Force Systems

Equivalent systems

Do we need this section?

It can be helpful at times to reduce the forces on a body and combine resultant moments and simplify the analysis. For example, you can replace a force couple with their equivalent couple moment in a system. These forces and moments will have the same external effect on the body, but the internal forces on the rigid body may be different.

Concentrated forces

Do we need this section? - from lecture 2

Concentrated forces are forces acting at a specific point on a body.

Fig: ConcentratedForce.PNG

A concentrated force applied to a bar at point A.

Distributed loads

Distributed loads (usually written as $ w(x) $) are forces applied over a length, volume, or area. The SI units for distributed loads are N/m. These loads, written as a function of length $ x $ can be simplified into an equivalent force, $ F_R $, which results in the same external loading on the rigid body. The magnitude of the equivalent force $ F_R $ is the area under the $ w(x) $ curve. The location of the resultant force $ F_R $ is the location of the centroid of the shape of the force curve.

Later, we can add a side comment about examples of distributed loads in real life. For example, books on a bookshelf, air pressure over a lake, the distribution of loads on on the ground over your foot when you take a step

Rectangular loading

For a constant distributed load of magnitude c applied to a bar over a length l, the resultant force magnitude is

$$ F_R = c*l\ $$

the resultant force location is at $ l/2 $.

Fig: DistLoad.PNG

Example distributed load (top) and its equivalent force system (bottom).

Triangular loading

For a triangular distributed load (a uniformly varying load), the magnitude of the load is equivalent to the area of the triangle.

$$ F_R = 0.5*c*l\ $$

The location of $ F_R $ is 1/3 from the base of the triangle.

Fig: TriangleLoad.PNG

Example triangular distributed load (top) and its equivalent force system (bottom).

Nonuniform distributed loads

More complex distributed loads can be reduced to two distributed loads of uniform shape, or the integral of the distributed load function can be taken to obtain the area under $ w(x) $, which is $ F_R $.

$$ F_R = \int_{0}^{L} w(x) \,dx \ $$

The location of $ F_R $ on the beam can be found by solving for the moment $ M_R $ about a point. In this case we will take the moment about the point $ x $=0 and setting it equivalent to the moment produced by the resultant force $ F_R $ at $ x $=0.

$$ M_R = \int_{0}^{L} x*w(x) \,dx = \bar{x}*F_R\ $$

Now we can solve for the center of mass of the distributed load, $ \bar{x} $, using

$$ \bar{x} = \dfrac{M_R}{F_R}\ $$

Fig: NonuniformLoad

Nonuniform load. Need to remake this in the same style as the other pictures.

TAM 2xx References