chapter 04 part 1
TRANSCRIPT
7/28/2019 Chapter 04 Part 1
http://slidepdf.com/reader/full/chapter-04-part-1 1/25
Force System Resultants4
Engineering Mechanics:
Statics in SI Units, 12e
Copyright © 2010 Pearson Education South Asia Pte Ltd
7/28/2019 Chapter 04 Part 1
http://slidepdf.com/reader/full/chapter-04-part-1 2/25
Copyright © 2010 Pearson Education South Asia Pte Ltd
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 forcehaving a specified location
7/28/2019 Chapter 04 Part 1
http://slidepdf.com/reader/full/chapter-04-part-1 3/25
Copyright © 2010 Pearson Education South Asia Pte Ltd
Chapter Outline
1. Moment of a Force – Scalar Formation
2. Cross Product
3. Moment of Force – Vector Formulation
4. Principle of Moments5. Moment of a Force about a Specified Axis
6. Moment of a Couple
7. Simplification of a Force and Couple System
8. Further Simplification of a Force and Couple System
9. Reduction of a Simple Distributed Loading
7/28/2019 Chapter 04 Part 1
http://slidepdf.com/reader/full/chapter-04-part-1 4/25
Copyright © 2010 Pearson Education South Asia Pte Ltd
4.1 Moment of a Force – Scalar Formation
• Moment of a force about a point or axis – a measure
of the tendency of the force to cause a body to rotate
about the point or axis
• Torque –
tendency of rotation caused by Fx or simplemoment (Mo) z
7/28/2019 Chapter 04 Part 1
http://slidepdf.com/reader/full/chapter-04-part-1 5/25
Copyright © 2010 Pearson Education South Asia Pte Ltd
4.1 Moment of a Force – Scalar Formation
Magnitude
• For magnitude of MO,
MO = Fd (Nm)
where d = perpendicular distancefrom O to its line of action of force
Direction• Direction using “right hand rule”
7/28/2019 Chapter 04 Part 1
http://slidepdf.com/reader/full/chapter-04-part-1 6/25
Copyright © 2010 Pearson Education South Asia Pte Ltd
4.1 Moment of a Force – Scalar Formation
Resultant Moment
• Resultant moment, MRo = moments of all the forces
MRo = ∑Fd
7/28/2019 Chapter 04 Part 1
http://slidepdf.com/reader/full/chapter-04-part-1 7/25Copyright © 2010 Pearson Education South Asia Pte Ltd
Example 4.1
For each case, determine the moment of the force about
point O.
7/28/2019 Chapter 04 Part 1
http://slidepdf.com/reader/full/chapter-04-part-1 8/25Copyright © 2010 Pearson Education South Asia Pte Ltd
Solution
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.
)(.200)2)(100()( CW mN mN M a o
7/28/2019 Chapter 04 Part 1
http://slidepdf.com/reader/full/chapter-04-part-1 9/25Copyright © 2010 Pearson Education South Asia Pte Ltd
Solution
)(.5.37)75.0)(50()( CW mN mN M b o
)(.229)30cos24)(40()( CW mN mmN M c o
7/28/2019 Chapter 04 Part 1
http://slidepdf.com/reader/full/chapter-04-part-1 10/25Copyright © 2010 Pearson Education South Asia Pte Ltd
Solution
)(.4.42)45sin1)(60()( CCW mN mN M d o
)(.0.21)14)(7()( CCW mkN mmkN M e o
7/28/2019 Chapter 04 Part 1
http://slidepdf.com/reader/full/chapter-04-part-1 11/25Copyright © 2010 Pearson Education South Asia Pte Ltd
4.2 Cross Product
• Cross product of two vectors A and B yields C, which
is written as
C = A X B
Magnitude• Magnitude of C is the product of
the magnitudes of A and B and
sine of angle θ
• For angle θ, 0° ≤ θ ≤ 180° C = AB sinθ
7/28/2019 Chapter 04 Part 1
http://slidepdf.com/reader/full/chapter-04-part-1 12/25Copyright © 2010 Pearson Education South Asia Pte Ltd
4.2 Cross Product
Direction
• Vector C has a direction that is perpendicular to the
plane containing A and B such that C is specified by
the right hand rule• Expressing vector C when
magnitude and direction are known
C = A X B = ( AB sinθ)uC
7/28/2019 Chapter 04 Part 1
http://slidepdf.com/reader/full/chapter-04-part-1 13/25
Copyright © 2010 Pearson Education South Asia Pte Ltd
4.2 Cross Product
Laws of Operations
1. Commutative law is not valid
A X B ≠ B X A
Rather,
A X B = - B X A
• Cross product B X A yields a
vector opposite in direction to C
B X A = -C
7/28/2019 Chapter 04 Part 1
http://slidepdf.com/reader/full/chapter-04-part-1 14/25
Copyright © 2010 Pearson Education South Asia Pte Ltd
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 )
• Proper order of the cross product must be maintained
since they are not commutative
7/28/2019 Chapter 04 Part 1
http://slidepdf.com/reader/full/chapter-04-part-1 15/25
Copyright © 2010 Pearson Education South Asia Pte Ltd
4.2 Cross Product
Cartesian Vector Formulation
• A more compact determinant in the form a special
utility for a cross-product in Cartesian vector form
z y x
z y x
B B B
A A A
k ji
B X A
7/28/2019 Chapter 04 Part 1
http://slidepdf.com/reader/full/chapter-04-part-1 16/25
Copyright © 2010 Pearson Education South Asia Pte Ltd
4.3 Moment of Force - Vector Formulation
• Moment of force F about point O can be expressed
using cross product
MO = r X F
Magnitude
• For magnitude of cross product,
M O = rF sinθ
7/28/2019 Chapter 04 Part 1
http://slidepdf.com/reader/full/chapter-04-part-1 17/25
Copyright © 2010 Pearson Education South Asia Pte Ltd
4.3 Moment of Force - Vector Formulation
Direction
• Direction and sense of MO are determined by right-
hand rule
*Note:- “curl” of the fingers indicates the sense of rotation
- Maintain proper order of r and F since cross product
is not commutative
7/28/2019 Chapter 04 Part 1
http://slidepdf.com/reader/full/chapter-04-part-1 18/25
Copyright © 2010 Pearson Education South Asia Pte Ltd
4.3 Moment of Force - Vector Formulation
Principle of Transmissibility
• For force F applied at any point along the line of action
of the force, moment about O is MO = r x F
• F has the properties of a sliding vector, thus
MO = r 1 X F = r 2 X F = r 3 X F
7/28/2019 Chapter 04 Part 1
http://slidepdf.com/reader/full/chapter-04-part-1 19/25
Copyright © 2010 Pearson Education South Asia Pte Ltd
4.3 Moment of Force - Vector Formulation
Cartesian Vector Formulation
• For force expressed in Cartesian form,
• With the determinant expended,
MO = (r y F z – r z F y )i
– (r x F z - r z F x ) j + (r x F y – y F x )k
z y x
z y xO
F F F
r r r
k ji
F X r M
7/28/2019 Chapter 04 Part 1
http://slidepdf.com/reader/full/chapter-04-part-1 20/25
Copyright © 2010 Pearson Education South Asia Pte Ltd
4.3 Moment of Force - Vector Formulation
Resultant Moment of a System of Forces
• Resultant moment of forces about point O can be
determined by vector addition
MRo = ∑(r x F)
7/28/2019 Chapter 04 Part 1
http://slidepdf.com/reader/full/chapter-04-part-1 21/25
Copyright © 2010 Pearson Education South Asia Pte Ltd
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.
7/28/2019 Chapter 04 Part 1
http://slidepdf.com/reader/full/chapter-04-part-1 22/25
Copyright © 2010 Pearson Education South Asia Pte Ltd
Solution
Position vectors are directed from point O to each force
as shown.
These vectors are
The resultant moment about O is
m254
m5
k jir jr
OB
OA
mkN604030
304080
254
204060
050
k ji
k jik ji
F r F r F r M OBOAO
7/28/2019 Chapter 04 Part 1
http://slidepdf.com/reader/full/chapter-04-part-1 23/25
Copyright © 2010 Pearson Education South Asia Pte Ltd
4.4 Principles of Moments
• Also known as Varignon’ s Theorem
“Moment of a force about a point is equal to the sum of
the moments of the forces’ components about the point”
• Since F = F1 + F2,
MO = r X F
= r X (F1 + F2)
= r X F1 + r X F2
7/28/2019 Chapter 04 Part 1
http://slidepdf.com/reader/full/chapter-04-part-1 24/25
Copyright © 2010 Pearson Education South Asia Pte Ltd
Example 4.5
Determine the moment of the force about point O.
7/28/2019 Chapter 04 Part 1
http://slidepdf.com/reader/full/chapter-04-part-1 25/25
Copyright © 2010 Pearson Education South Asia Pte Ltd
Solution
The moment arm d can be found from trigonometry,
Thus,
Since the force tends to rotate or orbit clockwise about
point O, the moment is directed into the page.
m898.275sin3 d
mkN5.14898.25 Fd M O