.60 Btu/lbm.°F) from 80°F to 140°F by hot water (Cp = 1.0 Btu/lbm-°F) that enters the thin-walled 0.5-in-diameter tubes at 175°F and leaves at 120°F. The total length of the tubes in the heat exchanger is 400 ft. The convection heat transfer coefficient is 4 Btu/h-ft²°F on the glycerin (shell) side and 70 Btu/h-ft²°F on the water (tube) side. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Determine the rate of heat transfer in the heat exchanger before any fouling occurs. Correction factor F 1.0 10 0.9 0.8 R=4.0 3.0 2.0.15 1.0 0.8.0.6 0.4 0.2 0.7 0.6 R= T1-T2 12-11 0.5 12-11 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 (a) One-shell pass and 2, 4, 6, etc. (any multiple of 2), tube passes P= T₁-11 The rate of heat transfer in the heat exchanger is Btu/h.arrow_forward! Required information Air at 25°C (cp=1006 J/kg.K) is to be heated to 58°C by hot oil at 80°C (cp = 2150 J/kg.K) in a cross-flow heat exchanger with air mixed and oil unmixed. The product of heat transfer surface area and the overall heat transfer coefficient is 750 W/K and the mass flow rate of air is twice that of oil. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Air Oil 80°C Determine the effectiveness of the heat exchanger.arrow_forwardIn an industrial facility, a counter-flow double-pipe heat exchanger uses superheated steam at a temperature of 155°C to heat feed water at 30°C. The superheated steam experiences a temperature drop of 70°C as it exits the heat exchanger. The water to be heated flows through the heat exchanger tube of negligible thickness at a constant rate of 3.47 kg/s. The convective heat transfer coefficient on the superheated steam and water side is 850 W/m²K and 1250 W/m²K, respectively. To account for the fouling due to chemical impurities that might be present in the feed water, assume a fouling factor of 0.00015 m²-K/W for the water side. The specific heat of water is determined at an average temperature of (30 +70)°C/2 = 50°C and is taken to be J/kg.K. Cp= 4181 Water Steam What would be the required heat exchanger area in case of parallel-flow arrangement? The required heat exchanger area in case of parallel-flow arrangement is 1m².arrow_forwardA single-pass crossflow heat exchanger is used to cool jacket water (cp = 1.0 Btu/lbm.°F) of a diesel engine from 190°F to 140°F, using air (Cp = 0.245 Btu/lbm.°F) at inlet temperature of 90°F. Both air flow and water flow are unmixed. If the water and air mass flow rates are 85500 lbm/h and 400,000 lbm/h, respectively, determine the log mean temperature difference for this heat exchanger. Assume the correction factor F to be 0.92. Air flow (unmixed) Water flow (unmixed) The log mean temperature difference of the heat exchanger is °F.arrow_forwardarrow_back_iosSEE MORE QUESTIONSarrow_forward_ios
- Welding: Principles and Applications (MindTap Cou...Mechanical EngineeringISBN:9781305494695Author:Larry JeffusPublisher:Cengage Learning
