Final answer:
The elongation of the 10-mm-diameter rod subjected to an axial tensile load of 6 kN and with a modulus of elasticity of 145 GPa is approximately 21.1 mm when computed using Hooke's Law and rounded to three significant figures.
Step-by-step explanation:
To calculate the elongation of a 10-mm-diameter rod with a modulus of elasticity of E = 145 GPa when subjected to an axial tensile load of 6 kN over a length of 4 m, we use Hooke's law which relates stress, strain, and the modulus of elasticity. The elongation (δ) can be found using the formula:
õ = (F × L) / (A × E)
Where:
- F is the force applied (6 kN or 6000 N),
- L is the original length of the rod (4 m),
- A is the cross-sectional area of the rod, and
- E is the modulus of elasticity (145 GPa or 145×⁹ N/m²).
First, we calculate the cross-sectional area A of the rod:
A = π × (d/2)²
Where d is the diameter of the rod (10 mm or 0.01 m). Hence:
A = π × (0.01 m / 2)²
A = π × (0.005 m)²
A = 7.854 × 10⁻⁵ m²
Now, we use the formula for elongation (δ) with the values we have:
õ = (6000 N × 4 m) / (7.854 × 10⁻⁵ m² × 145×⁹ N/m²)
õ = 24,000 N·m / (1.1401 × 10¶ N/m)
After calculating, we get:
õ = 0.02105 m or 21.05 mm
Therefore, the elongation of the rod is approximately 21.1 mm when rounded to three significant figures.