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Stress and Strain worksheet

Grades 9–10 · CCSS HSA.CED.A.4, HSN.Q.A.1

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

1.
Problem 1.
- Find: Elongation under a load (from E)
- Axial force (F): 5 kN
- Cross-section: Round bar (diameter)
- Diameter: 12 mm
- Original length (L₀): 2 m
- Material (Young's modulus E): Concrete: 30 GPa (typical)
- Show stress in: Millions of psi (Msi)
- Show elongation in: Inches (in)
Find: Elongation (ΔL)
Answer
2.
Problem 2.
- Find: Stress, strain and Young's modulus (from a test)
- Axial force (F): 1 kN
- Cross-section: Enter the area
- Cross-sectional area (A): 60 mm²
- Original length (L₀): 0.5 m
- Measured change in length (ΔL): 0.62 mm
- Show stress in: Gigapascals (GPa)
- Show modulus in: Gigapascals (GPa)
Find: Young's modulus (E)
Answer
3.
Problem 3.
- Find: Stress, strain and Young's modulus (from a test)
- Axial force (F): 5 kN
- Cross-section: Round bar (diameter)
- Diameter: 8 mm
- Original length (L₀): 2 m
- Measured change in length (ΔL): 0.65 mm
- Show stress in: Pounds per square inch (psi)
- Show modulus in: Kips per square inch (ksi)
Find: Young's modulus (E)
Answer
4.
Problem 4.
- Find: Stress, strain and Young's modulus (from a test)
- Axial force (F): 20 kN
- Cross-section: Enter the area
- Cross-sectional area (A): 62 mm²
- Original length (L₀): 0.5 m
- Measured change in length (ΔL): 0.625 mm
- Show stress in: Millions of psi (Msi)
- Show modulus in: Kilopascals (kPa)
Find: Young's modulus (E)
Answer
5.
Problem 5.
- Find: Stress, strain and Young's modulus (from a test)
- Axial force (F): 50 kN
- Cross-section: Enter the area
- Cross-sectional area (A): 75 mm²
- Original length (L₀): 3 m
- Measured change in length (ΔL): 0.9 mm
- Show stress in: Millions of psi (Msi)
- Show modulus in: Kilopascals (kPa)
Find: Young's modulus (E)
Answer
6.
Problem 6.
- Find: Elongation under a load (from E)
- Axial force (F): 50 kN
- Cross-section: Round bar (diameter)
- Diameter: 12 mm
- Original length (L₀): 3 m
- Material (Young's modulus E): Steel: 200 GPa (typical)
- Show stress in: Millions of psi (Msi)
- Show elongation in: Micrometers (µm)
Find: Elongation (ΔL)
Answer
7.
Problem 7.
- Find: Elongation under a load (from E)
- Axial force (F): 50 kN
- Cross-section: Round bar (diameter)
- Diameter: 6 mm
- Original length (L₀): 1 m
- Material (Young's modulus E): Wood (along the grain): 13 GPa (typical)
- Show stress in: Millions of psi (Msi)
- Show elongation in: Micrometers (µm)
Find: Elongation (ΔL)
Answer
8.
Problem 8.
- Find: Stress, strain and Young's modulus (from a test)
- Axial force (F): 10 kN
- Cross-section: Round bar (diameter)
- Diameter: 12 mm
- Original length (L₀): 0.5 m
- Measured change in length (ΔL): 0.74 mm
- Show stress in: Megapascals (N/mm²) (MPa)
- Show modulus in: Pounds per square inch (psi)
Find: Young's modulus (E)
Answer
9.
Problem 9.
- Find: Stress, strain and Young's modulus (from a test)
- Axial force (F): 20 kN
- Cross-section: Enter the area
- Cross-sectional area (A): 66 mm²
- Original length (L₀): 0.5 m
- Measured change in length (ΔL): 0.66 mm
- Show stress in: Megapascals (N/mm²) (MPa)
- Show modulus in: Kilopascals (kPa)
Find: Young's modulus (E)
Answer
10.
Problem 10.
- Find: Elongation under a load (from E)
- Axial force (F): 50 kN
- Cross-section: Enter the area
- Cross-sectional area (A): 90 mm²
- Original length (L₀): 0.5 m
- Material (Young's modulus E): Aluminum: 70 GPa (typical)
- Show stress in: Kips per square inch (ksi)
- Show elongation in: Micrometers (µm)
Find: Elongation (ΔL)
Answer

Answer key: Stress and Strain worksheet

1. Elongation (ΔL): 0.11604 in
1. Change the force to N, the standard unit the formula uses (1 kN = 1,000 N)
  $5 kN \times 1,000$ $= 5,000 N$
2. Change the diameter to m, the standard unit the formula uses (1 mm = 0.001 m)
  $12 mm \times 0.001$ $= 0.012 m$
3. Find the area of the bar's round end (A) from its diameter (d), because stress spreads over that area
  $\frac{3.14159265 \times {0.012}^{2}}{4}$ $= 0.0001130973 m²$
4. Divide the force by the area to get the stress, the load each square meter of the bar carries
  $\frac{5,000}{0.0001130973}$ $= 44,209,706.414415 Pa$
5. Use the typical stiffness (Young's modulus) for concrete
  $30 GPa$ $= 30,000,000,000 Pa$
6. Divide the stress by the stiffness to get the strain, since a stiffer material stretches less
  $\frac{44,209,706.414415}{30,000,000,000}$ $= 0.0014736569$
7. Multiply the strain by the original length to get the total stretch, because a longer bar stretches more
  $0.0014736569 \times 2$ $= 0.0029473138 m$
8. Convert to in, the unit you chose to show
  $\frac{0.0029473138}{0.0254}$ $= 0.11603597 in$
2. Young's modulus (E): 13.44 GPa
1. Change the force to N, the standard unit the formula uses (1 kN = 1,000 N)
  $1 kN \times 1,000$ $= 1,000 N$
2. Change the area to m², the standard unit the formula uses (1 mm² = 0.000001 m²)
  $60 mm² \times 0.000001$ $= 0.00006 m²$
3. Divide the force by the area to get the stress, the load each square meter of the bar carries
  $\frac{1,000}{0.00006}$ $= 16,666,666.666667 Pa$
4. Change the change in length to m, the standard unit the formula uses (1 mm = 0.001 m)
  $0.62 mm \times 0.001$ $= 0.00062 m$
5. Divide the stretch by the original length to get the strain, the stretch per meter of bar
  $\frac{0.00062}{0.5}$ $= 0.00124$
6. Divide the stress by the strain to get Young's modulus, a measure of how stiff the material is
  $\frac{16,666,666.666667}{0.00124}$ $= 13,440,860,215.053764 Pa$
7. Convert to GPa, the unit you chose to show
  $\frac{13,440,860,215.053764}{1,000,000,000}$ $= 13.44086 GPa$
3. Young's modulus (E): 44,391.29 ksi
1. Change the force to N, the standard unit the formula uses (1 kN = 1,000 N)
  $5 kN \times 1,000$ $= 5,000 N$
2. Change the diameter to m, the standard unit the formula uses (1 mm = 0.001 m)
  $8 mm \times 0.001$ $= 0.008 m$
3. Find the area of the bar's round end (A) from its diameter (d), because stress spreads over that area
  $\frac{3.14159265 \times {0.008}^{2}}{4}$ $= 0.0000502655 m²$
4. Divide the force by the area to get the stress, the load each square meter of the bar carries
  $\frac{5,000}{0.0000502655}$ $= 99,471,839.432435 Pa$
5. Change the change in length to m, the standard unit the formula uses (1 mm = 0.001 m)
  $0.65 mm \times 0.001$ $= 0.00065 m$
6. Divide the stretch by the original length to get the strain, the stretch per meter of bar
  $\frac{0.00065}{2}$ $= 0.000325$
7. Divide the stress by the strain to get Young's modulus, a measure of how stiff the material is
  $\frac{99,471,839.432435}{0.000325}$ $= 306,067,198,253.64484 Pa$
8. Convert to ksi, the unit you chose to show
  $\frac{306,067,198,253.64484}{6,894,757.29316836}$ $= 44,391.294028 ksi$
4. Young's modulus (E): 258,064,516.13 kPa
1. Change the force to N, the standard unit the formula uses (1 kN = 1,000 N)
  $20 kN \times 1,000$ $= 20,000 N$
2. Change the area to m², the standard unit the formula uses (1 mm² = 0.000001 m²)
  $62 mm² \times 0.000001$ $= 0.000062 m²$
3. Divide the force by the area to get the stress, the load each square meter of the bar carries
  $\frac{20,000}{0.000062}$ $= 322,580,645.16129 Pa$
4. Change the change in length to m, the standard unit the formula uses (1 mm = 0.001 m)
  $0.625 mm \times 0.001$ $= 0.000625 m$
5. Divide the stretch by the original length to get the strain, the stretch per meter of bar
  $\frac{0.000625}{0.5}$ $= 0.00125$
6. Divide the stress by the strain to get Young's modulus, a measure of how stiff the material is
  $\frac{322,580,645.16129}{0.00125}$ $= 258,064,516,129.03223 Pa$
7. Convert to kPa, the unit you chose to show
  $\frac{258,064,516,129.03223}{1,000}$ $= 258,064,516.129032 kPa$
5. Young's modulus (E): 2,222,222,222.22 kPa
1. Change the force to N, the standard unit the formula uses (1 kN = 1,000 N)
  $50 kN \times 1,000$ $= 50,000 N$
2. Change the area to m², the standard unit the formula uses (1 mm² = 0.000001 m²)
  $75 mm² \times 0.000001$ $= 0.000075 m²$
3. Divide the force by the area to get the stress, the load each square meter of the bar carries
  $\frac{50,000}{0.000075}$ $= 666,666,666.666667 Pa$
4. Change the change in length to m, the standard unit the formula uses (1 mm = 0.001 m)
  $0.9 mm \times 0.001$ $= 0.0009 m$
5. Divide the stretch by the original length to get the strain, the stretch per meter of bar
  $\frac{0.0009}{3}$ $= 0.0003$
6. Divide the stress by the strain to get Young's modulus, a measure of how stiff the material is
  $\frac{666,666,666.666667}{0.0003}$ $= 2,222,222,222,222.222 Pa$
7. Convert to kPa, the unit you chose to show
  $\frac{2,222,222,222,222.222}{1,000}$ $= 2,222,222,222.222222 kPa$
6. Elongation (ΔL): 6,631.5 µm
1. Change the force to N, the standard unit the formula uses (1 kN = 1,000 N)
  $50 kN \times 1,000$ $= 50,000 N$
2. Change the diameter to m, the standard unit the formula uses (1 mm = 0.001 m)
  $12 mm \times 0.001$ $= 0.012 m$
3. Find the area of the bar's round end (A) from its diameter (d), because stress spreads over that area
  $\frac{3.14159265 \times {0.012}^{2}}{4}$ $= 0.0001130973 m²$
4. Divide the force by the area to get the stress, the load each square meter of the bar carries
  $\frac{50,000}{0.0001130973}$ $= 442,097,064.144154 Pa$
5. Use the typical stiffness (Young's modulus) for steel
  $200 GPa$ $= 200,000,000,000 Pa$
6. Divide the stress by the stiffness to get the strain, since a stiffer material stretches less
  $\frac{442,097,064.144154}{200,000,000,000}$ $= 0.0022104853$
7. Multiply the strain by the original length to get the total stretch, because a longer bar stretches more
  $0.0022104853 \times 3$ $= 0.006631456 m$
8. Convert to µm, the unit you chose to show
  $\frac{0.006631456}{0.000001}$ $= 6,631.45596216 µm$
7. Elongation (ΔL): 136,029.9 µm
1. Change the force to N, the standard unit the formula uses (1 kN = 1,000 N)
  $50 kN \times 1,000$ $= 50,000 N$
2. Change the diameter to m, the standard unit the formula uses (1 mm = 0.001 m)
  $6 mm \times 0.001$ $= 0.006 m$
3. Find the area of the bar's round end (A) from its diameter (d), because stress spreads over that area
  $\frac{3.14159265 \times {0.006}^{2}}{4}$ $= 0.0000282743 m²$
4. Divide the force by the area to get the stress, the load each square meter of the bar carries
  $\frac{50,000}{0.0000282743}$ $= 1,768,388,256.576615 Pa$
5. Use the typical stiffness (Young's modulus) for wood (along the grain)
  $13 GPa$ $= 13,000,000,000 Pa$
6. Divide the stress by the stiffness to get the strain, since a stiffer material stretches less
  $\frac{1,768,388,256.576615}{13,000,000,000}$ $= 0.1360298659$
7. Multiply the strain by the original length to get the total stretch, because a longer bar stretches more
  $0.1360298659 \times 1$ $= 0.1360298659 m$
8. Convert to µm, the unit you chose to show
  $\frac{0.1360298659}{0.000001}$ $= 136,029.86589051 µm$
8. Young's modulus (E): 8,664,967.3 psi
1. Change the force to N, the standard unit the formula uses (1 kN = 1,000 N)
  $10 kN \times 1,000$ $= 10,000 N$
2. Change the diameter to m, the standard unit the formula uses (1 mm = 0.001 m)
  $12 mm \times 0.001$ $= 0.012 m$
3. Find the area of the bar's round end (A) from its diameter (d), because stress spreads over that area
  $\frac{3.14159265 \times {0.012}^{2}}{4}$ $= 0.0001130973 m²$
4. Divide the force by the area to get the stress, the load each square meter of the bar carries
  $\frac{10,000}{0.0001130973}$ $= 88,419,412.828831 Pa$
5. Change the change in length to m, the standard unit the formula uses (1 mm = 0.001 m)
  $0.74 mm \times 0.001$ $= 0.00074 m$
6. Divide the stretch by the original length to get the strain, the stretch per meter of bar
  $\frac{0.00074}{0.5}$ $= 0.00148$
7. Divide the stress by the strain to get Young's modulus, a measure of how stiff the material is
  $\frac{88,419,412.828831}{0.00148}$ $= 59,742,846,505.96672 Pa$
8. Convert to psi, the unit you chose to show
  $\frac{59,742,846,505.96672}{6,894.757293168361}$ $= 8,664,967.302789 psi$
9. Young's modulus (E): 229,568,411.39 kPa
1. Change the force to N, the standard unit the formula uses (1 kN = 1,000 N)
  $20 kN \times 1,000$ $= 20,000 N$
2. Change the area to m², the standard unit the formula uses (1 mm² = 0.000001 m²)
  $66 mm² \times 0.000001$ $= 0.000066 m²$
3. Divide the force by the area to get the stress, the load each square meter of the bar carries
  $\frac{20,000}{0.000066}$ $= 303,030,303.030303 Pa$
4. Change the change in length to m, the standard unit the formula uses (1 mm = 0.001 m)
  $0.66 mm \times 0.001$ $= 0.00066 m$
5. Divide the stretch by the original length to get the strain, the stretch per meter of bar
  $\frac{0.00066}{0.5}$ $= 0.00132$
6. Divide the stress by the strain to get Young's modulus, a measure of how stiff the material is
  $\frac{303,030,303.030303}{0.00132}$ $= 229,568,411,386.59323 Pa$
7. Convert to kPa, the unit you chose to show
  $\frac{229,568,411,386.59323}{1,000}$ $= 229,568,411.386593 kPa$
10. Elongation (ΔL): 3,968.3 µm
1. Change the force to N, the standard unit the formula uses (1 kN = 1,000 N)
  $50 kN \times 1,000$ $= 50,000 N$
2. Change the area to m², the standard unit the formula uses (1 mm² = 0.000001 m²)
  $90 mm² \times 0.000001$ $= 0.00009 m²$
3. Divide the force by the area to get the stress, the load each square meter of the bar carries
  $\frac{50,000}{0.00009}$ $= 555,555,555.555556 Pa$
4. Use the typical stiffness (Young's modulus) for aluminum
  $70 GPa$ $= 70,000,000,000 Pa$
5. Divide the stress by the stiffness to get the strain, since a stiffer material stretches less
  $\frac{555,555,555.555556}{70,000,000,000}$ $= 0.0079365079$
6. Multiply the strain by the original length to get the total stretch, because a longer bar stretches more
  $0.0079365079 \times 0.5$ $= 0.003968254 m$
7. Convert to µm, the unit you chose to show
  $\frac{0.003968254}{0.000001}$ $= 3,968.25396825 µm$