Miscellaneous (dimension, angle, etc.) - black [pencil] Use different colors in diagrams. Body outline - blue [black] Vector

Transcription

1 1. Sstems of orces & s Statics, 2011/2 Department of Mechanical Engineering, Chulalongkorn Uniersit bjecties Students must be able to Course bjectie Analze a sstem of forces and moments Chapter bjecties Describe the characteristics and properties of forces and moments Determine and manipulate 2D/3D forces and moments (about a point, a line and couple) in scalar & ector forms Determine the resultant of forces & couples and find an equialent sstem at another position educe a 2D resultant to a single force or couple sstem and reduce a 3D resultant to a wrench 2 Contents Manipulation Quantities (Position, orce, ) orce Sstem esultants Equialent orce Sstems Use different colors in diagrams Bod outline - blue [black] Load - red Miscellaneous (dimension, angle, etc.) - black [pencil] Maths #1 Sine Law/Cosine Law A B C = = sin α sin β sin γ 2 2 C = A + B 2ABcos γ 3 4 1

2 peration #1 Unit aa = ( a ) A A = Ae, e = unit ector of A 5 peration #2 ight-handed coordinate sstem Dot Product A B = AB θ = A B + A B + A B cos z z 6 peration #3 ight-handed coordinate sstem Cross Product A B = ( ABsin θ ) e C 7 Phsical Quantit s representing phsical quantities can be classified ied Its action is associated with a unique point of application Described b magnitude, direction & point of application Sliding Has a unique like of action in space but not a unique point of application Described b magnitude, direction & line of action ree Its action is not confined or associated with a unique line in space Described b magnitude & direction 8 2

3 Position Position r = = + + A ra Ai A j Azk r = = + + B rb Bi B j Bzk r = r r AB B A Coordinates s Position s orce orce Definition orce is a ector quantit. Staticall, force is the action of one bod on another. In dnamics, force is an action that tends to cause acceleration of an object. The SI unit of force magnitude is the Newton (N). ne Newton is equialent to one kilogram-meter per second squared (kg m/s 2 or kg m s 2 ) Eamples of mechanical force include the thrust of a rocket engine, the impetus that causes a car to speed up when ou step on the accelerator, and the pull of grait on our bod. orce can result from the action of electric fields, magnetic fields, and arious other phenomena orce orce eal-life Eamples & Phsical Meanings orce Classification Eternal s. Internal Surface/Contact s. Bod Applied s. eactie (Action s. eaction) Concentrated s. Distributed Internal orce Cables & Springs θ k = spring constant T T s Cable in tension = ks

4 orce orce orce epresentation quantit Magnitude Direction Point of application 10 N igid Bod A bod is considered rigid when the change in distance between an two points is negligible. igid s. Deformable bodies In Statics, bodies are considered rigid unless stated otherwise orce orce Graphical epresentation of tension and compression orce on igid Bod #1 orce on rigid bod Magnitude Direction Line of action orce on igid Bod #2 The force ma be applied at an point on its gien line of action without altering the resultant effects of the force eternal to the rigid bod on which it acts. Principle of Transmissibilit

5 Components 2D = + = i+ j = cos θ, = sin θ tan θ = = Components 3D #1 Cartesian Components = i + j + zk e = e z = i + j + k Components 3D #2 2 2 p = + = i+ j p p e p = = i + j P P P = e + k p p z Components 3D #3 cos θ = cosθ = Eleated Angle Directional Cosine z cos θ z =

6 Components 3D #4 Directional Cosine = = = cos cos cos θ θ θ z z cos θ + cos θ + cos θ = 1 z e = = e i+ e j+ e k z e = cos θ i+ cos θ j+ cos θ k z Eample Hibbeler E 2-8 #1 ind Cartesian components of force Eample Hibbeler E 2-8 #2 θ = α, θ = 60, θ = cos θ + cos θ + cos θ = cos α + cos 60 + cos 45 = 1 cos α =± 0.5 α z z cos ( 0.5) 120 or 60 1 = ± = rotate z about ais = cos θ + θ + θ i cos j cos zk = (200 N)cos60 i+ (200 N)cos60 j+ (200 N)cos 45 k = (100 i+ 100 j k) N = (100 i+ 100 j+ 141 k) N Ans orce Projection Along a Line Dot Product Angle between two ectors θ 1 A B = cos AB Components of a ector Scalar projection along a line a-a = + = cos θ = e = ( cos θ ) e = ( e ) e =

7 Eample Hibbeler E 2-17 #1 Determine the angle between the force and the pipe segment BA, and the magnitude of the component of the force (parallel and perpendicular) to BA. Eample Hibbeler E 2-17 #2 r = ( 2i 2j+ 1 k) ft BA r BA = 3 ft r ( 3 1 BC = j + k) ft r BC = 10 ft r ( ) ft ( 3 1 BA rbc i j + k j + k) ft cos θ = = = rba rbc (3 ft)( 10 ft) θ = = 42.5 Ans r ( 3 1 BC j + k) ft = = (80 lb) = ( j k ) lb r 10 ft BC Eample Hibbeler E 2-17 #3 r ( ) ft BA i j + k e BA = = = i j + k 3 ft BA ( lb) ( BA = eba = j + k i j + k) = lb = 59.0 lb Ans BA r = = lb = 54.0 lb Ans 2 2 BA Eample Hibbeler E 2-17 #4 BA = cos θ = (80 lb)cos = lb = 59.0 lb Ans = sin θ = (80 lb)sin = lb = 54.0 lb Ans

8 orce Parallelogram Summation Concurrent orces # Principle of Transmissibilit Lines of Action 29 orce Summation Concurrent orces #2 Triangular Law = Principle of Transmissibilit = 30 orce Summation Concurrent orces #3 Beware of incoming danger! 31 orce Summation Parallel orces #1 Cancellation irst Method Cancellation of Components 1 2 ind P such that = + ( P), = + P = 1+ 2 = 1+ P + 2 P = P 1 2 P

9 orce Summation Parallel orces #2 Cancellation irst Method: Cancellation of Components ind P such that = + ( P), = + P = 1+ 2 = 1+ P + 2 P = P P orce Summation Parallel orces #3 Second Method: Equialent 1 = a a Equialent about ais aa - = = + = orce Eample Hibbeler E 2-6 #1 Determine the magnitude and direction of the resultant force. 35 orce Eample Hibbeler E 2-6 #2 = (600cos30 i+ 600sin30 j) N = ( 400 sin 45 i+ 400cos 45 j) N

10 orce Eample Hibbeler E 2-6 #3 sstem II = sstem I = (600cos30 600sin30 ) N ( 400sin45 400cos45 = i + j + i + j) N = ( i j) N = + = ( N) + ( N) θ = θ = tan = N = 629 N Ans 1 tan ( ) 1 ( N N) θ = = 67.9 Ans 37 orce Eample Hibbeler E 2-6 #4 Scalar =, sstem II, sstem I = cos30 sin = (600 N)cos30 (400 N)sin45 = N =, sstem II, sstem I = sin30 + cos = (600 N)sin30 + (400 N)cos45 = N = ( N) ( N) = N = 629 N Ans θ = tan ( ) θ = tan ( N N) θ = = 67.9 Ans 2 38 orce orce Eample Hibbeler E 2-9 #1 Eample Hibbeler E 2-9 #2 Determine the magnitude and the coordinate direction angles of the resultant force acting on the ring

11 orce Eample Hibbeler E 2-9 #3 sstem II = sstem I = (60 80 ) lb ( = j + k + i j + k) lb = (50i 40j+ 180 k) lb (50 lb) ( 40 lb) (180 lb) lb 191 lb Ans = + + = = orce Eample Hibbeler E 2-9 #4 e = = (50 i 40 j+ 180 k) lb ( lb) = i j k cos θ = e = θ = 74.8 cos θ = e = θ = 102 cos θ = e = θ = 19.6 Ans z z z orce orce Eample Hibbeler E 2-11 #1 Specif the coordinate direction angles of 2 so that the resultant acts along the positie ais and has a magnitude of 800 N. Eample Hibbeler E 2-11 #2 1 = 1cos θ + θ + θ 1i 1cos 1j 1cos z1k = (300 N)cos 45 i+ (300 N)cos60 j+ (300 N)cos120 k = ( i+ 150 j 150 k) N = i+ j + k z

12 orce Eample Hibbeler E 2-11 #3 (800 = j) N sstem II = sstem I = (800 j N) = ( i+ 150 j 150 k) N + ( 2i + 2j + 2zk) N (800 j N) = ( N + ) i+ (150 N + ) j + ( 150 N + ) k z orce Eample Hibbeler E 2-11 #4 dir: 0 = N + dir: 800 = 150 N + 2 z dir: 0 = 150 N + = 212 N = 650 N = 150 N Ans 2 2 z z z dir: N = (700 N)cos α α = dir: 650 N = (700 N)cos β β = dir: 0 = (700 N)cos γ γ = 77.6 Ans orce Eample Hibbeler E 2-15 #1 The roof is supported b cables as shown. If the cables eert forces AB = 100 N and AC = 120 N on the wall hook at A as shown, determine the magnitude of the resultant force acting at A. orce Eample Hibbeler E 2-15 #2 r (4 0) (0 0) (0 4) m (4 4 AB = i j k + + = i k) m rab = rab + rab = (4 m) + ( 4 m) = m r (4 4 AB i k) m AB = AB ( ) = (100 N) r AB m = ( i k) N AB r (4 2 4 AC = i + j k) m, rac = 6 m r AC AC = AC ( ) r AC ( AC = i + j k) N

13 orce Eample Hibbeler E 2-15 #3 orce Eample Hibbeler E 2-15 #4 sstem II = sstem I = + AB AC ( ) N ( = i k + i + j k) N = ( i+ 40 j k) N = ( N) + (40 N) + ( N) = N = 217 N Ans orce esolution #1 A ector ma be resoled into two components. = A+ B orce esolution #

14 d Definition #1 M is the tendenc of force to rotate a bod about an ais. M = d Definition #2 M is the tendenc of force to rotate a bod about an ais. z z ( M ) z ( M ) d Definition #3 is a ector quantit. Magnitude Direction Ais of otation The unit of moment is N m The moment-arm d The right-hand rule determined b ector cross product Positie sign conention: 2D +k (CCW), 3D +e aes of a force or torque Cross Product z k = i j + ĵ k i j k A B = A A A B B B z z A B = ( AB ) ( ) ( ) z AB z i AB z AB z j+ AB AB k î ĵ î

15 j r k rz z About a Point #2 i M = r = r About a Point #1 58 Cartesian ormulae 57 Cartesian ormulae M = r = (r z rz )i ( r z rz ) j + ( r r )k r magnitude r A r d 59 M = ra = rb = rc 60 Principle of Transmissibilit M moment ais M = r sin θ = d About a Point #4 M = r A M moment ais About a Point #3 15

16 About a Point #5 Verignon Theorem 1 The moment of a force about an point is equal to the sum of the moments of the components of the force about the same point. 2 r = M = r 1 + r 2 = r ( 1+ 2) = r 61 Eample Hibbeler E 4-4 #1 Determine the moment about the support at A. 62 Eample Hibbeler E 4-4 #2 r (1 3 2 B = i + j + k) m r (3 4 C = i + j) m r = r r = ( 2i 1j+ 2 k) m CB B C r = + CB ( 2i 1j 2 k) m, rcb = 3 m r + CB ( 2i 1j 2 k) m 1 u = = = ( i j k) r CB 3 m 3 = (60 N) u = ( 40 i 20 j+ 40 k) N 63 Eample Hibbeler E 4-4 #3 i j k M = = 3 m 4 m 0 m = ( A rc i j k) N m 40 N 20 N 40 N M A = (160 N) + ( 120 N) + (100 N) = N = 224 N m Ans 64 16

17 Eample Hibbeler E 4-4 #4 MA = rc = rb i j k M = = 1 m 3 m 2 m = ( A rb i j k) N m 40 N 20 N 40 N Eample Hibbeler E 4-7 #1 Determine the moment of the force about point Eample Hibbeler E 4-7 #2 r = (400 sin30 i 400cos30 j) N = (200 i j) N r = (0.4 i 0.2 j) m i j k M = r = 0.4 m 0.2 m 0 m 200 N N 0 N M = k N m M = 98.6 k N m Ans 67 Eample Hibbeler E 4-7 #3 Scalar M = + M M = (400 N)sin 30(0.2 m) (400 N)cos 30(0.4 m) M = 40 N m N m = N m Ans 68 17

18 About an Ais #1 M a = ea ( r ) M = M e = e ( r ) e a a a a a Eample Hibbeler E 4-8 #1 Determine the moments of this force about the and a aes. B r = + + A ( 3i 4j 6 k) m M = i ( ra ) i j k M = 3 m 4 m 6 m 40 N 20 N 10 N M = + i ( 80 i 210 j 100 k) N m M = 80 N m M = Mi = 80 i N m Ans Eample Hibbeler E 4-8 #2 M r = + B ( 3i 4 j) m, rb = 5 m 3 4 u = = ( + a rb rb i j) 5 5 M = a ua ( ra ) a 3 4 = ( i + j) ( 80 i 210 j k) N m N m M a = 3 4 M = = ( 120 N m)( + ) = (72 96 a Maua i j i j) N m Ans 5 5 B 71 Eample Hibbeler E 4-9 #1 Determine the moment M AB produced b = ( 600i + 200j 300k) N, which tends to rotate the rod about the AB ais i

19 Eample Hibbeler E 4-9 #2 ( ) m, 0.2 m rb = i + j rb = r ( B i j) m u = = B r B 0.2 m u B = i j M = u ( r ) AB B r is directed from an point on the AB ais to an point on the line of action of the force. Eample Hibbeler E 4-9 #3 MA = rad M = (0.2 m) ( A j i j k) N M = ( 60i+ 120 k) N m A Tr point A MAB = MA ub M = ( AB i k N m) M = ( AB i j) M = N m AB M = M u M A AB B AB B = ( 48.0 i 24.0 j) N m Ans Eample Hibbeler E 4-9 #4 Tr point B tr r = (0.2 i 0.2 j+ 0.3 k) m BC r BC M = AB ub ( rb C ) M = + AB ( i j) M = + AB (0.2 i 0.2 j 0.3 k) m M = ( AB i j k) N M = ( A B i j) M = ( A B j k) N m M = N m A B Couple

20 Couple Definition #1 M = d = 0 The two forces that are equal, opposite, and colinear is called a couple. Couple is a ector Magnitude. Direction. of a couple is constant regardless of moment ais. Couple Definition #2 M = ra ( ) + rb = ( ra + rb) M = r AB A couple moment is a free ector A couple can act at an point since M depends onl upon the position ector r directed between the forces Couple Summation M = M1 + M2 M = ( r ) M 2 M M 2 P M 1 M 1 Eample Hibbeler E 4-13 #1 eplace the two couples acting on the pipe column b a resultant couple moment. M1 = d = (150 N)(0.4 m) = 60 N m M = (60 i ) N m (125 N) C = Ce C = j k 5 5 ( C = j k) N M2 = rdc C = (0.3 i m) (100 j 75 k) N = (22.5 j + 30 k) N m

21 Eample Hibbeler E 4-13 #2 M = M + M = (60 i j+ 30 k) N m Ans Concepts #1 eiew s can be manipulated b scalar multiplication, addition, subtraction, dot product, cross product and mied triple product. s representing can be classified into free, sliding and fied ectors. Position ectors describe the position of a point relatie to a reference point or the origin. Staticall, force is the action of one bod on another. In dnamics, force is an action that tends to cause acceleration of an object. To define a force on rigid bodies, the magnitude, direction and line of action are required. Thus, the principle of transmissibilit is applicable to forces on rigid bodies. 82 esultant Concepts #2 eiew To define a moment about a point, the magnitude, direction and the point are required. To define a moment about an ais, the magnitude, direction and the aes are required. To define a couple, the magnitude and direction are required. 83 esultant Definition The force-couple sstems or force sstems can be reduced to a single force and a single couple (together called resultant) that eert the same effects of Net force Tendenc to translate Net moment Tendenc to rotate Two force-couple sstems are equialent if their resultants are the same

22 esultant esultant orce Sstems #1 = = + sstem II sstem I 1 2 M,sstem II = M,sstem I M = MC + ( r1 1 ) + ( r2 2 ) =, =, M = M + M C esultant Eample Hibbeler E 4-14 #1 eplace the current sstem b an equialent resultant force and couple moment acting A esultant Eample Hibbeler E 4-14 #2 =,ss II,ss I cos θ = 100 N 400cos45 N cos θ = N =,ss II,ss I sin θ = 600 N 400sin45 N sin θ = N 87 esultant Eample Hibbeler E 4-14 #3 = ( cos θ) + ( sin θ) 2 2 = N = 962 N Ans tan θ θ = θ = tan 1 sin θ cos θ N N = = 247 Ans M = M + A,ss II A,ss I M M A A = (600 N)(0.4 m) (400 N)sin 45 (0.8 m) (400 N)cos 45 (0.3 m) = N m = 551 N m Ans 88 22

23 esultant esultant Eample Hibbeler E 4-15 #1 eplace the current sstem b an equialent resultant force and couple moment acting at its base (point ). Sstem I r r = + CB 0.15 i 0.1 j m CB = m 1 = 800 k N 2 = (300 N) u CB = (300 N)( rcb rcb ) = ( i j) N 4 3 M = Mu = (500) N m( + M j k) 5 5 = ( 400 j k) N m Eample Hibbeler E 4-15 #2 ss II = ss I = = ( 800 k) N + ( i j) N ( = i + j k) N = ( 250 i+ 166 j 800 k) N Ans Sstem II esultant Eample Hibbeler E 4-15 #3 M = M,ss II,ss I M = M + ( r 1) ( 2) C + rb M ( ) N m (1 m) ( 800 = j + k + k k N) + M ( ) m ( = i + j + k i + j) N M = ( 400 j+ 300 k) N m + (0) N m + M = ( j k) N m M = ( 166 i 650 j+ 300 k) N m Ans Equialent Sstem Equialent Sstem Definition Two force-couple sstems are equialent if the hae the same resultant. ( ) = ( ) ( M ) = ( M ) sstem I sstem II, sstem I, sstem II i

24 Equialent Sstem eduction Concurrent orce Sstems = Equialent Sstem eduction Coplanar orce Sstems #1 3 M 1 r 3 r 1 M 2 r = i M = M + ( r ) i i i M Equialent Sstem eduction Coplanar orce Sstems #2 M = d M d Equialent Sstem eduction Coplanar orce Sstems #3 A sstem of 2D forces and couples can be simplified to A sstem of a force through a point and a perpendicular couple A single resultant force

25 Equialent Sstem eduction Parallel orce Sstems #1 Equialent Sstem eduction Parallel orce Sstems #2 M 1 1 r 1 z r 2 r M 2 M z Equialent Sstem eduction Parallel orce Sstems #3 M z z d Equialent Sstem eduction Parallel orce Sstems #4 = = ( i i) = ( ) i i

26 Equialent Sstem Eample Hibbeler E 4-16 #1 Determine the magnitude, direction and location on the beam of a resultant force which is equialent to the sstem of forces measured from E. 101 Equialent Sstem Eample Hibbeler E 4-16 #2,ss II =,ss I cos θ = (500 N)cos60 + (100 N) = 350 N,ss II =,ss I sin θ = (500 N)sin60 + (200 N) = N = N = 420 N 102 Equialent Sstem Eample Hibbeler E 4-16 #3 sin θ N tan θ = = cos θ 350 N θ = = 33.7 = 33.7 CW Ans ME,ss II = M E,ss I sin θ ( d) = (500 N)sin60 (4 m) (100 N)(0.5 m) (200 N)(2.5 m) ( N)( d) = N d = m = 5.07 m Ans 103 Equialent Sstem Eample Hibbeler E 4-19 #1 Determine the magnitude and direction of a resultant equialent to the gien force sstem and locate its point of application P on the coer plate. Sstem I Sstem II

27 Equialent Sstem Eample Hibbeler E 4-19 #2 300 lb, 8 ft, 200 lb, 8 A = k ra = i B = k rb = j ft 150 lb, ( 8 sin 45 8cos 45 C = k rc = i + j) ft ss II = ss I A B C M = M = + + = 650 k lb Ans,ss II,ss I M = r + r A A B B M = (8 i ft) ( 300 k lb) + ( 8 j ft) ( 200 k lb) + M M M + r C = ( 8 sin 45 i+ 8cos 45 j) ft ( 150 k lb) = 2400 j i j lb ft = ( ) lb ft (1) C Equialent Sstem Eample Hibbeler E 4-19 #3 M = r = ( i + ) j ft ( 650 k lb) M = ( 650 i+ 650 j) lb ft (2) rom (1) = (2) ( 650 i+ 650 j) lb ft = ( i j) lb ft Equating i and j components lb ft = = ft = 2.39 ft Ans 650 lb = lb ft 650 lb = ft = 1.16 ft Ans Equialent Sstem eduction 3D Sstem to a Wrench #1 about & about & ( + ) M M M about & ( M + M ) about P & M Equialent Sstem eduction 3D Sstem to a Wrench #2 about P & M about P & M about P (free ector) i j i

28 Equialent Sstem Sign Conention eduction 3D Sstem to a Wrench #3 Positie wrench screw Negatie wrench unscrew 109 Equialent Sstem eduction of orce Sstems A sstem of 2D forces and couples can be simplified to A sstem of a force through a point and a perpendicular couple A single resultant force or If there is no resultant force component, a single couple A sstem of 3D forces and couples can be simplified to A sstem of a force through a point and a couple A single wrench 110 Equialent Sstem Eample Hibbeler E #1 The three forces acting on the block each hae a magnitude of 10 N. eplace this sstem b a wrench and specif the point where the wrench intersects the z ais, measured from point. Equialent Sstem Eample Hibbeler E #2 = sstem I sstem II = (10 N)(cos 45 i sin45 j) (10 j N) + (10 N)( cos 45 i+ sin 45 j) = = 10 j N

29 Equialent Sstem Eample Hibbeler E #3 = 10 j N M = M j N m = M M,sstem I,sstem II ( r + + = + B 1) ( ra 2) ( ra 3 ) ( rd ) M j ( r 1) + ( 2 ) = ( ) + AB ra rd M j ( 2 k m) (10 N)(cos 45 i sin45 j) + (6 j+ 2 k) m ( 10 j N) = ( dk ) ( 10 j N) + ( M j N m) 113 Equialent Sstem Eample Hibbeler E #4 ( 10 2i 10 2j+ 20 i) N m = (10 N) di+ M j -direction: ( 10 2) N m = M M = 10 2 N m = N m -direction: ( ) N m = (10 N) d d = m Wrench: = 10 j N M = 14.1 j N m d = m Ans 114 Equialent Sstem eduction Summar General force sstems Single force + single couple 2D force sstems Single force or single couple simplest sstems 3D force sstems Wrench 115 Concepts #3 eiew Two force sstems are equialent if their resultants are the same. A force sstem or force-couple sstem can be reduced to the resultant (a single force and a single couple) that eert the same effects of net force and net moment. orce sstem reduction A 2D force sstem can be reduced to a single force or, in case of no net force, a single couple. A 3D force sstem can be reduced to a wrench