Force Method for Temperature Changes

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Introduction

The Force Method was developed in 1864 by James C. Maxwell[1] to solve indeterminate structures. This method can be applied to beams, frames and trusses with various degrees of indeterminacy. This section will discuss how the Force Method can be applied to solve indeterminate trusses with a focus on temperature changes in some of the truss members.

Theory

The Force Method utilizes the relationship between the flexibility of a given member and its displacement due to applied loads to produce a compatibility equation.[1] This compatibility relationship provides an additional equation to solve for all the unknown forces in the structure. Flexibility is the amount of deflection a member undergoes per unit load applied to that member.[2] For a truss structure the method of [Virtual Work] is used to calculate the flexibility of a given member, which in turn allows us to solve for a redundant force. [1] Once all redundant forces are known, simple truss analysis with [Method of Joints] is used to solve for the remaining unknown forces.[2]

Important Notes in Analyzing Force Method in Truss Members

The redundant force that is chosen does not have or equate to any axial force value. However, when the member is defined as redundant, it does not mean that it ceases to exist, but the member still rather remains in the structure but is simply cut off from the calculations of the real system.

When the redundant force member is cut from the truss system, it will still exhibit certain tensile and compressive forces due to the external loading upon the truss member. This will eventually cause the cut sections to overlap.

Usually, in using the force method on truss members, when the virtual work analysis step is reached, the virtual force would act upon the corresponding redundant member. This would in turn make the calculations easier as the primary system with the redundant force would be identical to the virtual work system. (Sourced from CIVE3203 lectures posted on CuLearn)


General Method

When approaching an indeterminate truss with members affected by temperature change, the first step is to determine the degree of internal indeterminacy of the truss using the following equation:

    m + r > 2j  [1] 

Where:

  • "j" is the number of joints in the truss and
  • "m" is the number of members and
  • "r" is the number of reactions; for internal indeterminacy calculation r is always 3.

The difference between the two sides of the equation is the degree of internal indeterminacy. Next calculate the degree of external indeterminacy. This can be done by visually inspecting the number of unknown reaction forces in the truss. The degree of external indeterminacy is "k", where k is the total number of unknown reaction forces minus three.[3]


    Number of Total Unknown Reaction Forces - 3 = k (degree of external indeterminacy) [3]

So the total degree of indeterminacy is the sum of the internal and external indeterminacy. The total degree of indeterminacy indicates the number of redundant forces in the system.[2] Now to chose the redundant forces; at least one of the redundant forces must be an internal force since we are focusing on trusses with a member affected by temperature change. The redundant forces are forces within the original truss structure that can be removed to produce a determinate truss without jeopardizing its stability.[2] Furthermore, the number of internal and external redundant forces are equal to their respective degrees of indeterminacy.[2]

    Number of Internal Redundant Forces = Degree of Internal Indeterminacy 
    Number of External Redundant Forces = Degree of External Indeterminacy

The new determinate truss with the redundant forces removed is known as the primary structure.[1] Now using the chosen redundant forces, you can set up all the additional compatibility equations. The number of compatibility equations is equal to the total number of redundant forces (ie. an individual compatibility equation for each redundant force).The compatibility equation is the sum of the deflections at the location of a given redundant force due to all external loads plus the sum of the deflections at the location of a given redundant force due to each redundant force equals to zero.[1]

     Number of Compatibility Equations = Number of Internal Redundant forces + Number of External Redundant forces[3]
     $ \sum $ Deflection at Redundantn due to External Loads + $ \sum $ Deflection at Redundantn due to Redundant loads = Zero  [1]

The underlining assumption in the compatibility equation above is that the truss is in a state of equilibrium; therefore it does not experience any resulting deformation (i.e. the total deflection equals to 0).[2] In order to calculate the deflection of the truss at the location of a given redundant force due to all external loads, the primary structure is used with all the original external loads applied to it.[2][Virtual Work] is then applied to calculate the deflection at the location of each redundant force due to the external loads.

    $ \Delta $ = $  \sum_{i=1}^{m} \frac{1}{EA}  $ Force in member i in the primary system $ \times $ Force in member i in the virtual system $ \times $ Length of the member [1] 
    $  \Delta = \sum_{i=1}^{m} \frac{1}{EA} F \times F_v \times L  $ [1] 

Where:

  • m is the number of members in the system and
  • $ \Delta $ is total deflection due a unit load in the place of a redundant member


It should be noted that the temperature change in any affected members of the truss is treated like an external load but, it is not included as one of the external loads on the primary structure. Instead the deflection due to temperature change is calculated separately near the end of the method and is then added as a constant deflection to the left side of the compatibility equation for internal redundant forces ONLY.[3] The deflection due to temperature change for an external redundant is considered to be zero. [3]

Next the flexibility coefficients for each redundant force need to be calculated. This is also done through virtual work: multiplying the primary system without external loads with a virtual system. The primary systems for this step resemble the virtual systems from the previous step.[1] Therefore, the the flexibilities can be calculated by multiplying the general virtual systems by each other and by themselves. For the general case, if the truss is one degree indeterminate, only one flexibility term needs to be calculated. If the degree of indeterminacy is two, then three flexibility terms need to be calculated. The equations for these cases are shown below.

    (one degree of indeterminacy) $  \mathit{f} = \frac{F_v^2*L}{EA}  $[1]  

Where:

  • $ F_v $ is the member force in the virtual system
  • $ L $ is the length of the member
  • $ E $ is the Modulus of Elasticity of the member
  • $ A $ is the area of the member
    (two degrees of indeterminacy) $ \mathit{f}_{11} = \frac{F_v_1^2*L}{EA}  $ and $ \mathit{f}_{22} =\frac{F_v_2^2*L}{EA}  $ and $ \mathit{f}_{12} = \mathit{f}_{21} =\frac{ F_v_1*F_v_2*L}{EA}  $ by Maxwell's law of reciprocal deflections [1] 

Where:

  • $ F_v_1 $ is the member force in the virtual system for the first redundant force
  • $ F_v_2 $ is the member force in the virtual system for the second redundant force
  • $ L $ is the length of the member
  • $ E $ is the Modulus of Elasticity of the member
  • $ A $ is the area of the member

As mentioned before the deflection due to temperature change is calculated separately from the external loads. This is done using the equation where deflection due to temperature change equals to the product of the change in temperature times the coefficient of thermal expansion times the length of the effected truss member.[3]

    $ \delta $ = $  (\Delta T)\alpha L  $[1] 

Where:

  • $ \delta $ is the deformation due to temperature change for the affected member
  • $ \alpha $ is the coefficient of thermal expansion of the affected truss member and
  • L is the length of the affected truss member

Then the deformation due to temperature change of each affected member is multiplied by its corresponding force in the virtual system of the internal redundant force. These are then summed together to find the total deflection cause by the temperature change of the internal virtual system. [1]

    $ (\Delta_T)_n = \sum F_v \delta  $[1] 

So now that all required terms in the compatibility equation has been determined, we input them back into the equations and solve for the redundant forces. For each Redundant force the compatibility equation becomes:

     $  (\Delta_{T})_n + \Delta  + \;(\; \sum_{i=1}^{m}\; \ ( \; f_{ni} \times  F_i \;) = 0  $ [1] 

Where: $ \Delta_{T} $ = the deflection due to temperature change of the given redundant

$ \Delta $ = the deflection due to external loads of the given redundant

$ \ f_{ni} $ = the flexibility of the given redundant "n" due to a unit load at the location of redundant "i"

$ F_i $ = the unknown ith redundant force

$ m $ = the number of redundant force


When the total degree of indeterminacy is one; m = 1 in the above equation resulting in one redundant force. This permits the compatibility equation to be simply rearranged to solve for the unknown redundant force.[2] On the other hand, with multiple degrees of indeterminacy, the above equation is applied for each redundant force resulting in a series of equations. To solve for all the unknown redundant forces, one simply solves a system of equations.[2]

Finally, with all the redundant forces known in the original truss structure,we simply apply the [Equilibrium Equations] and[Method of Joints] to solve for all remaining external and internal forces in the original truss structure.[2]


Simplified General Procedure for Analysis of Truss Structure with Temperature Changes[1] [3]

1. Determine the Degree of Internal Indeterminacy and External Indeterminacy

2. Chose Redundant forces:

  • Number of Internal redundant forces = Degree of Internal Indeterminacy
  • Number of External reaction redundant forces = Degree of External Indeterminacy

3. Once Redundant forces are chosen, remove them from the original truss to create the primary structure

4. Apply the all the external loads to the primary structure (excluding temperature change)

5. Use [Virtual Work] to solve for the deflection at the location of each redundant due to external loads

6. Use [Virtual Work] to solve for flexibility of at a given redundant due to unit loads at each redundant

7. Solve deflection due to temperature change for each redundant force. Deflections due to temperature change for an external redundant force is zero.

8. Input all tabulated term back into the compatibility equations; solve for unknown redundant forces

9. Re-apply the redundant forces in the original truss; Solve the remainder of the truss using [Equilibrium Equations] and [Method of Joints]

Example Problems

Example with 1 Degree of Indeterminacy

This method has been adapted from Kassimali's textbook, Structural Analysis, pages 574 to 577.[1]


Given the truss shown in Figure 1.1, find the internal forces of the truss given that member BC experiences an increase in temperature of 50$ ^{\circ} $C and member FE experiences an increase in temperature of 75$ ^{\circ} $C.

Given that EA is constant (E = 200 GPa, A= 3000 mm$ ^2 $), and $ \alpha $ = 1.4 x10$ ^{-5} $.

Figure 1.1: Original Truss [4] [5]


Step 1: Finding the Degree of Indeterminacy

Check the degree of indeterminacy of the truss.

$ m + r > 2j $

For this truss:

$ 10 + 3 > 2(6) $

Since 13 is greater than 12 the truss is 1 degree internally indeterminate. From looking at the support reactions, the truss is externally determinate (3 reactions and 3 equilibrium equations). Since there are no external applied loads, the support reactions are equal to 0.

Step 2: Choose Redundant(s)

Choose redundant(s). Remember a redundant cannot affect the stability of the truss and the number of redundants should be equal to the total degree of indeterminacy.

For this example the member BE was chosen as the redundant because at joint B there would only be 1 additional vertical force which would allow the internal forces at the joint to be solved using [equilibrium]. It also maintains the stability of the truss.

Figure 1.2: Truss Without Redundant Member [4] [5]

Figure 1.2: Truss Without Redundant Member [4] [5]

Step 3: Primary System with External Loads

Create the primary system with external loads(remove redundant(s)). For this primary structure, all the internal forces would be zero-force members since there are no external loads acting on the truss. Therefore the primary system with external loads does not need to be added into the virtual work table as the deflection due to this system is 0.

See above Figure 1.2

Step 4: Virtual Work for Trusses

Create the [Virtual Work Table] for trusses to keep track of the values in the truss. The headings required are shown below.[1] (Note: recall that the primary system with external forces can be ignored for this analysis)

Member L (m) A (m$ ^2 $) $ \Delta $ T ($ ^{\circ} $C) F$ _v $ (kN/kN) $ \delta = \alpha\Delta $T L(m) $ \frac{F_v^2 L}{EA} $ (m/kN) $ F_v \delta $ (kN m)

Step 5: Applying Virtual Work to find Virtual System

Create the virtual system by applying a point load in place of the redundant on the primary truss. Using this force calculate the rest of the forces in the primary structure using the [method of joints].

Figure 1.3: Truss with Unit Loads Being Applied at Location of Redundant Members [4] [5]


Figure 1.3: Truss with Unit Loads Being Applied at Location of Redundant Members [4] [5]


From Rules of Zero Force Members and From Looking at the Truss:

Zero Force Members:
AF
AB
DE
DC

@ Joint B:

$ \sum F_y = 0 $
$ 0 = 1(\frac{5}{5.59}) - BF(\frac{5}{5.22}) $
$ BF = 0.93 \;\;kN/kN [C] $


$ \sum F_x = 0 $
$ 0 = -0.93(\frac{1.5}{5.22}) - 1(\frac{2.5}{5.59}) - BC $
$ BC = 0.18 \;\;kN/kN [C] $

@ Joint F:

$ \sum F_x = 0 $
$ 0 = -0.93(\frac{1.5}{5.22}) + 1(\frac{2.5}{5.59}) + FE $
$ FE = 0.72 \;\;kN/kN [C] $

The remaining forces can be found by using the symmetry of the truss.

Figure 1.4: Solved Virtual System [4] [5]

Figure 1.4: Solved Virtual System [4] [5]


Fill in the virtual work table with the forces found in the virtual system.

Member L (m) A (mm$ ^2 $) $ \Delta $ T ($ ^{\circ} $C) F$ _v $ (kN/kN) $ \delta = \alpha\Delta $T L(m) $ \frac{F_v^2 L}{EA} $ (m/kN) $ F_v \delta $ (kN m)
AB 5 3000 0 0
AF 8.2 3000 0 0
BC 4 3000 50 -0.18 [C]
BF 5.2 3000 0 -0.93 [C]
BE 5.6 3000 0 +1.0 [T]
CF 5.6 3000 0 +1.0 [T]
CE 5.2 3000 0 -0.93 [C]
FE 1 3000 75 -0.72 [C]
DC 5 3000 0 0
DE 8.2 3000 0 0

Step 6: Calculate the Flexibility

Calculate the flexibility for the redundant by filling in the second last column of the virtual work table. Multiply each force in the virtual system by itself and by the length of the member and then divide the product by the product of the modulus of elasticity and the area of the beam. Add these values to the table and find their sum. This sum is the flexibility of the internal redundant.

$ \mathit{f}_{BE,BE} = \frac{\sum F_v^2 L}{EA} $
Member L (m) A (mm$ ^2 $) $ \Delta $ T ($ ^{\circ} $C) F$ _v $ (kN/kN) $ \delta = \alpha\Delta $T L(m) $ \frac{F_v^2 L}{EA}(\times 10^{-6} $ m/kN) $ F_v \delta $ (kN m)
AB 5 3000 0 0 0
AF 8.2 3000 0 0 0
BC 4 3000 50 -0.18 [C] 0.21
BF 5.2 3000 0 -0.93 [C] 7.56
BE 5.6 3000 0 +1.0 [T] 9.33
CF 5.6 3000 0 +1.0 [T] 9.33
CE 5.2 3000 0 -0.93 [C] 7.56
FE 1 3000 75 -0.72 [C] 0.85
DC 5 3000 0 0 0
DE 8.2 3000 0 0 0
$ \sum $ - - - - - 34.85

Therefore:

$ \mathit{f}_{BE,BE} = 34.85 \times 10^{-6} m/kN $

Step 7: Find Change in Length Due to Temperature Changes

The change of length due to temperature changes are taken into account by multiplying the temperature change by the coefficient of thermal expansion and the length of the member and record the values in the appropriate column in the table. This process uses the following equation:

$ \delta = \alpha \Delta T L $

After deformation of each member due to temperature change ($ \delta $) has been calculated, the total displacement in the system can be calculated by multiplying the deformation for each member by the virtual force in the member, then taking the sum of these values. This is using the principle of virtual work and can be seen in the equation below.

$ 1 kN (\Delta_{T})_{BE} = \sum F_v \delta $
Member L (m) A (mm$ ^2 $) $ \Delta $ T ($ ^{\circ} $C) F$ _v $ (kN/kN) $ \delta = \alpha\Delta $T L(m) $ \frac{F_v^2 L}{EA}(\times 10^{-6} $ m/kN) $ F_v \delta $ (kN m)
AB 5 3000 0 0 0 0 0
AF 8.2 3000 0 0 0 0 0
BC 4 3000 50 -0.18 [C] 0.0028 0.21 -0.000501
BF 5.2 3000 0 -0.93 [C] 0 7.56 0
BE 5.6 3000 0 +1.0 [T] 0 9.33 0
CF 5.6 3000 0 +1.0 [T] 0 9.33 0
CE 5.2 3000 0 -0.93 [C] 0 7.56 0
FE 1 3000 75 -0.72 [C] 0.00105 0.85 -0.000751
DC 5 3000 0 0 0 0 0
DE 8.2 3000 0 0 0 0 0
$ \sum $ - - - - - 0.00385 34.85 -0.00125

Therefore:

$ (\Delta_{T})_{BE} = -0.00125 m $

Step 8: Using the Compatibility Equation

For this example since there is only 1 redundant there is only 1 compatibility equation and it is as follows:

$ (\Delta_{T})_{BE} + \mathit{f}_{BE,BE} \; F_{BE} = 0 $

Where $ F_{BE} $ is the unknown redundant force for which the compatibility equation will be solved for.

Subbing in the values which were just found from the table:

$ -0.00125 m + (34.85 \times 10^{-6} m/kN) \; F_{BE} = 0 $

Therefore:

$ F_{BE} = 35.9 kN $

Step 9: Finding the Real Forces in the Truss

Since there are no external loads, the virtual system has a 1:1 relationship with the real system. Since the force in BE was determined, the internal virtual force for each member can be mutiplied by this value to find the real system.

Figure 1.5: Solved Real System [4] [5]


Member F$ _v $ (kN/kN) F = F$ _v \; F_{BE} $ (kN)
AB 0 0
AF 0 0
BC -0.18 [C] -6.43 [C]
BF -0.93 [C] -33.6 [C]
BE +1.0 [T] 35.9 [T]
CF +1.0 [T] 35.9 [T]
CE -0.93 [C] -33.6 [C]
FE -0.72 [C] -25.7 [C]
DC 0 0
DE 0 0







----END OF PROBLEM----

Example with Multiple Degrees of Indeterminacy

This method has been adapted from Kharagpur's "Analysis of Indeterminate Structures by the Matric Force Method" Ex. 10.3 Pg 11. [3]

Determine all reaction forces and memeber forces in the truss shown in Figure 2.1 due to a temperature change of 65°C in members BC and CD as well as external loads at point A and H. For this problem assume EA is constant and the members are made out of structural steel, so:

Figure 2.1: Original Structure [4]

$ E = 200000 MPa $

$ A = 3000 mm^2 $

$ \alpha = 1.33e^-^5 / ^{\circ}C $

Step 1: Finding the Degree of Indeterminacy

$ m + r > 2 j $
$ m=18, r=4, j=10 $
Therefore,
$ m+r=22 $ and $ 2j=20 \Rightarrow 2^{\circ}S.I. $

Step 2: Choose Redundant(s)

Due to this problem being 2° statically indeterminate, two redundant forces must be selected. Since this structure is 1° externally indeterminate, an external reaction force must be used as one of the redundant forces. Furthermore, due to the structure being a truss with a change in temperature, the other redundant must be one of the internal members.

Figure 2.2: Structure with Redundant Forces [4]

Figure 2.2: Structure with Redundant Forces [4]

As shown in figure 2.2 , the reaction force $ B_x $ was chosen as the external redundant because it changes the pin support at B into a roller and in turn making the structure externally determinate. For the internal redundant, member FI was chosen because it is not a zero forces member and removing it will not result in an unstable structure.

Step 3: Create Compatibility Equation

The following are the compatibility equations for the internal redundant and the external redundant respectively.


(Equation 1) $ (\Delta_T)_1 + \Delta_{FI} + \mathit{f}_{FI,FI} F_{FI} + \mathit{f}_{FI,Bx} B_x = 0 $ and;
(Equation 2) $ (\Delta_T)_2 + \Delta_{Bx} + \mathit{f}_{Bx,Bx} B_x + \mathit{f}_{Bx,FI} F_{FI} = 0 $

Where for equation one:

  • $ (\Delta_T)_1 $ is the deflection at member FI due to temperature change in the truss
  • $ \Delta_{FI} $ is the deflection at member FI due to all applied external loads
  • $ \mathit{f}_{FI,FI} $ is the flexibility of member FI due to the redundant force FI
  • $ \mathit{f}_{FI,FI} $ is the flexibility of member FI due to the redundant force Bx

Note: The terms in Equation 2 follow similar notation as Equation 1

Step 4: Primary Structure with External Loads

Remove the redundant member/force and calculate all reactions and member forces due to the external loads on the now determinate structure.

Figure 2.3: Primary Structure with External Loads [4]

Figure 2.3: Primary Structure with External Loads [4]

Step 5: Virtual System with Internal Unit Load

Apply a 1 kN virtual unit on the interior redundant member and solve for all member forces. For this problem, forces due to virtual work on the internal member FI will be noted as $ F_v_1 $ and forces due to external virtual work at point B will be denoted as $ F_v_2 $.

Figure 2.4: Virtual System with Internal Unit Load [4]

Figure 2.4: Virtual System with Internal Unit Load [4]

Step 6: Virtual System with External Unit Load

Apply a 1 kN virtual unit load at the location of the exterior redundant force and calculate all member forces and reactions.

Figure 2.5: Virtual System with External Unit Load [4]

Figure 2.5: Virtual System with External Unit Load [4]

Step 7: Complete Virtual Work Table for Deflection Due to External Loads

When creating a primary structure with redundant forces the resulting structure and member forces are the same for both the real and virtual system. Therefore, $ (F=F_v) $[6] . Therefore, we are now able to tabulate all values needed to complete the compatibility conditions.

Member L (m) A (m$ ^2 $) $ \Delta $ T ($ ^{\circ} $C) F (kN) F$ _v_1 $ (kN/kN) F$ _v_2 $ (kN/kN) $ F * F_v_1*L (kNm) $ $ F * F_v_2*L (kNm) $ $ F_v_1^2*L (m) $ $ F_v_2^2*L (m) $ $ F_v_1*F_v_2*L (m) $
AB 6 0.003 0 -37.5 0 0 0 0 0 0 0
BC 6 0.003 65 -37.5 0 1 0 -225 0 6 0
CD 6 0.003 65 -75 0 1 0 -450 0 6 0
DE 6 0.003 0 0 0 0 0 0 0 0 0
AH 10 0.003 0 62.5 0 0 0 0 0 0 0
HI 6 0.003 0 -37.5 0 0 0 0 0 0 0
FG 12 0.003 0 37.5 -1.66 0 -747 0 33.07 0 0
FC 7.21 0.003 0 -45.06 1 0 -324.88 0 7.21 0 0
BF 4 0.003 0 -25 0 0 0 0 0 0 0
IJ 6 0.003 0 0 0 0 0 0 0 0 0
IG 7.21 0.003 0 -45.06 1 0 -324.88 0 7.21 0 0
CG 7.21 0.003 0 0 1 0 0 0 7.21 0 0
DG 4 0.003 0 -25 0 0 0 0 0 0 0
EJ 10 0.003 0 0 0 0 0 0 0 0 0
FH 4 0.003 0 -50 0 0 0 0 0 0 0
GJ 4 0.003 0 0 0 0 0 0 0 0 0
CI 8 0.003 0 25 -1.11 0 -222 0 9.86 0 0
$ \sum $ -1618.77/EA -675/EA 64.55/EA 12/EA 0

Note: Due to the formation of the truss members in this structure, $ \mathit{f}_{FI,Bx}=\mathit{f}_{Bx,FI}=0 $ but this is not always the case. [1]

Step 8: Calculate deflections due to Temperature Change for Interior and Exterior Redundant

$ (\Delta_T)_1 = \alpha T L = ( 1.33e^-^5 / ^{\circ}C ) ( 65 ^{\circ}C ) (12 m) = 10.37e^{-3} m $


$ (\Delta_T)_2 = \alpha T L = 0 \; \Rightarrow \; $ No deflections are produced at an external redundant due to a temperature change in a truss member

Step 9: Solve for Forces using Compatibility Equations

$ (\Delta_T)_1 + \Delta_{FI} + \mathit{f}_{FI,FI} F_{FI} + \mathit{f}_{FI,Bx} B_x = 0 \; \Rightarrow \; 10.37e^{-3} m \; + \; (-1618.77/EA) \; + (64.55/EA)F_{FI} \;+\;0 = \;0 $
Therefore, $ F_{FI} \; = \; -71.3 \; kN \; = \; 71.3 \; kN \; (C) $


$ \(\Delta_T)_2 + \Delta_{FI} + \mathit{f}_{Bx,Bx} B_x + \mathit{f}_{Bx,FI} F_{FI} = 0 \; \Rightarrow \; 0 \; + (-675/EA) \; + \; (12/EA)B_x \; + \; 0 \; = 0 $
Therefore, $ B_x \; = \; 56.25 \; kN $

Step 10: Solve for all Truss Members and Reactions

Substitute $ F_{FI} $ and $ B_x $ into the original structure and solve for all reaction forces and member forces using truss analysis.

Figure 2.6: Final Structure [4]
Member F (kN)
AB -37.5
BC 18.75
CD -18.75
DE 0
AH 62.5
HI -37.5
FI -71.31
FG 156.45
FC -116.37
BF -25
IJ 0
IG -116.37
CG -71.31
DG -25
EJ 0
FH -50
GJ 0
CI 104.04

Glossary

Compatibility Equation: An equation which combines the deformations caused by the primary system with the deformations caused by a point load in place of the redundant. [1] The deformations caused by the unit point load are written in terms of the member's flexibility and the unknown force in that member. [1] A typical compatibility equation for a truss with a single degree of indeterminacy is: $ \Delta_{Primary} + {f}_{Found\;from\;virtual\;system} \times F_{unknown} = 0 $.[1]

Flexibility: Defined as being the change in length of a member divided by the force applied to or existing in the member, $ \mathit{f} = \frac{\Delta}{P} $.[2]

Primary Structure: A stable determinate version of the structure which is created by removing the redundant member or force.[1]

Redundant Restraints: A redundant can be chosen as any force, moment or reaction which when removed from an indeterminate structure results in a stable, determinate structure (primary structure).[1]

Coefficient of Thermal Expansion: " the amount of expansion (or contraction) per unit length of a material resulting from one degree change in temperature" [7]

References

  1. 1.00 1.01 1.02 1.03 1.04 1.05 1.06 1.07 1.08 1.09 1.10 1.11 1.12 1.13 1.14 1.15 1.16 1.17 1.18 1.19 1.20 1.21 1.22 1.23 1.24 Kassimali, A., 'Structural Analysis', 4th Ed., Cengage Learning, 2011
  2. 2.00 2.01 2.02 2.03 2.04 2.05 2.06 2.07 2.08 2.09 2.10 Erochko, J., 'Lecture #16 Force Method for Structures with Multiple Degrees of Indeterminacy', 2013, [1]
  3. 3.0 3.1 3.2 3.3 3.4 3.5 3.6 3.7 Kharagpur, I., Structural Analysis,version 2 CE,2008[2]
  4. 4.00 4.01 4.02 4.03 4.04 4.05 4.06 4.07 4.08 4.09 4.10 4.11 4.12 4.13 4.14 4.15 4.16 4.17 Valiquette, B., Student, Carleton University, 2013
  5. 5.0 5.1 5.2 5.3 5.4 5.5 5.6 5.7 Arnold, P., Student, Carleton University, 2013
  6. Erochko J. Force Method for an Indeterminate Truss Using an Internal Redundant [Youtube Video] November 14, 2013
  7. 'The Free Dictionary', [3]