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

0 0 0

0 0 0

=Σ=Σ=Σ

=Σ=Σ=Σ

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

=++−=Σ==−=Σ

PmmPFPPPPF

personladdery

x

μμμ

μμ

μ +

+=

+

+=

1)(

1

)(21

gmmP

gmmP plpl

0)20(2

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

1763.0=μ

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

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

Load Analysis – 3-D

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.

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

GG HMmaF&=Σ

=Σ F=Force; m=mass, a= acceleration

MG=moment about the center of gravity

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

zzyyxx maFmaFmaF =Σ=Σ=Σ

kIjIiIH zzyyxxGˆˆˆ ωωω ++=

yxyxzzz

xzxzyyy

zyzyxxx

IIIM

IIIM

IIIM

ωωα

ωωα

ωωα

)(

)(

)(

−−=Σ

−−=Σ

−−=Σ

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

Euler’s Equations

For 2-D:

yyxx maFmaF =Σ=Σ zzz IM α=ΣWhere αz 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

∫∫==

δδ

Q axis-y therespect towith

Areaan ofMoment First

Q axis- x therespect towith

Areaan ofMoment First

y

x

AxAx

AyAy

A

A

⋅===

⋅===

∫

∫

δ

δ

IbVQ

=τ

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

=⋅

⋅⋅⎟⎠⎞

⎜⎝⎛

===

=⋅

⋅⋅⎟⎠⎞

⎜⎝⎛

===

∑∫

∑∫

δ

δ

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

δ

δ

Second Moment or Moment of Inertia of an Area

∫∫ ==A

yA

x AxIAyI δδ 22

∫ +==A

yxO IIAJ δρ 2

Rectangular moments of inertia

Polar moments of inertia

Radius of Gyration

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. φφ CosrxSinry ⋅=⋅= yCosrArea δφ ⋅⋅⋅= 2

δφφδ ⋅⋅= Cosry

2

4

42

44

2

2

4224

rIIJrI

rCosSinrI

yxzy

x

⋅=+=

⋅=

⋅=⋅⋅⋅⋅= ∫−

ππ

πδφφφπ

π

Parallel-Axis Theorem

2' yxx dAII ⋅+=

Example 4:

( )4

'

224

2'

4

2516

r4 4

4

rπ.I

rπ rπdAII

rI

x

yxx

x

⋅⋅=

⋅⋅⋅+⋅

=⋅+=

⋅=π

IMy

−=σ

Mass Moment of Inertia

It is the product of the element’s mass and the square of the element’s distance from the axis.

( )( )( )∫∫∫

⋅+=

⋅+=

⋅+=

amz

amy

amx

myxI

mzxI

mzyI

δ

δ

δ

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

Load Classification and Sign Convention

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

deflectionx

=> υδυδ 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

δυδ

δδ

δυδ

δδ

δυδ

δυδ

=⇒−=

=⇒=

=⇒=

Successive Integration Method

1

2

1

2600)(

600)(

)()(

CxxV

CxxxV

xxqxV

qxV

+−=

+−=

−=

−=

∫∫

δ

δδδ

2/3m

=650N1050N=

10 << x

NVthenmxFor

xxV

CxFor

CxxV

xFor

350)1(_____1__

6502

600)(

650__0__2

600)(

10__

21

1

2

==

+−=

==

+−=

<<

NVthenmxFor

xxV

CVxFor

CxxV

xFor

1050)2(_____2__

1502

600)(

150__150500350)1(__1__2

600)(

21__

22

2

2

==

+−=

=−=−==

+−=

<<

550)1(____1__0__0)0(____0__

650100)(

6502

600)()(

6502

600)(

10__

3

33

2

2

=====

++−=

⎟⎟⎠

⎞⎜⎜⎝

⎛+−==

+−=

<<

∫∫

MthenxForCMthenxFor

CxxxM

xxxxVxM

xxV

xFor

δδ

0)2(____2__500__550)1(____1__

150100)(

1502

600)()(

1502

600)(

21__

4

43

2

2

=====

++−=

⎟⎟⎠

⎞⎜⎜⎝

⎛+−==

+−=

<<

∫∫

MthenxForCMthenxFor

CxxxM

xxxxVxM

xxV

xFor

δδ

MMax=550N-m

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

A0: original cross-sectional area 0AF

=σ

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

Normal strain or tensile strain or Axial strain

00

0

ll

lll Δ=

−=ε

Hooke’s LawWhen strains are small, most of materials are linear elastic.

σ

ε

E

Normal: σ = Ε ε

Young’s modulus

EAlFl

o

o

⋅⋅

=Δ

Springs: the spring rate

o

o

lEA

lFk ⋅

=Δ

=

Torsion Loading resulting from the twist of a shaft.

lr

zrZ

θδδθγθ

⋅≅⋅=,

lrG

zrGG zz

θδδθγτ θθ

⋅⋅≅⋅⋅=⋅= ,,

G = Shear Modulus of Elasticity

Shear strain

Twist Moment or Torque

∫∫ ⋅⋅

=⋅⋅=A zA z Ar

lGArT δθδτ θθ

2,,

∫ ⋅=A

ArJ δ2Area Polar Moment of Inertia

JGlT

lJGT

⋅⋅

=⋅⋅

= θθ or

JrT

z⋅

≅,θτThus:JrT o

Max⋅

≅τAngular spring rate:

lGJTka⋅

==θ

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

Where:σ 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.τ 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

−=σ

Maximum Stresses in Beams

IbVQ

=τ

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 Hooke’s 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 transformation equations of plane stress or the Mohr’s circle to find the stresses at each point along the height of the beam or to describe how the principal stresses changes as we go from to top to the bottom of the beam.By investigating the stresses at many cross sections of the beam, we can determine how the principal stresses vary throughout the beam. Stress trajectory: Gives the directions of the principal stresses.Stress Contours: Curves connecting points of equal principal stress

Principal-stress trajectories for beams of rectangular cross section: (a) cantilever beam, and (b) simple beam. (Solid lines represent tensile principal stresses and dashed lines

represent compressive principal stresses.)

Example 8-3: A simple beam AB with a span length L = 6ft supports a concentrated load P = 10800lb acting a distance c = 2ft from the right-hand support (see figure below). The beam is made of steel and has a rectangular cross section (width b=2in and height h = 6in).Investigate the principal stresses and maximum shear stresses at cross section mnlocated at a distance x = 9in from the end A of the beam. (Consider only the in-plane stresses) Point y (in)

A 3B 2C 1D 0E -1F -2G -3

The reaction of the beam at support A is RA = P/3 = 3600lb, and therefore the bending moment and shear force at the section mn are

M = RA x = (3600lb)(9in) = 32400lb-inV = RA = 3600lb

Normal stress on cross section mn

Where y has units in inches and σx has units in psi. The stresses calculated are positive when in tension. Note that a positive value of y(upper half of the beam) gives a negative stress, as expected.

( )( )( )( )

yinin

yinlbbh

MyI

MyX 900

623240012

1233 −=

−−=−=−=σ

Solution

Shear stresses on cross section mnThe shear stresses are given by the shear formula

in which the first moment Q for a rectangular cross section is

IbVQ

=τ

( )( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛−=⎟⎟

⎠

⎞⎜⎜⎝

⎛−⎟

⎠⎞

⎜⎝⎛==

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛ −+⎟

⎠⎞

⎜⎝⎛ −=

22

32

2

3

22

46

4212

becomes formulashear the thus,422

22

yhbh

Vyhbbbh

VIb

VQτ

yhbyhyyhbQ

⎟⎟⎠

⎞⎜⎜⎝

⎛−=⋅⋅== ∫∫ 2

22

42 yh bδyby AyQ

h/

yA

x δ

AV

Max 23

=τ

The shear stresses τxy acting on the x face of the stress element are positive upwards, whereas the actual shear stresses τ act downward. Therefore

Substituting the numerical values into this equation gives

In which y has units of inches and τxy has units of psi

⎟⎟⎠

⎞⎜⎜⎝

⎛−−== 2

2

3 46 yhbh

VIb

VQτ

( )( )( )

( ) ( )222

3 9504

662

36006 yyinininlbτ XY −−=⎟⎟

⎠

⎞⎜⎜⎝

⎛−−=

Note: The maximum shear stress that occurs at the neutral axis in a rectangular section can be simplified by using the following equation:

AV

Max 23

=τ

Point y (in) σx (psi) τxy (psi)A 3 -2700 0B 2 -1800 -250C 1 -900 -400D 0 0 -450E -1 900 -400F -2 1800 -250G -3 2700 0

The normal stresses vary linearly from a compressive stress of -2700psi at the top of the beam (point A) to a tensile stress of 2700psi at the bottom of the beam (point G). The shear stresses have a parabolic distribution with a maximum stress at the neutral axis (point D).

Calculation of stresses on cross section mnWe divide the height of the beam into six equal intervals and label the corresponding points from A to G.

Principal Stresses and Maximum Shear StressesThe principal stresses and maximum shear stresses at each of the seven points A through G may be determined from the following equations

( )22

2,1 22 xyyxyx τ

σσσσσ +⎟⎟

⎠

⎞⎜⎜⎝

⎛ −±⎟⎟

⎠

⎞⎜⎜⎝

⎛ +=

( )22

2 xyyx

MAX τσσ

τ +⎟⎟⎠

⎞⎜⎜⎝

⎛ −=

Point y (in) σx (psi) τxy (psi) σ1 (psi) σ2 (psi) τmax (psi)-2700 1350

934602450602934

1350

-1834-1052-450-152-340

034152450

105218342700

A 3 -2700 0B 2 -1800 -250C 1 -900 -400D 0 0 -450E -1 900 -400F -2 1800 -250G -3 2700 0

( )22

2,1 22 xyxx τ

σσσ +⎟

⎠⎞

⎜⎝⎛±⎟

⎠⎞

⎜⎝⎛=( )2

2

2 xyx

MAX τσ

τ +⎟⎠⎞

⎜⎝⎛=

Since there is no normal stress in the y direction, this equation simplifies to:

Thus, by substituting the values of σx and τxy, we can calculate the principal stresses σ1 and σ2 and the maximum shear stress τmax.

The largest tensile stress anywhere in the beam is the normal stress at the bottom of the beam at the cross section of maximum bending moment (σtens)max = 14400psi).The largest shear stress occurs to the right of the load P (V = RB = 7200lb). Therefore, the largest value that occurs at the neutral axis is (τxy)max = 900psiThe largest shear stress anywhere in the beam occurs at 45o planes at either the top or bottom (τxy)max = 14400 / 2 = 7200psi