# Force Method for Multiple Degrees of Indeterminacy

## Contents

## Introduction

The *Force Method*, also known as *Method of Consistent Deformations* was introduced by James C. Maxwell in 1864 and later refined by Otto Mohr and Heinrich Müller-Breslau for the analysis of indeterminate structures^{[1]}. This method resolves structure deflections by analyzing its flexibility through taking a reaction(s) as redundant, and applying a small imaginary load (ie: 1kN) in place to determine its flexibility (in that direction). Applying Maxwell's Law we use that flexibility and the real forces to find our real "redundant" forces. This method was one of the first available for the analysis of statically indeterminate structures^{[1]}. The basic concept of this method requires choosing supports from the structure as redundants in order to satisfy the determinacy equation as determinate and solve for the beam using equilibrium. These equations, in turn, must satisfy compatibility and displacement conditions for a structure in order to determine the redundant forces.

For example, if a beam is supported by a pin support(r=2) and 4 roller supports (r=4), its degree of determinacy is *i=6-3=3*. In order for the beam to be analyzed using the force method, __3 support reactions__ must be selected as redundant.

Recall the determinacy equation:

- $ 3m + r = 3j + c $
^{[2]}

- $ 3m + r = 3j + c $

Once the supports are removed, we have a new structure called the *primary structure*. The removed restraints are now labelled as redundant along with their corresponding reactions and/or internal forces. These redundant forces are then applied as unknown loads (P) on our primary determinate structure, which are calculated each at a time using the following compatibility relationship:

- $ \Delta_{OriginalStructure}=\Delta_{PrimaryStructure} + \sum \Delta_{P} $
^{[2]}

- $ \Delta_{OriginalStructure}=\Delta_{PrimaryStructure} + \sum \Delta_{P} $

The analysis of a statically indeterminate structure is more involved than that of a statically determinate one.

**Several important reasons for considering an indeterminate structure for design**

• The stress and deflection of an indeterminate structure will be less than of a similar determinate structure.^{[1]}

• An indeterminate structure has a tendency to redistribute loads to its redundant supports/members in case of failure of the design or overloading. In this situation the indeterminate structure can maintain its stability and prevent collapse (good for earthquakes).^{[1]}

**Disadvantages include**

• Additional cost of fabrication and materials, the adding cost has to be compared to the added strength or necessary strength required.^{[1]}

• Also the force method may not be the most efficient method to use when solving indeterminate structures. There are many options including Slope-Deflection Method for Continuous Beams, Slope-Deflection Method for Frames without Sidesway, Slope-Deflection Method for Frames with Sidesway, Moment Distribution Method for Continuous Beams, etc.

## Important Variables

Concepts including External and internal determinacy, stability, and virtual work must be understood to preform an analysis of an indeterminate structure using the force method. Refer to Determinacy, Indeterminacy and Stability, Beam Virtual Work, Frame Virtual Work, Truss Virtual Work.

**Redundant**
A redundant force is a reaction/internal force or moment we choose to do a constant deformation analysis of and determine first. To analyze an indeterminate structure you have to select and analyze multiple redundant forces. The number of redundants is equal to the degree of static indeterminacy. ^{[3]}

**Maxwell's Law of reciprocal deflections**
This law states that the flexibility on either side of a reaction has to be equal.^{[1]}

## Summary

The Force Method for Multiple Degrees just builds on the Force Method for One Degree of Indeterminacy and also when using the Force method for a truss it is good to know Force Method Using Internal Forces as Redundants.^{[2]} For example in the figure below you can see that there is a beam with a pin (r=2) and 3 rollers (r=3).

**Steps for Solving:**

These steps are formed from reading the Structural Analysis textbook by Kassimali ^{[2]} and from Professor Erochko's lectures. ^{[4]}

**1)** Calculate the degree of indeterminacy and select that many unknown forces as **redundant**. Refer to Determinacy, Indeterminacy and Stability to determine how to calculate degrees of indeterminacy.

In this case when solving for the degree of indeterminacy we get that this beam is 2 degrees statically indeterminate (i=5-3). So we will select By and Cy as the redundant forces

**2)** Write the compatibility equations^{[2]}:

For our case: $ \Delta_{By,O} + f_{By,By}B_{y} + f_{By,Cy}C_{y} = 0 $ $ \Delta_{Cy,O} + f_{Cy,By}B_{y} + f_{Cy,Cy}C_{y} = 0 $

**3)** Now take out the **redundant** and solve for the reaction forces.

In this case take out By and Cy, and solve for Ax, Ay, and Dy. As seen in figure 2 below.

**4)** Next make as many of the same beams without the external loading as there are redundants. On one beam/frame/truss include a 1 kN force at one **redundant** and on the other beam/frame/truss include a 1 kN force at the other **redundant**. As seen in the figure below. You may have to add an internal redundant, Force Method Using Internal Forces as Redundants.

**5)** Solve for the reaction forces for the virtual beams/frames/trusses and make a moment diagram for all of the beams/frames/trusses. When working with a truss with an internal redundant, it is not necessary to make the moment diagrams, instead you need to find all internal member forces of the truss with external forces and the internal member forces of the truss with a 1 kN force at the redundant, Force Method Using Internal Forces as Redundants.

**6)** Solve for the deflections:

For our case: $ \Delta_{By,O} $ $ \Delta_{Cy,O} $

This is done by using virtual work. In our case to find the deflection use virtual work between the original moment diagram and the moment diagram for a 1kN force at By and once again between the original moment diagram and the moment diagram for the 1kN force at Cy. Review Beam Virtual Work, Frame Virtual Work, and Truss Virtual Work to know how to solve for the different kinds of virtual work problems.

**7)** Solve for the flexibility coefficients:

For our case: $ f_{By,By}, f_{By,Cy} $ $ f_{Cy,By}, f_{Cy,Cy} $

Using **Maxwell's law of reciprocal deflections** we can conclude that:

```
$ f_{By,Cy}=f_{Cy,By} $
```

These flexibility coefficients can also be found using the virtual work method. In this case to find $ f_{By,By} $, you would do virtual work with the moment diagram for the beam with a 1 kN force at By with itself. The same would be done to find $ f_{Cy,Cy} $, except you would use the moment diagram for Cy. To find $ f_{By,Cy}=f_{Cy,By} $ you would do virtual work with the moment diagram of Cy and By. Review Beam Virtual Work, Frame Virtual Work, and Truss Virtual Work to know how to solve for the different kinds of virtual work problems.

**8)** We have now found all the unknowns for the compatibility equations:

For our case: $ \Delta_{By,O} + f_{By,By}B_{y} + f_{By,Cy}C_{y} = 0 $ $ \Delta_{Cy,O} + f_{Cy,By}B_{y} + f_{Cy,Cy}C_{y} = 0 $

We can now solve for the unknown forces $ B_{y}, C_{y} $ because we have two equations and two unknowns.

**9)** Lastly put the unknown forces you have found, $ B_{y}, C_{y} $, on the original beam and solve for the other unknowns.

## Example Problem # 1 - Beam

In the following examples the value of IE is taken as constant and hence moment diagrams are equivalent to M/EI diagrams. The table integration method is used for calculations as per Beam Virtual Work and/or Frame Virtual Work in the preceding sections. Kindly note that all figures/diagrams used in the solution of these examples belong to the authors of this page.

This example illustrates how to apply the compatibility equations using the force method for structures that have more than one degree of indeterminacy.

In order to begin, you must determine how many redundant forces exist.

**So, how many degrees of indeterminacy does this structure have?**

Unknowns (Ma,Ax,Ay,By,Cx,Cy)-# of equilibrium reactions (X,Y,Moment)6 -3 = 3 redundant forces

Once you know __how many__ redundant forces need to be removed, you determine __which__ 3 forces are chosen to be *redundant*. The resulting beam is determinate, and referred to as the *primary beam*. In our example, we chose to replace the fixed end at point A with a free end. This will remove 3 redundant forces (Moment at A, Ax and Ay) therefore the new beam (with the fixed end removed) is our *primary beam* as shown in the figure below. Reaction supports are calculated through equilibrium and the shear and moment diagrams are constructed. The "real" redundant reactions will be found using virtual work and the integration table.

**The shear and moment diagrams for the Primary beam are presented in the following figure:**

**Solving for compatibility equations**

__Write out the 3 compatibility equations__

$ \Delta_{Ax,o} + f_{Ax,Ax}(A_{x}) + f_{Ax,Ay}(A_{y}) + f_{Ax,Ma}(M_{A})=0 $

$ \Delta_{Ay,o} + f_{Ay,Ax}(A_{x}) + f_{Ay,Ay}(A_{y}) + f_{Ay,Ma}(M_{A})=0 $

$ \Theta_{Ma,o} + f_{Ma,Ax}(A_{x}) + f_{My,Ay}(A_{y}) + f_{Ma,Ma}(M_{A}) = 0 $

**Solve for deflections**

The first step to solving for the unknown reactions in the compatibility equation $ \(F_A_x , F_A_y $ , and $ F_M_a) $ is to determine $ \Delta_{Ax,o} , $$ \Delta_{A_Yo} $ and $ \Theta_{M_ao} $ using virtual work (as discussed in the preceding section).

**Term****Image****Procedure**$ \Delta_{Ax,o} $ Redraw the primary beam with no loads except for an addition of 1kN where Ax would have been if the fixed end was there, in the positive direction. So you are replacing the redundant forces with 1 kN each at a time. Solve for the reactions and draw the moment diagram. $ \Delta_{Ay,o} $ Just like the procedure for Ax but this time the 1 kN is put where Ay would have been. $ \Theta_{Ma,o} $ Just like the procedure for Ax and Ay but this time the 1 kN is put where the moment would have been.

To calculate slope & deflections, using the Integration Table, multiply each redundant moment diagram by the primary beam moment diagram.

- $ \Theta_{Ma,o} = {\frac{38.71}{EI}} kN $

- $ \Delta_{Ax,o} = 0 kN $

- $ \Delta_{Ay,o} = {\frac{84.45}{EI}} kN $

**Solve for flexibility terms**
$ \Delta_{Ax,o} + f_{Ax,Ax}(A_{x}) + f_{Ax,Ay}(A_{y}) + f_{Ax,Ma}(M_{A})=0 $

$ \Delta_{Ay,o} + f_{Ay,Ax}(A_{x}) + f_{Ay,Ay}(A_{y}) + f_{Ay,Ma}(M_{A})=0 $

$ \Theta_{Ma,o} + f_{Ma,Ax}(A_{x}) + f_{Ma,Ay}(A_{y}) + f_{Ma,Ma}(M_{A}= 0 $

*Using Maxwell's Reciprocal Law we conclude:*

$ f_{Ax,Ma}=f_{Ma,Ax} $ $ f_{Ay,Ma}=f_{Ma,Ay} $ $ f_{Ax,Ay}=f_{Ay,Ax} $

Now we solve our flexibility unknowns through multiplying the virtual moment diagrams based on the subscripts. For example, $ f_{Ax,Ay} $ would be calculated by multiplying the virtual moment diagrams of Ax redundant and Ay redundant. This will result in:

$ f_{Ax,Ax} = 0 kN $

$ f_{Ax,Ay} = f_{Ay,Ax} = 0 $

$ f_{Ax,Ma} = f_{Ma,Ax} = 0 kN $

$ f_{Ay,Ma} = f_{My,Ay} = {\frac{3.33}{EI}} kN $

$ f_{Ma,Ma} = {\frac{3.67}{EI}} kN $

Subbing the newly obtained results back in our compatibility equations:

As expected, since we don't have a moment due to a force redundant in the x direction, there will be no load inducted in it and hence all x terms go to 0. This leaves us with two compatibility equations to solve:

```
$ {\frac{84.45}{EI}} + 0 + {\frac{9.34}{EI}}A_{y} + {\frac{3.33}{EI}}M_{A} = 0 $
```

```
$ {\frac{38.71}{EI}}} + 0 + {\frac{3.33}{EI}}A_{y} + {\frac{3.67}{EI}}M_{A} = 0 $
```

Solving simultaneously we get: $ A_{y} = - 7.81 kN $ and $ M_{A} = -3.46 kNm $

When we apply these forces back on the original beam, we find: $ B_{y} = 49.69 kN $ and $ C_{y} = 28.12 kN $

The figure to the right represents the original beam but with the new forces determined in place. The shear and moment diagram is included in the figure.

## Example Problem # 2 - Frame

Consider this real life situation where a stage set is held up by a structural frame. Each speaker weighs 20 N and the light set can be represented with a uniformly distributed load of 5 N/m. The moment diagram is required to observe the behavior of the loaded frame. The scenario is illustrated in terms of forces to the right.

The structure is 3 degrees statically indeterminate. B was chosen to be the redundant support; Bx,By and Mb are the redundant forces.

- The primary frame and its shear and moment diagrams are as follows:

The following table will show the diagram for each redundant force and will briefly discuss the procedure:

**Term****Image****Procedure**$ \Delta_{Bx,o} $ Redraw the primary frame with no loads except for an addition of 1k where Bx would have been if the fixed end was there, in the positive direction. So you are replacing the redundant forces with 1 N each at a time. Solve for the reactions and draw the moment diagram. $ \Delta_{By,o} $ Same as Bx but this time the 1 N is put where By would have been. $ \Theta_{Mb,o} $ Same as Bx and Yy but this time the 1 N is put where the moment would have been.

Similar to the previous beam example analysis, slope & deflections are calculated using the primary moment diagram times using the Integration Table, multiply each redundant moment diagram by the primary beam moment diagram.

- $ \Theta_{Mb,o} = {\frac{120}{EI}} N $

- $ \Delta_{Bx,o} = {\frac{-60}{EI}} N $

- $ \Delta_{By,o} = {\frac{-320}{EI}} N $

Then solving for the flexibility terms in the compatibility equations through the virtual moment diagrams:

$ \Delta_{Bx,o} + f_{Bx,Bx}(B_{x}) + f_{Bx,By}(B_{y}) + f_{Bx,Mb}(M_{b})=0 $

$ \Delta_{By,o} + f_{By,Bx}(B_{x}) + f_{By,By}(B_{y}) + f_{By,Mb}(M_{b})=0 $

$ \Theta_{Mb,o} + f_{Mb,Bx}(B_{x}) + f_{Mb,By}(B_{y}) + f_{Mb,Mb}(M_{b}= 0 $

*Using Maxwell's Reciprocal Law we conclude:*

$ f_{Bx,Mb}=f_{Mb,Bx} $ $ f_{By,Ma}=f_{Mb,By} $ $ f_{Bx,By}=f_{By,Bx} $

results:

$ f_{Bx,Bx} = {\frac{5.33}{EI}} N $

$ f_{Bx,By} = f_{By,Bx} = {\frac{-10}{EI}} N $

$ f_{Bx,Mb} = f_{Mb,Bx} = {\frac{-5}{EI}} N $

$ f_{By,By} = {\frac{37.33}{EI}} N $

$ f_{By,Mb} = f_{Mb,By} = {\frac{12}{EI}} N $

$ f_{Mb,Mb} = {\frac{6}{EI}} N $

Substituting the results in our compatibility equation and factoring out EI yields:

$ \ (5.33)(B_{x}) + (-10)(B_{y}) + (-5)(M_{b})=60 $

$ \ (-10)(B_{x}) + (37.33)(B_{y}) + (12)(M_{b})=320 $

$ \ (-5)(B_{x}) + (12)(B_{y}) + (6)(M_{b})=-120 $

Solving simultaneously for the unknown redundant reactions results:

$ B_{x} = - 34.4 N $ $ B_{y} = 42 N $ and $ M_{b} = -132.7 Nm $

These reaction forces are then applied on the original beam and equilibrium is used to solve for the other support reactions:

$ A_{x} = -34.4 N $ $ A_{y} = 102 N $ and $ M_{a} = -288 Nm $

The figure to the right represents the original beam with the real forces and resulting shear and moment diagrams.

## References

- ↑
^{1.0}^{1.1}^{1.2}^{1.3}^{1.4}^{1.5}Hibbeler. R.C. (2012).*Structural Analysis: 8th Edition)*. Pearson Education, Inc. Upper Saddle River, New Jersey. - ↑
^{2.0}^{2.1}^{2.2}^{2.3}^{2.4}Kassimali, A. (2011).*Structural Analysis: SI Edition (4th ed.)*. Stamford, CT: Cengage Learning. - ↑ http://www.engr.uky.edu/~gebland/CE%20582/Class%20Notes/Lecture%20-%20Force%20Method.pdf.
- ↑ Prof. Erochko, J.
*CIVE 3203 Lecture 16 Presentation*(Lecture, CIVE 3203 Structural Analysis, Ottawa, ON, October 25, 2013.)