Calculation of Reactions, Internal Shears
and 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
Calculation of reactions due to
concentrated loads
• 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.
Calculation of reactions due to
concentrated loads
• 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 of Figure 1, at a location x.
1.0
dR
y = 1.0 - x/L
Calculation of reactions due to
uniform loads
• As demonstrated earlier, such a load would
cause a differential left reaction of magnitude
dReaction = dP*(1-x/L)
• Thus 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) = w(1-x/L)dx = w(1x/L)dx
R = w(height of the influence line)dx
= w*(area under the influence line)
Calculation of reactions due to
uniform loads
• 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
Calculation of reactions due to
uniform loads
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
Calculation of maximum positive
shear at a point in a beam
• 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
L/4
w
w
0.75
0.25
0.25
0.50
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
Calculation of maximum negative
shear at a point in a beam
• 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
0.75
0.25
L/4
w
w
0.25
0.50
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
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
Calculation of maximum positive
moment at a point in a beam
• 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
L/4
w
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
Calculation of maximum negative
moment at a point in a beam
• 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
w
w