Engineering

152. Why is a surfacing weld different from the other weld types? 153. Why is it desirable to use energy sources for welding that have high heat densities? 154. What is the unit melting energy in welding, and what are the factors on which it depends? 155. Define and distinguish the two terms heat transfer factor and melting factor in welding. 156. What is the heat-affected zone in a fusion weld? 157. How do brazing and soldering differ from the fusion-welding processes? 158. How do brazing and soldering differ from the solid-state welding processes? 159. What is the technical difference between brazing and soldering? 160. Under what circumstances would brazing or soldering be preferred over welding? 161. What are the two joint types most commonly used in brazing?

EXPERT ANSWER 152. Why is a surfacing weld different from the other weld types? A surfacing weld is done to place the filler materials onto the surface of the base metal in the form of weld beads. It is different from other weld types because it is not made to join parts, only to add …

152. Why is a surfacing weld different from the other weld types? 153. Why is it desirable to use energy sources for welding that have high heat densities? 154. What is the unit melting energy in welding, and what are the factors on which it depends? 155. Define and distinguish the two terms heat transfer factor and melting factor in welding. 156. What is the heat-affected zone in a fusion weld? 157. How do brazing and soldering differ from the fusion-welding processes? 158. How do brazing and soldering differ from the solid-state welding processes? 159. What is the technical difference between brazing and soldering? 160. Under what circumstances would brazing or soldering be preferred over welding? 161. What are the two joint types most commonly used in brazing? Read More »

The welding process in this image is 1 point Solid wire electrode Shielding gas Current conductor Travel Wire guide and contact tube Nozzle Solidified weld metal Shielding gas Arc Base metal Molten weld metal SMAW GMAW O TIG O Oxy-Acetylene

EXPERT ANSWER 1.In Tig welding, external wire is used to feed or weld since Tungsten elecrtrode is non – consumable . So this is not TIG welding. 2. This is not oxy-acetylene welding, because in oxy-acetylene welding two cylinder is used for carrying oxygen and acetylene. 3. In SMAW, arc submerged in granular flux of …

The welding process in this image is 1 point Solid wire electrode Shielding gas Current conductor Travel Wire guide and contact tube Nozzle Solidified weld metal Shielding gas Arc Base metal Molten weld metal SMAW GMAW O TIG O Oxy-Acetylene Read More »

26.20 What is friction stir welding (FSW), and how is it different from friction welding?

26.20 What is friction stir welding (FSW), and how is it different from friction welding? EXPERT ANSWER Friction stir welding is fusion welding process in this friction stir tool is used to join the metals. The tool has a small probe projecting out from the shoulderd tool which sits at the joint line of two …

26.20 What is friction stir welding (FSW), and how is it different from friction welding? Read More »

Question 2 ‘Smoothtools’ manufacturing company want to enter the ‘woodworking tools’ market and one of the first products it has decided to manufacture is a general purpose smoothing plane. (See Figure Q2- parts view). After ‘reverse engineering’ the wood plane, the planning department have arrived at a build sequence and timings for each task which is shown in Table Q2(a). Furthermore, the marketing department at ‘Smoothtools’ have produced a sales forecast for the new planes which is given in Table Q2(b). The plant is expected to have a work pattern of 40 hours week, 48 weeks per year. Assuming a Product layout (Long thin) arrangement, determine the following information for the 1 st quarter of production: d) If a ‘short fat’ layout is used in the 1st quarter, how many people would be required to assemble the products? What is the balance loss and potential output for this arrangement? (6 Marks) e) Considering the two options. Which layout is best suited to meet the demand in 4 th quarter and why? (12 Marks) f) What are the advantages and disadvantages of both ‘Iong thin’ and ‘short fat’ layouts? (16 Marks) \begin{tabular}{|l|l|c|} \hline L & Make up box and wrap plane, pack and stock & 12 secs \\ \hline \end{tabular} Table Q2(a) Standard time for each task and build sequence Table Q2(b) Sales forecast Year 1

EXPERT ANSWER

A driver on a level two-lane highway observes a truck completely blocking the highway. The driver was able to stop her vehicle only 30 ft from the truck. If the driver was driving at 55 mi/h, how far was she from the truck when she first observed it (assume perceptionreaction time is 1.5 sec)? How far was she from the truck at the moment the brakes were applied (use a/g = 0.35)?

Problem A driver on a level two-lane highway observes a truck completely blocking the highway. The driver was able to stop her vehicle only 30 ft from the truck. If the driver was driving at 55 mi/h, how far was she from the truck when she first observed it (assume perceptionreaction time is 1.5 sec)? …

A driver on a level two-lane highway observes a truck completely blocking the highway. The driver was able to stop her vehicle only 30 ft from the truck. If the driver was driving at 55 mi/h, how far was she from the truck when she first observed it (assume perceptionreaction time is 1.5 sec)? How far was she from the truck at the moment the brakes were applied (use a/g = 0.35)? Read More »

a 2) Select a surface casing setting depth for the following data. Use Eaton’s chart for fracture gradients. Use a 0.5-lb/gal kick size. Intermediate setting depth = 11,000ft Original mud weight = 10.5 lb/gal a Select a surface casing setting depth for the following data. Use Eaton’s chart for fracture gradients. Use a 0.5-lb/gal kick size. Intermediate setting depth = 11,000ft Original mud weight = 10.5 lb/gal 2) Select a surface casing setting depth for the following data. Use Eaton’s chart for fracture gradients. Use a 0.5-lb/gal kick size. Intermediate setting depth = 11,000ft Original mud weight = 10.5 lb/gal 120 14 0 16.0 170 18.0 2 4 6 IN 13 13 15 12 19 19 Fracture pressure gradient, lb/gal

EXPERT ANSWER

(1) A coated carbide tool is being used during a turning operation on a brass workpiece (Ø 0.5 in). The operation is currently performed at a spindle speed of 2800 rpm yielding a tool life of 18 minutes. In order to reduce tooling expenditures, the company would like to extend the expected tool life to 30 minutes. a) Determine the new spindle speed at which the operation should be performed to achieve this, and b) In addition to changing the speed, suggest at least two other actions that the company can take to prolong tool life. (Hint: You will need information from textbook Ch. 22) (12 pts.) TABLE 22.2 n Representative values of n and C in the Taylor tool life equation, Equation (22.1), for selected tool materials Compiled from [4], [9], and other sources. с Nonsteel Cutting Steel Cutting Tool Material m/min (ft/min) m/min (ft/min) Plain carbon tool steel 0.1 70 (200) (60) High-speed steel 0.125 (350) (200) Cemented carbide 0.25 900 (2700) 500 (1500) Cermet 0.25 600 (2000) Coated carbide 0.25 700 (2200) Ceramic 0.6 3000 (10,000) 20 120 70

EXPERT ANSWER Given that the tool used = Coated carbide. Material cutting is brass which is non steel. spindle speed = 2800 RPM Cutting speed = * d * N / 1000 =  * 0.5 * 2800 / 1000 = 4.398 inches / min Taylors Eqn V Tn = C V1 = 4.398 T1 = 18 n=0.25 T2 …

(1) A coated carbide tool is being used during a turning operation on a brass workpiece (Ø 0.5 in). The operation is currently performed at a spindle speed of 2800 rpm yielding a tool life of 18 minutes. In order to reduce tooling expenditures, the company would like to extend the expected tool life to 30 minutes. a) Determine the new spindle speed at which the operation should be performed to achieve this, and b) In addition to changing the speed, suggest at least two other actions that the company can take to prolong tool life. (Hint: You will need information from textbook Ch. 22) (12 pts.) TABLE 22.2 n Representative values of n and C in the Taylor tool life equation, Equation (22.1), for selected tool materials Compiled from [4], [9], and other sources. с Nonsteel Cutting Steel Cutting Tool Material m/min (ft/min) m/min (ft/min) Plain carbon tool steel 0.1 70 (200) (60) High-speed steel 0.125 (350) (200) Cemented carbide 0.25 900 (2700) 500 (1500) Cermet 0.25 600 (2000) Coated carbide 0.25 700 (2200) Ceramic 0.6 3000 (10,000) 20 120 70 Read More »