# Beam Virtual Work

## Contents

## Introduction to Virtual Work in Beams

The Method of Virtual Work is a technique used to determine slope (θ) and deflection (Δ) of a structure subjected to loading. In beams, the slope and deflection at a given point can be obtained by analyzing the 'real system', then subsequently creating a 'virtual system' using a unit load or moment at the point of interest. By examining both the real and virtual systems, and applying the suitable Virtual Work equation, a value for deflection or slope at a point can be obtained. See the following sections for a detailed explanation of how the Virtual Work Method applies to beams, as well as relevant figures, terms, expressions and a worked example.

To see how the Virtual Work Method applies to other types of structures, see Truss Virtual Work and Frame Virtual Work.

## Important Terms

Below is a list of important terms relevant to the Method of Virtual Work in beams.

### Real System

The real system refers to actual loading on the beam as defined by a problem statement, including all external and internal moments and loads.^{[1]}

### Virtual System

The virtual system refers to the system created with the application of a unit load. When defining a virtual system, external moments and loads from the real system are ignored. ^{[1]}

### Unit Load/Moment

A unit load/moment is a arbitrary load or moment with a value of 1 (see Figure 1). It is applied to a structure in the virtual system to determine how the other members and supports will react to the force applied. When using a unit load/moment to determine the deflection or slope at a point, it must be applied at the point of interest on the beam (the point at which you are trying to find slope or deflection). ^{[2]}

### Deflection

Deflection (Δ) is the distance between a point on a structure when no load is applied to the same point’s new position when loading is applied. In Figure 2, the deflected shape of the beam due to the load is shown in red. The total deflection of any point on the beam is the distance from the original beam to the same point on the deflected beam. Deflection in a beam can be caused by many factors including loading, temperature and fabrication errors, however this page focuses exclusively on deflections due to loads. ^{[3]}

### Rotation and Slope

Rotation is an angle created when a beam is displaced as a result of a force. A tangent line can be drawn from the point of interest on the deflected beam to the point's original position as depicted in Figure 2. The value of $ \theta $ is the slope of the deflected shape at the position of interest. ^{[2]}

## Procedure for Analysis

The following is a step-by-step procedure to find defection and slope at a point using the Virtual Work Method.^{[4]}^{[2]} An understanding of Beam Analysis is recommended before undertaking this type of Virtual Work problem.

**Step 1:**Solve for the support reactions in the real system using Equilibrium and Compatibility.

**Step 2:**Create an expression(s) for moment in the real system in terms of 'x' distance. The number of expressions will be governed by the number of cuts that are needed to solve the structure.

**Step 3:**Create a virtual system by removing all forces acting on the beam, and applying a unit load to find the expression for deflection, and a unit moment to find an expression for slope. Repeat step 2 for the virtual system.

**Step 4:**Substitute moment values into the deflection expression (equation 1 below), and integrate to solve for the deflection at the point of interest.

**Step 5:**Substitute moment values into the slope expression (equation 2 below), and integrate to solve for the slope at the point of interest.

## Virtual Work Equations

#### Virtual Work Expression for Beam Deflections

- $ 1(\Delta)=\int_0^L {\frac{M_v M}{EI}dx } $ (
*Equation 1*)^{[2]}

#### Virtual Work Expression for Beam Slopes

- $ 1(\theta)= \int_0^L {\frac{M_v M}{EI}dx } $ (
*Equation 2*)^{[2]}

## Worked Example

**Problem Statement:** Find the slope and deflection at point C using the method of virtual work.

**Step 1:** Solve for the support reactions in the real system using equilibrium.

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

$ \Sigma F_y = 0 \Rightarrow A_y = 250 \textrm{ kN [up]} $

$ \Sigma M_A = 0 \Rightarrow M_A = 1662.5 \textrm{ kN*m [counter-clockwise]} $

**Step 2:** Find the two moment equations by cutting the beam twice. One cut will be through section CB, and the other will be through section CA, left of point B.

*Cut 1:* Member CB

$ \Sigma F_x = 0 \Rightarrow A_C = 0 \textrm{ kN [right]} $

$ \Sigma F_y = 0 \Rightarrow V_C = 75 \textrm{ kN [up]} $

$ \Sigma M_A = 0 \Rightarrow M_C = -75x \textrm{ kN*m [clockwise]} $

*Cut 2:* Member CA

$ \Sigma F_x = 0 \Rightarrow A_C = 0 \textrm{ kN [right]} $

$ \Sigma F_y = 0 \Rightarrow V_C = 25x-100 \textrm{ kN [up]} $

$ \Sigma M_A = 0 \Rightarrow M_C = -75x-{\frac{25(x-7)^2}{2}} \textrm{ kN*m [clockwise]} $

**Step 3:** Repeat step 2 for the virtual system. Using a unit load and a unit moment at point C, two moment equations must be found, one for deflection and one for slope.

*For Deflection:* Apply a 1 kN load at the point of deflection (C)

$ \Sigma F_x = 0 \Rightarrow A_C = 0 \textrm{ kN [right]} $

$ \Sigma F_y = 0 \Rightarrow V_C = 1 \textrm{ kN [up]} $

$ \Sigma M_A = 0 \Rightarrow M_C = -x \textrm{ kN*m [clockwise]} $

*For Slope:* Apply a 1 kN*m moment at the point where slope needs to be determined (C)

$ \Sigma F_x = 0 \Rightarrow A_C = 0 \textrm{ kN [right]} $

$ \Sigma F_y = 0 \Rightarrow V_C = 0 \textrm{ kN [up]} $

$ \Sigma M_A = 0 \Rightarrow M_C = -1 \textrm{ kN*m [clockwise]} $

**Deflection Summary:**

$ \ Segment $ $ \ Origin $ $ \ Limits (m) $ $ \ M \textrm{ (kN) } $ $ \ M_V \textrm{ (kN) } $ CB C 0-7 $ \ -75x $ $ \ -x $ BA C 7-14 $ \ -75x-{\frac{25(x-7)^2}{2}} $ $ \ -x $

**Slope Summary:**

$ \ Segment $ $ \ Origin $ $ \ Limits (m) $ $ \ M \textrm{ (kN) } $ $ \ M_V \textrm{ (kN) } $ CB C 0-7 $ \ -75x $ $ \ -1 $ BA C 7-14 $ \ -75x-{\frac{25(x-7)^2}{2}} $ $ \ -1 $

**Step 4:**Substitute values into the deflection expression, and integrate to solve for the deflection at point C.

$ 1(\Delta_C)=\int_0^L {\frac{M_v M}{EI}dx } $

$ 1(\Delta_C)={\frac{1}{EI}}[\int_0^7 (-75x)(-x)dx + \int_7^{14} -75x-{\frac{25(x-7)^2}{2}}(-x)dx] $

$ 1(\Delta_C)= {\frac{86107.3}\ \ kN^{2}*m^{2}}{EI} = {\frac{86107.3}{(70)(10^6)(2340)(10^{-6})}} = 0.526\ \ m $

$ 1(\Delta_C)= 526\ \ mm $

**Step 5:**Substitute values into the slope expression, and integrate to solve for the slope at point C.

$ 1(\theta_C)=\int_0^L {\frac{M_v M}{EI}dx } $

$ 1(\theta_C)={\frac{1}{EI}}[\int_0^7 (-75x)(-1)dx + \int_7^{14} -75x-{\frac{25(x-7)^2}{2}}(-1)dx] $

$ 1(\theta_C)= {\frac{8779.17}\ \ kN^{2}*m^{2}}{EI} = {\frac{8779.17}{(70)(10^6)(2340)(10^-6)} } = 0.0536\ \ rad $

$ 1(\theta_C)= 3.1^0 $

**FINAL ANSWER:**The slope at point C is 3.1° while the deflection is 526mm in the downward direction.

## Worked Example 2

**Problem Statement:** For the beam below, use virtual work to find the minimum moment of interia (*I*) such that it's deflection does not exceed L/360.

**Step 1:** Solve for the support reactions in the real system using equilibrium.

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

$ \Sigma M_A = 0 \Rightarrow D_y = 60 \textrm{ kN [up]} $

$ \Sigma F_y = 0 \Rightarrow A_y = 60\textrm{ kN [up]} $

**Step 2:** Draw shear and moment diagrams.

**Step 3:** Repeat step 1 and 2 for the virtual system. Using a unit load at mid-span. Note that maximum deflection is at middle of beam.

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

*Symmetrical:*
$ \Rightarrow A_y = D_y \ $

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

**Step 4:** Apply principle of virtual work.

Using the virtual work integration table [1]^{[5]} to find the appropriate equation for each section.

$ 1(\Delta)=\int_0^L{\frac{M_v M}{EI}dx } $ (*Equation 1*)^{[2]}

$ 1(\Delta_M)=({ {\frac{L M Q}{3E(2I)} } )*2} + ({{\frac{L M}{2EI} (Qa+Qb)}} + {{\frac{L M}{12EI} (3Qa+5Qb)}})*2 $

$ 1(\Delta_M)={ {\frac{(2)(2)(120)(1)}{6EI} } } + {{\frac{(2)(2)(120)}{2EI} (1+2)}} + {{\frac{(2)(2)(60)}{12EI} (3(1)+5(2))}} $

$ {\frac{1060\ \ kN*m^{3}*(10^{12})}{200000\ \ MPa * ( I ) } } \leq {\frac{8000\ \ mm}{360} } $

$ I \geq 239*10^{6}\ \ mm^{4} $

**FINAL ANSWER:** The minimum moment of inertia (*I*) is $ 239*10^{6}\ \ mm^{4} $

## Further Applications of the Virtual Work Method

Concepts from the Virtual Work Method can also be extended to other structural analysis techniques. The Force Method, for example, utilizes a virtual system and a similar method in order to solve indeterminate systems.

For simple applications of the Force Method, see Force Method for One Degree of Indeterminacy.

## References

- ↑
^{1.0}^{1.1}Professor Jeffrey Erochko. "Introduction to Structural Analysis, Lecture 8 (2013)". Carleton University - ↑
^{2.0}^{2.1}^{2.2}^{2.3}^{2.4}^{2.5}Kassimali, A. (2011). "Structural Analysis: SI Edition (4th ed.)" . Stamford, CT: Cengage Learning - ↑ Hibbeler, R.C. (2012). "Structural Analysis (8th ed.)". Upper Saddle River, NJ: Pearson Prentice Hall
- ↑ "Williams, Alan (2009). Structural Analysis - In Theory and Practice".Oxford, UK:Elsevier
- ↑ Professor Jeffrey Erochko. "Introduction to Structural Analysis, learnaboutstructures.com (Structural Engineering Resources)". Carleton University