Mechanics Tutorial 5 - Moments and Bridges

At its simplest, a bridge is a plank of wood placed between two supports.  As the plank of wood has a mass, it must also have a centre of mass.  At this level, we will assume that the plank is totally regular, and its centre of mass is exactly in the middle.

Notice that at each end there is a pivot, around which we will need to take moments.  Even if there were no actual pivot mechanism, we treat the bridge as if there were.  If the plank broke in the middle, the right hand half would turn about the right pivot, and the left half around the left pivot.

 


So let us have a look at the moments around each pivot:

Clockwise Moment = mg × d

Anticlockwise moment = mg × d

 

Since force = moment ÷ distance, it doesn’t take a genius to see that the force on each pivot, A and B, is given by:

F = mg/2

Question 1

A bridge is 8 metres long and has a mass of 20 000 kg.  Calculate:

a.       The weight of the bridge;                                                                          

b.      The moment about each end of the bridge;                                               

c.       The force acting on each end of the bridge.                                              

 

When we look at forces acting on a bridge with traffic on, we are usually asked to consider the force acting on each abutment (the pier at each end).  We need to take moments about each end.  Bridge engineers have to do this carefully, otherwise there would be a disaster. We will look at this by a worked example.

Now let’s put a load on the bridge of mass m1 kg, x metres from one end of the bridge:

The centre of mass is in the same place, but we now have an extra load of m1g on the bridge.  Taking moments about B:

Moment = (mg × d) + (m1g × x)

 Taking moments about A:

Moment = (mg × d) + (m1g × [2d – x])

Force on A can be worked out:

Force on A = Moment about B ÷ 2d = (mg × d) + (m1g × x)

                                                            2d

Similarly, the force on B can be worked out:

 

Force on B = moment about A ÷ 2d = (mg × d) + (m1g × [2d – x])

                                                                         2d

 

Worked Example

What is the force acting on each end of the bridge?

Moments about A = (10 000 × 9.8× 3) + (2000 × 9.8 × [62]) = 294 000 Nm + 78 400 Nm

Moments about A = 372 400 Nm

Moments about B = (10 000 × 9.8× 3) + (2000 × 9.8 × 2) = 294 000 Nm + 39 200 Nm

Moments about B = 333 200 Nm

Force on A = 333 200 ÷ 6 = 55 500 N

Force on B = 372 400 ÷ 6 = 62 000 N

Now let's do an example with two vehicles on.  The principles are the same.

Worked Example

On the bridge shown in the diagram above, what are the forces on the abutments A and B?

To work out the force on A, we need to take moments about B

  • Moment made by the centre of mass of the bridge = 200000 N × 8 m = 1 600 000 Nm

  • Moment made by the car = 11000 N × 6 m = 66000 Nm

  • Moment made by the lorry = 45000 N × 12 m = 540 000 Nm

  • Total moments = 1 600 000 Nm + 66000 Nm + 540 000 Nm = 2206000 Nm

  • Force on A = 2206000 Nm ÷ 16 m = 138 000 N

To work out the force on B, we need to take moments about A:

  • Moment made by the centre of mass of the bridge = 200000 N × 8 m = 1 600 000 Nm

  • Moment made by the car = 11000 N × 10 m = 110 000 Nm

  • Moment made by the lorry = 45000 N × 4 m = 180 000 Nm

  • Total moments = 1 600 000 Nm + 110 000 Nm + 180 000 Nm = 1890000 Nm

  • Force on B = 1890000 Nm ÷ 16 m = 118 000 N

Note that we have given our answer to 3 significant figures.  No more than this is needed.

 

Question 2

In the bridge below, a car of mass 1500 kg is 1.5 m from side A and a van of mass 2500 kg is 1.75 m from side B.  What forces are acting on sides A and B?

 

Ladders

A ladder uses the ideas of turning moments.