# Force Method Using Internal Forces as Redundants

## Contents

- 1
**Introduction to Force Method** - 2
**Internal force redundant** - 3
**Compatibility Conditions** - 4
**Compatibility Equations** - 5
**Key Terms** - 6
**Procedure for analyzing a truss using internal axial force as a redundant** - 7
**Procedure for analyzing a beam using internal moments as a redundant** - 8
**Example Problem 1** - 9
**Example Problem 2** - 10
**Table 1 for Example Problem 2** - 11
**References**

**Introduction to Force Method**

Force method is commonly used while analyzing an indeterminate structure. The basic concept of force method is to pick an appropriate redundant force(s) and remove it from the structure to make the structure statically determinate. Once that is done, we can look at the structure separately, one with all the external loads and the other with the just the redundant force as the external load. Then, with the help of compatibility conditions and equations of equilibrium, unknowns are calculated. When using the force method, any redundant force can be picked as long as the removal of the redundant force gives a determinate structure. This section focuses on applying the force method using only internal redundants.
^{[1]}

**Internal force redundant**

Both internal axial force and internal moments are considered to be an internal redundant. They can be used when analyzing any indeterminate structure as long as the removal of the internal force redundant results in a determinate and a stable structure. When analyzing a continuous beam, using internal moment as redundant is more convenient compared to any other method. However, only internal force redundant method is used when analyzing an externally determinate but internally indeterminate structure.^{[2]}.

**Compatibility Conditions**

**For Trusses when using Internal Axial Force:** After the redundant force member is selected and a cut is made, the two points from the cut in the two sides have to be still located at the same point. ^{[3]}

**For Internal Moments:** In the actual continuous beam, the slopes from the point of the hinge to the right and from the point of the hinge to the left side of the beam have to be the same.^{[4]}.

**Compatibility Equations **

Compatibility equations are significant while solving for internal forces. Compatibilty equations need to be set up in the right manner to find the forces.

**For Internal Moment Rotation:**

$ \theta_{CO,REL}+f_{c,REL} Mc=0 $

Above is the compatibility equation for an internal moment (hinge) as a redundant. When using the compatibility equation θCO,REL is the rotation caused by the external loads applied on the bean which is found by calculating the () . And fC,RELis the flexibility(rotation) caused by the internal moment hinge that is applied as a unit load. Mc is the total internal moment which can be calculated by equilibrium equations.
^{[5]}.

**For Internal Force deflection:**

$ \Delta_{AC} + f_{AC,}_A_C F_{AC} = 0 $

∆AC is the deflection between the members AC and is calculated by using . fAC, is the deflection caused by the unit load applied on that member to calculate the redundant force. FAC is the force in the member AC.
^{[6]}.

**Key Terms**

**Redundant Force:**Redundant forces are found in an indeterminate beam. It is the extra force(s) applied on the beams which when removed from the beam makes it statically determinate. The degree indeterminacy states the number of redundant forces in a beam. ^{[7]}.

**Indeterminate Beam:** When the number of reactions exceeds the number of equations of equilibrium, then it is an indeterminate beam. ^{[8]}

when
*m = # of members*
*r = reaction forces*
*j = # of joints*
*e = equation of condition*;

*Degree of Indeterminacy for Beam* = $ (3m+r) - (3j+ec) $ ^{[4]}.

*Degree of Indeterminacy for Truss* = $ (m+r)-2j $ ^{[4]}.
.

**Determinate Beam:** When the number of reactions is equal to the number of equations of equilibrium, then it is a determinate beam.

It is a determinate beam if * $ r = 3 + e $* ^{[4]}.

A truss is statically determinate if * $ m = 2j+3 $* ^{[4]}.

**Flexibility :** Flexibility is a coefficient for finding the force by force method. It is the deflection caused by the unit load applied to the structure.

**Deflection:** Deflection is caused by the external load applied to a structure.

**Procedure for analyzing a truss using internal axial force as a redundant **

1. Calculate the degree of indeterminancy. Make sure indeterminancy is 1.

2. Pick an appropriate redundant force from the given members of the truss as long as the resulting structure is stable and statically determinate.

3. Make a cut on the particular member that is picked as a redundant.

4. Set up the compatibility equation. For example, a sample equation is $ \Delta_{AC} + f_{AC,}_A_C F_{AC} = 0 $

5. Find the relative displacement of the two different ends of the cut ($ \Delta{AC} $) as well as relative displacement of the ends of the cut under the redundant load ($ f_{AC,AC} $).

6. Find relative displacement ($ \Delta{AC} $) and flexibility ($ f_{AC,}_A_C F_{AC} $ )using virtual work for truss.

7. Finally, the axial force can be calculated. The rest of the forces can then be calculated by equilibrium equations.
^{[9]}.

**Procedure for analyzing a beam using internal moments as a redundant **

1. Calculate the degree of indeterminancy. Make sure the degree of indeterminancy is equal to 1.

2. Insert an internal hinge in the middle of the beam in order to use internal moment as the redundant.

3. Set up the compatibility equation. For example a sample equation is $ \Delta_{CO,REL}+f_{C,REL}Mc=0 $

4. Find the total relative rotation between any sides (since both the sides are equal) of the hinge ($ \Delta{CO,REL} $ ) and flexibility ($ f_{C,REL} $ ) using the beam deflection formulas.

5. Finally, solve for $ M_c $ from the compatibility equation. The remaining forces can be calculated by equilibrium equations.
^{[9]}.

**Example Problem 1**

Looking at Figure 1 seen on the right, an Indeterminate beam is shown. This beam consists of a pin support at point A and roller supports at points B and C. The support at A has vertical and horizontal reaction forces, whereas the forces at B and C only have vertical reaction forces as they are rollers. This question contains more unknowns than equations, which means we would not be able to solve it in its current form. To solve the reaction forces, we must make the beam determinate, to do this we consider that there is a hinge at point C instead of a roller. This means the roller at B will actually be considered to be a hinge. (Please see Determinacy, Indeterminacy and Stability for a better understanding of indeterminate beams)

So, now that there is a hinge at C other than a roller, we know that by solving, we will have equal equations and equal unknowns which means the reactions at A and C can be found. To find the reactions at A and C, the moment around point C is taken into account: $ M_c --> A_y * (32) - 500(24) - 500 (8) = 0 --> A_y = 500 Kn $

To find the vertical reaction at C:

$ \sum {F_y} --> A_y - 500 - 500 + C_y =0 --> C_y = 500 Kn $

To find the horizontal reaction a A:

$ \sum {F_x} --> A_x + 0 = 0 --> A_x = 0 $

These forces can be found in the first Shear and Moment diagram labelled Figure 4.

The next step to finding all the reaction forces, is by finding the compatibility equation. To find the compatibility equation for the given beam, a unit force, also called the redundant force in this case, is applied to the hinge at C, this unit force may be of any magnitude desired. For simplicity, a magnitude of 1 KN is used as the unit force. This unit force will deflect the beam upwards as shown in figure 2 (Virtual System). As seen in figure 2, this 1 Kn unit force is added and all other forces (reaction forces and external forces) are not considered when drawing the deflected shape. A maximum displacement of $ F_{bb} $ is drawn. The first value in our compatibility equation is $ F_{bb} $. So now, the reactions at A and C must be found in accordance with the new 1Kn force on the beam. (Again, disregarding all other forces). The same procedure can be done: $ M_c --> -A_y * (32) + (1) (16) = 0 --> A_y = -0.5 Kn $

$ \sum {F_y} --> -A_y + 1 -C_y =0 --> C_y = -0.5 Kn $

$ \sum {F_x} --> A_x + 0 = 0 --> A_x = 0 $

The reaction forces at A and C are negative, which means they act downwards. This can be seen in the shear and moment diagrams in figure 5.

Now, the unit force used previously is removed, the external loads and reactions are taken into account. (We continue to consider the roller at C to be a hinge for this step as well) The reactions at A and C were found earlier in step 1. When the external forces are taken into account, the beam has a deflected shape of that seen in figure 3. The maximum displacement (downward) can be named as $ \Delta {B_o} $. This is the second part of the compatibility equation. The shear and moment diagrams can be seen to the right.

The two parts of the compatibility equation are : $ F_{bb} $ and $ \Delta {B_o} $. Now, the upward maximum displacement of the beam ( $ F_{bb} $ ) which was caused by the 1 Kn unit force, is equal to the downward displacement caused by the external forces. This concludes that our compatibility equation is:

```
$ F_{bb}*B_y + \Delta {B_o} = 0 $
```

( The reason at reaction a B ($ B_y $) is included in the equation is because that is the force at the reaction at B once we reconsider it as a roller instead of a hinge.)

The final step to solving the reaction forces, is to determine the magnitudes of $ F_{bb} $ and $ \Delta {B_o} $. To do so we will use the method of Beam Virtual Work. a virtual work integration table will be used:

( This is not the only way to find the magnitudes of $ F_{bb} $ and $ \Delta {B_o} $, another way could be to use the Conjugate Beam Method)

A full description of how to use this table can be found in Beam Virtual Work , Truss Virtual Work, Frame Virtual Work. A brief description of how to use this table: 1. Draw the moment diagrams for the real system and the virtual system. 2. Look at the shape of the real system moment diagram and compare it with the shape of the virtual system moment diagram. 3. Using the table, determine which equation should be used.

In this case, the moment diagrams are (Triangle + Triangle) + (Rectangle + Triangle). By looking that the integration table, it can be deduced that the equation will be :

$ F_{bb} = LMQ/3 $ + $ LMQ/2 $

$ \Delta {B_o} = ((8) (4000) (16) / 3) + ((4) (4000) (8) / 2) --> \Delta {B_o} = 234666.667/EI $ (Note: EI is a constant given in the question. All values on the moment diagram are divided by EI, since EI is constant, they are not shown and were taken as common factors.)

Now that $ \Delta {B_o} $ was found, the value of $ F_{bb} $ should be found. To do this, similar steps to how $ \Delta {B_o} $ was found are followed. Instead of using the real system moment diagram and the virtual system moment diagram, only the virtual system moment diagram will be used. Basically, it is as if both the moment diagrams are the same. We will use the same moment diagram twice (The virtual work moment diagram ofcourse, from figure 5). This means the integration table equation would be (Triangle + Triangle) + (Triangle + Triangle) From the integration table: $ F_{bb} = LMQ/3 + LMQ/3 --> F_{bb} = ((8)(8)(16)/3) + ((8)(8)(16)/3) = = 682.666667/EI $ Finally, our compatibility equation can now be used to find $ B_y $ (The vertical reaction at B)

$ F_{bb}*B_y + \Delta {B_o} = 0 --> (682.667/EI)B_y + 234666.667/EI = 0 --> B_y = 343.75/EI $ So the reaction at the roller at B is equal to 343.75 Kn.

**Example Problem 2**

^{[4]}.

Given the Truss structure in Figure 6, find the reactions and the forces in each member using the method of consistent deformations.

**Step #1:**

Using the indeterminacy test, it can be seen that the truss is externally determinate and internally indeterminate. Therefore, one of the internal members must be taken as a redundant, in our case, member BE, which will make the structure internally determinate. An easy way to imagine this is simply take out member BE and solve for the reactions and forces in the members using the method of joints as shown in figure 8.

**Step #2:**

In figure 7, the structure was once again considered but this time taking out all external reactions and supports and placing a unit load of 1KN(T) on member BE. Then again using method of joints the rest of the forces in the members were found.

**Step#3:**

After the forces in the members were found both using all external reactions without the redundant member and again with no external reactions or supports but with a force member of 1KN for member BE, a table was created ,Table 1.

**1st column**: showing the members.

**2nd column**: the lengths of each member.

**3rd column**: the real forces of the members while omitting member BE (redundant).

**4th column**: the forces of each member omitting all external reactions and supports while adding a 1KN force on member BE.

**5th column**: the product of the 2nd, 3rd, and 4th columns for each member alone.

**6th column**: taking the product of the 4th column multiplied by the lengths of each member.

**7th column**: showing the force in each member.

**Table 1 for Example Problem 2**

Members L(cm) $ \ F_D(KN) $ $ \ U_B_E(KN/KN) $ $ \ F_DU_B_EL(KN.m) $ $ \ U^2_B_EL(m) $ $ \ F = F_D+U_B_EF_B_E(KN) $ AB 5 125 -0.78 -487.5 3.042 272.42 BC 5 125 0 0 0 125 CD 6.4 -160 0 0 0 -160 DE 5 -437.3 -0.78 1705.47 3.042 -289.9 EA 4 0 -0.62 0 1.5376 117.18 AD 6.4 400 1 2560 6.4 211 BE 6.4 0 1 0 6.4 -189 BD 4 -150 -0.62 372 1.5376 -32.82

Finally, using the compatibality equation, $ \Delta_{BE} + f_{BE,}_B_E F_{BE} = 0 $,

$ \Delta_{BE} $ = $ \frac{1}{EA}} $ X $ \sum $ $ \ F_DU_B_EL(KN.m) $ = $ \ {\frac{4149.97(KN.m}{EA}} $^{[4]}.

$ \ f_{BE,}_B_E $ = $ \frac {1}{EA} $ X $ \sum $ $ \ U^2_B_EL(m) $ = $ \frac {21.9592(m)}{EA} $^{[4]}.

$ \ F_{BE} $ = $ \ {\frac{\Delta_{BE}}{f_{BE,}_B_E}} $ = -189 KN = 189 (C)^{[4]}.

**References**

- ↑ http://www.learningace.com/doc/637135/8c278793fb4c4dca9d58f38375e0f60f/chapter5-forcemethod
- ↑ Kassimali, A. (2011).
*Structural Analysis: SI Edition (4th ed.)*, Pg 531,535. Stamford, CT: Cengage Learning. - ↑ J. Erochko, Cive 3203, Video Note, Topic: "Force Method Example", Carleton University, Ottawa, Ontario, 07/11/13.
- ↑
^{4.0}^{4.1}^{4.2}^{4.3}^{4.4}^{4.5}^{4.6}^{4.7}^{4.8}Kassimali, A. (2011).*Structural Analysis: SI Edition (4th ed.)*. Stamford, CT: Cengage Learning. - ↑ Kassimali, A. (2011).
*Structural Analysis: SI Edition (4th ed.),Pg 534*. Stamford, CT: Cengage Learning. - ↑ Kassimali, A. (2011).
*Structural Analysis: SI Edition (4th ed.)*,Pg 537. Stamford, CT: Cengage Learning. - ↑ Kassimali, A. (2011).
*Structural Analysis: SI Edition (4th ed.), Pg 103*. Stamford, CT: Cengage Learning. - ↑ http://ocw.nthu.edu.tw/ocw/upload/8/259/Chapter_10-98.pdf
- ↑
^{9.0}^{9.1}Kassimali, A. (2011).*Structural Analysis: SI Edition (4th ed.)*, Pg 538,543. Stamford, CT: Cengage Learning.