2. Chapter Objectives
Concept of moment of a force in two and three dimensions
Method for finding the moment of a force about a specified axis.
Define the moment of a couple
Determine the resultants of non-concurrent force systems
Reduce a simple distributed loading to a resultant force having a specified location

3. Chapter Outline Moment of a Force – Scalar Formation Cross Product
Moment of Force – Vector Formulation
Principle of Moments
Moment of a Force about a Specified Axis
Moment of a Couple
Simplification of a Force and Couple System
Further Simplification of a Force and Couple System
Reduction of a Simple Distributed Loading
Moment is a combination of a physical quantity and a distance.
= the product of quantity (as a force) and the distance to a particular axis or point.

4. 4.1 Moment of a Force – Scalar Formation
Moment of a force about a specified point or axis is a quantity indicating an attempt to rotate an object about the point or axis.

5. 4.1 Moment of a Force – Scalar Formation
Magnitude
Magnitude of MO
(Moment about O from F) is
d = Perpendicular distance measured
from O to a line passing the force
Direction
Find the direction by right hand rule

6. 4.1 Moment of a Force – Scalar Formation
Resultant Moment
MRo = The sum of moments from all forces

7. Example 4.1
For each case, determine the moment of the force about point O.
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 curl.

8. Solution

10. Example 4.2
Determine the resultant moment of the forces about point O.
Assume positive moment as counter-clockwise;

11. 4.2 Cross Product
The cross product of the vectors A and B become the vector C
C = A X B
Magnitude
0≤ θ ≤ 180°, θ is the angle between A and B
C = AB sinθ

12. 4.2 Cross Product Direction
Vector C is perpendicular to the plane containting Vector A and B according to the right hand rule (sweep from A to B and the direction is indicated by the thumb)
Vector C can be writing in a vector form as
C = A X B = (AB sin θ)uC
uC is a unit vector of vector C

13. 4.2 Cross Product B X A = -C Laws of Operations
1. Commutative law not applied
A X B ≠ B X A
But
A X B = - (B X A)
Cross product B X A give a vector
that is opposite direction to vector C
B X A = -C

14. 4.2 Cross Product Laws of Operations 2. Multiplication by a Scalar
a( A X B ) = (aA) X B = A X (aB) = ( A X B )a
3. Distributive Law
A X (B + D) = (A X B) + (A X D)
And
(B + D) X A = (B X A) + (D X A)

15. 4.2 Cross Product
Cartesian Vector Formulation i j k

16. 4.2 Cross Product Cartesian Vector Formulation
A more compact determinant in the form as

17. 4.3 Moment of Force - Vector Formulation
Moment of a force F about O can be found by a cross product
MO = r X F
Magnitude
cross product, MO = rF sin θ
Consider if F or r can be moved along a line of action
(sliding vector) และ d = r sinθ, MO = rF sinθ = F (rsinθ) = Fd
Direction
Direction of MO can be found by the right hand rule
*Note:
- Sweep r to F and the thumb indiction the direction of the moment MO

18. 4.3 Moment of Force - Vector Formulation
Principle of Transmissibility
For a a force F acting at A, the moment about O is MO = rA x F
F can be moved along a line of action
MO = r1 X F = r2 X F = r3 X F

19. 4.3 Moment of Force - Vector Formulation
Cartesian Vector Formulation
For force expressed in Cartesian form,
หาค่า determinant,
MO = (ryFz – rzFy)i + (rzFx - rxFz)j + (rxFy – ryFx)k

20. 4.3 Moment of Force - Vector Formulation
The resultant moment of forces
The resultant moment about O is the sum of all moments about O
MRo = ∑(r x F)

21. Example 4.4
Two forces act on the rod. Determine the resultant moment they create about the flange at O. Express the result as a Cartesian vector.
A (0, 5, 0)
B (4, 5, -2)

22. Solution
Position vectors are directed from point O to each force as shown.
These vectors are
The resultant moment about O is

23. 4.4 Principles of Moments Varignon’s Theorem
“The moment of a resultant force is equivalent to the sum of moment from the compenent forces
From F = F1 + F2

24. Example 4.5
Determine the moment of the force about point O.

25. Solution The moment arm d can be found from trigonometry,
Since the force tends to rotate or orbit clockwise about point O, the moment is directed into the page. or

26. 4.5 Moment of a Force about a Specified Axis
Moment of a force about an axis is perpendicular to the plane containing the position vector and the force
The component moment about a specified axis can be found by Scalar notation or vector Cartesian vector
e.g. My from Mo

27. 4.5 Moment of a Force about a Specified Axis
Using scalar notation
From the right hand rule My is pointed in y direction
For any axis a
The force will not generate a moment if the force is parallel to the specified axis

28. 4.5 Moment of a Force about a Specified Axis
Using vector Cartesian notation
Find the moment about O
The component in y can be found from a projection on the axis y
For any axis a

29. 4.5 Moment of a Force about a Specified Axis
For a vector Cartesian notation
is the component of a unit vector
is the component of a position vector r
(โมเมนต์รอบแกนที่สนใจ ไม่ใช่รอบจุดที่สนใจ)

30. 4.5 Moment of a Force about a Specified Axis
In determinant form,
If the result is positive, the direction of a moment is in the same direction as the specified axis
Then

31. Example 4.8
Determine the moment produced by the force F which tends to rotate the rod about the AB axis.
A (0, 0, 0)
B (0.4, 0.2 0)

32. Solution
Unit vector defines the direction of the AB axis of the rod, where
rC or rD can be used
For simplicity, choose rD
The force is

33. Solution

34. Check for different position vector r
Using FD
F (-0.8, -0.4 ,0)
Using ED
D (0.6,0,0)
E (0.8, 0.4 ,0)

35. 4.6 Moment of a Couple Couple The forces are parallel
Identical magnitude but opposite direction
The distance between the forces is d
The resultant force = 0
Attempt to rotate in a direction
The moment of a couple is a sum of moments about a specified axis

36. 4.6 Moment of a Couple
Moment of a couple is a Free vector since it only depends on r, hence it can be applied anywhere and the result will be the same
(This is different to a moment of a force as the result will be different for difference specified points or axes)

37. 4.6 Moment of a Couple Magnitude of couple moment M = Fd
Direction and sense are determined by right hand rule
M acts perpendicular to plane containing the forces

38. 4.6 Moment of a Couple In a vector notation For couple moment,
M = r X F
If consider the moment about A, moment from –F will be zero

39. 4.6 Moment of a Couple Equivalent Couples
Two couple moments are equivalent if they are equal
Both couple moments are in the same planes

40. 4.6 Moment of a Couple The resultant moment of couples
Couple moments are free vector and can then be added using vector addition
MR = M1 + M2
In general, MR = ∑(r X F)

41. Example 4.12
Determine the couple moment acting on the pipe. Segment AB is directed 30° below the x–y plane.

42. SOLUTION I (VECTOR ANALYSIS)
Take moment about point O,
Take moment about point A

43. SOLUTION II (SCALAR ANALYSIS)
Take moment about point A or B,
Apply right hand rule, M acts in the –j direction

44. 4.7 Simplification of a Force and Couple System
Equivalent force and moment will give the same effect as when applying the original force and moment
Original system
+ Opposite actions
Equivalent system
Principle of Transmissibility

45. 4.7 Simplification of a Force and Couple System
Consider F1 F2 and couple moment M
Can move the forces to O and then generate additionl couples about MO as M1=r1xF1 and M2=r2xF2
The equivalent force about O and the result couple moments can be written as
For 2D in x–y plane and the couple moment perpendicular to this plane,

46. 4.7 Simplification of a Force and Couple System
Procedure for Analysis
Define origin at O that point to the interested direction
Sum forces
Sum moments

47. Example 4.16
A structural member is subjected to a couple moment M and forces F1 and F2. Replace this system with an equivalent resultant force and couple moment acting at its base, point O.

48. Solution
Express the forces and couple
moments as Cartesian vectors.

49. Solution
Force Summation.

50. 4.8 Further Simplification of a Force and Couple System
For equivalent force and moment, about point O, we can find one equivalent force and one moment O (FR) and (MR)O
Can further simplified to have only one force if FR and (MR)O are perpendicular to each other
This include
Forces with a common point
Coplanar forces
Parallel forces

51. 4.8 Further Simplification of a Force and Couple System
Concurrent Force System
All forces have a commom point O
There is no moment, hence we can find an equivalent force

52. 4.8 Further Simplification of a Force and Couple System
Coplanar Force System
All forces are in the same plane
Moments of each force are perpendicular to the plan
The resultant moment can be found by moving the resultant forces by d

53. 4.8 Further Simplification of a Force and Couple System
Parallel Force System
ตัวอย่าง แรงทั้งหมดขนานกับแกน z แรงลัพท์ก็ขนานกับแกนนี้
โมเมนต์ของแรงเหล่านั้น และโมเมนต์ลัพท์วางอยู่บนระนาบ
โมเมนต์ลัพท์สามารถแทนด้วยแรงที่เลื่อนตั้งฉากออกไปด้วยระยะ d
แนวของ FR และ (MR)O ตั้งฉากกัน

54. 4.8 Further Simplification of a Force and Couple System
Reduction to a Wrench
โดยทั่วไปแนวของ FR และ (MR)O ไม่ตั้งฉากกัน (ลดรูปต่อไม่ได้)
(a) แตก (MR)O ออกเป็นแนวตั้งฉากและขนานกับ FR ได้ (M┴ , M ║ )
(b) แทน M┴ ได้ด้วยการย้าย FR ไปจุด P ด้วยระยะ d
(c) ย้าย M ║ ไปจุด P ได้เนื่องจากเป็น Free vector
FR และ M ║ พยายามที่จะเลื่อนและหมุนวัตถุรอบแกน เรียกว่า Wrench หรือ Screw

55. Example 4.18
The jib crane is subjected to three coplanar forces. Replace this loading by an equivalent resultant force and specify where the resultant’s line of action intersects the column AB and boom BC.
1.75 kN

56. Solution
Force Summation

57. Solution For magnitude of resultant force,
For direction of resultant force,
Assume line of FR
(assume y or x)

58. Solution
Moment Summation
 Summation of moments about point A,

59. Solution
Moment Summation
 Principle of Transmissibility

60. 4.9 Reduction of a Simple Distributed Loading
For large areas, they are modelled as distributed loadings
It is the force per unit area, which is equivalent to pressure
Unit Pascal (Pa): 1 Pa = 1N/m2
Uniform Loading Along a Single Axis
This is the most common problem

61. 4.9 Reduction of a Simple Distributed Loading
Magnitude of Resultant Force
For an infinitestimal dF acting on an infinitestimal dA
For an entire length L, the resultant force is the integral over the length
Essentially, the resultant force is the area under the graph

62. 4.9 Reduction of a Simple Distributed Loading
Location of Resultant Force (for Centroid )
MR = ∑MO
dF causes the moment xdF = x w(x) dx about O
The the sum of moment,
Solving for

63. Ex. Determine the magnitude and location of the equivalent resultant force acting on the shaft.
Solution
For the small differential area element,
For resultant force

64. Solution
For location of line of action,
Checking,