Exam Monday April 14th

# Matplotlib compatibility patch for Pyodide
import matplotlib
if not hasattr(matplotlib.RcParams, "_get"):
    matplotlib.RcParams._get = dict.get

Exam Monday April 14th#

Today you’ll make the third exam assignment covering Statically indeterminate structures including its prerequisites and/or the third exam assignment covering Continuum mechanics including its prerequisites and/or the second exam assignment covering Buckling including its prerequisites. For more information about the exam see the assessment information in course information

Exam assignment 3 Statically indeterminate structures#

Your own submission and its grading can be found here: ANS exam assignment Statically indeterminate structures 3. The exam assignment was provided as follows:

Given is the structure as shown in the figure below.

../../_images/SO.svg

Calculate the normal force- and bending moment distribution using a force-, displacement- or hybrid- (‘hoekveranderingsvergelijkingen’ with moveable nodes) method.

Solution

Convert the structure into a statically determinate structure with a displacement or force condition.

For example when using the force method:

../../_images/SB.svg

Using forget-me-nots, the displacement in A can be described as:

  • \(w_\text{A} = \cfrac{36}{25} - {24}{625} A_v \)

  • \(u_\text{A} = 15 - \cfrac{3}{10} A_h\)

With the requirements of no displacement, this results in:

  • \(A_v = -37.5 \text{ kN}\)

  • \(A_h = -50 \text{ kN}\)

This gives the normal force and bending moment distribution:

../../_images/N-lijn.svg
../../_images/M-lijn.svg

Exam assignment 3 Continuum mechanics#

Your own submission and its grading will be available on ANS exam assignment Continuum mechanics 3 after the exam.

Given is the following structure:

../../_images/structure12.svg

  • Find the stresses alongside all edges of the stress shown element from the point E in the cross-section just to the left of D in element CD.

  • Evaluate whether the stress element fails according to Von Mises’ failure criterium.

Solution

In the cross-section, the bending moment is \(0\). The normal force is \(54 \text{ kN}\) and the shear force \(72 \text{ kN}\)

Relevant cross-sectional properties are:

  • \(A = 10800 \text{ mm}^2\)

  • From top fibre downwards, the normal force centre is at \(200 \text{ mm}\)

  • \(I_{zz} = 108043200 \text{ mm}^4\)

This gives in E:

  • \(\sigma = 5 \text{ MPa}\)

  • \(\tau \approx 10 \text{ MPa}\)

The stresses on the stress element can be calculated using the transformation formulas or using Mohr

../../_images/Mohr.svg

This leads to:

  • \(\sigma_1 = 12.8 \text{ MPa}\)

  • \(\sigma_2 = -7.80 \text{ MPa}\)

and the following stresses on the stress-element:

../../_images/stress_state.svg

The required yield stress results according to Von Mises is: \(\sqrt{\cfrac{1}{2}\left( \left(12.8+7.8 \right)^2 + 12.8^2 + \left(-7.8\right)^2 \right) } = 12.7 \text{ MPa}\), so the point does not fail.

Exam assignment 2 Buckling#

The exam assignment was provided as shown here

Solution

The solution is shown here

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