Influence Line:
Assume,
Here AB is a simply-support beam. 1 kip load is
applied downward on the point A.We will get an upward reaction value (noted as
RA) at Point A because of applied 1 kip downward load.
Now when that
1 kip downward load is moved from Point A to Point B then this load (1 kip) will
act all points on the beam AB & we will get a different reaction value of RA
each time while 1 kip load will move on the different points of beam AB (Point
A to Point B). By adding these all reaction values of RA, we will
get a straight line. This straight line will known to us as Influence Line for
RA that is IL for RA.
Now by the same way, we apply that 1 kip vertically
downward load at Point B at first. We will get an upward reaction value (noted
as RB) at point B. Now move this load from Point B to Point A. We
will get a different reaction value of RB each time while 1 kip load
will move on the different points of beam AB (Point B to Point A). By adding
these all reaction values of RB, we will get a straight line. This straight
line will known to us as Influence Line for RB that is IL for RB.
Example-01:
Draw Influence Line for RA and RB.
Solution:
IL of RA:
Now at first put 1 kip load
downward at point A and then move this 1 kip load from Point A to Point B, we
get,
When 1 kip load at first on the Point A then
reaction of A that is RA=1, RB=0, because total applied load is
resisted by RA.
Now when 1 kip is started to move from Point A to
Point B and at last will rest on Point B then RA=0 and RB=1.
Now if we want to draw the influence line for RA,
then we get the following structure,
When 1 kip load rest upon Point A then RA=1
and when 1 kip load rest upon Point B then RA=0.
But as the
definition of Influence Line, we need to add all reaction values which we will
get from all points upon AB, during moving of 1 kip downward load from Point A
to Point B. But here we have added reaction values only when 1 kip load rests
at starting point and Ending Point.
Now we will calculate different reaction values of RA
from different points on beam AB.
At first consider the 1 kip load rests at Point C
which is located at a distance 2 feet from point A.
Then,
∑MA= 1*2 – RB*10 = 0
So, RB = 0.2 kip,
∑FY= RA – 1 + 0.2 = 0
So, RA= 0.8 kip.
Now consider the 1 kip load rests at Point D which
is located at a distance 5 feet from point A.
Then,
∑MA= 1*5 – RB*10 = 0
So, RB = 0.5 kip,
∑FY= RA – 1 + 0.5 = 0
So, RA= 0.5 kip.
Though 1 kip load is applied at middle point of AB
then RA will be half of applied load and RB will be another
half of applied load that means the summation of RA & RB
will be the equal value of the applied load.
Now consider the 1 kip load rests at Point E which
is located at a distance 8 feet from point A.
Then,
∑MA= 1*8 – RB*10 = 0
So, RB = 0.8 kip,
∑FY= RA – 1 + 0.5 = 0
So, RA= 0.2 kip.
Now put all these reaction values of RA and
plot them on a Plain Graph Paper, which values we have already calculated by applying 1 kip
load at Point C, D & E.
After plotting them on a Plain Graph Paper we get IL
of RA which is similar with the previous, by this experiment we can
take decision that,
To find IL of any Simply Supported Beam we need to
find only two reaction values at Starting Point & at Ending Point, we need
not find any more reaction values at any other point on the BEAM.
IL for RB :
When 1 kip load at first on the Point B then
reaction of B that is RB=1, RA=0, because total applied
load is resisted by RB.
Now when 1 kip is started to move from Point B to
Point A and at last will rest on Point A then RB=0 and RA=1.
So IL of RB will be as follows,
We need not calculate different reaction values of RB
from different points on beam AB, because IL will be same (it is already proved
for RA)