
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.
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Chapter 24 Solutions
Welding: Principles and Applications (MindTap Course List)
- Drawing the section view for the following multiview drawing AutoCAD you see the section pratice I need to show how to autocadarrow_forwardA boiler with 80% efficiency produces steam at 40bar and 500 C at a rate of 1.128kg/s. The temperature of the feed water is raised from 25 C to 125 C in the economizer and the ambient air is drawn to the boiler at a rate of 2.70 kg/s at 16 C. The flue gases leave the chimney at rate of 3 kg/s at 150 C with specific heat of 1.01 kJ/kg.K. The dryness fraction of steam collected in the steam drum is 0.95. 1- Determine the heat value of the fuel. 2- The equivalence evaporation. 3- Draw the heat balance sheet.arrow_forwardA rotating shaft is made of 42 mm by 4 mm thick cold-drawn round steel tubing and has a 6 mm diameter hole drilled transversely through it. The shaft is subjected to a pulsating torque fluctuating from 20 to 160 Nm and a completely reversed bending moment of 200 Nm. The steel tubing has a minimum strength of Sut = 410 MPa (60 ksi). The static stress-concentration factor for the hole is 2.4 for bending and 1.9 for torsion. The maximum operating temperature is 400˚C and a reliability of 99.9% is to be assumed. Find the factor of safety for infinite life using the modified Goodman failure criterion.arrow_forward
- I need help with a MATLAB code. This code just keeps running and does not give me any plots. I even reduced the tolerance from 1e-9 to 1e-6. Can you help me fix this? Please make sure your solution runs. % Initial Conditions rev = 0:0.001:2; g1 = deg2rad(1); g2 = deg2rad(3); g3 = deg2rad(6); g4 = deg2rad(30); g0 = deg2rad(0); Z0 = 0; w0 = [0; Z0*cos(g0); -Z0*sin(g0)]; Z1 = 5; w1 = [0; Z1*cos(g1); -Z1*sin(g1)]; Z2 = 11; w2 = [0; Z2*cos(g2); -Z2*sin(g2)]; [v3, psi3, eta3] = Nut_angle(Z2, g2, w2); plot(v3, psi3) function dwedt = K_DDE(~, w_en) % Extracting the initial condtions to a variable % Extracting the initial condtions to a variable w = w_en(1:3); e = w_en(4:7); Z = w_en(8); I = 0.060214; J = 0.015707; x = (J/I) - 1; y = Z - 1; s = Z; % Kinematic Differential Equations dedt = zeros(4,1); dedt(1) = pi*(e(3)*(s-w(2)-1) + e(2)*w(3) + e(4)*w(1)); dedt(2) = pi*(e(4)*(w(2)-1-s) + e(3)*w(1) - e(1)*w(3)); dedt(3) = pi*(-e(1)*(s-w(2)-1) - e(2)*w(1) + e(4)*w(3));…arrow_forwardalpha 1 is not zero alpha 1 can equal alpha 2 use velocity triangle to solve for alpha 1 USE MATLAB ONLY provide typed code solve for velocity triangle and dont provide copied answer Turbomachienery . GIven: vx = 185 m/s, flow angle = 60 degrees, (leaving a stator in axial flow) R = 0.5, U = 150 m/s, b2 = -a3, a2 = -b3 Find: velocity triangle , a. magnitude of abs vel leaving rotor (m/s) b. flow absolute angles (a1, a2, a3) 3. flow rel angles (b2, b3) d. specific work done e. use code to draw vel. diagram Use this code for plot % plots Velocity Tri. in Ch4 function plotveltri(al1,al2,al3,b2,b3) S1L = [0 1]; V1x = [0 0]; V1s = [0 1*tand(al3)]; S2L = [2 3]; V2x = [0 0]; V2s = [0 1*tand(al2)]; W2s = [0 1*tand(b2)]; U2x = [3 3]; U2y = [1*tand(b2) 1*tand(al2)]; S3L = [4 5]; V3x = [0 0]; V3r = [0 1*tand(al3)]; W3r = [0 1*tand(b3)]; U3x = [5 5]; U3y = [1*tand(b3) 1*tand(al3)]; plot(S1L,V1x,'k',S1L,V1s,'r',... S2L,V2x,'k',S2L,V2s,'r',S2L,W2s,'b',U2x,U2y,'g',...…arrow_forward3. Find a basis of eigenvectors and diagonalize. 4 0 -19 7 a. b. 1-42 16 12-20 [21-61arrow_forward
- 2. Find the eigenvalues. Find the corresponding eigenvectors. 6 2 -21 [0 -3 1 3 31 a. 2 5 0 b. 3 0 -6 C. 1 1 0 -2 0 7 L6 6 0 1 1 2. (Hint: λ = = 3)arrow_forwardUSE MATLAB ONLY provide typed code solve for velocity triangle and dont provide copied answer Turbomachienery . GIven: vx = 185 m/s, flow angle = 60 degrees, (leaving a stator in axial flow) R = 0.5, U = 150 m/s, b2 = -a3, a2 = -b3 Find: velocity triangle , a. magnitude of abs vel leaving rotor (m/s) b. flow absolute angles (a1, a2, a3) 3. flow rel angles (b2, b3) d. specific work done e. use code to draw vel. diagram Use this code for plot % plots Velocity Tri. in Ch4 function plotveltri(al1,al2,al3,b2,b3) S1L = [0 1]; V1x = [0 0]; V1s = [0 1*tand(al3)]; S2L = [2 3]; V2x = [0 0]; V2s = [0 1*tand(al2)]; W2s = [0 1*tand(b2)]; U2x = [3 3]; U2y = [1*tand(b2) 1*tand(al2)]; S3L = [4 5]; V3x = [0 0]; V3r = [0 1*tand(al3)]; W3r = [0 1*tand(b3)]; U3x = [5 5]; U3y = [1*tand(b3) 1*tand(al3)]; plot(S1L,V1x,'k',S1L,V1s,'r',... S2L,V2x,'k',S2L,V2s,'r',S2L,W2s,'b',U2x,U2y,'g',... S3L,V3x,'k',S3L,V3r,'r',S3L,W3r,'b',U3x,U3y,'g',...... 'LineWidth',2,'MarkerSize',10),...…arrow_forwardUSE MATLAB ONLY provide typed code solve for velocity triangle and dont provide copied answer Turbomachienery . GIven: vx = 185 m/s, flow angle = 60 degrees, R = 0.5, U = 150 m/s, b2 = -a3, a2 = -b3 Find: velocity triangle , a. magnitude of abs vel leaving rotor (m/s) b. flow absolute angles (a1, a2, a3) 3. flow rel angles (b2, b3) d. specific work done e. use code to draw vel. diagram Use this code for plot % plots Velocity Tri. in Ch4 function plotveltri(al1,al2,al3,b2,b3) S1L = [0 1]; V1x = [0 0]; V1s = [0 1*tand(al3)]; S2L = [2 3]; V2x = [0 0]; V2s = [0 1*tand(al2)]; W2s = [0 1*tand(b2)]; U2x = [3 3]; U2y = [1*tand(b2) 1*tand(al2)]; S3L = [4 5]; V3x = [0 0]; V3r = [0 1*tand(al3)]; W3r = [0 1*tand(b3)]; U3x = [5 5]; U3y = [1*tand(b3) 1*tand(al3)]; plot(S1L,V1x,'k',S1L,V1s,'r',... S2L,V2x,'k',S2L,V2s,'r',S2L,W2s,'b',U2x,U2y,'g',... S3L,V3x,'k',S3L,V3r,'r',S3L,W3r,'b',U3x,U3y,'g',...... 'LineWidth',2,'MarkerSize',10),... axis([-1 6 -4 4]), ...…arrow_forward
- The answer should equal to 1157. Please sent me the solution. Thank you!arrow_forwardBONUS: If the volume of the 8cm x 6.5cm x 6cm Block of Aluminum was 312cm3 before machining, find how much material was removed when the fixture below was machined. +2 2.00 cm 6.00 cm 2.50 cm 6.50 cm 1.00 cm 2.50 cm 11.00 cm 8.00 cm 30 CP 9411 FL.4) (m² 1157 Area of triangle = 1/2*B*H Area of circle = лR² Circumference of a circle = 2πR 6.00 cm 6.50 cm 1.50 cm Radius 1.50 cm 1.00 cmarrow_forwardConsider a 5m by 5m wet concret patio with an average water film thickness of .2mm. Now wind at 50 km/h is blowing over the surface. If the air is at 1 atm, 15oC and 35 percent relative humidity, determine how long it will take for the patio to completely dry.arrow_forward
- Welding: Principles and Applications (MindTap Cou...Mechanical EngineeringISBN:9781305494695Author:Larry JeffusPublisher:Cengage Learning
