Download George Washington Bridge: Reactions

REACTIONS
INTRODUCTORY PRINCIPLES
MAIN SPAN REACTIONS
SIDE SPAN REACTIONS
The loads on the bridge will be transferred through the cables to the tower supports and finally to the anchor
supports. This analysis will not investigate the reactions at the bases of the towers because the cables are the
topic here. Therefore, we will study the reactions at the ends of the cables (in the anchors) and at the tops of the
tower. Because of the idealization of symmetry, the reactions only need to be found at one tower and one
anchor.
Because of their flexibility, cables can change their shape so that any applied load will only create tensile forces
in them. Thus the cables develop the graceful curvature that is so characteristic of suspension bridges. The
curvature inclines the cables at the point where they are supported. The tensile force that is acting along the
axis of a cable must be resisted by both a horizontal and a vertical reaction.
This is the same at the anchor where the cables come in at an angle.
The reaction forces can be found by taking advantage of the cables’ inability to resist bending moment at any
point, i.e., the bending moment will be zero all along the cable. Since the bending moment is zero, one can find
the force that would create moments around any point of the cable and set the sum of them equal to zero. For
example, the bending moment at the hinge shown is zero, and, therefore, if the member is to be in equilibrium
(immobile) the bending moments acting on it must balance. If the bending moment of each force is found, one
will see that they cancel because they are trying to turn the member in opposite directions.
3 kips  4 ft  12 ft  k
6 kips  2 ft  12 ft  k
The sum of the moments about a hinge is zero ( M  0 ). This can be seen in the equation:
M  0
 3 kips  4 ft  6 kips  2ft
 12 ft  k  12 ft  k
0
MAIN SPAN REACTIONS
This same principle can be applied to the cables. Concentrating first on the main span, the reactions in the
tower supports can be found. If the endpoint of the main span (A) is picked for convenience, the forces acting
on the cables will balance around it because M  0 .
The horizontal reaction at point B creates no bending moment around point A. This is because it is acting along
the imaginary line that would connect points A and B. The vertical reaction, however, would create a bending
moment around point A. The magnitude of this bending moment is equal to the distance multiplied by the force:
M  VB1  l
 VB1  3500 ft
The other force created by a bending moment around point A is the load on the cables – both dead and live
loads. This uniform load will be idealized by a point load which will act at the midpoint of the span. The
magnitude of this point load is:
q  l   qd ,m  ql   l
 164,500 kips
This force will act a distance of
l
from point A.
2
l
 1750 ft
2
From M A  0 the magnitude of the vertical reaction can be found by:
M A  0  VB1  l 
-
q l  
l
2
The total load and vertical reaction act in opposite directions, so one must be considered negative. Substituting
in the known values of  q  l  and l and rearranging the equation, one finds:
l
2
VB1  82, 250 kips
VB1  l   q  l  
The vertical reaction at the left-hand tower is found through a vertical force balance, i.e., the forces acting
upward (reactions) must equal the forces acting downwards (the loads).
VA1  VB1  q  l
VA1  164,500 kips  82, 250 kips
 82, 250 kips
This result shows that the vertical reactions at the ends of a symmetrical span are equal.
Now that the vertical reactions are known, the horizontal reactions from the main span can be found. The same
principle of a moment balance can be applied to find the horizontal reactions, this time finding the moments
about the centerpoint of the cable, M C  0 . Working with one side of the cables, one finds the following set
up of forces.
 q  l  , is found as 82,250 k and acts at a distance
l
 875 ft from point C. The bending moments
4
2
around the center can be set equal to zero and then the horizontal reactions can be found:
The load,
l
M C  0  VB1  
2
 H B1  d 
-
-
 q  l  l 

 
 2  4 
q l
2
Rearranging, we can solve for H B1 :
where: VB1 
q  l  l   q  l  l 
   
 
 2  2   2  4 
 H B1  d   
q l2
8d
 220, 000 k.
H B1 
The values of the horizontal and vertical reactions can be used to find the slope of the cables where they meet
the supporting tower. (This angle will be necessary in subsequent calculations of internal force).
tan  
VB1
82, 250 k.

 0.374
H B1 220, 000 k.
  tan 1  0.374   20.5
SIDE SPAN REACTIONS
The reactions in the side spans will be investigated next. The value of the load on the spans is found from qd , s
and ql . The horizontal reaction at the tower is known because the analysis has assumed that no horizontal force
will be exerted on the tower; therefore, the horizontal reaction to the loads on the side span is equal and
opposite to the horizontal reaction already found.
H B1  H B 2  220,000 k.
To find the remaining reactions in the side spans, that portion of the bridge will be isolated and the horizontal
reaction retained as a force. To create a horizontal equilibrium, i.e., have all the horizontal forces balance, the
horizontal reaction in the anchor must have this same value H = 220,000 k.
The vertical reaction at the tower can be found by applying the moment balance principle around the anchor
point, D. Forces that will create moment are both the vertical and horizontal reactions at the top of the tower
and the loads suspended from the cables. The vertical reaction is the unknown force that will be found from
this balance. The load from qd and ql is found over the side span length of 650 feet:
qs  ls   qd ,s  ql   ls
 31, 200 k.
This force acts a distance of
ls
 325 feet from point D.
2
M D  0 
 HB2  d 
 qs  ls  
ls 

2
+
-
VB 2  ls 
Therefore:
 qs  ls 2 

 2 
 H B2  d   
VB 2 
l
 143, 200 k.
This value is added to the value of VB1 which is the vertical reaction developed by the load on the main span.
Together the values show the total vertical reaction at the top of the tower.
Vtotal  VB1  VB 2
 225, 450 k.
To find the vertical reaction at the anchor, the principle of vertical equilibrium is used – all the vertical forces
on the side spans must balance.
VD  VB 2  qs  ls
 112, 000 k.
This reaction is unlike any vertical reaction encountered thus far in the analysis. Usually a vertical reaction is
reacting against compressive forces – forces that would push the structure into the ground. At the anchor,
however, the forces are tensile and are trying to pull the cable out of the anchor block of concrete.
The side spans meet the tower at an angle of 33.1 and the anchor at an angle of 27.0 as found from the
following calculations:
143, 200
   33.1
220, 000
112, 000
tan  
   27.0
220, 000
tan  
This analysis has idealized the bridge as symmetrical, so the tower and anchor reactions, as well as the angles
on the left-hand side will be equal and opposite to those found on the right-hand side. Now all the reactions,
both horizontal and vertical, have been found in all four supports of the bridge.
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