1. Structure I Course Code: ARCH 208 Dr. Aeid A. Abdulrazeg
Statics:The Next Generation Mehta & Danielson Lecture Notes for Sections

2. a) understand and define moment, and,
MOMENT OF A FORCE
Objectives :
We will be able to:
a) understand and define moment, and,
b) determine moments of a force in 2-D and 3-D cases.
Applications
Moment in 2-D
Concept quiz
Group Problem Solving
Attention quiz
Moment of a
force

3. What is the net effect of the two forces on the wheel?
APPLICATIONS
What is the net effect of the two forces on the wheel?
Statics:The Next Generation Mehta & Danielson Lecture Notes for Sections

4. APPLICATIONS (continued)
What is the effect of the 30 N force on the lug nut?
Statics:The Next Generation Mehta & Danielson Lecture Notes for Sections

5. MOMENT IN 2-D
The moment of a force about a point provides a measure of the tendency for rotation (sometimes called a torque).
Statics:The Next Generation Mehta & Danielson Lecture Notes for Sections

6. MOMENT IN 2-D (continued)
In the 2-D case, the magnitude of the moment is Mo = F d
As shown, d is the perpendicular distance from point O to the line of action of the force.
In 2-D, the direction of MO is either clockwise or
counter-clockwise depending on the tendency for rotation.
Statics:The Next Generation Mehta & Danielson Lecture Notes for Sections

7. MOMENT IN 2-D (continued)F a b d O
For example, MO = F d and the direction is counter-clockwise.
Often it is easier to determine MO by using the components of F as shown. a b O F
F x
F y
Using this approach, MO = (FY a) – (FX b). Note the different signs on the terms! The typical sign convention for a moment in 2-D is that counter-clockwise is considered positive. We can determine the direction of rotation by imagining the body pinned at O and deciding which way the body would rotate because of the force.
Statics:The Next Generation Mehta & Danielson Lecture Notes for Sections

8. The sense of M depends on the direction in which F tends to rotate the body. The right-hand rule is used to identify this sense. We represent the moment of F about O-O as a vector pointing in the direction of the thumb, with the fingers curled in the direction of the rotational tendency.
Statics:The Next Generation Mehta & Danielson Lecture Notes for Sections

9. EXAMPLE 1
Given: A 400 N force is applied to the frame and  = 20°.
Find: The moment of the force at A.
Plan:
1) Resolve the force along x and y axes.
2) Determine MA using scalar analysis.

10. EXAMPLE 1 (continued)
Solution
+  Fy = cos 20° N
+  Fx = sin 20° N
+ MA = {(400 cos 20°)(2) + (400 sin 20°)(3)} N·m
= N·m

11. 2. If M = r  F, then what will be the value of M • r ? A) 0 B) 1
CONCEPT QUIZ
1. If a force of magnitude F can be applied in four different 2-D configurations (P,Q,R, & S), select the cases resulting in the maximum and minimum torque values on the nut. (Max, Min).
A) (Q, P) B) (R, S)
C) (P, R) D) (Q, S)
Answers:
1. D
2. A
2. If M = r  F, then what will be the value of M • r ?
A) 0 B) 1
C) r 2 F D) None of the above.
Statics:The Next Generation Mehta & Danielson Lecture Notes for Sections

12. Given: A 40 N force is applied to the wrench.
GROUP PROBLEM SOLVING
Given: A 40 N force is applied to the wrench.
Find: The moment of the force at O.
Plan: 1) Resolve the force along x and y axes.
2) Determine MO using scalar analysis.
Solution: +  Fy = - 40 cos 20° N
+  Fx = - 40 sin 20° N
+ MO = {-(40 cos 20°)(200) + (40 sin 20°)(30)}N·mm
= N·mm = N·m
Statics:The Next Generation Mehta & Danielson Lecture Notes for Sections

13. ATTENTION QUIZ
10 N
5 N
3 m P m
1. Using the CCW direction as positive, the net moment of the two forces about point P is
A) 10 N ·m B) 20 N ·m C) N ·m
D) 40 N ·m E) N ·m
Answers:
1. B
2. A
Statics:The Next Generation Mehta & Danielson Lecture Notes for Sections

14. EXAMPLE 2
Calculate the magnitude of the moment about the base point O of the 600-N force

15. (I) The moment arm to the 600-N force is
d = 4 cos 40o+ 2 sin 40o = 4.35 m
By M = Fd the moment is clockwise and has the magnitude
MO = 600(4.35) = 2610 N .m
(II) Replace the force by its rectangular components at A
F1 = 600 cos 40o = 460 N,
F2 = 600 sin 40o = 386 N
By Varignon’s theorem, the moment becomes
MO = 460(4) + 386(2) =2610 N .m

16. MOMENT OF A COUPLE
Today’s Objectives:
Students will be able to
a) define a couple, and,
b) determine the moment of a couple.
In-Class activities:
Check homework, if any
Reading quiz
Applications
Moment of a Couple
Concept quiz
Group problem solving
Attention quiz

17. READING QUIZ
1. In statics, a couple is defined as __________ separated by a perpendicular distance.
A) two forces in the same direction.
B) two forces of equal magnitude.
C) two forces of equal magnitude acting in the same direction.
D) two forces of equal magnitude acting in opposite directions.
2. The moment of a couple is called a _________ vector.
A) free B) spin
C) romantic D) sliding

18. APPLICATIONS
A torque or moment of 12 N · m is required to rotate the wheel. Which one of the two grips of the wheel above will require less force to rotate the wheel?

19. APPLICATIONS (continued)
The crossbar lug wrench is being used to loosen a lug net. What is the effect of changing dimensions a, b, or c on the force that must be applied?

20. MOMENT OF A COUPLE
A couple is defined as two parallel forces with the same magnitude but opposite in direction separated by a perpendicular distance d.
The moment of a couple is defined as
MO = F d (using a scalar analysis) or as
MO = r  F (using a vector analysis).
Here r is any position vector from the line of action of –F to the line of action of F.

21. MOMENT OF A COUPLE (continued)
The net external effect of a couple is that the net force equals zero and the magnitude of the net moment equals F d
Since the moment of a couple depends only on the distance between the forces, the moment of a couple is a free vector. It can be moved anywhere on the body and have the same external effect on the body.
Moments due to couples can be added using the same rules as adding any vectors.

22. Given: Two couples act on the beam and d equals 8 ft.
EXAMPLE
Given: Two couples act on the beam and d equals 8 ft.
Find: The resultant couple
Plan:
1) Resolve the forces in x and y directions so they can be treated as couples.
2) Determine the net moment due to the two couples.
Statics:The Next Generation Mehta & Danielson Lecture Notes for Sections

23. Solution:
The x and y components of the top 60 lb force are:
(4/5)(60 lb) = 48 lb vertically up
(3/5)(60 lb) = 36 lb to the left
Similarly for the top 40 lb force:
(40 lb) (sin 30°) up
(40 lb) (cos 30°) to the left
The net moment equals to
+ M = -(48 lb)(4 ft) + (40 lb)(cos 30º)(8ft)
= = ft·lb

24. GROUP PROBLEM SOLVING - SCALAR
Given: Two couples act on the beam. The resultant couple is zero.
Find: The magnitudes of the forces P and F and the distance d.
PLAN:
1) Use definition of a couple to find P and F.
2) Resolve the 300 N force in x and y directions.
3) Determine the net moment.
4) Equate the net moment to zero to find d.

25. Solution:
From the definition of a couple
P = 500 N and
F = 300 N.
Resolve the 300 N force into vertical and horizontal components. The vertical component is (300 cos 30º) N and the horizontal component is (300 sin 30º) N.
It was given that the net moment equals zero. So
+ M = - (500)(2) + (300 cos 30º)(d) + (300 sin 30º)(0.2) = 0
Now solve this equation for d.
d = (1000 – 60 sin 30º) / (300 cos 30º) = 3.96 m

26. EXAMPLE
The rigid structural member is subjected to a couple consisting of the two 100-N forces. Replace this couple by an equivalent couple consisting of the two forces (P) and (-P), each of which has a magnitude of 400 N. Determine the proper angle Ɵ.

27. The forces P and -P produce a counterclockwise couple
The original couple is counterclockwise when the plane of the forces is viewed from above, and its magnitude is
M = 100(0.1) =10 N. m
The forces P and -P produce a counterclockwise couple
M = 400(0.040) Cos Ɵ
10 = (400)(0.040) Cos Ɵ

28. . Its moment and force equals _____ N·m. &_____ N.
ATTENTION QUIZ
1. A couple is applied to the beam as shown. Its moment equals _____ N·m.
A) 50 B) 60
C) 80 D) 100
50 N 1m 2m 5 3 4
2. Replace the horizontal 80-lb force acting on the lever by an equivalent system consisting of a force at O and a couple.
. Its moment and force equals _____ N·m. &_____ N.
A) 624, 50 B) 600, 80
C) 624, 80 D) 100, 20