Exam Friday

Today some studnets make an additional exam assignment covering static indeterminate structures, continuum mechanics or stability 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 will be available on after the exam.

Given is the following structure:

• \(EA_{\rm{AC}} = 240 \, \rm{MN}\)

• \(EI_{\rm{AC}} = 1200 \, \rm{MNm}^2\)

• \(EA_{\rm{BC}}, EI_{\rm{BC}} \gg EA_{\rm{AC}}, EI_{\rm{AC}}\)

Fig. 241  

Exercise

Show that this structure is statically indeterminate to the second degree.

Solution

Fig. 242  

For this structure, the external static indeterminacy is equal to the internal static indeterminacy.

There are 5 unknowns and 3 equilibrium equations, making the structure statically indeterminate to the second degree.

Exercise

Provide three valid variants to make this structure statically determinate for the purpose of the force method. Provide three different variants: over all variants, you must have adjusted at least each node and element once. For each of the variants, provide the necessary compatibility equation(s) to determine the statically indeterminate force(s).

Solution

Potential alternatives could be:

Fig. 243  

Fig. 244  

Fig. 245  

Exercise

Find the displacement of \(\rm{C}\) using the force method.

Solution

The following statically indeterminate structure is chosen:

• \(EA_{\rm{AC}} = 240 \, \rm{kN}\)

• \(EI_{\rm{AC}} = 1200 \, \rm{kNm}^2\)

• \(EA_{\rm{BC}}, EI_{\rm{BC}} \gg EA_{\rm{AC}}, EI_{\rm{AC}}\)

Fig. 246  

First, the internal forces in \(\rm{C}\) are determined:

Fig. 247  

This gives:

• \(V_{\rm{C}}^{\rm{AC}} = B_{\rm{h}}\)

• \(N_{\rm{C}}^{\rm{AC}} = B_{\rm{v}}\)

• \(M_{\rm{C}}^{\rm{AC}} = 3 \cdot B_{\rm{h}} + 4 \cdot B_{\rm{v}}\)

Now, the displacements of node \(\rm{C}\) can be determined based on the elongation of a bar and forget-me-nots for part \(\rm{AC}\):

Fig. 248  

This gives:

• \(u_{\rm{C,h}} = \cfrac{21}{200} B_{\rm{h}} + \cfrac{3}{50} B_{\rm{v}}\) (→)

• \(u_{\rm{C,v}} = \cfrac{1}{40} B_{\rm{v}} + \cfrac{243}{200}\) (↓)

• \(\theta_{\rm{C}} = \cfrac{3}{100} B_{\rm{h}} + \cfrac{1}{50} B_{\rm{v}}\) (↻)

As \(\rm{CB}\) doesn’t deform but can rotate, this leads to:

• \(u_{\rm{B,h}} = u_{\rm{C,h}} + \theta_{\rm{C}} \cdot 3 = \cfrac{39}{200} B_{\rm{h}} + \cfrac{3}{25} B_{\rm{v}}\) (→)

• \(u_{\rm{B,v}} = u_{\rm{C,v}} + \theta_{\rm{C}} \cdot 4 = \cfrac{3}{25} B_{\rm{h}} + \cfrac{21}{200} B_{\rm{v}} + \cfrac{243}{200}\) (↓)

Equating the displacements of node \(\rm{B}\) to zero gives:

• \(B_{\rm{h}} = 24 \, \rm{kN}\)

• \(B_{\rm{v}} = - 39 \, \rm{kN}\)

This gives:

• \(u_{\rm{C,h}} = 0.18 \, \rm{m}\) (→)

• \(u_{\rm{C,v}} = 0.24 \, \rm{m}\) (↓)

Exercise

Draw the displaced statically indeterminate structure for which you indicate the displacements of node \(\rm{C}\) and the location of the point of inflection (the point for which the curvature changes sign).

Solution

As the bending moment in \(\rm{C}\) is \(3 \cdot B_{\rm{h}} + 4 \cdot B_{\rm{v}} = - 84 \, \rm{kNm}\) with a shear force of \(B_{\rm{h}} = 24 \, \rm{kN}\) (which equates the slope of the bending moment diagram), the bending moment is \(0\) at \(\cfrac{84}{24} = 3.5 \, \rm{m}\) to the bottom of node \(\rm{C}\), which is the point of inflection.

For a more correct representation of the deformed structure, the rotation of node \(\rm{C}\) can be taken into account, which is \(-0.06 \, \rm{rad}\) (↺).

This gives:

Fig. 249  

Exam assignment 3 Continuum mechanics

Your own submission and its grading will be available on after the exam.

Given is the following structure:

Fig. 250  

Forces and support reactions are acting on the normal force centre. The cross-section can be regarded as thin-walled

Fig. 251  

Exercise

Show that \(A = 150 \sqrt{2} \, \rm{cm}^2\), \(\bar z_{\rm{N.C.}} = 0 \, \rm{mm}\) and \(I_{zz} = 31250 \sqrt{2} \, \rm{cm}^4\).

Solution

The area of the cross section is:

\[ A = 2 \cdot 50 \cdot \sqrt{2} \cdot 1.5 = 150 \sqrt{2} \, \rm{cm}^2 \]

There is as much material above as below the \(y\)-axis, which gives \(\bar z_{\rm{N.C.}} = 0 \, \rm{mm}\).

The second moment of area \(I_{zz}\) is (taking into account the projected horizontal thickness of the parts):

\[I_{zz} = 2 \cdot \cfrac{1}{12} \cdot 1.5 \sqrt{2} \cdot 50^3 = 31250 \cdot \sqrt{2} \, \rm{cm}^4\]

Exercise

Show that \(V_{\rm{B}}^{\rm{SB}} = -60 \, \rm{kN}\) and \(M_{\rm{B}} = -240 \, \rm{kNm}\).

Solution

Using equilibrium equation you can find:

• \( V_{\rm{B}}^{\rm{SB}} = -60 \, \rm{kN}\)

• \( M_{\rm{B}} = -240 \, \rm{kNm}\)

Exercise

Determine the 3D stress tensor on a positive cross section just left of \(\rm{B}\) in point \(\rm{E}\). Indicate the directions of this stress tensor.

Solution

To find the shear stress in \(\rm{E}\), the part below \(\rm{E}\) can be seen as sliding off, this gives:

\[\tau = \cfrac{\left| -60000 \cdot \left( 0.125+0.\cfrac{125}{2} \right) \cdot 0.015 \cdot \sqrt{2} \cdot \cfrac{0.125}{2} \right|}{0.015 \cdot 31250 \cdot \sqrt{2} \cdot 10^{-8}} = 4.5 \cdot 10^6 \, \rm{Pa} = 4.5 \, \rm{MPa}\]

As the shear force acts upwards on a positive cross section, the shear stress acts in in top right direction in \(\rm{E}\)

The normal force can be found with:

\[\sigma = \cfrac{ -240 \cdot 10^3\cdot 0.125}{31250 \cdot \sqrt{2} \cdot 10^{-8}} = -48 \cdot 10^6 \cdot \sqrt{2} \, \rm{Pa} \approx -67.88 \, \rm{MPa}\]

So the 3D stress tensor is:

\[\begin{split}\sigma = \begin{bmatrix} -67.88 & 0 & -4.5 \\ 0 & 0 & 0 \\ -4.5 & 0 & 0 \end{bmatrix} \, \rm{MPa}\end{split}\]

With the \(x\)-axis in the original \(x\)-direction, the \(z\)-axis along the edge to the bottom left and the \(y\)-axis perpendicular to the edge to the top left:

Fig. 252  

Exercise

Determine the deviatoric stress tensor on a positive cross section just left of \(\rm{B}\) in point \(\rm{E}\). Include the direction of the coordinate system of this deviatoric stress tensor.

Solution

The principle stresses are:

• \(\sigma_1 = \frac{1}{2} \cdot -67.88 - \frac{1}{2} \cdot \sqrt{ \left(-67.88 \right)^2 + 4 \cdot 4.5^2} \approx -68.18 \, \rm{MPa}\)

• \(\sigma_2 = \frac{1}{2} \cdot -67.88 + \frac{1}{2} \cdot \sqrt{ \left(-67.88 \right)^2 + 4 \cdot 4.5^2} \approx 0.297 \, \rm{MPa}\)

• \(\sigma_3 = 0 \, \rm{MPa}\)

This gives the isotropic stress:

\[\sigma_{\rm{o}} = \cfrac{-68.18 + 0.297 + 0}{3} \approx -22.63 \, \rm{MPa}\]

This gives the deviatoric stress tensor:

\[\begin{split}\sigma_{\rm{d}} = \begin{bmatrix} -67.88 - \left(-22.63\right) \\ 0.297 - \left(-22.63\right) \\ 0 - \left(-22.63\right) \end{bmatrix} = \begin{bmatrix} 45.55 \\ -22.92 \\ -22.63 \end{bmatrix} \, \rm{MPa}\end{split}\]

This stress tensor is coordinated in the principle stress directions, which are not the same as the original stress tensor.

The coordination system of the deviatoric stress tensor is the principle stress direction. This is the original coordinate system rotated over an angle of \(\theta = \frac{1}{2} \cdot \arctan \left( \cfrac{4.5}{\frac{1}{2} \left(67.88 - 0\right)} \right) \approx 3.74^{\circ}\) around the \(y\)-axis of the 3D stress tensor in the direction from \(z\) to \(x\).