PortlandJib

TEACHERS – are you looking for a curriculum that will get your students interested in engineering? Will the bridge deck crashing down do the trick?

I use this model of a cable stayed bridge to cover force vectors, torque, center of gravity, applied trig and overturning moment.

And as an added bonus, the kids can take the bridge apart and build:

a boom crane—a tower crane–a suspension bridge—an elevator—a cantilever bridge or anything they can dream up.

Changing the angle of the cable will change the Tension. How much tension will the cable support before it breaks?

There are lots of variables here.

size and placement of the load

angle of the supporting cable

thickness of the wood dowel.

In a cable-stayed bridge -with one or 2 towers- the cable stays that support then deck, extend from the tower directly to the deck

I’ve designed a curriculum that teaches the kids about force vectors, torque, center of gravity, applied trig and overturning moment. I use a model of a cable stayed bridge with just a single tower.

Now to the reason behind this new curriculum- how to compute the amount of Tension in the cable and figuring out when it will break and the deck will fall.

The 2×4 sits on the ground- or rather 1 end rests on the ground and the other end is held up by the string.

The 2×4 weighs 6#. One end rests on the ground (I use a small triangle to denote this) – so the ground is supporting half the 6# load.

This means that Ben only needs 3# of force to lift the other end.

Now when Ben applies a force to lift the 2×4 at an angle of 30 deg. he needs 6# of force. We have to find a way to compute the force necessary at different angles.

Easy first we turn the forces into a right triangle.

The vertical leg is 3#- that represents the original force when the lifting was vertical -90 deg.

The hypotenuse is the 6# force needed at 30 deg.

The complement to the 30 deg angle on the 2×4 is 60 deg. That makes the upper angle 30 deg.

The ratio of the opposite side ( the side opposite the angle we’re talking about- the 30 deg. angle)—3#

over the hypotenuse—-6# is 3/6 or 1/2 or .5

For those of you who don’t already know basic trig—this is SINE—– opposite divided by the hyp.

We write it like this— sin 30 = 3/6

sin 30 =.5

so substituting

.5 = 3/6

A triangle only has 3 sides and 3 angles so there are only so many ways to juggle the pieces-so how hard could it be?

Now we have the basic formula we can use to compute the tension in the cable at different angles.

What would the Tension(force) be if the string was at a 20 deg. angle?

The vertical leg is still 3# and F is the unknown force required.

Sine will give us the answer.

sin 20 deg = 3#/F and sin 20 deg. = .342

now .342 = 3/F

substituting we get F= 8.77pounds of force

As we expected when we lowered the angle, the force (Tension) needed increased from 6# to 8.77#

I encourage my students to forget using their calculators to find the trig values. Instead use the trig tables like this one.

This way the kids can see how the sine values change as the angle changes.

All this was for the 2×4 but we need to use it for the cable of the bridge.

The deck of the bridge weighs 3.4#, and just as half of the weight of the 2×4 was supported by the floor, half of the deck (1.7#) is supported by the steel pivot pin mounted in the tower.

And the remaining 1.7# is the vertical leg of the triangle

Here’s the formula:

sin 35 deg = 1.7/F

sine 35 = .573

with substituting we get

.573 = 1.7/F and finally

F = 2.9 pounds of force just to hold up the deck.

What is the Tension in the cable with a 2 brick load?

The entire 9.4# load is supported by the vertical leg of the triangle.

Here’s the set -up.

sin 35 = 9.4/F

sin 35 = .573

.573 = 9.4/F

F = 16.4#

Now we add the Tension in the cable from the weight of the deck (2.9#) and get

19.3#

Next we lower the cable to 15 deg.

First the Tension in the cable just for the deck. – Scroll up and review the formula

sin 15 = 1.7 /F

sin 15 = .258

.258 = 1.7/F

F = 6.58#

that’s just for the deck.

Now the 4.7# brick

sin 15 = 4.7/F

sin 15 = .258

.258 = 4.7/F

F= 18.2# then we add the Tension from the deck (6.58#) and get
24.78#

much more than the 19.3# with 2 bricks @35deg.

I worked with a middle school team at Franciscan Montessori school. They managed to place 5 bricks

on the deck with a 35deg angle on the cable.

Did you notice the windlass? The blue wheel has an axle connected to it. The line (string) wraps around the axle, so when the wheel turns, the line is gets pulled and the deck/load is raised.

Now the girls lowered the angle of the cable and added the bricks.