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Mechanics of Materials

Stress State: Dive into the World of Forces and Stresses!

Are you ready for an exciting journey into the world of physics? Then buckle up and discover the secrets of the stress state with us!

What is stress? Imagine you are building a giant Lego structure. The individual blocks push and pull against each other - that's exactly what stress is! In this course, you will learn how to calculate and understand these forces.

Stress components: Break down stress into its individual parts and discover how they interact. Just as a puzzle consists of many pieces, stress is also made up of different components.

Calculation: Crack the code of stress calculation! With a few clever formulas and tools, you can determine the forces in any component.

Transformation: Stresses change depending on the perspective. Learn how to transform them into different cutting planes and thus make the whole story of the load in the component visible.

Maximum stresses: Where does the greatest danger lurk? Find out where the stresses are highest in the component and how you can minimize them.

Mohr's circle of stress: This ingenious tool helps you to visualize stresses and to grasp important information at a glance.

Discover the fascination of the stress state! In this course you will not only learn dry knowledge, but also immerse yourself in the world of engineering. With good explanations and exciting application examples, the stress state becomes child's play.

Together we are strong! We will accompany you on your journey and help you to understand the complex concepts of the stress state. With our support you will master every challenge and become an expert for stable constructions.

So what are you waiting for? Start your journey into the world of stress now!

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Practice Exercises
Table of Contents

2.1 Normal Stress in Uniaxial Loading

The Stress-Free Guide

Ditch the boring intro and dive straight into the action! Uniaxial loading? Think of a poor rod being stretched or squished by a mean force. That's it!

What happens?

The poor rod gets deformed, and that's what we care about. It reveals how much stress is lurking inside the material.

Uniaxial stress state due to tensile force.
Fig. 1.2.1: Uniaxial state of stress

Meet our hero: Normal stress. It describes the force per unit area acting on the poor rod.

  • Positive force? It's tensile stress, pulling the rod like a taffy.
  • Negative force? It's compressive stress, pushing the rod like a bully.
The formula for normal stress is a piece of cake:
$$ \begin{aligned} \sigma_x = \dfrac{N_x}{A} = \dfrac{F}{A} \end{aligned} $$

(1.3)

Symbol Name Description Unit
\(\sigma_x\) Sigma Normal stress in the x-direction N/mm2 = MPa
\(N_x\) Normal force in the x-direction N
\(F\) Externally applied tensile or compressive force N
\(A\) Area mm2
But wait, there's more!
  • The force must be perpendicular to the area.
  • The area must be the cross-sectional area of the rod.
Example time!

Imagine a 10 cm2 rod with a 100 N load.

$$ \begin{flalign} \text{Stress} = \dfrac{100~\mathrm{N}}{10~\mathrm{cm^2}} = 10~\frac{\mathrm{N}}{\mathrm{cm^2}} \end{flalign} $$
What does that mean?

Each square centimeter of the rod is under 10 N of force. That's quite a lot, huh?

Where's the shear stress?

Relax, it's not here. The stress vector \(\vec{t}\) is perpendicular to the area, so the shear stress component is zero. No need to worry about it for now.

Hanging Cone - Stress times variable

Things are about to get even more exciting! What happens when the force or the cross-sectional area isn't constant?

The stress isn't constant either! Take a hanging cone for example.

Suspended conical shape with dimensions.
Fig. 1.2.2: Suspended conical shape
Quiz time:

Where is the stress in the hanging cone smallest?

  1. At the pointy tip
  2. At the big, round top
  3. Neither

The answer is 1, of course! The stress is smallest at the pointy tip.

Why?

It's not because the cross-sectional area is smallest there. It's because there's no force acting at the tip! The cone hangs from the top, so the entire weight pulls down on the top area. The further down you go, the smaller the pulling force. And where there's no force, there's no stress!

In our cone, the normal force and the diameter change depending on the location. So the stress also changes:

$$ \begin{aligned} \sigma(x) = \dfrac{N(x)}{A(x)} \end{aligned} $$

(1.3b)

Conclusion:

Uniaxial loading is a breeze! With a bit of math and imagination, you can calculate the stress in any material.

Now what?

Study the example and test your knowledge! Calculate the stress in a rod of your choice. Have fun!

Example 1.1: Tapered Rod

A conical rod with a circular cross-section and length \(l = 250~\mathrm{mm}\) is loaded as depicted in Figure E1 by a compressive force \(F=10~\mathrm{kN}\) along the axis of the rod. The diameter at \(x=0\) is twice as large as the diameter at \(x=l\) with \(2d_0 = 150~\mathrm{mm}\) and \(d_0 = 75~\mathrm{mm}\), respectively.

Tapered rod subjected to compressive force.
Fig. E1: Tapered Rod
  1. What is the formula for calculating the normal stress \(\sigma\) at any location x in a section perpendicular to the rod's axis?
  2. What is the magnitude of the normal stress \(\sigma\) at the location \(x=200~\mathrm{mm}\) in a section perpendicular to the rod's axis?
Solution

Considerations

First question: What is the subject area here, actually? Can you justify it?

The rod is loaded by a force \(F\) whose line of action is the axis of the rod. This means: We have a uniaxial stress state.

Which formulas should we dig out to nail this problem?

We are looking for the normal stress in a section perpendicular to the rod's axis. The general formula is:

$$ \begin{aligned} \sigma_x = \dfrac{N_x}{A} \end{aligned} $$

(1.3)

The difficulty of this task lies in the fact that the cross-sectional area \(A\) of the rod is conical and therefore different for each position \(x\). Hence, \(A=A(x)\). Accordingly, we need formula 1.3b to solve this task:

$$ \begin{aligned} \sigma(x) = \dfrac{N(x)}{A(x)} \end{aligned} $$

(1.3b)

  1. What is the formula for calculating the normal stress \(\sigma\) at any location x in a section perpendicular to the rod's axis?

Step 1: Determine Normal Force

Imagine cutting the rod at any point \(x\). At the intersection, two forces act, opposite and equal in magnitude: the normal force N in duplicate. This is the only way this intersection can be in equilibrium.

As a reminder: On the positive, left side of the cut, we apply \(N\) in a positive direction (towards the positive x-axis), while on the negative, right side of the cut, we apply it in a negative direction (against the positive x-axis).

Cut conical rod with normal force at the cut.
Fig. E2: Determine Normal Force

Since the rod is in equilibrium (i.e., it does not move), the forces in the x-direction must be equal.

If we apply the horizontal equilibrium condition for, say, our left cut, we can determine the normal force \(N\):

$$ \begin{align} \tag{1} \rightarrow: F + N &= 0\\[7pt] \tag{2} N &= \underline{-F} \label{(1)} \end{align} $$

So: N = -F. This means that the normal force is negative because it counteracts the compressive force. It acts exactly opposite to how we have drawn it in Fig. E2.

Thus, we obtain a constant (independent of \(x\)), negative, internal normal force, i.e., we have a compressive force, and according to Equation 1.3/1.3b, a compressive stress. This must be the case, as the area is positive and negative force through positive area results in negative stress (= compressive stress).

How do we know that the normal force is constant over the entire rod?

Quite simply: At the beginning of the rod, an external force acts on our rod. Since there is no change in this force due to additional external forces until the end of the rod, the normal force remains constant over the entire rod.

Step 2: Determine Area as a Function of \(x\)

Since it is a conical rod, the cross-sectional area in the perpendicular cut to the axis of the rod is always circular. The formula for the area of a circle is:

$$ \begin{align} \tag{3} A = \dfrac{d^2 \cdot \pi}{4}=r^2 \cdot \pi \end{align} $$

As mentioned above, the difficulty of this task lies in the fact that the cross-sectional area \(A\) of the rod is conical and therefore varies for each position \(x\). When we examine the problem in a graph, we quickly realize that we need to represent the radius \(r\) as a function of the position \(x\) in order to solve this task.

Conical rod in side view.
Fig. E3: Sketch of the conical rod in side view, upper part

So we are looking for \(r(x)\) in order to describe the area as follows:

$$ \begin{align} \tag{4} A(x) = r(x)^2 \cdot \pi \end{align} $$

We can solve such a problem in at least two different ways. Either we apply the 2nd similarity theorem or we establish the functional equation of the line \(r(x)\).

Both approaches lead to the exact same result. The choice of which one to use depends on personal preference and familiarity. Therefore, let's take a closer look at both options.

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