Mektek vs Kbh

7/28/2019 Mektek vs Kbh

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CHINE DESIGN - An Integrated Approach, 2ed by Robert L. Norton, Prentice-Hall 2000

ading Loading

ear Shear

oment Moment

ope Slope

eflection Deflection

yFa

EI

a lmax = ( )2

6

3 yw

EI

l a l amax = + ( )24

3 44 3 4

when :a ywl

EImax= = 0

8

4

M Mw

l amax = = ( )1 2 22

when :a Mwl

max= =02

2

when :a l M Flmax= =

R F1 = R w l a1 = ( )

M Famax =

= + ( )FEI

ax x x a2

22 2

M M R x F x a

F a x x a

= +

= + ( )1 1

1

1

= +

1 2

2

11 2

2EI

M xR

x

Fx a

yEI

Mx

Rx

Fx a

F

EIx ax x a

= +

= ( )

1

2 6 6

63

1 2 1 3 3

3 2 3

when :a l yFl

EImax= =

3

3

M M R x w x a

wl a x l a x a

= +

= ( ) ( ) [ ]1 1

2

2 2 2

22 2

= ( ) ( ) ( )wEI

l a x l a x x a6

3 32 2 2 3

= +

1 2

6

11 2

3EI

M xR

x

wx a

yEI

Mx

Rx

wx a

w

EIl a x l a x x a

= +

= ( ) ( ) ( )

1

2 6 24

244 6

1 2 1 3 4

3 2 2 2 4

M

w

l a1

2 2

2= ( )M Fa

1 =

q M x R x F x a= + 12

1

1 1q M x R x w x a= + 1

2

1

1 0

V M x R F x a

F x a

= +

= ( )

1

1

1

0

0

1

V M x R w x a

w l a x a

= +

= ( ) [ ]

1

1

1

1

1

V R Fmax = =1 V R w l amax = = ( )1

G U R E D - 1

tilever Beams with Concentrated or Distributed Loading. Note: < > Denotes a Singularity Function

) Cantilever beam with concentrated loading (b) Cantilever beam with uniformly distributed loading

Fx a 1

x

R1

l

a wxa 0

x

M1

R1

l

a

x

Vmax

x

Mmax

x

max

x

ymax

x0

Vmax

V

x0

MMmax

x0

max

x0

yymax


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(b) Simply supported beam with uniformly distributed loadingSimply supported beam with concentrated loading

pe Slope

Momentoment

ear Shear

ading Loading

flection Deflection

q R x w x a R x l= + 11 0

21

Rw

ll a1

2

2= ( )

Rw

ll a22 2

2= ( )

V R Rmax = MAX( , )1 2

when = :a Mwx

l x02

= ( )

when = :

a

ywx

EIlx x l

0

242

2 3 3= ( )

R Fa

l1 1=

R Fa

l2 =

V R Rmax = MAX( , )1 2

q R x F x a R x l= + 11 1

21

V R F x a R x l

Fa

lx a

= +

=

10

20

01

M R x F x a R x l

Fa

lx x a

= +

=

11

21

11

yF

EIa

a

llamax =

32

34

2

M Faa

lmax =

1

when = :al

MFl

max2 4

=

=

+ + ( )

F

EI

a

lx x a

a

la al l

2

1

33 2

2 2

2 2

yF

EI

a

lx x a

a

la al l x

=

+ + ( )

6

1

3 2

3 3

2 2

=( )

+ ( ) ( )[ ]

w

EI

x

ll a x a

ll a l l a

24

64

12

22 3

4 2 2

M R xw

x a R x l

w x

ll a x a

= +

= ( )

12

21

2 2

2

2

V R w x a R x l

wl

l a x a

= +

= ( )

11

20

2 11

2

yw

EI

xl

l a x a

x

ll a l l a

=( )

+ ( ) ( )[ ]

24

2

2

3

2 4

4 2 2

U R E D - 2

ly Supported Beams with Concentrated or Distributed Loading. Note: < > Denotes a Singularity Function

x

x

ymax

x0

V

x0

y

ymax

l

a

x

R1 R2

wxa0Fxa 1l

a

x

R1 R2

x

Mmax

x

max

x0

max

x0

M Mmax


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ope Slope

Momentoment

ear Shear

ading Loading

eflection Deflection

q R x w x w x a R x b= + + 11 0 0

2

R Fa

b2 =

q R x F x a R x b= + 11 1

2

1

V Fb a

b xa

b x b x a=

+ 0 0 0

Rwa

b2

2

2=

Mw

aa

bx x

x aa

bx b

=

+ +

2

2

22

22

1

=

+

+

+ +

w

EI

a b ab bb

b a

a a

bx x

a

bx b x a

1

242 4

1

2 4

1

6

4

1

6

2 2 3 4

22 3

22 3

yw

EI

a b ab bb

b a x

aa

bx x

x aa

bx b

=

+

+

+ +

24

2 41

42

2

2 2 3 4

23 4

42

3

=

+

+ ( )

F

EI

b a

bx

a

bx b x a

ba b

2

3

2 2 2

V w a x x a

a

b x b= + + ( )

1

20

21

R waa

b1 1

2=

M Fb a

bx

a

bx b x a=

+

1 1 1

R Fb a

b1 =

yF

EI

b a

bx

a

bx b x a

b a b x

=

+

+ ( )

6

3 3 3

G U R E D - 3

rhung Beams with Concentrated or Distributed Loading. Note: < > Denotes a Singularity Function

x

x

ymax

wxa 0

x

Mmax

R1

l

x

R2

b

a

Fxa1

R1

l

x

R2

b

a

Overhung beam with concentrated loading (b) Overhung beam with uniformly distributed loading

x0

V

x0

yymax

x

0

max

min

0

M

Mmax

x

max


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Biaxial stresses

Locate critical point (max stresses)

P

a

b

M=Pb

T=Pa

=Mc/I

=

Td/J

x

xz

x

xz

Mohrs circle

Determine maximum shear stress

(x,

xz)

12

max)

max)

2

How to Predict Failure


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Maximum Shear Stress Theory (MSST) (A.k.a Tresca)

Tension test specimenyields whensy

(0,0)

max= -sy/2

(sy,0) 2

max

ys=

In real applications,2

21max

= 2 1

max

So yielding occurs when ys= 21

Safety factor is21

=y

s

sn

3-D Mohr Max. Shear Stress Theory (MSST)

Be careful when z

= 0 is outside 2-D Mohrs circle.

2 1

max

3

Always order principalnormal stresses according to

321 >>

22

31max

==

ys Now yielding occurs when

max

Safety factor is

31 =

y

s

sn

ys= 31

Graphical representation

ys= 21

2

1

sy

sy-sy

-sy

yieldor 21 >> yy ss

yieldor 21


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DET vs. MSST

2

1

sy

sy-sy

-syShear diagonal

2 1

Different especially when 1 = -2(e.g. pure torsional loading)

3-D Distortion Energy Theory (DET)

Be careful when z

= 0 is outside 2-D Mohrs circle.

2 13

Now von Mises stress is

( ) ( ) ( )[ ]12232132122

1 ++=e

Brittle Materials

Brittle materials dont yield, they fracture.

Strength in compression >> Strength in tension Three theories presented

Material strength

5 Sut= Ultimate (fracture) strength in tension.

5 Suc = Ultimate (fracture) strength in compression.

5 Strengths are always positive numbers

5 Stresses 1, 2, and 3 can be negative or positive.

Maximum Normal Stress Theory (MNST)

whichever is smaller

fractureor 21 >> yy ss

fractureor 21


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Internal Friction Theory (IFT)

fractureor 21 >> yy ss

fractureor 21 > yy ss

fractureor 21


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Iron is taken from the earth and copper is smelted from ore.Man puts an end to the darkness;

he searches the farthest recesses for ore in the darkness.The Bible (Job 28:2-3)

Image: Iron flows from a blast furnace. Source:American Iron and Steel Institute.

Figure 3.5 Stress-strain diagram for a ductile material.text reference: Figure 3.5, , page 96

A

P

P

(


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text reference: Figure 3.6, page 97

Figure 3.1 Ductile material from a standardtensile test apparatus. (a) Necking; (b) failure.


0

0

=

l

llEL

fr

%EL

Manifest

danger

stress concentrations


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Figure 3.2 Failure of a brittlematerial from a standard

tesiletest apparatus.


%EL%EL



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Figure 3.10 Stress-strain diagram for polymer below, at, and above its glass transitiontemperature Tg.



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=E

uniaxial

linear

Esteel

Ealum


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yx yx

torsion

=G

yx

x

y

steel

alum v

rubber vG E

transverse

axial

=

( )+=

EG


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yield

Allowable building

yallowy SS

yallow S=

yallowy SS

yallow S=

=

y

dUr

0rupture


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Class Members Short nameEnginering alloys(themetals and alloys ofengineering)

AluminumalloysCopper alloysLead alloysMagnesiumalloysMolybdenumalloysNickel alloysSteels

Tin alloysTitaniumalloys

Tungsten alloysZinc alloys

Al alloysCu alloysLead alloysMg alloysMo alloysNi alloysSteels

Tin alloysTi alloys

W alloysZn alloysEngineering polymers(thethermoplastics andthermosets of engineering)

EpoxiesMelaminesPolycarbonatePolyesterPolyethylene, high densityPolyethylene, low densityPolyformaldehydePolymethylmethacrylatePolypropylenePolytetrafluoroethylenePolyvinyl chloride

EPMELPCPESTHDPELDPEPFPMMAPPPTFEPVC

Engineering ceramics(fineceramics capableofload-bearing application)

AluminaDiamondSialonsSilicon carbideSilicon nitrideZirconia

Al2O3CSialonsSiCSi3N4ZrO2

Table 3.7 Material classesand members and shortnames of each member.[From Ashby (1992)].

text reference: Table 3.7, page 123

RA RB RC

SU HB HBSU HB

HB


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Charpy Izod


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quenching

critical temperature

below critical temperatureand


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carburizing

hardness

strength

hardness toughness


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Expensive!

y

y


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Brass zinc


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Figure 3.4 Cross section of fiber reinforcedcomposite material.


Figure33 Strength/density for variousmaterials

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Mektek vs Kbh