STRESSES
IN PRESTRESSED CYLINDER
IN PRESTRESSED CYLINDER
A steel
ring having an internal diameter of 8.99 in (228.346 mm) and a thickness of %
in
ring having an internal diameter of 8.99 in (228.346 mm) and a thickness of %
in
(6.35
mm) is heated and allowed to shrink over an aluminum cylinder having an
external
mm) is heated and allowed to shrink over an aluminum cylinder having an
external
diameter
of 9.00 in (228.6 mm) and a thickness of 1A
in (12.7 mm). After the steel cools,
of 9.00 in (228.6 mm) and a thickness of 1A
in (12.7 mm). After the steel cools,
the
cylinder is subjected to an internal pressure of 800 lb/in2
(5516
kPa). Find the stresses
cylinder is subjected to an internal pressure of 800 lb/in2
(5516
kPa). Find the stresses
in the two materials.
For aluminum, E = 10 x 106 lb/in2 (6.895 x 107
kPa).
For aluminum, E = 10 x 106 lb/in2 (6.895 x 107
kPa).
Calculation
Procedure:
Procedure:
1.
Compute the radial pressure caused by prestressing
Compute the radial pressure caused by prestressing
Use the
relation;? = 2ϕD/{LD2[l/(taEa)
+ l/tsEs)]},
where/? = radial pressure resulting
relation;? = 2ϕD/{LD2[l/(taEa)
+ l/tsEs)]},
where/? = radial pressure resulting
from
prestressing, lb/in2
(kPa),
with other symbols the same as in the previous calculation
prestressing, lb/in2
(kPa),
with other symbols the same as in the previous calculation
procedure
and the subscripts a and s referring to aluminum and steel,
respectively. Thus,
and the subscripts a and s referring to aluminum and steel,
respectively. Thus,
p
= 2(0.01)/{92[1/(0.5
x l 0 x 106) +
1/(0.25 >( 30 x 106)]) =
741 lb/in2 (5109.2 kPa).
= 2(0.01)/{92[1/(0.5
x l 0 x 106) +
1/(0.25 >( 30 x 106)]) =
741 lb/in2 (5109.2 kPa).
2.
Compute the corresponding prestresses
Compute the corresponding prestresses
Using
the subscripts 1 and 2 to denote the stresses caused by prestressing and
internal
the subscripts 1 and 2 to denote the stresses caused by prestressing and
internal
pressure,
respectively, we find sa1
= pD/(2ta),
where the symbols are the same as in the
respectively, we find sa1
= pD/(2ta),
where the symbols are the same as in the
previous
calculation procedure. Thus, sa1
= 741(9)/[2(0.5)]
= 6670-lb/in2 (45,989.7-kPa)
calculation procedure. Thus, sa1
= 741(9)/[2(0.5)]
= 6670-lb/in2 (45,989.7-kPa)
compression.
Likewise, ss1 =
741(9)/[2(0.25)] = 13,340-lb/in2
(91,979-kPa)
tension.
Likewise, ss1 =
741(9)/[2(0.25)] = 13,340-lb/in2
(91,979-kPa)
tension.
3.
Compute the stresses caused by internal pressure
Compute the stresses caused by internal pressure
Use the
relation s s2lsa2
= Es/Ea
or, for this cylinder, ss2lsa2
= (30 x 106)/(10
x 106) = 3.
relation s s2lsa2
= Es/Ea
or, for this cylinder, ss2lsa2
= (30 x 106)/(10
x 106) = 3.
Next,
compute sa2
from ta2
tsSs2
= pD/2, or sa2
= 800(9)/[2(0.5
+ 0.25 x 3)] = 2880-
compute sa2
from ta2
tsSs2
= pD/2, or sa2
= 800(9)/[2(0.5
+ 0.25 x 3)] = 2880-
lb/in2
(19,857.6-kPa)
tension. Also, ss2
= 3(2880)
– 8640-lb/in2 (59,572.8-kPa)
tension.
(19,857.6-kPa)
tension. Also, ss2
= 3(2880)
– 8640-lb/in2 (59,572.8-kPa)
tension.
4.
Compute the final stresses
Compute the final stresses
Sum the
results in steps 2 and 3 to obtain the final stresses: sa3
= 6670 – 2880 = 3790-
results in steps 2 and 3 to obtain the final stresses: sa3
= 6670 – 2880 = 3790-
lb/in2
(26,132.1-kPa)
compression; ss3 =
13,340 + 8640 = 21,980-lb/in2
(151,552.1-kPa)
(26,132.1-kPa)
compression; ss3 =
13,340 + 8640 = 21,980-lb/in2
(151,552.1-kPa)
tension.
5.
Check the accuracy of the results
Check the accuracy of the results
Ascertain
whether the final diameters of the steel ring and aluminum cylinder are equal.
whether the final diameters of the steel ring and aluminum cylinder are equal.
Thus,
setting s’ = 0 in ϕ/E> = (D/E)(s – vs’) we find ϕDa
=
-3790(9)7(10 x 106)
= -0.0034
setting s’ = 0 in ϕ/E> = (D/E)(s – vs’) we find ϕDa
=
-3790(9)7(10 x 106)
= -0.0034
in
(-0.0864 mm), D0 =
9.0000 – 0.0034 = 8.9966 in (228.51 mm). Likewise, ϕDs
=
(-0.0864 mm), D0 =
9.0000 – 0.0034 = 8.9966 in (228.51 mm). Likewise, ϕDs
=
21,980(9)7(30
x io6) = 0.0066 in (0.1676 mm), Ds
= 8.99 + 0.0066 = 8.9966 in (228.51
x io6) = 0.0066 in (0.1676 mm), Ds
= 8.99 + 0.0066 = 8.9966 in (228.51
mm).
Since the computed diameters are equal, the results are valid.
Since the computed diameters are equal, the results are valid.
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