08-01Chap Gere

# 08-01Chap Gere - 8 Applications of Plane Stress (Pressure...

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Spherical Pressure Vessels When solving the problems for Section 8.2, assume that the given radius or diameter is an inside dimension and that all internal pressures are gage pressures. Problem 8.2-1 A large spherical tank (see figure) contains gas at a pressure of 400 psi. The tank is 45 ft in diameter and is constructed of high-strength steel having a yield stress in tension of 80 ksi. Determine the required thickness (to the nearest 1/4 inch) of the wall of the tank if a factor of safety of 3.5 with respect to yielding is required. Solution 8.2-1 Spherical tank 8 Applications of Plane Stress (Pressure Vessels, Beams, and Combined Loadings) C ROSS SECTION Radius: Internal pressure: p 5 400 psi r 5 1 2 (45 ft) 5 270 in. Yield stress: s y 5 80 ksi (steel) Factor of safety: n 5 3.5 M INIMUM WALL THICKNESS t min From Eq. (8-1): or Use the next higher 1 / 4 inch: t min 5 2.50 in. t 5 prn 2 s y 5 2.36 in. s y n 5 pr 2 t s max 5 pr 2 t Problem 8.2-2 Solve the preceding problem if the internal pressure is 3.5 MPa, the diameter is 18 m, the yield stress is 550 MPa, and the factor of safety is 3.0. Determine the required thickness to the nearest millimeter. Solution 8.2-2 Spherical tank C ROSS SECTION Radius: Internal pressure: p 5 3.5 MPa r 5 1 2 (18 m) 5 9 m Yield stress: s y 5 550 MPa Factor of safety: n 5 3.0 M INIMUM WALL THICKNESS t min From Eq. (8-1): or Use the next higher millimeter: t min 5 86 mm t 5 prn 2 s y 5 85.9 mm s x n 5 pr 2 t s max 5 pr 2 t p p

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Problem 8.2-3 A hemispherical window (or viewport ) in a decompression chamber (see figure) is subjected to an internal air pressure of 80 psi. The port is attached to the wall of the chamber by 18 bolts. Find the tensile force F in each bolt and the tensile stress s in the viewport if the radius of the hemisphere is 7.0 in. and its thickness is 1.0 in. Solution 8.2-3 Hemispherical viewport 498 CHAPTER 8 Applications of Plane Stress F REE - BODY DIAGRAM Radius: r 5 7.0 in. Internal pressure: p 5 80 psi Wall thickness: t 5 1.0 in. 18 bolts T 5 total tensile force in 18 bolts HORIZ 5 T 2 pA 5 0 T 5 pA 5 p ( p r 2 ) F 5 force in one bolt T ENSILE STRESS IN VIEWPORT (E Q . 8-1) s 5 pr 2 t 5 280 psi F 5 T 18 5 1 18 ( p pr 2 ) 5 684 lb a p Problem 8.2-4 A rubber ball (see figure) is inflated to a pressure of 60 kPa. At that pressure the diameter of the ball is 230 mm and the wall thickness is 1.2 mm. The rubber has modulus of elasticity E 5 3.5 MPa and Poisson’s ratio n 5 0.45. Determine the maximum stress and strain in the ball. Solution 8.2-4 Rubber ball C ROSS - SECTION Radius: r 5 (230 mm) / 2 5 115 mm Internal pressure: p 5 60 kPa Wall thickness: t 5 1.2 mm Modulus of elasticity: E 5 3.5 MPa (rubber) Poisson’s ratio: n5 0.45 (rubber) M AXIMUM STRESS (E Q . 8-1) 5 2.88 MPa M AXIMUM STRAIN (E Q . 8-4) 5 0.452 e max 5 pr 2 tE (1 2 n ) 5 (60 kPa)(115 mm) 2(1.2 mm)(3.5 MPa) (0.55) s max 5 pr 2 t 5 (60 kPa)(115 mm) 2(1.2 mm) p
Problem 8.2-5 Solve the preceding problem if the pressure is 9.0 psi, the diameter is 9.0 in., the wall thickness is 0.05 in., the modulus of elasticity is 500 psi, and Poisson’s ratio is 0.45.

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## This note was uploaded on 03/19/2008 for the course E M 316 taught by Professor Korkolis during the Spring '08 term at University of Texas.

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08-01Chap Gere - 8 Applications of Plane Stress (Pressure...

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