Engineering Mechanics: Statics in SI Units, 12E

3/15/2015

Engineering Mechanics: Statics in SI Units, 12e

44 Force System Resultants

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 force having a specified location

1 3/15/2015

Chapter Outline

1. Moment of a Force – Scalar Formation 2. Cross Product 3. Moment of Force – Vector Formulation 4. Principle of Moments 5. 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

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 simple moment ( Mo) z

2 3/15/2015

4.1 Moment of a Force – Scalar Formation

Magnitude

• For magnitude of MO,

MO = Fd (N.m) or (Ib.ft) where d = perpendicular distance from O to its line of action of force

Direction • Direction using “right hand rule ”

4.1 Moment of a Force – Scalar Formation

Resultant Moment

• Resultant moment, MRo = moments of all the forces

MRo = ∑Fd

3 3/15/2015

Example 4.1

For each case, determine the moment of the force about point O.

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.

a)( Mo = (100N 2)( m) = 200N m(. CW )

4 3/15/2015

Solution

b)( Mo = (50N .0)( 75m) = 37 5. N.m(CW )

o c)( Mo = 40( N 4)( m + 2cos30 m) = 229N m(. CW )

Solution

o d)( Mo = 60( N 1)( sin45 m) = 4.42 N m(. CCW )

e)( Mo = 7( kN 4)( m −1m) = 0.21 kN m(. CCW )

5 3/15/2015

6 3/15/2015

4.2 Cross Product

• Cross product of two vectors A and B yields C, which is written as C = A X B and is read C equals A cross B Magnitude • Magnitude of C is the product of the magnitudes of A and B • For angle θ, 0 ≤ θ ≤ 180 °

C = AB sinθ

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

Magnitude of C Direction of C

7 3/15/2015

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 A X B yields a vector opposite in direction to C

B X A = -C

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

8 3/15/2015

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 r r r i j k r r AXB = Ax Ay Az

Bx By Bz

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

9 3/15/2015

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

MO = rF sinθ • Treat r as a sliding vector. Since d = r sinθ,

MO = rF sinθ = F (rsinθ) = Fd

4.3 Moment of Force - Vector Formulation

Principle of Transmissibility • 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

10 3/15/2015

4.3 Moment of Force - Vector Formulation

Cartesian Vector Formulation • For force expressed in Cartesian form, r r r i j k r r r M O = FXr = rx ry rz

Fx Fy Fz • With the determinant expended,

MO = ( ryFz – rzFy)i – (rxFz - rzFx)j + ( rxFy – ryFx)k

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)

11 3/15/2015

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.

Solution

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 = 0 5 0 + 4 5 − 2 − 60 40 20 80 40 − 30 = {}30i − 40 j + 60k kN ⋅m

12 3/15/2015

13 3/15/2015

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

Example 4.5

Determine the moment of the force about point O.

14 3/15/2015

Solution

The moment arm d can be found from trigonometry,

d = (3)sin 75 =° .2 898 m

Thus,

M O = Fd = (5)( .2 898) = 5.14 kN ⋅m

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

15 3/15/2015

Examples and Problems

• Resolve the following problems for Homework: 4-29, 4-37 and 4-49

4.5 Moment of a Force about a Specified Axis

• 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

dy = d cosθ

M y = Fd y

16 3/15/2015

4.5 Moment of a Force about a Specified Axis

Scalar Analysis

• According to the right-hand rule, M y is directed along the positive y axis • For any axis, the moment is

M a = Fda

• Force will not contribute a moment if force line of action is parallel or passes through the axis

4.5 Moment of a Force about a Specified Axis

Vector Analysis

• For magnitude of MA,

MA= MO·ua

where ua = unit vector

• In determinant form, u u u r r r r ax ay az M a = uax ⋅( FXr ) = rx ry rz

Fx Fy Fz

17 3/15/2015

Example

Example 4.8

Determine the moment produced by the force F which tends to rotate the rod about the AB axis.

18 3/15/2015

Solution

Unit vector defines the direction of the AB axis of the rod, where r r { 4.0 i + 2.0 j} u = B = = .0 8944i + .0 4472 j B 2 2 r B 4.0 + 2.0

For simplicity, choose rD

rD = { 6.0 i}m The force is F = {− 300k} N

Solution

r r { 4.0 i + 2.0 j} u = B = = .0 8944i + .0 4472 j B 2 2 r = { 6.0 i}m F = {− 300k} N r B 4.0 + 2.0 D

19 3/15/2015

Examples and Problems

• Resolve the following problems for Homework: 4-51, 4-57 and 4-63

20 3/15/2015

4.6 Moment of a Couple

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

Page148 Slide 85

4.6 Moment of a Couple

Scalar Formulation • Magnitude of couple moment M = Fd • Direction and sense are determined by right hand rule • M acts perpendicular to plane containing the forces

21 3/15/2015

4.6 Moment of a Couple

Vector Formulation • For couple moment, M = r X F • 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 • 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

22 3/15/2015

4.6 Moment of a Couple

Resultant Couple Moment • 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)

23 3/15/2015

Example 4.12

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 (-250 k) + rB X (250 k) = (0.8j) X (-250 k) + (0.6cos30 ºi + 0.8j – 0.6sin30 ºk) X (250 k) = {-130 j}N.cm

Take moment about point A

M = rAB X (250 k) = (0.6cos30 °i – 0.6sin30 °k) X (250 k) = {-130 j}N.cm

24 3/15/2015

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 = {-130 j}N.cm

4.7 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

25 3/15/2015

4.7 Simplification of a Force and Couple System

• Equivalent resultant force acting at point O and a resultant couple moment is expressed as

FR = ∑ F

()M 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 = ∑ Fy

()M R O = ∑ M O + ∑ M

4.7 Simplification of a Force and Couple System

Procedure for Analysis 1. Establish the coordinate axes with the origin located at point O and the axes having a selected orientation 2. Force Summation 3. Moment Summation

26 3/15/2015

27 3/15/2015

Example 4.16

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.

28 3/15/2015

Solution

Express the forces and couple moments as Cartesian vectors. r r F1 {−= 800 }Nk r r r  r  F = 300( )uN = 300( N) rCB  2 CB    rCB  r r  − 15.0 i + 1.0 j  r r = 300  {−= 6.249 i + }4.166 Nj 2 2  )15.0( + )1.0(   4  r  3  r r r M −= 500  j + 500 k {−= 400 j + 300 .} mNk  5   5 

Solution

 4  r  3  r r r M −= 500   j + 500   k { −= 400 j + 300 .} mNk  5   5  Forcer r Summation. F Σ= F; rR r r r r r F = F + F −= 800k − 249 6. i +166 4. j R 1 r2 r r = {−249 6. i +166 4. j −800k}N

r r r r r r r r M Σ= M Σ+ M = M + FXr + FXr Ro C O C 1 B 2r r r i j k r r r r = (−400 j + 300k + k X ()1() −800k ) + − 15.0 1.0 1 − 249 6. 166 4. 0 r r r = {−166i − 650 j + 300k N.} m

29 3/15/2015

4.8 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

4.8 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

30 3/15/2015

4.8 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

4.8 Further Simplification of a Force and Couple System

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

31 3/15/2015

Example 4.18

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 beam BC.

32 3/15/2015

Solution

Force Summation

+ → FRx = ΣFx; 3 FRx = − 5.2 kN  − 75.1 kN 5 = − 25.3 kN = 25.3 kN ←

+ → FRy = ΣFy;  4 FRy = − 5.2 N  − 6.0 kN  5 = −2.60kN = 2.60N ↓

Solution

For magnitude of resultant force,

2 2 2 2 FR = (FRx ) + (FRy ) = )25.3( + )60.2( = 16.4 kN

For direction of resultant force,

−1 FRy ÷ −1 60.2  θ = tan  ÷= tan  ÷  FRx   25.3  = 7.38 o

33 3/15/2015

Solution

Moment Summation Ï Summation of moments about point A,

M RA = ΣM A; 25.3 kN y)( + 60.2 kN )0( = 75.1 kn m)1( − 6.0 kN 6.0( m) 3  4 + 50.2 kN  2.2( m) − 50.2 kN  6.1( m) 5  5 y = .0 458m

Solution

Moment Summation Ï Principle of Transmissibility

M RA = ΣM A; 25.3 kN 2.2( m) − 60.2 kN x)( = 75.1 kn m)1( − 6.0 kN 6.0( m)  3   4  + 50.2 kN  2.2( m) − 50.2 kN  6.1( m)  5   5  x = 2.177m

34 3/15/2015

4.9 Reduction of a Simple Distributed Loading

• Large surface area of a body may be subjected to distributed loadings • Loadings on the surface is defined as pressure • Pressure is measured in Pascal (Pa): 1 Pa = 1N/m 2

Uniform Loading Along a Single Axis • Most common type of distributed loading is uniform along a single axis

4.9 Reduction of a Simple Distributed Loading

Magnitude of Resultant Force • Magnitude of dF is determined from differential area dA under the loading curve. • For length L,

F = w x dx = dA = A R ∫ ( ) ∫ L A • Magnitude of the resultant force is equal to the total area A under the loading diagram .

35 3/15/2015

4.9 Reduction of a Simple Distributed Loading

Location of Resultant Force

• MR = ∑MO • dF produces a moment of xdF = x w(x) dx about O • For the entire plate,

M = ΣM Fx = xw x)( dx Ro O R ∫ L • Solving for x ∫ xw x)( dx ∫ xdA x = L = A ∫ w x)( dx ∫ dA L A

Example 4.21

Determine the magnitude and location of the equivalent resultant force acting on the shaft.

36 3/15/2015

Solution

For the colored differential area element, dA = wdx = 60x2dx For resultant force

FR = ΣF; 2 F = dA = 60x2dx R ∫ ∫ A 0 2  x3  23 03  = 60  = 60 −   3 0  3 3  =160N

Solution

For location of line of action, 2 2 4 4 4 2  x  2 0  xdA x 60( x )dx 60  60 −  ∫ ∫  4  4 4 x = A = 0 = 0 =   ∫ dA 160 160 160 A = 5.1 m Checking, ab 2m(240N / m) A = = =160 3 3 3 3 x = a = 2( m) = 5.1 m 4 4

37 3/15/2015

38 3/15/2015

QUIZ

1. What is the moment of the 10 N force about point A

(M A)? F = 12 N A) 3 N·m B) 36 N·m C) 12 N·m D) (12/3) N·m E) 7 N·m d = 3 m • A

2. The moment of force F about point O is defined as MO = ______. A) r x F B) F x r C) r • F D) r * F

39 3/15/2015

QUIZ

3. If a force of magnitude F can be applied in 4 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) S C) (P, R) D) (Q, S) R PQ 4. If M = r ××× F, then what will be the value of M • r ? A) 0 B) 1 C) r 2F D) None of the above.

QUIZ

5. 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) - 20 N ·m D) 40 N ·m E) - 40 N ·m

10 N 3 m P 2 m 5 N

40 3/15/2015

QUIZ

7. When determining the moment of a force about a specified axis, the axis must be along ______. A) the x axis B) the y axis C) the z axis D) any line in 3-D space E) any line in the x-y plane

8. The triple scalar product u • ( r ××× F ) results in A) a scalar quantity ( + or - ) B) a vector quantity. C) zero. D) a unit vector. E) an imaginary number. Copyright © 2010 Pearson Education South Asia Pte Ltd

QUIZ

9. The vector operation ( P ××× Q) • R equals A) P ××× (Q • R). B) R • (P ××× Q). C) ( P • R) ××× (Q • R). D) ( P ××× R) • (Q ××× R ). 10. The force F is acting along DC. Using the triple product to determine the moment of F about the bar BA, you could use any of the following position vectors except

A) rBC B) rAD C) rAC

D) rDB E) rBD

41 3/15/2015

QUIZ

11. For finding the moment of the force F about the x-axis, the position vector in the triple scalar product should be ___ .

A) rAC B) rBA

C) rAB D) rBC

12. If r = {1 i + 2 j} m and F = {10 i + 20 j + 30 k} N, then the moment of F about the y-axis is ____ N·m. A) 10 B) -30 C) -40 D) None of the above.

QUIZ

13. 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

14. The moment of a couple is called a _____ vector. A) Free B) Spin C) Romantic D) Sliding

42 3/15/2015

QUIZ

15. F1 and F2 form a couple. The moment of the couple is given by ____ . F1

A) r1 ××× F1 B) r2 ××× F1 r1 C) F2 ××× r1 D) r2 ××× F2 r2 F2 16. If three couples act on a body, the overall result is that A) The net force is not equal to 0. B) The net force and net moment are equal to 0. C) The net moment equals 0 but the net force is not necessarily equal to 0. D) The net force equals 0 but the net moment is not necessarily equal to 0 .

QUIZ

17. A general system of forces and couple moments acting on a rigid body can be reduced to a ___ . A) single force B) single moment C) single force and two moments D) single force and a single moment 18. The original force and couple system and an equivalent force-couple system have the same _____ effect on a body. A) internal B) external C) internal and external D) microscopic

43 3/15/2015

QUIZ

18. The forces on the pole can be reduced to a single force and a single moment at point ____ . Z S A) P B) Q C) R R D) S E) Any of these points. Q P Y X 19. Consider two couples acting on a body. The simplest possible equivalent system at any arbitrary point on the body will have A) One force and one couple moment. B) One force. C) One couple moment. D) Two couple moments.

QUIZ

20. Consider three couples acting on a body. Equivalent systems will be ______at different points on the body. A) Different when located B) The same even when located C) Zero when located D) None of the above.

44 3/15/2015

QUIZ

21. The resultant force (FR) due to a distributed load is equivalent to the _____ under the distributed loading curve, w = w(x). A) Centroid B) Arc length y Distributed load curve C) Area D) Volume w

22. The line of action of the distributed load’s equivalent force passes through the ______of the distributed load. A) Centroid B) Mid-point C) Left edge D) Right edge

QUIZ

23. What is the location of F R, i.e., the distance d? A) 2 m B) 3 m C) 4 m FR D) 5 m E) 6 m A B A B 3 m 3 m d 24. If F1 = 1 N, x1 = 1 m, F2 = 2 N and x2 = 2 m, what is the location of FR, i.e., the distance x. A) 1 m B) 1.33 m C) 1.5 m F x2 1 x FR D) 1.67 m E) 2 m F2

45 3/15/2015

QUIZ

25. F R = ______A) 12 N B) 100 N 100 N/m C) 600 N D) 1200 N 12 m

26. x = ______. F A) 3 m B) 4 m R C) 6 m D) 8 m