1 material strengthkkkr3644 part ii - principles of material strength ~ students

8/12/2019 1 Material StrengthKKKR3644 Part II - Principles of Material Strength ~ Students

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2/59

CLASSIFICATION OFPRESSURE VESSELS v e n o wo c asses epen ng on e

ratio of the wall thickness to vesseldiameter t D :

and allied industries are classified as thin-walled vessels.

Thick-walled vessels are used for P.


3/59

PRINCIPAL STRESSES The state of stress at a point in a

structural member under a complexsystem of loading is described by themagnitude and direction of the principle

stresses. Principle stresses = maximum values of

the normal stresses at the point; whichact on planes on which the shear stressis zero.

In a two-dimensional stress system, the

principal stresses at any point arerelated to the normal stress in the x and

y irections, x an y, an t e s earstress, xy at the point of the followingeqn.:

1

The maximum shear stress at the pointis equal to:

21 42,stresses,Pr nc paxyxyxy ++=

( )212

1stressshearMaximum =


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MATERIAL STRENGTH Example of principal stress at vessel wall:

General example of symmetrical vessel at an axis:


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PRINCIPAL STRESSES IN PV WALL

OOElement abcd

(1) 1 = meridional / longitudinalstress, acting along a meridian0-0 axis.

a

b

d

c

(2) 2

= circumferential/tangential/hoop stress, acting

along parallel to 0-0 axis. 3 ,

normal to 0-0 axis. For thin wall,

3


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THEORIES OF FAILURE - unidirectional

Generally stress 2:(i) Compressive stresses are -ve;

(ii) Tensile stresses are +ve. For a structure under unidirectional stress (tensile

or com ressive failure will occur if > thetensile strength of the material.

The failure point in a simple tension is taken as theyield-point stress, e.


7/59

THEORIES OF FAILURETHEORIES OF FAILURE combined stressescombined stresses

For components subjected to combined stresses (normal orshear stresses), failure analysis becomes more complicated.

en ng momen s ress

Longitudinal

Shear stress

Circumferentialstress

3 commonly used theories to analyze failure under combinedstresses:


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the principal stresses reaches the, e

(unidirectional).,

1=e or 2=e or 3=e


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Maximum Shear Stress Theory

Assuming the failure will occur in a complex stress system when

the maximum shear stress reaches the value of the shear

The shear stress at whichthe material fails under

stress at failure in simple tension.

tensile test:

2

'

e

e

=

For system of combinedstresses, 3 shear stresses:

2;

2;

213

332

221

1 ===

eee === 321 oror


10/59

Maximum Strain Energy Theory

Assuming failure will occur in a complex stresssystem when the total strain energy per unit

volume reaches the value at which failure occurs insimple tension.

materials under complex loading.

Most design codes uses Maximum Shear Stresseory an ax mum ra n nergy eory.


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Under certain loading conditions failure of a

,

by buckling or wrinkling: Buckling results in a gross and sudden change of shape

, Plastic failure, the structure retains the same basic

shape.

This mode of failure will occur when the structureis not elastically stable, lacks of sufficient stiffness,rigidity to withstand the load.

The stiffness of a structural material is dependent

not on the basic strength of the material but on:


12/59

Stress - Strain Stress is defined as:

F==

Load

Strain is defined as:l1l2

oareasect ona-

Strain due to elon ation of sam le:

ol==

lengthOriginal

engoange

Strain due to compression of1

12

l

lle

=

sample:

1

21

d

ddc

= ForceF

o sson s ra o s e ne as:

e

c

=

d1


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Stress () Strain () CurveStress ()

u

eb

a

The slope for straight line (elastic area):

Strain ()a b

)(ModulussYoung'/ElasticityofModulus Ea

a

b

b

ab

ab ====


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Summary of Stress () and Strain ()StressStress

((==FF//AAoo N/mN/m22))

Shear stressShear stress Tensile stress Compressive-

StrainStrain

(( unitlessunitless))

onga on s ra nonga on s ra n ee (+(+veve)) ompress ve s ra nompress ve s ra ncc(+(+veve))

Poissons ratio (+Poissons ratio (+veve))

((vv==cc// ee uunitlessnitless))


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Under tensile stress: Under compressive stress:

o

oc

o

oe

Dl == and11

444444 3444444 21 o

oc

o

oe

lD

=

= and 11

444444 3444444 21

e

cv

=ratio,sPoisson'

e

cv

=ratio,sPoisson'


16/59


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Stress Analysis in Shell of Revolution

directions:1. Towards the walls of the shell of revolution

membrane stress at the walls.2. Top-bottom direction due to the longitudinal stress.


18/59

(1) Analysis of Membrane Stress

b

a

a

c

d


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(1) Analysis of Membrane Stress Symbols:

= n erna pressuret= thickness of shell1 = longitudinal / meridional stress2 = circumferential / tangential stress

r1 = mer ona ra us o curva ure

r2 =circumferential radius of curvature Values of r1 and r2 are determined by the

shapes of the shells: y n er: r1= an r2= ameter

Sphere: r1 = r2 = D/2

b

a

cd


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dS2

r2

d2b

a 1c

d

2

dS

r1

t

b

t

b1

dS1 dS2

2 c 1 a


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t

b

SidesSides ababoror cdcd

b

a 1F1

d

2

1

2tdS1d2/2

a2

d2/2 x

b


22/59

SidesSides adadoror bcbct

ab

a 1F2

d

2

2 d1/21tdS2

r1d1/2 y

d


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Total Force due to Membrane Stresses (Total Force due to Membrane Stresses (FFnn))

+

=+=

2

sin

2

sin222 1212

1221

ddS

ddStFFFn


24/59

Force due to Internal Pressure (Force due to Internal Pressure (FFPP))


25/59

=

==

2sin2

2sin2

elementofArea

22

11

dr

drPF

bcabPabcdPF

P

P

+

=+=

2sin

2sin222 121

21221

ddS

ddStFFFn

Simplified using:

(1)2

2

1

1

rrt

P +=

22222

11111

2211

dand

22sinand

22sin

dSddrS

dSddrdS ====

21 rr


26/59

(2) Analysis of longitudinal / meridional stress


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Analysis of longitudinal / meridional stress

r2

r2sinP

Areas = (r2sin)2


28/59

Perimeter = 2(r2sin)

Area = t2(r2sin)

=1sin

tr2sin

1

1

2r2sin


29/59

Balanced

= 1t2(r2sin)sin= P(r2sin)2

FP=Fn


30/59

Shapes of Shells of Revolution


31/59

Cylinder A cylinder is swept out by the rotation of a line

parallel to the axis of revolution: r1 = and r2 = D/2 (D = diameter of cylinder)

Pr P

21

t=

21 rrt=

PD1= 2

21

Dt

P +

=

t

PD

22=


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Sphere For sphere:

r1 = r2 = D/2 (D = diameter of sphere)

P Pr

21 rrt=

21

t=

PD1= 224

2

DDt

PD

t

P +=

t

PD

42=


33/59

Cone cone s swept out y a stra g t ne nc ne at an

angle ato the axis: r1 = and r2 = r/cos a(D2=diameter of circle)

P Pr

21 rrt=

21

t=

Pr1= art

P

cos21 +

=

Maximum value

at

Pr

cos

2=for 1 and 2when r=D2/2


34/59

Ellipsoid Ellipsoid is commonly used as head section or cover. Ellipsoid has 2 axes:

1. Minor axis (vertical)2. Major axis (horizontal)

The relationship between the meridional radius ofcurvature, r1 and circumferential radius of curvature r2with the minor and ma or axes:


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23

2 brr =(2)2

1

Pr=

23

2 brr =

P

a

P

a

21 rrt

=

21 rrt

=

=

2

2

4

22

2 br

ar

t

P

==

211and

aPaPa tb

Pa

2

2

21 ==

tt


36/59

Torus Torus is formed by rotating a circle of radius r2 about an

axis. The meridional radius of torus at d:

sin21

rRRr o

+==

s ns n

d


37/59

sin

sin

sin

21

rRRr o

+==

(1)21P

+=21

( ) +

=

sin2

s n1

2

222

rR

r

t

r

o

Pr

t2=

+

+=2

222 2

2 rRrR

tPr

o

o

=

2

222

2

2 rR

rR

t

Pr

o

o


38/59

Torispherical Heads A torispherical shape is often used as the end closure of

cylindrical vessels. It is formed from part of a torus and part of a sphere.

cheaper to fabricate. Symbols: R

k= knuckle radius (radius of torus), R

c=

crown radius (radius of sphere). For the spherical portion:

t

PRc

221 ==

For the torus:PR

t21

=

and is a function ofR

cand R

k.


39/59

R l h b l d l d f lR l h b l d l d f l


40/59

Relationships between longitudinal and circumferentialRelationships between longitudinal and circumferentialstresses with the wall thickness for shells of revolutionstresses with the wall thickness for shells of revolution

Shape r1 r2 1 2

Centre =0r1= sinsin

21r

o==t22

1=( )

+

=

sin2

1

2

222

rRto

Pr2=

Outer =/2 sin=1t

+

+=

2

222

2

2 rR

rR

t

Pr

o

o

Inner =3/2 sin=-1

=

2

222

2

2 rR

rR

t

Pr

o

o

Torispherical:Sphere

tPRc

221 ==

Torus

locationt

PRk

21=


41/59

FLAT PLATES Flat plates are used as covers for manholes, as blind flanges,and for the ends of small diameter and low pressure vessels.

,its original shape.


42/59

Types of Flat Plates

(i) Circular plate supported atits edges (clamped edges)

(ii) Simply supported circularflat plate

P

w

U if l l d d i l l t


43/59

Uniformly loaded circular plate,slope, at any radiusxis given by:

21

31 CxCPxdw

216 xDdx==

44444 344444 21

32

2

1

4

ln

464

CxCxC

D

Pxw +=

with,P=pressure, =slope of plate,w=deflectionx= radial distance to point of interest,

= exua r g y o p a et= plate thicknessv= Poissons ratio for the material

= o u us o e asticity o t e materia oung s o u usC1,C2,C3 = constants of integration which can be obtained

from the boundary conditions at the edge of the plate

CLAMPED EDGES FLAT PLATE


44/59

CLAMPED EDGES FLAT PLATE When the edge of the plate is rigidly clamped, not free torotate heavy flange or strong joint:


45/59

32

21

4NINTEGRATIO21

3

ln464216

1CxC

xC

D

Pxw

x

CxCPx

Ddx

dw+= ++==

00

0 22 === CC

3

224

CxPaPxw +=

11

3xCPx

+= Pa4

216D

2

D643

DC 81=

( )222

64 axDw =

( )2216

xa

D

Px=

D

Paw

64

4

max =

St i Ci l Fl t Pl t


46/59

Stress in Circular Flat Plate ,

external side plate will undergo elongation strain and theinternal side of plate will undergo compressive strain.

A y z

r

y

yy

xx

A

xx


47/59

( )[ ]2

3

21

1

dxdy

dxyd

r +

=

2

21

dx

yd

r

=r

yx =

2yd

=dx

x

Plate with tensile stress at the plate wall in the


48/59

Plate with tensile stress at the plate wall in thedirections of x (x) and z (z)

Tensile stress x will result inelongation strain in the x

direction ( ) and compressiveTensile stress z will result in

elongation strain in the zdirection (ez) and compressive

strain in the zdirection (cz): strain in the xdirection (cx):excz

xex v

E == and ezcx

zez v

E == and

If both xand zact simultaneously, hence thetotal strains at the respective xand zdirections

Ev

Ev

E

zxez

xcxexx

+=+=+=

xzz

EEE

exczezz

21 v

E xx

=

2

2

dx

ydyx =

xz v =

=2

2

2

1 dx

ydy

v

Ex


49/59

21 v

E xx

=

x

y= z

y=

=

=

=

=

EyEEE 11

x zezcx

zxezexcxexxx

rrvvvv 1111 2222

EyEEE 11

xz

cxczezczzz

rrvvvv 1111

2222

z

zr

=x

xr

=cxez =


50/59

=2

2

2

ydy

Ex =

2t

xydAM

876

222 2 tt

2t

t/2

=

=

2

2

22

2

2

2

2 11tt

dAydxv

dAdx

yv

M dA

dy x xCompressive

{

===2

32

2

2

12tt

tdyydAyI

22 121 dx

y

vM

=43421

{

===2

32

2

2

12tt

tdyydAyI

-t/2x=1 unit

For a stripe with 1 unitwidth (zdirection) and

thickness t, A =t X 1,= =

2

2

dx

ydDM=

Mis the total moment if plate

deflects in the ydirection-

along z-axis can be neglected

From the top view of plate


51/59

From the top view of plateCircular flat plate that

x-axisx

-

Side view of platex

-y

Law of similar triangle

dxdy

xrdxdy

rx

zz

11 ==

An element with width d zandy - axis


52/59

Mx

= bending moment at x-axis per unit lengthd

x-axis

t/2

=2

2

t

t

xydAM -t/2 dz Bending moment of

element with

}2t

c ness yanwidth dzat y-axis

+

=

zx

xr

vrv

Ey 11

1 2

=2t

xx dydzydzM

447

( )

+=

+= zxzx

xr

vr

Dr

vrv

EtM 1111112 2

3

Bending moment ofelement with

+=

xz

z

r

v

r

DM11

z= bending moment at

y-axis per unit length

dyvyd 21121 yd

=


53/59

+=

+=

dx

dy

x

v

dx

ydD

rv

rDM

zx

x 2

112

dxrx=

+=

+=2

2111

dx

ydv

dx

dyD

rv

rDMz

dy11=

For plate with clampededges, the obtained

xzz

( ) ( )[ ]vxvaPMx ++= 311622

e ec on w w=-y:

( )222

64

ax

D

Pw =

( ) ( )[ ]vxvaPMz 31116

22 ++=

8

2Pa

Mx =PaxPaPx

4224

+=

D

xPa

D

Px

dx

dy

DDD

1616

64326423

+=8

2

PavMz =

D

Pa

D

Px

dx

yd

1616

3 22

2

2

+= zx MMv1 >< yr

E

r

yEE

r

y xxxx

==== and

EIE x

I

tt

====

=876

2

2

22Pa

Mx = tyrrtt

x

22

32t

t

2y=

122AyI

t

==22 tIx

==32

MM x ==

2

min

2

16

3

16

3

=

=

x

x

Pdt

t

dP

SIMPLY SUPPORTED FLAT PLATE


55/59

SIMPLY SUPPORTED FLAT PLATE

For simply supported plate, the edge / boundary conditions:

w


56/59

32

21

4NINTEGRATIO21

3

ln464216

1CxC

xC

D

Pxwy

x

CxCPx

Ddx

dw+++== ++==

DIFFERENTIATION

00

0 22 === CC 216

1

22

1

3

x

CPx

Ddx

dy+=

2

1

4

CxCPx

w ++==

2161

2 Ddx+=

2

464D

0=C

+=dxxdx

DMx 2

CPxdw 1 13

( ) ( ) +++= vC

vD

Px

DMx 12316

1

Ddx 216

( )( )vD

vPaC

+

+=

18

32

1

( )( )

vPaC

+=

32

1


57/59

Pdyvyd 222

( )vD +18

1

( ) ( )[ ]vxvaPdx

ydv

dx

dy

xDM

dxxdx

z

x

31316

1

16

22

2

2

2

++=

+=

216

1 13

xCPx

Ddx

dy+=

216

31 12

2

2CPx

Ddx

yd+=

vPa

MM +== 32

16

Pd2

x ==64

6MMtt

32

2 tdAyI

t

== 2

2 tI

x

2

2t

( ) ( )2

min2

2

332

33

32

3

+=

+=

x

x

Pvdt

t

Pdv

Comparison between clamped edges and simply supported plates


58/59

24NINTEGRATIO

31 xCPxCxCPxdw32 n464216 xDwyxDdx +++== ++==

Clamped edge plate Simply supported plate

w=0 @ x=a

=0 @ x=a=0 @ x=0

Mx=0 @ x=a

w=0 @ x=a=0 @ x=0

( )( )

vxvaP

M

vxaP

Mx

313

316

22

22

++=

+=( ) ( )[ ]

P

vxvaP

Mx 3116

22

22

=

++=

z16

( ) 0at32

=+== xvPa

MMx

z16

PavM

PaM zx

== and

22

MMv zx >

1 material strengthkkkr3644 part ii - principles of material strength ~ students

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