Questions: A 2.2 -m-long steel rod must not stretch more than 1.2 mm when it is subjected to a 5.0-kN tension force. It is given that E= 200 GPa. Determine the smallest diameter rod that should be used. The smallest diameter rod that should be used is mm.

Solution Steps

We need to find the smallest diameter of a steel rod that will not stretch more than 1.2 mm under a tension force of 5.0 kN. The length of the rod is 2.2 m, and the modulus of elasticity \( E \) is 200 GPa.

The formula for the elongation \(\Delta L\) of a rod under tension is given by:

\[ \Delta L = \frac{F \cdot L}{A \cdot E} \]

where:

• \( F \) is the force applied (5.0 kN = 5000 N),
• \( L \) is the original length of the rod (2.2 m),
• \( A \) is the cross-sectional area,
• \( E \) is the modulus of elasticity (200 GPa = \(200 \times 10^9 \, \text{Pa}\)).

Rearrange the formula to solve for the cross-sectional area \( A \):

\[ A = \frac{F \cdot L}{\Delta L \cdot E} \]

Substitute the given values:

\[ A = \frac{5000 \, \text{N} \times 2.2 \, \text{m}}{0.0012 \, \text{m} \times 200 \times 10^9 \, \text{Pa}} \]

Calculate \( A \):

\[ A = \frac{11000}{240000000} = 4.5833 \times 10^{-5} \, \text{m}^2 \]

The cross-sectional area \( A \) of a circular rod is given by:

\[ A = \frac{\pi d^2}{4} \]

Solve for the diameter \( d \):

\[ d^2 = \frac{4A}{\pi} \]

\[ d = \sqrt{\frac{4 \times 4.5833 \times 10^{-5}}{\pi}} \]

Calculate \( d \):

\[ d = \sqrt{\frac{1.8333 \times 10^{-4}}{\pi}} \approx 0.0153 \, \text{m} = 15.3 \, \text{mm} \]

Final Answer