STRESSES IN PRESTRESSED CYLINDER
A steel 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
diameter 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
in the two materials. For aluminum, E = 10 x 106 lb/in2 (6.895 x 107 kPa).
Calculation Procedure:
1. Compute the radial pressure caused by prestressing
Use the 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
procedure 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. Compute the corresponding prestresses
Using 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
previous 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.
3. 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.
Next, compute sa2 from ta2tsSs2= 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.
4. Compute the final stresses
Sum the 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)
tension.
5. Check the accuracy of the results
Ascertain 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
in (-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
mm). Since the computed diameters are equal, the results are valid.