1. CHAPTER 3: BASIC CONCEPT OF MOMENTS
STATICS AND DYNAMICS
PDT 101/3
CHAPTER 3: BASIC CONCEPT OF MOMENTS
PREPARED BY:
TAN SOO JIN

2. CHAPTER OUTLINE Moment of a Force – Scalar Formation
Moment of a Force about a Specified Axis
Moment of a Couple
Simplification of a Force
Reduction of A Simple Distributed Loading

3. 3.1 Moment of a Force – Scalar Formation
The ability of a force to cause a body to rotate is measured by a quantity called the moment of the force.

4. 3.1 Moment of a Force – Scalar Formation
The turning effect of the force is measured by the product of F and d.

5. 3.1 Moment of a Force – Scalar Formation
Resultant Moment
- Resultant moment, MRo = moments of all the forces
MRo = ∑Fd

6. 3.1 Moment of a Force – Scalar Formation
Direction of a moments.
The two forces P and Q causes the lever to rotate about point O in opposite direction.
– The force P causes a counterclockwise rotation (positive
sign)
– The force Q causes a clockwise rotation (negative sign)

7. EXAMPLE 1
For each case, determine the moment of the force about point O.

8. SOLUTION 1
Line of action is extended as a dashed line to establish moment arm d.
Tendency to rotate is indicated and the orbit is shown as a colored curl.

11. EXAMPLE 2
Determine the moment of the force about point O

12. SOLUTION 2 + MO = {– 100 (3/5)N (5 m) – (100)(4/5)N (2 m)} N·m

13. EXERCISE 1
A 500 N force is applied to the end of a lever pivoted at point O. determine the moment of the force about O if θ= 30°.
(Ans: -50N.m)

14. EXERCISE Given: A 100 N force is applied to the hammer.
Find: The moment of the force at A.
(Ans: N.m)

15. 3.2 Moment of a Force About A Specified Axis
For moment of a force about a point, the moment and its axis is always perpendicular to the plane
A scalar or vector analysis is used to find the component of the moment along a specified axis that passes through the point

16. 3.2 Moment of a Force About A Specified Axis
Scalar Analysis
According to the right-hand rule, My is directed along the positive y axis
For any axis, the moment is
Force will not contribute a moment if force line of action is parallel or passes through the axis

17. EXERCISE 2
A force F of magnitude 60 N is applied to the gear. Determine the moment of F about point O.
(Ans: N.m)

18. 3.3 Moment of A Couple Couple Equivalent Couples
- Two parallel forces that have the same magnitude
but opposite directions and separated by a
perpendicular distance, d.
Equivalent Couples
- 2 couples are equivalent if they produce the same moment
- Forces of equal couples lie on the same plane or plane parallel to one another

19. 3.4 Simplification of a Force
Equivalent resultant force acting at point O and a resultant couple moment is expressed as
If force system lies in the x–y plane and couple
moments are perpendicular to this plane,

20. EXERCISE 3
The caster unit is subjected to the pair of 400-N forces shown. Determine the moment associated with these forces on the metal nut.
(Ans: -14N.m)

21. EXERCISE 4
Two couples act on the beam as shown. Determine the magnitude of F so that the resultant couple moment is 300 N · m counterclockwise. Take moment about point A (Ans: N) A

22. EXERCISE 5
Determine the moments of the 800 N force acting on the frame about points A and B.
(Ans: MC=0, MD=400Nm)
MA = 800 N (2.5)
= 2000 N
MB = 800 N (1.5)
= 1200 N
How about point C and D?

23. EXERCISE 6
The force F acts at the end of the angle bracket shown. Determine the moment of the force about point O.
(Ans: Nm)

24. SOLUTION + MO = (400sin30o N) (0.2m) – (400cos30o N)(0.4m)

25. EXERCISE 7 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.

26. SOLUTION + FRx= 50(sin 30) + 100(3/5) = 85 kN
+  FRy = (cos 30) – 100(4/5)
= kN
+ MRA = 200 (3) + 50 (cos 30) (9)
– 100 (4/5) 6 = kN.m FR
FR = ( )1/2 = 184 kN
 = tan-1 ( 163.3/85) = 62.5°
The equivalent single force FR can be located at a distance d measured from A.
d = MRA/FRy = / = m

27. EXERCISE 8 Given: A 2-D force and couple system as shown.
Find: The equivalent resultant force and couple moment acting at A.

28. SOLUTION + Fx = 450 (cos 60) – 700 (sin 30) = – 125 N
+  Fy = – 450 (sin 60) – 300 – 700 (cos 30)
= – 1296 N
Now find the magnitude and direction of the resultant.
FRA = ( )1/2 = 1302 N and  = tan-1 (1296 /125)
= 84.5°
+ MRA = 450 (sin 60) (2) (6) (cos 30) (9)
= Nm

29. 3.5 Reduction of A Simple Distributed Loading
There is a bundle (called a bunk) of 50 mm x 100 mm boards stored on a storage rack. This lumber places a distributed load (due to the weight of the wood) on the beams holding the lumber.
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?

30. Distribution 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.

31. 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).

32. Location of 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

33. 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).

34. EXAMPLE
Until you learn more about centroids, we will consider only rectangular and triangular loading diagrams.
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.

35. Solution
Now let’s complete the calculations to find the concentrated loads (which is a common name for the resultant of the distributed load).
The rectangular load: FR = 10  5 = 50 kN and = 0.5 x 5 = 2.5 m. x
The triangular loading:
FR = (0.5) (600) (6) = 1,800 N and = 6 – (1/3) 6 = 4 m.
Please note that the centroid of a right triangle is at a distance one third the width of the triangle as measured from its base. x

36. EXERCISE 9 Given: The loading on the beam as shown.
Find: The equivalent force and its location from point A.

37. SOLUTION