Influence Lines for Distributed Loads

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Introduction

To familiarize yourself with the topics covered in this section, we recommend that you refer to Beam and Frame Influence Lines.

Summary

Introduced by E. Winkler in 1867 [1], the concept of influence lines is an important analytical method in the analysis of structures. All structures (typically buildings), when designed, consider the implications of dead or fixed loads which do not move (self weight, equipment, etc). However, live loads (people, etc) move across a certain distance and need to be considered, especially for an engineer designing bridges. This is why influence lines are used. The definition of an influence line is given as:

An influence line is a graph of a response function of a structure, as a function of the position of a downward unit load moving across the structure.[1]

Essentially, beams need to be designed based on the maximum stress loading which they receive, as each member may be different. An influence line represents how shears, moments, and reactions change as a load moves across a member. So when created, an influence line will easily indicate this maximum and beams can be properly designed in order to prevent failure.

Influence Lines for Distributed Loads can then be simplified down to:

  • Take 1

Using the shear influence line at any point on a beam, given some sort of load combination of a single or set of distributed loads (i.e. Live, Dead, Snow, etc.), the shear at a desired point is equal to the area under the distributed load of the shear influence line diagram multiplied by the magnitude of the uniformly distributed load for that entire area. This can be applied for both the shear and moment influence line diagrams at any point by simply substituting moment for shear in the above sentence.

  • Take 2

This method seems quite simple, yet the major aspect to take into account is the change in sign that may occur throughout a typical influence line diagram. Therefore, it can be said that, in any case (and keeping the signs of the influence line in check), the shear or moment at a point is the sum of all the areas in the influence line multiplied by the magnitudes of the uniformly distributed loads above each respective area.

Therefore, a pseudo equation for this topic can be defined as (for both shear and moment at a point):

$ M_{x}|V_{x} = \sum A_{i} \cdot DUDL + sum A_{i} \cdot LUDL + \sum P \cdot PL\ $

Where:

  • $ M_{x} \ $ - Moment at point x ($ kN \cdot m \ $)
  • $ V_{x} \ $ - Shear at a point x ($ kN \ $)
  • P - value of influence diagram at a point load
  • PL - point load magnitude ($ kN\ $)
  • DUDL - dead uniformly distributed load ($ kN/m\ $)
  • LUDL - live uniformly distributed load ($ kN/m\ $)
  • $ \sum A_{i}\ $ - sum of the area below a single UDL ($ m^{2} \ $)
Figure 1 - Simple Example - Uniformly Distributed Live Load Only

For this simple example of a moment influence diagram of a point between two reactions 'point A', with a LUDL of $ 15 kN/m \ $ between $ 4m \leq x \leq 8m\ $, the above equation can then be written as:

$ M_{A} = 0 + (\frac{1}{2} \cdot (4m) \cdot (2) \cdot (15 kN/m)) + 0\ $
$ M_{A} = 60 kN \cdot m \ $

Therefore, when a LUDL of $ 15 kN/m \ $ is placed between $ 4m \leq x \leq 8m\ $ of the beam, it will result in an internal moment at point A of $ 60 kN \cdot m \ $ .






Worked Example

For the following beam:

Figure 2 - Hinged Continuous Beam




Find the maximum positive and negative shear and moment at point B, given the following load combination:

  • Concentrated live load of 126 kN
  • Uniformly distributed live load of 20 kN/m
  • Uniformly distributed dead load of 12 kN/m




For the solution of this problem, steps 1 to 6 are similar to the above mentioned section on Influence Lines for Beams and Frames, yet the question is worked through nonetheless, to demonstrate the use of all the various influence diagrams. The following solution is broken into the following steps [1] [2] [3] [4]:

Step One

A cut is made at the hinge and then the moments about D between D and E are found (all moments about any of the points are taken in the counter clockwise direction being positive).

For $ (0m \leq x \leq 12m) $ :

$ \sum M_{DE} = 0\ $
$ E_{y}(4) = 0\ $
$ E_{y} =0 \ $
Step Two

Take the moment about D again but now include the unit force from a distance to the right of D.

For $ (12m \leq x \leq 16m) $ :

$ \sum M_{DE} = 0\ $
$ -1(x-12) + E_{y}(4) = 0\ $
$ E_{y} = \frac{x}{4} - 3\ $
Step Three

To find the reaction at Cy, take the moment about A. This gives the following equations:

$ \sum M_{AE} = 0\ $
$ -1(x) + C_{y}(8) + E_{y}(16) = 0\ $
$ C_{y} = \frac{x}{8} - (2)E_{y}\ $

Therefore, subbing in the two conditions for $ E_{y}\ $:

For $ (0m \leq x \leq 12m) $ :

$ C_{y} = \frac{x}{8} - (0)E_{y}\ $
$ C_{y} = \frac{x}{8}\ $

For $ (12m \leq x \leq 16m) $ :

$ C_{y} = \frac{x}{8} - (2) - (2)(\frac{x-12}{4})\ $
$ C_{y} = - \frac{3x}{8} + 6\ $
Step Four

Now that two of the reactions are found, use sum of the forces in the y direction to find the reaction at A (all forces are taken in the upwards positive direction).

$ \sum F_{y} = 0\ $
$ A_{y} + E_{y} + C_{y} - 1 = 0\ $
$ A_{y} = 1 - E_{y} - C_{y}\ $

Therefore, subbing in the two conditions for both $ E_{y}\ $ and $ C_{y}\ $:

For $ (0m \leq x \leq 12m) $ :

$ A_{y} = 1 - (0) - (\frac{x}{8})\ $
$ A_{y} = - \frac{x}{8} + 1\ $

For $ (12m \leq x \leq 16m) $ :

$ A_{y} = 1 - (\frac{x}{4} - 3) - (6 - \frac{3x}{8})\ $
$ A_{y} = \frac{x}{8} - 2\ $
Step Five

To find the shear at B, first, imagine placing a roller at point B which can allow beam AB and BE to move up and down.

So to find the shear, the beam is cut at point B. The sum of the forces in the y direction is then found at point B.

$ \sum F_{y} = 0\ $

For $ (0m \leq x \leq 4m) $ (point load is to the left of B) :

$ -B_{y} + A_{y} - 1 = 0\ $
$ -B_{y} + (1 - \frac{x}{8}) - 1 = 0\ $
$ B_{y} = -\frac{x}{8}\ $

For $ (4m \leq x \leq 12m) $ (point load is to the right of B, before the hinge) :

$ -B_{y} + A_{y} = 0\ $
$ -B_{y} + (1 - \frac{x}{8}) = 0 \ $
$ B_{y} = - \frac{x}{8} + 1\ $

For $ (12m \leq x \leq 16m) $ (point load is to the right of B, after the hinge) :

$ -B_{y} + A_{y} = 0\ $
$ -B_{y} + \frac{x}{8} - 2 = 0\ $
$ B_{y} = \frac{x}{8} - 2\ $
Step Six

To find the moment, the sum of the moments is taken at point B of the same cut at B.

$ \sum M_{E} = 0\ $

For $ (0m \leq x \leq 4m) $ (point load is to the left of B) :

$ -M_{E} + A_{y}(4) - 1(4-x) = 0\ $
$ -M_{E} + (1 - \frac{x}{8}) - 1(4-x) = 0\ $
$ M_{E} = \frac{x}{2}\ $

For $ (4m \leq x \leq 12m) $ (point load is to the right of B, before the hinge) :

$ -M_{E} + A_{y}(4) = 0\ $
$ -M_{E} + (1 - \frac{x}{8})(4) = 0 \ $
$ M_{E} = - \frac{x}{2} + 4\ $

For $ (12m \leq x \leq 16m) $ (point load is to the right of B, after the hinge) :

$ -M_{E} + A_{y}(4) = 0\ $
$ -M_{E} + (\frac{x}{8} - 2)(4) = 0\ $
$ M_{E} = \frac{x}{2} - 8\ $

Therefore, the following influence diagrams are found from using the calculated equations for each section of the beam, for each reaction and including the shear and moment for point B.


Figure 3 - Influence diagrams for all support reactions including shear and moment influence lines for point B










































Step Seven

The dead and live uniformly distributed loads, with the live point load, are placed to find both maximum positive and negative shear and moment at point B. The following diagram illustrates the load combinations needed for each case:

Figure 4 - Load Combinations for Maximum Positive and Negative Shear and Moment at point B




































Now, by simply multiplying the loads by the specified areas under the specific influence line diagrams:

For this section:

  • P - value of influence diagram at point B
  • LL - point live load magnitude (kN)
  • DUDL - dead uniformly distributed load (kN/m)
  • LUDL - live uniformly distributed load (kN/m)
Maximum Shear at point B
$ S_{B} = \sum A_{i}*DUDL + \sum A_{i}*LUDL + P*LL\ $
Maximum Positive Shear

Due to the DUDL:

$ S_{B,D} = (\frac{1}{2})((-0.5)(4 m)(12 kN/m) + (\frac{1}{2})((0.5)(4 m)(12 kN/m) + (2)[(\frac{1}{2})((-0.5)(4 m)(12 kN/m) + (\frac{1}{2})((-0.5)(4 m)(12 kN/m)]\ $
$ S_{B,D} = -24 kN\ $

Due to the LUDL:

$ S_{B,L} = (\frac{1}{2})((0.5)(4 m)(20 kN/m)\ $
$ S_{B,L} = 20 kN\ $

Due to the LL:

$ S_{B,P} = (0.5)(126 kN)\ $
$ S_{B,P} = 63 kN\ $

Therefore,

$ S_{B} = (-24 kN) + (20 kN) + (63 kN)\ $
$ S_{B} = 59 kN $
Maximum Negative Shear

Due to the DUDL: This is the same as before for maximum positive shear:

$ S_{B,D} = -24 kN\ $

Due to the LUDL:

$ S_{B,L} = (\frac{1}{2})((-0.5)(4 m)(20 kN/m) + (2)[(\frac{1}{2})((-0.5)(4 m)(20 kN/m) + (\frac{1}{2})((-0.5)(4 m)(20 kN/m)]\ $
$ S_{B,L} = -60 kN\ $

Due to the LL:

$ S_{B,P} = (-0.5)(126 kN)\ $
$ S_{B,P} = -63 kN\ $

Therefore,

$ S_{B} = (-24 kN) + (-60 kN) + (-63 kN)\ $
$ S_{B} = -147 kN\ $
Maximum Moment at point B
$ M_{B} = \sum A_{i}*DUDL + \sum A_{i}*LUDL + P*LL\ $
Maximum Positive Moment

Due to the DUDL:

$ M_{B,D} = (2)(4 m)(12 kN/m) + (-2)(4 m)(12 kN/m)\ $
$ M_{B,D} = 0 kN\ $

Due to the LUDL:

$ M_{B,L} = ((2)(4 m)(20 kN/m)\ $
$ M_{B,L} = 160 kN m\ $

Due to the LL:

$ M_{B,P} = (2)(126 kN)\ $
$ M_{B,P} = 252 kN m\ $

Therefore,

$ M_{B} = (0 kN m) + (160 kN m) + (252 kN m)\ $
$ M_{B} = 412 kN m\ $
Maximum Negative Moment

Due to the DUDL: This is the same as before for maximum positive shear:

$ M_{B,D} = 0 kN m\ $

Due to the LUDL:

$ M_{B,L} = (-2)(4 m)(20 kN/m)\ $
$ M_{B,L} = -160 kN m\ $

Due to the LL:

$ M_{B,P} = (-2)(126 kN)\ $
$ M_{B,P} = -252 kN m\ $

Therefore,

$ M_{B} = (0 kN m) + (-160 kN m) + (-252 kN m)\ $
$ M_{B} = -412 kN m\ $

References

  1. 1.0 1.1 1.2 Kassimali, A. (2011). Structural Analysis: SI Edition (4th ed.). Stanford, CT: Cengage Learning.
  2. http://www.youtube.com/watch?v=Fj5SVD0Rmy0
  3. http://www.engineeringwiki.org/wiki/Beam_and_Frame_Influence_Lines
  4. Hibbeler, R. C. (2010). Engineering Mechanics: Statics and Dynamics (12th ed.). Prentice Hall.