A cantilever beam is anchored at one end only- to a wall, a tower or ???
while the other
end is free and unsupported.
Most cantilever bridges are "balanced cantilevers" with a central tower from which
arm" and a "cantilever arm" extend out in opposing directions.
Here's the Coos Bay Bridge under construction in the 1930s
They built outward from the middle of the tower, keeping the 2
sides equal and balanced. They
continued to add more members
on the left, making things UN-balanced, so a pier was added as
Eventually the 2 arms got closer to each other so they were
"bridged" by a steel beam
connecting the bottom chord
and the top chord.
The bridge was later named in honor of the designer-engineer
The Conde McCullough
Many cantilever bridges will have a suspended section between a pair of balanced cantilevers.
way the main span( the distance between 2 piers), can grow without increasing
the size and expense of longer cantilever
The Firth of Forth Bridge (built 1890) has 3 immense balanced cantilevers
with much smaller suspended
sections (2) hung between
In most truss bridges the top chord is in Compression while the bottom chord
is in Tension. see the Truss
Bridge page-coming soon.
These forces are reversed in a cantilever bridge.
We'll use just 1 of the balanced cantilever towers for this exercise.
The load (and the deck it sits on) is held up from below by the angled strut . This combined
pushes down on the strut putting it in Compression.
The strut is part of the bottom chord.
At the same time
the strings connecting the deck with the top of the tower
are supporting the load from above, so they're being pulled
tight and are in Tension.
In other words the top chord is in Tension and the bottom chord is in Compression-
opposite of most trusses.
Baker and Fowler (the engineers responsible for the Firth of Forth bridge) wanted to
demonstrate to the public how a cantilever bridge worked so they popularized this photo-
The 2 guys in chairs represent the 2 towers. Their arms ( top chords)are being
pulled down by
the weight of the suspended section (Mr. Watanbe) on one side
and the anchors (the pile of bricks) on the other side,
while the angled sticks -the struts,
are being pushed down by the same forces.
That means these bottom chords are
What would happen if the rope connected to the 2 piles of bricks was cut?
weight would cause both men and their chairs to tip to the middle.
We replaced Mr. Watanabe with 6 bricks. Notice there are 3 bricks in each of the anchors
Each kid ( tower) is balanced with that 3 brick load on 1 side and half the suspended
section (half of 6) on the other side.
In this cantilever model the left tower is anchored to the table-it's tied
to a steel weight.
The other tower is anchored by Jessie.
We used the model below to find the forces at work.
The load (3 bricks) and the weight of the deck and the strut are trying to tip the
in a Counter Clockwise direction. With no counter-weight present
the only thing preventing this tip-over are the strings
attached to the anchor end
that are pulling the tower down in a Clockwise direction.
How much counter-weight (ct.wt.) would we need to keep the tower balanced
and not to have use the strings?
2 bricks (148 oz.) are 32" away from the pivot so
CCW moment = F x D
or 148oz. x 32" = 4736 inch ounces
and the opposite moment is
CW moment = F x D 4 bricks
(296 oz.) x 18" = 5328 inch ounces
Not even close but we didn't take into account the weight of
That would get us much closer but the purpose of this exercise is to
compute the force required
to keep the bridge level using a rope
(the red line) tied to the top of the tower.
For now we'll ignore the weight of the tower, deck etc.
so we can concentrate on the math.
3 brick load (222 oz.) is 32" from the pivot so
M = F x D or 222 x 32" = 7104
inch ounces.That's the CCW moment.
To balance that we have to pull on the red line to create a CW moment
7104- then the sum of the moments around the pivot
point will be equal to zero.
Remember that red line is the LINE OF FORCE
and the shortest distance from the line of force to
point is the perpendicular distance....28" So
CW moment = F x D or
7104 = F x 28" then
F = 7104/28 = 253.714 oz. or 15.87 pounds
about 2 lbs. more than the weight of the
How much force would we need if we lowered the angle to 60 deg.?
In this new rt. triangle the shortest distance from the line of force
to the pivot is the
line opposite the 60 deg angle. so we'll use sine.
sin 60 = distance/ 28" and sin
60 = .866
and we get distance =
.866 x 28 = 24.248"
Now we got back to M = F
7104 = F x 24.248 and finally we get
F = 292.972 oz. or 18.31 lbs.
As the angle gets smaller the distance also decreases
but that means the force
How much force would we need if the angle was 20deg.?