Influence Line
Influence lines show how internal forces or reactions at a specific location vary as a unit load moves across a structure.
They are particularly useful in determining the critical positions of live loads.
Two Common Methods:
- Tabular Method (Analytical)
- Müller-Breslau Principle (Qualitative)
1. Tabular Method (Table-Based Analytical Influence Lines)
This method is best suited for statically determinate structures. It involves placing a unit load at key points and solving for the response of the desired function (e.g., support reaction, shear, or moment).
Steps:
- Select the function you want to analyze: a support reaction, shear at a section, or moment at a section.
- Divide the beam into key load positions — usually at supports and points of interest.
- Place a unit load (1.0 kN or 1.0 unit) at one position at a time.
- Using equilibrium, solve for the value of the function of interest at each load position.
- Plot or tabulate the values to create the influence line diagram.
Example Table Format:
| Unit Load Position |
Support Reaction $R_A$ |
Shear at Section X |
Moment at Section X |
| At A | 1.0 | ... | ... |
| Midspan | ... | ... | ... |
| At B | 0.0 | ... | ... |
The tabular method is exact and useful for numerical influence line diagrams, especially for statically determinate systems.
2. Müller-Breslau Principle (Qualitative Influence Lines)
The Müller-Breslau Principle provides a fast, visual way to sketch the shape of an influence line, even for indeterminate structures.
It states:
“The influence line for a response function (reaction, shear, or moment) has the same shape as the deflected shape of the structure when the corresponding restraint is released and given a unit displacement in the direction of the function.”
Steps:
- Identify the function you want to draw an influence line for (e.g., $R_A$, $V_x$, $M_x$).
- Modify the structure by releasing the restraint or cut at the location of the function.
- Apply a unit displacement in the direction of the desired function:
- Vertical displacement for vertical reactions and shear
- Rotation for moments
- Sketch the deflected shape — this is the qualitative influence line.
- Positive displacement direction = positive influence value. The relative height of the sketch indicates the magnitude.
The Müller-Breslau Principle is particularly useful for indeterminate structures or quick checks of influence line shapes.
After using the Müller-Breslau sketch, you can optionally apply the conjugate beam method or unit load method to compute exact ordinates if needed.
Exam Generator Problems
Additional board-style practice items for this topic.
Question Bank: q5
PSAD - Structural Theory / Influence Line / Engr. Janclyde Espinosa (Clidez)
CE Board May 2010
A simply supported beam 10 meters long has an overhang of 2m at the left support. If a highway uniform load of 9.35kN/m and a concentrated load of 116kN passes through the beam, compute the following based on the influence line for maximum shear at midspan.
Determine the length of the beam where the uniform load could produce maximum positive shear at the midspan.
- 7m
- 2m
- 5m
- 10m
Determine the length of the beam where the uniform load could produce maximum negative shear at the midspan.
- 5m
- 2m
- 7m
- 10m
If the concentrated load will be placed at the end of the overhang, compute the maximum shear at the midspan.
- 23.2kN
- 58.4kN
- 30kN
- 27.9kN
Solution pending in psadquestions/q5.json.
Question Bank: q138
PSAD - Structural Theory / Influence Line / Engr. Janclyde Espinosa (Clidez)
Refer ot the compound bridge girder shown. Analyze the beam using the influence line diagram. The components of the moving load are the following:
Concentrated load = 120kN
Uniform live load = 10kN/m
Compute the ordinate at E considering the inluence line of the maximum reaction at F.
- 1.20
- 0
- 1.00
- 1.50
Calculate the maximum reaction at F in kN.
- 474
- 378
- 447
- 456
Calculate the maximum reaction at B in kN.
- 324
- 307
- 286
- 344
Solution pending in psadquestions/q138.json.
Question Bank: q566
PSAD - Structural Theory / Influence Line / Engr. Deguma
For the beam shown, a=25m, b=5m, and c=15m.
Uniformly Distributed Load, wLL=20kN/m
Concentrated Load = 60kN
For maximum reaction at the support at A, solve for the beam length,
which should be loaded with the uniformly distributed load?
- 25m
- 15m
- 10m
- 5m
For maximum reaction at the support at D, solve for the beam length,
which should be loaded with the uniformly distributed load?
- 15m
- 25m
- 5m
- 10m
Which of the following most nearly gives the maximum reaction (kN) at the
support at B due to the given loads?
- 612
- 645
- 588
- 540
Solution pending in psadquestions/q566.json.
Question Bank: q626
PSAD - Structural Theory / Influence Line / Mastermatician
A 12-meter-long beam is simply supported at the right
end and at 2 meters from the left end. It is subjected to a highway
load consisting of 9.35 kN/m uniformly distributed load and 116 kN
concentrated load. Based on the influence line for maximum
moment at midspan,
What is the total length in meters of the beam which should be
subjected to a uniformly distributed load to obtain the
maximum positive moment at midspan?
- 10
- 12
- 2
- 5
What is the total length in meters of the beam which should be
subjected to the uniformly distributed load to obtain the
maximum negative moment at the midspan?
- 2
- 5
- 10
- 12
What would be the critical positive moment (kN-m) at midspan
due to the highway lane load?
- 407
- 214
- 325
- 154
Solution pending in psadquestions/q626.json.
Question Bank: q628
PSAD - Structural Theory / Influence Line / Mastermatician
A 10 m long beam is simply supported at the left end and at 2
m from the right end. It is to be analyzed for uniformly
distributed moving load. For maximum reaction at the right
support, what is the total beam length (m) which should be
subjected to the moving load?
- 10m
- 8m
- 12m
- 6m
Solution pending in psadquestions/q628.json.
Question Bank: q629
PSAD - Structural Theory / Influence Line / Mastermatician
The beam in the figure supports a distributed load of
1.5 kN/m and a single concentrated load of 8 kN. The dead load
is 2 kN/m.
Determine the maximum positive moment at C, in kN-m.
- 13
- 23
- 12
- 18
Determine the maximum positive shear at C, in kN.
- 6.50
- 3.45
- 4.78
- 4.34
Solution pending in psadquestions/q629.json.