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                Applications of Influence Line Diagram  

Applications of Influence Line Diagram

From the influence line we can determine the response (reaction, shear, moment etc. ) of a structure due to multiple point loads as well as due to uniform load. Influence line also helps in finding the maximum value of a response function.

Response due to Point Load

Response due to single point load = magnitude of load x ordinate of ILD at the position of point load.
Response due to multiple point loads = equal to the algebraic sum of the product of load and the corresponding ordinate of ILD.

Illustrative Example: Consider A simply supported beam (figure 1) having a span of 7 m.
Influence line diagram application
The reaction at support B due to 10 kN load will be equal to (10) x (value of ordinate at 2m from the ILD for RB) = 10 x (2/7) = 20/7 kN.

Similarly the reaction at B due to 5 kN load would be equal to (5) x (value of ordinate at 5 m from the ILD for
) = 5 x (5/7) = 25/7 kN
Reaction at support B due to both the loads = 10 x (2/7) + 5 x (5/7)
                           = 20/7 + 25/7 = 45/7 kN = 6.43 kN

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Response due to Uniform Load

Response due to uniform load is equal to the intensity of load multiplied by the area of ILD under the load.

Consider a beam of span 5m shown in figure 2 and loaded with a 3 m long uniform load of intensity 7 kN/m as shown.

influence line application

The value of reaction at B due to this uniform load =

 (intensity of uniform load) x (area of ILD between C and D).

area of ILD between C and D = area of trapezium

=0.5(ordinate at C+ordinate at D) x (length of uniform load)

                      = 0.5 (2/7 + 5/7) (3) = 0.5 (7/7) (3) = 1.5

therefore, Reaction at B = (7) (1.5) = 10.5 kN

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Last updated on Wednesday October 23, 2013