# Truss Influence Lines

## Introduction

In many cases, when analyzing a structure, the applied loads may not always be a constant force located at a fixed location. These loads are referred to as live loads.

Trusses, which are used commonly in bridge applications (as well as other applications) are frequently, if not constantly, exposed to these live loads. When analyzing trusses, the applied loading (including live loads) on the different members of the truss must be calculated in such a way that would account for any combination and/or position of the live loads, preferably in an easy and efficient manner. Similar to live loading on beams and frames, influence lines can be constructed for the truss members in question (if not all the members).

For the purpose of this analysis, the following trusses are both internally and externally determinate.

## Method for Analysis

Suppose that the influence lines of members CD, DH, HI, EF, FK, and KL of the Howe truss[1] below are required.

### Step I: Support Reaction Analysis

For analysis purposes, the truss will be subjected to a unit load of 1 kN at any arbitrary distance 'x' which is to be measured from point 'A'. To begin, the influence lines of the support reactions must be determined using equilibrium.

 (1) Solving for $A_{x}$ $+ {\rightarrow} \sum F_{x}=0$ $A_{x}=0$ (2) Solving for $K_{y}$ $\sum\ M_{A}=0$ $0=-1(x) + K_{y}(20)$ $K_{y}=\frac{x}{20}\$ (3) Solving for $A_{y}$ $+ {\uparrow} \sum F_{y}=0$ $0=-1+K_{y}+A_{y}$ $A_{y}=1-K_{y}$ $A_{y}=1- \frac{x}{20}\$

The influence lines for the reactions $A_{y}$ and $K_{y}$ can now be drawn according to the equations.

The influence lines drawn above illustrate the value of each reaction according to where the unit load is placed.

NOTE: Notice that when the unit load is placed at $A$, reaction $K_{y}$ is zero while $A_{y}$ is equal to the unit load, i.e. $A_{y}$ is solely supporting the unit load. The same goes for when the unit load is at $K$, $A_{y}=0$ while $K_{y}=1.0$.

LESSON: When the load is at a reaction point, that reaction force solely supports the unit load while all the other reactions are zero at that point.

### Step II: Member Analysis for Cut a-a

In order to find the influence lines of all the members, the 'method of sections' must be utilized. To find the influence lines of members CD, DH, and HI, the truss must be cut along the axis 'a-a'.

All members are drawn assuming they are in tension.

In order to simplify the analysis, the moment will be taken about point H thus eliminating $F_{DH}$ and $F_{HI}$ from the equation. The remaining force, $F_{CD}$ can then be isolated. To further simplify the process, when the unit load is acting along Section A, $x <= 8m$, Section B will be analyzed. When the unit load is acting along Section B, $x >= 12m$, Section A will be analysed. This technique will allow us to not consider the unit load for that particular section since it is in fact only acting on the other section[2].

 Section B Analysis: Unit load acting along Section A $x<=8m$, Assuming counter clockwise rotation as positive: $\sum M_{H}=0$ $0=5F_{CD}+K_{y}(4+8)$ $F_{CD}= \frac{-12}{5}$$\frac{x}{20}$$ = \frac{-3}{25}\ x$ (for $x<=8m$) Section A Analysis: Unit load is acting along Section B $x>=12m$ Assuming counter clockwise rotation as positive: $\sum M_{H}=0$ $0=-5F_{CD}-8A_{y}$ $F_{CD}= \frac{-8}{5}\ A_{y}$ $F_{CD}= \frac{-8}{5}$$1-\frac{x}{20}$$ = \frac{2}{25}\ x - \frac{8}{5}\$ (for $x>=12m$)

Now that $F_{CD}$ is known with respect to $x$ the influence diagram can be created. Since influence lines are always linear [3], the portion of $8m<=x<=12m$ can be drawn by simply connecting the two portions:

Using equilibrium the remaining member forces of the cut can be found with respect to $x$. Similar to the previous method for finding force member, $F_{CD}$, the unit load is placed on the opposite side of the truss from where it is being analysed. Again, two equations for each force are found and used to create the influence diagrams for $F_{DH}$ and $F_{HI}$.

 Section B Analysis: Unit load acting along Section A $x<=8m$, $+ {\uparrow} \sum F_{y}=0$ $0=K_{y}- \frac{5}{6.4}\ F_{DH}$ $F_{DH}=\frac{6.4}{5}\ K_{y}$ $F_{DH}=\frac{6.4}{5}\ \frac{x}{20}\ = \frac{6.4}{100}\ x$ for $x<=8m$ Section A Analysis: Unit load is acting along Section B $x>=12m$ $+ {\uparrow} \sum F_{y}=0$ $0=A_{y}+\frac{5}{6.4}F_{DH}$ $F_{DH}=\frac{-6.4}{5}\ (A_{y})$ $F_{DH}=\frac{6.4}{5}\ (1-x/20)$ $F_{DH}=\frac{6.4}{100}\ x - \frac{6.4}{5}\$ for $x>=12m$

The influence line for the portion $8m<=x<=12m$ can be drawn by connecting the other two portions[3].

When solving for $F_{HI}$, the unit load is not acting in the x-direction, therefore any of the two sections can be used while using $\sum F_{x}=0$. Using Section A, this knowledge is exploited to determine the following relationship: $0=F_{CD}+F_{HI}+\frac{4}{6.4}\ F_{DH}$. However, when substituting for $F_{CD}$ and $F_{DH}$ the influence line equation will be obtained by substituting the respective equations of $F_{CD}$ and $F_{DH}$ for when $x<=8m$ and $x>=12m$.

 For x >= 12m: $0=F_{CD}+F_{HI}+\frac{4}{6.4}\ F_{DH}$ $0=(\frac{2}{25}\ x -\frac{8}{5}\) + F_{HI}+ (\frac{6.4}{100}\ x - \frac{6.4}{5}\) (\frac{4}{6.4}\)$ $F_{HI}= -\frac{12}{100}\ x + \frac{12}{5}\$ For x <= 8m: $0=F_{CD}+F_{HI}+\frac{4}{6.4}\ F_{DH}$ $0=\frac{-3}{25}\ x + F_{HI}+ \frac{6.4}{100}\ * \frac{4}{6.4}\$ $F_{HI}=\frac{2}{25}\ x$

### Step III: Member Analysis for Cut b-b

The influence lines for members EF, FK, and KL follow the same procedure for analysis, but this time cutting the structure along axis 'b-b'.

 For $x<=20m$ (Using Section B) Assuming counter clockwise rotation as positive: $\sum M_{K} = 0$ $0 = 5F_{EF}$ $F_{EF} = 0$ Note: This makes sense since when $x<=20m$, EF, FK, and KL are zero force members For $x = 24m$ (Using Section A) Assuming counter clockwise rotation as positive: $\sum M_{K} = 0$ $0 = -20A_{y}-5F_{EF}$ $F_{EF} = \frac{-20}{5}A_{y} = -4(1-\frac{x}{20}\) = \frac{x}{5}-4$
 for $x<=20m$ (Using Section B) $+ {\uparrow} \sum F_{y}=0$ $0 = -F_{FK}$ $F_{FK}=0$ for $x=24m$ (Using Section A) $+{\uparrow} \sum F_{y} = 0$ $0=A_{y}+K_{y}+F_{FK}$ $F_{FK} = -A_{y} - K_{y}$ $F_{FK} = -(1-\frac{x}{20})-\frac{x}{20}$ $F_{FK}=-1$
 for $x<=20m$ $+ {\rightarrow} \sum F_{x}=0$ $0 = F_{KL}+F_{EF}$ $F_{KL} = 0$ for $x=24m$ $+ {\rightarrow} \sum F_{x}=0$ $0=F_{KL}+F_{EF}$ $0 = F_{KL}+(\frac{x}{5}-4)$ $F_{KL}=-\frac{x}{5}+4$

## Example: Warren Truss

Determine the influence lines of the indicated members of the Warren truss[4] below.

 Solving for $G_x$ $+ \rightarrow \sum{F_x}=0$ $G_x=0$ Solving for $J_y$ $\sum{M_G}=0$ $0 = 1(6-x) + 18J_y$ $J_y= \frac{x}{18}-\frac{1}{3}$ Solving for $G_y$ $+\uparrow \sum{F_y}= 0$ $0 = J_y+G_y-1$ $G_y =-\frac{x}{18}+\frac{4}{3}$

Solving for the reaction force in member $F_{DI}$

 For $x<=12$ (Using Section B) $+ \uparrow \sum{F_y} = 0$ $0 = J_y+\frac{4}{5}F_{DI}$ $F_{DI} = -\frac{5}{4}(\frac{x}{18}-\frac{1}{3})$ For $x>=18$ (Using section A) $+ \uparrow \sum{F_y} = 0$ $0 = G_y -\frac{4}{5}F_{DI}$ $F_{DI} = \frac{5}{4}(\frac{4}{3}-\frac{x}{18})$

Solving for the reaction force in member $F_{DE}$

 For $x<=12m$ (Using Section B) $\sum{M_I} = 0$ $0= 6J_y+4F_{DE}$ $F_{DE} = \frac{1}{2}-\frac{x}{12}$ For $x>=18m$ (Using Section A) $\sum{M_I}=0$ $0 =-12G_y-4F_{DE}$ $F_{DE} = \frac{x}{6}-4$

Solving for the reaction force in member $F_{HI}$ and using Section B:

 For $x<=12m$ $+\rightarrow \sum{F_x} = 0$ $0 = -F_{HI} -\frac{3}{5}F_{DI}-F_{DE}$ $F_{HI} = \frac{-3}{5}\frac{-5}{4}(\frac{x}{18} - \frac{1}{3}) - (\frac{1}{2}-\frac{x}{12})$ $F_{HI} = \frac{x}{8} - \frac{3}{4}$ For $x>=18$ $+ \rightarrow \sum{F_x} = 0$ $0 = -F_{HI} -\frac{3}{5}F_{DI}-F_{DE}$ $F_{HI} = -\frac{3}{5}\frac{5}{4}(\frac{4}{3}-\frac{x}{18}) - (\frac{x}{6} - 4)$ $F_{HI} = 3 - \frac{x}{8}$

## Example

Determine the influence lines for the members AF and CF of the following truss

We have to make a cut that goes through CF so we can find the force at the member.

First we must find the influence lines of the reactions at A and E.

$\sum{M_E}=0$

$80A_y - (80 - x)1.0 = 0$

$A_y = 1 - \frac{x}{80}$

$\sum{M_A}=0$

$80E_y - (x)(1.0) = 0$

$E_y =\frac{x}{80}$

With the point load being at A, Ay is 1.0 and Ey is zero. At B, with x = 20, we replace the value of x in the Ey equation, which gives us By = 0.25.

Influence line for reaction at A

With the point load now being at E, Ey is 1.0 and Ay is zero. At B, with x = 20, we replace the value of x in the Ay equation, which gives us By = 0.75. Same thing at C, with x = 40, we get Cy = 0.5.

Influence line for reaction at E
Inluence line of AF
$+\uparrow\sum{F_y}=0$

$A_y -1 + \frac{3}{5}F_{AF} = 0$

at x = 0, the 1 KN load is at joint A, with $A_y = 0$ we get:

$F_{AF}=0$

at 20<x<80 when the 1 KN load is located to the right of joint B we get:

$+\uparrow\sum{F_y}=0$

$A_y + \frac{3}{5}F_{AF} = 0$

$F_{AF} = -1.67A_y$

For the influence line of AF, with x = 20, $F_{AF}$ will equal 0.25. And with x = 80, $F_{AF}$ will equal to 0.

Influence line for AF
Influence line of CF
To find the force of member CF we connect members FG and BC at point I. The moment at point I will cancel all the forces and leave us with one force at member CF.

when x is between 0 and 20 we get:

$\sum{M_I}=0$

$\frac{3}{5}F_{CF}(80) + E_y(120)=0$

$F_{CF} =-2.5E_y$

at x is between 40 and 80 we get:

$\sum{M_I}=0$

$A_y(40) - \frac{4}{5}F_{CF}(15) - \frac{3}{5}F_{CF}(60)=0$

$F_{CF} =0.83A_y$

Influence line for CF

## References

1. Kassimali, Aslam. Structural Analysis. 4th ed. Stamford: Cengage Learning, 92. eBook.
2. Erochko, Jeffrey. Personal Communication, Oct. 16, 2013.
3. Hibbeler, Russ. Structural Analysis. 8th ed. Upper Saddle River: Pearson Prentice Hall, 2012. 232. eBook.
4. "Heritage Documentation Programs." National Park Service. Historic American Engineering Record, n.d. Web. 16 Nov 2013. <http://www.nps.gov/history/hdp/samples/HAER/truss.htm>.