- Scale the diagram
Scale OA and OB to obtain/max = 10,020 lb/in^2 (69,087.9 kPa);/min = 380 lb/in^2 (2620.1
kPa). Both stresses are tension. - Determine the stress angle
Scale angle BCG and measure it as 48°22'. The angle between the x axis, on which the
maximum stress exists, and the side AD of the prism is one-half of BCG. - Construct the x and y axes
In Fig. 2Ia, draw the x axis, making a counterclockwise angle of 240 11' with AD. Draw
the y axis perpendicular thereto. - Verify the locations of the principal planes
Consider ADJ as a free body. Set the length AD equal to unity. In Fig. 2 Ic, since there is
no shearing stress on AJ, %FH = T cos 6 - 8400 - 3600 tan O = O; T cos B = 8400 +
3600(0.45) = 10,020 lb/in^2 (69,087.9 kPa). The stress on AJ = TIAJ = T cos O = 10,020
lb/in^2 (69,087.9 kPa).
HOOP STRESS IN THIN-WALLED CYLINDER
UNDER PRESSURE
A steel pipe 5 ft (1.5 m) in diameter and^3 /s in (9.53 mm) thick sustains a fluid pressure of
180 lb/in^2 (1241.1 kPa). Determine the hoop stress, the longitudinal stress, and the in-
crease in diameter of this pipe. Use 0.25 for Poisson's ratio.
Calculation Procedure:- Compute the hoop stress
Use the relation s = pD/(2i), where s = hoop or tangential stress, lb/in
2
(kPa); p = radial
pressure, lb/in^2 (kPa); D = internal diameter of cylinder, in (mm); t = cylinder wall thick-
ness, in (mm). Thus, for this cylinder, s = 180(60)/[2(^3 / 8 )] = 14,400 lb/in^2 (99,288.0 kPa). - Compute the longitudinal stress
Use the relation s' =pD/(4t), where s' = longitudinal stress, i.e., the stress parallel to the
longitudinal axis of the cylinder, lb/in^2 (kPa), with other symbols as before. Substituting
yields s' = 7200 lb/in^2 (49,644.0 kPa). - Compute the increase in the cylinder diameter
Use the relation AD = (D/E)(s - vs'\ where v = Poisson's ratio. Thus AD = 60(14,400 -
0.25 x 7200)/(30 x 106 ) = 0.0252 in (0.6401 mm).
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^1 A in (12.7 mm). After the steel cools,
the cylinder is subjected to an internal pressure of 800 lb/in^2 (5516 kPa). Find the stresses
in the two materials. For aluminum, E = 10 x 106 lb/in^2 (6.895 x 107 kPa).