LESSON NOTE ON STRESSES IN PRESTRESSED CYLINDER

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/(t_aE_a) + l/t_sE_s)]}, 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 s_a1= pD/(2t_a), where the symbols are the same as in the

previous calculation procedure. Thus, s_a1= 741(9)/[2(0.5)] = 6670-lb/in² (45,989.7-kPa)

compression. Likewise, s_s1 = 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= E_s/Ea or, for this cylinder, s_s²ls_a²= (30 x 10⁶)/(10 x 10⁶) = 3.

Next, compute s_a² from t_a²t_sS_s²= pD/2, or sa²= 800(9)/[2(0.5 + 0.25 x 3)] = 2880-

lb/in2 (19,857.6-kPa) tension. Also, s_s² = 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.