# Muller Breslau Principle

## Contents

## Müller-Breslau's principle

### Background/Biography

The founder/creator of the Müller-Breslau's principle is Heinrich Müller-Breslau, who was born on the 13th of May, 1851, in Breslau, Germany (now known as Wroclaw, Poland). Heinrich later decided to add Breslau to the end of his name, after his hometown.

Müller began his career in the engineering field by joining the Prussian Army's engineering sector. After the war he then pursued further education in Berlin, Germany by taking an array of courses involving engineering and mathematics. Müller never graduated, yet began as a engineering consultant. Müller's main focus was on iron structures including bridges.

After a few years, Müller had published a series of articles which led to him working as a professor of bridge design at the Polytechnic Institute of Hannover. Müller continued to publish and consult, which then led to being appointed to the Chair of Structural Engineering at the Berlin-Charlottenburg Institute of Technology.

Müller's designs are influential to modern day structural engineering practice, and has personally designed many structures in Germany, Russia, etc.

- all information is that of the writer of the source and all accreditation for such information rests with them
^{[2]}

### Important Previous Knowledge

In order to fully understand Müller-Breslau's principle we need to understand some other parts of structural analysis. As we will be constructing Influence lines, having a good understanding of them would be invaluable.

Beam and Frame Influence Lines

In order to construct these influence lines we need to understand deflected shapes. If you require further information please view the page below.

### Description of Principle

The Müller-Breslau principle is a method which is used to determine the influence lines of a simply supported beam/structure. This method removes a support and uses a unit force to determine the reaction at the removed support.

As per Kassimali text ^{[3]}

"The influence line for a force (or moment) response function is given by the deflected shape of the released structure obtained by removing the restraint corresponding to the response function from the original structure and by giving the released structure a unit displacement (or rotation) at the location and in the direction of the response function, so that only the response function and the unit load perform external work."

The Müller-Breslau principle is just a simplified way of finding the influence lines of members. It involves modifying the member for our use and then finding the deflected shape of the modified member. To modify the member we release the restraint at the point which we are going to create the influence line about.

ORIGINAL | MODIFIDED |

Roller | Point Load |

Hinge | Released Hinge |

Pin | Roller + Point Load* |

Fixed | (Roller + Point Load)* or Hinge** |

Continuous Beam | Cut Beam (finding internal shear) |

Continuous Beam | Hinge (finding internal moment) |

*the point load is positive and applied in the direction of the force the we are drawing the influence line for. **apply a hinge when the moment is what we are drawing the influence line for.

We then think about what the deflected shape would look like if a positive force or moment were applied at that point. We are basically including a qualitative deflection for a force and a qualitative rotation for a moment.

We have completed the qualitative influence line diagram.

For the quantitative diagram the geometry will be very helpful to find the maxima and minima. To do this we need to find the value of the influence line at one point. We can use a unit load on the point of interest. Using simple equilibrium along with equations of condition we will find the value at that point and use geometry for the rest of the values.

## General Procedure

As per Kassimali text ^{[4]}

### Step 1

Firstly, we need to ensure that the point to which the influence line is being drawn will not restrict the movement of the unit load that will be applied. This means we must remove the restraint and replace it with a support reaction that will allow the beam to move in the desired direction.

### Step 2

Secondly we will need to select a unit load that will move across the beam creating a deflected shape, typically we will choose 1 kN.

### Step 3

Next we apply that unit load at the point being evaluated and draw the resulting deflected shape. The influence line at that point will be the same as the deflected shape. ^{[5]}

### Step 4

Similar triangles will be used to solve for the magnitudes of the influence line at various significant points along the beam.

### Step 5

To find the influence line at a point on the beam that doesn't have a support reaction we will solve for the shear force. We will place the unit load on point being evaluated. We will then use equilibrium to solve for the support reactions due to the applied unit load. We then cut the beam at the point being sheared and use equilibrium to solve for the values of the influence line just to the left and just to the right of the point in question. The sum of these values at the point being evaluated should sum to equal our unit load.

### Step 6

To solve for the influence line for bending moments we must ensure that the point we wish to evaluate is free to rotate. We apply our unit moment at the chosen point and draw the resulting deflected shape. This deflected shape diagram is the same as the influence diagram. We solve for the values of the influence line due to the point moment by applying a the unit load on the point in question. We then solve for the reaction forces like before. With the reaction forces known, we can then cut the beam where the unit load was applied and solve for all the unknown moment values on the influence line by equilibrium.

## Example

### Question:

Using Müller-Breslau's Principle for finding influence lines, determine the vertical reactions at supports A, C, E, and F, the shear at points C', D' and E' (midpoints of section CD, DE, and EF respectively) and the internal moments at the listed midpoints C', D' and E' for the simple supported beam shown in *Fig. 1*.

### Preparation:

For this solution, we need to make use of similar triangles and two main equations which simplify geometric relations to solve for displacements.

**1.**The first of these equations relates the cut length (a) and the total length of the section (L) to produce a value for the magnitude of the force and is given by

$ \[\triangle_{x} = \frac{a}{L}\] $

**2.**The second equation relates the interior angles of triangle and its side lengths to produce the magnitude of the moment. The equation is simplified to

$ \[\triangle{x} = a(1 - \frac{a}{L})\] $

### Solution:

Before solving any equations, we first must apply Müller-Breslau's Principle to find the influence lines themselves. This is done by removing the reaction support or adding the shear or moment conditions and displacing the beam by a unit load ( 1 kN ) or moment ( 1 kN m) at said location. This will represent the reaction, shear or moment at the specified location as a load moves across the beam.

Following, you will find pre-drawn influence lines for all specified locations at which we would like to analyze.

**Fig. 1 (a) represents the influence lines, for the support reactions, for a unit load moving across the beam. the first beam represents the influence diagram for the reaction at support A and the successive beams are for the reactions at supports C, E and F respectively.**

Support A | Support C | Support E | Support F |

There are no calculations needed for support A since the beam displaces by 1 kN and returns to its original at the first hinge |
$ \[\frac{1}{3}=\frac{x_{B}}{5.5} \\x_{B}=1.83 \] $ |
$ \[\frac{1}{4}=\frac{x_{D}}{7} \\x_{D}= 1.75 \\\ \\\frac{1.75}{3}=\frac{x_{tot}}{5.5} \\x_{tot}= 3.208 \\x_{B}= x_{tot}-x_{D} \\x_{B}= 1.46\] $ |
$ \[\frac{1}{4}=\frac{x_{tot}}{7} \\x_{tot}= 1.75 \\x_{D}= x_{tot}-x_{F} \\x_{D}= 0.75 \\\frac{0.75}{3}=\frac{x_{tot2}}{5.5} \\x_{tot2}= 1.375 \\x_{B}= x_{tot2}-x{D} \\x_{B}= 0.625\] $ |

The table above represents the calculations made to determine the magnitudes of the influence lines at various locations as the 1 kN load travels across the beam. When the unit load is placed at support A, the beam faces no deflection after the hinge at point B and as mentioned in the table, no calculations are needed as there are no other deflections. When we place the load at support C however, the beam is deflected from support A to support C where the maximum deflection occurs at hinge B. To find the magnitude of this deflection at point B we use similar triangles as demonstrated in the table. Next we move our unit load to support E where we see a deflected shape throughout the entire beam. We must now calculate the magnitude of the influence line using similar triangles for points B and D just like we did previously. Finally we place out 1kN load at the support F which results in a deflected shape throughout the whole beam. Once again, we calculate the magnitude of the influence line with similar triangles. We have now completed the influence line diagrams for all support points along the beam.

**Fig. 1 (b) represents the influence diagram for the internal shear at points C', D' and E' respectively.**

Midpoint C' | Midpoint D' | Midpoint E' |

$ \[x_{C'+} = \frac{1.5}{3} \\x_{C'+}= 0.5 \\x_{C'-} = 1- x_{1} \\x_{C'-} = 0.5 \\\frac{0.5}{1.5} = \frac{x_{tot}}{4} \\x_{tot} = 1.33 \\x_{B} = x_{tot}-x_{C'-} \\x_{B}=0.833\] $ |
$ \[\frac{1}{3} = \frac{x_{tot}}{5.5} \\x_{tot}= 1.83 \\x_B=x_{tot} -x_D \\x_B=0.83\] $ |
$ \[\frac{0.5}{2} = \frac{x_{tot}}{5} \\x_{tot}= 1.25 \\x_D=x_{tot} -x_{E'} \\x_D=0.75 \\\frac{0.75}{3} = \frac{x_{tot}}{5.5} \\x_{tot}= 1.375 \\x_B=x_{tot} -x_{D} \\x_B=0.625\] $ |

**NOTE:* Since the cuts are at the midpoint of each section, we can assume that it will have 0.5 kN up and 0.5 kN down. Otherwise we would have to use equation 1 to solve for the displacements*

In the table above we are solving for the influence line diagrams due to a unit shear placed at the mid point between support C and hinge D (C'), hinge D and support E (D'), and support E and support F (E'). Starting with C' and recognizing that the shear is placed at the midpoint of the section of the beam and that the shear from the left and right of point C' is equal with opposite directions, we can say that C' from the left is equal to C' from the right which is half of the unit load. Then we cut can cut the beam at point C' and use similar triangles to find the magnitude of the influence line for the total cut beam. We then subtract our value for C' from the total cut beam and result with the magnitude of the influence line at point B. The same process is done to solve for D' and finally E'.

**Fig. 1 (c) represents the influence diagram for the internal moment at points C', D' and E' respectively. For internal moments, we will make use of equation 2, stated above.**

Midpoint C' | Midpoint D' | Midpoint E' |

$ \[\triangle_{C'}=1.5(1-\frac{1.5}{3}) \\\triangle_{C'} = 0.75 \\\frac{0.75}{1.5}=\frac{\triangle_{tot}}{4} \\\triangle_{tot}=2 \\\triangle_{B}=\triangle_{tot}-\triangle_{C'} \\\triangle_B=1.25\] $ |
$ \[\triangle_{D}=1.5(1-\frac{1.5}{3}) \\\triangle_{D} = 0.75 \\\frac{0.75}{3}=\frac{\triangle_{tot}}{5.5} \\\triangle_{tot}=1.375 \\\triangle_{B}=\triangle_{tot}-\triangle_D \\\triangle_B=0.625\] $ |
$ \[\triangle_{E'}=2(1-\frac{2}{4}) \\\triangle_{E'} = 1 \\\frac{1}{2}=\frac{\triangle_{tot}}{5} \\\triangle_{tot}=2.5 \\\triangle_{D}=\triangle_{tot}-\triangle_{E'} \\\triangle_D=1.5\] $ $ \[\frac{1.5}{3}=\frac{\triangle_{tot2}}{5.5} \\\triangle_{tot2}=2.75 \\\triangle_{B}=\triangle_{tot2}-\triangle_D \\\triangle_B=1.25\] $ |

**NOTE:* For internal moments, influence diagrams only give displacements. Displacements must be then multiplied by the distance to the original point to find magnitude of the internal moment due to a unit moment placed some point along the beam*

The next step is very similar to solving influence lines due to unit loads and shears, only now, we have applied a unit moment of 1kN. We will apply this moment in the same locations, C', D' and E'. Starting at C', the point moment will cause the beam to rise between support C and hinge D and descend between supports A and C. From this, we can draw the deflected shape which is also the influence diagram, and use equation 2 to solve for the displacement at C'. Then we can use similar triangles to solve for the total beam displacement from support A to hinge D. We may then subtract the displacement at C' from the total displacement to solve for the displacement at B. The same process is repeated as we move our point moment along the beam to points D' and E' as shown in the table above.

**Fig. 1 (d): The Completed Influence Line Diagrams for All Cases Calculated Above.**

## Influence Diagram Reference Guide

*The following is a sample reference guide made to illustrate how some common elements of a simply supported beam can influence your Influence diagram. This table can be used as a guide to solving Müller-Breslau problems.*

Reaction, Shear and Moment Influence Lines |
Hinged Influence lines |

## References

- ↑ National Secretariat of the Scientific-Disciplinary Sector ICAR/08 Construction Science. (copyright 2009). "Heinrich Müller-Breslau" [Online]. Available: http://www.scienzadellecostruzioni-segreteria.it/passato.php >.
- ↑ Charles Scribner's Sons. (copyright 2008). "Müller-Breslau, Heinrich" [Online]. Available:http://www.encyclopedia.com/doc/1G2-2830903084.html >.
- ↑ Kassimali, A. (2011).
*Structural Analysis: SI Edition (4th ed.)*. Stamford, CT: Cengage Learning. - ↑ Kassimali, A. (2011).
*Structural Analysis: SI Edition (4th ed.)*. Stamford, CT: Cengage Learning. - ↑ Fanous, F. (2000, April 21).
*Introductory Problems in Structural Analysis.*Qualitative Influence Lines using the Muller Breslau Principle. Retrieved on November 23, 2013, from http://www.public.iastate.edu/~fanous/ce332/influence/simplecantenvelope.html.