2 d equilibrium-split

Engineering Mechanics
Equilibrium of Rigid Bodies

Main Text
• Major chunk of lectures based on:
• Vector Mechanics for Engineers, Beer, Johnston et
al., 10th ed., McGraw-Hill.
• Referred to as BJ10. Indian Edition available.
• Also, some problems from Beer and Johnston 3rd
and 8th editions, BJ3 and BJ8.
• Dynamics will 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:
• sagar.divekar@mheducation.com Sagar Divekar
santosh.joshi@mheducation.com Santosh Joshi

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

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

Application
4 – 5 BJ10
Engineers
designing this
crane will need to
determine the
forces that act on
this body under
various conditions.
Slide from BJ10

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

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!

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

Equilibrium
• System is in
equilibrium if
and only if the
sum of all the
forces and
moment (about
any point)
equals zero.
If true for one point O, true
w.r.t any other point! Convince
yourself

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.

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.

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 direction are 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 DOF then there is no reaction
from the support in that direction.

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

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

• The third dimension is very small as
compared to the other two and loads are
coplanar.

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

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

Structure
4 – 18 BJ10
a force of unknown
direction and
magnitude.

Structure
4 – 18 BJ10
a force of unknown
direction and
magnitude.
• Reactions equivalent
to a force of unknown
direction and
magnitude and a
couple of unknown
magnitude.

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

Slot Connection
http://oli.cmu.edu

Simple Examples
Roller Support Fixed Support

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

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.

Simple examples
• Copyright, Dr. Romberg
FBD
FBD

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.

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

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.

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?

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?

Equations of equilibrium in 2D
• Three equations per free body.
• Writing more than three equations 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

Problem 1
• Determine the tension in cable ABD and
reaction at support C.

Categories of Equilibrium in 2D
MK5

Problem 2 (BJ10)
• A 70 kg (W) overhead garage door consists of a uniform rectangular
panel AC 2100 mm high (h), supported by the cable AE attached at the
middle of the upper edge of the door and by two sets of frictionless
rollers at A and B. Each set consists of two rollers one either side of the
door. The rollers A are free to move in horizontal channels, while rollers
B are guided by vertical channels. If the door is held in the position for
which BD=1050 mm, determine (a) the tension in the cable AE, (2) the
reaction at each of the four rollers. Assume a = 1050 mm, b = 700mm

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

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

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

Body
• Assuming that their lines of action intersect,
the moment of F1 and F2 about the point of
intersection represented by D is zero.
Slide from BJ10

Body
• Since the rigid body is in equilibrium, the sum
of the moments of F1, F2, and F3 about any axis
must be zero. It follows that the moment of F3
about D must be zero as well and that the line
of action of F3 must pass through D.
Slide from BJ10

Body
• Since the rigid body is in equilibrium, the sum
of the moments of F1, F2, and F3 about any axis
must be zero. It follows that the moment of F3
about D must be zero as well and that the line
of action of F3 must pass through D.
• The lines of action of the three forces must be
concurrent or parallel.
Slide from BJ10

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 T in the
rope and the reaction at A.

4 – 38 BJ10
a rope.
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.

4 – 38 BJ10
a rope.
SOLUTION:
• The three forces must be concurrent
for static equilibrium. Therefore, the
reaction R must pass through the
intersection of the lines of action of
the weight and rope forces.
Determine the direction of the
reaction R.

4 – 38 BJ10
a rope.
SOLUTION:
• The three forces must be concurrent
for static equilibrium. Therefore, the
reaction R must 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.

Sample Problem 4.6
4 – 39 BJ10
joist.

Sample Problem 4.6
4 – 39 BJ10
joist.
• Determine the direction of the
reaction R.
 
 
 
6361
4141
3132
tan
m2.313m51508282
m515020tanm4141)2545(cot
m4141
m828245cosm445cos
ooo
2
1
oo
.
.
.
AE
CE
..BDBFCE
..CDBD
.AFAECD
.ABBF







6.58

Sample Problem 4.6
4 – 40 BJ10
• Determine the magnitude of the
reaction R.

38.6sin
N1.98
110sin4.31sin

RT
N8.147
N9.81


R
T

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 A and a rocker
at B. The center of gravity of
the crane is located at G.
Determine the components of
the reactions at A and B.

SOLUTION:
• Create a free-body diagram for the
crane.

SOLUTION:
crane.
• Determine the reactions at B by
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.

SOLUTION:
crane.
• Determine the reactions at A by
solving the equations for the
sum of all horizontal force
components and all vertical
force components.

SOLUTION:
crane.
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.

4 - 42
• Create the free-body
diagram.

4 - 42
diagram.
• Determine B by solving the equation
for the sum of the moments of all
forces about A.
   
  0m6kN523
m2kN819m510


.
..B:MA
kN1107.B 

4 - 42
diagram.
forces about A.
   
  0m6kN523
m2kN819m510


.
..B:MA
kN1107.B 
solving the equations for the sum of all
horizontal forces and all vertical forces.
00  BA:F xx
kN1.107xA

4 - 42
diagram.
forces about A.
   
  0m6kN523
m2kN819m510


.
..B:MA
kN1107.B 
00  BA:F xx
kN1.107xA
0kN523kN8190  ..A:F yy
kN333.Ay 

4 - 42
diagram.
• Check the values obtained.
forces about A.
   
  0m6kN523
m2kN819m510


.
..B:MA
kN1107.B 
00  BA:F xx
kN1.107xA
0kN523kN8190  ..A:F yy
kN333.Ay 

Hydraulic Cylinder
show Mathematica demo on this

System constrained to
various degrees

Questions?
Tutorial after tea-break

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.

Problem 1 (BJ10)
4 – 47 BJ10
fixed end E.
SOLUTION:
- Discuss with a neighbor the
steps for solving this problem.

Problem 1 (BJ10)
4 – 47 BJ10
fixed end E.
SOLUTION:
• Create a free-body diagram
for the frame and cable.

Problem 1 (BJ10)
4 – 47 BJ10
fixed end E.
SOLUTION:
• Apply the equilibrium
equations for the reaction
force components and couple
at E.
• Create a free-body diagram
for the frame and cable.

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.

Problem-1 BJ10
exercise.
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

Problem-1 BJ10
exercise.
to zero.
  0kN150
5.7
5.4
:0  xx EF
kN0.90xE
  0kN150936cos:0 o
 .EF xx
  0Nk150
57
6
:0 
.
EF xx
A.
B.
C.
D.
  0kN150936sin:0 o
 .EF xx
E.   0kN150936sin:0 o
 .EF xx
kN0.90xE

Problem-1 BJ10
exercise.
to zero.
  0kN150
5.7
5.4
:0  xx EF
kN0.90xE
  0kN150936cos:0 o
 .EF xx
  0Nk150
57
6
:0 
.
EF xx
A.
B.
C.
D.
  0kN150936sin:0 o
 .EF xx
E.   0kN150936sin:0 o
 .EF xx
kN0.90xE
• What does the negative sign
signify?
• why the others are incorrect?

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

Problem 1 BJ10
4 - 49
• Now apply the
of forces in the y-
zero.
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

Problem 1 BJ10
4 - 49
• Now apply the
of forces in the y-
zero.
kN200yE
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

Ey  200 kN
    0kN150936coskN204:0 o
 .EF yy

Problem 1 BJ10
4 - 49
• Now apply the
of forces in the y-
zero.
kN200yE
A.
B.
C.
D.
E.
• What does the positive sign
signify?
• Discuss why the others are
    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

Ey  200 kN
    0kN150936coskN204:0 o
 .EF yy

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.

Problem 1 BJ10
must equal zero.
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.

Problem 1 BJ10
must equal zero.
condition, and why.
• Assume that you choose point E to
apply the sum-of-moments
condition.

Problem 1 BJ10
must equal zero.
condition, and why.
  :0EM    
   
  0m5.4kN150
5.7
6
m8.1kN20m6.3kN20
m4.5kN20m7.2kN20



EM
mkN0.180 EM
• Assume that you choose point E to
apply the sum-of-moments
condition.

Problem 2
MK2
The uniform beam has
an overall length of 6m
and a mass of 300kg. The
force P applied to the
hoisting cable is slowly
increased to raise the
ring C, the two 4-m ropes
AC and BC, and the
beam. Compute the
tensions in the ropes at A
and B when the beam is
clear of its supports and
the force P is equal to the
weight of the beam

A light rod AD supports a
150N vertical load and is
attached to collars B and C,
which may slide freely on the
rods shown. Knowing that the
wire attached at A forms an
angle α = 300 with the
horizontal, determine
a) The tension in the wire
b) The reaction at B and C
Problem 3 (BJ3)

The device shown in
section can support the
load L at various
heights by resetting the
pawl C in another tooth
at the desired height on
the fixed vertical
column D. Determine
the distance b at which
the load should be
positioned in order for
the two rollers A and
B to support equal
forces. The weight of
the device is negligible
compared with L.
Problem 4
MK5, 3.111

Problem 5
A semi-circular rod
ABCD is supported
by a roller at D and
rests on two
frictionless cylinders
B and C. Find the
maximum angle,
force P can make
with the vertical if
applied at point A
and the rod remains
in equilibrium.

A uniform 400-kg drum is
mounted on a line of rollers at A
and a line of rollers at B. An 80-
kg man moves slowly a distance
of 700 mm from the vertical
centerline before the drum
begins to rotate. All rollers are
perfectly free to rotate except
one of them at B which must
overcome appreciable friction in
its bearing. Calculate the friction
force F exerted by that one
roller tangent to the drum and
find the magnitude R of the force
exerted by all rollers at A on the
drum for this condition
Problem 6
MK5, 3.57

Problem 7 MK5
A special jig is designed to position
large concrete pipe sections and
consists of a 80 Mg sector mounted on
a line of rollers at B. One of the
rollers at B is a gear which meshes
with the a ring of gear teeth on the
sector as to turn the sector about its
geometric center O. When α = 00 , a
counterclock wise torque of 2460 Nm
must be applied to the gear at B to
keep the assembly form rotating.
When α = 300 , a clock wise torque of
4680 Nm is needed to prevent
rotation. Locate the mass center G of
the jig by calculating r and θ. Note
that the mass center of the pipe
section is at O.
Dia = 480
mm

2 d equilibrium-split

Recommended

Recommended

More Related Content

What's hot

What's hot (20)

Viewers also liked

Viewers also liked (17)

Similar to 2 d equilibrium-split

Similar to 2 d equilibrium-split (20)

More from sharancm2009

More from sharancm2009 (13)

Recently uploaded

Recently uploaded (20)

2 d equilibrium-split