INTERNAL FORCES
FORCES FROM VERTICAL LOADS
FORCES FROM HORIZONTAL LOADS
FORCES
TOTAL AXIAL
The internal forces in the columns are found using the reactions, the loads and the principles of equilibrium.
The simplest internal forces are the axial ones, which result from the vertical loads and reactions. They reach a
maximum at the base of the Tower.
FORCES FROM VERTICAL LOADS
The axial force in the columns is the inclined force, N, shown acting along the axis of one column. This force is
equal to the combination of a vertical and a horizontal component. The gravity loads and reactions are vertical
and create the vertical component; the horizontal component develops because the vertical forces are carried in
an inclined column, i.e., for the columns to be in equilibrium there must be a horizontal component of the axial
force.
These three forces create a ‘force polygon’ where, drawn head to tail, the scaled components are equal to the
axial force and three form a closed shape. The magnitudes of the horizontal components and the axial force can
be found from the vertical force and the angle of the column axis with trigonometry. The chart shown in the
geometry section gives the slope of the columns at the base as   18.4 . The simple trigonometric relationship
V
 cos 
N
can be rearranged to find N:
V
cos 
11,140 k

cos18.4
 11, 730 k
N
This is the axial force that is developed in both columns at their bases. The horizontal components at the base
can be found from the trigonometric formula:
H
 tan 
V
or
H  V tan 
 11,140 k  tan18.4 
 3700 k
These horizontal forces tend to push the Tower apart but are resisted by the connection all along the Tower and
the ground. This can be visualized if one imagines a ladder propped against a building. If there was no friction
between the ground and ladder base, or if the base was on wheels, the ladder would slide away from the wall.
In the same manner, the columns of the Tower would slide apart if there were no connections between them.
These connections, which have been idealized as continuous, experience tension force equal to the horizontal
forces they are resisting.
At the base, the ground will resist with horizontal reactions of 3700 kips acting inward.
The axial force will decrease with height as the vertical load and angle of inclination decrease. As an example,
the axial force at the second platform will be found. The entire top and middle parts are above this point,
therefore, the vertical load is:
Q  Q1  Q2
 1100 k  2200 k
 3300 k
The chart given in the Geometry section shows that at this height of 380 feet, the slope of the elements is 11.6 .
The trigonometric formula:
N
V 2
cos 
is used to find:
1650 k
cos11.6
 1685 k
N
The value at the base is almost seven times this value! The compression forces from the dead load will be
identical in both of the columns because they are each taking half of the vertical load. (The cross-sectional area
could be decreased at the higher point because the smaller compression force would create a small compressive
stress. This is indeed what Eiffel did, as shown in the earlier figure; this is neglected in the simplification.) The
horizontal component of the axial force at the second platform is:
V 
H     tan  
2
 1650 k  tan11.6 
 339 k
This force will be resisted by the connection made by the second platform itself.
FORCES FROM HORIZONTAL LOADS
The internal forces created by the overall bending moment are investigated next. The bending moment from the
wind load can be found at any point if the entire structure is treated as a single unit rather than as two columns.
The bending moment is found around the tower axis, whereas the axial forces were found along the columns’
axes. The overall bending moment from the horizontal wind load will produce tension (T) in one column and
compression (C) in the other. Exaggerated action under wind load is shown:
The values of the bending moment all along the height of the Tower must be known to find the resulting tension
and compression forces in the columns. A bending moment diagram illustrates the continuous values. The
horizontal force creating the bending moment is the wind, simplified as a point load acting at the mid-point of
the area under consideration.
At a point a distance ‘x’ from the bottom as shown in the diagram below, the bending moment is found using
 l  x  as the distance.
p  l  x  as the magnitude of the point load and
2
The equation for the moment at any point along the Tower uses these two values:
M  p l  x  

p l  x 
l  x 
2
2
2
For the Eiffel Tower, l = 984 feet and p = 2.6 k/ft; the reaction calculation was made where x = 0 to find M =
1,260,000 ft-k.
The reaction formula used earlier, M  P  l 2 , is a simplified version of this formula where x = 0 and where
P  p l .
The graph of the equation M  p  l  x  2 with changing values of x is a diagram of the continuous bending
moment along the Tower. The shape is also a parabola, only the magnitude is different.
2
The shape of the Eiffel Tower roughly imitates the shape of this diagram; the idealization imitates it exactly, but
the actual tower is more sharply curved as shown earlier.
At any point along the height of the Tower, the overall bending moment, M, and the width of the structure, d,
are used to determine the value of both the tension (T) and compression (C) forces (which are equal) from the
formula:
T C 
M
d
This formula was justified in the Monument analysis where it was shown that the tension or compression forces
result from the bending moment acting over the depth (or width here) of the structure.
This calculation uses the bending moments found around the tower axis to find axial forces that act along the
column axes.
At the base of the Tower the moment force is M = 1,260,000 ft-k (the reaction force), and the width, d, is 328
feet. Therefore,
M
d
1, 260, 000 ft  k

328 ft
 3,850 k
T C 
where T = +3,850 k and C = -3,850 k because compression forces are designated as negative. Both of these
forces are vertical reactions to the wind.
The axial forces in the inclined elements resulting from the wind load are found from the vertical reactions and
the horizontal reactions to the wind found earlier.
On the leeward side, where the wind creates a vertical compression reaction, Vc and H contribute part of their
forces along the axis of the column. The compression force, Vc, contributes a component of  Vc cos   to the
axial force, Nc and H contributes   H sin   to Nc. Nc can be found from Vc, H and the angle at the base of the
column,   18.4 . Negative signs are used because the axial force is compressive.
N c   Vc cos     H sin  
  3,850 k  cos18.4   1, 280 k  sin18.4 
 4, 050 k
The windward side will have a positive value of N because Vt and H create axial tension:
N t  Vt cos   H sin 
  3,850 k  cos18.4   1, 280 k  sin18.4 
 4, 050 k
The axial forces produced by the horizontal load of the wind are 4, 050 k .
To further illustrate this idea, the axial forces produced by the wind load at the second platform will be
investigated. The horizontal and bending moment reactions from the wind on the top and middle parts of the
Tower are found with a free-body diagram cut at the second platform.
H m  p   l1  l2 
  2.6 k ft    604 ft 
 1,570 k
where P  p   l1  l2 
Mm
l  l 
l  l 
 P 1 2  p 1 2
2
2
2
 604 ft 
 1,570 k   

 2 
 474, 000 ft  k
The chart shows that d = 123 ft at the second platform, so the tension and compression are:
Vt  Vc 
 M 474, 000 ft  k

 3,850 k
d
123 ft
Vt  3,850 k
Vc  3,850 k
Hm is found as 1,570 k  2  785 k in each column. With these values Nc and Nt can be found using the
formulas introduced earlier:
N c   Vc cos     H sin  
  3,850 k  cos11.6    785 k  sin11.6 
 3,930 k
N t  Vt cos     H sin  
  3,850 k  cos11.6    785 k  sin11.6 
 3,930 k
The axial forces and the vertical components are nearly equal. This illustrates that at a small angle these
calculations hardly need be performed, especially considering the simplicity of this analysis. Generally at an
angle less than 15 , one can consider the axial force and vertical force equal. The vertical forces, Vt and Vc, at
this height are roughly equal to the vertical tension and compression found at the base; they are all roughly
4,000 kips. This shows that if the shape of a structure imitates the shape of its bending moment diagram, the
vertical forces created by the bending moment will be constant all along its height. The width of the base is
equally capable of resisting the bending moment at the base as the width at the second platform is capable of
resisting the bending moment at that point.
TOTAL AXIAL FORCES
Now that the axial forces from both vertical and horizontal loads have been calculated, the total axial forces in
the column supports are found by simply adding the axial force found from the two loads. Compression forces
are negative and tension forces are positive. This is shown in the first diagram, below. The second diagram
shows the corresponding forces at the second platform.
At this point (the second platform) the windward column experiences tension forces rather than the compression
found near the base. This creates no structural problems because wrought iron resists tension just as well as
compression. Next, the total axial forces are used to find the internal stress. First, however, a quick
investigation of the shear force in the columns further illustrates the efficient shape of the Tower. Since the
shape of the Tower nearly imitates the shape of the moment diagram, and since the vertical compression force
has been found to be constant all along the Tower’s height, there should, theoretically, be no shear force in the
column from the horizontal wind load. This is investigated using the vertical compression and horizontal forces
found earlier. (See the diagram at the top of the next page.) Where Sc  C sin  and Sh  H cos  , the total
shear force is found by the formula:
Stot  Sh  Sc
 H cos   C sin 
 1280 k  cos18.4    3850 k  sin18.4 
 1215 k  1215 k
0
This result verifies the concept that, from the wind load, the parabolic shape of the Tower will allow only
uniform vertical forces, and axial forces that decrease with the angle of the columns.
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