To determine the total elongation of the steel rod, we need to use Hooke's Law for elongation due to a vertical load, which is given by the formula:

$\Delta L = \frac{P L}{A E}$

Where:

• $\Delta L$ = elongation of the rod (in meters)
• $P$ = applied force (in newtons)
• $L$ = original length of the rod (in meters)
• $A$ = cross-sectional area of the rod (in square meters)
• $E$ = Young's Modulus of the material (in pascals or N/m²)

Given:

• Cross-sectional area $A = 400 \, \text{mm}^2 = 400 \times 10^{-6} \, \text{m}^2$
• Length $L = 3 \, \text{m}$
• Young's Modulus $E = 200 \, \text{GPa} = 200 \times 10^9 \, \text{N/m}^2$
• Force $P = 15 \, \text{kN} = 15 \times 10^3 \, \text{N}$

Now, substitute the given values into the formula:

$\Delta L = \frac{(15 \times 10^3) \times 3}{(400 \times 10^{-6}) \times (200 \times 10^9)}$

Let’s calculate this:

$\Delta L = \frac{45,000}{(400 \times 10^{-6}) \times (200 \times 10^9)} = \frac{45,000}{80 \times 10^3} = 0.5625 \, \text{mm}$

Total elongation of the rod = 0.5625 mm

Would you like more details or further clarification on any step?

Here are some related questions you might find useful:

1. How does Young’s Modulus affect the elongation of materials?
2. What would happen to the elongation if the cross-sectional area were doubled?
3. How does the material’s density affect its elongation under the same force?
4. What if the rod’s length were doubled, but all other factors remained the same?
5. What is the relationship between stress and strain in the context of material deformation?

Tip: Ensure that the units are consistent when using formulas, especially converting from mm² to m² or kN to N.