Chap. 4 Force System Resultants

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

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

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

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

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

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 + +

4-1-8 Cross Product Cartesian Vector Formulation ),, ( ),, ( z y x z y x B B B B 0 0 0 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 (+) (-)

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

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 4-1-10

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 4-1-11

Moment of a Force-Vector Formulation Resultant Moment of a System of Forces M O Σ( r F) R 4-1-12

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) 4-1-13

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 5 0 + 4 5 40 20 80 40 { 30i 40 j + 60k} kn m r B F j k 2 30 4-1-14

Principle of Moments Varignon s Theorem MO r F 1 + r F 2 + r F 3 + r F 4 +... 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 4-1-15

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? 4-1-16

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. 4-1-17

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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. 4-1-19

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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. 4-1-21

Chap. 4-2 Force System Resultants

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

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

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. 4-2-4

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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. 4-2-7

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Moment of a Couple 力偶 M O r (r r - r r F (-F) + B B F r B ) F (F) 4-2-10

Moment of a Couple Scalar Formulation MFd Vector Formulation Mr x F 4-2-11

Moment of a Couple Equivalent Couples 等效 Resultant Couple Moment( 合力偶力偶合 ) 4-2-12

Example 4.12 Determine the couple moment acting on the pipe. Segment B is directed 30 below the x y plane. 4-2-13

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 4-2-14

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 4-2-15

p.155 Problem 76 Determine the required magnitude of force F if the resultant couple moment on the frame is 200 N M, clockwise. 4-2-16

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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. 4-2-18

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p.159 Problem 4-102 If F 1 500 N and F 2 1000 N, determine the magnitude and coordinate direction angles of the resultant couple moment. 4-2-20

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Chap. 4-3 Force System Resultants

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

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

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

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) + 0.1 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 0 1 + r B F 2 i j k (0,-400,300) + 0.15 0.1 1 249.6 166.4 0 (-166.4,-650,300)N m 4-2-5

p. 169, 4-116 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. 4-2-6

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

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

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 4-2-10

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 4-2-11

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 4-2-12

p.180, 4-124. 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. 4-2-13

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p.182, 4-140. 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. 4-2-16

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Reduction of a Simple Distributed Loading (x, y) 形心 centroid 集中力 concentrated force 4-2-19

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 4-2-20

p.189, 4-144. Replace the distributed loading by an equivalent resultant force and specify its location, measured from point. 4-2-21

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p.191, 4-152. 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 10 0 1 5 x 10 10000N 10000 1250 1 2 10 0 1 4 x 8 1250N x 1 2 4 x dx 10 0 3 ns m 8.00 m dx 10 0 ns 4-2-23

p.192, 4-160. The distributed load acts on the beam as shown. Determine the magnitude of the equivalent resultant force and specify its location, measured from point. 4-2-24

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