Under a tensile load of 240 kN, what is the stress in a round bar that stretches 1.2 mm?

To determine the correct stress in the round bar under a tensile load, we start by recalling the definition of stress, which is calculated as the force applied divided by the cross-sectional area over which the force is distributed.

In this scenario, the tensile load is 240 kN, which translates to 240,000 N. Stress (σ) is defined mathematically as:

[

\sigma = \frac{F}{A}

]

where:

• ( F ) is the force applied (240,000 N),

• ( A ) is the cross-sectional area of the round bar.

To find the stress, we also need the diameter of the round bar to calculate its cross-sectional area (A). For a round bar, the area can be computed using the formula:

[

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

]

where ( d ) is the diameter of the bar. However, without additional information about the diameter, we can use properties of materials and general assumptions about stress and strain relationships. Given that we've specified the amount of extension (1.2 mm), material properties such as Young's modulus can be utilized if necessary to relate stress to strain.

From the previous calculations, if we had