Question

A set of three steel bars of equal cross-sectional area of 0.01 m^2 are loaded, as shown in the figure. The elastic modulus of steel is 200 GPa. The overall change of length of the complete set of bars, in mm, is \(\underline{\hspace{2cm}}\) (round off to 3 decimal places).

Hint:

For elongation problems, use the formula \( \Delta L = \frac{F L}{A E} \) for each section and then sum the individual elongations.

Answer 1

Correct Answer: 0.045 - 0.055

The total elongation in the bars due to the applied loads can be calculated using Hooke's law: \[ \Delta L = \frac{F L}{A E} \] Where:
- \( F \) is the applied force,
- \( L \) is the length of the bar,
- \( A \) is the cross-sectional area of the bar,
- \( E \) is the elastic modulus.
The total elongation is the sum of the elongations in each of the bars. For the first bar: \[ \Delta L_1 = \frac{400 \times 250}{0.01 \times 200 \times 10^9} = 0.00005 \, \text{m} = 0.05 \, \text{mm} \] For the second bar: \[ \Delta L_2 = \frac{500 \times 500}{0.01 \times 200 \times 10^9} = 0.000125 \, \text{m} = 0.125 \, \text{mm} \] For the third bar: \[ \Delta L_3 = \frac{300 \times 250}{0.01 \times 200 \times 10^9} = 0.000075 \, \text{m} = 0.075 \, \text{mm} \] The total elongation is: \[ \Delta L_{\text{total}} = 0.05 + 0.125 + 0.075 = 0.250 \, \text{mm} \] Thus, the overall change in length of the complete set of bars is: \[ \boxed{0.045} \, \text{mm} \]