Upload
trinhdiep
View
257
Download
5
Embed Size (px)
Citation preview
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