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Rigid Frame & Truss Bridges
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1
Rigid Frame & Truss Bridges
Background
Rigid Frame Bridges
A rigid frame bridge is one in which the piers and the girder are one solid
structure. The cross sections of the beams in a rigid frame bridge are usually I
shaped or box shaped. The styles us
ed almost exclusively are the pi

shaped
frame, the batter post frame, and the V shaped frame. The batter post rigid
frame bridge is particularly well suited for river and valley crossings because
piers tilted at an angle can straddle the crossing more eff
ectively without
requiring the construction of foundations in the middle of the river or piers in deep
parts of a valley.
Figure 1. Batter post rigid frame bridge.
V

shaped frames make effective use of foundations. Each V

shaped pier
provides two
supports to the girder, reducing the number of foundations and
creating a less cluttered profile.
Figure 2. V

shaped frame bridge.
Pi shaped rigid frame structures are used frequently as the piers and supports for
inner city highways. The frame su
pports the raised highway and at the same time
allows traffic to run directly under the bridge.
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Figure 3. Pi shaped rigid frame bridge.
Some Examples…
Engineering
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Truss Bridges
The truss is a simple skeletal structure. In design theory, the individual members
of a simple truss are only subje
ct to tension and compression forces and not
bending forces. Thus, for the most part, all beams in a truss bridge are straight.
Trusses are comprised of many small beams that together can support a large
amount of weight and span great distances. In most
cases the design,
fabrication, and erection of trusses is relatively simple. However, once
assembled trusses take up a greater amount of space and, in more complex
structures, can serve as a distraction to drivers.
There are both simple and continuous tru
sses. The small size of the individual
parts of a truss make it the ideal bridge for places where large parts or sections
cannot be shipped or where large cranes and heavy equipment cannot be used
during erection. Because the truss is a hollow skeletal str
ucture, the roadway
may pass over (Figure 4) or even through (Figure 5) the structure allowing for
clearance below the bridge often not possible with other bridge types.
Figure 4. Roadway passing over the truss structure.
Figure 5. Roadw
ay passing through the truss structure.
Engineering
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GatorTRAX
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Rigid Frame & Truss Bridges
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Trusses are also classified by the basic design used. The most representative
trusses are the Warren truss, the Pratt truss, and the Howe truss. The Warren
truss is perhaps the most common truss for both simple and
continuous trusses.
For smaller spans, no vertical members are used lending the structure a simple
look (Figure 5). For longer spans vertical members are added providing extra
strength (Figure 4). Warren trusses are typically used in spans of between 50

10
0m.
The Pratt truss (Figure 6) is identified by its diagonal members which, except for
the very end ones, all slant down and in toward the center of the span. Except for
those diagonal members near the center, all the diagonal members are subject to
tensio
n forces only while the shorter vertical members handle the compressive
forces. This allows for thinner diagonal members resulting in a more economic
design.
Figure 6. Pratt truss bridge.
The Howe truss (Figure 7) is the opposite of the Pratt tr
uss. The diagonal
members face in the opposite direction and handle compressive forces. This
makes it very uneconomic design for steel bridges and its use is rarely seen.
Figure 7. How
e
truss bridge.
Engineering
University of Florida
GatorTRAX
College of Engineering
Rigid Frame & Truss Bridges
Copyright 2004 by Gator
Trax
. Created, Designed, and Edited for Gator
Trax
programs
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Some example…
Engineering
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Mathematical Concepts
Quick Review
–
Simple Algebra
A x B = C x D
A= (C x D) / B
Sample Problem Set 1
. Solving for the given variable.
A x 3 = 2 x 6
A=_________
A x 4 = 12 x 5
A=_________
8 x B = 3 x 16
B=_________
B x 9 = 3 x 6
B=_________
11 x 4 = C x 2
C=_________
45 x 9 = 5 x C
C=________
_
100 x 20 = 50 x D
D=_________
55 x 23 = D x 11
D=_________
Sample Problem Set 2
. Solve for the given variable (include units)
A x 3 in. = 2 lbs x 6in.
A=_________
A x 4 lbs = 12in. x 5lbs
A=_________
8N x B = 3m x 16N
B=_________
B x 9mm = 3km
x 6mm
B=_________
11m x 4N = C x 2N
C=_________
45tons x 9yd = 5tons x C
C=_________
100in x 20lbs = 50in x D
D=_________
55cm x 23N = D x 11N
D=_________
Units of Force
F = m x a
F= Force (Newtons, Pound Force)
m = mass (kilogram, slug)
a = acce
leration (m/s², ft/s², in/s²)
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Pythagorean Theorem
Force Balancing
The same algebraic expressions can be used to solve for an unknown variable.
Sample Problem Set 3
.
A 25 Newton block is 5 meters away from center of beam. A 100 Newton man
wants to lift the block and balance the beam. How far from the center must the
man stand to lift the block?
ANSWER:________________________
A 75 lb student wants to balance a
beam. A 30 lb block is 4 ft from the center.
Where should the student stand to balance the beam?
ANSWER:________________________
a
2
+ b
2
= c
2
B
A
D
C
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Point Loads and Moments
In a bridge, the sum of all the forces acting on all the members will equal zero.
Therefore F
a
+ F
b
+ F
c
= 0
To Find Pin Forces (A,B) use the following:
B = (C x D)/ (D+E)
A = (C x E)/(D+E)
D
E
A
C
B
A
300 lbs
B
70 ft
30 ft
Engineering
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Sample Problem Set 4
.
From figure above, find the forces at A and B.
Force A = _
________________
Force B = _________________
Statically Determinate Trusses
The Pratt and Howe Trusses are statically determinate. A truss structure is
statically determinate if the member forces are determined entirely by the
conditions of static
equilibrium at the joints. This means that the sum of all the
forces at a joint is equal to zero.
Figure 8. Bridge Joint.
Method of Joints
To balance th
e forces in a truss structure using the method of
joints, we begin with a joint at a reaction force and isolate that joint. We then
sum the horizontal and the vertical forces on it. The sum of the horizontal forces
should equal zero. The sum of the vert
ical forces should also equal zero. We
can now solve for any unknown forces by proceeding joint by joint through the
bridge only picking up two new unknowns at each joint until we reach the other
reaction.
Let us solve the following example using the m
ethod of joints.
Engineering
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GatorTRAX
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Rigid Frame & Truss Bridges
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10
Let us assume that
W
, the weight of the truss, is 15N. The length of side
a
is 4m
and the length of side
b
is 3m. Angle ABC is equal to the angle formed by line
BC and the line of the force F
2
.
First, let us solve for the length of
side c.
We use the Pythagorean theorem to obtain the following equation:
a
2
+ b
2
= c
2
4
2
+ 3
2
= c
2
16 + 9 = c
2
25 = c
2
c = 5
Before we begin examining the joints, let us go over how we indicate direction of
forces. A force acts alon
g a line or member. The member will either be in
tension or compression. The following illustrate the direction of the force for a
member in tension and a member in compression.
We can think of a member in tension as something that i
s being pulled apart.
The force is equal on both ends, but in opposite directions.
Engineering
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We can think of a member in compression as something that is being pushed, or
compressed together from each end. The force is equal on both ends, but,
again, in opposite directions.
When examining the forces, just as when we graph on the coordinate plane, we
have to keep in mind their positive and negative directions. Up and to the right is
positive. Down and to the left is negative. We have to indi
cate their signs when
placing them into algebraic equations so that we can solve them correctly.
Now, let us examine Joint A.
We will begin by summing the forces in the vertical direction:
F
V
= 0 = 12

F
AB
(4/5)
12 = F
AB
(4/5)
12(5/4) = F
AB
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F
AB
= 15
Now we will sum the forces in the horizontal direction:
F
H
= 0 =

F
AC

F
AB
(3/5)
F
AB
= 15
(from vertical Joint A analysis)
F
AC
=

15(3/5)
F
AC
=

9
Next we will move on to Joint C.
Again, we will begin by summing the force
s in the horizontal direction:
F
H
= 0 = F
AC
–
F
3
F
AC
=

9
(from joint A analysis)

9 =
–
F
3
F
3
= 9N
Next, we sum the forces in the vertical direction:
F
V
= 0 = F
BC
–
W
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W = 15
(we were given this)
0 = F
BC
–
15
F
BC
= 15
Finally, we move to Joint B.
From previous analyses, we know that F
AB
= 15 and that F
BC
= 15. We will
redraw the diagram to include what we know.
Engineering
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GatorTRAX
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Now, let us sum the forces in the vertical direction:
F
V
= 0 =

15(4/5)
–
15
–
F
2
(4/5)
F
2
(4/5) =
15(4/5)
–
15
F
2
=

3.75 N
(a negative sign in front of the force means that we drew it in
the wrong direction)
F
2
= 3.75 N
Finally, we will sum the forces in the horizontal direction keeping F
2
drawn in its
original directio
n:
F
H
= 0 = 15(3/5) + F
2
(3/5)
–
F
1
F
1
= 15(3/5) + F
2
(3/5)
F
2
=

3.75 N
(from vertical Joint B analysis)
F
1
= 15(3/5)
–
3.75(3/5)
F
1
= 9
–
2.25
F
1
= 6.75N
Engineering
University of Florida
GatorTRAX
College of Engineering
Rigid Frame & Truss Bridges
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Trax
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15
We have now used the method of joints to solve for all the unknown forces in this
truss stru
cture.
Sample Problem Set 5
. Try solving for the unknown forces in the following
diagram using the method of joints.
Let us assume that
W
, the weight of the truss, is 20N. The length of side
a
is 8m
and the length of side
b
is 6m. Angle ABC is equa
l to the angle formed by line
BC and the line of the force F
2
. The 10N reaction force is along the same line as
line AB.
Method of Sections
Using the method of sections, we consider the forces acting
on some larger part of the bridge. In the following
diagram, we would analyze
the forces acting on triangle ABC, the entire shaded region, rather than the
forces acting on each individual joint, as before.
Engineering
University of Florida
GatorTRAX
College of Engineering
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Trax
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16
Let us assume again that
W
, the weight of the truss, is 15N. The length of side
a
is 4m
and the l
ength of side
b
is 3m. The length of side
c
is as we determined earlier, 5m.
Angle ABC is equal to the angle formed by line BC and the line of the force F
2
.
We begin by summing the moments about B.
M
B
= 0 =

F
3
(4) + 12(3)
F
3
(4) = 12(3)
F
3
= 9N
Next, we sum the forces in the vertical direction:
F
V
= 0 = 12
–
15
–
F
2
(4/5)
F
2
(4/5) = 12

15
F
2
=

3(5/4)
F
2
=

3.75
(again, a negative sign in front of the force means that we
drew it in the wrong
direction)
F
2
= 3.75 N
Engineering
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College of Engineering
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Finally, let us sum the forces in the horizontal direction:
F
H
= 0 = F
3
–
F
1
+ F
2
(3/5)
F
3
= 9N
(from moment summation)
F
2
=

3.75
(from vertical force summation)
F
1
= F
3
+ F
2
(3/5)
F
1
= 9
–
3.75(3/5)
F
1
= 6.75N
As, you can see, we come up with the same results using the method of joints
and the method of sections.
Engineering
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Sample Problem Set 6
. Try solving for the unknown forces in the last sample
problem using the method of sections. Check to see that you
obtain the same
results as before.
Again, we will assume that
W
, the weight of the truss, is 20N. The length of side
a
is 8m and the length of side
b
is 6m. Angle ABC is equal to the angle formed
by line BC and the line of the force F
2
. The 10N rea
ction force is along the same
line as line AB.
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