Calculation of Reactions, Internal Shears and Internal Moments Using
Influence Lines
Objectives of the materials covered: Given an influence line and the optimal load
positioning, students should be able to compute the resulting magnitude of reactions,
shears, and moments caused by any combination of loads placed on the structure.
Calculation of reactions due to concentrated loads: Once the influence line has been
developed, and the critical placement of dead and live loads has been determined, the
influence line can then be used to quickly determine the value of the resulting reaction,
shear, or moment.
For example, to determine the maximum possible value for the left reaction on a simple
beam, the influence line is first generated and the loads placed such that they cause the
greatest possible influence. This was accomplished in the previous section, and is
repeated here:
1.0
0.75
Figure 1
It is now possible to calculate the maximum left reaction that the truck wheels could
cause by simply multiplying the magnitudes of the wheel loadings times the
corresponding heights of the influence line at the locations of these loads:
Reaction at A = 20,000# x 1.0 + 10,000# x 0.75 = 27,500#
Note that by using a dimensionless unit load, the resulting influence line also has no
units, and the answer obtained by multiplying the height of the influence line times the
applied loads gives the proper units for the reaction.
The influence line could also be expressed by the following equation, where y is the
height of the influence line at a distance x from the left reaction:
y = 1.0 – x/L
Thus if you wanted to know the magnitude of the left reaction corresponding to the
application of a load at x = 0.75L, it could be found from
Left Reaction = Load*y = Load*(1.0 – x/L) = Load*(1.0 – 0.75L/L) = 0.25*Load
If you wanted the magnitude of the left reaction due to the application of a 5000N force
at x = 0.875L, it could be found as
Left Reaction = 5000N * (1.0 – x/L) = 5000N * (1.0 – 0.875L/L) = 625N
Note that this is why we used a dimensionless unit load, so that when a real load with
units of kips or Newtons is applied to the beam, the result would take on the units of the
applied loads.
Calculation of reactions due to uniform loads:
Results for uniform live loads can also be determined as follows. Assume that a
differentially small load dP were applied to the beam, at a location x.
1.0
y = 1.0 - x/L
dR
As demonstrated earlier, such a load would cause a differential left reaction of magnitude
dR = dP*(1-x/L) where dP = w*dx
Then the total reaction caused by a continuous uniform load across the beam could be
found by integrating this incremental load across the beam, wherever the uniform load is
applied:
R = dP(1-x/L) = wdx(1-x/L) = w(1-x/L)dx
R = w(height of the influence line)dx = w*(area under the influence line)
Thus the maximum possible reaction on a simply supported beam due to a uniformly
distributed load of any length can be determined using the influence line of Figure 1.
First, the influence line tells us where to place the uniform load for maximum positive
effect, namely across the entire length of the beam where the influence line is positive.
W
x
Next, the value of the reaction can be computed by multiplying the magnitude of the
uniform load w by the area under the influence line over which the uniform load is
applied:
Ra = Height of uniform load * Area under the influence line triangle
Ra = w kips/ft * (L ft * (1.0+0)/2) = wL/2 kips
Now admittedly this result is rather obvious, and most people would not create an
influence line for such a simple case. However, we will later show that determining
where to place these loads for maximum effect can be quite confusing unless influence
lines are available.
Calculation of maximum positive shear at a point in a beam: Previously we showed
the generation of an influence line for shear at the quarter point in a simply supported
overhanging beam, with critical placement of a set of concentrated and uniform loads,
which will be repeated here. Note that for maximum positive shear, the uniform live load
is placed only in the ranges where the influence line is positive.
L/4
w
w
0.75
0.25
0.25
0.50
Assuming a uniform load of 2,000 pounds/foot, and L = 10 feet, we can calculate the
maximum positive shear at the quarter point by:
V ¼ positive =
Concentrated Load*height + Concentrated Load*height
+ Uniform Load*Area under diagram
+ Uniform Load*Area under diagram
V ¼ positive =
20k * 0.75 + 10k * (2/3)*0.75
+ 2k/ft * 0.25 * (10ft/4)/2 + 2k/ft * 0.75 * (3*10ft/4)/2
26.25 kips
=
Calculation of maximum negative shear at a point in a beam: Previously we showed
the generation of an influence line for shear at the quarter point in a simply supported
overhanging beam, with critical placement of a set of concentrated and uniform loads,
which will be repeated here:
0.75
0.25
L/4
w
w
0.25
0.50
Assuming a uniform load of 2,000 pounds/foot, and L = 10 feet, we can calculate the
maximum negative shear at the quarter point by:
V ¼ negative =
Concentrated Load*height + Concentrated Load*height
+ Uniform Load*Area under diagram
+ Uniform Load*Area under diagram
V ¼ negative =
20k *(- 0.50) + 10k * (1/2)*(-0.50)
+ 2k/ft *(-0.25) * (10ft/4)/2 + 2k/ft * 0.50 * (10ft/2)/2
15.625 kips
=
Calculation of maximum positive moment at a point in a beam: Previously we
showed the generation of an influence line for moment at the quarter point in a simply
supported overhanging beam, with critical placement of a set of concentrated and
uniform loads, which will be repeated here. Note again that loads are placed only in
regions where the influence line is positive, the large wheel load is placed over the largest
positive value, and the smaller wheel load is placed to the right, or left, of the large wheel
load, depending on which side results in the highest value.
L/4
w
Assuming a uniform live load of 2,000 pounds/foot, and L = 10 feet, we can calculate the
maximum positive moment at the quarter point by:
M ¼ positive =
=
=
Concentrated Load*height + Concentrated Load*height
+ Uniform Load*Area under influence line
+ Uniform Load*Area under influence line
20kip * 0.1875*10ft + 10kip * 0.1875*10ft*2/3
+ 2kip/ft * (0.1875*10ft*(10ft/2))
406.25 kip ft
Calculation of maximum negative moment at a point in a beam: Previously we
showed the generation of an influence line for moment at the quarter point in a simply
supported overhanging beam, with critical placement of a set of concentrated and
uniform loads, which will be repeated here:
w
w
Assuming a uniform load of 2,000 pounds/foot, and L = 10 feet, we can calculate the
maximum negative moment at the quarter point by:
M ¼ negative =
Concentrated Load*height + Concentrated Load*height
+ Uniform Load*Area under influence line
+ Uniform Load*Area under influence line
=
=
20kip * -0.375*10ft + 10kip * (-0.375*10ft)/2
+ 2kip/ft * (-0.375*10ft * (10ft/2)/2)
+ 2kip/ft * (-0.03125*10ft)*(10ft/8)/2
131.64