# Deflection Influence Lines

## Contents

## Introduction

Influence lines are used to analyze determinate structures that carry variable loads such as live loads. Influence lines are graphic representations of the magnitude of a section force at a fixed location due to a single point load with variable positions along the structure.^{[1]}
This means that influence lines are drawn in order to indicate where a load is going to create the maximum effect of the reaction force, shear force, moment or deflection on the structure.^{[2]}

This section will only cover influence lines for deflection, so it will explain how to construct the influence line for deflection at a certain point using one of the methods used for computing deflection such as, Direct integration method, superposition method, moment-area method, conjugate beam method or virtual work method.

This method is beneficial because you can find the deflection at any length in the beam by substituting the value (X) that you are looking for. In other methods you are normally looking for the deflection at a certain point, if you want to find it at a different point on the same beam you will have to do the same process over with different restrictions on the point of deflection you are looking for. Using the equation of the deflection curve with respect to X followed by Maxwell’s Law saves time when finding deflections at any point on the beam.

## Maxwell's Law of Reciprocal Deflections

When finding deflection influence lines one of the main concepts that we use is Maxwell's law of reciprocal deflection, this law only applies to linearly elastic structures^{[1]}.

Maxwell's law of reciprocal deflection creates a relationship between two points along a beam in order to find the influence line for a deflection at one of those two points. A unit load (1 kN) at any point on the structure X with respect to the deflection at point B is equal to the deflection at any point X with a unit load at point B.^{[1]} This is the basic theory of Maxwell's equation shown below.

__Maxwell's law of reciprocal deflections:__^{[1]}
- $ \text{\it f_{\it XB} =\it f_{\it BX} } $
$ \text{\it f_{\it XB}} $ :the deflection at X due to the unit load at B. |

The law of reciprocal deflection can be used for different structures such as beams, trusses, and rarely frames.

For both beams and trusses we can imagine that the 1kN load is traveling across the structure.^{[3]} It is valuable to fully understand what is equal to 1kN . We know that 1kN = 101.97 kg = 224.81lb from this we can conclude that 1kN is not a substantial force acting on the structure, therefore the diagrams ( figure 1 and 2 ) below exaggerates the size of deflection for better understanding of the material.

With that knowledge, imagine two high school students walking side-by-side across the structure with the reference point (X) being the center of the structure. When the students reach one fourth of the way, the force they are applying will cause a deflection at point X which we call ƒSX . When the students reach the middle of the structure the deflection at the point they were earlier (one fourth of the way on the structure) which we can call ƒXS , will be equal to the deflection calculated at the centre. This is the basics of Maxwell’s law.

### Beam

### Trusses

### Frames

Most frames are not linearly elastic and therefore Maxwell's law of reciprocal deflection can not be applied.

## Method for Determining Influence Line Deflection

According to the limitation of Maxwell's Law (linearly elastic structures only), the deflection shape of the structures, which used influence line method to find deflection, can be express as an equation of x which is the distance from origin to where the unit load placed. Therefore, only Maxwell's Law and the equations of the structure deflection curve is needed for the method of using influence line to find the deflection of a structure. This section shows by steps of the influence line method to determine the deflection of beams and trusses.

__Beams__

[Steps can be refer to *Figure 3(a)*]

__Step 1:__Place a unit load (1 kN) at the desired point.

__Step 2:__Determine the deflection at any point (X) on the beam ΔXB by using Moment Area Theorem,Conjugate Beam Method, Beam Virtual Work or Integrating to Find Deflections.

__Step 3:__$ f_{XB} = \triangle_{XB} $

__Step 4:__Apply Maxwell’s Law to find deflection at B. Therefore, $ f_{BX} = f_{XB} $

For certain beams there are simplified equations of elastic curve that are calculated. ^{[4]}

__Trusses__

__Step 1:__Place a unit load (1 kN) at the desired point.

__Step 2:__Determine the deflection at any point (X) on the beam ΔXB by using Truss Virtual Work.

__Step 3:__$ f_{XB} = \triangle_{XB} $

__Step 4:__Apply Maxwell’s Law to find deflection at B. Therefore, $ f_{BX} = f_{XB} $

## Example

__Problem:__

Find the influence line for the deflection of the beam at point B.

EI=constant

This solution follows the steps outlined in the *Method for Determining Influence Lines Deflections* section, as well as an adapted version of a moment area theorem analysis which can be found in the Kassimali, A. (2011). Structural Analysis: SI Edition (4th ed.). Stamford, CT: Cengage Learning^{[1]} textbook.

-Find support reactions at points A and C for unit load at a distance x from A using the Equilibrium Equations for a 2D structure^{[1]}.

$ \sum M_A=0 $ (counterclockwise rotation as positive)

$ \sum M_A= -1(x) + C_y(7) $

$ -1(x) + C_y(7)=0 $

$ C_y= \frac {x}{7} $

$ \sum F_y=0 $ (up as positive)

$ A_y-1+\frac{x}{7}=0 $

$ A_y=1-\frac{x}{7} $

$ \sum F_x=0 $ (right as positive)

$ A_x=0 $

- Find the shear, moment and curvature diagrams if the unit load was applied at the point B (if you dont know how to do that you can refer to the Kassimali, A. (2011). Structural Analysis: SI Edition (4th ed.). Stamford, CT: Cengage Learning^{[1]}).

So we get the following diagrams:

__Shear diagram (in KN)__

B is at 5m

$ A_y=1-\frac{5}{7}=\frac{2}{7} $

$ C_y=\frac{5}{7} $

__Moment diagram (in KN*m)__

$ \frac{2}{7}*5=\frac{10}{7} $

$ \frac{-5}{7}*2=\frac{-10}{7} $

__Curvature diagram (M/EI)__

To find the curvature diagram, we need to divide our moment diagram by our EI ^{[1]}. Since in our case the EI is a constant, our curvature diagram will be the same shape as our moment diagram.

- Now using the Moment Area Theorem we can find $ \Delta_{XB}=A_1\bar{x} _1+A_2\bar{x} _2+A_3\bar{x} _3+A_4\bar{x} _4 $^{[1]} by multiplying the area of each shape by the distance of their centroid to B.

As mentioned in the *Method for determining Influence Line Deflections*, the result will be equal to $ f_{XB} $

Now in this case, because of the shape of the curvature diagram, we will have to come up with 2 equations since we do not have one continuous line or curve. One equations for when x is smaller than 5m, and another for when x is bigger than 5m.

Now if we look at area 1 in our figure and divide it into 3 sections:

- We can see that using the similar triangle theorem^{[5]}, we have:

$ \frac{W}{\frac{10}{7EI}}=\frac{x}{5} $

$ W=\frac{10x}{35EI} $

And now we can calculate $ \Delta_{XB} $ for the first part of our figure (the sum of areas 1.2 and 1.3 multiplied by the distance from their centroids to B) :

$ \Delta_{XB}=\frac{10x}{35EI}\frac{5-x}{2}+(\frac{10}{7EI}-\frac{10x}{35EI})(\frac{5-x}{2})(\frac{5-x}{3}) $

Which simplifies to:

$ \Delta_{XB}=\frac{1}{EI}[\frac{-x^3+7x^2-55x+125}{21}] $ for x smaller or equal to 5m

And as mentioned in the *Method for Determining Influence Line Deflections* section above, $ \Delta_{XB}=f_{XB} $, therefore:

$ f_{XB}=\frac{1}{EI}[\frac{-x^3+7x^2-55x+125}{21}] $ for x smaller or equal to 5m

- Now we need to find an equation for when x is bigger than 5m:

- We can once again apply the similar triangle theorem^{[5]} to determine the equation for W, and then calculate the areas of the shapes 2.2 and 2.3 and the distance from their respective centroid to B.

- Using the same reasoning as before we end up with:

$ W=\frac{10(7-x)}{14EI} $

- Now we can find the equation desired:

$ \Delta_{XB}=[(x-5)(\frac{10(7-x)}{14EI})(\frac{x-5}{2})]+[\frac{10}{7EI}-(\frac{10(7-x)}{14EI})(x-5)(\frac{1}{2})(\frac{x-5}{3})] $

- And this equation simplifies to give:

$ \Delta_{XB}=f_{XB}=\frac{1}{21EI}(-5x^3+85x^2-475x+905) $ for when x is bigger or equal to 5m

- And now, using *Maxwell's Law of Reciprocal Deflection*^{[1]} explained above, we get $ f_{XB}=f_{BX} $

Therefore we now have:

For x smaller or equal to 5m

$ f_{BX}=\frac{1}{EI}[\frac{-x^3+7x^2-55x+125}{21}] $

And for x bigger or equal to 5m

$ f_{BX}=\frac{1}{21EI}(-5x^3+85x^2-475x+905) $

And these are our deflection influence lines for point B.

## References

- ↑
^{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}Kassimali, A. (2011).*Structural Analysis: SI Edition (4th ed.)*. Stamford, CT: Cengage Learning. - ↑ http://www.ce.memphis.edu/3121/notes/notes_06a.pdf.
- ↑
^{3.0}^{3.1}^{3.2}^{3.3}http://www.cdeep.iitb.ac.in/nptel/Civil%20Engineering/Structural%20Mechanic%20II/Course_home_4.6.html - ↑ http://www.engineering.com/Library/ArticlesPage/tabid/85/ArticleID/136/Beam-Deflections-and-Slope.aspx
- ↑
^{5.0}^{5.1}Maths Revision.*Similar Triangles*[Online].Available FTP:http://www.mathsrevision.net/gcse-maths-revision/trigonometry/similar-triangles