Displacement method – method und example

I hope you have read the guide displacement method – introduction. Guides with this method will not be separated into an algorithm and example, which will allow me to easily present the mechanism of DP operation and the theory regarding the displacement method.

Exercise.
Draw graphs of the internal forces of the statically indeterminate scheme using the displacement method, having the data and scheme shown below.

  1. Static schema

Data.
EJ1=1870kNm^{2}
EJ2=2900kNm^{2}
q=10kN/m
M=16kNm
P=10kN

  1. Determination of kinematic degree of indeterminability of the system (DKI)

DKI = \Sigma \varphi + \Sigma \psi

\Sigma \varphi – total number of unknown turns of rigid nodes (number of rigid nodes).
In this example, \Sigma \varphi=1 and this is node “3”. The rigid joint has a minimum of two statically indeterminate rods.

\Sigma \psi – the total number of unknown, independent shifts of rigid nodes, equal to the degree of kinematic variation of the articulation system.
Kinematic chain (articulated system)

\Sigma \psi = 3 * t - w = 3 * 3 - (21 + 22 + 23 + 24) = 9 - 8 = 1

DKI = 1 + 1 = 2

The system is kinematically indeterminate twice, the rotation of node 3 - \varphi 3 = Z1 is unknown and the travel of node3 - \psi 3 = Z2 is unknown.

  1. The basic system and the canonical equation.

The basic system must be kinematically determinable, i.e. the rotations and displacements of all rigid nodes must be known. We create it by adding fictitious links to directions of unknown turnover.
1 – turnover blocking,
2 – blocking the shift.

Assumption.
\Delta l = 0 -> \Delta l23 = 0 – that is, we omit changes in length.
I wrote about it in the introduction to the displacement method.
The canonical equation.

\left\{\begin{matrix}k11 * Z1 + k12 * Z2 + k1p = 0 & \\ k21 * Z1 + k22 * Z2 + k2p = 0 & \end{matrix}\right.

k_{ij} – reaction from a generalized relocation of ties;
k_{ip} – reaction from external impact (strength, temperature, subsidence);

  1. Calculation of the k_{ij} reaction in fictitious bonds.

a) State of deformation Z1 = \varphi 3 = 1 – nodal moments (at the ends of the cut bars) from transformation patterns.

b) State of deformation Z2 = \psi 3 = 1 – calculations of the angles of rotation of chords from kinematic chain equations and trigonometric relations.

Equation of the kinematic chain in the general form.

According to the X axis : -\Sigma L_{ij}^{x} * \psi _{ij} = V_{k} - V_{o}
According to the Y axis : -\Sigma L_{ij}^{y} * \psi _{ij} = U_{k} - U_{o}

where:
V_{k}\: \: and\: \: U_{k} – horizontal and vertical displacement of the final chain node, these quantities must be known
V_{o}\: \: and\: \: U_{o} – vertical and horizontal displacements of the initial node of the chain, these quantities must be known

Chain 1 – 2 – 3 – 4

According to the X axis: -L_{12}^{x} * \psi _{12} - L_{23}^{x} * \psi _{23} - L_{34}^{x} * \psi _{34} = V_{k} - V_{o}
According to the Y axis: L_{12}^{y} * \psi _{12} + L_{23}^{y} * \psi _{23} + L_{34}^{y} * \psi _{34} = U_{k} - U_{o}

According to the X axis: -0 * 0,50 - 5,00 * \psi 23 - 3,00 * \psi 34 = 0 - 0
According to the Y axis:  4,00 * (- 0,25) + 0,00 * \psi 23 - 4,00 * \psi 23 = 0 - 0

\psi 12 = -0,25

\psi 23 = \frac{3}{20}

\psi 34 = 0,25

c) Nodal moments at the ends of cut rods from the transformation formula

d) Reaction k11

K_{11} – from the equilibrium conditions of the cut node “3” loaded with node moments from the graph M.
Turn like \varphi 3.

\Sigma M = 0
k_{11} - M_{34} - M_{32} = 0
k_{11} = 1496 + 1740
k_{11} = 3236

e) Reaction K12 = K21

K12 = K21 – from the condition of the balance of the cut node “3” loaded with node moments.

\Sigma M = 0
k_{12} - M_{34} + M_{32} = 0
k_{12} = 561 - 261
k_{12} = 300

f) Reaction K22
K22 – we assume that the reaction of k22 must be positive.

Virtual work equation.

k_{22} * U - M_{12} * \psi _{12} - M_{23} * \psi _{23} - M_{34} * \psi _{43} = 0
k_{22} * 1 = 351 * (\frac{1}4{}) + 261 * (\frac{3}{20}) + 561 * (\frac{1}{4}) + 561 * \frac{1}{4}
k_{22} = 407,30

  1. Calculation of k_{ip} reaction in fictitious bonds.

a) State P – nodal moments (at the ends of the cut bars).

M_{12} = M_{34} = M_{43} = 0 (rods not loaded)
M_{21} = M_{12} (joint)
M_{32} = \frac{(P * L_{23}^{2})}{8}= \frac{(10 * 5^{2})}{8} = 31,25 kNm

b) Reaction K1p

K1p – from the equilibrium condition of the cut node “3” loaded with node moments from the Mp chart and the external moment applied in this node.

\Sigma M = 0
-k_{1p} -16,00kNm+31,25kNm=0
k_{1p} = 15,25kNm

c) Reaction K2p
We calculate how K22, taking into account the work of external concentrated forces applied in nodes.
The kinematic chain.

\delta a = \psi 23 * \frac{L_{23}}{2} = \frac{3}{20} * 2,50 = \frac{15}{40}
\delta p = \psi_{34} * L_{34}^{y} = \frac{1}{4} * 4 = 1

Remember that the work of continuous load is equal to the work of the resultant force Q!
Virtual work equation.

k_{2p} * u + Q * \delta _{p} + M_{32} * \psi _{23} - P * \delta_{p} = 0
k_{2p} * 1,00 = -50,00 * \frac{15}{40} - 31,25 * \frac{3}{20} + 10,00 * 1,00
k_{2p} = - 13.44kN

  1. The solution of the system of canonical equations. Calculation of the actual displacement of node “3”.

\left\{\begin{matrix} 32,36 * Z_{1} + 300,00 * Z_{2} + 15,25 = 0 & \\ 300,00 * Z_{1} + 407,30 * Z_{2} - 13,44 = 0 & \end{matrix}\right.

Z_{1} = - 0,008341 [rad] \varphi 3 – turnover

Z_{2} = 0,03913 [m] \psi 3 – shift

  1. Diagram of bending moments in the real system from superposition patterns.

M = M_{1} * Z_{1} + M_{2} * Z_{2} + M_{P}

  1. Graph of transverse (cutting) forces from the balance conditions of the cut bars, loaded at the ends with moments from the “M” graph and the span load.

We cut individual rods and calculate cutting forces.

  1. Graph of normal (axial) forces from the equilibrium conditions of cut nodes loaded with forces from the “T” graph and external concentrated forces applied in these nodes.

Having cutting forces, we cut nodes and calculate normal forces.

That would be enough. I hope that the above example will help you understand what the displacement method is. As I wrote earlier, the method of displacement is twice as difficult as the method of forces I described in previous guides, which is why you also need twice as many reworked examples to learn this method well.