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12-1

Chapter 12:Static Equilibrium and Elasticity

.

12.2 Centre of Gravity

12.3 Rigid Objects in Static Equilibrium

12.4 Elastic Pro erties of Solids

12-2

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Over 13 Weeks (39 hrs):

I. Vectors and D namics22 hrs

II . Static Equilibrium and Elasticity (6 hrs)

Chapter 12: Static Equilibrium and Elasticity

III. F ui Mec anics3 rs

IV. Thermod namics8 hrs

12-3

CONTENTS of PART II LECTURES

L23 Conditions for Equilibrium

L24 Equivalent Force Systems, Centre of Gravity

L25 Rigid Objects in Static Equilibrium (1)

L27 Analysis of Trusses.

g ec s n a c qu r um

L28 Elastic Properties of Solids: Youngs Modulus, ShearModulus and Bulk Modulus

12-4

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an

Textbook:

R. A. Serway and J W Jewett, Jr, Physics for

Scientists and Engineers, 6th Ed, Vol 1, Thomson,2004.

R. C. Hibbeler, Engineering Mechanics Statics, 2nd

, , .

A. Gupta and C. K. Soh, Intelligent Interactive Tutoring

12-5

- , , .

12.1 Conditions for E uilibrium

Equilibrium - General

Equilibrium Equations End Supports

12-6

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Static E uil ibrium

qu r um mp es t e o ect s at rest stat c

or its center of massmoves with a constant

Static equilibrium is a common situation in

Principles involved are of particular interest

eng neer ng

to civil engineers, architects, and mechanical

engineers

12-7

.

In this chapter, we will concentrate on Static

Equilibrium

The object will not be moving

D namic E uilibrium is also ossible The object would be translating with a constant

velocity and/or rotating with a constant angular

velocity

In either case F = 0 and = 0

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Conditions for Equilibrium

Fi = 0 e o ec s mo e e as apar c e,

then this is the only condition that must

Net torque (or moment) equals zero

i = i =

This is needed if the object cannot be

12-9

. .

= r F (or = r F) -

rule to determine the

direction of the torque The tendency of the force

to cause a rotation about

moment arm d

12-10

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Vector Re resentation of Moment 2

12-11

Moment of a Force

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Vector Representation of Moment

12-13

Cross Product of Two Vectors

12-14

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12-15

Coordinate Independence of Couples

F

12

FA

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Need the angular acceleration

(vs.translational acceleration aof the ob ect to

be zero

= For rotational equilibrium,

i = i = This must be true for any axis of rotation

12-17

Ax is of Rotation for Torque Equation

e ne orque s a ou an ax s roug anypoint in the xyplane

If an object is in translational equilibrium and

,the net torque must be zero about any otheraxis i.e.

Fx = 0

Fy = 0

12-18

z = 0 (or Mz = 0), about any other axis

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Axis of Rotation for Torque Equation, 2

Given that: F = 0 an

O = 0 (or MO = 0)022110 =++= LFrFr

( ) ( )

( )'

''

212211

2211'0

++++=

++=LL

L

FFrFrFr

FrrFrr

0=

--- If an object is in translational equilibrium and the

12-19

net torque is zero a out one axis, t en t e nettorque must be zero about any other axis

E uil ibrium E uations

.

Translational equilibrium => resultant external

Fi = 0

be zero when viewed from an inertial frame of reference

Rotational equilibrium => resultant external

torque about anyaxis must be zero:

i = 0 (or Mi = 0 )

12-20

The angular acceleration must equal zero

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,

We will restrict the applications to situations

in which all the forces lie in the xyplane

These are called coplanarforces since they lie in

the same plane

here are three resulting equations

Fx = 0

Fy = 0

z = 0 or Mz = 0

12-21

12-22

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MRy

-

Can move Can rotate in Fixed

12-23

horizontally androtate in the plane

the plane

Support Provided by a Smooth oroug ur ace

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Support Reactions: Simply Supported Beam

for finding Reactions

RAxRAx

y

y RAx

y

y

12-25

FX = 0 RAx=0

Princi le of Transmissibil it

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E uivalent Force S stems

body form a force system

Two force systems (applied to the same rigid

),,,( 121

1

1

2

1

1 Lvv

Lvv

MMFF ),,,( 222

1

2

2

2

1 Lvv

Lvv

MMFF

body) are equivalent when:

he resultant forces of the two force s stems are

the same;

he net torques of the two force systems (about

=j

j

i

i FF21vv

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the same point) are the same )()( 21 OO = vv

E uivalent Force S stem

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Force and Moment Vectors

12-29

Simplification of Force Systems

equivalent to a force applied at the given

. ., .

=

iA FF

1vv

= )(AMMAvv

(or, any force system can be simplified to be

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12.2 Center of Gravit

force applied to a rigid body (or.

Find a point (i.e., A) such that therA

r

A

simplified moment is just 0.

( ) 0)( === WdrrAMM AAvvvvv

V

WrWdrWdr AV

A

V

vvvvvv==

12-31

--- e str ute grav tat ona orce s equ va ent to t e

concentrated simplified force applied at the point A.

t e var ous grav tat onaforces acting on all the

equivalent to a single

ravitational force actinthrough a single pointcalled the Center of Gravity(CG)

L++= 222111

xgmxgmx ii

xm=

12-32

L++ 2211 gmgm i

m

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Center of Gravit 2

e orque ue o e grav a ona orce on

an object of mass Mis the force Mgacting at

Ifg is uniform over the object, then the

en er o rav y o e o ec co nc es

with its Center of Mass(CM)

t e o ect s omogeneous an symmetr ca ,

the Center of Gravity(CG) coincides with its

12-33

eome r c en er

Centre of Gravity and Mass for Particles

~~~

~~~

for CG===

i

ii

i

ii

i

ii

W

Wzz

W

Wyy

W

Wxx

for CM

===i

ii

i

ii

i

ii

mz

m

myy

m

mxx

:coordinates of

each particleyx ~,~,~

coor na es o ecentre of gravity

G of particles

:,, yx

12-34

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=

=

dmxMx

m

~=

=

dWxWx ~

= dmyMy~

~= dWyWy~

~,

= mz=

12-35

-

( ) WxdWxxdWWx ==

=

12-36

( ) WydWyydWWy ==

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Ex 12 .1: Locate the CM of theara o c o , x = y

2

( ) dyydy

xdydydxdL 22 211 +=

+=+=

m479.1411

21

=+== dyydLL

( ) 21

0

2m8484.021 =+== dyyydLyydm

m574.0== Mydmy

:density per length:

( ) ( ) 21

0

222 m6063.02121 =+=+==

dyyydyyxdLxxdm

12-37

m410.0== Mxdmx

Ex: Locate the Center of Geometry of

y2 2

1

2= - -

x

Parts Ai xi Aixi yi Aiyi2.14Ax

1 12 2 24 1.5 18

2 -3.14 1 -3.14 2 -6.28

07.286.6

.===

iAx

51.110.39

=== ii Ay

12-38

3 -2 3.33 -6.66 0.67 -1.33

6.86 --- 14.2 --- 10.39

6.86iA

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12.3 Rigid Objects in Stat ic Equilibrium

Problem-Solving Strategy

External and Internal Forces

Support Provided by a Surface

x . : e g e an xamp e

Ex 12.3: Standing on a Horizontal Beam

Ex 12.4: Leaning Ladder

Ex 12.5: Negotiating a Curb

12-39

Truss

Problem-Solving Strategy

Draw a diagram of the system

Draw a free-body diagram(FBD)

object

Indicate the locations ofallthe forces

For systems with multiple objects, draw a

separate free-body diagramfor each object

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Problem-Solving Strategy, 2

Find the components of the forces along the two axes = Be careful of signs

torque (moment) on the object

Remember that the choice of the axis is arbitrary

Choose an origin that simplifies the calculations as

much as possible

12-41

A force that acts along a line passing through the

origin produces a zero torque

Problem-Solving Strategy, 3

Apply the second condition for equilibrium,

i = 0 (or Mi = 0 ) The two conditions of equilibrium will give a

Solve the simultaneous equations

,

the direction opposite what you drew in the free-

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External and Internal Forces

12-43

External and Internal Forces 2

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External and Internal Forces 3

section.

12-45

Ex 12.2: Weighted Hand Example

Model the forearm as arigid bar

e weig t o t e orearmis ignored

the x-direction

conditions from the free-body diagram

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Ex 12.2: Weighted Hand Example, 2

o ve or t e un nown

forces (F and R) pp y e con on or

torque equilibrium using the

rotation (o =0)mgo =

0)35)(50()3(F ==

12-47

N583F =

Ex 12.2: Weighted Hand Example, 3

Apply the condition for forceequilibrium (Fy = 0)

mgy =

0N0.50RF ==

50FR =

N53350583 ==

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Ex 12.3: Standing on a Horizontal Beam

uniform => the center of ravit

is at the geometric centerof the beam

he person of 600 N isstanding on the beam

a are e ens on

in the cable and the

12-49

wall on the beam?

Ex 12.3: Standing on a Horizontal Beam, 2

Draw a free-bodydiagram

Use the pivot in theproblem (at the wall)as e p vo

This will generally be

Note there are threeunknowns TR

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Ex 12.3: Standing on a Horizontal Beam, 3

resolved into componentsin the free-bod dia ram

Apply the two conditionsof e uilibrium to obtainthree equations

Solve for the unknowns

026004200853sin

==

== TA=

=

N580

N313

R

T

12-51

020060053sinsin =+= TRFyx

= 1.71

he ladder is uniform

=> the weight of the ladder

center (its center of gravity)

here is static friction(s=0.4) between the

ladder and the ground

Find the minimum anglemin at which the ladder

12-52

oes not s ip

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Ex 12.4: Leaning Ladder, 2

raw a ree- o y agram or

the ladder

The frictional force fis:

=s , s .static friction

Let Obe the axis of rotation

Apply the equations for the

two conditions of equilibrium

12-53

Solve the equations

Ex 12.4: Leanin Ladder 3

0cos2

sin == mgPo l

l

0

0

==

==

PfF

mgnF

x

y

from 1st and 2nd eqs:

mgnmg

P == andcot

from 3rd eq:

mgnmg

Pf ss === cot2

12-54

== 3.5125.12

1tan

min

s

Q

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Ex 12.4: Leaning Ladder, Extended

at a distance dfrom the

The higher the personclimbs the lar er theangle at the base needsto be in order to remainin equilibrium

Eventually, the ladder

12-55

may s ip

Ex 12 .5: Negotiating a Curb

(A) n magn u e oforce F to be

(B) Magnitude and

reaction R

mg=700 N

r= 30 cm

h = 10 cm

12-56

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Ex 12 .5: Negotiat ing a Curb, 2

be applied to roll up==

( ) 2 222 == hrhhrrd

N3132

=

=hr

mgdF

direction of reaction R

( )

==

=+=

66

N767

1

22

mgtan

FmgR

12-57

, ,form a triangle

F

Roof Truss

12-58

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12-59

.(2). Connected by joints;(3). Loads are applied to joints.

Simpletruss:

(1). Start from a triangle;(2). Add every 2 members to form a new joint;

. .

12-60

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(1). Find reactions;

A simple truss can be solved by Joint Method with 3 steps:

(2). Number the joints (from 0 to n);

(3). Solve the joints (from n to 0).

(a) Free body diagram of joint with tension convention;

(b) Fx = 0; Fy = 0;

(c) + is tension force, - indicates compression force.

12-61

(1)Statically Determinate Reactions

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Determine forcesactin at the ins

Rockers / rollers atsu orts A &E

Fg= 7200 N

Note that FAB=FBAB D

FBA F

FBD FDB

F FDE

L = 50 mA C E

FAB

FAC FCA

FCB FCD

FCE

FED

FEC

12-63

nA nEFg

Anal sis of a Truss 2

B DF

FBD FDB

A C E

FAB

FAC FCA

FCB FCD

FCE

BC DC DE

FED

FECequilibrium

nA nEFg

( ) AEgEAgEAy nnFLLnMFnnF =====+= N360002,0

on)(compressiN7200030sin ==+= ABABAy FFnF

12-64

(tension)N6235030cos ==+= ACABACx FFFF

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12.4 Elasticit

Deformation

Youngs Modulus

Shear Modulus

Moduli and Types of Materials

12-65

Deformation

So far we have assumed that objectsremain rigid when external forces act ont em

Except springs

ctua y, o ects are e orma e

It is possible to change the size and/or shape

Internal forces resist the deformation

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Stress and Strain

Stress Is proportional to the force causing the

deformation

It is the external force acting on the object perun area

Strain

Is a measure of the degree of deformation

12-67

between stress and strain For sufficiently small stresses, strain is directly proportional

o s ress

It depends on the material being deformed

It also depends on the nature of the deformation Elastic modulus in general relates what is done to a

solid object to how that object responds

strain

s ressmoduluselastic

12-68

ar ous ypes o e orma on ave un que e as cmoduli

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Youngs Modulus

Measures the resistance of a solid to a change in

Shear Modulus

within a solid parallelto each other

Bulk Modulus

Measures the resistance of solids or liquids to

changes in their volume

12-69

he bar is stretchedby an amount L

force F

ratio of the externalforce to the cross-sectional area A, i.e.

AF

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Youn s Modulus 2

ens on stra n s t e rat o o t e c ange n

length to the original length, i.e. LL oung s mo u us, , s e ra o o ose wo

ratios:

tensile stress

tensile strainY

L =

Units are: N / m2

(Pa)

i

12-71

Stress vs. Strain Curve

Ex eriments show that for certainstresses, the stress is directlyproportional to the strain

the curve

The elastic limitis themaximum stress that can beapplied to the substance before itbecomes ermanentl deformed

When the stress exceeds the elastic limit, the substance willbe permanently deformed

12-72

The curve is no longer a straight line

With additional stress, the material ultimately breaks

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deformation occurs whena force acts parallel to oneof its faces while theopposite face is held fixedy ano er orce

This is called a shears ress

For small deformations, no change in volume occurs

12-73

A good first approximation

Shear Modulus, 2

Shear stressis / A Fis the tangential force

A is the area of the face being sheared

x x is the horizontal distance the sheared face moves h is the height of the object

Shear modulus is the ratio of the shear stress tothe shear strain:

shear stress

shear strainS

x= =

12-74

Units are: N / m2 (Pa)

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occurs when a force of uniformma nitude is a liedperpendicularly over the entiresurface of the object.

The object will undergo achange in volume, but not ins ape.

Volume stressis defined as the ratio of the magnitude

12-75

, , A, of the surface.

Bulk Modulus 2

s s a so ca e epressure.

Volume strainis the ratio of the change in.

Bulk modulus is the ratio of the volume stress

to the volume strain:

volume stressF

PAB

= = =

The negative sign indicates that an increase in

i iV V

12-76

pressure wi resu t in a ecrease in vo ume.

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Compressibility is the inverse of the bulkmodulus.

It is often used instead of the bulk modulus.

12-77

Both solids and liquids have a bulk modulus

Liquids cannot sustain a shearing stress or

a ens e s ress

If a shearing force or a tensile force is applied

to a liquid, the liquid simply flows in response.

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Moduli Values

12-79

A rigid body is supported by 3 deformable rods, find

reactions N , N and N (all the self wei hts are ne li ible).

x

yP= 100 N

o u us: = m

Rod length: L=1 m

Cross area: A=0.1 m2

NA N N

Distances between rods: 1 m

01000 =++= CBAy NNNF :Forces are in equilibrium:

=+= .: CBA

Deformations are compatible: Points A, B and C are on the same line

12-80

+AE

LNA A10,:

+AE

LNB B11,:

+AE

LNC C12,:

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Thank You !Thank You !

12-81