2 d equilibrium-split

mechanics

1. Engineering Mechanics
Equilibrium of Rigid Bodies

2. Main Text
•Major chunk of lectures based on:
•Vector Mechanics for Engineers, Beer, Johnston et
al., 10
th
ed., McGraw-Hill.
•Referred to as BJ10. Indian Edition available.
•Also, some problems from Beer and Johnston 3
rd
and 8
th
editions, BJ3 and BJ8.
•Dynamicswill be exclusively taught from BJ10.
Many wonderful new resources available in BJ10.
Become more clear as we proceed.
•Many slide contents in our lectures from BJ10
Instructor resources.
•Attractive online features available for Instructors
http://highered.mcgrawhill.com/sites/1259062910/information_center_view0/
•For purchase/other details kindly contact:
•[email protected]
[email protected]

3. Secondary Text
•Many really interesting and challenging
problems from:
–Engineering Mechanics: Statics/Dynamics,
Meriamand Kraige, Eds. 2, 5, 7. (MK3,
MK5, MK7).

4. Online resources
•Nice demonstrations from Wolfram (look under
the mechanics section). Will show some of them.
–http://demonstrations.wolfram.com/
•Beautiful lectures notes by Prof. Allan Bower at
Brown University
–http://www.brown.edu/Departments/Engineering/Courses/En4/Notes/notes.html
•Nice general lectures on Dynamics on youtube:
–http://www.youtube.com/user/mellenstei
•Nice animations to textbook problems:
–http://wps.prenhall.com/wps/media/objects/3076/3149958/studypak/index_st.h
tml

5. Application
4 –5 BJ10
Engineers
designing this
cranewillneedto
determine the
forcesthatacton
thisbodyunder
variousconditions.
Slide from BJ10

6. Previously discussed
•Vector mechanics
•Definition of force and moment/torque
•Equivalent systems
•Distributed loads
•Centroid, Moment of Inertia etc.

7. What’s a rigid body?
•Mathematically a body is rigid means:
–Distance between any two points of the body
does not change in the course of motion
•In reality, no body is fully rigid! Rigidity
only implies that the deformations are very
small as compared to the body dimensions.
•Many real life structures can be idealized
as rigid!

8. Examples
IIT main building Crane
http://www.yjcrane.com/img/cranes/crawler-crane.jpg

9. Equilibrium
•Systemisin
equilibriumif
andonlyifthe
sumofallthe
forces and
moment(about
any point)
equalszero.
If true for one point O, true
w.r.tany other point! Convince
yourself

10. Supports and Equilibrium
•Any structure is made of many
components.
•The components are the be connected by
linkages.
•Other wise the structure will lose its
integrity.
•Different component of structure talk to
each other via linkages.
•The structure should be globally
supported to prevent it from falling over.

11. Different Structural Supports
•Supports are required to maintain
system in equilibrium.
•Too few supports makes system unstable
general loading
•Too many supports make the system
over-rigid.

12. Constraints and Reactions
•There is an intricate relationship between
kinematics (motion) and reactions (forces).
•Always note that in the case of supports
displacement (rotation)and force (torque)in
any given directionare complementary.
•If a support rigidly constrains a given degree
of freedom (DOF) for a rigid body then it
gives rise to a reaction corresponding to that
DOF.
•Similarly if a support freely allows motion of
particular DOFthen there is no reaction
from the support in that direction.

13. What are 2D structures?
•No real life structure is 2D!
•So what’s the deal with 2D?

14. What are 2D structures?
•Symmetry in the structure and loading about a
plane. The problem can then be simplified to a
2D problem! Convince yourself.
BJ10

15. What are 2D structures?
•The third dimension is very small as
compared to the other two and loads are
coplanar.

16. Non-Symmetrical but bodies
connected by pin are very close to
each other

17. Reactions at Supports and Connections for a Two-Dimensional
Structure
4 -17
•Reactions equivalent to
a force with known line
of action.
Slide from BJ10

18. Reactions at Supports and Connections for a Two-Dimensional
Structure
4 –18 BJ10
•Reactions equivalent to
a force of unknown
direction and
magnitude.

19. Reactions at Supports and Connections for a Two-Dimensional
Structure
4 –18 BJ10
•Reactions equivalent to
a force of unknown
direction and
magnitude.
•Reactions equivalent
to a force of unknown
direction and
magnitude and a
couple of unknown
magnitude.

20. Summary
•Pin support
•Pin connection
http://oli.cmu.edu

21. Roller
http://oli.cmu.edu

22. Slot Connection
http://oli.cmu.edu

23. Simple Examples
Roller Support
Fixed Support

24. Need for different types of
supports
http://fastestlaps.com/articles/wads_on_sportscars_audi_rs6.html
www.howstuffworks.com

25. Free Body Diagram (FBD)
The Heart of mechanics
•Single most important concept in
engineering mechanics.
•Zoom in on a given component of a structure.
•Means replace supports (connections) with the
corresponding reactions.
•Replace kinematic constraints with
corresponding reactions.
•Concepts will get more clear as we proceed
further.

26. Simple examples
•Copyright, Dr. Romberg
FBD
FBD

27. More
Examples
of FBD

28. Practice (BJ10)
4 –26 BJ10
The frame shown supports part of the
roof of a small building. Your goal is to
draw the free body diagram (FBD) for
the problem. The tension in the cable
BDF is 150kN.
On the following page, you will choose
the most correct FBD for this problem.
First, you should draw your own FBD.

29. Practice
4 –27 BJ10
Choose the most
correct FBD for the
original problem.

30. Practice
4 –27 BJ10
A B
C D
150 kN
150 kN
150 kN
150 kN
Choose the most
correct FBD for the
original problem.

31. Practice
4 –27 BJ10
A B
C D
150 kN
150 kN
150 kN
150 kN
Choose the most
correct FBD for the
original problem.
B is the most correct, though C is
also correct. A & D are incorrect;
why?

32. Practice
4 –27 BJ10
A B
C D
150 kN
150 kN
150 kN
150 kN
Choose the most
correct FBD for the
original problem.
B is the most correct, though C is
also correct. A & D are incorrect;
why?
why each choice is
correct or incorrect?

33. Equations of equilibrium in 2D
•Threeequations per free body.
•Writing more than threeequations per free body is punishable
by law (at least it should be).
We can also use equations like this
or like this where A, B, C are not in a
straight line
•C

34. Problem 1
•Determine the tension in cable ABDand
reaction at support C.

36. Categories of Equilibrium in 2D
MK5

37. Adequacy of
Constraints
MK5

38. Problem 2 (BJ10)
•A70kg(W)overheadgaragedoorconsistsofauniformrectangular
panelAC2100mmhigh(h),supportedbythecableAEattachedatthe
middleoftheupperedgeofthedoorandbytwosetsoffrictionless
rollersatAandB.Eachsetconsistsoftworollersoneeithersideofthe
door.TherollersAarefreetomoveinhorizontalchannels,whilerollers
Bareguidedbyverticalchannels.Ifthedoorisheldinthepositionfor
whichBD=1050mm,determine(a)thetensioninthecableAE,(2)the
reactionateachofthefourrollers.Assumea=1050mm,b=700mm

40. Link: Two-Force Member
•Member with negligible weight and arbitrary
shape connected to other members by pins

41. Two Force member
http://oli.cmu.edu

42. Equilibrium of a Three-Force
Body
•Consider a rigid body subjected to forces
acting at only 3 points.
Slide from BJ10

43. Equilibrium of a Three-Force
Body
•Consider a rigid body subjected to forces
acting at only 3 points.
•Assuming that their lines of action intersect,
the moment of F
1and F
2about the point of
intersection represented by Dis zero.
Slide from BJ10

44. Equilibrium of a Three-Force
Body
•Consider a rigid body subjected to forces
acting at only 3 points.
•Assuming that their lines of action intersect,
the moment of F
1and F
2about the point of
intersection represented by Dis zero.
•Since the rigid body is in equilibrium, the sum
of the moments of F
1, F
2, and F
3about any axis
must be zero. It follows that the moment of F
3
about Dmust be zero as well and that the line
of action of F
3must pass through D.
Slide from BJ10

45. Equilibrium of a Three-Force
Body
•Consider a rigid body subjected to forces
acting at only 3 points.
•Assuming that their lines of action intersect,
the moment of F
1and F
2about the point of
intersection represented by Dis zero.
•Since the rigid body is in equilibrium, the sum
of the moments of F
1, F
2, and F
3about any axis
must be zero. It follows that the moment of F
3
about Dmust be zero as well and that the line
of action of F
3must pass through D.
•The lines of action of the three forces must be
concurrent or parallel.
Slide from BJ10

46. Sample Problem 4.6 BJ-10
4 –38 BJ10
A man raises a 10-kg joist,
of length 4 m, by pulling on
a rope.
Find the tension Tin the
rope and the reaction at A.

47. Sample Problem 4.6 BJ-10
4 –38 BJ10
A man raises a 10-kg joist,
of length 4 m, by pulling on
a rope.
Find the tension Tin the
rope and the reaction at A.
SOLUTION:
•Create a free-body diagram of the
joist. Note that the joist is a 3 force
body acted upon by the rope, its
weight, and the reaction at A.

48. Sample Problem 4.6 BJ-10
4 –38 BJ10
A man raises a 10-kg joist,
of length 4 m, by pulling on
a rope.
Find the tension Tin the
rope and the reaction at A.
SOLUTION:
•Create a free-body diagram of the
joist. Note that the joist is a 3 force
body acted upon by the rope, its
weight, and the reaction at A.
•The three forces must be concurrent
for static equilibrium. Therefore, the
reaction Rmust pass through the
intersection of the lines of action of
the weight and rope forces.
Determine the direction of the
reaction R.

49. Sample Problem 4.6 BJ-10
4 –38 BJ10
A man raises a 10-kg joist,
of length 4 m, by pulling on
a rope.
Find the tension Tin the
rope and the reaction at A.
SOLUTION:
•Create a free-body diagram of the
joist. Note that the joist is a 3 force
body acted upon by the rope, its
weight, and the reaction at A.
•The three forces must be concurrent
for static equilibrium. Therefore, the
reaction Rmust pass through the
intersection of the lines of action of
the weight and rope forces.
Determine the direction of the
reaction R.
•Utilize a force triangle to determine
the magnitude of the reaction R.

50. Sample Problem 4.6
4 –39 BJ10
•Create a free-body diagram of the
joist.

51. Sample Problem 4.6
4 –39 BJ10
•Create a free-body diagram of the
joist.
•Determine the direction of the
reaction R.
 
 
6361
4141
3132
tan
m2.313 m51508282
m515020tan m4141)2545(cot
m4141
m828245cosm445cos
ooo
2
1
oo
.
.
.
AE
CE
..BDBFCE
..CDBD
.AFAECD
.ABBF





 
6.58

52. Sample Problem 4.6
4 –40 BJ10
•Determine the magnitude of the
reaction R.
38.6sin
N 1.98
110sin4.31sin

RT N 8.147
N9.81


R
T

53. Sample Problem 4.1 BJ10
A fixed crane has a mass of
1000 kg and is used to lift a
2400-kg crate. It is held in
place by a pin at Aand a rocker
at B. The center of gravity of
the crane is located at G.
Determine the components of
the reactions at Aand B.

54. Sample Problem 4.1 BJ10
A fixed crane has a mass of
1000 kg and is used to lift a
2400-kg crate. It is held in
place by a pin at Aand a rocker
at B. The center of gravity of
the crane is located at G.
Determine the components of
the reactions at Aand B.
SOLUTION:
•Create a free-body diagram for the
crane.

55. Sample Problem 4.1 BJ10
A fixed crane has a mass of
1000 kg and is used to lift a
2400-kg crate. It is held in
place by a pin at Aand a rocker
at B. The center of gravity of
the crane is located at G.
Determine the components of
the reactions at Aand B.
SOLUTION:
•Create a free-body diagram for the
crane.
•Determine the reactions at Bby
solving the equation for the sum of
the moments of all forces about A.
Note there will be no contribution
from the unknown reactions at A.

56. Sample Problem 4.1 BJ10
A fixed crane has a mass of
1000 kg and is used to lift a
2400-kg crate. It is held in
place by a pin at Aand a rocker
at B. The center of gravity of
the crane is located at G.
Determine the components of
the reactions at Aand B.
SOLUTION:
•Create a free-body diagram for the
crane.
•Determine the reactions at Bby
solving the equation for the sum of
the moments of all forces about A.
Note there will be no contribution
from the unknown reactions at A.
•Determine the reactions at Aby
solving the equations for the
sum of all horizontal force
components and all vertical
force components.

57. Sample Problem 4.1 BJ10
A fixed crane has a mass of
1000 kg and is used to lift a
2400-kg crate. It is held in
place by a pin at Aand a rocker
at B. The center of gravity of
the crane is located at G.
Determine the components of
the reactions at Aand B.
SOLUTION:
•Create a free-body diagram for the
crane.
•Determine the reactions at Bby
solving the equation for the sum of
the moments of all forces about A.
Note there will be no contribution
from the unknown reactions at A.
•Determine the reactions at Aby
solving the equations for the
sum of all horizontal force
components and all vertical
force components.
•Check the values obtained for
the reactions by verifying that
the sum of the moments about B
of all forces is zero.

58. Sample Problem 4.1 BJ10
4 -42
•Create the free-body
diagram.

59. Sample Problem 4.1 BJ10
4 -42
•Create the free-body
diagram.
•Determine Bby solving the equation
for the sum of the moments of all
forces about A.  
0m6kN523
m2kN819m510


.
..B:M
A kN1107.B

60. Sample Problem 4.1 BJ10
4 -42
•Create the free-body
diagram.
•Determine Bby solving the equation
for the sum of the moments of all
forces about A.  
0m6kN523
m2kN819m510


.
..B:M
A kN1107.B
•Determine the reactions at Aby
solving the equations for the sum of all
horizontal forces and all vertical forces.00  BA:F
xx kN1.107
x
A

61. Sample Problem 4.1 BJ10
4 -42
•Create the free-body
diagram.
•Determine Bby solving the equation
for the sum of the moments of all
forces about A.  
0m6kN523
m2kN819m510


.
..B:M
A kN1107.B
•Determine the reactions at Aby
solving the equations for the sum of all
horizontal forces and all vertical forces.00  BA:F
xx kN1.107
x
A 0kN523kN8190  ..A:F
yy kN333.A
y


62. Sample Problem 4.1 BJ10
4 -42
•Create the free-body
diagram.
•Check the values obtained.
•Determine Bby solving the equation
for the sum of the moments of all
forces about A.  
0m6kN523
m2kN819m510


.
..B:M
A kN1107.B
•Determine the reactions at Aby
solving the equations for the sum of all
horizontal forces and all vertical forces.00  BA:F
xx kN1.107
x
A 0kN523kN8190  ..A:F
yy kN333.A
y


63. Hydraulic Cylinder
show Mathematicademo on this

64. System constrained to
various degrees

65. Questions?
Tutorial after tea-break

66. Tutorial on 2D equilibrium

67. Problem 1 (BJ10)
4 –47 BJ10
The frame supports part of the
roof of a small building. The
tension in the cable is 150 kN.
Determine the reactions at the
fixed end E.

68. Problem 1 (BJ10)
4 –47 BJ10
The frame supports part of the
roof of a small building. The
tension in the cable is 150 kN.
Determine the reactions at the
fixed end E.
SOLUTION:
-Discuss with a neighbor the
steps for solving this problem.

69. Problem 1 (BJ10)
4 –47 BJ10
The frame supports part of the
roof of a small building. The
tension in the cable is 150 kN.
Determine the reactions at the
fixed end E.
SOLUTION:
-Discuss with a neighbor the
steps for solving this problem.
•Create a free-body diagram
for the frame and cable.

70. Problem 1 (BJ10)
4 –47 BJ10
The frame supports part of the
roof of a small building. The
tension in the cable is 150 kN.
Determine the reactions at the
fixed end E.
SOLUTION:
-Discuss with a neighbor the
steps for solving this problem.
•Apply the equilibrium
equations for the reaction
force components and couple
at E.
•Create a free-body diagram
for the frame and cable.

71. Problem-1 BJ10
•The free-body diagram was
created in an earlier
exercise.
•Apply one of the three
equilibrium equations.
Try using the condition
that the sum of forces in
the x-direction must sum
to zero.

72. Problem-1 BJ10
•The free-body diagram was
created in an earlier
exercise.
•Apply one of the three
equilibrium equations.
Try using the condition
that the sum of forces in
the x-direction must sum
to zero.  0kN150
5.7
5.4
:0  xx
EF  0kN150936cos:0
o
 .EF
xx
•Which equation is correct? 0Nk150
57
6
:0 
.
EF
xx
A.
B.
C.
D. 0kN150936sin:0
o
 .EF
xx
E. 0kN150936sin:0
o
 .EF
xx

73. Problem-1 BJ10
•The free-body diagram was
created in an earlier
exercise.
•Apply one of the three
equilibrium equations.
Try using the condition
that the sum of forces in
the x-direction must sum
to zero.  0kN150
5.7
5.4
:0  xx
EF kN 0.90
x
E  0kN150936cos:0
o
 .EF
xx
•Which equation is correct? 0Nk150
57
6
:0 
.
EF
xx
A.
B.
C.
D. 0kN150936sin:0
o
 .EF
xx
E. 0kN150936sin:0
o
 .EF
xx kN 0.90
x
E

74. Problem-1 BJ10
•The free-body diagram was
created in an earlier
exercise.
•Apply one of the three
equilibrium equations.
Try using the condition
that the sum of forces in
the x-direction must sum
to zero.  0kN150
5.7
5.4
:0  xx
EF kN 0.90
x
E  0kN150936cos:0
o
 .EF
xx
•Which equation is correct? 0Nk150
57
6
:0 
.
EF
xx
A.
B.
C.
D. 0kN150936sin:0
o
 .EF
xx
E. 0kN150936sin:0
o
 .EF
xx kN 0.90
x
E
•What does the negative sign
signify?
•why the others are incorrect?

75. Problem 1 BJ10
4 -49
•Now apply the
condition that the sum
of forces in the y-
direction must sum to
zero.

76. Problem 1 BJ10
4 -49
•Now apply the
condition that the sum
of forces in the y-
direction must sum to
zero.
•Which equation is correct?
A.
B.
C.
D.
E.  0kN150936sinkN204:0
o
 .EF
yy    0kN150
57
6
kN204:0 
.
EF
yy    0kN150
57
6
kN204:0 
.
EF
yy    0kN150
57
6
kN204:0 
.
EF
yy    0kN150936coskN204:0
o
 .EF
yy

77. Problem 1 BJ10
4 -49
•Now apply the
condition that the sum
of forces in the y-
direction must sum to
zero. kN200
y
E
•Which equation is correct?
A.
B.
C.
D.
E.  0kN150936sinkN204:0
o
 .EF
yy    0kN150
57
6
kN204:0 
.
EF
yy    0kN150
57
6
kN204:0 
.
EF
yy    0kN150
57
6
kN204:0 
.
EF
yy 
E
y200 kN    0kN150936coskN204:0
o
 .EF
yy

78. Problem 1 BJ10
4 -49
•Now apply the
condition that the sum
of forces in the y-
direction must sum to
zero. kN200
y
E
•Which equation is correct?
A.
B.
C.
D.
E.
•What does the positive sign
signify?
•Discuss why the others are
incorrect.  0kN150936sinkN204:0
o
 .EF
yy    0kN150
57
6
kN204:0 
.
EF
yy    0kN150
57
6
kN204:0 
.
EF
yy    0kN150
57
6
kN204:0 
.
EF
yy 
E
y200 kN    0kN150936coskN204:0
o
 .EF
yy

79. Problem 1 BJ10
•Finally, apply the
condition that the sum of
moments about any point
must equal zero.
•Discuss with a neighbor
which point is the best for
applying this equilibrium
condition, and why.

80. Problem 1 BJ10
•Finally, apply the
condition that the sum of
moments about any point
must equal zero.
•Discuss with a neighbor
which point is the best for
applying this equilibrium
condition, and why.
•Three good points are D, E, and F.
Discuss what advantage each
point has over the others, or
perhaps why each is equally good.

81. Problem 1 BJ10
•Finally, apply the
condition that the sum of
moments about any point
must equal zero.
•Discuss with a neighbor
which point is the best for
applying this equilibrium
condition, and why.
•Three good points are D, E, and F.
Discuss what advantage each
point has over the others, or
perhaps why each is equally good.
•Assume that you choose point Eto
apply the sum-of-moments
condition.

82. Problem 1 BJ10
•Finally, apply the
condition that the sum of
moments about any point
must equal zero.
•Discuss with a neighbor
which point is the best for
applying this equilibrium
condition, and why.
•Three good points are D, E, and F.
Discuss what advantage each
point has over the others, or
perhaps why each is equally good. :0
EM    
  
  0m5.4kN150
5.7
6
m8.1kN20m6.3kN20
m4.5kN20m7.2kN20



E
M mkN0.180 
EM
•Assume that you choose point Eto
apply the sum-of-moments
condition.

83. Problem 2
MK2
Theuniformbeamhas
anoveralllengthof6m
andamassof300kg.The
forcePappliedtothe
hoistingcableisslowly
increasedtoraisethe
ringC,thetwo4-mropes
ACandBC,andthe
beam.Compute the
tensionsintheropesatA
andBwhenthebeamis
clearofitssupportsand
theforcePisequaltothe
weightofthebeam

85. AlightrodADsupportsa
150Nverticalloadandis
attachedtocollarsBandC,
whichmayslidefreelyonthe
rodsshown.Knowingthatthe
wireattachedatAformsan
angleα=30
0
withthe
horizontal,determine
a)Thetensioninthewire
b)ThereactionatBandC
Problem 3 (BJ3)

87. Thedeviceshownin
sectioncansupportthe
loadLatvarious
heightsbyresettingthe
pawlCinanothertooth
atthedesiredheighton
thefixedvertical
columnD.Determine
thedistancebatwhich
theloadshouldbe
positionedinorderfor
thetworollersAand
Btosupportequal
forces.Theweightof
thedeviceisnegligible
comparedwithL.
Problem 4
MK5, 3.111

89. Problem 5
Asemi-circularrod
ABCDissupported
byarolleratDand
rests on two
frictionlesscylinders
BandC.Findthe
maximum angle,
forcePcanmake
withtheverticalif
appliedatpointA
andtherodremains
inequilibrium.

91. Auniform400-kgdrumis
mountedonalineofrollersatA
andalineofrollersatB.An80-
kgmanmovesslowlyadistance
of700mmfromthevertical
centerlinebeforethedrum
beginstorotate.Allrollersare
perfectlyfreetorotateexcept
oneofthematBwhichmust
overcomeappreciablefrictionin
itsbearing.Calculatethefriction
forceFexertedbythatone
rollertangenttothedrumand
findthemagnitudeRoftheforce
exertedbyallrollersatAonthe
drumforthiscondition
Problem 6
MK5, 3.57

93. Problem 7 MK5
Aspecialjigisdesignedtoposition
largeconcretepipesectionsand
consistsofa80Mgsectormountedon
alineofrollersatB.Oneofthe
rollersatBisagearwhichmeshes
withthearingofgearteethonthe
sectorastoturnthesectoraboutits
geometriccenterO.Whenα=0
0
,a
counterclockwisetorqueof2460Nm
mustbeappliedtothegearatBto
keeptheassemblyformrotating.
Whenα=30
0
,aclockwisetorqueof
4680Nmisneededtoprevent
rotation.LocatethemasscenterGof
thejigbycalculatingrandθ.Note
thatthemasscenterofthepipe
sectionisatO.
Dia= 480
mm

96. Point Connections