Force method – example

In this guide, we will solve the statically indeterminate system using the force method. It is somewhat complementary to the force method guide – an algorithm, because in that guide there was only theory. Here we will solve a sample task that you can get during a test or exam.

Exercise.
Determine the distribution of internal forces M, T, N in a statically indeterminate system using the force method.

  1. Real system
  2. Calculation of the degree of static indeterminate

DSI = 3(A) + 2(D) - 3 * 1 = 5 - 3 = 2DSI = 2

  1. Construction of the basic layout
    We reject both ties at point D.
  1. Arranging a system of canonical equations\left\{\begin{matrix} &0=\delta _{1P}+\delta _{11}*X_{1}+\delta _{12}*X_{2} \\ &0=\delta _{2P}+\delta _{21}*X_{1}+\delta _{22}*X_{2} \end{matrix}\right.
  2. Determination of displacements from the Maxwell-Mohr patterns (we omit the influence of T and N, because their impact on the displacement is insignificant).
    \delta _{ij}=\int \frac{M_{i}*M_{i}}{EJ}\, ds              \delta _{ip}=\int \frac{M_{i}*M_{p}}{EJ}\, ds

5.1. Graphs of internal forces from hyperstatic load X_{1} = 1.
State X1

5.2. Graphs of internal forces from hyperstatic load X_{2} = 1.
State X2

5.3. Graphs of internal forces from external impact P.
Stan P

 

5.4 Calculation of displacements δij: (I will not multiply graphs here by myself, because we can already do it, but if you do not know what’s going on, I invite you to read: calculating displacements).

\delta _{11} = 0,0107 (we multiply the X1 status graphs)
\delta _{12} = \delta _{21} = 0,024 (multiplication of state X1 by X2)
\delta _{22} = 0,108 (we multiply the X2 status graphs)

5.5 Calculation of δip displacement.

\delta _{1p} = -0,096 (multiplication of state X1 by state P)

\delta _{2p} = -0,378 (multiplication of state X2 by state P)

  1. The solution of the system of canonical equations.

\left\{\begin{matrix} 0,0107 * X_{1} + 0,024 * X_{2} = 0,096 & \\ 0,0240 * X_{1} + 0,108 * X_{2} = 0,378 & \end{matrix}\right.
X_{1} = 2,25kN \: \: \: \: \: \: \: X_{2} = 3,00 kN

  1. Determining the diagram of internal forces in the real system from superposition patterns.

7.1. Bending moment       
M = M_{P} + M_{i} * X_{i}
M_{ab} = -24,00+4,00*2,25+6,00*3,00=3kNm
M_{ba} = -24,00+0,00*2,25+6,00*3,00=-6,00kNm
M_{bc} = -24,00+0,00+6,00*3,00=-6,00kNm
M_{cb} = 0,00+0,00+3,00*3,00=9,00kNm
M_{cd} = 0,00+0,00+3,00*3,00=9,00kNm
M_{cd} = 0,00+0,00+0,00=0,00kNm

7.2. Shear forces
T = T_{P} + T_{i} * X_{i}
T_{ab} = 0,00 + 0,00-1,00 *2,25=-2,25kN
T_{ba} = 0,00 + 0,00-1,00 *2,25=-2,25kN
T_{bc} = 8,00-1,00*3,00=5,00kN
T_{cb} = 8,00-1,00*3,00=5,00kN
T_{cd} = -1,00*3,00=-3,00kN
T_{dc} = -1,00*3,00=-3,00kN

7.3. Normal forces
N = N_{P} + N_{i} * X_{i}
N_{ab} = -8,00+1,00*3,00=-5,00kN
N_{ba} = -8,00+1,00*3,00=-5,00kN
N_{bc} = 0,00+0,00-1,00*2,25=-2,25kN
N_{cb} = 0,00+0,00-1,00*2,25=-2,25kN
N_{cd} = -2,25kN
N_{dc} = -2,25kN

  1. Determination of supporting reactions from external force diagrams.

Checking the equilibrium conditions.

\Sigma P_{X} = H_{a} - H_{d} = 0,00 kN - ok!
\Sigma P_{Y} = V_{a} - P + V_{d} = 0,00kN - ok!
\Sigma M_{a} = M_{a} + P * 3,00 - H_{d} * 4,00 - V_{d} * 6,00 = 0,00 kNm - ok!

  1. Static check of correctness of the solution in nodes (node B).

All forces are balanced. Static check is correct!

  1. Kinematic check
    It is necessary to determine the displacement of the cross-section in the real system at the point whose value we know, for example: from the support ties. We know that there must be a zero displacement on the support, when such displacement comes out, it means that we have calculated the task correctly.