A bar with a free length of 160 mm and a cross section of 25 mm2 is pulled with a force of 500 Newtons.  The bar extends one-tenth of a millimeter.  What is the axial stress applied to the bar in MPa (Mega Pascal)?




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.

A bar with a free length of 160 mm and a cross section of 25 mm² is pulled with a force of 500 Newtons. The bar extends one-tenth of a millimeter. What is the axial stress applied to the bar in MPa (Mega Pascal)?

Calculate the axial stress applied to a bar with a length of 160 mm, a cross-sectional area of 25 mm², and a force of 500 N. Learn to apply the stress formula σ = F / A and understand the principles of material deformation. Suitable for Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation