2DEquilibrium 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 BJ10Instructor resources.

• Attractive online features available for Instructorshttp://highered.mcgrawhill.com/sites/1259062910/information_center_view0/

• For purchase/other details kindly contact:• [email protected] Sagar Divekar

[email protected] Santosh Joshi

mailto:[email protected]

mailto:[email protected]

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

http://demonstrations.wolfram.com/

http://www.brown.edu/Departments/Engineering/Courses/En4/Notes/notes.html

http://www.youtube.com/user/mellenstei

http://wps.prenhall.com/wps/media/objects/3076/3149958/studypak/index_st.html

http://wps.prenhall.com/wps/media/objects/3076/3149958/studypak/index_st.html

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

Roller

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

More

Examples

of 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 kN150 kN

150 kN

Choose the most

correct FBD for the

original problem.

Practice

4 – 27 BJ10

A B

C D

150 kN

150 kN150 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 kN150 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

Adequacy of

Constraints

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 themiddle of the upper edge of the door and by two sets of frictionlessrollers at A and B. Each set consists of two rollers one either side of thedoor. The rollers A are free to move in horizontal channels, while rollersB are guided by vertical channels. If the door is held in the position forwhich BD=1050 mm, determine (a) the tension in the cable AE, (2) thereaction 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




• 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







• 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







• 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.

63614141

3132tan

m2.313 m51508282

m515020tan m4141)2545(cot

m4141

m828245cosm445cos

ooo

21

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

N 1.98

110sin4.31sin

RT

N 8.147

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:







• Determine the reactions at A by

solving the equations for the

sum of all horizontal force

components and all vertical

force components.










SOLUTION:








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:M A

kN1107.B


4 - 42


diagram.



forces about A.

0m6kN523

m2kN819m510

.

..B:M A

kN1107.B


solving the equations for the sum of all

horizontal forces and all vertical forces.

00 BA:F xx

kN1.107xA


4 - 42


diagram.



forces about A.

0m6kN523

m2kN819m510

.

..B:M A

kN1107.B




00 BA:F xx

kN1.107xA

0kN523kN8190 ..A:F yy

kN333.Ay


4 - 42


diagram.

• Check the values obtained.



forces about A.

0m6kN523

m2kN819m510

.

..B:M A

kN1107.B




00 BA:F xx

kN1.107xA

0kN523kN8190 ..A:F yy

kN333.Ay

Hydraulic Cylinder

show Mathematica demo on this

System constrained to

various degrees

Questions?

Tutorial after tea-break

Tutorial on 2D equilibrium

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.

0kN1505.7

5.4:0 xx EF

0kN150936cos:0 o .EF xx

• Which equation is correct?

0Nk15057

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.

0kN1505.7

5.4:0 xx EF

kN 0.90xE



0Nk15057

6:0

.EF xx

A.

B.

C.

D.



kN 0.90xE

Problem-1 BJ10



exercise.






to zero.

0kN1505.7

5.4:0 xx EF

kN 0.90xE



0Nk15057

6:0

.EF xx

A.

B.

C.

D.



kN 0.90xE

• 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

0kN15057

6kN204:0

.EF yy

0kN15057

6kN204:0

.EF yy

0kN15057

6kN204:0

.EF yy

0kN150936coskN204:0 o .EF yy

Problem 1 BJ10

4 - 49

• Now apply the


of forces in the y-


zero.

kN200yE


A.

B.

C.

D.

E.


0kN15057

6kN204:0

.EF yy

0kN15057

6kN204:0

.EF yy

0kN15057

6kN204:0

.EF yy

Ey 200 kN


Problem 1 BJ10

4 - 49

• Now apply the


of forces in the y-


zero.

kN200yE


A.

B.

C.

D.

E.

• What does the positive sign

signify?

• Discuss why the others are

incorrect.


0kN15057

6kN204:0

.EF yy

0kN15057

6kN204:0

.EF yy

0kN15057

6kN204:0

.EF yy

Ey 200 kN


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

0m5.4kN1505.7

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 hasan overall length of 6mand a mass of 300kg. Theforce P applied to thehoisting cable is slowlyincreased to raise thering C, the two 4-m ropesAC and BC, and thebeam. Compute thetensions in the ropes at Aand B when the beam isclear of its supports andthe force P is equal to theweight 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 ismounted on a line of rollers at Aand a line of rollers at B. An 80-kg man moves slowly a distanceof 700 mm from the verticalcenterline before the drumbegins to rotate. All rollers areperfectly free to rotate exceptone of them at B which mustovercome appreciable friction inits bearing. Calculate the frictionforce F exerted by that oneroller tangent to the drum andfind the magnitude R of the forceexerted by all rollers at A on thedrum 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

Point Connections

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2DEquilibrium Split