Page 51 - Handbook of Civil Engineering Calculations, Second Edition

1.34 STRUCTURAL STEEL ENGINEERING AND DESIGN

5. Scale the diagram
Scale OA and OB to obtain f max 10,020 lb/sq.in. (69,087.9 kPa); f min 380 lb/sq.in.
(2620.1 kPa). Both stresses are tension.
6. 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.
7. Construct the x and y axes
In Fig. 21a, draw the x axis, making a counterclockwise angle of 24°11
with AD. Draw
the y axis perpendicular thereto.
8. Verify the locations of the principal planes
Consider ADJ as a free body. Set the length AD equal to unity. In Fig. 21c, since there is
no shearing stress on AJ, F H T cos 8400 3600 tan 0; T cos 8400
3600(0.45) 10,020 lb/sq.in. (69,087.9 kPa). The stress on AJ T/AJ T cos
10,020 lb/sq.in. (69,087.9 kPa).

HOOP STRESS IN THIN-WALLED CYLINDER
UNDER PRESSURE

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A steel pipe 5 ft (1.5 m) in diameter and /5 in. (9.53 mm) thick sustains a fluid pressure of
180 lb/sq.in. (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:
1. Compute the hoop stress
Use the relation s pD/(2t), where s hoop or tangential stress, lb/sq.in. (kPa); p ra-
dial pressure, lb/sq.in. (kPa); D internal diameter of cylinder, in (mm); t cylinder
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wall thickness, in. (mm). Thus, for this cylinder, s 180(60)/[2( /8)] 14,400 lb/sq.in.
(99,288.0 kPa).
2. 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/sq.in. (kPa), with other symbols as before. Substitut-
ing yields s
7200 lb/sq.in. (49,644.0 kPa).
3. Compute the increase in the cylinder diameter
Use the relation D (D/E)(s vs
), where v Poisson’s ratio. Thus D 60(14,400
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0.25 7200)/(30 10 ) 0.0252 in. (0.6401 mm).

STRESSES IN PRESTRESSED CYLINDER

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A steel ring having an internal diameter of 8.99 in. (228.346 mm) and a thickness of /4 in.
(6.35 mm) is heated and allowed to shrink over an aluminum cylinder having an external
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diameter of 9.00 in. (228.6 mm) and a thickness of /2 in. (12.7 mm). After the steel cools,
the cylinder is subjected to an internal pressure of 800 lb/sq.in. (5516 kPa). Find the
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stresses in the two materials. For aluminum, E 10 10 lb/sq.in. (6.895 10 kPa).