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
5-2
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)
5-4
(continued with ch5a.ppt)
5-5
5-6
Free-Body Diagrams
Example 5.1
Draw the free-body diagram of the uniform
beam. The beam has a mass of 100kg.
5-7
Free-Body Diagrams
Solution
Free-Body Diagram
5-8
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
5-10
p.216, 5-9. Draw the free-body diagram of the beam, which is pinconnected at A and rocker-supported at B.
5-11
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
5-14
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.
5-17
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
5-20
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
5-21
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.
5-22
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.
5-27
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.
5-31
120 kN
1.2 m
1.5 m
5-32
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.
5-33
5-34
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.
5-35
5-36
5-37
Free Body Diagrams (Three Dimension)
5-38
5-39
5-40
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.
5-41
Solution
5-42
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
5-44
Constraints for a Rigid Body
„
Improper Constraints-intersect points on a
common axis
同軸
5-45
Constraints for a Rigid Body
„
Improper Constraints-intersect points on a
common axis
2-D reaction force 共點
5-46
Constraints for a Rigid Body
„
Improper Constraints-reactive force are all parallel
5-47
Constraints for a Rigid Body
„
Partially Constrained 部分限制
5-48
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.
5-49
RE
∑MX =0
∑ MY = 0
5-50
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.
5-51
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.
5-52
d’
d’
(< r(1+W/P) )
5-53
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.
5-54
5-55