Lesson Monday September 22th

During today’s lesson it’s demonstrated how you to use the force method for statically indeterminate problems which only involve extension

Demonstration

Given a structure as shown below:

Fig. 34 Structure

We can find the normal force distribution using the force method. Therefore, we need to know the degree of statically determinacy. The free-body-diagram of the full structure is as follows:

Fig. 35 Free-body-diagram full structure

There are 4 unknown support reactions. With only 3 equilibrium equations for this self-contained structure, it can be concluded that this structure is a first-order statically determinant structure.

To apply the force method we need to replace a part of the structure by a statically indeterminate force, which turns the structure into a statically determinate structure. To solve the statically indeterminate structure we need to include a compatibility condition.

Define statically determinate structure with compatibility condition

In this structure, we choose to replace the right support with a rolling hinged support. This leads to a horizontal statically indeterminate force \(B_\rm{v}\) with the compatibility conditions \(u_{h,\rm{B}} = 0\):

Fig. 36 Statically determinate structure with compatibility condition

Solve displacements statically determinate structure in terms of statically indeterminate force

The force distribution and displacements in this statically determinate structure is now solved for in terms of \(B_\rm{h}\).

Support reactions

First, the support reactions are solved for

Fig. 37 Free-body-diagram full structure

\[\begin{split} \begin{array}{c} \sum {{{\left. T \right|}_{\rm{A}}} = 0} \to {B_{\rm{v}}} = 5 \ {\rm{ kN}}\\ \sum {{F_{\rm{v}}} = 0} \to {A_{\rm{v}}} = 15 \ {\rm{ kN}}\\ \sum {{F_{\rm{h}}} = 0} \to {A_{\rm{h}}} = {B_{\rm{h}}} \end{array} \end{split}\]

Leading to:

Fig. 38 Free-body-diagram full structure with resulting support reactions

Section forces

The section forces are solved for, starting with the forces in \(\rm{BE}\) and \(\rm{BD}\):

Fig. 39 Free-body-diagram joint \(\rm{B}\)

\[\begin{split} \begin{array}{c} \sum {{F_{\rm{v}}} = 0} \to {N_{{\rm{BE}}}} = -6.25 \ {\rm{ kN}}\\ \sum {{F_{\rm{h}}} = 0} \to {N_{{\rm{BD}}}} = 3.75 - {B_{\rm{h}}} \end{array} \end{split}\]

Fig. 40 Free-body-diagram joint \(\rm{B}\) with resulting sections forces

Now, let’s continue with a section through beams \(\rm{AD}\), \(\rm{CD}\) and \(\rm{CE}\):

Fig. 41 Free-body-diagram part \(\rm{AC}\)

\[\begin{split} \begin{array}{c} \sum {{F_{\rm{v}}} = 0} \to {N_{{\rm{CD}}}} = - 6.25 \ {\rm{ kN}}\\ {\sum {\left. T \right|} _{\rm{D}}} = 0 \to {N_{CE}} = - 7.5 \ {\rm{ kN}}\\ \sum {{F_{\rm{h}}} = 0} \to {N_{{\rm{AD}}}} = 11.25 - {B_{\rm{h}}} \end{array} \end{split}\]

Fig. 42 Free-body-diagram part \(\rm{AC}\) with resulting section forces

Thirdly, let’s continue with the joint \(\rm{D}\):

Fig. 43 Free-body-diagram joint \(\rm{D}\)

\[\sum {{F_{\rm{v}}} = 0} \to {N_{{\rm{DE}}}} = 6.25 \ {\rm{ kN}}\]

Fig. 44 Free-body-diagram joint \(\rm{D}\)

And finally joint \(\rm{C}\):

Fig. 45 Free-body-diagram joint \(\rm{D}\)

\[\sum {{F_{\rm{v}}} = 0} \to {N_{{\rm{AC}}}} = - 18.75 \ {\rm{ kN}}\]

Fig. 46 Free-body-diagram joint \(\rm{D}\)

Shortening/lengthening of elements

Now, for each element the shortening / lengthening can be calculated:

\[\begin{split}\Delta L = \frac{{NL}}{{EA}} \to \begin{array}{c} {\Delta {L_{{\rm{AC}}}} = - 0.025 \ {\rm{ m}}}\\ {\Delta {L_{{\rm{CE}}}} = - 0.012 \ {\rm{ m}}}\\ {\Delta {L_{\rm{BE}}} = \cfrac{1}{{120}} \approx - 0.00833 \ {\rm{ m}}}\\ {\Delta {L_{{\rm{CD}}}} = \cfrac{1}{{120}} \approx - 0.00833 \ {\rm{ m}}}\\ {\Delta {L_{{\rm{DE}}}} = \cfrac{1}{{120}} \approx 0.00833 \ {\rm{ m}}}\\ {\Delta {L_{{\rm{AD}}}} = 0.018 - \cfrac{1}{{625}}{B_{\rm{h}}} = 0.018 - 0.0016{B_{\rm{h}}}{\rm{ m}}}\\ {\Delta {L_{{\rm{DB}}}} = 0.006 - 0.0016{B_{\rm{h}}}{\rm{ m}}} \end{array}\end{split}\]

Displacement structure due to \(20 \rm{ kN}\)

For now we’ll ignore the shortening/lengethening due to \(B_\rm{h}\):

Fig. 47 Shortening/lengthening of elements

This gives the following Williot-diagram with a fixed \(\rm{AC}\):

Fig. 48 Williot diagram with fixed \(\rm{AC}\)

Leading to the following displacements if \(\rm{AC}\) doesn’t rotate:

joint	Displacement due to \(20 \ \rm{ kN}\) with fixed \(\rm{AC}\) in horizontal direction → \(\left( \rm{mm}\right)\)	Displacement due to \(20 \ \rm{ kN}\) with fixed \(\rm{AC}\) = in vertical direction ↓ \(\left( \rm{mm}\right)\)
\(\rm{A}\)	\(0\)	\(0\)
\(\rm{C}\)	\(-15\)	\(20\)
\(\rm{D}\)	\(18\)	\(-15.1667\)
\(\rm{E}\)	\(-27\)	\(-59.333\)
\(\rm{B}\)	\(24\)	\(-108\)

\(\rm{B}\) shouldn’t move vertically, so this structure has to be rotated back with \(\theta \approx \cfrac{{108}}{{12000}} = 9 \cdot {10^{ - 3}}{\rm{ rad}}\) ⟳, leading to:

joint	Displacement due to \(\theta\) in horizontal direction → \(\left( \rm{mm}\right)\)	Displacement due to \(\theta\) in vertical direction ↓ \(\left( \rm{mm}\right)\)
\(\rm{A}\)	\(0\)	\(0\)
\(\rm{C}\)	\(36\)	\(27\)
\(\rm{D}\)	\(0\)	\(54\)
\(\rm{E}\)	\(36\)	\(81\)
\(\rm{B}\)	\(0\)	\(108\)

Resulting in total displacements of:

joint	Displacement due to in horizontal direction → \(\left( \rm{mm}\right)\)	Displacement in vertical direction ↓ \(\left( \rm{mm}\right)\)
\(\rm{A}\)	\(0\)	\(0\)
\(\rm{C}\)	\(21\)	\(47\)
\(\rm{D}\)	\(18\)	\(38.833\)
\(\rm{E}\)	\(9\)	\(21.67\)
\(\rm{B}\)	\(24\)	\(0\)

Fig. 49 Displaced structure

Displacement structure due to \(B_\rm{h}\)

For the displacement due to \(B_\rm{h}\) in \(\rm{kN}\), the following Williot-diagram with a fixed \(\rm{AD}\) can be drawn:

Fig. 50 Williot diagram with fixed \(\rm{AD}\)

Leading to the following displacements if \(\rm{AD}\) doesn’t rotate:

joint	Displacement due to \(B_\rm{h}\) in horizontal direction →	Displacement due to \(B_\rm{h}\) in vertical direction ↓
\(\rm{A}\)	\(0\)	\(0\)
\(\rm{C}\)	\(-0.8{B_{\rm{h}}}\)	\(-0.6{B_{\rm{h}}}\)
\(\rm{D}\)	\(-1.6{B_{\rm{h}}}\)	\(0\)
\(\rm{E}\)	\(-0.8{B_{\rm{h}}}\)	\(0.6{B_{\rm{h}}}\)
\(\rm{B}\)	\(-3.2{B_{\rm{h}}}\)	\(2.4{B_{\rm{h}}}\)

Again, \(\rm{B}\) Shouldn’t move vertically, so this structure has to be rotated back with \(\theta = \cfrac{2.4{B_{\rm{h}}}}{{12000}} = 0.0002{B_{\rm{h}}} \ {\rm{ rad}}\) ⟲, leading to:

joint	Displacement due to \(\theta\) in horizontal direction → \(\left( \rm{mm}\right)\)	Displacement due to \(\theta\) in vertical direction ↓ \(\left( \rm{mm}\right)\)
\(\rm{A}\)	\(0\)	\(0\)
\(\rm{C}\)	\(-0.8 B_\rm{h}\)	\(-0.6 B_\rm{h}\)
\(\rm{D}\)	\(0\)	\(-1.2B_\rm{h}\)
\(\rm{E}\)	\(-0.8 B_\rm{h}\)	\(-1.8B_\rm{h}\)
\(\rm{B}\)	\(0\)	\(-2.4 B_\rm{h}\)

Resulting in total displacements of:

joint	Displacement due to in horizontal direction → \(\left( \rm{mm}\right)\)	Displacement in vertical direction ↓ \(\left( \rm{mm}\right)\)
\(\rm{A}\)	\(0\)	\(0\)
\(\rm{C}\)	\(-1.6B_\rm{h}\)	\(-1.2B_\rm{h}\)
\(\rm{D}\)	\(-1.6{B_{\rm{h}}}\)	\(-1.2B_\rm{h}\)
\(\rm{E}\)	\(-1.6B_\rm{h}\)	\(-1.2B_\rm{h}\)
\(\rm{B}\)	\(-3.2B_\rm{h}\)	\(0\)

Fig. 51 Displaced structure

Solve statically indeterminate structure with compatibility conditions

Now, we can fill in the compatibility conditions:

\[{u_{{\rm{B,h}}}} = 0 \to 0.024 - 0.0032{B_{\rm{h}}} = 0 \to {B_{\rm{h}}} = 7.5 \ {\rm{ kN}}\]

Section forces statically indeterminate structure

The section forces can be calculated by filling in the resulting \(B_\rm{h}\) in our previous expressions:

Element	Normal force \(\rm{kN}\)
\(\rm{AC}\)	-18.75
\(\rm{CE}\)	-7.5
\(\rm{BE}\)	-6.25
\(\rm{CD}\)	-6.25
\(\rm{DE}\)	6.25
\(\rm{AD}\)	3.75
\(\rm{DB}\)	-3.75

Fig. 52 Normal force distribution

Displacements statically indeterminate structure

Now, the displacements can be found as well by filling in the resulting \(B_\rm{h}\) in our previous expressions:

joint	Displacement in horizontal direction → \(\left( \rm{mm}\right)\)	Displacement in vertical direction ↓ \(\left( \rm{mm}\right)\)
\(\rm{A}\)	\(0\)	\(0\)
\(\rm{C}\)	\(9\)	\(38\)
\(\rm{D}\)	\(6\)	\(29.833\)
\(\rm{E}\)	\(-3\)	\(12.66\)
\(\rm{B}\)	\(0\)	\(0\)

Fig. 53 Displaced structure