
Concept explainers
Why is usage of the term GWW preferable to MIG for gas metal arc welding?

Reason of using the term GMAW over MIG for gas metal arc welding.
Answer to Problem 1R
GMAW is preferred over MIG for gas metal arc welding as other gases apart from argon are used in metal arc welding.
Explanation of Solution
The term MIG stands for Metal inert gas welding, an inert gas is used to protect the weld from impurity. Initially, Argon as an inert gas was used forshielding.
Later, after developments in the technology associated with welding, carbon dioxide
Argon was used for shielding the weld. It was termed as MIG welding and was named as Gas Metal Arc Welding (GMAW) by American Welding society because other gases can also be used for shielding.
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Chapter 10 Solutions
Welding: Principles and Applications (MindTap Course List)
- A 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_forwardI 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_forward
- 3. Find a basis of eigenvectors and diagonalize. 4 0 -19 7 a. b. 1-42 16 12-20 [21-61arrow_forward2. 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_forward
- 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, 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_forwardThe 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_forward
- Consider 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_forward70. Compute the number of cubic centimeters of iron required for the cast-iron plate shown. The plate is 3.50 centimeters thick. Round the answer to the nearest cubic centimeter. 50.0 cm 40.0 cm Radius 150° 115.0 cm- 81.0 cmarrow_forwardLaw of Sines Solve the following problems using the Law of Sin 7. Find side x. All dimensions are in inches. -°-67°-37° 81° x Sin A 8.820 X 67°00' 32°00' a sin A b C sin B sin Carrow_forward
- Welding: Principles and Applications (MindTap Cou...Mechanical EngineeringISBN:9781305494695Author:Larry JeffusPublisher:Cengage Learning
