Reduction of a Force & Couple System | Civil Engineering Optional Notes for UPSC PDF Download

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