1. SIMPLIFICATION OF FORCE AND COUPLE SYSTEMS & FURTHER SIMPLIFICATION OF A FORCE AND COUPLE SYSTEM
Objectives:
Determine the effect of moving a force.
b) Find an equivalent force-couple system for a system of forces and couples.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections

2. APPLICATIONS
What are the resultant effects on the person’s hand when the force is applied in these four different ways?
Why is understanding these difference important when designing various load-bearing structures?
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections

3. APPLICATIONS (continued)
Several forces and a couple moment are acting on this vertical section of an I-beam.
For the process of designing the I-beam, it would be very helpful if you could replace the various forces and moment just one force and one couple moment at point O with the same external effect? How will you do that?
| | ??
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections

4. SIMPLIFICATION OF FORCE AND COUPLE SYSTEM (Section 4.7)
When a number of forces and couple moments are acting on a body, it is easier to understand their overall effect on the body if they are combined into a single force and couple moment having the same external effect.
The two force and couple systems are called equivalent systems since they have the same external effect on the body.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections

5. MOVING A FORCE ON ITS LINE OF ACTION
Moving a force from A to B, when both points are on the vector’s line of action, does not change the external effect.
Hence, a force vector is called a sliding vector. (But the internal effect of the force on the body does depend on where the force is applied).
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections

6. MOVING A FORCE OFF OF ITS LINE OF ACTIONB
When a force is moved, but not along its line of action, there is a change in its external effect!
Essentially, moving a force from point A to B (as shown above) requires creating an additional couple moment. So moving a force means you have to “add” a new couple.
Since this new couple moment is a “free” vector, it can be applied at any point on the body.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections

7. SIMPLIFICATION OF A FORCE AND COUPLE SYSTEM
When several forces and couple moments act on a body, you can move each force and its associated couple moment to a common point O.
Now you can add all the forces and couple moments together and find one resultant force-couple moment pair.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections

8. SIMPLIFICATION OF A FORCE AND COUPLE SYSTEM (continued)
If the force system lies in the x-y plane (a 2-D case), then the reduced equivalent system can be obtained using the following three scalar equations.
WR = W1 + W2
(MR)o = W1 d1 + W2 d2
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections

9. FURTHER SIMPLIFICATION OF A FORCE AND COUPLE SYSTEM (Section 4.8)
If FR and MRO are perpendicular to each other, then the system can be further reduced to a single force, FR , by simply moving FR 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.

10. Given: A 2-D force system with geometry as shown.
EXAMPLE #1
Given: A 2-D force system with geometry as shown.
Find: The equivalent resultant force and couple moment acting at A and then the equivalent single force location measured from A.
Plan:
1) Sum all the x and y components of the forces to find FRA.
2) Find and sum all the moments resulting from moving each force component to A.
3) Shift FRA to a distance d such that d = MRA/FRy
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections

11. EXAMPLE #1 (continued) +  FRx = 150 (3/5) + 50 –100 (4/5) = 60 lb
+  FRy = 150 (4/5) (3/5)
= 180 lb
+ MRA = 100 (4/5) 1 – 100 (3/5) 6
– 150(4/5) 3 = – 640 lb·ft FR
FR = ( )1/2 = 190 lb
 = tan-1 ( 180/60) = 71.6 °
The equivalent single force FR can be located at a distance d measured from A.
d = MRA/FRy = 640 / 180 = ft.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections

12. Given: The slab is subjected to three parallel forces.
EXAMPLE #2
Given: The slab is subjected to three parallel forces.
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 FRO = Fi = FRzo k
2) Find MRO =  (ri  Fi) = MRxO i + MRyO j
3) The location of the single equivalent resultant force is given as x = – MRyO/FRzO and y = MRxO/FRzO

13. EXAMPLE #2 (continued)
FRO = {100 k – 500 k – 400 k} = – 800 k N
MRO = (3 i)  (100 k) + (4 i + 4 j)  (-500 k)
+ (4 j)  (-400 k)
= {–300 j j – 2000 i – 1600 i}
= { – 3600 i j }N·m
The location of the single equivalent resultant force is given as,
x = – MRyo / FRzo = (–1700) / (–800) = 2.13 m
y = MRxo / FRzo = (–3600) / (–800) = 4.5 m

14. Given: A 2-D force and couple system as shown.
GROUP PROBLEM SOLVING
Given: A 2-D force and couple system as shown.
Find: The equivalent resultant force and couple moment acting at A.
Plan:
1) Sum all the x and y components of the two forces to find FRA.
2) Find and sum all the moments resulting from moving each force to A and add them to the 45 kN m free moment to find the resultant MRA .
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections

15. GROUP PROBLEM SOLVING (continued)
Summing the force components:
+  Fx = (5/13) 26 – 30 sin 30°
= – 5 kN
+  Fy = – (12/13) 26 – 30 cos 30°
= – kN
Now find the magnitude and direction of the resultant.
FRA = ( )1/2 = 50.2 kN and  = tan-1 (49.98/5) = 84.3 °
+ MRA = {30 sin 30° (0.3m) – 30 cos 30° (2m) – (5/13) 26 (0.3m)
– (12/13) 26 (6m) – 45 } = – 239 kN m
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections

16. GROUP PROBLEM SOLVING (continued)
Given: Forces F1 and F2 are applied to the pipe.
Find: An equivalent resultant force and couple moment at point O.
Plan:
a) Find FRO =  Fi = F1 + F2
b) Find MRO =  MC +  ( ri  Fi )
where,
MC are any free couple moments (none in this example).
ri are the position vectors from the point O to any point on the line of action of Fi .
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections

17. GROUP PROBLEM SOLVING (continued)
F1 = {– 20 i –10 j + 25 k} lb
F2 = {–10 i + 25 j + 20 k} lb
FRO = {–30 i + 15 j + 45 k} lb
r1 = {1.5 i + 2 j} ft
r2 = {1.5 i + 4 j + 2 k} ft
Then, MRO =  ( ri  Fi ) = r1  F1 + r2  F2
= {(50 i – 37.5 j + 25 k ) + (30 i – 50 j k )} lb·ft
= {80 i – 87.5 j k} lb·ft
MRO = {
} lb·ft
i j k +
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections

18. REDUCTION OF A SIMPLE DISTRIBUTED LOADING=
Objective:
Determine an equivalent force for a distributed load.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 4.10

19. APPLICATIONS
There is a bundle (called a bunk) of 2” x 4” boards stored on a storage rack. This lumber places a distributed load (due to the weight of the wood) on the beams holding the bunk.
To analyze the load’s effect on the steel beams, it is often helpful to reduce this distributed load to a single force How would you do this?
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 4.10

20. APPLICATIONS (continued)
The uniform wind pressure is acting on a triangular sign (shown in light brown).
To be able to design the joint between the sign and the sign post, we need to determine a single equivalent resultant force and its location.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 4.10

21. In such cases, w is a function of x and has units of force per length.
DISTRIBUTED LOADING
In many situations, a surface area of a body is subjected to a distributed load. Such forces are caused by winds, fluids, or the weight of items on the body’s surface.
We will analyze the most common case of a distributed pressure loading. This is a uniform load along one axis of a flat rectangular body.
In such cases, w is a function of x and has units of force per length.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 4.10

22. MAGNITUDE OF RESULTANT FORCE
Consider an element of length dx.
The force magnitude dF acting on it is given as
dF = w(x) dx
The net force on the beam is given by
+  FR = L dF = L w(x) dx = A
Here A is the area under the loading curve w(x).
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 4.10

23. LOCATION OF THE RESULTANT FORCE
The force dF will produce a moment of (x)(dF) about point O.
The total moment about point O is given as
+ MRO = L x dF = L x w(x) dx
Assuming that FR acts at , it will produce the moment about point O as
+ MRO = ( ) (FR) = L w(x) dx
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 4.10

24. LOCATION OF THE RESULTANT FORCE (continued)
Comparing the last two equations, we get
You will learn more detail later, but FR acts through a point “C,” which is called the geometric center or centroid of the area under the loading curve w(x).
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 4.10

25. EXAMPLES
Until you learn more about centroids, we will consider only rectangular and triangular loading diagrams whose centroids are well defined and shown on the inside back cover of your textbook.
Look at the inside back cover of your textbook. You should find the rectangle and triangle cases. Finding the area of a rectangle and its centroid is easy!
Note that triangle presents a bit of a challenge but still is pretty straightforward.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 4.10

26. The rectangular load: FR = 400  10 = 4,000 lb and = 5 ft. x
EXAMPLES
Now lets complete the calculations to find the concentrated loads (which is a common name for the resultant of the distributed load).
The rectangular load: FR = 400  10 = 4,000 lb and = 5 ft. x
The triangular loading:
FR = (0.5) (600) (6) = 1,800 N and = 6 – (1/3) 6 = 4 m.
Please note that the centroid in a right triangle is at a distance one third the width of the triangle as measured from its base. x
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 4.10

27. GROUP PROBLEM SOLVING Plan:
Given: The loading on the beam as shown.
Find: The equivalent force and its location from point A.
Plan:
The distributed loading can be divided into three parts. (one rectangular loading and two triangular loadings).
2) Find FR and its location for each of these three distributed loads.
3) Determine the overall FR of the three point loadings and its location.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 4.10

28. x x GROUP PROBLEM SOLVING (continued)
For the left triangular loading of height 8 kN/m and width 3 m,
FR1 = (0.5) 8 kN/m  3 m = 12 kN
x1 = (2/3)(3m) = 2 m from A
For the top right triangular loading of height 4 kN/m and width 3 m,
FR2 = (0.5) (4 kN/m) (3 m) = 6 kN
and its line of action is at = (1/3)(3m) + 3 = 4 m from A x 2
For the rectangular loading of height 4 kN/m and width 3 m,
FR3 = (4 kN/m) (3 m) = kN
and its line of action is at = (1/2)(3m) + 3 = 4.5 m from A x 3
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 4.10

29. GROUP PROBLEM SOLVING (continued)
For the combined loading of the three forces, add them.
FR = 12 kN + 6 kN kN = 30 kN
+ MRA = (2) (12) (6) + (4.5) 12 = kN • m
Now, (FR x) has to equal MRA = kN • m
So solve for x to find the equivalent force’s location. Hence, x = (102 kN • m) / (30 kN) = m from A.
Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Section 4.10