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### Transcript of Static Equilibrium - · PDF fileStatic equilibrium: Sum of vertical and horizontal forces must...

• Static EquilibriumStatic equilibrium requires a balance of forces and a balance of moments.

0 0 0

0 0 0

===

===

zyx

zyx

MMM

FFF

Example 1:

A painter stands on a ladder that leans against the wall of a house at an angle of 20o. Assume that the painter is a midheight of the ladder. Calculate the minimum coefficient of friction for static equilibrium.

• 0)( 0

21

1221

=++====

PmmPFPPPPF

x

++

=+

+=

1)(

1

)(21

gmmP

gmmP plpl

0)20(2

)()20()20( 21 =+ SinLgmmLCosPLSinP pl

1763.0=

Static equilibrium: Sum of vertical and horizontal forces must be zero.

Taking the moment about point O and setting it equals to zero:

Newton First Law: A body at rest tend to remain at rest and a body in motion at constant velocity will tend to maintain that velocity unless acted upon by an external force.

Newton Second Law: The time rate of change of momentum of a body is equal to the magnitude of applied force and acts in the direction of the force.

Newton Third Law: when two particles interact, a pair of equal and opposite reaction forces will exist at their contact point. This force pair will have the same magnitude and act along the same direction line but have opposite sense.

• Newtons second law can be written for a rigid body in two forms, one for linear forces and one for moments or torques

GG HMmaF&=

= F=Force; m=mass, a= accelerationMG=moment about the center of gravity

GH& =time rate of change of the moment or the angular momentum about the CG

zzyyxx maFmaFmaF ===

kIjIiIH zzyyxxG ++=

yxyxzzz

xzxzyyy

zyzyxxx

IIIM

IIIM

IIIM

)(

)(

)(

=

=

=

If the x, y and z axes are chosen to coincide with the principal axes of inertia of the body. Where Ix, Iy and Iz are the principal centroidal mass moments of inertia (second moments of mass). In 3-D

Eulers Equations

• For 2-D:

yyxx maFmaF == zzz IM =Where z is the angular acceleration

Moment of a Force about an axis

• 3-D Equilibrium Example:Two transmission belts pass over sheaves welded to an axle supported by bearings at B and D. The radius at A = 2.5 and at C = 2. The axle rotates at constant speed. Find T and the reaction forces at B, D. Assume that bearing at D exerts no axial thrust and neglect weights of sheaves and axle.

• 8in

6in6in

• x

y

A B C D33.75lb 33.75lb67.5lb

33.75lb

33.75lb

202.5lb-in

• x

z

A B C D70lb 28lb42lb

28lb

42lb

336lb-in

• DefinitionsCentroid of Area (Center of Gravity of an area): Point that defines the geometric center of the area

( )yx,

A

Axx

A

Ayy AA

==

Q axis-y therespect towith

Areaan ofMoment First

Q axis- x therespect towith

Areaan ofMoment First

y

x

AxAx

AyAy

A

A

===

===

IbVQ

=

• Example 2:

Calculate the center of gravity of the rectangle:

(a)Without a hole

(b)With a hole of dimensions cand d

a) Without a hole

22

22

1

1

aba

baa

xA

Axx

A

Axx

bba

bab

yA

Ayy

A

Ayy

n

iiA

n

iiA

=

===

=

===

• b) With a hole

dcba

dccebaa

xA

Axx

A

Axx

dcba

dcdfbab

yA

Ayy

A

Ayy

n

iiA

n

iiA

===

+

===

)2

(2

)2

(2

1

1

Second Moment or Moment of Inertia of an Area

==A

yA

x AxIAyI 22

+==A

yxO IIAJ 2

Rectangular moments of inertia

Polar moments of inertia

2222

2

2

yxOOO

yy

xx

rrrArJ

ArI

ArI

+==

=

=

Example 3:Find the moment of inertia of the circular area about the x and y axes, the polar moment of inertia and the radius of gyration about the x and y axes. CosrxSinry == yCosrArea = 2

= Cosry

2

4

42

44

2

2

4224

rIIJrI

rCosSinrI

yxzy

x

=+=

=

==

• Parallel-Axis Theorem

2' yxx dAII +=

Example 4:

( )4

'

224

2'

4

2516

r4 4

4

r.I

r rdAII

rI

x

yxx

x

=

+

=+=

=

• IMy

=

• Mass Moment of Inertia

It is the product of the elements mass and the square of the elements distance from the axis.

( )( )( )

+=

+=

+=

amz

amy

amx

myxI

mzxI

mzyI

22

22

22

• Classification with respect to the method of application:

(a)Normal tensile

(b)Normal compressive

(c)Shear

(d)Bending

(e)Torsion

(f) Combined

• The sign convention that will be used here is as follows:

deflectionx

=> 02

2

• Distributed loads, Shear Force and

Bending Moment in Beams

• 4

4

3

3

2

2

2

2

xEI-qq

xV

xEIVV

xM

xEIM

EIM

x

==

==

==

• Successive Integration Method

1

2

1

2600)(

600)(

)()(

CxxV

CxxxV

xxqxV

qxV

+=

+=

=

=

2/3m

=650N1050N=

10

• NVthenmxFor

xxV

CVxFor

CxxV

xFor

1050)2(_____2__

1502

600)(

150__150500350)1(__1__2

600)(

21__

22

2

2

==

+=

====

+=

• 0)2(____2__500__550)1(____1__

150100)(

1502

600)()(

1502

600)(

21__

4

43

2

2

=====

++=

+==

+=

• Normal Stress and Strain:Where F: force, normal to the cross-sectional area,

A0: original cross-sectional area 0AF

=

1Pa = 1 N.m-2; 1MPa = 106Pa; 1GPa=109Pa

Normal strain or tensile strain or Axial strain

00

0

ll

lll =

=

• Hookes LawWhen strains are small, most of materials are linear elastic.

E

Normal: =

Youngs modulus

EAlFl

o

o

=

Springs: the spring rate

o

o

lEA

lFk =

=

lr

zrZ

=,

lrG

zrGG zz

== ,,

G = Shear Modulus of Elasticity

Shear strain

Twist Moment or Torque

==A zA z

Arl

GArT 2,,

= A ArJ 2Area Polar Moment

of InertiaJGlT

lJGT

=

= or

JrT

z

,Thus: JrT o

Max

Angular spring rate:

lGJTka

==

• We can obtain the normal and shear stresses from flexure and shear formulas

Where: is the normal stress acting on the cross section, M is the bending moment, y is the distance from the neutral axis and I is the moment of inertia of the cross sectional area with respect to

the neutral axis. is the shear stress at any point in the cross section, V is the shear force,Q is the first moment of the cross sectional area outside of the point in the cross section where the stress is being found, and b is the width of the cross section.

IMy

=

Maximum Stresses in Beams

IbVQ

=

• The normal stresses obtained from the flexure formula have their maximum values at the farthest distance from the neutral axis. The normal stresses are calculated at the cross section of maximum bending moment.

The shear stress obtained from the shear formula usually have their highest value at the neutral axis. The shear stresses are calculated at the cross section of maximum shear force. In most circumstances, these are the only stresses that are needed for design purposes. However to obtain a more complete picture of the stresses, we will need to determine the principal stresses and maximum shear stresses at various points in the beam.

• Consider a simple rectangular beam below, and a cross section to the left of the load. Points A and E are at the top and bottom of the beam. Point C is in the midheight of the beam and points B and Dare in between.If Hookes law applies (linear elasticity), the normal and shear stresses at each of these five points can be readily calculated from the flexure and shear formulas.All the elements of vertical and horizontal faces, are in plane stress, because there is no stresses acting perpendicular to the plane of the figure.

Beams of Rectangular Cross Section

• Stresses in a beam of rectangular cross section:

(a) simple beam with points A, B, C, D, and E on the side of the beam;

(b) normal and shear stresses acting on stress elements at points A, B, C, D, and E;

(c) principal stress; and (d) maximum shear stresses.

Points A and E elements are in uniaxialcompressive and tensile stresses respectively. Point C (neutral axis) element is in pure shear. Points B and Delements have both normal and shear stresses.

• We may use either the