Chap. 5 Equilibrium of a Rigid Body

Chapter Outline „ „ „ „ „

Conditions for Rigid-Body Equilibrium Free-Body Diagrams Equations of Equilibrium Two and Three-Force Members Constraints and Statical Determinacy

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Conditions for Rigid-Body Equilibrium „

The equilibrium of a body is expressed as F = ∑F = 0 (M ) = ∑ M = 0 R

R O

„

O

Consider summing moments about some other point, such as point A, we require

∑M

A

= r × FR + (M R )O = 0 5-3

FREE BODY DIAGRAMS (TWO DIMENSION) (constraint)

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(continued with ch5a.ppt)

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Free-Body Diagrams Example 5.1 Draw the free-body diagram of the uniform beam. The beam has a mass of 100kg.

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Free-Body Diagrams Solution Free-Body Diagram

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Free-Body Diagrams Example 5.3 Two smooth pipes, each having a mass of 300kg, are supported by the forks of the tractor. Draw the free-body diagrams for each pipe and both pipes together. 5-9

Free-Body Diagrams Solution „ For idealized models, „

Free-Body Diagram of pipe A

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p.216, 5-9. Draw the free-body diagram of the beam, which is pinconnected at A and rocker-supported at B.

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5.3 Equations of Equilibrium „

„

„

For equilibrium of a rigid body in 2D, ∑Fx = 0; ∑Fy = 0; ∑MO = 0 ∑Fx and ∑Fy represent the algebraic sums of the x and y components of all the forces acting on the body ∑MO represents the algebraic sum of the couple moments and moments of the force components about an axis perpendicular to x-y plane and passing through arbitrary point O, which may lie on or off the body 5-12

5.3 Equations of Equilibrium Alternative Sets of Equilibrium Equations „

„

For coplanar equilibrium problems, ∑Fx = 0; ∑Fy = 0; ∑MO = 0 can be used Two alternative sets of three independent equilibrium equations may also be used ∑Fa = 0; ∑MA = 0; ∑MB = 0 When applying these equations, it is required that a line passing through points A and B is not perpendicular to the a axis 5-13

5.3 Equations of Equilibrium Alternative Sets of Equilibrium Equations „ A second set of alternative equations is ∑MA = 0; ∑MB = 0; ∑MC = 0 „ Points A, B and C do not lie on the same line

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5.3 Equations of Equilibrium Procedure for Analysis Free-Body Diagram „ Establish the x, y, z coordinates axes in any suitable orientation „ Draw an outlined shape of the body „ Show all the forces and couple moments acting on the body „ Label all the loadings and specify their directions relative to the x, y axes 5-15

5.3 Equations of Equilibrium Procedure for Analysis Free-Body Diagram „ The sense of a force or couple moment having an unknown magnitude but known line of action can be assumed „ Indicate the dimensions of the body necessary for computing the moments of forces „ Equations of Equilibrium 5-16

5.3 Equations of Equilibrium Example 5.5 Determine the horizontal and vertical components of reaction for the beam loaded. Neglect the weight of the beam in the calculations.

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5.3 Equations of Equilibrium Solution FBD „ 600N force is represented by its x and y components „ 200N force acts on the beam at B and is independent of the force components Bx and By, which represent the effect of the pin on the beam 5-18

5.3 Equations of Equilibrium Solution Equations of Equilibrium + → ∑ Fx = 0; 600 cos 45o N − Bx = 0 Bx = 424 N „

„

A direct solution of Ay can be obtained by applying ∑MB = 0 about point B Forces 200N, Bx and By all create zero moment about B 5-19

5.3 Equations of Equilibrium Solution ∑ M B = 0; 100 N ( 2 m ) + ( 600 sin 45 o N )( 5 m ) − ( 600 cos 45 o N )( 0 . 2 m ) − A y ( 7 m ) = 0 A y = 319 N + ↑ ∑ F y = 0; 319 N − 600 sin 45 o N − 100 N − 200 N + B y = 0 B y = 405 N

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5.3 Equations of Equilibrium Solution Checking, ∑M

A

= 0;

− ( 600 sin 45 o N )( 2 m ) − ( 600 cos 45 o N )( 0 . 2 m ) − (100 N )( 5 m ) − ( 200 N )( 7 m ) + B y ( 7 m ) = 0 B y = 405 N

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5.3 Equations of Equilibrium Example 5.6 The cord supports a force of 500N and wraps over the frictionless pulley. Determine the tension in the cord at C and the horizontal and vertical components at pin A.

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5.3 Equations of Equilibrium Solution FBD of the cord and pulley „ Principle of action: equal but opposite reaction observed in the FBD „ Cord exerts an unknown load distribution p along part of the pulley’s surface „ Pulley exerts an equal but opposite effect on the cord 5-23

5.3 Equations of Equilibrium Solution FBD of the cord and pulley „ Easier to combine the FBD of the pulley and contracting portion of the cord so that the distributed load becomes internal to the system and is eliminated from the analysis 5-24

5.3 Equations of Equilibrium Solution Equations of Equilibrium ∑M

A

= 0;

500 N ( 0 . 2 m ) + T ( 0 . 2 m ) = 0 T = 500 N

Tension remains constant as cord passes over the pulley (true for any angle at which the cord is directed and for any radius of the pulley 5-25

Two- and Three-Force Members „

大小相等、反向、作用線相等 Slender rod (beams)

平行力

共點力

F3一定要通過 F1和F2的交點 5-26

5.4 Two- and Three-Force Members Example 5.13 The lever ABC is pin-supported at A and connected to a short link BD. If the weight of the members are negligible, determine the force of the pin on the lever at A.

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5.4 Two- and Three-Force Members Solution FBD „ Short link BD is a two-force member, so the resultant forces at pins D and B must be equal, opposite and collinear „ Magnitude of the force is unknown but line of action known as it passes through B and D „ Lever ABC is a three-force member 5-28

5.4 Two- and Three-Force Members Solution FBD „ For moment equilibrium, three non-parallel forces acting on it must be concurrent at O „ Force F on the lever at B is equal but opposite to the force F acting at B on the link „ Distance CO must be 0.5m since lines of action of F and the 400N force are known 5-29

5.4 Two- and Three-Force Members Solution Equations of Equilibriumθ = tan −1 ⎛⎜ 0.7 ⎞⎟ = 60.3o ⎝ 0.4 ⎠ + → ∑ Fx = 0;

FA cos 60.3o − F cos 45o + 400 N = 0 + ↑ ∑ Fy = 0; FA sin 60.3o − F sin 45o = 0

Solving,

FA = 1.07 kN F = 1.32kN 5-30

p.227, Problem 5-20 The train car has a weight of 120 kN and a center of gravity at G. It is suspended from its front and rear on the track by six tires located at A, B, and C. Determine the normal reactions on these tires if the track is assumed to be a smooth surface and an equal portion of the load is supported at both the front and rear tires.

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120 kN

1.2 m

1.5 m

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p.230, Problem 5-32 The jib crane is supported by a pin at C and rod AB. If the load has a mass of 2 Mg with its center of mass located at G, determine the horizontal and vertical components of reaction at the pin C and the force developed in rod AB on the crane when x = 5 m.

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p.235, 5-60. The uniform rod has a length l and weight W. It is supported at one end A by a smooth wall and the other end by a cord of length s which is attached to the wall as shown. Show that for equilibrium it is required that h = [(s2 - l2)>3]1>2.

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Free Body Diagrams (Three Dimension)

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Example 5.14 Several examples of objects along with their associated free-body diagrams are shown. In all cases, the x, y and z axes are established and the unknown reaction components are indicated in the positive sense. The weight of the objects is neglected.

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Solution

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Equation of Equilibrium „

Vector

ΣF = 0 ΣM O = 0 „

Scalar

ΣFX = ΣFY = ΣFZ = 0 Σ M X = ΣM Y = Σ M Z = 0 5-43

Constraints for a Rigid Body „

Redundant constraints 靜不定

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Constraints for a Rigid Body „

Improper Constraints-intersect points on a common axis 同軸

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Constraints for a Rigid Body „

Improper Constraints-intersect points on a common axis 2-D reaction force 共點

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Constraints for a Rigid Body „

Improper Constraints-reactive force are all parallel

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Constraints for a Rigid Body „

Partially Constrained 部分限制

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p.253, 5-67. Due to an unequal distribution of fuel in the wing tanks, the centers of gravity for the airplane fuselage A and wings B and C are located as shown. If these components have weights WA = 225 kN, WB = 40 kN, and WC = 30 kN, determine the normal reactions of the wheels D, E, and F on the ground.

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RE

∑MX =0 ∑ MY = 0

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p.255, 5-73. Determine the force components acting on the ball-and-socket at A, the reaction at the roller B and the tension on the cord CD needed for equilibrium of the quarter circular plate.

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p.257, 5-85. The circular plate has a weight W and center of gravity at its center. If it is supported by three vertical cords tied to its edge, determine the largest distance d from the center to where any vertical force P can be applied so as not to cause the force in any one of the cables to become zero.

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d’

d’

(< r(1+W/P) )

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p. 261, prob. 5-96 The symmetrical shelf is subjected to a uniform load of 4 kPa. Support is provided by a bolt (or pin) located at each end A and A’ and by the symmetrical brace arms, which bear against the smooth wall on both sides at B and B’. Determine the force resisted by each bolt at the wall and the normal force at B for equilibrium.

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