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Weld Metal Weight worksheet

Grades 7–9 · CCSS 7.G.B.6, HSG.SRT.C.8

Solve each problem and show your work. Round answers to 2 decimal places where needed.

1.
Problem 1.
- Units: Millimeters, meters, kilograms
- Weld type: Fillet weld
- Leg 1: 12 mm
- Leg 2: 4 mm
- Weld length: 3 m
- Allowance for reinforcement and convexity: 10%
- Weld metal: Aluminum (2.70 g/cm³)
- Process (deposition efficiency): Other (enter efficiency)
- Deposition efficiency: 76%
Find: Weld metal deposited
Answer
2.
Problem 2.
- Units: Inches, feet, pounds
- Weld type: Single-V groove (butt)
- Plate thickness: 0.75 in
- Root gap: 0.0625 in
- Root face (land): 0 in
- Included groove angle: 45 °
- Weld length: 11 ft
- Allowance for reinforcement and convexity: 0%
- Weld metal: Carbon or low-alloy steel (7.85 g/cm³)
- Process (deposition efficiency): Solid or metal-cored wire, GMAW (about 95%)
Find: Weld metal deposited
Answer
3.
Problem 3.
- Units: Millimeters, meters, kilograms
- Weld type: Double-V groove (butt)
- Plate thickness: 20 mm
- Root gap: 1 mm
- Root face (land): 2 mm
- Included groove angle: 70 °
- Weld length: 3 m
- Allowance for reinforcement and convexity: 10%
- Weld metal: Other (enter density)
- Density: 7 g/cm³
- Process (deposition efficiency): Stick, SMAW (about 55%)
Find: Weld metal deposited
Answer
4.
Problem 4.
- Units: Inches, feet, pounds
- Weld type: Single-V groove (butt)
- Plate thickness: 1 in
- Root gap: 0 in
- Root face (land): 0.125 in
- Included groove angle: 60 °
- Weld length: 10 ft
- Allowance for reinforcement and convexity: 20%
- Weld metal: Carbon or low-alloy steel (7.85 g/cm³)
- Process (deposition efficiency): Submerged arc, SAW (about 100%)
Find: Weld metal deposited
Answer
5.
Problem 5.
- Units: Inches, feet, pounds
- Weld type: Single-V groove (butt)
- Plate thickness: 0.5 in
- Root gap: 0.125 in
- Root face (land): 0 in
- Included groove angle: 70 °
- Weld length: 9 ft
- Allowance for reinforcement and convexity: 20%
- Weld metal: Austenitic stainless steel (8.0 g/cm³)
- Process (deposition efficiency): Submerged arc, SAW (about 100%)
Find: Weld metal deposited
Answer
6.
Problem 6.
- Units: Inches, feet, pounds
- Weld type: Fillet weld
- Leg 1: 0.1875 in
- Leg 2: 0.125 in
- Weld length: 7 ft
- Allowance for reinforcement and convexity: 15%
- Weld metal: Aluminum (2.70 g/cm³)
- Process (deposition efficiency): Flux-cored wire, FCAW (about 90%)
Find: Weld metal deposited
Answer
7.
Problem 7.
- Units: Inches, feet, pounds
- Weld type: Double-V groove (butt)
- Plate thickness: 0.75 in
- Root gap: 0.0625 in
- Root face (land): 0.125 in
- Included groove angle: 70 °
- Weld length: 9 ft
- Allowance for reinforcement and convexity: 20%
- Weld metal: Aluminum (2.70 g/cm³)
- Process (deposition efficiency): Solid or metal-cored wire, GMAW (about 95%)
Find: Weld metal deposited
Answer
8.
Problem 8.
- Units: Inches, feet, pounds
- Weld type: Double-V groove (butt)
- Plate thickness: 0.375 in
- Root gap: 0.25 in
- Root face (land): 0.125 in
- Included groove angle: 70 °
- Weld length: 7 ft
- Allowance for reinforcement and convexity: 15%
- Weld metal: Other (enter density)
- Density: 8 g/cm³
- Process (deposition efficiency): Stick, SMAW (about 55%)
Find: Weld metal deposited
Answer
9.
Problem 9.
- Units: Millimeters, meters, kilograms
- Weld type: Double-V groove (butt)
- Plate thickness: 10 mm
- Root gap: 1 mm
- Root face (land): 0 mm
- Included groove angle: 50 °
- Weld length: 4 m
- Allowance for reinforcement and convexity: 0%
- Weld metal: Aluminum (2.70 g/cm³)
- Process (deposition efficiency): Solid or metal-cored wire, GMAW (about 95%)
Find: Weld metal deposited
Answer
10.
Problem 10.
- Units: Inches, feet, pounds
- Weld type: Square groove (butt, gap only)
- Plate thickness: 0.25 in
- Root gap: 0.25 in
- Weld length: 11 ft
- Allowance for reinforcement and convexity: 15%
- Weld metal: Austenitic stainless steel (8.0 g/cm³)
- Process (deposition efficiency): Solid or metal-cored wire, GMAW (about 95%)
Find: Weld metal deposited
Answer

Answer key: Weld Metal Weight worksheet

1. Weld metal deposited: 0.214 kg
1. A fillet is a right triangle, so its cross-section is leg 1 × leg 2 ÷ 2
  $\frac{12 \times 4}{2}$ $= 24 mm²$
2. Add 10% for reinforcement, since real welds bulge above the flat shape
  $24 \times 1 + \frac{10}{100}$ $= 26.4 mm²$
3. Volume = area × length. Change meters to mm (× 1,000), then mm³ to cm³ (÷ 1,000)
  $\frac{26.4 \times 3,000}{1,000}$ $= 79.2 cm³$
4. Weight = volume × density (how heavy each cm³ is), ÷ 1,000 to go from grams to kg
  $\frac{79.2 \times 2.7}{1,000}$ $= 0.21384 kg$
5. Filler to buy = weld metal ÷ deposition efficiency, because some filler is lost to stubs, spatter and slag
  $\frac{0.21384}{\frac{76}{100}}$ $= 0.281368 kg$
2. Weld metal deposited: 10.477 lb
1. Find the V's area. It depends on the beveled depth (thickness − root face) and on half the groove angle
  ${(0.75 - 0)}^{2} \times tan 22.5 °$ $= 0.232995 in²$
2. Add the strip under the root gap (gap × thickness) to the V's area
  $0.0625 \times 0.75 + 0.232995$ $= 0.27987 in²$
3. Add 0% for reinforcement, since real welds bulge above the flat shape
  $0.27987 \times 1 + \frac{0}{100}$ $= 0.27987 in²$
4. Volume = area × length. Change feet to inches first (× 12)
  $0.27987 \times 132$ $= 36.942857 in³$
5. Change the density into pounds per cubic inch, to match the inch measurements
  $\frac{7.85 \times 16.387064}{453.59237}$ $= 0.283599 lb/in³$
6. Weight = volume × density
  $36.942857 \times 0.283599$ $= 10.476966 lb$
7. Filler to buy = weld metal ÷ deposition efficiency, because some filler is lost to stubs, spatter and slag
  $\frac{10.476966}{\frac{95}{100}}$ $= 11.028386 lb$
3. Weld metal deposited: 3.082 kg
1. Find both V areas. Each V is half the beveled depth (thickness − root face), so together they need less metal than one big V
  $2 \times {(\frac{20 - 2}{2})}^{2} \times tan 35 °$ $= 113.433621 mm²$
2. Add the strip under the root gap (gap × thickness) to both V areas
  $1 \times 20 + 113.433621$ $= 133.433621 mm²$
3. Add 10% for reinforcement, since real welds bulge above the flat shape
  $133.433621 \times 1 + \frac{10}{100}$ $= 146.776983 mm²$
4. Volume = area × length. Change meters to mm (× 1,000), then mm³ to cm³ (÷ 1,000)
  $\frac{146.776983 \times 3,000}{1,000}$ $= 440.33095 cm³$
5. Weight = volume × density (how heavy each cm³ is), ÷ 1,000 to go from grams to kg
  $\frac{440.33095 \times 7}{1,000}$ $= 3.082317 kg$
6. Filler to buy = weld metal ÷ deposition efficiency, because some filler is lost to stubs, spatter and slag
  $\frac{3.082317}{\frac{55}{100}}$ $= 5.604212 kg$
4. Weld metal deposited: 18.052 lb
1. Find the V's area. It depends on the beveled depth (thickness − root face) and on half the groove angle
  ${(1 - 0.125)}^{2} \times tan 30 °$ $= 0.442034 in²$
2. Add the strip under the root gap (gap × thickness) to the V's area
  $0 \times 1 + 0.442034$ $= 0.442034 in²$
3. Add 20% for reinforcement, since real welds bulge above the flat shape
  $0.442034 \times 1 + \frac{20}{100}$ $= 0.530441 in²$
4. Volume = area × length. Change feet to inches first (× 12)
  $0.530441 \times 120$ $= 63.652867 in³$
5. Change the density into pounds per cubic inch, to match the inch measurements
  $\frac{7.85 \times 16.387064}{453.59237}$ $= 0.283599 lb/in³$
6. Weight = volume × density
  $63.652867 \times 0.283599$ $= 18.051905 lb$
7. Filler to buy = weld metal ÷ deposition efficiency, because some filler is lost to stubs, spatter and slag
  $\frac{18.051905}{\frac{100}{100}}$ $= 18.051905 lb$
5. Weld metal deposited: 8.898 lb
1. Find the V's area. It depends on the beveled depth (thickness − root face) and on half the groove angle
  ${(0.5 - 0)}^{2} \times tan 35 °$ $= 0.175052 in²$
2. Add the strip under the root gap (gap × thickness) to the V's area
  $0.125 \times 0.5 + 0.175052$ $= 0.237552 in²$
3. Add 20% for reinforcement, since real welds bulge above the flat shape
  $0.237552 \times 1 + \frac{20}{100}$ $= 0.285062 in²$
4. Volume = area × length. Change feet to inches first (× 12)
  $0.285062 \times 108$ $= 30.786724 in³$
5. Change the density into pounds per cubic inch, to match the inch measurements
  $\frac{8 \times 16.387064}{453.59237}$ $= 0.289018 lb/in³$
6. Weight = volume × density
  $30.786724 \times 0.289018$ $= 8.897928 lb$
7. Filler to buy = weld metal ÷ deposition efficiency, because some filler is lost to stubs, spatter and slag
  $\frac{8.897928}{\frac{100}{100}}$ $= 8.897928 lb$
6. Weld metal deposited: 0.11 lb
1. A fillet is a right triangle, so its cross-section is leg 1 × leg 2 ÷ 2
  $\frac{0.1875 \times 0.125}{2}$ $= 0.011719 in²$
2. Add 15% for reinforcement, since real welds bulge above the flat shape
  $0.011719 \times 1 + \frac{15}{100}$ $= 0.013477 in²$
3. Volume = area × length. Change feet to inches first (× 12)
  $0.013477 \times 84$ $= 1.132031 in³$
4. Change the density into pounds per cubic inch, to match the inch measurements
  $\frac{2.7 \times 16.387064}{453.59237}$ $= 0.097544 lb/in³$
5. Weight = volume × density
  $1.132031 \times 0.097544$ $= 0.110423 lb$
6. Filler to buy = weld metal ÷ deposition efficiency, because some filler is lost to stubs, spatter and slag
  $\frac{0.110423}{\frac{90}{100}}$ $= 0.122692 lb$
7. Weld metal deposited: 2.321 lb
1. Find both V areas. Each V is half the beveled depth (thickness − root face), so together they need less metal than one big V
  $2 \times {(\frac{0.75 - 0.125}{2})}^{2} \times tan 35 °$ $= 0.136759 in²$
2. Add the strip under the root gap (gap × thickness) to both V areas
  $0.0625 \times 0.75 + 0.136759$ $= 0.183634 in²$
3. Add 20% for reinforcement, since real welds bulge above the flat shape
  $0.183634 \times 1 + \frac{20}{100}$ $= 0.220361 in²$
4. Volume = area × length. Change feet to inches first (× 12)
  $0.220361 \times 108$ $= 23.799003 in³$
5. Change the density into pounds per cubic inch, to match the inch measurements
  $\frac{2.7 \times 16.387064}{453.59237}$ $= 0.097544 lb/in³$
6. Weight = volume × density
  $23.799003 \times 0.097544$ $= 2.321443 lb$
7. Filler to buy = weld metal ÷ deposition efficiency, because some filler is lost to stubs, spatter and slag
  $\frac{2.321443}{\frac{95}{100}}$ $= 2.443624 lb$
8. Weld metal deposited: 3.228 lb
1. Find both V areas. Each V is half the beveled depth (thickness − root face), so together they need less metal than one big V
  $2 \times {(\frac{0.375 - 0.125}{2})}^{2} \times tan 35 °$ $= 0.021881 in²$
2. Add the strip under the root gap (gap × thickness) to both V areas
  $0.25 \times 0.375 + 0.021881$ $= 0.115631 in²$
3. Add 15% for reinforcement, since real welds bulge above the flat shape
  $0.115631 \times 1 + \frac{15}{100}$ $= 0.132976 in²$
4. Volume = area × length. Change feet to inches first (× 12)
  $0.132976 \times 84$ $= 11.170002 in³$
5. Change the density into pounds per cubic inch, to match the inch measurements
  $\frac{8 \times 16.387064}{453.59237}$ $= 0.289018 lb/in³$
6. Weight = volume × density
  $11.170002 \times 0.289018$ $= 3.228335 lb$
7. Filler to buy = weld metal ÷ deposition efficiency, because some filler is lost to stubs, spatter and slag
  $\frac{3.228335}{\frac{55}{100}}$ $= 5.8697 lb$
9. Weld metal deposited: 0.36 kg
1. Find both V areas. Each V is half the beveled depth (thickness − root face), so together they need less metal than one big V
  $2 \times {(\frac{10 - 0}{2})}^{2} \times tan 25 °$ $= 23.315383 mm²$
2. Add the strip under the root gap (gap × thickness) to both V areas
  $1 \times 10 + 23.315383$ $= 33.315383 mm²$
3. Add 0% for reinforcement, since real welds bulge above the flat shape
  $33.315383 \times 1 + \frac{0}{100}$ $= 33.315383 mm²$
4. Volume = area × length. Change meters to mm (× 1,000), then mm³ to cm³ (÷ 1,000)
  $\frac{33.315383 \times 4,000}{1,000}$ $= 133.261532 cm³$
5. Weight = volume × density (how heavy each cm³ is), ÷ 1,000 to go from grams to kg
  $\frac{133.261532 \times 2.7}{1,000}$ $= 0.359806 kg$
6. Filler to buy = weld metal ÷ deposition efficiency, because some filler is lost to stubs, spatter and slag
  $\frac{0.359806}{\frac{95}{100}}$ $= 0.378743 kg$
10. Weld metal deposited: 2.742 lb
1. A square groove is a plain rectangle: root gap × thickness
  $0.25 \times 0.25$ $= 0.0625 in²$
2. Add 15% for reinforcement, since real welds bulge above the flat shape
  $0.0625 \times 1 + \frac{15}{100}$ $= 0.071875 in²$
3. Volume = area × length. Change feet to inches first (× 12)
  $0.071875 \times 132$ $= 9.4875 in³$
4. Change the density into pounds per cubic inch, to match the inch measurements
  $\frac{8 \times 16.387064}{453.59237}$ $= 0.289018 lb/in³$
5. Weight = volume × density
  $9.4875 \times 0.289018$ $= 2.742061 lb$
6. Filler to buy = weld metal ÷ deposition efficiency, because some filler is lost to stubs, spatter and slag
  $\frac{2.742061}{\frac{95}{100}}$ $= 2.88638 lb$