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This document from the spring 2009 ce 331 course introduces the concept of influence lines, which are used to calculate member forces in structures due to moving loads. Examples and formulas for calculating reactions, moments, and influence diagrams. Students will learn how to identify critical locations, construct influence lines, and use them to determine maximum moments and reactions.
What you will learn
Typology: Schemes and Mind Maps
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First of all, what is an influence line? An influence line is a plot of a particular member force at a particular location (say moment at the middle of Span 1) due to a moving load, P(x).
ft P
x
a (^) ,
Influence line for moment at “a” (M (^) a) due to unit load at “x”, Px
x, ft M (^) a, k-ft
5 2.
14 -0.
25 0.
Influence lines are useful for calculating member forces due to moving loads, for example on crane rails or bridges. A critical location is first identified (at midspan say) and the influence line is constructed by calculating the moment at midspan due to a unit point load at each location (x) along the bridge. The influence line can then be used to:
Example: Calculate the Ma due to the truck axle loads shown below.
k ft a
k ft k ft a
k ft k ft a
−
− −
=
a
x
Px
a
9 ft
30 k^10 k
X, ft
10 20 30
Influence diagrams can be constructed by applying a unit deformation associated with the type of member force.
reaction
shear
moment
1 rad
1
1
Draw the influence line for moment at E by “breaking” the beam at E and rotating the right end 1 radian^ relative to the left end, as shown. Since the beam segments are both 12 feet on either end of the break, the angles of each end are equal and equal to one half of 1.0rad^ = 0.5 rad^. The ordinate of the influence line at E is calculated from the following equation:
tan
tan
rad a 12'
a
forsmalldisplacements
a
The other ordinates are calculated using similar triangles.
The moment at E due to the concentrated load is maximum when the load is placed at E. The moment due to the 18 k^ load is then calculated from:
k ft k ft E
k M^18 = ( 18 )( 6 )= 108 −
The moment at E due to the uniform load is maximum when the load is placed at on the beam where the influence diagram is positive. The moment due to the 0.64klf^ load is then calculated from:
k ft k ft k ft E
klf ft klf ft ft k ft E
klf
− − −
−
ft P
M x
E (^) ,
B
5
1 rad
0.5 rad
a=
5 12’^ 12’^3