Chap. 4
Force System Resultants
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
4-1-2
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
simple moment (Mo) z
4-1-3
Moment of a Force – Scalar
Formation
Magnitude
„ For magnitude of MO,
MO = Fd (N⋅m)
where d = perpendicular distance
from O to its line of action of force
Direction
„ Direction using “right hand rule”
4-1-4
Moment of a Force – Scalar
Formation
Resultant Moment
„ Resultant moment, MRo = moments of all the
forces
MRo = ∑Fd
4-1-5
Cross Product
„
Cross Product (vector product)
C=A×B
„
Magnitude
C = AB sinθ
„
0° ≤θ ≤180°
Direction
C = ( AB sin θ )u C
C⊥ A
C⊥B
4-1-6
Cross Product
„
Laws of Operation
A× B = −B × A
α ( A × B) = (α A) × B = A × (α B) = ( A × B)α
A × ( B + D) = ( A × B) + ( A × D)
A × ( D + B) = ( A × D) + ( A × B)
4-1-7
Cross Product
„
Cartesian Vector Formulation
i×i = 0
j × i = −k
k ×i = j
i× j = k i×k = − j
j× j = 0
j×k = i
k × j = −i k × k = 0
A × B = ( Ax , Ay , Az ) × ( Bx , B y , Bz )
= ( Ay Bz − Az B y )i + ( Az Bx − Ax Bz ) j + ( Ax B y − Ay Bx )k
or
i
j
A × B = Ax
Bx
k
i
j
k i
j
Ay
Az = Ax
Ay
Az Ax
Ay
By
Bz
By
Bz B x
By
Bx
(+)
(-)
4-1-8
Moment of a Force-Vector
Formulation
MO = r × F (=rB × F)
MO = r Fsinθ= F(r sinθ) = Fd
4-1-9
Moment of a Force-Vector
Formulation
„
Transmissibility of a Force 傳遞性
MO = rA × F = rB × F = rC × F
A,B,C在F的作用線上
4-1-10
Moment of a Force-Vector
Formulation
„
Vector Formulation
i
j
k
M O = r × F = rx
ry
rz
Fx
Fy
Fz
4-1-11
Moment of a Force-Vector
Formulation
„
Resultant Moment of a System of Forces
M RO = Σ(r × F )
4-1-12
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)
4-1-13
Position vectors are directed from point O to each force
as shown.
These vectors are
rA = {5 j} m
rB = {4i + 5 j − 2k } m
The resultant moment about O is
r
M O = ∑ (r × F ) = rA × F + rB × F
i
j
k
i
j
k
5 0 + 4 5 −2
= 0
− 60 40 20 80 40 − 30
= {30i − 40 j + 60k } kN ⋅ m
4-1-14
Principle of Moments –
Varignon’s Theorem
MO = r × F1 + r × F2 + r × F3 + r × F4 + ...
= r × ( F1 + F2 + … )
=r×F
*合力對固定點的力矩
等於各分力對固定點的力矩之總合
4-1-15
p. 133, 4–4. Two men exert forces of F = 400 N and P = 250 N
on the ropes. Determine the moment of each force about A.
Which way will the pole rotate, clockwise or counterclockwise?
4-1-16
4–24. In order to raise the lamp post from the position shown,
force F is applied to the cable. If F = 1000 N, determine the
moment produced by F about point A.
4-1-17
4-1-18
p. 137, 4-36 The wheelbarrow and its contents have a center of
mass at G. If F = 100 N, and the resultant moment produced by
force F and the weight about the axle at A is zero, determine the
mass of the wheelbarrow and its contents.
4-1-19
4-1-20
p. 138, 4-47 The force F = {6i + 8j +10k} N creates a moment about point O
of MO = {-14i + 8j + 2k} N⋅m. If the force passes through a point having an x
coordinate of 1 m, determine the y and z coordinates of the point. Also,
realizing that MO = Fd, determine the perpendicular distance d from point O to
the line of action of F.
4-1-21
Chap. 4-2
Force System Resultants
Moment of a Force About a
Specified Axis
„
1.
2.
3.
My為F對特定軸(y-axis)
的moment，也是Mo在y
軸上的投影(projection)
O為y軸上任意一點，先
求Mo(=r×F)
再以特定軸的單位向量
(此例為j)和Mo 的dot
product求My(= Mo⋅ j)
My = [j ⋅(r×F)] j
4-2-2
Moment of a Force About a
Specified Axis
a-a’ is an arbitrary axis
M a = M O cos θ = M O ⋅ u a
= u a ⋅ (r × F)
u ax
u ay
u az
= rx
ry
rz
Fx
Fy
Fz
Triple scalar product
純量三倍積
(平行六面體的體積)
M a = M a u a = [u a ⋅ (r × F)]u a
4-2-3
p.145 Problem 4-56
Determine the moment produced by force F about segment AB of
the pipe assembly. Express the result as a Cartesian vector.
4-2-4
4-2-5
4-2-6
p.147 Problem 4-65
The A-frame is being hoisted into an upright position by the
vertical force of F = 400 N. Determine the moment of this force
about the y axis when the frame is in the position shown.
4-2-7
4-2-8
4-2-9
Moment of a Couple 力偶
M O = r A × (-F) + r B × (F)
= (r B - r A ) × F
= r AB × F
=r×F
4-2-10
Moment of a Couple
„
Scalar Formulation
M=Fd
„
Vector Formulation
M=r x F
4-2-11
Moment of a Couple
„
Equivalent Couples
等效
„
Resultant Couple
Moment(合力偶 力偶合)
4-2-12
Example 4.12
Determine the couple moment acting on the pipe.
Segment AB is directed 30° below the x–y
plane.
4-2-13
SOLUTION I (VECTOR ANALYSIS)
Take moment about point O,
M = rA × (-250k) + rB × (250k)
= (0.8j) × (-250k) + (0.6cos30ºi
+ 0.8j – 0.6sin30ºk) × (250k)
= {-130j}N.cm
Take moment about point A
M = rAB × (250k)
= (0.6cos30i – 0.6sin30k)
× (250k)
= {-130j}N.cm
4-2-14
SOLUTION II (SCALAR ANALYSIS)
Take moment about point A or B,
M = Fd = 250N(0.5196m)
= 129.9N.cm
Apply right hand rule, M acts in the –j direction
M = {-130j}N.cm
4-2-15
p.155 Problem 76
Determine the required magnitude of force F if the resultant
couple moment on the frame is 200 N⋅M, clockwise.
4-2-16
4-2-17
p.158 Problem 4-92
Determine the required magnitude of couple moments so that the
resultant couple moment is MR = {-300i + 450j - 600k} N⋅m.
4-2-18
4-2-19
p.159 Problem 4-102
If F1 = 500 N and F2 = 1000 N, determine the magnitude and
coordinate direction angles of the resultant couple moment.
4-2-20
4-2-21
4-2-22
Chap. 4-3
Force System Resultants
Simplification of a Force and Couple
System
„
„
„
An equivalent system is when the external effects are
the same as those caused by the original force and
couple moment system
External effects of a system is the translating and
rotating motion of the body
Or refers to the reactive forces at the supports if the
body is held fixed
4-2-2
Equivalent System
„
Point O is On the Line of Action
„
Point O Is Not On the Line of Action
4-2-3
Equivalent System
„
Resultant of a Force and Couple
System
Force Summation : FR = ΣF
Moment Summation : MRO = ΣMC+ΣMo
4-2-4
Ex. 4-15, replace the system by an equivalent resultant force and couple
moment acting on point O.
F1 = −800kN
u CB =
(−0.15,0.1,0)
(−0.15) 2 + 0.12
F 2 = 300 ⋅ u CB = ( −249.6,166.4,0) N
M = (0,−400,300) N ⋅ m
Σ F = ( - 250, 166.4, - 800 ) N
0
M RO = M + r C × F1 + r B × F2
i
j
k
= (0,-400,300) + − 0.15
0.1 1
− 249.6 166.4 0
= (-166.4,-650,300)N ⋅ m
4-2-5
p. 169, 4-116 Replace the force system acting on the pipe
assembly by a resultant force and couple moment at point O.
Express the results in Cartesian vector form.
4-2-6
4-2-7
Further Reduction of a Force and
Couple System
M RO = FR d
4-2-8
Further Simplification of a Force and
Couple System
Concurrent Force System
„ A concurrent force system is where lines of
action of all the forces intersect at a common
point O
FR = ∑ F
4-2-9
Further Simplification of a Force and
Couple System
Coplanar Force System
„ Lines of action of all the forces lie in the same
plane
„ Resultant force of this system also lies in this
plane
4-2-10
Further Simplification of a Force and
Couple System
Parallel Force System
„ Consists of forces that are all parallel to the z
axis
„ Resultant force at point O must also be
parallel to this axis
4-2-11
Further Simplification of a Force and
Couple System
„
Reduction to a Wrench (扳手)
M || (在 F R 方向)
d=
M⊥
FR
M RO = M || + M ⊥
4-2-12
p.180, 4-124. Replace the force and couple moment
system acting on the overhang beam by a resultant force,
and specify its location along AB measured from point A.
4-2-13
4-2-14
4-2-15
p.182, 4-140. Replace the three forces acting on the
plate by a wrench. Specify the magnitude of the force
and couple moment for the wrench and the point P(y, z)
where its line of action intersects the plate.
4-2-16
4-2-17
4-2-18
Reduction of a Simple Distributed
Loading
( x , y)
集中力
形心
centroid concentrated force
4-2-19
Reduction of a Simple Distributed
Loading
„
Magnitude
FR = ∫ w ( x )dx = ∫ dA = A
L
„
A
Location
Σ M O = M RO :
∫ xw ( x )dx = x F R
∫ A xdA ∫ xdA
⇒ x=
=
A
∫ A dA
求形心
first moment
4-2-20
p.189, 4-144. Replace the distributed loading by an
equivalent resultant force and specify its location,
measured from point A.
4-2-21
4-2-22
p.191, 4-152. Wind has blown sand over a platform such that the intensity
3
of the load can be approximated by the function w=(0.5x ) N/m. Simplify
this distributed loading to an equivalent resultant force and specify the
magnitude and location of the force, measured from A.
dA = wdx
FR = ∫ dA =
10
∫
0
1 3
x dx
2
10
⎡1 ⎤
= ⎢ x4⎥
⎣8 ⎦0
= 1250N
FR = 1.25kN
∫ xdA =
10
∫
0
Ans
1 4
x dx
2
10
⎡1
⎤
= ⎢ x5⎥
⎣ 10
⎦0
= 10000N ⋅ m
10000
x=
= 8 . 00 m Ans
1250
4-2-23
p.192, 4-160. The distributed load acts on the beam as shown. Determine
the magnitude of the equivalent resultant force and specify its location,
measured from point A.
4-2-24
4-2-25