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For spot welding of two steel sheets (base metal) each of 3 mm thickness, welding current of 10000 A is applied for 0.2s. The heat dissipated to the base metal is 1000 J. Assuming that the heat required for melting 1 mm3 volume of steel is 20 J and interfacial contact resistance between sheets is 0.000 2? , the volume (in mm3) of weld nugget is _______
(Important - Enter only the numerical value in the answer)
Correct answer is between '140,160'. Can you explain this answer?
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For spot welding of two steel sheets (base metal) each of 3 mm thickne...
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Given data:
- Thickness of each steel sheet (base metal) = 3 mm
- Welding current = 10000 A
- Welding time = 0.2 s
- Heat dissipated to the base metal = 1000 J
- Heat required for melting 1 mm^3 volume of steel = 20 J
- Interfacial contact resistance between sheets = 0.0002
Calculating the volume of the weld nugget:
- The heat dissipated to the base metal is equal to the heat required for melting the volume of the weld nugget.
- Let's assume the volume of the weld nugget is V mm^3.
Heat dissipated to the base metal = Heat required for melting the volume of the weld nugget
1000 J = 20 J/mm^3 * V mm^3
Solving the equation:
V = 1000 J / (20 J/mm^3)
V = 50 mm^3
Calculating the effective resistance:
- The effective resistance (Reff) is the sum of the contact resistance (Rc) and the resistance of the base metal (Rbm).
- Rc = 0.0002 (given)
- Rbm = (resistivity of steel * base metal length) / (base metal area)
Calculating the resistivity of steel:
- The resistivity of steel (ρ) is the inverse of the conductivity (σ) of steel.
- The conductivity of steel is typically around 6.2 x 10^6 S/m.
ρ = 1 / σ
ρ = 1 / (6.2 x 10^6 S/m)
ρ ≈ 1.6 x 10^-7 Ω.m
Calculating the resistance of the base metal:
- The base metal length is equal to the welding current multiplied by the welding time.
- The base metal area is equal to the product of the thickness of each steel sheet and the welding spot diameter (which can be assumed to be equal to the thickness of the steel sheets).
Base metal length = 10000 A * 0.2 s
Base metal length = 2000 m
Base metal area = 2 * 3 mm * 3 mm
Base metal area = 18 mm^2
Rbm = (1.6 x 10^-7 Ω.m * 2000 m) / (18 mm^2)
Rbm ≈ 1.8 x 10^-6 Ω
Calculating the effective resistance:
Reff = Rc + Rbm
Reff = 0.0002 + 1.8 x 10^-6 Ω
Reff ≈ 1.82 x 10^-6 Ω
Calculating the welding current:
- The welding current (I) is equal to the heat dissipated to the base metal divided by the product of the effective resistance and the welding time.
I = 1000 J / (1.82 x 10^-6 Ω * 0.2 s)
I ≈ 2.75 x 10^6 A
Calculating the volume of the weld nugget:
- The volume of the weld nugget is directly proportional to the
- Thickness of each steel sheet (base metal) = 3 mm
- Welding current = 10000 A
- Welding time = 0.2 s
- Heat dissipated to the base metal = 1000 J
- Heat required for melting 1 mm^3 volume of steel = 20 J
- Interfacial contact resistance between sheets = 0.0002
Calculating the volume of the weld nugget:
- The heat dissipated to the base metal is equal to the heat required for melting the volume of the weld nugget.
- Let's assume the volume of the weld nugget is V mm^3.
Heat dissipated to the base metal = Heat required for melting the volume of the weld nugget
1000 J = 20 J/mm^3 * V mm^3
Solving the equation:
V = 1000 J / (20 J/mm^3)
V = 50 mm^3
Calculating the effective resistance:
- The effective resistance (Reff) is the sum of the contact resistance (Rc) and the resistance of the base metal (Rbm).
- Rc = 0.0002 (given)
- Rbm = (resistivity of steel * base metal length) / (base metal area)
Calculating the resistivity of steel:
- The resistivity of steel (ρ) is the inverse of the conductivity (σ) of steel.
- The conductivity of steel is typically around 6.2 x 10^6 S/m.
ρ = 1 / σ
ρ = 1 / (6.2 x 10^6 S/m)
ρ ≈ 1.6 x 10^-7 Ω.m
Calculating the resistance of the base metal:
- The base metal length is equal to the welding current multiplied by the welding time.
- The base metal area is equal to the product of the thickness of each steel sheet and the welding spot diameter (which can be assumed to be equal to the thickness of the steel sheets).
Base metal length = 10000 A * 0.2 s
Base metal length = 2000 m
Base metal area = 2 * 3 mm * 3 mm
Base metal area = 18 mm^2
Rbm = (1.6 x 10^-7 Ω.m * 2000 m) / (18 mm^2)
Rbm ≈ 1.8 x 10^-6 Ω
Calculating the effective resistance:
Reff = Rc + Rbm
Reff = 0.0002 + 1.8 x 10^-6 Ω
Reff ≈ 1.82 x 10^-6 Ω
Calculating the welding current:
- The welding current (I) is equal to the heat dissipated to the base metal divided by the product of the effective resistance and the welding time.
I = 1000 J / (1.82 x 10^-6 Ω * 0.2 s)
I ≈ 2.75 x 10^6 A
Calculating the volume of the weld nugget:
- The volume of the weld nugget is directly proportional to the
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For spot welding of two steel sheets (base metal) each of 3 mm thickness, welding current of10000 A is applied for 0.2s. The heat dissipated to the base metal is 1000 J. Assuming thatthe heat required for melting 1 mm3 volume of steel is 20 J and interfacial contact resistancebetween sheets is 0.000 2? , the volume (in mm3) of weld nugget is _______(Important - Enter only the numerical value in the answer)Correct answer is between '140,160'. Can you explain this answer?
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For spot welding of two steel sheets (base metal) each of 3 mm thickness, welding current of10000 A is applied for 0.2s. The heat dissipated to the base metal is 1000 J. Assuming thatthe heat required for melting 1 mm3 volume of steel is 20 J and interfacial contact resistancebetween sheets is 0.000 2? , the volume (in mm3) of weld nugget is _______(Important - Enter only the numerical value in the answer)Correct answer is between '140,160'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about For spot welding of two steel sheets (base metal) each of 3 mm thickness, welding current of10000 A is applied for 0.2s. The heat dissipated to the base metal is 1000 J. Assuming thatthe heat required for melting 1 mm3 volume of steel is 20 J and interfacial contact resistancebetween sheets is 0.000 2? , the volume (in mm3) of weld nugget is _______(Important - Enter only the numerical value in the answer)Correct answer is between '140,160'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for For spot welding of two steel sheets (base metal) each of 3 mm thickness, welding current of10000 A is applied for 0.2s. The heat dissipated to the base metal is 1000 J. Assuming thatthe heat required for melting 1 mm3 volume of steel is 20 J and interfacial contact resistancebetween sheets is 0.000 2? , the volume (in mm3) of weld nugget is _______(Important - Enter only the numerical value in the answer)Correct answer is between '140,160'. Can you explain this answer?.
For spot welding of two steel sheets (base metal) each of 3 mm thickness, welding current of10000 A is applied for 0.2s. The heat dissipated to the base metal is 1000 J. Assuming thatthe heat required for melting 1 mm3 volume of steel is 20 J and interfacial contact resistancebetween sheets is 0.000 2? , the volume (in mm3) of weld nugget is _______(Important - Enter only the numerical value in the answer)Correct answer is between '140,160'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about For spot welding of two steel sheets (base metal) each of 3 mm thickness, welding current of10000 A is applied for 0.2s. The heat dissipated to the base metal is 1000 J. Assuming thatthe heat required for melting 1 mm3 volume of steel is 20 J and interfacial contact resistancebetween sheets is 0.000 2? , the volume (in mm3) of weld nugget is _______(Important - Enter only the numerical value in the answer)Correct answer is between '140,160'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for For spot welding of two steel sheets (base metal) each of 3 mm thickness, welding current of10000 A is applied for 0.2s. The heat dissipated to the base metal is 1000 J. Assuming thatthe heat required for melting 1 mm3 volume of steel is 20 J and interfacial contact resistancebetween sheets is 0.000 2? , the volume (in mm3) of weld nugget is _______(Important - Enter only the numerical value in the answer)Correct answer is between '140,160'. Can you explain this answer?.
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