Post on 12-Dec-2015
description
Lecture 9Lecture 9
Stresses in thick walled cylinders:Cylindrical pressure vessels, hydraulic cylinders, gun barrels, and pipes carrying fluids at high pressures develop both radial and tangential stresses with values that depend upon the radius of the element under consideration. In determining the radial stress σ
rand the tangential
stress σt.
Consider the element shown in Figure in a pressure cylinder loaded with internal and external pressure Pi &Powith internal and external pressure Pi &Po
The element force balance will be a) ΣFr = 0 in the radial direction for unit length
)1(0.
0))(.(2...2..
LL====−−−−−−−−
====++++++++−−−−++++
dr
dr
ddrrdrdr
dddrdr
rrt
rrtr
σσσσσσσσσσσσ
φφφφσσσσ
σσσσφφφφ
σσσσφφφφσσσσ
b) Strain
).(1
)3(..2
..2).(.2
)2(
2 trr
t
tt
r
u
E
r
u
r
rur
L
dr
du
εεεεµµµµεεεεσσσσ
ππππππππππππ
εεεε
εεεε
++++−−−−
====
====−−−−++++
====∆∆∆∆
====
====
LL
LL
3,2
).(1
)4()..(1
)4().(1
3,2
2
22
2
2
2
LL
LL
++++−−−−
====
++++−−−−−−−−
====
++++−−−−
====
from
u
E
dr
du
r
u
r
u
dr
ud
u
E
dr
d
r
u
dr
du
u
E
from
rtt
r
r
εεεεµµµµεεεεσσσσ
µµµµσσσσ
µµµµσσσσ
)6(0.1
15,4
)5().(1
3,2
22
2
2
LL
LL
====−−−−++++
++++−−−−
====
r
u
dr
du
rdr
ud
infrom
dr
du
r
u
u
E
from
t µµµµσσσσ
Equation “6” is a second degree differential equation has a general solution:
)(1
5,4
)7(
.
22
122
12
22
1
21
LL
r
CC
r
CC
E
from
r
CC
dr
du
r
CrCu
r
−−
++++++++−−−−
−−−−====
−−−−====
++++====
µµ
µµµµµµµµ
σσσσ
)9()1(
)1(1
)(1
)8()1(
)1(1
22212
22
122
12
22212
LL
LL
r
BA
rCC
E
r
CC
r
CC
E
r
BA
rCC
E
t
t
r
++++====
−−−−++++++++
−−−−====
−−−−++++++++
−−−−====
−−−−====
−−−−−−−−++++
−−−−====
µµµµµµµµ
µµµµσσσσ
µµµµµµµµ
σσσσ
µµµµµµµµ
µµµµσσσσ
A & B from boundary conditions:At r = a (inner radius) …… σr = -Pi & at r = b (outer radius) …… σr = -Po
- Pi = A – B/a2 ……… - Po = A – B/b2
Solving for A & B:
22
2
2222
22
22
22
22
).(.
..
)(..
..
ab
PPr
abbPaP
ab
PPabB
ab
bPaPA
iooi
r
iooi
−−−−
−−−−−−−−−−−−====
−−−−
−−−−====
−−−−
−−−−====
σσσσ
LL
22
2
2222
22
2
2222
22
).(.
..
).(.
..
ab
PPr
abbPaP
or
ab
PPr
abbPaP
ab
iooi
rort
iooi
t
r
−−−−
−−−−±±±±−−−−====
−−−−
−−−−++++−−−−====
−−−−
σσσσ
σσσσ
Case of internal pressure only:Po = 0
aratabP
r
b
ab
aP
r
b
ab
aP
it
it
ir
====++++
====
++++−−−−
====
−−−−−−−−
====
LL).(
)1.()(
.
)1.()(
.
22
max
2
2
22
2
2
2
22
2
σσσσ
σσσσ
σσσσ
aratab
bP
aratab
irt
it
====−−−−
====−−−−
====
====−−−−
====
LL
LL
).(2
)(
)(
22
2
max
22max
σσσσσσσσττττ
σσσσ
In thin wall cylinder b – a = t (thickness)b2 – a2 = (a + b).(a – b) ≅ 2 a . t & b2 + a2 ≅ 2 a2
σt = Pi . a/t
Case of external pressure only:Pi = 0
aratab
bP
r
a
ab
bP
r
a
ab
bP
ot
ot
or
====−−−−
−−−−====
++++−−−−
−−−−====
−−−−−−−−
−−−−====
LL22
2
max
2
2
22
2
2
2
22
2
.2.
)1.(.
)1.(.
σσσσ
σσσσ
σσσσ
aratab
t ====−−−−
−−−−==== LL22maxσσσσ
Case of Solid shaft:a = 0σr = σt = - Po
DeformationIf σσσσr and σσσσt are known:
.
.
r
rtt
trr
radiusinincreaseu
EE
EE
σσσσµµµµ
σσσσεεεε
σσσσµµµµ
σσσσεεεε
====∆∆∆∆====
−−−−====
−−−−====
)..(.
..2
..2).(.2
rtt
t
r
E
rru
r
u
r
rur
radiusinincreaseu
σσσσµµµµσσσσεεεε
ππππππππππππ
εεεε
−−−−========
====−−−−++++
====
====∆∆∆∆====
Stresses produced by Shrink-Fits
Using Superposition method for both cylinders σσσσt = σσσσt (created from shrink fit) + σσσσt (created from internal pressure)
ba
rt
cb
rt
oi
E
b
E
b
bacylUcbcylU
−−−−−−−−
−−−−++++
−−−−====
−−−−++++−−−−====
)..()..(
)2()1(
22
11
σσσσµµµµσσσσσσσσµµµµσσσσδδδδ
δδδδ LL
222
22
2122
22
1
222
22
222
22
2122
22
1
....
....
ab
ab
E
bP
bc
bc
E
bP
ab
ab
ab
ab
E
bP
bc
bc
E
bP
shsh
shsh
−−−−
−−−−
++++++++
++++
−−−−
++++====
⟩⟩⟩⟩⟩⟩⟩⟩−−−−
++++
++++
−−−−
++++−−−−++++
++++
−−−−
++++====
µµµµµµµµδδδδ
µµµµ
µµµµµµµµδδδδ
2
22
222
2222
21
.2
)(.
.
0
).(.2
)).((.
2&1.&
c
bc
b
EP
ashaftsolidfor
acb
bcab
b
EP
cylforsametheareEif
abEbcE
sh
sh
−−−−====
====
−−−−
−−−−−−−−====
−−−− −−−−
δδδδ
δδδδ
µµµµ
Maximum torque that can be transmitted by shrink-fit connection
T = f . Psh .π . d2 . L /2
f = coefficient of frictionL = length of contact aread = shaft diameter
Force required to assemble the two members
F = f . Psh .π . d . L
Design considerations: If shaft radius has maximum tolerance + s1 and minimum + s2 , and the maximum disc tolerance +d1 and minimum +d2
Where s1 > s2 > d1 > d2
maximum interference δmax = maximum shaft radius – minimum disc radiusδmax = (b + s1) – (b + d2) = s1 – d2
where b = nominal radius of the assembly Minimum interference δmin = (b + s2) – (b + d1) = s2 – d1
The joint must transmit torque in case of minimum interference, and to have safe stress in case of maximum interference.
22min
min)(
.. bcE
Psh−−−−
====δδδδ
T = f . Psh min .π . 2 . b2 . L
2min
min.2
.cb
Psh ====
b
E
bc
cP
c
bc
b
EP
sh
sh
.2
.
).(
.2
)(.
.
maxmax
22
2
maxmax
2
22max
max
δδδδττττ
ττττ
δδδδ
====
−−−−====
−−−−====