CHAP 5 Equilibrium of a Rigid body
5.1 Conditions for Rigid body equilibrium
Consider a rigid body which is at rest or moving with x y z reference at constant velocity
x
2F
3F
4F
ij
y
z
1F
rigid body
Free body diagram of ith particle of the body
ijf
iF
ir
x
y
z
i
j
External force (外力 ) : gravitational, electrical, magnetic or contact force
iF
Internal force (內力 )if
if
n
j 1
ijf ij
Force equilibrium equation for particle i
0ii fF (Newton first law)
0 ii fF
0if
0FFi
Force equilibrium equation for the whole body
(Newton’s 3rd law 作用力與反作用力 )
Moment of the forces action on the ith particle about pt. O
iiijo fFrM
0 iiii frFr
Moment equilibrium equation for the body
0
0
0
0
jjo
jj
jjjj
jjjjo
FrM
fr
frFr
fFrM
Equations of equilibrium for a rigid body are
0 F
0 oM
力平衡
力矩平衡
1. Free-body Diagram (1) F.B.D A sketch of the outlined shape of the body represents it as being
isolated or “free” from its surrounding , i.e ., a free body”.
5.2 Equilibrium in Two Dimensions
(2) Support Reactions
A .Type of support : see Table 5-1
B . General rules for support reaction:
If a support prevents the translation of a body in a given direction, then a force is developed on the body in that direction . Likewise, if rotation is prevented, a couple moment is exerted on the body.
(a) roller or cylinder support
AyF
ByF
xF
Fy
xF
Fy
(b) pin support
Examples:
(c) Fixed support
M
FAy
FAx
(3) External and Internal forces
A. Internal force
Not represented on the F.B.D. became their net effect on the body is zero.
B. External force
Must be shown on the F.B.D.
(a) “Applied” loadings
(b) Reaction forces 反作用力(c) Body weights 重力
(4) Weight and the center of gravity
The force resultant from the gravitational field is referred as the weight of the body, and the location of its point of application is the center of gravity G.
w
body
平行力 等效力系
(ch4)
w
P=G( 重心 )
2. Equations of Equilibrium for 2D rigid body
y
x Couple moment
1F
2F
3F
1M
2M
0M0F 0
0M
0F
0F
0
y
x
Here: Fx
Fy
0M
algebraic sum of x components of all force on the body.
algebraic sum of y components of all force on the body.
algebraic sum of couple moments and moments of allthe force components about an axis xy plane and⊥passing 0.
(1) Conditions of equilibrium
(2) Alternative equilibrium equation
0M
0M
0F
B
A
a
When the moment points A and B do not
lie on a line that is “perpendicular” to the axis a.
(A)
ARA
R
MM
FF
aaFF
MM
Ra
RAA
0
00 0/ RRBAB FoFrM
0M
0M
0M
C
B
A
a
a
A
B
C AF
AM
(B)
Points A, B and C do not lie on the same line
ARA
R
MM
FF
0
00
BR
RAA
MBAcollinearF
MM0/ RRCAC FoFrM
600N
200N
2m 3m 2m
100N
y
xBxA B
ByAy
3 unknown Ax, Bx, By
Equations of equilibrium
0.0245sin600545sin600721000
020010045sin6000
045cos6000
AyM
ByAyFy
BxFx
B
3 equations for 3 unknowns
(3) Example
BAR FFF 0
0Mo
5.3 Two-and Three-Force Members
1. Two-Force member
A member subject to no couple moments and forces applied at only two points on the member.
A
B
A
B
FA
FBEquations of Equilibrium
2. Three-Force member
0F 0F 0Mo 0Mo
A member subject to only three forces, which are either concurrent or parallel if the member is in equilibrium.
(1)Concurrent (2)parallel(3力交於O 點 ) (3力相交無限遠處 )
o
F1F2
F3
F1 F2
F3
5.4 Equilibrium in Three Dimensional Rigid Body
1. Free Body Diagrams
(1)F.B.D
Same as 2D equilibrium problems
(2)Support Reactions
A. Types of support:see Table 5-2
B. General rules for reaction
Same as two-dimensional case
Examples:
(a) Ball and Socket joint
No translation along any direction
Rotate freely about any axis
Fx
Fy
Fz
3 reaction forces
(b) single journal bearing
Rotate freely about its longitudinal axis
Translate along its longitudinal direction
two unkown forces
and couple momentsx
yMx
Mz
z
Fz
Fx (c) single pin
Only allow to rotate about a specific axis.
x
y
z
Mz
My
FxFy
Fz
Three unkown forces
and two couple moments
A. Vector equations of equilibrium
B. Scalar equations of equilibrium
0F 0Mo
0Fx
0Fy
0Fz
0Mx
0My
0Mz
2. Equations of Equilibrium
1. Redundant constraints
(1) Redundant constraints
Redundant supports are more than necessarily to hold a body in equilibrium.
2KN-m 500N
5.6 Constraints for a rigid body
Ex:
Equation of motion=3
5 unknown reactions >3 equation of motion
there are two support reactions which are redundant supports and more than necessarily.
(2) Statically indeterminate 靜不定 There are more unknown loadings on the body than equations of equilibrium available for the solution.
F.B.D of above example
x
y
A BC
2KN-m
Ay By Cy
MA
Ax
500N
unknown loadings AX,AY,MA,BY,CY;5
Equations of equilibrium ΣFX=0,ΣFY=0,ΣMA =0;3
5>3
Statically indeterminate structure
(3)Solutions for statically indeterminate structure
Additional equations are needed ,which are obtained from the deformation condition at the points of redundant support based on the mechanics of deformation, such as mechanics of materials.
Equations of Equilibrium for above example are
+ ΣFX=0 AX=0
+ΣFY=0 500-AY-BY-CY=0
+ΣMA=0 MA-2-DYBY-DCCY=0
Need two more equations to solve the five unknown forces.
(1) Reaction force = equations of equilibrium
If this kind of improper constraint occurs then system is instable
A. The lines of action of the reactive forces intersect points on a common axis (concurrent).
2.Improper Constraints
A
B
C
0.2m
100N
Ao
B
C100NF.B.D FC
FAFB
body will rotate about Z-axis or point O0
0
o
yx
M
FF
o
FC
A
B C
FB
FA
100N
F.B.D
Body will translate along x direction.
B. The reactive forces are all parallel
100NA
B C
0 xF
(2) Reaction forces < equations of equilibrium
If the body is partially constrained then it is in instable condition
100N
Stable?
F.B.D
0
0
0
o
y
x
M
F
F
FA FB
100N
o
Not in equilibrium
Top Related