Normal Stress and Strain in Axially Loaded Members

Hello! Welcome to the first lesson of our second module, Mechanics of Materials.

In our previous module on Statics, we mastered the analysis of external forces acting on rigid bodies to ensure they remained in equilibrium. As we concluded, our focus now shifts from the external to the internal. Real-world components are not perfectly rigid; they deform under load. These external forces create internal forces distributed within the material, which we call stress, and cause the material to deform, which we call strain.

Today, we will build the foundation for this new topic. Our learning outcome is to calculate normal stress and strain in axially loaded members. Understanding these concepts is the first and most critical step in analyzing the strength, performance, and reliability of any mechanical component, from a simple bolt to the complex struts in an aircraft's landing gear.

1. Defining Normal Stress

When a straight member, like a bar or a cable, is pulled or pushed along its axis, we say it is under axial load. This load creates internal forces that resist the pulling (tension) or pushing (compression). Stress is the measure of the intensity of these internal forces.

To get a clear conceptual picture, let's start with a short video.

As the video explained, normal stress is the internal force per unit area that acts perpendicular (or "normal") to a surface. We denote it with the Greek letter sigma ($\sigma$).

The fundamental formula for normal stress is:
$\sigma =\frac{P}{A}$
Where:

• $\sigma$ is the normal stress.
• $P$ is the internal axial force at the cross-section.
• $A$ is the cross-sectional area perpendicular to the force.

Sign Convention and Units

In engineering, we use a standard sign convention:

• Tensile stress (pulling the material apart) is positive (+).
• Compressive stress (squashing the material) is negative (-).

Since stress is force divided by area, its units are crucial. Your background in electronics engineering makes you very familiar with the importance of consistent units, and it's just as critical here.

• In the SI system, the standard unit is the Pascal (Pa), which is one Newton per square meter ($N/m^2$). Since this is a very small unit, we more commonly use megapascals ($1\text{ MPa}=10^6\text{ Pa}$) and gigapascals ($1\text{ GPa}=10^9\text{ Pa}$). A useful identity is that $1\text{ MPa}=1{\text{ N/mm}}^2$.
• In the U.S. Customary system, units are pounds per square inch (psi) or kips per square inch (ksi), where 1 kip = 1000 lbs.

Test your understanding!

Consider a solid circular rod with a diameter of 20 mm. It is being pulled by a tensile force of 50 kN. A second rod, with a square cross-section of 20 mm x 20 mm, is pulled with the same 50 kN force. Which rod experiences higher stress, and why?

Show answer

The circular rod experiences higher stress. Both rods have the same internal force $P=50\text{ kN}$. The key is the area, $A$.

• Area of circular rod: $A_{circle}=\pi r^2=\pi (10\text{ mm}{)}^2\approx 314{\text{ mm}}^2$
• Area of square rod: $A_{square}=(20\text{ mm}{)}^2=400{\text{ mm}}^2$

Since stress is inversely proportional to area ($\sigma =P/A$), the circular rod with the smaller cross-sectional area will have the higher stress.

2. Finding the Internal Force, $P$

A common point of confusion is identifying the correct value for $P$. The force $P$ in the stress formula is the internal resultant force acting on the cross-section you are analyzing. This is where our skills from Statics come into play. To find this internal force, we use the method of sections:

1. Make an imaginary "cut" through the member at the location of interest.
2. Draw a free-body diagram (FBD) of one of the two segments you've created.
3. The internal force $P$ is the force required to keep that segment in equilibrium.

For a bar with multiple forces, the internal force $P$ will change along its length. The diagram below shows a bar subjected to several forces and the resulting internal axial force diagram. This diagram is analogous to the shear force diagrams you've encountered before; it plots the value of the internal force $P$ at every point along the bar's axis.

3. Defining Normal Strain

When we apply a stress to a material, it deforms. Normal strain, denoted by the Greek letter epsilon ($ϵ$), is the measure of this deformation. It quantifies how much the member stretches or shortens relative to its original size.

Let's return to the video for a quick explanation.

As the video showed, the formula for average normal strain is:
$ϵ=\frac{\delta }{L_0}$
Where:

• $ϵ$ is the normal strain.
• $\delta$ (delta) is the total change in length (elongation).
• $L_0$ is the original, undeformed length of the member.

Strain is a dimensionless quantity because it is a ratio of length to length (e.g., m/m or in/in). It is often expressed as a percentage or in terms of "microstrain" ($\mu ϵ$), where $1\mu ϵ=1\times 10^{-6}$. The sign convention is the same as for stress: tensile strain (elongation) is positive (+), and compressive strain (shortening) is negative (-).

4. Worked Example: Stress Calculation in a Real System

Now, let's put these concepts together. The following video solves a problem that requires you to first use statics to find the forces in two cables and then use those forces to calculate the normal stress in each cable. This is a perfect example of how Statics and Mechanics of Materials are interconnected.

The example highlights the standard workflow:

1. Statics Analysis: Determine the internal force P at the desired location.
2. Geometry: Calculate the cross-sectional area A.
3. Stress Calculation: Apply the formula $\sigma =P/A$.

For more practice with the direct application of the formula, including rearranging it to find a required area (a common design task), the following resource is useful.

Conclusion

In this lesson, we have taken our first step into the world of Mechanics of Materials. We've moved beyond treating bodies as perfectly rigid and started to quantify the internal effects of external loads.

Key Takeaways:

• Normal Stress ($\sigma =P/A$) is a measure of the intensity of the internal force perpendicular to a cross-section. Tension is positive, compression is negative.
• The force $P$ is the internal axial force, found using the method of sections from Statics.
• Normal Strain ($ϵ=\delta /L_0$) is a measure of the intensity of deformation, or the fractional change in length. Elongation is positive, shortening is negative.
• The standard engineering workflow is to first solve the statics problem to find internal forces, then use those forces to calculate stress.

Next Lesson Preview:

So far, we have defined stress and strain as two separate concepts. However, for any given material, they are intrinsically linked. Applying a stress causes a strain. The nature of this relationship defines a material's mechanical properties, such as its stiffness and strength. In our next lesson, we will explore this crucial link by interpreting stress-strain curves and applying Hooke's Law, which mathematically connects stress and strain. This will unlock our ability to predict how much a component will deform under a given load.