Skip to main content Skip to page footer

Exercise F-1.1.4

Practice Exercise in Technical Mechanics 2, Mechanics of Materials

Topic: Uniaxial Stress State.

Practice Exercise F-1.1.4

Uniaxial Stress State: Normal and Shear Stress at an Arbitrary Section Angle

Problem Statement

A clamped beam with a square cross-section (side length \(d=20~\mathrm{mm}\)) is subjected to a tensile force \(F=10~\mathrm{kN}\) along the beam axis.

The illustration shows a beam with a square cross-section, fixed on the left side. In the drawing, three sectional planes a, b, and c are indicated. Sectional plane a is perpendicular to the beam axis. Sectional plane b runs from the upper left to the lower right, with the given angle Beta describing the smaller angle at the top. Sectional plane c runs from the lower left to the upper right, with the given angle Gamma describing the smaller angle at the bottom. The side length of the cross-section is denoted as d. A tensile force F acts at the right end of the beam at the center of the cross-section.
Fig. 1: Clamped beam

Determine the average normal stress and the average shear stress...

  1. ...acting in cross-sectional plane a.
  2. ...acting in cross-sectional plane b (\(\beta = 50°\)).
  3. ...acting in cross-sectional plane c (\(\gamma = 40°\)).
Short Solution
  1. Determine the average normal stress and the average shear stress acting in cross-sectional plane a.
$$ \begin{aligned} \sigma_a &= 25~\dfrac{\mathrm{N}}{\mathrm{mm}^2 } = 25~\mathrm{MPa}\\ \tau_a &= 0~\dfrac{\mathrm{N}}{\mathrm{mm}^2 } = 0~\mathrm{MPa} \end{aligned} $$
  1. Determine the average normal stress and the average shear stress acting in cross-sectional plane b (\(\beta = 50°\)).
$$ \begin{aligned} \sigma_b &= 14,671~\dfrac{\mathrm{N}}{\mathrm{mm}^2 } = 14,671~\mathrm{MPa}~(\nearrow)\\ \tau_b &= -12,310~\dfrac{\mathrm{N}}{\mathrm{mm}^2 } = -12,310~\mathrm{MPa}~(\nwarrow) \end{aligned} $$
  1. Determine the average normal stress and the average shear stress acting in cross-sectional plane c (\(\gamma = 40°\)).
$$ \begin{aligned} \sigma_c &= 10,329~\dfrac{\mathrm{N}}{\mathrm{mm}^2 } = 10,329~\mathrm{MPa}~(\searrow)\\ \tau_c &= 12,310~\dfrac{\mathrm{N}}{\mathrm{mm}^2 } = 12,310~\mathrm{MPa}~(\nearrow) \end{aligned} $$
Comprehensive Solution

Preliminary Considerations

Ready to crack the stress state for different cutting angles in a beam? Cool stuff! Don't worry, it's a piece of cake.

First, take a deep breath and relax. The beam is loaded by a force \(F\) acting along its axis. This means we're dealing with a uniaxial stress state – easy peasy, right?

Now, it's time to dig out your formulas.

You'll need the following:

  • For a perpendicular cut to the beam axis:
    $$ \begin{aligned} \sigma_x = \dfrac{N_x}{A} \end{aligned} $$

    (1.3)

  • For an arbitrary cutting angle:
    $$ \begin{aligned} \sigma_\xi &= \dfrac{\sigma_x}{2}\bigl(1+\cos(2\varphi)\bigr) \end{aligned} $$

    (1.4)

    $$ \begin{aligned} \tau_{\xi\eta} &= -\dfrac{\sigma_x}{2}\bigl(\sin(2\varphi)\bigr) \end{aligned} $$

    (1.5xy)

    $$ \begin{aligned} \tau_{\xi\eta} &= \dfrac{\sigma_x}{2}\bigl(\sin(2\varphi)\bigr) \end{aligned} $$

    (1.5xz)

Before we dive into the calculations, let's clarify a few important things:
  • What type of loading is present? (Tensile, compressive, ...)
  • What is the magnitude of the force?
  • What is the cross-sectional area?
  • At what angle is the cut made?
That's it for the theory!

In the next step, we'll move on to the practical application.

So, get to work and good luck!
  1. Determine the average normal stress and the average shear stress acting in the section plane a.

We are looking for the normal stress and shear stress in a perpendicular section to the beam axis. The general formula for normal stress is:

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

(1.3)

The shear stress must be \(0\) since the external force \(F\) in the perpendicular section can only cause normal force and not shear force.

Step 1: Determine the normal force

To calculate the normal force, we cut the beam at section a and represent the external force \(F\) and the internal normal force \(N\). As a reminder: On the positive, left side of the section, we represent \(N\) in the positive direction (towards the positive x-axis), while on the negative, right side of the section, we represent \(N\) in the negative direction (opposite to the direction of the positive x-axis).

This graphic shows the beam cut at section a with the normal force reaction.
Fig. 2: Determine the normal force at section a

To save ourselves from calculating the support reactions (even though it's obvious that only one support reaction on the beam axis in the opposite direction of the x-axis and of magnitude \(F\) must act), we set up the horizontal equilibrium condition for the right section. This allows us to determine the normal force \(N\):

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

Continue with TechMechAcademy+

Everything. Always. Everywhere.

With TechMechAcademy+ full access to all content.


For Beginners: TechMechAcademy+. Full access to all content for one day.

Overview of the benefits:

  • Most cost-effective offer.
  • Ideal if you need access to already identified content for a short period.
  • Unlimited access to all existing and newly created content throughout the entire premium membership.
  • Guaranteed premium membership for 24 hours. Access automatically ends at 0:00 CET (Central European Time) on the following day. No cancellation necessary.

€3.99

TechMechAcademy+. Full access to all content for one week.

Overview of the benefits:

  • Cost-effective offer.
  • Ideal for short-term exam preparation with the content of TechMechAcademy.
  • Unrestricted entry to every piece of content, both existing and newly generated, across the entirety of the premium membership.
  • Guaranteed premium membership for one week. Access starts on the day of activation and ends automatically at 0:00 CET (Central European Time) on the following week's day. No cancellation necessary.

€9.99

Bestseller: TechMechAcademy+. Full access to all content for one month.

Overview of the benefits:

  • Benefit from the bestseller.
  • Ideal for effective exam preparation with the content of TechMechAcademy.
  • Complete access to all content, both existing and newly produced, under the umbrella of the premium membership.
  • Guaranteed premium membership for one month. Access starts on the day of activation and ends automatically at 0:00 CET (Central European Time) on the following month's day. No cancellation necessary.

€14.99

TechMechAcademy+. Full access to all content for three months.

Overview of the benefits:

  • Cost-effective due to extended duration.
  • Ideal for working alongside the semester with the content of TechMechAcademy.
  • Unrestricted admission to all content, both pre-existing and recently generated, within the scope of the premium membership.
  • Guaranteed premium membership for three months. Access starts on the day of activation and ends automatically at 0:00 CET (Central European Time) on the following day of the third month from the start date. No cancellation necessary.

€29.99

TechMechAcademy+. Full access to all content for six months.

Overview of the benefits:

  • Cost-effective due to extended duration.
  • Ideal for semester preparation, concurrent semester work, and/or post-semester work with the content of TechMechAcademy.
  • Unrestricted entry to a comprehensive collection of both existing and freshly generated content as part of the premium membership.
  • Guaranteed premium membership for six months. Access starts on the day of activation and ends automatically at 0:00 CET (Central European Time) on the following day of the sixth month from the start date. No cancellation necessary.

€49.99

Are you already a TechMechAcademy+ premium member? Then please log in here to enjoy full access to all content.