Chap. 4 Force System Resultants

Transcription

1 Chap. 4 Force System Resultants

2 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 xis 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

3 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 F x or simple moment (M o ) z 4-1-3

4 Moment of a Force Scalar Formation Magnitude For magnitude of M O, M O 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

5 Moment of a Force Scalar Formation Resultant Moment Resultant moment, M Ro moments of all the forces M Ro Fd 4-1-5

6 Cross Product Cross Product (vector product) Magnitude C B Direction C B sinθ 0 θ 180 C ( Bsinθ )u C C B C 4-1-6

7 4-1-7 Cross Product Laws of Operation ) ( ) ( ) ( ) ( ) ( ) ( ) ( D B D B B B B B B B + + α α α α ) ( ) ( ) ( B D B D + +

8 4-1-8 Cross Product Cartesian Vector Formulation ),, ( ),, ( z y x z y x B B B B k k i j k j i k i k j j j k i j j k i k j i i i k B B j B B i B B x y y x z x x z y z z y ) ( ) ( ) ( + + y x y x z y x z y x z y x z y x B B j i B B B k j i B B B k j i B or (+) (-)

9 Moment of a Force-Vector Formulation M O r F (r B F) M O r Fsinθ F(r sinθ) Fd 4-1-9

10 Moment of a Force-Vector Formulation Transmissibility of a Force 傳遞性 M O r F r B F r C F,B,C 在 F 的作用線上

11 Moment of a Force-Vector Formulation Vector Formulation M r F O i r F x x r j F y y k r F z z

12 Moment of a Force-Vector Formulation Resultant Moment of a System of Forces M O Σ( r F) R

13 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. (0, 5, 0), B(4, 5, -2)

14 Position vectors are directed from point O to each force as shown. These vectors are r r B { 5 j} m { 4i + 5 j 2k}m The resultant moment about O is r M O i 0 60 ( r F ) j k F + r i { 30i 40 j + 60k} kn m r B F j k

15 Principle of Moments Varignon s Theorem MO r F 1 + r F 2 + r F 3 + r F r ( F 1 + F 2 + ) 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

16 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. Which way will the pole rotate, clockwise or counterclockwise?

17 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

18 4-1-18

19 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 is zero, determine the mass of the wheelbarrow and its contents

20 4-1-20

21 p. 138, 4-47 The force F {6i + 8j +10k} N creates a moment about point O of M O {-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. lso, realizing that M O Fd, determine the perpendicular distance d from point O to the line of action of F

22 Chap. 4-2 Force System Resultants

23 Moment of a Force bout a Specified xis M y 為 F 對特定軸 (y-axis) 的 moment, 也是 M o 在 y 軸上的投影 (projection) 1. O 為 y 軸上任意一點, 先求 M o (r F) 2. 再以特定軸的單位向量 ( 此例為 j) 和 M o 的 dot product 求 M y ( M o j) 3. M y [j (r F)] j 4-2-2

24 Moment of a Force bout a Specified xis a-a is an arbitrary axis M a M cosθ O M O u a a a a u a u r F ax x x (r F) u r F [ u a (r F) ] u a M M u ay y y u rz F az Triple scalar product 純量三倍積 ( 平行六面體的體積 ) z 4-2-3

25 p.145 Problem 4-56 Determine the moment produced by force F about segment B of the pipe assembly. Express the result as a Cartesian vector

26 4-2-5

27 4-2-6

28 p.147 Problem 4-65 The -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

29 4-2-8

30 4-2-9

31 Moment of a Couple 力偶 M O r (r r - r r F (-F) + B B F r B ) F (F)

32 Moment of a Couple Scalar Formulation MFd Vector Formulation Mr x F

33 Moment of a Couple Equivalent Couples 等效 Resultant Couple Moment( 合力偶力偶合 )

34 Example 4.12 Determine the couple moment acting on the pipe. Segment B is directed 30 below the x y plane

35 SOLUTION I (VECTOR NLYSIS) Take moment about point O, M r (-250k) + r B (250k) (0.8j) (-250k) + (0.6cos30ºi + 0.8j 0.6sin30ºk) (250k) {-130j}N.cm Take moment about point M r B (250k) (0.6cos30i 0.6sin30k) (250k) {-130j}N.cm

36 SOLUTION II (SCLR NLYSIS) Take moment about point or B, M Fd 250N(0.5196m) 129.9N.cm pply right hand rule, M acts in the j direction M {-130j}N.cm

37 p.155 Problem 76 Determine the required magnitude of force F if the resultant couple moment on the frame is 200 N M, clockwise

38 4-2-17

39 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

40 4-2-19

41 p.159 Problem If F N and F N, determine the magnitude and coordinate direction angles of the resultant couple moment

42 4-2-21

43 4-2-22

44 Chap. 4-3 Force System Resultants

45 Simplification of a Force and Couple System n 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

46 Equivalent System Point O is On the Line of ction Point O Is Not On the Line of ction 4-2-3

47 Equivalent System Resultant of a Force and Couple System Force Summation : FR ΣF Moment Summation : MRO ΣMC +ΣMo 4-2-4

48 Ex. 4-15, replace the system by an equivalent resultant force and couple moment acting on point O. F u 1 CB 800kN ( 0.15,0.1,0) 2 ( 0.15) F2 300 u CB ( 249.6,166.4,0)N M (0, 400,300)N m 2 ΣF ( - 250,166.4, -800 ) N M RO M + r C F r B F 2 i j k (0,-400,300) (-166.4,-650,300)N m 4-2-5

49 p. 169, 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

50 4-2-7

51 Further Reduction of a Force and Couple System M RO F R d 4-2-8

52 Further Simplification of a Force and Couple System Concurrent Force System concurrent force system is where lines of action of all the forces intersect at a common point O F R F 4-2-9

53 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

54 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

55 Further Simplification of a Force and Couple System Reduction to a Wrench ( 扳手 ) M ( 在 FR方向 ) d M F R M RO M + M

56 p.180, Replace the force and couple moment system acting on the overhang beam by a resultant force, and specify its location along B measured from point

57 4-2-14

58 4-2-15

59 p.182, 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

60 4-2-17

61 4-2-18

62 Reduction of a Simple Distributed Loading (x, y) 形心 centroid 集中力 concentrated force

63 Reduction of a Simple Distributed Loading Magnitude F w(x)dx d R L Location ΣM O M xw ( x) dx RO : xf R x xd d xd 求形心 first moment

64 p.189, Replace the distributed loading by an equivalent resultant force and specify its location, measured from point

65 4-2-22

66 p.191, Wind has blown sand over a platform such that the intensity of the load can be approximated by the function w(0.5x 3 ) N/m. Simplify this distributed loading to an equivalent resultant force and specify the magnitude and location of the force, measured from. d F F x R R 1.25kN xd wdx d x N x N x x dx ns m 8.00 m dx 10 0 ns

67 p.192, The distributed load acts on the beam as shown. Determine the magnitude of the equivalent resultant force and specify its location, measured from point

68 4-2-25