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1 P4 Stress and Strain Dr. A.B. Zavatsky HT08 Lecture 7 Further Development of Theory and Applications Hooke’s law for plane stress. Relationship between the elastic constants. Volume change and bulk modulus. Spherical and cylindrical pressure vessels.

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### Transcript of Lecture 7 Further Development of Theory and Applications

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P4 Stress and Strain Dr. A.B. ZavatskyHT08

Lecture 7

Further Development of Theory and Applications

Hooke’s law for plane stress. Relationship between the elastic constants.Volume change and bulk modulus. Spherical and cylindrical pressure vessels.

2

Generalized Hooke’s Law

σxσx

σy

σy

σz

σz

Apply σx, get εx, εy = -νεx, εz = -νεxApply σy, get εy, εx = -νεy, εz = -νεyApply σz, get εz, εx = -νεz, εy = -νεz

For an isotropic linearly elastic material, ε = σ / E in the x, y, and z directions.

( )

( )

( )yxzz

zxyy

zyxx

E

E

E

σνσνσε

σνσνσε

σνσνσε

−−=

−−=

−−=

1

1

1The resulting equations are:Use superposition to get εx :

( )zyxx

zyxx

zyx

x

E

EEE

E

σνσνσε

σνσ

νσε

νενεσε

−−=

⎟⎠⎞

⎜⎝⎛−⎟⎟

⎞⎜⎜⎝

⎛−=

−−=

1 etc.

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Hooke’s Law for Plane Stress

For plane stress, substitute σz = 0 into the generalized Hooke’s Law equations to get:

( )

( )

( )yxz

xyy

yxx

E

E

E

σσνε

σνσε

σνσε

+−

=

−=

−=

1

1

Gxy

xyτ

γ =Remember also the shear strain:

HLT, page 108

4

These equations can be re-written in terms of stresses:

( )

( )01

1

2

2

=

+−

=

+−

=

z

xyy

yxx

E

E

σ

ενεν

σ

ενεν

σ

xyxy Gγτ =

HLT, page 108

These equations contain three material constants: E, G, and ν.We can show that these constants are related by the equation:

)1(2 ν+=

EG HLT, page 110

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τ = Gγ

Consider a plane stress element in “pure shear” and relate the shear strains and stresses to the strains and stresses along the θ = 45° direction.

a b

d c

Element before shear is applied.

ab

dc

Element after shear applied.bd lengthens, ac shortens

γπ−

2

Angle changes related to γ

h

h

hLbd 2=

γπ+

2

6

?2 =′= bdbd LhL

( )( )bdbd

bdbdbdbdbd

bdbdbdbd

bdbd

hL

LLLL

LLLL

ε

ε

εε

+=′

+=Δ+=′

=ΔΔ

=

12

1

Using geometry, the normal strain εbd can be related to the shear strain γ.

d

ba

h

h L’bd

γπ+

2 24γπ

( )( )[ ] [ ]

( )γεε

γπε

γπε

γπ

sin1)(21

)2/cos(11

)2/cos(1212

)2/cos()(2

:rule cosine theUsing

2

2

22

222

+=++

+−=+

+−=+

+−+=′

bdbd

bd

bd

bd

hh

hhhhL

2/sin0)(

small, are and Since2

γεγγε

γε

=≈≈

bd

bd

bd

7

Now use Mohr’s circle and Hooke’s law to relate strains to stresses. Find the stress along the θ = 45° direction :

τσ

τA

A

BB

2θ1 2θ1 = 90°θ1 = 45°

σ1σ2

σ1 = + τσ2 = - τ

σ2 = - τ

σ1 = + τ

θ = 45°

The strain in the σ1 direction is:

( )ντε

ντε

σνσε

τ

+=

−=

−=

11

)(1

211

E

E

EE

E

= εbd

8

( )

( ) ( )

)1(2

112

12/

ν

νγντγ

ντεγε

+=

+=+=

+==

EG

EG

E

Ebdbd

Data from HLT, page 41

Mild Steel: E = 210 GPa,ν = 0.27-0.30, G = 81 GPaAluminium 2024: E = 72 GPa,ν = 0.33, G = 28 GPa

Mild Steel: G = 210 / [2(1+0.27)] = 83 GPaG = 210 / [2(1+0.30)] = 81 GPa

Aluminium 2024: G = 72 / [2(1+0.33)] = 27 GPa

Using the equation just derived:

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This equation is valid for both large and small strains. If strains are assumed to be small, the product terms all tend to 0.

Volume Change and Bulk ModulusNormal stresses produce changes in volume, whereas shearstresses produce changes in shape.Normal stress causes a change in length dL of each face. Since ε = dL / L, each dL = εL.

a

x

y

z

c

b

aεx

bεy

cεz

)1(

)1)(1)(1(

)1)(1)(1(

))()((

1

1

1

1

zyxzxzyyxzyxo

zyxo

zyx

zyx

o

VV

VV

abcV

ccbbaaV

abcV

εεεεεεεεεεεε

εεε

εεε

εεε

+++++++=

+++=

+++=

+++=

= Original volume VoNew volume V1

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The unit volume change (“dilatation”,“volumetric strain”) is defined as:

zyx

o

zyxo

oe

VV

VVe

εεε

εεε

++=

++=

Δ=

)(

Consider next two situations: spherical stress and uniaxial stress.

)1(1 zyxoVV εεε +++=

Assuming that the strains are small,

)(

)1(1

zyxo

ozyxo

o

VV

VVVVVV

εεε

εεε

++=Δ

−+++=Δ−=Δ

The change in volume ΔV is then:

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Spherical stress is defined as σx = σy = σz = σo.

( ) ( )

( ) zyo

x

ooozyxx

E

EE

εενσε

νσνσσνσνσσε

==−=

−−=−−=

21

11(Hooke’s Law)

= εo

Define the bulk modulus of elasticity:)21(3 ν−

=EK

o

oo

o

eK

Ke

εσσ

σ

3strain volumetricstress spherical

===

=

HLT,page 110

These equations also hold for hydrostatic stress (σx = σy = σz = - σo)

Ee

e

oo

zyx

)21(33 νσε

εεε

−==

++= (unit volume change)

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Data from HLT, page 41

Mild Steel: E = 210 GPa,ν = 0.27-0.30, K = 160-170 GPaAluminium 2024: E = 72 GPa,ν = 0.33, K = 75 GPa

Mild Steel: K = 210 / { 3 [1-2(0.27)] } = 152 GPaK = 210 / { 3 [1-2(0.30)] } = 175 GPa

Aluminium: K = 72 / { 3 [1-2(0.33)] } = 71 GPa

Using the equation for K just derived:

If ν = 1/2, K → ∞.This corresponds to a rigid material having no change in volume (that is, the material is incompressible).

If ν = 1/3, K = E.If ν = 0, K = E / 3.

)21(3 ν−=

EK

13

Next, consider the unit volume change for uniaxial stress, σy = σz = 0.

zyxe εεε ++=

E

E

E

xz

xy

xx

νσε

νσε

σε

−=

−=

=

Substituting σy = σz = 0 into Hooke’s Law gives:

( )νσ

νσνσσ

21−=

−−=

Ee

EEEe

x

xxx

The volumetric strain is then:

Note here that the maximum possible value of Poisson’s ratio for common materials is 0.5, because a larger value means that the volume would decrease when the material is in tension, which is contrary to ordinary physical behaviour.

14

Example (based on Gere, 6th ed, p 537, 7.6-10)

A solid steel sphere (E = 210 GPa, ν = 0.3) is subjected to hydrostatic pressure p such that its volume is reduced by 0.4%.

Calculate: (a) the bulk modulus of elasticity K for the steel (b) the pressure p(c) the strain energy stored in the sphere if its diameter d = 150

mm.

Solution:

(a) The bulk modulus of elasticity is found using the equation for K derived earlier.

GPa175Pa10175))3.0(21(3

10210)21(3

99

=×=−

×=

−=

νEK

15

(b)

MPa700Pa10700)004.0)(10175(

004.0%4.0

69 =×=×==

==Δ

=

eK

VVe

o

o

σ

σο is the same as the pressure p.

(c) The strain energy density u (strain energy per unit volume) is given by the area under the σο = K e curve (which is linear).

36

9

262

mJ1040.1

)10175(2)10700(

221

×=×

×===

Keu o

oσσ

The strain energy density U is then simply

J 2478J)1077.1)(1040.1( 36 =××== −uVU

The volume V of the sphere is

33333 m1077.1)1075(34

34 −− ×=×== ππ rV

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Pressure Vessels

• Closed structures containing liquids or gases under pressure. Examples are tanks, pipes, pressurized cabins in aircraft, etc.

• Pressure vessels are considered to be thin-walled when the ratio of radius r to wall thickness t is greater than 10.

• Assume that the internal pressure pin exceeds the pressure poutacting on the outside of the vessel (usually atmospheric pressure). If pout > pin, the vessel could collapse inward due to buckling.

• We will derive equations based on the “net” or “gauge” pressure p, where p = pin – pout.

• We are interested in the stresses and strains that develop in the walls of pressure vessels.

17

Spherical Pressure VesselsTo determine the stresses in the (thin) walls of a spherical pressure vessel with inner radius r and wall thickness t, first cut through the sphere on a vertical diametral plane.

fluid pressure p

tensile stress σin the walls

Fluid force in horizontal directionP = p (πr 2)

Tensile force in horizontal directionT = σ (2πrmt) where rm= r + t/2

Next, isolate half of the sphere and its fluid contents as a single free body.

r

t

p

18

For equilibrium, forces in the horizontal direction must balance.

trrp

rptrF

m

m

2

0)()2(2

2

=

=−=Σ

σ

ππσ

For thin-walled vessels, r ≅ rm and the equation becomes

trp

2=σ * Note that using r instead of rm actually gives a result

closer to the theoretically “exact” result.

Since the same equation for tensile stresses would result from any slice through the centre of the sphere, we conclude that the wall of a spherical pressure vessel is subjected to uniform tensile stress in all directions.

σ

σ

Sometimes called “membrane stresses”

19

Stresses at the Outer Surface

σx

σy= σ

σx= σ

σy

σx = σy = σ = pr / 2tσz = 0τxy = 0

When new stress elements on the sphere are obtained from rotating this element about the z axis, the normal stresses remain the same and there are no shear stresses. So, every plane tangent to the sphere is a principal plane, and every direction a principal direction.

The principal stresses are σ1 = σ2 = pr / 2t, σ3 = 0.

The maximum (out-of-plane) shear stress is τmax = (σ1 – σ3) / 2 = pr / 4t

Draw the three Mohr’s circles to convince yourself of this.

20

1. The wall thickness must be small in comparison to the other dimensions. The ratio r/t should be 10 or more.

2. The internal pressure must exceed the external pressure to avoid inward buckling.

3. The analysis is based only on the effects of (net) internal pressure. The effects of external loads, reactions, the weight of the contents, and the weight of the structure are not included.

4. The formulae derived are valid throughout the (thin) wall of the vessel, except near points of stress concentration.

21

Cylindrical Pressure Vessels

The thin-walled cylindrical tank is subjected to a net internal pressure p.

A Bm

n

p

q

bσx

σy

A m

n

Longitudinal Stress

fluid pressure p

tensile stress σLin the walls

Fluid force in horizontal directionP = p (πr 2)

Tensile force in horizontal directionT = σL (2πr t)

22

For equilibrium, forces in the horizontal direction must balance.

m

n

tr

p

q

b

Circumferential (or Hoop) Stress

The longitudinal stress is then:trp

L 2=σ

fluid pressure p Fluid forceP = p (2br)

tensile stress in the walls σh

Tensile forceT = σh (2bt)

For equilibrium, forces in the horizontal direction must balance.

The circumferential (or hoop) stress is then:trp

h =σ

23

The normal stresses σx and σy are principal stresses since no shear stresses are acting.

A Bm

n

p

q

bσx

σy

Circumferential (Hoop) Direction

trp

hy === σσσ 1

Longitudinal Direction

trp

Lx 22 === σσσ

An obvious discontinuity exists at the ends of the cylinder, where the ends (usually plates or hemispheres) are attached, because the geometry of the structure changes abruptly.

24

Stresses at the Outer Surface

σx

σy= σ1

σx= σ2

σy

σx = σ2 = pr / 2tσy = σ1 = pr / tσz = σ3 = 0τxy = 0

The maximum (out-of-plane) shear stresses are

The maximum (in-plane) shear stress is

tprtprtpr42

)2/()/(2

21)2,1max( =

−=

−=

σστ

tprtpr22

0)/(2

31)3,1max( =

−=

−=

σστ

tprtpr42

0)2/(2

32)3,2max( =

−=

−=

σστ

overall max shear stress

25

ExampleA spherical pressure vessel having 450 mm inside diameter and 6 mm wall thickness is to be constructed by welding together two aluminium hemispheres. From tests, it is found that the ultimate and yield stresses in tension at the weld are 165 MPa and 110 MPa, respectively. The tank must have a factor of safety of 2.1 with respect to the ultimate stress and 1.5 with respect to the yield stress. What is the maximum permissible pressure in the tank? (Gere and Timoshenko, 3rd ed, p 413)

r = d/2 = 0.450/2 = 0.225 m t = 0.006 mr / t = 37.5 (>10, so thin-walled assumption okay)

The allowable stress based on the ultimate stress is:σallow = σult / n = 165/2.1 = 78.6 MPa

The allowable stress based on the yield stress is:σallow = σy / n = 110/1.5 = 73.3 MPa

The latter is lower, so it is the most critical and governs the design.

26

Tension at the weld for the spherical vessel is σ = pr / 2t

p = 2 t σallow / r = 2 (0.006)(73.3 x 106) / 0.225p = 3.91 x 106 Pa = 3.91 MPa

So, the maximum allowable pressure is 3.9 MPa.(Note that for safety reasons we have rounded down here, not up.)

27

ExampleA cylindrical pressure vessel is constructed with a helical weld that makes an angle of 55° with the longitudinal axis. The tank has inside radius 1.8 m and wall thickness 8 mm. The maximum internal pressure is 600 kPa.

Find the circumferential and longitundal stresses, the absolute maximum shear stress, and the normal and shear stresses acting perpendicular and parallel to the weld. (Gere and Timoshenko, 3rd ed, p 414)

r / t = 1.8/0.008 = 225 > 10 (so thin-walled assumption okay)

55°

Circumferential and longitudinal stressesσx = σL = pr / 2t = (600 x 103)(1.8)/2(0.008) = 67.5 MPaσy = σh = pr / t = 135 MPa

28

Absolute maximum shear stressσ1 = σh, σ2 = σL, σ3 = 0 (at the outer surface)τmax = (σ1 - σ3 )/2 = 67.5 MPa

135

67.5

Need stresses perpendicular and parallel to the weld. Consider the stress element below and use either the transformation equations or Mohr’s circle with θ = 35°(why not 55°?) to find σx1 = 89.7 MPa, σy1 = 112.8 MPa, τx1y1 = 31.7 MPa.