Idea Transcript
SKAA 1213 ‐ Engineering Mechanics TOPIC 5
M Moment t and d Couple C l Lecturers: Rosli Anang Dr. Mohd Yunus Ishak Dr. Tan Cher Siang
Moment of a Force • Moment of a force about a point/axis the tendency of the force to cause the body to tendency of the force to cause the body to rotate about the point/axis. • Moment is a vector quantity
Moment of a Force
Moment axis
Mo
1. Scalar Formulation of Moment F d
M = Fd Fd O
Where d is the perpendicular distance from the axis of point O to the action of the force F).
Direction of force : specified by using the right hand rule. rule
O
Moment Arm F
F θ
d l
l
θ
O
O
M = Fd Fd O
d
Note: d = l sin θ F
θ=
0o l
O
θ = 90o
F
d=0
l d=l O
d
Example 1 Determine the moment of the 70N force about point A. [ Answer : (a)M 2800Nmm (b) M =2704.6Nmm ] p A
(a)
A
(b)
Example 2 Example 2 Determine the moment of the 70N and 60N 3611 4 Nm ] forces about point A [ Answer : M = 3611.4 forces about point A. A
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Example 3 Determine the moment of each the force about Determine the moment of each the force about point O. [ Answer : MO1 = 200kNm, MO2 =70kNm , MO3 = 50kNm, MO4 = 70kNm , MO5 = 125kNm ]
Example 4 Determine the moments of the 40kN force about points A, B, C and D. [ Answer : MA = 0 MB= 48kNm MC= 20kNm MD= 20kNm ]
Resultant Moment of Coplanar Forces • determined by total up the moments of all the forces algebraically the forces algebraically.
+ MRO = Fd z
d2
F2
F1 d3
F3
O d 1
The counterclockwise curl written along the written along the equation indicates that, the moment of any force will be positive if it is will be positive if it is directed along the +z axis.
Example 5 Determine the moment of the three forces about point O. [Answer : M = -120kNm ] O
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Example 6 Determine the moments of the three forces 85kNm , M = 125Nm ] about point B and C [Answer : M = -85kNm about point B and C. B
C
Cross Product
C = A x B
cross product of two vectors A and is written as C = A and is written as =AxB
C = AB sin θ
θ
Magnitude of C = AB sin θ Magnitude of C AB sin θ
A B
The direction of vector C The direction of vector C is is perpendicular to the plane A & B such that C & B such that C is specified by is specified by the right‐hand rule.
C uC
A
θ
C = AB sin θ
B
Cross Product - Laws of Operations
Commutative law: A B ≠ B x A x B A A x B = ‐B x A Multiplication by a scalar: ) ( ) B = A x ((aB) = (A x ) ( B)a ) a ((A x B) = (aA) x Distributive law: A x (B + D) = (A x B) + (A x D) D) (A B) (A D)
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Cross Product of the Cartesian unit vectors In a similar manner In a similar manner,
z k = i x j i x
i x j = k
i x k = ‐j
i x i = 0
j x k = i k i k x i = j
j x i = ‐kk k x j = ‐i
jj x j = 0 xj 0 k x k = 0
i
j y
+ j
‐
This diagram is helpful for obtaining the result of the result of k cross products of unit vectors unit vectors
Tips : Apply right hand rule 14
C Cross product of two vectors d t ft t A and A d B. B D t Determinant form: i tf i
A x B =
j k
Ax Ay Az Bx By Bz
Moment of a Force 2. Vector Formulation of Moment p • The moment of F about point O discussed earlier, can be expressed using the vector cross‐product; MO = r x F Where r represent the position vector drawn from O to any point lying on the line of action of F.
Magnitude: MO = rF sin θ = F(r sin θ) = Fd
F
Moment axis
MO
r
O
Direction: Apply Apply right‐hand rule at the right hand rule at the intersection point of the tails of extended r and F
Moment axis
Note that the moment axis is perpendicular to the plane containing r and F
MO
r
θ r
θ
d O
F
r is treated as a sliding vector t
Example 7
A force F=90N directed from C to D. Determine the magnitude of the moment created about pp p , the support at point A, and their coordinate direction angles. [ Answer :M = 253.4Nm α = 126.9 β = 81 = 141.6 A
rC
rD
rD= 3j + 0.6k
o
o
o
]
Resultant moment of a System of Forces F 3
F2
z MRO r2
r3
y
O r1 F1
x
MRO = ∑ = ∑ (r x F)
Example Three forces acting on the rod. Determine the Three forces acting on the rod Determine the resultant moment about O and the coordinate di direction angles. Given i l Gi F1= 20i 20i + 80k, F 80k F2= 40i 40i + 30j ‐ 25k and F3= ‐35i + 50j ‐ 15k. [ Answer :MRO= 410Nm α = 120.8o β = 117o = 43o ]
Principle of Moments (Varignon’s theorem) The moment of a force about a point is equal to the summation of the moments of the force’s f th t f th f ’ components about the point. Proof: M = r x F1 + r x F2 = r x (F1 + F2) =rxF = r x
F1 F
F2 r
O
O F = F1 + F2
Useful to determine the moment arms of the force’s Useful to determine the moment arms of the force’s components than the moment arm of the force itself.
Moment of a Force about a Specified Axis • Can be solved by scalar Can be solved by scalar or vector or vector analysis. analysis z b MO
5 3
O
4
My
0.5 m
y 0.3 m
0.4 m
x
F = 20 N
In some situations we need the component of the moment along a specified axis that passes through the point. Let say component of MO about y axis, My .
1. Scalar Analysis z b MO
5 3
O
4
My
05m 0.5
y 0.3 m
0.4 m
x
F = 20 N
a) Mo = 20(0.5) = 10 Nm (direction defined by RH Rule about the Ob axis) Rule about the Ob My = (3/5)(10) = 6 Nm (component method) b) My = 20(0.3) = 6 Nm (direct method)
• If the line of action of a force F is perpendicular to any specified axis aa, then; Ma= Fd Fda
where da is the perpendicular distance where is the perpendicular distance from the force line of action to the axis.
• The The direction is determined from the thumb direction is determined from the thumb of the Right Hand when the fingers are curled in accordance with the direction of curled in accordance with the direction of rotation. • A A force will NOT force will NOT contribute a moment about a specified contribute a moment about a specified axis if the force line of action is parallel to the axis or its line of action passes through the axis.
Example What are the values moment about the x,y, z axes. [Answer : MOx=13 Nm MOy== 59 Nm MOz= -32 32 Nm] z
6
10
x
7
y
Example Example Find the Moment about a specified axis using Scalar notation method [Answer Scalar notation method. [A : M = 48Nm 48N M = 0Nm 0N , M = 0Nm] 0N ] Ox
(0,0,8)
O (0,6,0)
oy
oz
Example Example Find the Moment about a specified axis using 0Nm M = 50 Nm ] Scalar notation method [Answer : M = 0Nm , M = 0Nm, Scalar notation method. Ox
O
oy
oz
(0,5,0)
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2. Vector Analysis z b MO O
x
rA
My
ua = j y 0.3 m
MO = rA x F = (0.3i + 0.4j) X (‐20k) = (‐8i + 6j) Nm
0.4 m
F = (-20k) N
The component of this moment along the y axis is then determined from the dot product i i th d t i d f th d t d t Since the unit vector of this axis is ua=j, then My = MO ∙ua = (‐8i + 6j) ∙ j = 6 Nm
Vector analysis is advantages to find moment of Vector analysis is advantages to find moment of force about an axis when the force components or the moment arms are difficult to determine. a b MO = r x F
θ
Ma
y
O b’ ua
H t How to get M t Ma ?
r
1. Find MO = r x F 2. Ma = MO cos θ = MO∙ua Ma = (r x F) ∙ u = (r x F) ∙ ua = u = ua∙ ( r x F) ∙ ( r x F)
a’ F
In vector algebra , combination of dot and cross product yielding the scalar d d t i ldi th l Ma is called the triple scalar product.
The triple scalar product : i j k Ma = (u = (uax i + u + uay j + u + uaz k) ∙ k) ∙ rx ry rz Fx Fy Fz
uax uay uaz = r r y rz x Fx Fy Fz Once Ma is determined, M Once M is determined Ma as a Cartesian vector ; as a Cartesian vector ; Ma = Maua = [ua∙ (r x F)]ua
To find the resultant of a series of forces about the axis aa’,, the moment components of each force are axis aa the moment components of each force are added together algebraically, since the component lies along the same axis. Ma = ∑[ua∙(r x F)] = ua∙ ∑(r x F)
Example The force F The force F= ‐35i 35i + 50j + 50j ‐ 15k acts at C. Determine acts at C Determine the moment of this force about x and a axes. [ Answer : Mx= -210Nm 210Nm Ma = -161Nm 161Nm ]
Couples and Couple Moments • Definition : two parallel forces that have the g but opposite direction, and pp , same magnitude separated by a perpendicular distance, d. • The moment produced is called couple moment. Th t d di ll d l t ∑F = 0, the only effect is tendency of rota , y y on. •∑ d
B
-F F
-F r
A
F F
2D
rB
rA O
3D
Moment of a Couple Determination of moments of couple forces Determination of moments of couple forces about any point : ‐F B
r A
about A: M = r X F
F
about O: M = rB X (F) + rA X (‐F)
rB
rA 3D O
This indicates that a couple moment is a free vector. It This indicates that a couple moment is a free vector. It can act at any point since M only depends upon the position vector r, not rA and rB.
2D
Scalar Formulation: M =Fd
‐F
d
F
Vector Formulation: M = r x F
3D B
‐F
r A
F
Properties of Moment of a Couple 1. The couple moment is unaffected by the pivot location *Couples at the same position for example below.
MA=20(0.3)+20(1.7) = 40Nm 40N
MA=20(6)‐20(4) = 40Nm
2. A couple can be shifted and still have the same 2 A couple can be shifted and still have the same moment about a given point.
Couple at different position & moments calculated at the same point. MA=30(2) = 60Nm
• Equivalent Couples Equivalent Couples Two couples which produce the same moment lie either in the same plane or in planes parallel lie either in the same plane or in planes parallel to each other. The direction of the couple moments is the same and is perpendicular to the moments is the same and is perpendicular to the parallel planes.
• Resultant Couple Moment Since couple moments are free vectors Since couple moments are free vectors they can they can be applied at any point on a body and added vertically. vertically
M1 M2
Two set of couple forces MR
M1
M2 P
Two couple moments
Moved to any arbitrary point and added to obtain point and added to obtain resultant couple moment MR= M1 + M2
Example Replace the forces acting on the structure by an equivalent resultant force and couple moment at A. [ Answer: M = = -46.6Nm () ] RA
Example Example Determine the moment of the couple on the [ Answer : M= -221.5Nm ] member shown. b h
d
Equivalent System
Replacing system of forces and couple moments acting on a body by a single force and couple acting on a specified point O that produce the same external effects of translation and rotation. C Case 1: 1 Point O Is On the Line of Action P i t O I O th Li f A ti A O
F F
=
O -F
A
F
A F
=
O
Case 2: Point O Is Not On the Line of Action C 2 P i t O I N t O th Li f A ti M=rxF F O
F A
F
F
=
A
O
=
A
O P
-F F
*Note: Since couple is a free vector , it may be applied at any point
Resultants of a Force System MRO = ∑M ∑ C + ∑M ∑ O FR = ∑F M2 = r2 x F2
F1
F2 r2
r1 O
=
F2 MC
MC
F1 O
MRO
=
M1 = r1 x F1
FR = F1 + F2 θ O
NOTE: • Both the magnitude & direction of FR are independent of the location of O, however, independent of the location of O however • MRO depends on the location of O since the moment M1 & M & 2 are determined by using the d db h position vectors r1 & r2. • MRO is a free vector and can act at any point on the body.
Example Determine the magnitude, direction and Determine the magnitude direction and location of a resultant force which is equivalent to the given system of forces measured to the given system of forces measured horizontally from A. [ Answer : F = 272N() θ = 68.4 d = 0.18m ] R
o
Example p Determine the magnitude and direction of a resultant force equivalent force system and locate its point of application. [ Answer : FR = -1190N () ,y = 2.84m x = 1.24m]
Simplification to a Single Force System Consider a special case for which the system of forces and couple moments reduces at point O of the resultant force FR and couple MR which are perpendicular to each other. M1
F2 r2
O r3 M2
FR
F1
r1
b
a
=
O a
a
b
MRO
= F3
FR
b
dd
P O
a
b d =
MRO FR
If the system of forces is either concurrent, If th t ff i ith t coplanar, or parallel, it can be reduced (as in the above case), to a single resultant force F b ) t i l lt t f FR. This is because in each of these cases FR and MR will always be perpendicular to each other when y p yp point. the force system is simplified at any
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1 Concurrent Force System 1.Concurrent Force System
F1
FR =∑F F2
= F3
2.Coplanar Force System y F3 Couple p moments are ┴ to plane of forces
r3 r1 F1
y M2
F2
x
O M1
MRO = ∑M + ∑r x F
=
r2
y d= M /F RO R x
O
d
= O
FR= ∑F
FR= ∑F
x
3.Parallel Force System ll l z Couple p moments are ┴ to the forces M1
F2
F1
x
r1
FR= ∑F
z r2
O r3
MRO = ∑M + ∑r x F
F3
y
FR= ∑F
= O M2
x
z
y
=
O x
y
Example Determine the magnitude and location of the equivalent resultant force acting on the equivalent resultant force acting on the beam. beam. _ [ Answer : FR = -1190N () , x = 3m ]
dA = w dx = 50x2 dx
Example Determine the magnitude and location of the Determine the magnitude and location of the equivalent resultant force acting on the beam. _ [ Answer : FR = 140KN , x = 1.86m ]
Reduction of a simple Distributed p Loading p x
p=p(x)
L y a/2
a/2
Uniform pressure along one axis on a flat rectangular axis on a flat rectangular surface. The load intensity is of the load represented by o t e oad ep ese ted by the arrows form a system of parallel forces, infinite in p , numbers, each acting on a separate differential area. p
w=w(x) w w(x)
dF dA
x
dx L
Load function, p = p(x) [pressure uniform in y [p f y axis] Multiply pyp p=p(x) p( ) with the width a, we obtain; p( ) w= p(x) a = w x
This loading function is a measure of load di t ib ti distribution along the line y=0 which is the plane l th li 0 hi h i th l of symmetry of the loading. Note: it is load per unit length. it l th 55
FR
A
C O x L
In a system of coplanar parallel forces, the load intensity can be represented by w = w(x) This system of forces can be This system of forces can be simplified to a single force FR and its location x can be and its location x can be specified.
Magnitude of Resultant Force w=w(x)
For an elemental F l t l length l th dx d as shown in the diagram, the force acting is; dF = w(x) dx = dA [shaded area]
dF dA
x
L
dx
For entire length; ∑F: FR = ∫w(x) ∫ ( ) dx = ∫ dA = A +↓FR = ∑
Hence, the magnitude of the resultant force is equal to the total area A under the loading equal to the total area A under the loading diagram w = w(x).
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Location of Resultant Force MRO = MO Equating the moment of the F Equating the moment of the FR and the force distribution about O.
dA O x
dx L
dF produces d a momentt off x dF =x w(x) dx about O.
+ MRO = SMO: x FR = ∫ x w(x) dx L Solving for x; Solving for x;
FR
A
∫L x w(x) dx
x = ∫ w(x) dx L
∫A x dA
= ∫ dA A
w=w(x)
dF
C O x L
Location of Resultant Force w=w(x)
dF
This eqn represents the x coordinate for the geometric di f h i center (centroid) of the area under the distributed loading d h di ib d l di diagram w(x). The resultant force has a line of p g action which passes through the centroid C fo the area defined by the distributed loading diagram g g w(x).
dA O x
dx L
dF produces d a momentt off x dF =x w(x) dx about O.
FR
A
C O x L