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 8.00 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.   

State what assumption you made.

Use g = 9.81 N kg-1.

Answer

 

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

A bridge has a mass of 10 000 kg and is made of a uniform beam AB 6.0 m long as shown.  A 2000 kg mass is place on the beam 2.0 m from the end B as shown

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

Use g = 9.8 N kg-1.

Answer

Moments about A = (10 000 kg × 9.8 N kg-1 × 3) + (2000 kg × 9.8 N kg-1 × [6.0 m2.0 m]) = 294 000 N m + 78 400 N m

Moments about A = 372 400 N m

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

Moments about B = 333 200 N m

Force on A = 333 200 N m ÷ 6.0 m = 55 533 N = 53 000 N (2 s.f.)

Force on B = 372 400 N m ÷ 6.0 m = 62 067 N = 62 000 N (2 s.f.)

 

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?

Use g = 9.8  N kg-1.

Answer

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

  • The lorry is 4.0 m from A, while the car is 16 m - 6.0 m = 10 m from A

  • The lorry is 16 m - 4.0 m = 12 m from B, while the car is 6.0 m from B

  • Moment made by the centre of mass of the bridge = 20000 kg × 9.8 N kg-1 × 8.0 m = 1 568 000 N m

  • Moment made by the car = 1100 kg × 9.8 N kg-1 × 6.0 m = 64680 N m

  • Moment made by the lorry = 4500 kg × 9.8 N kg-1 × 12 m = 529 200 N m

  • Total moments = 1 568 000 N m + 64680 N m + 529 200 N m = 2 161 880 N m

  • Force on A = 2 161 880 N m N m ÷ 16 m = 135 117 N = 140 000 N (2 s.f. as the data are to 2 s.f.)

 

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

  • Moment made by the centre of mass of the bridge = 20000 kg × 9.8 N kg-1 × 8.0 m = 1 568 000 N m

  • Moment made by the car = 1100 kg × 9.8 N kg-1 × 10 m = 107 800 N m

  • Moment made by the lorry = 4500 kg × 9.8 N kg-1 × 4.0 m = 176 400 N m

  • Total moments = 1 568 000 N m + 107 800 N m + 176 400 N m = 1 852 200 N m

  • Force on B = 1 852 200 N m ÷ 16 m = 115 763 N = 120 000 N (2 s.f.)

Note that we have given our answer to 2 significant figures, as the data are to 2 s.f.

 

Question 2

In the bridge below which is 6.00 m in length, a car of mass 1500 kg is 1.50 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?

 

Give your answer to an appropriate number of significant figures.

Use g = 9.8 N kg-1.

Answer

 

 Many other problems involving moments can be solved easily by modelling them as bridge problems, for example: