Force Method for Support Settlement

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When foundations are not fully compacted, soil settlement occurs. When a structure is grounded on such a foundation, the change in height at different points under the structure creates stress within it. The force method for support settlement was created to solve the reactions of an indeterminate structure that is subjected to foundation shifts. Although the method is very similar to the force method used for structures with multiple degrees of indeterminacy, there are some differences.

Theory for Support Settlement


When analyzing an indeterminate structure that has support settlements, the methodology does not change much. The main difference is that the compatibility equations are set equal to the settlement relating to that point rather than zero. To begin, determine the degree of indeterminacy and choose your redundants. Remove the supports at the redundant locations and determine the deflections at each point(s) using a beam deflection method. To find the flexibility coefficients, apply unit loads separately to each point to determine the deflection. Depending on the amount of redundants more flexibility coefficients will have to be determined. [1]

The Chord is the straight line drawn between the two end supports of a structure. Figure i shows the chord as the dashed line. The sign convention for support settlement is positive if the settlement is in the same direction as the redundant forces. [2]For the example in Figure i there would be a negative settlement at C and a positive settlement at E.

Figure i: Example Structure with Support Settlement.

Figure i

- For more information in regards to just the Force Method visit the Force Method for Multiple Degrees of Indeterminacy wiki.

Procedure for Support Settlement

1. Find the degree of indeterminacy of the structure and select redundants. The amount of redundants must equal the degree of indeterminacy

2. Remove the chosen redundants so that you are left with a determinate structure. Leave all other applied loading the way it is.

3. From the locations that you removed your redundants, calculate the deflections. These will be the first terms in your compatibility equations.

4. In order to find the flexibility coefficients of the compatibility equations, we must setup our beam so that a unit load is applied to where the removed support was. You must apply the unit loads at the redundant position one at time. This means that if you have two redundants you need to make two different primary beams loaded with the unit load for each redundant. Even though the primary beam is only loaded with one unit load, the deflections at all redundant positions need to be calculated.

5. With the values from steps 3 and 4, we can substitute and solve for the redundant forces using the compatibility equations. It is always good to check your answers with the equations of equilibrium when you are finished.

To demonstrate the procedure used to solve an indeterminate structure with support settlements the following example is provided.

Example Problem

Solve for the reactions in the indeterminate structure shown in figure 1. There are Support Settlements at C and E of 18 mm and 22 mm, respectively. $ E = 200 GPa , I = 3000 x (10)^6 mm^4 $

This structure is 2 degrees indeterminate since there are 5 reactions and only 3 equations of equilibrium. Therefore we need to select two redundant forces. For this example we will choose Cy and Ey to be our redundants.

Figure 1: Indeterminate Structure.

Figure 1

Figure 2: Support Settlements.

Figure 2

The Support Settlements are $ \Delta_{C} = 0.018 m $ and $ \Delta_{E} = 0.022 m $.

We notice that the settlement of C and E relative to the chord is in the same direction as the redundants so they are positive in magnitude.

Figure 3: the primary structure with the supports removed.

Figure 3

The Compatibility equations are then written as:

$ \Delta_{CO} + f_{CC}C_y + f_{CE}E_y = \Delta_{C} $

$ \Delta_{EO} + f_{EC}C_y + f_{EE}E_y = \Delta_{E} $

Note: the compatibility equations for the beam remain the same as the one used for the force method, except that the right-hand side is equal to the settlements. The flexibility coefficients are shown as deflections from the redundants in Figure 4.

Figure 4: the structure with each redundant being applied to it separately.

Figure 4

To solve for the deflections we create the shear and bending moment diagrams for the primary beam and the redundant combinations as shown in Figure 5 and Figure 6.

Figure 5: The Shear and Bending Moment Diagrams for Primary Loading.

Figure 5

By using the beam deflection formula, we get :

$ \Delta_{CO} = \int_{A-C} \frac{M}{EI}(x)dx $

From $ \frac{M}{EI} $ diagram,

$ \Delta_{CO} = (5)(\frac{1}{2})(\frac{825}{(200)(3000)})(5 + \frac{5}{3}) + (5)(\frac{5}{2})(\frac{825}{(200)(3000)}) + (5)(\frac{5}{3})(\frac{1376-825}{(2)(200)(3000)})(\frac{1}{2}) $

$ \Delta_{CO} = 0.042 m = 42 mm $

Due to the symmetry in Figure 3 we can say that,

$ \Delta_{EO} = 0.042 m = 42 mm $

Figure 6: Shear and Bending Moment Diagrams for Redundants.

Figure 6

$ f_{CC} = (\frac{1}{2})(10)(\frac{1}{3})(10)(\frac{6.67}{(200)(3000)}) $

$ f_{CC} = 0.0001853 m $

$ f_{EC} = (\frac{1}{2})(\frac{3.33}{(200)(3000)})(10)(\frac{10}{3}) $

$ f_{EC} = 0.0000925 m $

Due to symmetry it follows that,

$ f_{EE} = 0.0001853 m $

$ f_{CE} = 0.0000925 m $

Substitution into our compatibility equations

$ \Delta_{CO} + f_{CC}C_y + f_{CE}E_y = \Delta_{C} $

$ 0.042 + (0.0001853)C_y + (0.0000925)E_y = 0.018 $

$ \Delta_{EO} + f_{EC}C_y + f_{EE}E_y = \Delta_{E} $

$ 0.042 + (0.0000925)C_y + (0.0001853)E_y = 0.022 $

Combining equations yields:

$ C_y = 100.7 kN $

$ E_y = 57.6 kN $

Applying equations of equilibrium, we find Gy to be 93.0 kN and Ay to be 78.7 kN.

Figure 7: FBD with final reaction forces.

Figure 7


  1. Caprani, Collin. “Structural Analysis III, Moment Distribution”, 2008. Internet: [Accessed November 24th, 2013]
  2. Kassimali, A. (2011). Structural Analysis: SI Edition (4th ed.). Stamford, CT: Cengage Learning.