Lesson November 10th

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

Results first exam assignment on Statically indeterminate structures

Discussed based on frequently made mistakes as shown in Exam Friday November 7th

Demonstration transforming stresses

Given the following structure and cross section.

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

Internal forces

First, let’s find the internal forces:

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

Cross-sectional properties

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

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

Normal and shear stresses

The normal stresses can be calculated as:

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

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

Leading to the following diagram:

The shear forces can be calculated as:

\[\sigma_{\rm{xm,max,flange}} = - \cfrac{16000 \cdot 12 \cdot 100 \cdot -150}{2 \cdot 12 \cdot 72.0288 \cdot 10^{-6}} \approx 1.7 \cdot 10^6 \ \rm{ Pa}\]

\[\sigma_{\rm{xm,min,web}} = - \cfrac{16000 \cdot 12 \cdot 100 \cdot -150}{2 \cdot 4 \cdot 72.0288 \cdot 10^{-6}} \approx 5.0 \cdot 10^6 \ \rm{ Pa}\]

\[\sigma_{\rm{xm,E}} = - \cfrac{16000 \cdot \left(12 \cdot 100 \cdot -150 + 2\cdot 4\cdot 75 \cdot -112.5 \right)}{2 \cdot 4 \cdot 72.0288 \cdot 10^{-6}} \approx 6.9 \cdot 10^6 \ \rm{ Pa}\]

\[\sigma_{\rm{xm,max,web}} = - \cfrac{16000 \cdot \left(12 \cdot 100 \cdot -150 + 2\cdot 4\cdot 150 \cdot -75 \right)}{2 \cdot 4 \cdot 72.0288 \cdot 10^{-6}} \approx 7.5 \cdot 10^6 \ \rm{ Pa}\]

Leading to the following diagram:

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

Find maximum stress

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

Leading to the tensor \(\sigma\):

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

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

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

With a corresponding angle \(\alpha\):

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