### 3 EQUILIBRIUM

```FBD: 3D Force Reaction
1
FBD :
3D Force Reaction
1 unknown
3 unknown
2 unknown
No moment support? 2
3 unknowns
6 unknowns
6. smooth pin
5 unknown
3
7. bearing
journal-bearing support
4 unknown
thrust-bearing support
5 unknown
For some problem, the couples in both case should be
treated as zero to provide statical determinacy
4
If thrust barring
If glue or friction exist
6
3D Equilibrium Supports #2
7
3D Equilibrium Supports #3
8
3D Equilibrium Supports #4
9
3D Equilibrium Supports #5
Comparison with 2D supports
10
Equilibrium in 3D
R  F  0
M  M  0
or
or
Fx= 0
Fy= 0
Fz= 0
Mx= 0
My= 0
Mz= 0
any but only
three independent
(perpendicular) axis
any but only one point
Body
(bodies)
in
Equilibrium
OR
any but only three
independent axis
The 3 axises need not
interesect at one point
y
x
z
R  0
y
x
M
position (where the axis pass)
also matters.
Direction of moment
axis has a affect on
solving problem.
z
O
0
12
Equilibrium in 3D
or
R  F  0
M  M  0
or
Fx= 0
Fy= 0
Fz= 0
Mx= 0
My= 0
Mz= 0
at most 6 unknowns may be found.
any but only
three independent
(perpendicular) axis
any but only one point
Body
(bodies)
in
Equilibrium
OR
any but only three
independent axis
vector approach may be easier
- Each of the equation may be applied independently; e.g., an accelerating car on a
flat surface may be treated as in equilibrium in the vertical direction. Same for the
moment equations.
ax  0
z  0
- Not in this class, but be careful about the moment equations, things get very
complicated if the body is not spinning in a single plane!
13
nˆ

M

F
r
r
X
position vector:
from any point on line  to
any point on tline of action of
the force.
A
d
Y
Z
nˆ
F5

M F ,  { ( r  F )  nˆ }nˆ
F3
F1
 M  ?
{
5
{(r  F )  nˆ }
i
}nˆ
i=1
F4
F2
 { (r  F4 )  nˆ }nˆ
Forces which interest or parallel with axis,
do not cause the moment in that axis
15
Example 3/5
The uniform 7m steel shaft has a mass of 200kg and is
supported by a ball-and-socket joint at A in the
horizontal floor. The ball end B rests against the
smooth vertical walls. Find the force exerted by the
walls and the floor on the ends of the shaft.
this is not FBD.
17
FBD: frequent mistake
Correct FBD
System Isolation
Don’t forget axis
Write force name
Use (at least)
3 different colors
No axis
No system isolation
(surrounding still exists)
Caution:
In some slides using in this class, FBD may be drawn wrongly
according to the rule introduced, DO NOT IMITATE this style
18
W  mg  200(9.81)
 1962 N
h  7  (6  2 )  3
2
2
2
Fx  0
Fy  0
Fz  0
M
M
M
x
0
y
0
z
0
any point but only one point
A= (2,6,0)
B=(0,0,3)
G=(1,3,1.5)
  M A  0


iˆ
ˆj
1
rAG  rAB  iˆ  3 ˆj  1.5kˆ
2
iˆ
ˆj
kˆ
Vector Cross Product is
useful in 3D Problem
(3By  5890)iˆ  (3Bx  1962) ˆj  ( 2 By  6Bx )kˆ  0
2 6 3  1 3 1.5  0
Bx By 0
0 0 1962
  F  0


Independent
Eq. = 2
rAB  2iˆ  6 ˆj  3kˆ
rAB  ( Bx  Bx )  rAG W  0
kˆ
Use point A
Bx  654 N
By  1962 N
Ans
(654  Ax )iˆ  (1962  Ay ) ˆj  ( 1962  Az )kˆ  0
Ax  654 N
A y  1962 N
A y  1962 N
| A | 654  1962  1962  2850 N
2
2
2
Ans
19
The homogeneous plate shown
has a mass of 100 kg and is
subjected to a force and couple
moment along its edges. If it is
supported in the horizontal plane
by means of a roller at A, a balland-socket joint at B, and a cord at
C, determine the components of
reaction at the supports.
200 N-m
300 N
z
y
x
TC
Hibbeler Ex 5-15
[ Fx  0]
Bx  0
[ Fy  0]
By  0
981 N
By
Az
Bx
Bz
[ Fz  0]
Az  Bz  TC  300  981  0
20
z
TC
200 N-m
300 N
y
x
[ Fx  0]
Bx  0
[ Fy  0]
By  0
3m
[ Fz  0]
981 N
M  0
By
2m
Az
Bx
Bz
[ M x  0] 2TC  2Bz  (1)981  0
[ M y  0]
Az  Bz  TC  300  981  0
Selection of moment axis
OR
(1.5)300  (1.5)981
OR
[ M x  0]
[ M y  0] +3TC  (1.5)300
 3Bx  3 Az  200  0
Az  790N
Bz  -217N
 2 Az  (2)300  (1)981  0
 (1.5)981  200  0
TC  707N
21
Boom AB lies in vertical y-z
plane, is supported by a balland-socket joint at B and by 2
cables at A. Calculate tension
in each cable resulting from
the 20-kN force acting in the
horizontal plane at the
midpoint M of the boom.
Neglect the weight of the
boom.
22
z
3m
T1
A
20
M
4m
P




 4 iˆ  5 ˆj  10kˆ

T2  T2 
 42  52  102






T1
( 4 iˆ  5 ˆj  10kˆ )
11.87

T2
(4 iˆ  5 ˆj  10kˆ )
11.87
problem
Bz
By
B
D
  4 iˆ  5 ˆj  10kˆ

T1  T1 
 42  52  102

T2
10m
F

P  20(sin 20 iˆ  cos20 ˆj )
2m
y
There are 5 unknowns:
Bx
E
4m
C
use moment at B:
x
T1
T2
Bx
By Bz

 MB  0
(need only 2 components (eq.) from 3)
rBA  3 ˆj  10kˆ
rBM
1 ˆ
 (3 j  10kˆ)
2
M
B
 rBA  T1  rBA  T2  rBM  P  0
component x , component z
23
Technique: Finding unknown in 1 equation
The bent rod is supported at A
by a journal bearing, at D by a
ball-and-socket joint,
at B by means of cable BC.
Use only one equilibrium
equation to obtain a direct
solution for the tension in cable
BC. Assume that the bearing at
A is capable of exerting force
components only in the z and y
directions (since it is properly
aligned on the shaft).
Hibbeler Ex 5-19
30
Find T only in 1 scalar Equation.
TB  TB
rB  1 ˆj
Az
1
(0.2iˆ  0.3 ˆj  0.6kˆ)
0.7
Ay
TB
uˆ DA
rB
uˆ DA 

rE
W  100(9.81)kˆ
rE  0.5 ˆj
rDA
rDA
1 ˆ 1 ˆ
i
j
2
2
Dy
W
Dx
Dz
Scalar Equation
M
DA
{ uˆDA  ( ri  Fi ) }  0
TB W Dx Dy Dz Ay Az
i
MDA  0
uˆDA  (rB TB  rE  W )  0
31
uˆDA  rB  TB  rE  W   0
 MDA  0
TB  TB
1
(0.2iˆ  0.3 ˆj  0.6kˆ)
0.7
uˆDA  (rB  TB )  uˆDA  (rE  W )  0
ˆj

1 ˆ 1 ˆ  ˆ TB
ˆ
ˆ
ˆ
(
i
j )  ( j ) 
(0.2i  0.3 j  0.6k )  ( )  ( W kˆ)   0
0.7
2
2
2


1 ˆ 1 ˆ  TB
1

ˆ
ˆ
(
i
j )   (2k  6i )  ( )  (W iˆ)   0
2
2
2
7

6TB W
 0
7
2
You can use other rE and rB
(Solve as 2D problem)
TB  572.25 N
Az
TB
(TB ) z
6
 TB
7
1
2
1
1
(TB ) z
W
0
2
2 2
(TB ) z  W
1
0
2
6
1
(TB )  W  0
7
2
Ay
uˆ DA
1
2 2
rB
W
rE
Dx
Dy
Dz
32
3/90
Sign has a mass of 100 kg, center
of mass at the center of the sign.
- Support at C can be treated as a ball-andsocket joint.
- At D, support is provided in y-direction only
- Find the tensions, T1 and T2
- Find total force supported at C and the
lateral force R supported at D.
( 4iˆ  1.5 ˆj  2.5kˆ)
T1  T1
L1
L1  4 2  1.52  2.52  4.95
( 2.5iˆ  1.5 ˆj  2.5kˆ)
T2  T2
L2
L2  2.52  1.52  2.52  3.84
34
this is not a FBD, why?
L1  42  1.52  2.52  4.95
T1 
T1
( 4iˆ  1.5 ˆj  2.5kˆ)
L1
L2  2.52  1.52  2.52  3.84
T2 
T2
(2.5iˆ  1.5 ˆj  2.5kˆ)
L2
M
R
E
Cx
Cy
M
x
F
y
0
F
z
0
2.5
2.5
T1 
T2  Cz  0
L1
L2
 Cy (1)  0
0
 Cx (3.5)  100(9.81)(2)  0
z
0

x
0

AB
M
Cz
1.5
1.5
 T1 
T2  R  0
L1
L2
0
F
Get R and C
(1.5T1 )
(1.5T2 )
(4) 
(2.5)  0
L1
L2
4
2.5
T1 
T2  Cx  0
L1
L2
Get T1 and T2
35
Properly Alignment and Moment Support
A,B,C:
journal
bearings
“Properly Alignment”
If the line of pipe is initially
placed without pre-torsion, the 6
couple reactions are not necessary
to maintain equilibrium, and then
can be considered as not existing.
The 6 force reactions developed by the
bearings are sufficient for maintaining the
equilibrium since they prevent the shaft from
rotating about any of the coordinate axes.
6 unknown reactions with
6 independent equations.
37
Properly Alignment and Moment Support
“initially
properly
aligned”
Even though the line of pipe is
initially placed without pretorsion or properly aligned.
The 4 force reactions developed at
bearing A and tension T are not sufficient
to maintain the equilibrium of the body.
The couple reactions are developed to make the object in equilibrium
38
Properly Alignment and Moment Support
“properly
aligned”
“properly
aligned”
The 6 force reactions developed by the
bearing, hinge and ball are sufficient for
maintaining the equilibrium, so the
couple reactions will not be developed at
the bearing and hinge.
39
thrust-bearing
“initially
properly
aligned”
40
Sample 3/6
A 200-N force is applied to the handle of the hoist in the direction shown.
The bearing A supports the thrust (force in the direction of the shaft axis),
Determine the mass m which can be supported and the total radial force
exerted on the shaft by each bearing. Each bearing is initially properly
align.
42
P  200(cos60sin 45iˆ  sin60 ˆj  cos60cos45kˆ)
By
 (70.7iˆ  173.2 ˆj  70.7kˆ)
Bx
Ay
P
Ax
F  0 F  0 F  0
x
M
x
y
0
M
at same point
z
y
0
M
z
0
No need to be
the same point
Az
m(9.81)100 – 173.2(250) = 0
-Bx(150)-70.7(175)+70.7(250)= 0
2D FBD
Ay(150)+m(9.81)200-173.2(325)=0
Ax+Bx-70.7=0
Az-70.7=0
By+Ay-m(9.81)100-173.2=0
2D View
convenient when almost forces are othorgonal.43
3/83
The shaft, level and handle
are welded together and
constitute a single rigid body.
Their combined mass is 28 kg
with mass center at G. The
assembly is mounted in A and
B and rotation is prevented by
exerted on the shaft by bearing
A and B, while the 30 N-m
couple is applied to the handle
as shown.
BearingA,B : Journal bearing
45
ignore
2 force member
T // plane x-y
2 force member
Most forces are
parallel with
rectangular axis
z
Ay
By
T cos 
y
T sin 
T
z
Ax
Bearing A,B :
Journal bearing
mg
Bx
y

M  30
mg
Ax  Bx
46
| A |
Ax2  Ay2
| B | Bx2  By2
T=251N
Ax=11.74N
30-Tsin36.9(0.6)+28(9.81)(0.22)=0
28(9.81)(0.3)-Tsin36.9(0.5)-Ax(0.6)=0
Bx=112.2N
Ax+Bx+Tsin36.9-28(9.81)=0
300
300
600
Ay
T cos 
200
z
By
Ay(0.6)-Tcos36.9(0.5)=0
Ay+By-Tcos36.9=0
Ay=167.5N
By=33.5N
200
y
T sin 
T
100
200
Ax
mg
300
z
Bx
y

380
M  30
220
mg
Ax  Bx
47
Bearing A,B :
Journal bearing
30-Tsin36.9(0.6)+28(9.81)(0.22)=0
28(9.81)(0.3)-Tsin36.9(0.5)-Ax(0.6)=0
Ax+Bx+Tsin36.9-28(9.81)=0
Ay(0.6)-Tcos36.9(0.5)=0
Ay+By-Tcos36.9=0
Bearing A,B :
thrust bearing
Az+Bz=0
T=251N
Ax=11.74N
Bx=112.2N
Ay=167.5N
By=33.5N
Cannot be solved
to get exact value.
7 unknowns, 6 equations
You will get exact value for some
unknowns but not for all of them.
48
Statically determinate (SD) problems
the problem that can be analyzed by
using the equilibrium condition alone.

number of
unknowns
F5
F4
For one obj, usually
 F  0 3 scalar eqs (2D)
 M  0 or 6 scalar eqs (3D)
number of independent
equilibrium equations
maybe > (3 or 6)
(many FBDs)
y
x
F6
z

F3

F1

F2
maybe < ( 3 or 6 )
FE
(some specific problems
i.e. point equilibrium)
54
Categories of Equilibrium (3D)

F2
1) Concurrent at a point
y

F3
O

F1
x
0
F
x
0
M
F
dependant  M
y
0
F
y
0
M
0
z
z
0

     
M

r
 A  F1  r  F2  r  F3

 r  ( F1  F2  F3 )  0
y
x
2) Concurrent with a line
z
Independent Equations:

F3

F1
Independent Equations:
z

F4

F2
x
x
0
F
x
0
M
F
M
y
0
F
y
0
M
z
z
0
0
All couples occuring when
moving force |_ this direction
M | _ r ( M  r  F)
55
Categories of Equilibrium (3D)

F4
y
x
3) Parallel
Independent Equations:

F2

F3

F1

F1

F4
y
x

F3
x
0
F
x
0
M
F
M
All couples occuring when
moving force // the plane

M

F2
z
z
y
0
F
y
0
M
z
z
0
0
M | _ F ( M  r  F)
4) General
Independent Equations:
x
0
F
x
0
M
F
M
y
0
F
y
0
M
z
z
Constraints and statical determinacy: see the book
0
0
56
SD/SI
1) SD
Type of Statics Problem
(statically determinate)
the problem that can be solved
for all unknown with only static
equilibrium equations.
2) SI-redundant
mass m
bar
Unknown 3 ,
Independent eq 3
(redundant statically indeterminate)
the problem that can not be
solved for all unknown with only
static equilibrium equations.
Unknown 4 ,
Independent eq 3
3) SI-Improper
(improper statically indeterminate)
the problem that can not be
solved due to improper support.
unknown 3 ,
independent 2
SD/SI
Statically Indeterminacy
Statically indeterminate (SI) problems
the problem that can not be analyzed by
using the equilibrium condition alone.
SI-redundant
SI-improper
Problem with
redundant supports
Problem with imporper supports
- Not able to maintain equilibrium
No solution: (cant maintain moment)
number of
unknowns

number of independent
equilibrium equation
unknown 3 ,
independent 2
59
3/91
The window is temporarily held
open in the position shown. If
a=0.8 m and b = 1.2 m and the
mass of window is 50 kg with
mass center at is geometric center,
determine the compressive force
F(CD) in the prop and all the
component of the forces exerted
by the hinges B on the windows.
Assume that A is a thrust-bearing
hinge but the hinge B is not.
61
D : (0.8sin50 , 0.8cos50 ,0)  (0.613, 0.514,0) C : (0,0.8,0.3)
By
y
1.2
Ay
Ax
Az
z

F  FnˆCD  F (0.828iˆ  0.286ˆj  0.406kˆ)
x
( M A ) z  0 :
Bx
mg
0.8
F solved
 50(9.81)(0.4 sin 50)  (0.828F )(0.8)  0
( M A ) y  0 :
Bx solved
 Bx (1.2)  (0.828F )(0.3)  0
50
C
( M A ) x  0 :
D
By solved
By (1.2)  50(9.81)(0.6)
Ans
 (0.406F )(0.8)  (0.386F )(0.3)  0
 Fx  0 :
F
 0:
F
 0:
y
z
Ax  Bx  0.828F  0
Ax solved
Ay  By  0.386F  50(9.81)  0 Ay solved
Az  0.406F  0
Az solved
62
Basic concept:
Since all forces are given in orthogonal system, it is easier to solve the
problem with orthogonal projections as sample problem 3/6.
63
Cx is the key for determining Ax and Bx
On x-z plane: take moment at line AB to
reduce unknowns
M
AB
  W  0.150 Cx  3
0  W  0.150 Cx  3
Cx  W  0.050  176.6 N
On x-y plane: take moment at point B to
reduce unknowns
y
x A
x
z
C
x
Az
yW
Bx
M
Az +Bz
z
Cx
FBD
x W
Ax  3  Cx  2.4
0  Ax  3  Cx  2.4
Ax +Bx
B
B
z
Ax  Cx  0.8  141.3N
Ans
Then sum forces on x-axis
F
x
Ax  Cx  Bx
0  Ax  Cx  Bx
Ans 64
Bx   Ax  Cx  35.3N
Recommended Problem
3/73 3/82 3/92 3/93
65
Equilibrium
Here ends the most important
chapter of the subject.
66
Review Quiz #4
Review
• Classify problems in equilibrium into SD and SI
categories
– How can I recognize the SD and SI problems?
– Why do we need to differentiate between SD and
SI problems?
– What does it mean when an object has redundant
supports? How about the improper supports?
– What is the degree of redundancy?
67
Concepts
Review
• When a body is in equilibrium, the resultant force
and couple about any point O are both zero.
Problems can be analysed using free body
diagrams (FBDs).
• Statically determinate (SD) problems can be solved
using the equilibrium conditions alone, while the
statically indeterminate (SI) problems cannot due to
too few or too many supports or constraints.
68
Chapter Objectives Descriptions #1
• Analyse objects (particles and rigid bodies) in
equilibriums
– Specify the condition of equilibrium
– Describe objects in equilibrium in physical and
idealized worlds
– Choose appropriate bodies/parts of bodies for the
analyses
– Specify support reactions from physical worlds
and vice versa
– Specify loads from physical worlds and vice versa
– Draw free-body diagrams
– Calculate the support reactions from condition of
equilibrium
– Identify 2-force and 3-force members
69
Chapter Objectives Descriptions #2
• Classify problems in equilibrium into SD and SI
categories
– Specify types and conditions of SD and SI
problems
– Describe physical meanings of SI problems
– Use FBDs differentiate between SD and SI
problems
– Obtain the degree of redundancy (when
applicable)
70
Review Quiz #1
Review
• Analyse objects (particles and rigid bodies) in
equilibriums
– What are the conditions of equilibrium for each
type of bodies?
– What are the physical meanings of 2D contact
with cable, spring, smooth surface, contact with
rough surface, roller, pin, fixed and slider
supports?
– What are the physical meanings of 3D contact
with cable, smooth surface support, roller, ball &
socket, journal bearing, thrust bearing, smooth
pin, hinge and fixed support?
– What are the differences between 2D and 3D
FBDs?
71
Review Quiz #2
Review
• Analyse objects (particles and rigid bodies) in
equilibriums
– Can we analyze objects in equilibrium without
FBDs? Why or why not?
– What are the steps in the sketching of an FBD?
– In drawing the FBDs, how did we choose the
directions of support reactions?
– Why do we have to delete the physical supports,
or isolate objects, before adding the support
reactions?
– How many independent equilibrium equations can
you obtain from 2D and 3D diagrams? What are
they?
72
Review Quiz #3
Review
• Analyse objects (particles and rigid bodies) in
equilibriums
– Although we can choose the point about which the
moments are evaluated arbitrarily, do you have
some guidelines of choosing for simpler analyses?
– Why it is worthwhile to recognize that an object is
a two-force member? What about the three-force
member? How can I identify 2-force and 3-force
members in 2D and 3D problems?
73
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