In the textbook (Machine Elements in Mechanical Design 6th Edition by Robert L. Mott, Edward M. Vavrek, and Jyhwen Wang) chapter 20 question 11. Im having trouble trying to understand and solve it with the information given to me. Image below. **Welded Joint Design Exercise****Instruction:**For Problems 7–12, design a welded joint to join the two members shown in each figure. Specify the weld pattern, the type of electrode to be used, and the size of the weld. - **Problems 7–9:** The members are made from ASTM A36 steel.- **Problems 10–12:** The members are made from ASTM A441 steel.Use the method of treating the joint as a line. Refer to Table 20–3 for allowable forces per inch of leg in building-type structures.**Problem References:**7. Figure P20–18. Figure P20–49. Figure P20–510. Figure P20–611. Figure P20–11**Figure Explanation:****Figure P20–11 (Problems 11 and 12):**This figure depicts a welded joint design where both members are subjected to forces. The dimensions are as follows:- A horizontal member is 21.5 inches long.- A vertical plate is attached perpendicularly to one end.- Force \( P_1 \) is 3000 lb, applied vertically at 6.5 inches from the edge along the member.- Force \( P_2 \) is 1200 lb, applied vertically at 9 inches from the same edge.- There is an overhang of 12 inches beyond the application of force \( P_1 \).- The total width of the horizontal member is 4.5 inches.This setup is used to evaluate the design and strength of the welded joint to ensure it sustains the applied forces effectively.

Given: To find: Design the welded jointWeld pattern: C- Shape Weld Type of Electrode: E70 Size of the weld: Weld properties can be found from Fig 20-8 y¯=d2b+2d=6.529+2×6.5=1.9 in Aw=b+2d=9+2×6.5=22 in Sw=2bd+d23=2×9×6.5+6.523=53.1 in2 Jw=b+2d312-d2b+d2b+2d=425.9 in3Shear force: fs=PAw=3000+120022=190.9 lb/in Bending force: fb=MSw=3000+1200×21.553.1=1700.6 lb/in Twisting force: ft=TcJw=1200×16.5×1.9425.9=88.3 lb/in Resultant: f2t+f2b+f2s=88.32+1700.62+190.92=1713.5 lb/in Using allowable force w=1713.511200=0.153 in Hence size of weld should be: 0.153 in