Beam and Frame Influence Lines

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Introduction

The method of influence lines on beams and frames is used for the analysis of live loads moving along a structure.[1] With this analysis we are able to calculate the forces and moments due to these traveling live loads as they move along a structure. The influence lines at each support/reaction can be determined. This information presents us with the effect of the live load on a certain support/reaction as the load moves along a structure, as well as the maximum load and its location.[2] The method to draw these influence lines will be explained in the next section, followed by a solved example.

This page only explains the general method of influence lines, for another approach see the Muller Breslau Principle. You can also refer to Truss Influence Lines, Deflection Influence Lines, Influence Lines for Distributed loads, or Influence Lines for a Series of Moving Loads for a further understanding.

Theory

Beam with respective Influence Line Diagrams

Explanation

Influence lines play a crucial role in the design and analysis of structures. It shows the variation of moment or force at a specific point of a structure as a live load moves across it. When the influence lines are drawn, the maximum force or moment along the structure can be found. This is very useful in the design of bridges, buildings, construction equipment, etc. since these parameters are typical design requirements or limiting factors. [3]

Why use a live unit load to find influence line diagrams?

For a convenient analysis, a unit load is being used when drawing influence lines. This makes it easier to do calculations and also easy to determine any reactions/moments influence on a structure at any desired location. The influence of a reaction/moment can be found, by simply multiplying the value on the influence line at the desired location by the reaction/moment value.


Influence lines are found by applying the equilibrium method/equations. By applying the equilibrium equations, the travelling live load can be incorporated as a function of its position. These equations can be applied to find the live load influence lines at any point along the entire structure. [2]

$ \sum F_{x} = 0 $

$ \sum F_{y} = 0 $

$ \sum M_{Q} = 0 $

Method of Analysis (inspired by Kassimali[2])

  • As an example, a problem is asking for you to find the influence lines for some support, reaction or bending moment.
  • A direction must be chosen, in which the live load will move across the beam or frame. This will be the reference point for x on the diagram. Once a reference position is chosen, the influence line diagrams can be solved. (i.e on the right diagram, the left of the diagram is chosen to be the reference).
  • The next step would be solving with one of the reactions/moments in terms of the variable distance x. The answer is a function of x called a response function. [4]
  • The load should always be placed so that the response function is easier to solve for. This is not necessarily the closest point to the reference point.
  • Find all the desired reactions/moments in terms of x using equations of equilibrium. There should be at least one response function for every desired reaction or moment.
  • Plug in all the corresponding location points of interest x to find the value of the response function when the unit load is at a location x.
  • Once these values are obtained, plot the values on a line representing the beam length. Each value from the influence line diagram will represent the magnitude of the analyzed reaction when the unit load is placed at a point x on the beam or frame.
  • Repeat the steps for all desired supports, reactions or bending moments.
  • It is important to note that when a unit load is placed directly over a single support, the reaction will be equal and opposite. Therefore the reaction influence line for any given support will have a value of 1 directly over that support. This can be used as a check for the above method.

Solved examples

Example 1:

Problem: Draw the influence line diagrams for the vertical support reactions $ A_y, C_y, $ and $ E_y $ and for the internal moment at D, $ M_D $

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  • In this example there are no axial forces, therefore they are not accounted for.
  • By checking the Determinacy of the beam, it is found to be a determinate beam. Therefore, all of the reactions and moments can be solved for.
  • The hinge gives 1 degree of freedom, therefore the beam needs to be "cut" at the hinge.


$ 0 \leq x \leq 2 $ $ 2 \leq x \leq 6 $
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$ \sum M_B^{AB} =0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1(2-x)-2A_y=0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Rightarrow \ A_y = \frac{2-x}{2} $ $ A_y=0 $


  • As the unit load moves across the beam 2 cases should be accounted for:
  1. The unit load is within the beam section AB
  2. The unit load is beyond the beam section AB


The reaction $ A_y $ is now determined, so the analysis can be continued on the other beam supports.


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To calculate the reaction $ C_y $ use the moment at point E:


$ \sum M_E =0 : \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1(6-x)-6A_y-2C_y=0 $


$ C_y= \frac{6-x}{2} - 3A_y \ \ \ \ \ \ \ \Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ A_y \ \ \ \ \ \ \ \ \ could \ be \ zero \ or \ \ \ \ \ \ \frac{2-x}{2} $


Substituting the $ A_y $ values into the above equation gives:


$ C_y= \frac{6-x}{2} - 3(\frac{2-x}{2}) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{6-x}{2} - \frac{6-3x}{2} = x = C_y $


$ C_y=\begin{cases}x & \ \ \ \ \ \ \ 0 \leq x \leq 2 \\ \\ \\ \\ \\ \frac{6-x}{2} & \ \ \ \ \ \ \ 2 \leq x \leq 6 \end{cases} $


Finally, the only unknown reaction is $ E_y $. It can be found by the summation of forces in the y direction:


$ +\uparrow \sum F_y = 0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ A_y + C_y+ E_y-1=0 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \Rightarrow \ \ \ \ \ \ \ \ E_y=1-C_y-A_y $


Substituting the $ C_y $ and $ A_y $ values into the above equation gives:


$ E_y=\begin{cases} 1-x- \frac{2-x}{2} \ \ \ \Rightarrow \ \ \ \frac{-x}{2} & \ \ \ \ 0 \leq x \leq 2 \\ \\ 1-\frac{6-x}{2}-0 \ \ \Rightarrow \ \ \ \frac{x}{2}-2 & \ \ \ \ 2 \leq x \leq 6 \end{cases} $


To calculate an internal moment at a point, the beam must be cut at this point's location. This results in two cases for point D.


Similar to the case of finding the reaction, $ A_y $ 2 cases must be accounted for:

  1. The unit load is before the cut section DE.
  2. The unit load is within the cut section DE.


For internal moment
$ 0 \leq x \leq 5 $ $ 5 \leq x \leq 6 $
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$ \sum M_D =0 : \ \ \ \ \ \ \ \ -M+E_y(1)=0 \ \ \ \ \ \ \ \ \ \Rightarrow $ $ \sum M_D =0 : \ \ \ \ \ \ \ \ -M-1(x-5)+E_y(1)=0 $
$ M=E_y $ $ M=E_y - x + 5 $


Substituting the $ E_y $ values into the above equations gives:


$ M_D=\begin{cases} \frac{-x}{2} & \ \ \ 0 \leq x \leq 2 \\ \\ \\ \frac{x}{2} - 2 & \ \ \ 2 \leq x \leq 5 \\ \\ \\ \frac{x}{2}-2-x+5 \ \Rightarrow \ 3-\frac{x}{2} & \ \ \ 5 \leq x \leq 6 \end{cases} $


The final step is to substitute the x values into the equations determined above, and to draw the influence line diagrams.


$ A_y=\begin{cases} \frac{2-x}{2} & \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0 \leq x \leq 2 \\ \\ \\ \\ \\ \ 0 & \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2 \leq x \leq 6 \end{cases} $
at $ x=0 \ \ \ \ \ \ \ \ \ \ \ A_y=\frac{2-0}{2} = 1 $
at $ x=2 \ \ \ \ \ \ \ \ \ \ \ A_y=\frac{2-2}{2} = 0 $
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$ C_y=\begin{cases} x & \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0 \leq x \leq 2 \\ \\ \\ \\ \\ \ \frac{6-x}{2} & \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2 \leq x \leq 6 \end{cases} $
at $ x=0 \ \ \ \ \ \ \ \ \ \ \ C_y=0 $
at $ x=2 \ \ \ \ \ \ \ \ \ \ \ C_y=\frac{6-2}{2} = 2 $
at $ x=6 \ \ \ \ \ \ \ \ \ \ \ C_y=\frac{6-6}{2} = 0 $
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$ E_y=\begin{cases} \frac{-x}{2} & \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0 \leq x \leq 2 \\ \\ \\ \\ \\ \ \frac{x}{2} -2 & \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2 \leq x \leq 6 \end{cases} $
at $ x=0 \ \ \ \ \ \ \ \ \ \ \ E_y=0 $
at $ x=2 \ \ \ \ \ \ \ \ \ \ \ E_y=\frac{-2}{2} = -1 $
at $ x=6 \ \ \ \ \ \ \ \ \ \ \ E_y=\frac{6}{2} -2 = 1 $
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$ M_D=\begin{cases} \frac{-x}{2} & \ \ \ \ \ 0 \leq x \leq 2 \\ \\ \\ \\ \\ \ \frac{x}{2} -2 & \ \ \ \ \ 2 \leq x \leq 5 \\ \\ \\ \\ \\ \ 3- \frac{x}{2} & \ \ \ \ \ \ 5 \leq x \leq 6 \end{cases} $
at $ x=0 \ \ \ \ \ \ \ \ \ \ \ M_D=0 $
at $ x=2 \ \ \ \ \ \ \ \ \ \ \ M_D=\frac{-2}{2} = -1 $
at $ x=5 \ \ \ \ \ \ \ \ \ \ \ M_D= 3- \frac{5}{2} = 0.5 $
at $ x=6 \ \ \ \ \ \ \ \ \ \ \ M_D= 3- \frac{6}{2} = 0 $
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As shown at the transition point $ x=2 $, substituting $ x<2 $ in both equations results in the same answer. This is a helpful way to check the validity of the equations constructed. A similar case is when $ x=5 $ for $ M_D $.

Example 2:

Problem: Draw the influence line diagrams for the vertical support reactions $ B_y $ , $ C_y $ and for the internal moment $ M_d $

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  • In this example we have a simply supported beam with zero degrees of freedom (no hinges) which means the analysis will be simpler since we don't have to cut the beam,and there are no external forces in the axial direction so the axial component of the pin $ B_x $ is equal to zero.
  • By checking the Determinacy of the beam, it is found to be a determinate beam. Therefore, all of the reactions and moments can be solved for using equations of equilibrium.
  • We start off by defining our reference point at point A (as shown in the figure below) and then use moment equilibrium to calculate the reactions at the supports as a point point load is moving from the reference point till the end of the beam, the reactions will be written as a function of X which is our distance from the reference point.

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  • To write an equation for the reaction $ B_y $ as the point load moves along the beam we use moment equilibrium at roller C as the following

$ \sum M_C =0 : \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1(6-x)-4B_y=0 $


$ B_y= \frac{6-x}{4} $
  • Now to draw the influence line for the reaction $ B_y $ we simply substitute x by different values of distance from the reference point to get how much the reaction of $ B_y $ would be when the point load is in that position X

Influence line for $ B_y $ :

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  • Now we use moment equilibrium at pin B to get an equation for the vertical reaction $ C_y $

$ \sum M_B =0 : \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 4C_y+1(2-x)=0 $

$ C_y= \frac{x-2}{4} $
  • Note that when the point load is at a position after x = 2m the equation for $ M_b $ would change because the point load would generate a negative moment according the positive sign convection, the equation would be as the following:

$ \sum M_B =0 : \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 4C_y-1(x-2)=0 $

  • Since the beam is continuous and the influence lines are linear we get the same value for $ C_y $
$ C_y= \frac{x-2}{4} $
  • Now we draw the influence line for $ C_y $ by substituting values of x

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  • Note that we can find the value of the influence line at any point along the beam using similar triangles since it has the same slope
  • Now to find the influence line for internal moment $ M_d $ we cut the beam at point D and solve for the internal moment using either side, in this case we will use the right segment since we have only one support reaction and it will be easier to analyze
  • After cutting the beam we apply the unit load before point D and after it, and we write an equation of the influence line as follows:
  • unit load applied before point D (x<4):

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$ \sum M_d = 2C_y \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ M_d = 2 ( \frac{x-2}{4} ) $

  • when load is after point D (x>4)

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$ \sum M_d = 2C_y - 1(x-4) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ M_d = 2 ( \frac{x-2}{4} )-1(x-4) $

  • Now we simply substitute values of x for both cases and draw the influence line for the internal moment

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Example 3:

Problem: Draw the influence line diagrams for the vertical support reactions $ A_y, M_A, $. notice that the length of the beam=L Fixebeam1.png

  • to find the influence lines for reaction at A , we should apply a moving load of 1 KN into different positions to come up with an equation for the value of $ A_y $ at different positions of moving load.

Fixedbeam2.png

$ \ A_y-(1) = 0 $.

$ \ A_y = 1 $

therefore the influence line for $ A_y $ would look like this:

Fixedbeam3a.png

  • Influence line for $ M_A $ :

the equation that we come up with for moment at point A is: $ \ M_A-(1)(x)=0 $.

$ \ M_A = -x $

so the influence line for $ M_A, $ would look like this:

Fixedbeam5new.png

Validity check

  • For reaction $ B_y $ influence line we can validate our answer by substituting x by 2 which is the position of the support B and we should get the answer 1 because when the load is applied right on top of the support it should take all the load, same thing for reaction $ C_y $ the value we should get when we substitute x by 6 is 1. Another thing is that we should also have a value of zero at all other support reactions because when the load is on top of another support it should take all the load and we should have a value of zero for the reaction we're drawing the influence line for.
  • For the internal moment $ M_d $ the shape of the influence line should look similar to the deflected shape of the beam when the moment we're solving for is removed, or in other words to make sure our influence line is drawn correctly for an internal moment we simply introduce a hinge at the point we're solving for and apply a unit rotation in the positive direction, the influence line should look like the deflected shape.
  • Note that the influence diagrams are not values of different reactions at different parts of the beam but it is the value of one reaction when the point load is in that position along the beam.


These examples were solved using the General method. For a more detailed example please watch this video done by Professor Jeffrey Erochko: Solving Influence Lines using General Method

In this part of the wiki, only the general method is explained but you can watch the same example being solved by the same professor with Muller Breslau principle : Solving Influence Lines using Muller Breslau Principle

References

  1. Kharagpur. "Structural Analysis.pdf, Version 2 CE IIT". 7 August 2008. Accessed on 26 November 2010.
  2. 2.0 2.1 2.2 Kassimali, A (2011). Structural Analysis: SI edition (4th ed.) Stanford, CT: cengage Learning.
  3. Department of Civil Engineering, lecture on influence lines, The University of Memphis, <http://www.ce.memphis.edu/3121/notes/notes_06a.pdf>
  4. Dr. George E. Blandford's lecture, University of Kentucky, <http://www.engr.uky.edu/~gebland/CE%20382/CE%20382%20Four%20Slides%20per%20Page/L8%20-%20Influence%20Line%20Diagrams.pdf>