Lesson November 11th

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

Lesson November 11th#

Today we’ll discuss the results of the first exam assignment on Statically indeterminate structures.

Results first exam assignment on Statically indeterminate structures#

Approaches taken during the exam assignment:

  • Mostly applied displacement method with displacements at \(\text{D}\)

  • One person applied force method for horizontal support reactions at \(\text{C}\), another person for \(N_\text{BD}\). Both of them abandoned that a

  • Not explicitly clear which statically determinate structure is solved for

  • Displacement of \(30 \text{ mm}\) used incorrectly as full total elongation of bar \(\text{BD}\)

Demonstration transforming stresses and Circle of Mohr#

Given the following structure and cross section.

../../_images/structure1.svg

We’ll find the maximum stresses in point \(\text{E}\) in cross-section \(\text{A}\) and its direction.

Internal forces#

First, let’s find the internal forces:

../../_images/Mline.svg
../../_images/Vline.svg

At cross-section \(\text{A}\) this gives a moment of \(-14.6 \text{ kNm}\) and shear force of \(+16 \text{ kN}\).

Cross-sectional properties#

For this thin-walled cross-section, the second moment of area of the cross-section can be calculated with:

\[ I_\text{zz} = 2 \cdot \cfrac{1}{12} \cdot 4 \cdot 300 ^3 + 2 \cdot \left(\cfrac{1}{12} \cdot 100 \cdot 12^3 + 12 \cdot 100 \cdot 150^2 \right) = 72.0288 \cdot 10^6 \text{ mm}^4\]

Normal and shear stresses#

The normal stresses can be calculated as:

\[\sigma_\text{E} = \cfrac{-14600 \cdot -0.075}{72.0288 \cdot 10^{-6}} \approx +15 \cdot 10^6 \text{ Pa}\]
\[\sigma_\text{max} = \cfrac{-14600 \cdot -0.150}{72.0288 \cdot 10^{-6}} \approx +30 \cdot 10^6 \text{ Pa}\]

Leading to the following diagram:

../../_images/normall_stress.svg

The shear forces can be calculated as:

\[\sigma_\text{xm,max flange} = - \cfrac{16000 \cdot 12 \cdot 100 \cdot 150}{2 \cdot 12 \cdot 72.0288 \cdot 10^{-6}} \approx 1.7 \cdot 10^6 \text{ Pa}\]
\[\sigma_\text{xm,min web} = - \cfrac{16000 \cdot 12 \cdot 100 \cdot 150}{2 \cdot 4 \cdot 72.0288 \cdot 10^{-6}} \approx 5.0 \cdot 10^6 \text{ Pa}\]
\[\sigma_\text{xm,E} = - \cfrac{16000 \cdot \left(12 \cdot 100 \cdot 150 + 2\cdot 4\cdot 75 \cdot 112.5 \right)}{2 \cdot 4 \cdot 72.0288 \cdot 10^{-6}} \approx 6.9 \cdot 10^6 \text{ Pa}\]
\[\sigma_\text{xm,max web} = - \cfrac{16000 \cdot \left(12 \cdot 100 \cdot 150 + 2\cdot 4\cdot 150 \cdot 75 \right)}{2 \cdot 4 \cdot 72.0288 \cdot 10^{-6}} \approx 7.5 \cdot 10^6 \text{ Pa}\]

Leading to the following diagram:

../../_images/shear_stresses.svg

The stress state for a rectangular element at point \(\text{E}\) is therefore:

../../_images/element.svg

Find maximum stress#

As the stress can be represented as a tensor. Therefore, a \(x,y\)-coordinate system is introduced:

../../_images/element2.svg

Leading to the tensor \(\sigma\):

\[\begin{split}\sigma = \left[ \begin{array}{} {15}&{-6.9}\\ {-6.9}&{0} \end{array} \right]{\text{ MPa}}\end{split}\]

The maximum stress can be found by applying the transformation rules:

\[\sigma_1 = \frac{1}{2} \left(15 + 0\right) + \sqrt{\left(\frac{1}{2}\left(15-0\right)\right)^2 + \left(-6.9\right)^2} \approx 17 \text{ MPa}\]

With a corresponding angle \(\alpha\):

\[\alpha = \frac{1}{2} \arctan\left(\cfrac{-6.9}{\frac{1}{2}\left(15-0\right)}\right) \approx 21^\text{o}\]

This can also be found using a circle of Mohr:

First, the two points \(\left(15,-6.9\right)\) and \(\left(0,-6/9\right)\) are added:

../../_images/mohr1.svg

Then, Mohr’s circle can be drawn

../../_images/mohr2.svg

Given the same maximum of \(\sigma_1 \approx 17 \text{ MPa}\). Finally, the corresponding rotation can also be found:

../../_images/mohr3.svg

Which gives the same \(\alpha \approx 31^\text{o}\).

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