material shape.ppt

7/30/2019 material shape.ppt

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MECH4301 2008 Lecture 6 (1/3) Shape Factors 1/23

Material and Shape:

Textbook Chapters 11 and 12

Lecture 6 (1/3)

Efficient?

Materials for efficient structures

MECH4301 2008Materials Selection in Mechanical Design

To c reate a deformation work -stress chart for foams,

use (densif icat ion strain * yield s trength) as y-axis and

yield strength as x-axis.)See Announcement in Bb.


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Shape and mechanical efficiency

Section shape becomes important when materials are loaded

in bending, in torsion, or are used as slender columns.

Examples of Shape:

Shapes to which a material can be formed are limited by the

material itself.

Shapes from: http://www.efunda.com/math/areas/RolledSteelBeamsS.cfm

Is shape

important fortie rods?


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Shape and mode of loading

Standard structural members

Loading:

tension/compression

Area A and shape

IXX, IYY matter

Area A and shape J

matter

Area A and shape Imin

matter

Area A matters,

not shape

Loading: bending

Loading: torsion

Loading: axial

compression


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Certain materials can be made to certain

shapes: what is the best material/shape

combination (for each loading mode)?


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Shape efficiency: bending stiffness pp. 289-290

1212

24

oo Ab I

b

b

Area A is

constant

Area Ao = b2

modulus E

unchanged

Neutralreferencesection

Shaped

sections

Define a standard reference section: a solid square, area A = b2

3

0

L

IECSo

3

L

IECS

3

1

L

EICFS

Moments of Sections; p 477 12

3bhI

221212

AAE

E

S

S

oooo

e

II

I

I

I

I

Ao = A

Define shape factor forelastic bending,

measuring efficiency, as


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A shaped beam of shape factor forelastic bending,

e= 10, is 10 times

stiffer than a solid square section

beam of similar cross section area.

212

AE

E

S

S

ooo

e

I

I

I

I

I

bending stiffness


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I-sections

Properties of the shape factor The shape factor is dimensionless -- a pure number.

It characterises shape, regardless of size.

Circular tubes

10e

10e

These sections are e

times stiffer in bending than a solid

square section of the same cross-sectional area

Increasing size at constant shape = constant SF

Rectangular

Sections e= 2


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Define a standard reference section: a solid square, area A = b2

Shape efficiency: bending strength p. 294/5

66

2/3

0

3

AbyIZ

m

o

o

** Z

y

IM

m

f

maxy

ZI

b

b

Area A is

constant

Area A = b2

yield strength

unchanged

*

Neutralreferencesection

Moments of Sections; p 477my

IZ,modulussection

Define shape factor for the onset of

plasticity (failure), measuring efficiency, as

** Z

y

IM

m

f

*

0

*0

0 Z

y

IM

m

f

2/3*

*6A

Z

Z

Z

Z

Z

M

M

oofo

f

f

A = Ao


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**

Zy

IM

m

f


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A shaped beam of shape factor forbending strength, f= 10, is 10 times

stronger than a solid square section

beam of similar cross section area.

2/3*

*

6 A

Z

Z

Z

Z

Z

M

M

oofo

f

f

bending strength


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Section shape Area Am

Secondmoment I, m

4

Elastic shapefactor

bh 12

hb 3

b

h

abba

4

3

b

a3

tr2

)rr( 2

i

2o

tr

)rr(4

3

4

i

4

o

)tr(

t

r3

)tb,h(

)bh(t2

)

h

b31(th

6

1 3 tb,h(

)h/b1(

)h/b31(

t

h

2

12

)tb,h(

tb2

)hh(bio

2o

3i3o

htb2

1

)hh(12b

)tb,h(

tbh

23

2o

)tb,h(

)bh(t2

)

h

b31(th

6

1 3 )tb,h(

)h/b1(

)h/b31(

t

h

2

12

h2t

b

h

b

2a

2b

h

b

b

hohi

2ro2ri

Tabulation of shape factors (elastic bending) p. 292/3

tr

tr

rttr

Ao

e 3

)2(1212 2

3

2

I

I

I

bh

hbbh

Ao

e

22

3

2 1121212

I

I

I

A2 = Ao2

Second moment of

section, I


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Comparison of shapes done so far at constant

material (E, y) and given cross section area, A

How to compare different materials and

different shapes at:

Constant structural stiffness, S ?

Constant failure moment, Mf ?

Material substitution at constant stiffness or

strength allowing for differences in shape


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L

mA

A

Ae

o

e

12I12

I

I 2

2

I

m = mass

A = area

L = length

= density

b = edge length

S = stiffness

I = second moment of area

E = Youngs Modulus

Beam (shaped section).

Bending stiffness of the beam S:

Trick to bring the Shape Factor in ?

Eliminating A from the eq. for the mass gives:

3L

IECS

2/1

2/1512

EC

LSm

e

LAm

Chose materials with largest

2/1

Ee

Minimise mass, m, where:

Function

Objective

ConstraintL

FArea A

Shape factorpart of thematerial index

Indices that include shape (1): minimise mass at constant stiffness p. 310


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Indices that include shape (2): minimise mass at constant strength p . 311

L

mA

A

A

f

o

f 6

ZZ

6Z

Z2/3

2/3

m = mass

A = area

L = length

= density

Mf= bending strength

I = second moment of area

E = Youngs Modulus

Z = section modulus

Beam (shaped section).

Bending strength of the beam Mf:

Trick to bring the Shape Factor in ?

Eliminating A from the equation for m gives:

** Z

y

IM

mf

3/2*

3/26

f

f LMm

LAm

Chose materials with largest

3/2*

f

Minimise mass, m, where:

Function

Objective

ConstraintL

FArea A

Shape factorpart of thematerial index


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From Lecture 4: Demystifying Material Indices (elastic bending)

2/1

1

1

2/15

1

12

EC

LSm

2/1

2

2

2/15

2

12

EC

LSm

2

1

1

2/1

1

2/1

2

2

1

2

M

ME

Em

m

For given shape, the reduction in mass atconstant bending stiffness is determined by

the ratio of material indices.

Same conclusion applies to bending strength.

Unshaped mass,Material 1

Unshaped massMaterial 2


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Demystifying Shape Factors (elastic bending)

2/11

1

2/15

12

EC

LSmo

2/1

1

1

2/1512

EC

LSm

e

s

2/11

2/11

2/11

1 1

)(

E

Em

m

o

s

Shaping (material fixed) at constant bending

stiffness reduces the mass of the

component in proportion to e

-1/2.

Optimum approach: simultaneously

maximise both M and .

Unshapedmass

Shaped mass, samematerial, same S

Q: Is the cross section

area constant when going

from mo to ms?


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Demystifying Shape Factors (failure of beams)

3/2*

3/26

LMm fo

3/2*3/2

6

f

fs LMm

3/23/2*3/2*

1ffo

s

mm

Unshapedmass

Shaped mass,same material,same Mf

Shaping (material fixed) at constant

bending strength reduces the mass of thecomponent in proportion to

f-2/3.

Optimum approach: simultaneously

maximise both M and .

EXAM QUESTION: Is the

cross section area constant

when going from mo to ms?


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Material , Mg/m3 E, GPa e,max

1020 Steel 7.85 205 65 1.8 14.7

6061 Al 2.70 70 44 3.1 20.4

GFRP 1.75 28 39 2.9 18.9

Wood (oak) 0.9 13 8 4 11.4

/2/1E /E 1/2maxe,

Practical examples of material-shape combinations

/

2/1

E

Materials for stiff beams of minimum weight

Fixed shape (e fixed): choose materials with greatest Shape e a variable: choose materials with greatest

Same shape for all (up to e = 8): wood is bestMaximum shape factor (e = e,max): Al-alloy is bestSteel recovers some performance through high e,max

/E 1/2maxe,


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Density (typical) (Mg/m^3)0.01 0.1 1 10

Y

oung'sModulus(typical)(GPa)

1e-004

1e-003

0.01

0.1

1

10

100

1000

Concrete

Titanium

Cork

PP

Flexible Polymer Foams

Rigid Polymer Foams

Tungsten Carbides

Steels Nickel alloys

Copper alloys

Zinc alloys

Lead alloys

Silicon Carbide

AluminaBoron Carbide

Silicon

Al alloys

Mg alloys

CFRP

GFRPBamboo

Wood

PlywoodPET

PTFE

PE

PUR

PVC

EVA

Silicone

Polyurethane

Neoprene

Butyl Rubber

Polyisoprene

CE

2/1

Note that new material with

Shape on selection charts: stiffness p. 312/3

Al: e = 44

Al: e = 1

Density (Mg/m3)

Yo

ungsmodulus(G

Pa)

e

e

e

e

e

eE

EE

/

/

/

2/1

2/1

2/1

e

s

/

e

s EE /




20/29


Shape on selection charts: stiffness p. 314

Density (Mg/m^3)0.1 1 10

Young'sModulus(GPa)

1

10

100

1000

Al Sf=44

Bamboo SF=1

Steel SF = 65

Bamboo SF =5.6

steel SF=1

Al SF =1

Drag the

labelsalong lines

of slope 1

Selection line

of slope 2

UnshapedSteel SF =1

UnshapedAluminium

UnshapedBamboo SF= 1

ShapedaluminiumSF = 44

Shaping

makes

Steel

competitive

with Al and

Bamboo

Shaped BambooSF=5.6

Shaped steel

SF=65


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Note that new material with

2

3/22*

2

2

3/2*

3/2*

/

/

/f

f

f

f

f

f

2** / fs

2/ fs

Shape on selection charts: strength p. 314

3/2*

100 2 f

steel

1 2 f

steel




22/29


Shape on selection charts: strength p. 314

Density (Mg/m^3)0.01 0.1 1 10

TensileStrength(MPa)

10

100

1000

Bamboo SF =1

Al 2024, SF=10 SF^2=100

steels SF=7 SF^2=49

bamboo SF = 2 SF^2=4

steel SF =1

Al Sf =1

selection line slope 1.5

Selection line ofslope 1.5

Shaped SteelSF=7; (SF)2=49

ShapedBamboo SF=2(SF)2=4

Shaping

makes

Steel

competitive

with Al and

Bamboo

Shaped Aluminium

SF=10; (SF)2

=100


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Shaping at constant cross section A increases the

bending stiffness or strength by at constant mass.This stems from the definition of shape factor

e = S/So= I/Io f= M/Mo = Z/Zo

Dragging the labels in the CES charts is equivalent

to shaping at constant bending stiffness or strength,

so the mass is reduced by 1/e1/2 (stiffness) orby 1/f2/3 (strength).

Exam question: (to get everybody confused!)


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Examples of indices including shape p. 318

Same as elastic

bending


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This afternoon: solve Exercises E8.1,8.8, 8.9 and 8.12.

Leave 8.6 / 8.7 for next sessions.

-Tutorial 3 (E8. Materials and Shape)(6 Exercises). Solve in this

order: E8.1; E8.8; E8.9; E8.12; (solve either E8.6 or E8.7) (seehints and instructions in BB).

Exercise #6 for Tute 3: Show that the shape factors of Table 12.5 (p.

325) are a factor 4/3 = 1.33 too large.


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Example using CES: dragging labels


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End of Lecture 6two more lectures re. shape factors to follow


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Shape factors for twisting and buckling

2T2

0

o

T

AK14.714.0K

KK

SS ATo

T

3/2AQ8.4

QQ

o

fT

Failure under torsion p. 296

Buckling p. 296

212

AE

E

S

S

oo

e

I

I

I Same as elastic

bending

Elastic twisting p. 294


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Moments of Sections; p 477

material shape.ppt

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