# Truss Virtual Work

## Contents

- 1 Introduction
- 2 Principle of Virtual Work
- 3 Virtual Work Expression
- 4 Procedure for Determining Truss Deflection
- 4.1 1. Determine Whether the Truss is Determinate, Indeterminate, or Unstable
- 4.2 2. Find the Support Reactions for the Truss
- 4.3 3. Find Real Support and Internal Forces, $ F $
- 4.4 4. Find Virtual Forces, $ F_v $
- 4.5 5. Find the length of each truss member
- 4.6 6. Find the cross-sectional area, $ A $ and the modulus of elasticity, $ E $.
- 4.7 7. Find the real displacement, $ \Delta $

- 5 Procedure for Determining Deformations Due to Temperature and Defects
- 6 Example 1
- 7 Example 2
- 8 Example 3
- 9 Example 4 (Warning many errors in this example)
- 10 Reference

## Introduction

The method of virtual work in trusses is used to determine the deflection of a joint in a truss. In order to determine this deflection, a real and virtual system are developed. In this section we will look at finding vertical or horizontal displacements in determinate trusses caused by external forces. Axial deformations caused by fabrication errors and temperate changes will also be addressed.

## Principle of Virtual Work

Virtual work is a method where a "virtual" load (an imaginary load that does not exist as a part of the real loadings)^{[1]} is applied to a structure at a point where the value of the slope or deflection is desired. This applied virtual load creates a virtual system which, through the principles of work and energy, is related to the real system. The relationship can then be used to solve for the deflection or slope at any point of the real system.

## Virtual Work Expression

### Truss Deflection - Basic Equation

The basic equation for determining the deflection in a truss is as follows:

$ 1 * \Delta = \sum F_v * \delta $^{[2]}

Where: $ 1 $ is a virtual, unit force; $ \Delta $ is an unknown, real displacement; $ F_v $ is a set of all virtual internal forces arising from and in equilibrium with $ 1 $; $ \delta $ is a set of internal displacements arising from $ F_v $. The internal displacement $ \delta $ can be expanded to $ \delta = FL/AE $, where $ F $ is the real external applied force; $ L $ the length of the truss member; $ A $ is the cross-sectional area of the member; and $ E $ is the modulus of elasticity. ^{[2]}

## Procedure for Determining Truss Deflection

### 1. Determine Whether the Truss is Determinate, Indeterminate, or Unstable

Determine whether the truss is determinate, indeterminate, or unstable using the equations and methods described in the Determinacy, Indeterminacy and Stability section.[1]

### 2. Find the Support Reactions for the Truss

The support reactions for the truss need to be found first by analyzing the truss with the three equations of equilibrium which are...

$ \sum F_x = 0 } $

$ \sum M_p = 0 } $

$ \sum F_y = 0 } $

### 3. Find Real Support and Internal Forces, $ F $

Calculate the real internal forces, $ F $ in each truss member caused by the real external loads placed on the truss. These internal forces can be determined by the Method of Joints.[2]

### 4. Find Virtual Forces, $ F_v $

Remove all real, external loads on the truss. Place the virtual unit load, $ 1 $, in the same direction as the displacement to be determined. If the vertical displacement is to be determined at a joint $ a $, apply the virtual load in the vertical direction at joint $ a $. Generally the unit load $ 1 $ is in the same units as the external load applied. In the case of a 5 KN external load, the virtual unit load would be 1 KN. Calculate the internal forces in each truss member with the $ 1 $ unit force applied. These forces are the virtual forces, $ F_v $. ^{[2]}

### 5. Find the length of each truss member

Calculate the length of each truss member. As the length and height of a truss are generally given the individual member lengths can be solved with basic trigonometry.

### 6. Find the cross-sectional area, $ A $ and the modulus of elasticity, $ E $.

These properties of the truss are generally given. Make sure to pay attention to the units of this given values.

### 7. Find the real displacement, $ \Delta $

Create a table with calculated values for $ F $, $ F_v $, $ L $, and $ \frac{F_v*F*L}{A*E} $. If $ A $ and $ E $ are uniform throughout the truss they can be removed and multiplied with the final summation. ^{[2]}

##### Virtual Work Table

member member length, $ L $ real internal force, $ F $ virtual internal force, $ F_v $ $ F_vFL $ AB BC CA $ \sum F_vFL $

Finally, for uniform $ A $ and $ E $, the displacement of the truss can be calculated:

$ \Delta = \frac{1}{1}*\frac{1}{AE}*\sum F_vFL $ ^{[2]}

## Procedure for Determining Deformations Due to Temperature and Defects

### Deformations Due to Temperature Change

For temperature change the axial deformation is expressed as $ \delta=\alpha*\Delta T*L $ ^{[2]}

Where $ \alpha $ is coefficient of thermal expansion, $ \Delta T $ is the change in temperature, and $ L $ is the length of the truss member.

To get the truss deformation $ \Delta $, the axial deformation, $ \delta $ is subbed into the equation $ 1 \Delta = \sum F_v \delta $ resulting in: $ \Delta = \frac{1}{1}* \sum F_v \alpha \Delta TL $, where $ F_v $ is the virtual internal force. ^{[2]}

### Deformations Due to Manufacturing Defects

For finding deformations due to manufacturing defects, the axial deformation, $ \delta $ (negative or positive) is known.

The equation for truss deformation simply becomes $ \Delta = \frac{1}{1}*\sum F_v*\delta $ ^{[2]}

## Example 1

Determine horizontal displacement at Point B, assuming all members have constant cross-sectional area $ A = 300 \textrm{mm}^2 $ and modulus of elasticity $ E = 250 \textrm{GPa} $.

We begin our solution by noting that this truss is a statically determinate structure and we proceed by drawing the free body diagram and solving for the support reactions using the equations of static equilibrium.

### Solve for Supports in Real System

$ \sum F_x = 0 \Rightarrow G_x = 0 \textrm{kN [right]} $

$ \sum M_D = 0 \Rightarrow G_y = 100 \textrm{kN [down]} $

$ \sum F_y = 0 \Rightarrow D_y = 300 \textrm{kN [up]} $

### Find Internal Forces in Real System

Having obtained the support reactions, we proceed with a joint analysis of the truss to determine forces acting on each member.

#### Joint A

$ \sum F_y = 0 \Rightarrow -200 \textrm{kN} + \frac{4}{5} F_{AB}= 0 \Rightarrow F_{AB}=250 \textrm{kN [T]} $

$ \sum F_x = 0 \Rightarrow F_{AD} = \frac{-3}{5} F_{AB} \Rightarrow F_{AD} = - 150 \textrm{kN [C]} $

#### Joint D

$ \sum F_x = 0 \Rightarrow F_{DE} = F_{AD} \Rightarrow F_{DE} = - 150 \textrm{kN [C]} $

$ \sum F_y = 0 \Rightarrow F_{BD}= - 300 \textrm{kN [C]} $

#### Joint G

$ \sum F_y = 0 \Rightarrow -100 \textrm{kN} + \frac{4}{5} F_{CG}= 0 \Rightarrow F_{CG}=125 \textrm{kN [T]} $

$ \sum F_x = 0 \Rightarrow F_{EG} = \frac{-3}{5} F_{CG} \Rightarrow F_{EG} = - 75 \textrm{kN [C]} $

#### Joint B

$ \sum F_y = 0 \Rightarrow \frac{-4}{5} F_{BE} + F_{BD} + \frac{-4}{5} F_{AB}= 0 \Rightarrow F_{BE}=125 \textrm{kN [T]} $

$ \sum F_x = 0 \Rightarrow F_{BC} + \frac{-3}{5} F_{AB} + \frac{3}{5} F_{BE} = 0 \Rightarrow F_{BC}= 75\textrm{kN [T]} $

#### Joint C

$ \sum F_y = 0 \Rightarrow F_{CE} = \frac{-4}{5} F_{CG} \Rightarrow F_{CE} = -100 \textrm{kN [C]} $

After obtaining both the support reactions and all the member forces of the real truss the free body diagram of the real truss can be drawn with the forces included.

### Construct Virtual System

Since we are required to find the horizontal displacement at Point B, we apply a leftward, horizontal 1kN virtual load to point B which can be seen in the free body diagram of the virtual system above.

#### Solve for Supports in Virtual System

We find the virtual reactions in the same way as we found the real reactions; for brevity we just state the results.

$ D_y^\prime = \frac{2}{3} \textrm{kN [up]} $

$ G_x^\prime = 1 \textrm{kN [right]} $

$ G_y^\prime = \frac{2}{3} \textrm{kN [down]} $

We note the existence of zero-force members $ F_{AB} $, $ F_{AD} $, and $ F_{DE} $ and simplify our joint analysis accordingly. Again, the intermediate steps have been omitted.

#### Joint G

$ F_{CG}^\prime = \frac{5}{6} \textrm{kN [T]} $

$ F_{EG}^\prime = \frac{1}{2} \textrm{kN [T]} $

#### Joint D

$ F_{BD}^\prime = \frac{-2}{3} \textrm{kN [C]} $

#### Joint B

$ F_{BE}^\prime = \frac{5}{6} \textrm{kN [T]} $

$ F_{BC}^\prime = \frac{1}{2} \textrm{kN [T]} $

#### Joint C

$ F_{CE}^\prime = \frac{-2}{3} \textrm{kN [C]} $

After obtaining both the support reactions and all the member forces of the virtual system the free body diagram of the virtual system can be drawn with the forces included.

#### Virtual Work Table

member $ L $ $ F $ $ F_v $ $ F_vFL $ AB 5 250 0 0 BC 3 75 1/2 112.5 CG 5 125 5/6 520.83 BD 4 -300 -2/3 800 CE 4 -100 -2/3 266.67 BE 5 125 5/6 520.83 AD 3 -150 0 0 DE 3 -150 0 0 EG 3 -75 1/2 -112.5 $ \sum F_vFL $ 2108.33$ \textrm{kN}^2\cdot\textrm{m} $

#### Find the Deflection

$ \Delta = \frac{1}{\textrm{kN}} \cdot \frac{1}{(300\textrm{mm}^2)\cdot (250\textrm{GPa})} \cdot 2108.33 \textrm{kN}^2 \cdot \textrm{m}} $

$ \Rightarrow \Delta = 28 \textrm{mm [left]} $

Thus the expected horizontal deflection at B will be 28 mm.

## Example 2

Determine vertical displacement at Point A due to temperature change in the beams of the truss displayed below. Assume all members have a constant coefficient of thermal expansion $ \alpha = 1.0 \times 10^{-5} ^{\circ} \textrm{C} $

Begin the analysis by applying a virtual load at the joint A in a downward direction and solve for the resulting support reactions.

### Solve for Supports in Virtual System

$ \sum F_x = 0 \Rightarrow E_x = 0 \textrm{kN [left]} $

$ \sum M_C=0 \Rightarrow E_y = 1 \textrm{kN [down]} $

$ \sum F_y=0 \Rightarrow C_y = 2 \textrm{kN [up]} $

### Find Internal Forces in Virtual System

Now that the support reactions for the virtual system have been solved for, we proceed with a joint analysis of the truss to determine forces acting on each member.

#### Joint A

$ \sum F_y = 0 \Rightarrow -1 \textrm{kN} + \frac{3}{5} F_{AB} = 0 \Rightarrow F_{AB}=1.67 \textrm{kN [T]} $

$ \sum F_x = 0 \Rightarrow -F_{AC} + \frac{4}{5} F_{AB} = 0 \Rightarrow F_{AC} = 1.34 \textrm{kN [C]} $

#### Joint B

$ \sum F_y = 0 \Rightarrow F_{BC} + \frac{-3}{5} F_{BA}= 0 \Rightarrow F_{BC} = 1 \textrm{kN [C]} $

$ \sum F_x = 0 \Rightarrow F_{BD} + \frac{-4}{5} F_{BA} = 0 \Rightarrow F_{BD} = 1.34 \textrm{kN [T]} $

#### Joint C

$ \sum F_y = 0 \Rightarrow 2 \textrm{kN} - F_{CB} - \frac{3}{5} F_{CD}= 0 \Rightarrow F_{CD}=1.67 \textrm{kN [C]} $

$ \sum F_x = 0 \Rightarrow F_{CA} - F_{CE} + \frac{-4}{5} F_{CD} = 0 \Rightarrow F_{CE} = 0 \textrm{kN} $

#### Joint D

$ \sum F_x = 0 \Rightarrow - F_{DB} + \frac{4}{5} F_{DC} + \frac{4}{5} F_{DG} = 0 \Rightarrow F_{DG}= 0\textrm{kN} $

$ \sum F_y = 0 \Rightarrow \frac{3}{5} F_{DC} - F_{DE} = 0 \Rightarrow F_{DE}=1 \textrm{kN [T]} $

Zero-force members in the system are members CE, EG, and DG.

Now that the support reactions and the member forces of the virtual system have been solved for, a free body diagram displaying these values will be created.

#### Virtual Work Table

Member $ L $ $ \Delta\textrm{T} $ $ F_v $ $ F_v \Delta TL $ AB 5 -15 1.67 -125.25 AC 4 10 -1.34 -53.6 BC 3 -15 -1 45 BD 4 -15 1.34 -80.4 CD 5 10 -1.67 -83.5 ED 3 10 1 30 $ \sum F_v\cdot\Delta\textrm{T}\cdot\textrm{L} $ -267.75$ \textrm{kN}\cdot^{\circ}\textrm{C}\cdot\textrm{m} $

#### Find the Deflection

$ \Delta = \frac{1}{1}* \sum F_v \alpha \Delta TL $

$ \Delta = \frac{1}{\textrm{kN}} \cdot (1.0\times10^{-5} / ^{\circ}\textrm{C})\cdot(-267.75 \textrm{kN} \cdot^{\circ}\textrm{C} \cdot\textrm{m}) $

$ \Delta = -0.0026775\textrm{m} $

$ \Delta = 2.68 \textrm{mm[up]} $

## Example 3

Determine vertical displacement at Point C due to fabrication errors in the beams of the truss displayed below.

Begin the analysis by applying a virtual load at the joint C in a downward direction as in the figure below and solve for the resulting support reactions.

### Solve for Supports in Virtual System

Due to symmetry the two vertical support reactions will be the same value therefore: $ A_y=D_y $

$ \sum F_y=0 \Rightarrow A_y + D_y = 1 \textrm{kN} \Rightarrow A_y=D_y=0.5\textrm{kN [up]} $

$ \sum F_x = 0 \Rightarrow A_x = 0 \textrm{kN [left]} $

### Find Member Forces in Virtual System

Now that the support reactions for the virtual system have been solved for, we proceed with a joint analysis of the truss to determine forces acting on each member.

#### Joint D

$ \sum F_y = 0 \Rightarrow 0.5 \textrm{kN} + \frac{4}{5} F_{DB} = 0 \Rightarrow F_{DB}=0.625 \textrm{kN [C]} $

$ \sum F_x = 0 \Rightarrow -F_{DC} + \frac{3}{5} F_{DB} = 0 \Rightarrow F_{DC} = 0.375 \textrm{kN [T]} $

#### Joint C

$ \sum F_y = 0 \Rightarrow F_{CB} - 1\textrm{kN}= 0 \Rightarrow F_{CB} = 1 \textrm{kN [T]} $

$ \sum F_x = 0 \Rightarrow F_{CD} - F_{CA} = 0 \Rightarrow F_{CA} = 0.375 \textrm{kN [T]} $

The truss is symmetrical so the rest of the member forces can be easily determined.

Now that the support reactions and the member forces of the virtual system have been solved for, a free body diagram displaying these values will be created.

#### Virtual Work Table

The members that are not affected by fabrication error will be neglected from the table. The values for $ F_v\delta (\textrm{kN}\cdot\textrm{mm}) $ of those members will result in zero.

This leaves only beam BD and AC to be added in the table.

Member $ \delta\textrm{(mm)} $ $ F_v\textrm{(kN)} $ $ F_v\delta \textrm{(kN}\cdot\textrm{mm)} $ BD 20 -0.625 -12.5 AC -10 0.375 -3.75 $ \sum F_v\delta $ 16.25

#### Find the Deflection

$ \Delta = \frac{1}{1}*\sum F_v*\delta $

$ \Delta = \frac{1}{\textrm{kN}} \cdot (16.25 \textrm{kN}\cdot\textrm{mm}) $

$ \Delta = 16.25 \textrm{mm [down]} $

## Example 4 (Warning many errors in this example)

For the truss shown below, use virtual work to determine the minimum cross- sectional area (A) in $ \textrm{mm}^2 $ such that the vertical deflection at point E $ (\Delta E ) $ does not exceed 40 mm. Where E = 200 KPA and the thermal coefficient of expansion $ (\alpha) = 12 \times 10^{-6} $.

First off we start by determining the stability and determinacy of the proposed truss, In our case the structure is found to be a statically determinate structure through calculations, Therefore we can proceed into solving the problem.

We will start by solving for the support reactions for the real system using the equations of static equilibrium :

### Solve for Supports in Real System

$ \sum M_A = 0 \Rightarrow B_x = 2.33 \textrm{kN [right]} $

$ \sum F_x = 0 \Rightarrow A_x = 3.33 \textrm{kN [left]} $

$ \sum F_y = 0 \Rightarrow A_y = 1 \textrm{kN [up]} $

### Analyze the Truss with Current Loading

Having obtained the support reactions, we analyze the truss by finding the internal forces in each member of the structure by using joint analysis.

#### Joint A

$ \sum F_y = 0 \Rightarrow 1 \textrm{kN} + F_{AB}= 0 \Rightarrow F_{AB}= -1 \textrm{kN} = 1 \textrm{kN [C]} $

$ \sum F_x = 0 \Rightarrow F_{AD} - 3.33 = 0 \Rightarrow F_{AD} = 3.33 \textrm{kN [T]} $

#### Joint B

$ \sum F_y = 0 \Rightarrow F_{AB} \textrm{kN} - F_{BD} sin(36.87) = 0 \Rightarrow F_{BD}= 1.67 \textrm{kN [T]} $

$ \sum F_x = 0 \Rightarrow F_{BC} + F_{BD} cos (36.87) + 2.33 = 0 \Rightarrow F_{BC} = -3.67 \textrm{kN} = 3.67 \textrm{kN [C]} $

#### Joint C

$ \sum F_y = 0 \Rightarrow 1 \textrm{kN} + 3.67 \textrm{kN} + F_{CE} cos(45) = 0 \Rightarrow F_{CE}= -6.60 \textrm{kN} = 6.60 \textrm{kN [C]} $

$ \sum F_x = 0 \Rightarrow - F_{CD} - (-6.60) sin(45) = 0 \Rightarrow F_{CD} = 4.67 \textrm{kN [T]} $

#### Joint D

$ \sum F_x = 0 \Rightarrow F_{DE} - 3.33 - (1.67) cos(36.87) = 0 \Rightarrow F_{DE} = 4.67 \textrm{kN [T]} $

#### Joint E

- check

$ \sum F_x = 0 \Rightarrow -4.67 + 6.6 cos(45) = 0 \Rightarrow -4.67 + 4.67 = 0 \textrm{kN [T]} $

- Note: the existence of zero-force members is not present in our structure.

Having solved for our reaction forces and successfully analyzed the real structure by finding all the internal forces of the members, a free body diagram of the real truss can be drawn with the forces included as seen below:

### Construct Virtual System

Since we are asked to find the cross- sectional area (A) of the members, we will start off by finding the deflection at point E (\delta E ) by applying a downward virtual unit load ( 1* KN ) in the direction and location of the deflection at point E.

#### Solve for Support Reaction Forces in Virtual System

We find the virtual reactions in the same way as we found the real reactions:

$ \sum M_A = 0 \Rightarrow B_x = 2.33 \textrm{kN [left]} $

$ \sum F_x = 0 \Rightarrow A_x = 2.33 \textrm{kN [right]} $

$ \sum F_y = 0 \Rightarrow A_y = 1 \textrm{kN [up]} $

### Analyze The Truss With Virtual Unit Load

We analyze the truss with the virtual unit load by finding the internal forces in the members, as we have before:

#### Joint A

$ \sum F_y = 0 \Rightarrow 1 \textrm{kN} + F_{AB}= 0 \Rightarrow F_{AB}= -1 \textrm{kN} = 1 \textrm{kN [C]} $

$ \sum F_x = 0 \Rightarrow F_{AD} + 2.33 = 0 \Rightarrow F_{AD} = -2.33 \textrm{kN} = 2.33 \textrm{kN [C]} $

#### Joint B

$ \sum F_y = 0 \Rightarrow 1 \textrm{kN} - F_{BD} sin(36.87) = 0 \Rightarrow F_{BD}= 1.67 \textrm{kN [T]} $

$ \sum F_x = 0 \Rightarrow F_{BC} + F_{BD} cos (36.87) - 2.33 = 0 \Rightarrow F_{BC} = 1 \textrm{kN [T]} $

#### Joint C

$ \sum F_x = 0 \Rightarrow F_{CE} cos(45) - 1 = 0 \Rightarrow F_{CE} = 1.41\textrm{kN [T]} $

$ \sum F_y = 0 \Rightarrow -F_{CD} \textrm{kN} - F_{CE} sin(45)\textrm{kN} = 0 \Rightarrow F_{CD}= -1 \textrm{kN} = 1 \textrm{kN [C]} $

#### Joint D

$ \sum F_x = 0 \Rightarrow F_{DE} + 2.33 - (1.67) cos(36.87) = 0 \Rightarrow F_{DE} = -1 \textrm{kN} = 1 \textrm{kN [C]} $

#### Joint E

- check

$ \sum F_x = 0 \Rightarrow -1 + 1.41 sin(45) = 0 \Rightarrow -1 + 1 = 0 $

- Note: the existence of zero-force members is not present in our structure.

After obtaining both the support reactions and all the member forces of the virtual system the free body diagram of the virtual system can be drawn with the forces included.

#### Virtual Work Table

We construct the values conducted from our calculation in the table below for easier calculations ahead:

member $ L $ $ F $ $ F_v $ $ F_vFL $ AB 3m -1 -1 3 BC 4m -3.67 1 -14.68 BD 5m 1.67 1.67 13.945 AD 4m 3.33 -2.33 -31.04 CE 4.24m -6.6 1.41 -39.5 CD 3m 4.67 -1 -14.01 DE 3m 4.67 -1 -14.01 $ \sum F_vFL $ 96.32 $ \textrm{kN}^2\cdot\textrm{m} $

Sample calculations :

- Member BD : $ L = \sqrt{3^{2}+ 4^{2}} = 5 m $

- Member CE: $ L = \sqrt{3^{2}+ 3^{2}} = 4.24 m $

- $ L = 4.24 m + \alpha\times \Delta T \times L = 4.24 + 12 \times 10^{-6} \times 20 \times 4.24 = 4.241 $

#### Find the Deflection

$ 1* . \Delta = \frac {\sum F_v . F . L} {E.A} $ < 40 mm

$ = \frac {96.32}{E.A} < 40mm $ A > 12.04 $ \textrm{mm}^2 $

**Therefore the minimum cross- sectional area (A) shall be taken as 12.04 $ \textrm{mm}^2 $**

## Reference

- ↑ Hibbeler, R.C. (2012).
*Structural Analysis (8th ed.)*. New Jersey, Pearson Education inc. - ↑
^{2.0}^{2.1}^{2.2}^{2.3}^{2.4}^{2.5}^{2.6}^{2.7}Kassimali, A. (2011).*Structural Analysis: SI Edition (4th ed.)*. Stamford, CT: Cengage Learning.