36548021-Deflection-of-Beams

8/7/2019 36548021-Deflection-of-Beams

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ByATIRAKE SINGH

09011D08019


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It is internal resistance which the body offerss tomeet with the load is called STRESS.

Force per unit area

Units: MPa, psi, ksi

Types: bearing, shearing , axial

P

A=

Load

Area=


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Classification of Stresses

1)Simple or direct stress

I)Tension

II)Compression

III)Shera

2)Indirect stress

I) Bending

II)Torsion

3)Combined stress

Any possible combination of types 1 and 2


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Ratio of deformation caused by load to theoriginal length of material

Units: Dimensionless

Change in Length

Original Length=

LL

=


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Stiffness = stress/strain =

For elastic materials :

oModulus of Elasticity

o

Elastic Modulus

oYoungs Modulus

Stre

ss,

Strain,

E

1


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The ratio of lateral strain to linear strain is known asPoissons ratio.

=lateral strainLateral strain

Where m is a constant and its value varies between 3 and 4 for

different material.


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Center of gravity of the body may be defined as thepoint through which the whole weight of a body maybe assumed act.


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Common shapes are

I Angle Channel

Common materials are steel and wood

Source: Load & Resistance Factor Design (First Edition), AISC


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Introduction to Beams

The parallel portions on an I-beam or H-beam arereferred to as the flanges. The portion that connects the

flanges is referred to as the web.

FlangesWeb

Flanges

Web


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Beams are supported in structures viadifferent configurations

Source: Statics (Fifth Edition), Meriam and Kraige, Wiley


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Beams are designed to support various types ofloads and forces

Concentrated Load Distributed Load

Source: Statics (Fifth Edition), Meriam and Kraige, Wiley


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Beam Theory

Consider a simply supported beam oflength, L. The cross section is

rectangular, with width, b, and depth, h.

L

b

h


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Beam Theory

An area has a centroid, which is similar to a center ofgravity of a solid body.

The centroid of a symmetric cross section can be easily

found by inspection. X and Y axes intersect at the

centroid of a symmetric cross section, as shown on therectangular cross section.

h/2

h/2

b/2 b/2

X - Axis

Y - Axis


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Beam Theory

An important variable in beam design is the moment of

inertia of the cross section, denoted by I. Inertia is a measure of a bodys ability to resist rotation.

Moment of inertia is a measure of the stiffness of the

beam with respect to the cross section and the ability of

the beam to resist bending.

As I increases, bending and deflection will decrease.

Units are (LENGTH)4, e.g. in4, ft4, cm4


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Beam Theory

I can be derived for any common area using calculus.

However, moment of inertia equations for common crosssections (e.g., rectangular, circular, triangular) are readily

available in math and engineering textbooks.

For a rectangular cross section,

b is the dimension parallel to the bending axis. h is the

dimension perpendicular to the bending axis.

12

bh3

x =Ib

h

X-axis (passingthrough centroid)


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Beam Theory

Example: Calculate the moment of inertia about the X-axis for a yardstick that is 1 high and thick.

12bh

3

x =I

b = 0.25

h = 1.00

( ) ( )

12

in1.00in0.253

x

=I

4x in0.02083=I

X-Axis

Y-Axis


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Beam Theory

Example: Calculate the moment of inertia about the Y-axis for a yardstick that is 1 high and thick.

h = 0.25

b = 1.00

X-Axis

Y-Axis

12bh

3

y =I

( ) ( )

12

in0.25in1.003

y

=I

4y in0.00130=I


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Beam Theory

Suppose a concentrated load, P, isapplied to the center of the simply

supported beam.

P

L


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20

Beam Theory

The beam will bend downward as a resultof the load P.

P


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21

Beam Theory

The deflection () is the verticaldisplacement of the of the beam as a

result of the load P.

Deflection,

L


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22

Beam Theory

The deflection () of a simply supported, center loadedbeam can be calculated from the following formula:

I48E

PL

3

=

where,

P = concentrated load (lbs.)

L = span length of beam (in.)

E = modulus of elasticity

(lbs./in.2)

I = moment of inertia of axis

perpendicular to load P (in.4)L

P


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23

Beam Theory

Modulus of elasticity, E, is a property that indicates thestiffness and rigidity of the beam material. For example,

steel has a much larger modulus of elasticity than wood.

Values of E for many materials are readily available in

tables in textbooks. Some common values are

Material Modulus of Elasticity (psi)

Steel 30 x 106

Aluminum 10 x 106

Wood ~ 2 x 106


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24

Beam Theory

Example: Calculate the deflection in the steel beamsupporting a 500 lb load shown below.

P = 500 lb

L = 36

b = 3

h = 2

12

bh3=I

I48E

PL

3

=


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25

Beam Theory

Step 1: Calculate the moment of inertia, I.

12

bh3=I

( )( )12

in2in3 3=I

4

in2=I


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26

Beam Theory

Step 2: Calculate the deflection, .

I48E

PL

3

=

( )( )

( )42

6

3

in2in

lb10x3048

in36lb500

=

( ) ( )( )4

2

6

3

in2in

lb10x3048

in46656lb500

=

in0.0081 =


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27

Beam Theory

These calculations are very simple for a solid, symmetriccross section.

Now consider slightly more complex symmetric cross

sections, e.g. hollow box beams. Calculating the

moment of inertia takes a little more effort. Consider a hollow box beam as shown below:

4 in.

6 in.

0.25 in.


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28

Beam Theory

The same equation for moment of inertia, I = bh3/12, canbe used.

Treat the outer dimensions as a positive area and the

inner dimensions as a negative area, as the centroids of

both are about the same X-axis.

Positive Area Negative Area

X-axisX-axis


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29

Beam Theory

Calculate the moment of inertia about the X-axis for thepositive area and the negative area using I = bh3/12.

The outer dimensions will be denoted with subscript o

and the inner dimensions will be denoted with subscript

i.

bi = 3.5 in.

ho = 6 in.

hi = 5.5 in.

bo = 4 in.

X-axis


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30

Beam Theory

bi = 3.5 in.

ho = 6 in.

hi = 5.5 in.

bo = 4 in.

X-axis

12hb

3

oopos =I 12hb

3

iineg =I

( ) ( )

12

in6in43

pos =I( )( )

12

in5.5in3.53

neg =I


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31

Beam Theory

Simply subtract Ineg from Ipos to calculate the moment ofinertia of the box beam, Ibox

4 in.

6 in.

0.25 in.

negposbox - III =

( ) ( ) ( ) ( )

12

in5.5in3.5

12

in6in433

box =I

4

box

in23.5=I

12

hb

12

hb 3ii3

oobox =I

( ) ( ) ( ) ( )12

in166.4in3.5

12

in216in4 33

box =I


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32

Beam Theory

The moment of inertia of an I-beam can be calculated ina similar manner.


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33

Beam Theory

Identify the positive and negative areas

Positive Area Negative Area


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34

Beam Theory

and calculate the moment of inertia similar to the boxbeam (note the negative area dimensions and that it is

multiplied by 2).

bo

hi

bi bi

ho

12

hb2

12

hb3

ii

3

oo =beamII


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35

Beam Theory

The moment of inertia of an H-beam can be calculated ina similar manner


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36

Beam Theory

The moment of inertia of an H-beam can be calculated ina similar manner


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37

Beam Theory

however, the H-beam is divided into three positiveareas.

b2

b1

h2 h1

b1

h1

12

hb

12

hb

12

hb3

11

3

22

3

11beam-H ++=I

12

hb

12

hb23

22

3

11

beam-H

+=I


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38

Beam Theory

Example: Calculate the deflection in the I-beam shownbelow. The I-beam is composed of three x 4 steel

plates welded together.

L = 8 ft

P = 5000 lbf

x 4 steel plate (typ.)


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39

Beam Theory

First, calculate the moment of inertia for an I-beam aspreviously shown, i.e. divide the cross section of the

beam into positive and negative areas.

bo = 4 in. bi = bi

12

hb2

12

hb3

ii

3

oo =beamII

ho = 5 in. hi = 4 in.


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40

Beam Theory

First, calculate the moment of inertia for an I-beam aspreviously shown, i.e. divide the cross section of the

beam into positive and negative areas.

bo = 4 in. bi =1.75in bi

( ) ( ) ( ) ( )

12

4in1.75in2

12

5in4in33

=beamII

ho = 5 in. hi = 4 in.

4

in23.0=beamII


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41

Beam Theory

Next, calculate the deflection (Esteel = 30 x 106

psi).

I48E

PL

3

=

L = 8 ft

P = 5000 lbf


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42

Beam Theory

Calculate the deflection, .

I48E

PL

3

=

( ) ( )

( )42f6

3

in23in

lb10x3048

in96lb5000

=

( ) ( )( )4

2

6

3

in23in

lb10x3048

in884736lb5000

=

in0.134=


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43

Beam Theory

Example: Calculate the volume and mass of the beam ifthe density of steel is 490 lbm/ft

3.

L = 8 ft x 4 steel plate (typ.)


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44

Beam Theory

Volume = (Area) x (Length)

( ) ( ) ( )8ft4in0.5in3V =

ALV =

( )( )in962.0in3V 2=3in576V =


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45

Beam Theory

Convert to cubic feet3

3

12in

1ftin576V

=

=3

33

1728in

1ft576inV

3ft0.333V =


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46

Beam Theory

Calculate mass of the beam Mass = Density x Volume

Vm=

( )33m 0.333ft

ft

lb490m

=

mlb163.3m=


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Questions??

Question ? ?


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Thank u

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36548021-Deflection-of-Beams