P V l dPressure Vessel and Combined LoadingMAE 314 – Solid MechanicsY. Zhu

Slide 1

Thin-Walled Pressure Vessels

• Assumptions

Cylindrical vessel with capped ends Spherical vessel

– Constant gage pressure, p = internal pressure – external pressure– Thickness much less than radius (t << r, t / r < 0.1)– Internal radius = r

Point of calculation far away from ends (St Venant’s principle)

Slide 2

– Point of calculation far away from ends (St. Venant s principle)

Cylindrical Pressure Vessel

Circumferential (Hoop) Stress: σ1

Sum forces in the vertical direction.

( ) 0)2(2 1 =Δ−Δ xrpxtσ

prtpr

=1σLongitudinal stress: σ2

Sum forces in the horizontal direction:

( ) 0)(2 22 =− rprt ππσ

tpr22 =σ

Slide 3

Cylindrical Pressure Vessel cont’d

• There is also a radial component of stress because the gage pressure must be balanced by a stress perpendicular to the surface.

σ p– σr = p– However σr << σ1 and σ2 , so we assume that σr = 0 and consider this a

case of plane stress.• Mohr’s circle for a cylindrical pressure vessel:• Mohr s circle for a cylindrical pressure vessel:

– Maximum shear stress (in-plane)

pr2 ==στ

– Maximum shear stress (out-of-plane)

t42max ==τ

tpr22max == στ

Slide 4

Spherical Pressure Vessel

Sum forces in the horizontal direction:

( ) 0)(2 22 =− rprt ππσ

In-plane Mohr’s circle is just a point

tpr221 == σσ

tpr42

1max ==

στ

Slide 5

Example Problem 1

When filled to capacity, the unpressurized storage tank shown containswater to a height of 48 ft above its base. Knowing that the lower portionof the tank has a wall thickness of 0.625 in, determine the maximumnormal stress and the maximum shearing stress in the tank. (Specificweight of water = 62.4 lb/ft3)g )

Slide 6

Slide 7

Stress Concentration

• The stresses near the points of application of concentrated loads can reach values much larger than the average value of the stress in the member.

• Stress concentration factor K = σ /σ• Stress concentration factor, K = σmax/σave

Slide 8 --- Uniaxial Loading ---

Slide 9 --- Uniaxial Loading ---

Example

• Determine the largest axial load P that can safely supported by a flat steel bar consisting of two portions, both 10 mm thick and, respectively, 40 and 60 mm wide connected by fillets of radius r = 8 mm assume an allowable mm wide, connected by fillets of radius r = 8 mm. assume an allowable normal stress of 165 MPa. (Example 2.12 in Beer’s book)

Slide 10 --- Uniaxial Loading ---

Poisson’s Ratio

• When an axial force is applied to a bar, the bar not only elongates but also shortens in the other two orthogonal directions.

• Poisson’s ratio (ν) is the ratio

lateral strain

Poisson s ratio (ν) is the ratio of lateral strain to axial strain.

zystrainlateral εευ −=−=−=

axial strainxxstrainaxial εε

υ −=−=−=

Minus sign needed to obtain a positive value – all engineering materials have opposite signs for axial and lateral strains

ν is a material specific property and is dimensionless.

pp g

Slide 11

Generalized Hooke’s Law

• Let’s generalize Hooke’s Law (σ=Eε).– Assumptions: linear elastic material, small deformationsp

EEEEEEzyx

yzyx

xνσσνσενσνσσε −+−=−−=

EEEzyx

zσνσνσε +−−=

• So, for the case of a homogenous isotropic bar that is axially loaded along the x-axis (σy=0 and σz=0), we get

νσσEE

xzy

xx

νσεεσε −=== Even though the stress in the y and z axes are zero, the strain is not!

Slide 12

Poisson’s Ratio cont’d

• What are the limits on ν? We know that ν > 0.– Consider a cube with side lengths = 1 PConsider a cube with side lengths 1

– Apply hydrostatic pressure to the cube P PPzyx −=== σσσ– Can write an expression for the change

in volume of the cubeP

P

P

( )( )( )( )( )( )zxyxzyx

zyxV

εεεεεεε

εεε

++++++−=

−+++=Δ

11

1111

00

00

zyxzy εεεεε +εx, εy, εz are very small, so

we can neglect the terms of order ε2 or ε3

Slide 13

order ε or ε

Poisson’s Ratio cont’d

• ∆V simplifies to• Plug σ=P/A=P into our generalized equations for strain.

zyxV εεε ++≅Δg g q

( )12 −=++−=−−= ννννσνσσε

PPPPEP

EP

EP

EP

EEEzyx

x

( )

( )12

12

=+=+=

−=+−=−+−=

νννσνσνσε

ννννσσνσε

PPPPEP

EP

EP

EP

EEEzyx

zyxy

• Plug these values into the expression for ∆V.

( )12 −=−+=+−−= νεEEEEEEEz

( )123−≅Δ ν

EPV

Slide 14

Poisson’s Ratio cont’d

• Since the cube is compressed, we know ΔV must be less than zero.

P( ) 0123P

P P( )

012

0123

<−

<−

ν

νEP

PP

P1210 <<ν

Slide 15

Shear Strain

• Recall that – Normal stresses produce a change in volume of the elementp g– Shear stresses produce a change in shape of the element

• Shear strain (γ) is an angle• Shear strain (γ) is an anglemeasured in degrees or radians(dimensionless)

• Sign convention is the same asfor shear stress (τ)for shear stress (τ)

Slide 16

Shear Strain cont’d

• There are two equivalent ways to visualize shear strain.

• Hooke’s Law for shear stress is defined as

yzyzxzxzxyxy GGG γτγτγτ ===– G = shear modulus (or modulus of rigidity)– G is a material specific property with the same units as E (psi or Pa).

Slide 17

Relation Among E, ν, and G

• An axially loaded slender bar will elongate in the axial direction and contract in the transverse directionscontract in the transverse directions.

• An initially cubic element oriented as in top figure will deform into a rectangular parallelepiped The axial load produces a

• If the cubic element is oriented as in the b tt fi it ill d f i t

parallelepiped. The axial load produces a normal strain.

C f l d h i

bottom figure, it will deform into a rhombus. Axial load also results in a shear strain.

( )ν+= 1E

• Components of normal and shear strain are related,

Slide 18

( )ν+12G

Example Problem 1

• A vibration isolation unit consists of two blocks of hard rubber bonded to a plate AB and to rigid supports as shown. Knowing that a force of magnitude P = 6 kips causes a deflection of δ=0 0625 in offorce of magnitude P = 6 kips causes a deflection of δ=0.0625 in. of plate AB, determine the modulus of rigidity of the rubber used.

Slide 19

Example Problem 1 Solution

Slide 20

Example Problem

A circle of diameter d = 9 in. is scribed on an unstressed aluminum plate of thickness t = 3/4 in. Forces acting in the plane of the plate later cause normal stresses σx = 12 ksi and σz = 20 ksi.

For E = 10x106 psi and ν = 1/3, determine the change in:

a) the length of diameter ABa) the length of diameter AB,

b) the length of diameter CD,

c) the thickness of the plate andc) the thickness of the plate, and

d) the volume of the plate.

(sample problem 2.5 in Beer’s book)

Slide 21

( p p )

Example Problem 2 Solution

• Apply the generalized Hooke’s Law to find the three components of normal strain.

• Evaluate the deformation components.

( )( )in.9in./in.10533.0 3−×+== dxAB εδ

3

( ) ( )ksi2010ksi121⎥⎤

⎢⎡ −−=

−−+=EEE

zyxx

νσνσσε( )( )in.9in./in.10600.1 3−×+== dzDC εδ

in.108.4 3−×+=ABδ

3( ) ( )

in./in.10533.0

ksi203

0ksi12psi1010

3

6

−×+=

⎥⎦⎢⎣×=

σ

( )( )in.75.0in./in.10067.1 3−×−== tyt εδ

in.104.14 3−×+=DCδ

in./in.10067.1 3−×−=

−+−=EEE

zyxy

νσσνσε in.10800.0 3−×−=tδ

• Find the change in volume

in./in.10600.1 3−×+=

+−−=EEEzyx

zσνσνσε

• Find the change in volume

( ) 33

333

in75.0151510067.1

/inin10067.1

×××==Δ

×=++=

−

−

eVV

e zyx εεε

Slide 22

in./in.10600.1 ×+ ( )3in187.0+=ΔV