Exam Friday November 8th

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Exam Friday November 8th#

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

Exam assignment#

Your own submission and its grading can be found here: ANS exam assignment Statically indeterminate structures 1. The exam assignment was provided as follows:

Given is the structure as shown in the figure below.

../../_images/structure.svg

Calculate the displacement of \(\text{D}\) and the normal forces in all the members using a force-, displacement- or hybrid- (‘hoekveranderingsvergelijkingen’ with moveable nodes) method.

Solution assignment 1

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

For example when using the force method:

../../_images/statically_determinate.svg

This gives:

  • \({N_{{\text{AD}}}} = - \cfrac{4}{5}{B_{\text{v}}}\)

  • \({N_{{\text{BD}}}} = + {B_{\text{v}}}\)

  • \({N_{{\text{CD}}}} = - \cfrac{3}{4}{B_{\text{v}}}\)

This gives elongations:

  • \(\Delta {L_{{\text{BD}}}} = \cfrac{{{B_{\text{v}}}}}{{7500}}\)

  • \(\Delta {L_{{\text{AD}}}} = - \cfrac{{{B_{\text{v}}}}}{{4800}}\)

  • \(\Delta {L_{{\text{CD}}}} = - \cfrac{{{B_{\text{v}}}}}{{10000}}\)

Resulting in a williot like this (keeping the unknown value of $\(B_\text{v}\)$ constant for all elongations):

../../_images/williot.svg

This gives a vertical displacement of \(\text{B}\) of \(\cfrac{{3{B_{\text{v}}}}}{{6400}}\).

With the requirements of \(30 \text{ mm}\) this leads to \(B_\text{v} = 64 \text{ kN}\), resulting in:

  • \(N_\text{BD} = +64 \text{ kN}\)

  • \(N_\text{AD} = -80 \text{ kN}\)

  • \(N_\text{CD} = -48 \text{ kN}\)

  • \(u_{\text{D,h}} = 6.4 \text { mm} \left( \to \right)\)

  • \(u_{\text{D,v}} = 21.4667 \text{ mm} \left( \downarrow \right)\)

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