Force System Analysis Resultants, Reduction and Distributed Loads - Engineering

The basic idea: Two force systems are equivalent if they result in the same resultant force and the same resultant moment.

Moving a force along its line of action: Moving a force along its line of action results in a new force system which is equivalent to the original force system.

Moving a force off its line of action: If a force is moved off its line of action, a couple must be added to the force system so that the new system generates the same moment as the old system.

The resultant of a force and couple system: For any point O, every force and couple system can be made equivalent to a single force passing through O and a single couple. The single force passing through O is equal to the resultant force of the original system, and the couple is equal to the resultant moment of the original system around point O.

When can one reduce a force and couple system to a single force?: For a force and couple system if the resultant force and the resultant couple are perpendicular, then one can find an equivalent system with a single force and no couple. To obtain this system, move the resultant force a distance d along the line perpendicular to the plane of the resultant force and resultant couple until the resultant force creates a moment equivalent to the resultant couple.

Note: All 2-D force systems can be reduced to a single force. To find the line of action of the force, the moment of the original system must be forced to be the same as the system with the single force.

Further Reduction of a Force and Couple System                                    

(Section 4.9)

If F_R and M_RO are perpendicular to each other, then the system can be further reduced to a single force, F_R , by simply moving F_R from O to P.

In three special cases, concurrent, coplanar, and parallel systems of forces, the system can always be reduced to a single force.

Example #1

Given: A 2­D force and couple system as shown.

Find: The equivalent resultant force and couple moment acting at A and then the equivalent single force location along the beam AB.

Plan:

1. Sum all the x and y components of the forces to find F_RA.
2. Find and sum all the moments resulting from moving each force to A.
3. Shift the F_RA to a distance d such that d = M_RA/F_Ry

Example #1

+ → Σ FR_x  =  25 + 35 sin 30°   = 42.5 lb

+  ↓ Σ F_Ry  =  20 + 35 cos 30°   = 50.31 lb

+  MRA   =  35 cos30° (2)  + 20(6)  – 25(3) = 105.6  lb∙ft

FR = ( 42.52 + 50.312 )^1/2 = 65.9 lb

θ = tan^-­1 ( 50.31/42.5) = 49.8 °

The equivalent single force F_R can be located on the beam AB at a distance d measured from A.

d = M_RA/F_Ry = 105.6/50.31 = 2.10 ft.

Example #2

Given: The building slab has four columns. F₁ and F₂ = 0.

Find:    The equivalent resultant force and couple moment at the origin O. Also find the location (x,y) of the single equivalent resultant force.

Plan:

1. Find F_RO = ∑F_i = F_Rzo k
2. Find M_RO = ∑ (r_i × F_i) = M_RxO i + M_RyO j
3. The location of the single equivalent resultant force is given as x = ­M_RyO/F_RzO and y = M_RxO/F_RzO

Example #2  

F_RO = {­50 k – 20 k} = {­70 k} kN

M_RO = (10 i) × (­20 k) + (4 i + 3 j)x(­50 k)

= {200 j + 200 j – 150 i} kN∙m

= {­150 i + 400 j } kN∙m

The location of the single equivalent resultant force is given as,

x = ­M_Ryo/F_Rzo = ­400/(­70) = 5.71 m

y = M_Rxo/F_Rzo = (­150)/(­70) = 2.14 m

Distributed Loads

Wind and water loads, cars on a bridge, and people on a crowded walkway often generate loads that are approximated as a pressure (force per unit area) or a distributed load (force per unit length). In order to utilize our equilibrium equations, however, we need forces and moments. Consequently, it is often necessary to replace a pressure or distributed load with a single force. First, consider a simple example. We will apply a uniform load to a beam that is 3 m long and the space, a between the wall and the beginning of the applied load is 0.5 m. It should be easy to see that, if we want to replace this with a single force, it must be a 250 N load placed in the middle of the loaded region (i.e.  1:75 m).

Next, take the system shown below, a cantilevered beam with an increasing, triangular distributed load which peaks at w₀.

The distributed load has units of force per unit length (N/m or lbs./ft.) and, in this case, can be written as,

          (11.1)

In order to ensure that the forces in the two systems are equivalent, we require

      (11.2)

which yields,

We also want it to have the same moment about the origin. In order to do this we require that,

  (11.3)

which yields,

and the position of the force,   must be 2/3 L

Example Here is another distributed load acting on a beam. Replace the distributed load with a single force.