thickness of the nugget is
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- Two 1.2 mm thick sheets of steel are joined by double-sided spot welding using a current of 6000A and current flow time of t = 0.15 sec. The electrodes are 3 mm in diameter and the weld spot thickness is approximately one third of the total thickness of the metal sheets. Assuming that the effective resistance in this operation is 200 micro resistance, estimate the heat generated and its distribution in the weld zone. (Take density of steel = 8600 Kg/m3 and its latent heat capacity, L =1400 J/g)A welding operation on an aluminum alloy makes a groove weld. The cross-sectional area of the weld is 30.0 mm2. The welding velocity is 4.0 mm/sec. The heat transfer factor is 0.92 and the melting factor is 0.48. The melting temperature of the aluminum alloy is 650°C. Determine the rate of heat generation required at the welding source to accomplish this weld.An electron-beam welding operation will join two pieces of steel plate together. The plates are 2.5 cm thick. The unit melting energy is 8050 J/cm3. The diameter of the work area focus of the beam is 0.15 cm, hence the width of the weld will be 0.15 cm. The accelerating voltage is 30 kV and the beam current is 35 milliamp. The heat transfer factor is 0.70 and the melting factor is 0.55. If the beam moves at a speed of 125 cm/min, will the beam penetrate the full thickness of the plates
- An RSW operation is used to join two pieces of sheet steel having a unit melting energy of 8369.85 J/cm3. The sheet steel has a thickness of 0.3125 cm. The weld duration will be set at 0.25 sec with a current of 11,000 amp. Based on the electrode diameter, the weld nugget will have a diameter of 0.75 cm. Experience has shown that 40% of the supplied heat melts the nugget and the rest is dissipated by the metal. If the electrical resistance between the surfaces is 130 micro-ohms, what is the thickness of the weld nugget assuming it has a uniform thickness?Two 1.2 mm thick flat copper sheets are being spot welded using a current of 6000A and a current flow time of 0.18 s. The electrodes are 5 mm in diameter. Estimate the generated in the weld zone. Take effective resistance as 150 ohm. also calculate temperature rise assuming that the heat generated is confined to the volume ogf materials directly between two electrodes and temperature is uniformly distributed.The welding parameters to be used to butt weld a metal alloy are being evaluated. In order to set up the welding machine, the melting factor for the metal (f2) has to be determined. The welding machine can deliver 4000 watts of power and the heat transfer factor for the welding process (f1) is 0.9. The final weld bead consists of 8 0% filler metal and 40% parent metal. The melting temperature of the filler metal and the parent metal is the same 2540 deg F. The cross-sectional area of the weld bead is 0.5 sq inches. The rate at which filler metal is added to the weld is 1 cu inch per min. Conversion factor 1 Btu = 1055 J and 0 deg F = 460 deg R Calculate the melting factor.
- What is staircase effect? A welding operation on an aluminum alloy makes a groove weld.The cross-sectional area of the weld is 30.0 mm2. The welding velocity is 4.0 mm/sec. Theheat transfer factor is [0.92+{AAA/(20+YY)}] and the melting factor is [0.48+{AAA/(20+YY)}], where AAA are the last three digits and YY are the first two digits of yourstudent ID. The melting temperature of the aluminum alloy is 667°C. Determine the rate ofheat generation required at the welding source.A GTAW operation is performed on low carbon steel, whose unit melting energy is 10.3 J/mm3. The welding voltage is 22 volts and the current is 135 amps. The heat transfer factor is 0.7 and the melting factor is 0.65. If filler metal wire of 3.5 mm diameter is added to the operation, the final weld bead is composed of 60% volume of filler and 40% volume base metal. If the travel speed in the operation is 5 mm/sec, determine (a) cross-sectional area of the weld bead, and (b) the feed rate (mm/sec) at which the filler wire must be supplied.A nozzle having I.D. 400 mm is fabricated from S.S. 316 plate. It is attached by welding to a vessel having I.D. 1500 mm. Internal design pressure is 10 kgf/cm2 and design temperature is 300oC. Maximum allowable stress at design temperature is 612.4 kgf/cm2. Joint efficiency is 0.85 and corrosion allowance is 1.5 mm for both shell and nozzle. Density of material of fabrication is 7830 kg/m3. Calculate thickness and weight of reinforcement pad
- The unit melting energy for a certain sheet metal is 10J/mm^3 .The thickness of each of the two sheets to be spot-welded is = mm. To achive required strength, it is desired to form a weldnugget that is =7 mm in diameter and 3.5 mm thick. The weld duration will be set at 44 sec .If the electrical resistance between the two surfaces is 144 micro-ohms and only one half the electricity generated will be used to form the weld nugget Determine the minium current level required in this operationwhy is arc shielding requried during arc welding ? how is the arc shielding accomplshed? provide the detailed explanationTe - Effective throat thickness well in mm = 6 sin45° Consider the following drawing in which plate 1 is welded onto plate 2 as shown for tensile and shear loading. The plates are fillet welded with a weld thickness of 6 mm as shown for a length of 150 mm. The plates have an ultimate strength of 275 MPa and the welding was done with F70 welding electrodes, which possesses an ultimate strength of 483 MPa. Consider a partial factor of safety of 1.7 for the welded joint and determine the design strength of the weld joint, both in tension (Tdw) and in shear (Vdw) due to the forces acting on it. a) Determine the design strength of the following weld in tension ( ) due to vertical forces. b) Determine the design strength of the weld in shear ( ) due to horizontal forces. Attached pic is shear equation for calculation Tension equation for calculation is: Tdw = FyLwte /Ymw