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

an "anchor
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

support.

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
Bridge

Many cantilever bridges will have a suspended section between a pair of balanced cantilevers.

In this
way the main span( the distance between 2 piers), can grow without increasing

the size and expense of longer cantilever
arms.

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

weight
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-

the
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
in Compression.

What would happen if the rope connected to the 2 piles of bricks was cut?

Mr. Watanabe's
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

resting on
the ground.

Each kid ( tower) is balanced with that 3 brick load on 1 side and half the suspended

load
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

tower over
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

the tower,deck,
strut etc.

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.

The
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

of
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
the pivot

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
3 bricks.

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

so
.866 =distance/28"

and we get distance =
.866 x 28 = 24.248"

Now we got back to M = F
x D

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
required INCREASES.

How much force would we need if the angle was 20deg.?