Exam Tuesday December 10th

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

Exam Tuesday December 10th#

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

Exam assignment 2 Statically indeterminate structures#

Your own submission and its grading will be available on ANS exam assignment Statically indeterminate structures 2 after the exam.

Given is the following structure:

../../_images/structure7.svg

Calculate the displacement of \(\rm{E}\) and the moments in all the members 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/statically_determinate1.svg

Using forget-me-nots, this gives a rotation of \(\rm{B}\) of: \(\varphi_B = \cfrac{5 \cdot \left(- 4 \cdot A_v + 2.5 \cdot 64 - 4 \cdot 65\right)}{3 \cdot \cfrac{100}{3} \cdot 10^{3}} = - 2.0 \cdot 10^{-4} A_{v} - 5.0 \cdot 10^{-3} \left(↺\right)\)

Again using forget-me-nots, this give a vertical displacement in \(\rm{A}\) of: \(w_{v,A} = \varphi_B \cdot 5 - \cfrac{A_v \cdot \cfrac{4}{5} \cdot 5 ^ 3}{3 \cdot \cfrac{100}{3} \cdot 10^{3}} + \cfrac{F \cdot 2.5 ^ 3}{3 \cdot \cfrac{100}{3} \cdot 10^{3}} + \cfrac{F \cdot 2.5 ^ 2}{2 \cdot \cfrac{100}{3} \cdot 10^{3}} \cdot 2.5 = -2.0 \cdot 10^{-3} \cdot A_v \)

Using \(w_{v,A} = 0\) gives \(A_v = = 0 \ \rm{ kN}\)

Again using forget-me-nots for both rotation of \(\rm{B}\), bending of \(\rm{BE}\) and bending of \(\rm{CD}\) a displacement of \(8.9518 \cdot 10^{-2} \ \rm{ m}\) is found.

The bending moment diagram results in:

../../_images/mline.svg

Exam assignment 1 Continuum mechanics#

Your own submission and its grading will be available on ANS exam assignment Continuum mechanics 1 after the exam.

Given is the following structure:

../../_images/structure23.svg

Find the stresses alongside all edges of the shown element from cross-section \(\rm{A}\) point \(\rm{C}\), and determine the required yield stress according to the Von Mises failure criterium.

Solution

In \(\rm{C}\) the shear stress due to bending is \(0\).

The normal stress due to bending is: \(\sigma = \cfrac{4 \cdot 4000 \cdot \cfrac{1}{2} \cdot \cfrac{0.250}{\sqrt{\pi}}}{\pi \cdot \left(\cfrac{1}{2} \cdot \cfrac{0.250}{\sqrt{\pi}}\right)^3 \cdot 0.010} = 102.4 \cdot 10^{6} \ \rm{ Pa}\). Which is a tension stress.

The shear stress due to torsion is: \(\tau = \cfrac{12000 \cdot \cfrac{1}{2} \cdot \cfrac{0.250}{\sqrt{\pi}}}{2 \cdot \pi \cdot \left(\cfrac{1}{2} \cdot \cfrac{0.250}{\sqrt{\pi}}\right)^3 \cdot 0.010} = 38.4 \cdot 10^{6} \ \rm{ Pa}\). This shear stress works clockwise on a positive \(x\)-plane and anticlockwise on a negative \(x\)-plane (corresponding of the top plane of the shown specimen).

The stress at the bottom plane of the shown specimen can be found by rotating the right plane \(-\cfrac{1}{4}\pi \ \rm{ rad}\):

  • \(\sigma_{\bar x \bar x} = 102.4 \sin \left(-\cfrac{1}{4}\pi\right)^2 + 2 \cdot -38.4 \sin\left(-\cfrac{1}{4}\pi\right) \cos \left( -\cfrac{1}{4}\pi \right)^2= 89.6 \ \rm{ MPa} \)

  • \(\tau_{\bar x \bar y} = 102.4 \sin \left( -\cfrac{1}{4}\pi\right) \cos \left(-\cfrac{1}{4}\pi\right) -38.4 \cos \left(-\cfrac{1}{4}\pi\right)^2 +38.4 \sin \left(-\cfrac{1}{4}\pi \right)^2= -51.2 \ \rm{ MPa}\)

This can also be one using Mohr’s circle:

../../_images/mohr.svg

This results in:

../../_images/stresses3.svg

The principal stresses can be found using:

  • \(\sigma_1 = \cfrac{102.4}{2}+\sqrt{\left(\cfrac{102.4}{2}\right)^2+38.4^2} = 115.2 \ \rm{ MPa} \)

  • \(\sigma_2 = \cfrac{102.4}{2}-\sqrt{\left(\cfrac{102.4}{2}\right)^2+38.4^2} = -12.8 \ \rm{ MPa} \)

Now the Von Mises failure criterion can be evaluated: \(f_y > \sqrt{\cfrac{1}{2} \left(\left(115.2+12.8\right)^2+\left(-12.8-0\right)^2+\left(0-115.2\right)^2\right)} = 12\sqrt{91} \ \rm{ MPa} \approx 122 \ \rm{ MPa}\) or \(f_y > \sqrt{102.4^2+3\cdot \left(38.4\right)^2} = 12\sqrt{91} \ \rm{ MPa} \approx 122 \ \rm{ MPa}\) ​

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