M

Engineering Mechanics: Statics in SI Units, 12e 4

Force System Resultants

Copyright © 2010 Pearson Education South Asia Pte Ltd

Force system resultants 力系合成 Chapter Outline 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9

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 Moment of a Force – Scalar Formation 一力的力矩 -- 純量法 •

•

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 Moment of a Force – Scalar Formation Magnitude •

For magnitude of MO, MO = Fd (Nm) where d = perpendicular distance from O to its line of action of force

Direction •

Direction using “right hand rule”


Example (4-1) For each case, determine the moment of the force about point O.


Solution ( a ) M o  (100 N )( 2 m )  200 N .m ( CW )


Solution ( b )M o  ( 50 N )( 0 . 75 m )  37 . 5 N .m (CW )


Solution 

( c )M o  ( 40 N )( 4 m  2 cos 30 m )  229 N .m (CW )


Solution 

( d )M o  ( 60 N )( 1 sin 45 m )  42 . 4 N .m (CCW )


Solution ( e )M o  ( 7 kN )( 4 m  1m )  21 . 0 kN .m (CCW )


4.1 Moment of a Force – Scalar Formation Resultant Moment 力矩合成（合力矩） •

Resultant moment, MRo = moments of all the forces

MRo = ∑Fd


TEST (4-1) • Determine the resultant moment (合力矩) produced by the forces about point O.


4.2 Cross Product 向量積（叉積） • Cross product of two vectors A and B yields C, which is written as

C=AXB


4.2 Cross Product Magnitude • Magnitude of C is the product of the magnitudes of A and B • For angle θ, 0° ≤ θ ≤ 180° C = AB sinθ

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 Copyright © 2010 Pearson Education South Asia Pte Ltd

4.2 Cross Product • Expressing vector C when magnitude and direction are known

C = A X B = (AB sinθ)uC


4.2 Cross Product Laws of Operations 1. Commutative law is not valid AXB≠BXA Rather, AXB=-BXA Cross product A X B yields a vector opposite in direction to C B X A = -C 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 AX(B+D)=(AXB)+(AXD) • Proper order of the cross product must be maintained since they are not commutative


4.2 Cross Product • Cartesian vector formulation

i j  k

ik  j

jxi=

jxj =

jxk=i

kxi=

kxj =

kxk=

ii  0


4.2 Cross Product • Cartesian vector formulation

ii  0

i j  k

j x i = -k

jxj =0

jxk=i

kxi=j

k x j = -i

kxk=0

ik  j


4.2 Cross Product • A = Ax i + Ay j + Az k • B = Bx i + By j + Bz k • AxB= • (Ay Bz- Az By) i - (Ax Bz- Az Bx) j + (Ax By- Ay Bx) k


4.2 Cross Product Cartesian Vector Formulation • Use C = AB sinθ on pair of Cartesian unit vectors • A more compact determinant in the form as    i j k   A  B  Ax A y Az Bx

By

Bz

A = Ax i + Ay j + Az k B = Bx i + By j + Bz k Copyright © 2010 Pearson Education South Asia Pte Ltd

4.2 Cross Product • Each component can be determined using 2  2 determinants.


TEST (4-2) • A= 1i+3j+ 2k • B = -40 i -20 j + 40 k

• AxB=?


TEST (4-2) Ans

  A B 

  A  B  Ax

 j

 k

Ay

Az

Bx

By

Bz

 i

• A= 1i+3j+ 2k • B = -40 i -20 j + 40 k  i

 j

 k

1

3

2

 40

 20

40

 [ 3 ( 40 )  2 (  20 )] i  [1( 40 )  2 (  40 )] j  [1(  20 )  3 (  40 )] k  160 i  120 j  100 k


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, MO = rF sinθ • Treat r as a sliding vector. Since d = r sinθ, MO = rF sinθ = F (rsinθ) = Fd 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 注意 r 跟 F 順序不可調換


4.3 Moment of Force - Vector Formulation Principle of Transmissibility 可傳遞性原理 • For force F applied at any point A, moment created about O is MO = rA x F • F has the properties of a sliding vector, thus MO = r1 X F = r2 X F = r3 X F


4.3 Moment of Force - Vector Formulation Cartesian Vector Formulation 卡式座標向量公式 • For force expressed in Cartesian form,    M O  r x F  rx

 j

 k

ry

rz

Fx

Fy

Fz

 i

• With the determinant expended, MO = (ryFz - rzFy) i - (rxFz - rzFx) j + (rxFy - ryFx) k 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)


Example (4-3) Two forces act on the rod. Determine the resultant moment they create about the flange at O. Express the result as a Cartesian vector.


Solution Position vectors are directed from point O to each force as shown. These vectors are r A  5 j  m rB  4 i  5 j  2 k  m

The resultant moment about O is  MO  

 r  F   r

A

 F1  rB  F 2

i

j

k

i

j

k

0

5

0  4

5

2

 60

40

20

40

 30

80

 30 i  40 j  60 k  kN  m Copyright © 2010 Pearson Education South Asia Pte Ltd

TEST (4-3) • Determine the moment of force F about point O. Express the resultant as a Cartesian vector.





Force system resultants 力系合成 Chapter Outline 4.1 Moment of a Force – Scalar Formation 4.2 Cross Product 4.3 Moment of Force – Vector Formulation

4.4 Principle of Moments 4.5 Moment of a Force about a Specified Axis 4.6 Moment of a Couple 4.7 Simplification of a Force and Couple System 4.8 Further Simplification of a Force and Couple System 4.9 Reduction of a Simple Distributed Loading Copyright © 2010 Pearson Education South Asia Pte Ltd

4.4 Principles of Moments • Also known as Varignon’s Theorem 瓦希農定理 “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


Example (4-4) Determine the moment of the force about point O.


Solution The moment arm d can be found from trigonometry, d  3  sin 75   2 . 898 m

Thus,

M O  Fd  5  2 . 898   14 . 5 kN  m

Since the force tends to rotate or orbit clockwise about point O, the moment is directed into the page.


Solution II Considering counterclockwise moments as positive, and applying the principle of moments, we have

(CCW)＋MO ＝－Fxdy－Fydx ＝ – (5 cos45 kN) (3 sin30 m) – (5 sin45 kN) (3 cos30 m) ＝ – 14.5 kN‧m ＝ 14.5 kN‧m (CW) Copyright © 2010 Pearson Education South Asia Pte Ltd

Solution III The x and y axes can be set parallel and perpendicular to the rod’s axis. Here Fx produces no moment about point O since its line of action passes through this point. Therefore,

(CCW)＋MO ＝－Fydx ＝－(5 sin75 kN)(3 m) ＝－14.5 kN‧m ＝14.5 kN‧m (CW)


TEST (4-4) • Determine the moment of the force about point O.

F4-6 Copyright © 2010 Pearson Education South Asia Pte Ltd

TEST (4-4) Ans •

+ Mo = + 500 sin45o (3+3 cos45o) - 500 cos45o (3 sin45o) = 1060.7 (N.m) Fy= 500 sin45o

Fx= 500 cos45o dy= 3 sin45o

dx= 3+3 cos45o Copyright © 2010 Pearson Education South Asia Pte Ltd

4.5 Moment of a Force about a Specified Axis 對特定軸的力矩 • For moment of a force about a point, the moment and its axis is always perpendicular to the plane

• A scalar or vector analysis is used to find the component of the moment along a specified axis that passes through the point


4.5 Moment of a Force about a Specified Axis Scalar Analysis 純量法 • According to the right-hand rule, My is directed along the positive y axis • For any axis, the moment is M

y

 Fd

y

• Force will not contribute a moment if force line of action is parallel or passes through the axis Copyright © 2010 Pearson Education South Asia Pte Ltd

4.5 Moment of a Force about a Specified Axis Vector Analysis 向量法 • For magnitude of MA, MA = MO cosθ = MO·ua where ua = unit vector • In determinant form,  M

u ax

a

    u a  ( r  F )  rx

Fx

u ay

u az

ry

rz

Fy

Fz


Example (4-5) Determine the moment produced by the force F which tends to rotate the rod about the AB axis.


Solution Unit vector defines the direction of the AB axis of the rod, where uB 

 rB



0 . 4 i  0 . 2 j  0 .4  0 .2 2

rB

 0 . 8944 i  0 . 4472 j

2

For simplicity, choose rD rD  0 . 6 i m

The force is F   300 k  N


Solution -- Example 4.8


TEST (4-5) • Determine the magnitude of the moment of the 200-N force about the x axis.


4.6 Moment of a Couple 力偶矩 • Couple 力偶 – two parallel forces – same magnitude but opposite direction – separated by perpendicular distance d • Resultant force = 0 • Tendency to rotate in specified direction • Couple moment 力偶矩 = sum of moments of both couple forces about any arbitrary point


4.6 Moment of a Couple Scalar Formulation 純量公式 • Magnitude of couple moment M = Fd • Direction and sense are determined by right hand rule • M acts perpendicular to plane containing the forces


4.6 Moment of a Couple Vector Formulation 向量公式 •

For couple moment, M=rXF

• If moments are taken about point A, moment of –F is zero about this point • r is crossed with the force to which it is directed


4.6 Moment of a Couple Equivalent Couples 等效力偶 • 2 couples are equivalent if they produce the same moment • Forces of equal couples lie on the same plane or plane parallel to one another


4.6 Moment of a Couple Resultant Couple Moment 合力偶矩 • Couple moments are free vectors and may be applied to any point P and added vectorially • For resultant moment of two couples at point P, MR = M1 + M2 • For more than 2 moments, MR = ∑(r X F)


Example (4-6) Determine the couple moment acting on the pipe. Segment AB is directed 30° below the x–y plane.


SOLUTION I (VECTOR ANALYSIS) Take moment about point O, M = rA X (-250k) + rB X (250k) = (0.8j) X (-250k) + (0.6 cos30ºi+ 0.8j – 0.6 sin30ºk) X (250k) = {-130j} N.m Take moment about point A M = rAB X (250k) = (0.6cos30° i – 0.6sin30° k) X (250k) = {-130j}N.m Copyright © 2010 Pearson Education South Asia Pte Ltd

SOLUTION II (SCALAR ANALYSIS) Take moment about point A or B, M = Fd = 250N (0.5196m) = 129.9 N.m Apply right hand rule, M acts in the –j direction M = {-130j} N.m


TEST (4-6) • Determine the couple moment acting on the pipe assembly and express the result as a Cartesian vector.





Force system resultants 力系合成 Chapter Outline 4.1 4.2 4.3 4.4 4.5 4.6

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

4.7 Simplification of a Force and Couple System 4.8 Further Simplification of a Force and Couple System 4.9 Reduction of a Simple Distributed Loading Copyright © 2010 Pearson Education South Asia Pte Ltd

4.7 Simplification of a Force and Couple System • An equivalent system (等效系統) is when the external effects (外部效應) 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 (平移及旋轉運動) of the body – Or refers to the reactive forces (反作用力) at the supports if the body is held fixed


4.7 Simplification of a Force and Couple System • Equivalent resultant force acting at point O (作用於O點) and a resultant couple moment is expressed as FR 

M 

F

R O



M

O



M

• If force system lies in the x–y plane and couple moments are perpendicular to this plane,  FR x   Fx

FR  y M R

F  M 

O

y

O



M


Example (4-7) 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.


Solution Express the forces and couple moments as Cartesian vectors.   F1  {  800 k } N    rCB   F2  ( 300 N ) u CB  ( 300 N )     rCB      0 . 15 i  0 . 1 j     300    {  249 . 6 i  166 . 4 j } N 2 2  ( 0 . 15 )  ( 0 . 1)    4 3 M   500   j  500   k  {  400 j  300 k } N .m 5 5


Solution Force Summation.

  FR   F ;       F R  F1  F 2   800 k  249 . 6 i  166 . 4 j     {  249 . 6 i  166 . 4 j  800 k } N

 M

Moment Summation.

         M C   M O  M  rC  F1  rB  F 2 Ro  i      (  400 j  300 k )  (1k )  (  800 k )   0 . 15  249 . 6      (  400 j  300 k )  ( 0 )  (  166 . 4 i  249 . 6 j )     {  166 i  650 j  300 k } N .m

 j

 k

0 .1

1

166 . 4

0


TEST (4-7) • Replace the loading system by an equivalent resultant force and couple moment acting at point O.


4.8 Further Simplification of a Force and Couple System 由簡化成作用於指定點O → 簡化成單一合力

Concurrent Force System 共交點力系 •

A concurrent force system is where lines of action of all the forces intersect at a common point O

FR 


F

4.8 Further Simplification of a Force and Couple System

Coplanar Force System 共平面力系 • •

Lines of action of all the forces lie in the same plane Resultant force of this system also lies in this plane



Parallel Force System 平行力系 • •

Consists of forces that are all parallel to the z axis Resultant force at point O must also be parallel to this axis



Reduction to a Wrench 化簡為一扳鉗力 • •

3-D force and couple moment system have an equivalent resultant force acting at point O Resultant couple moment not perpendicular to one another


Example (4-8) 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.


Solution Force Summation   F Rx   F x ; 3 F Rx   2 . 5 kN    1 . 75 kN 5   3 . 25 kN  3 . 25 kN    F Ry   F y ; F Ry

4   2 . 5 N    0 . 6 kN 5   2 . 60 kN  2 . 60 N  Copyright © 2010 Pearson Education South Asia Pte Ltd

Solution For magnitude of resultant force, FR  ( FRx )2  ( FRy )2  (3.25)2  (2.60)2  4.16kN

For direction of resultant force,  FRy 

 tan 1 2.60   3.25   FRx 

q  tan 1  38.7


Solution Moment Summation  Summation of moments about point A, M

RA

 M A;

3 . 25 kN ( y )  2 . 60 kN ( 0 )  1 . 75 kN (1m )  0 . 6 kN ( 0 . 6 m ) 3 4  2 . 50 kN   ( 2 . 2 m )  2 . 50 kN   (1 . 6 m ) 5 5 y  0 . 458 m Copyright © 2010 Pearson Education South Asia Pte Ltd

Solution Moment Summation  Principle of Transmissibility M

RA

 M A;

3 . 25 kN ( 2 . 2 m )  2 . 60 kN ( x )  1 . 75 kN (1m )  0 . 6 kN ( 0 . 6 m ) 3 4  2 . 50 kN   ( 2 . 2 m )  2 . 50 kN   (1 . 6 m ) 5 5 x  2 . 177 m


TEST (4-8) • Replace the loading system by an equivalent single resultant force and specify the x and y coordinates of its line of action.


4.9 Reduction of a Simple Distributed Loading 簡單分佈負載的簡化 • Large surface area of a body may be subjected to distributed loadings (分佈負載) • Loadings on the surface is defined as pressure • Pressure is measured in Pascal (Pa): 1 Pa = 1 N/m2

Uniform Loading Along a Single Axis • Most common type of distributed loading is uniform along a single axis 單一軸均勻負載 p(x) [=] N/m2 共平面分佈負載 w(x)=p(x)b [=] N/m Copyright © 2010 Pearson Education South Asia Pte Ltd

4.9 Reduction of a Simple Distributed Loading Magnitude of Resultant Force 合力的大小 • Magnitude of dF is determined from differential area dA under the loading curve. • For length L, dF=w(x)dx

FR 

 w  x dx   dA  L

A

A

• Magnitude of the resultant force is equal to the total area A under the loading diagram.


4.9 Reduction of a Simple Distributed Loading Location of Resultant Force 合力的位置 • MR = ∑MO • dF produces a moment of xdF = x w(x) dx about O • For the entire plate, M

 M O

Ro

x FR 

 xw ( x ) dx L

• Solving for x x 

 xw ( x ) dx L

 w ( x ) dx L



 xdA A

 dA A


Example (4-9) Determine the magnitude and location of the equivalent resultant force acting on the shaft (軸承).


Solution For the colored differential area element, dA  wdx  60 x dx 2

For resultant force FR   F ; 2

FR 

 dA 

 60 x dx

A

0

2

2

x   23 03   60    60    3 3 3  0   3

 160 N Copyright © 2010 Pearson Education South Asia Pte Ltd

Solution For location of line of action, 2

2

 xdA

x 

A

 dA



 x ( 60 x

2

) dx 

0

160

 x4  60    4 0 160

 24 04  60    4 4    160

A

 1 .5 m

Checking, A

ab



2 m ( 240 N / m )

3 x 

3 4

a

 160

3 3

( 2 m )  1 .5 m

4 Copyright © 2010 Pearson Education South Asia Pte Ltd

TEST (4-9) • Determine the resultant force and specify where it acts on the beam measured from A.


The End of CHAPTER 4
