The axial stress ($\sigma$) applied to a bar is given by the formula:

$\sigma = \frac{F}{A}$

where:

• $F$ is the applied force (in Newtons),
• $A$ is the cross-sectional area (in square millimeters).

Given:

• Force, $F = 500 \, \text{N}$,
• Cross-sectional area, $A = 25 \, \text{mm}^2$.

Now, applying the formula:

$\sigma = \frac{500}{25} = 20 \, \text{MPa}$

The axial stress applied to the bar is 20 MPa.

Would you like to explore further details or have any questions?

Here are 5 related questions for further practice:

1. How would the stress change if the force is doubled to 1000 N?
2. What would be the axial stress if the cross-sectional area were halved?
3. How is axial strain related to the stress applied to the bar?
4. Can you compute the Young's modulus of the material given the extension data?
5. How does axial stress differ from shear stress in material deformation?

Tip: Stress is always force divided by area, so changing either force or area directly affects the stress.