INFORMATIO

# Walls are or hoop stress xial stress or hoop strain

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walls are: or hoop stress xial stress or hoop strain ain try to serve as boilers ade is subjected to a aving an inner radius and inner radius r as ontaining gas or fluid,

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5 e. Change in length The change in length of the cylinder may be determined from the longitudinal strain. Change in length=longitudinal strain×original length L=ε 2 L= ) ( 1 1 2 E L L= ) 2 1 ( 2 tE pr L f. Change in diameter The change in diameter may be found from the circumferential change. Change in diameter=diametral strain×original diameter Diametral strain=circumferential strain d= ε 1 d= ) ( 1 2 1 E d d= ) 2 ( 2 tE pr d g. Change in internal volume Volumetric strain=longitudinal strain+2diametral strain ε v = ε 2 +2 ε 1 = ) ( 1 1 2 E +2 ) ( 1 2 1 E diametral strain ε v = ) 2 2 ( 1 2 1 1 2 E longitudinal strain = ) 2 2 ( 1 t pr t pr t pr t pr E diametral strain ε v = ) 4 5 ( 2 tE pr change in internal volume=volumetric strain×original volume v= ε v v v= ) 4 5 ( 2 tE pr v
2. Spherical Vessels: Because of the symme pressure will be two mutua value and a radial stress. 1 (2 rt)-p( r 2 )=0 t pr 2 1 2= t pr 2 1 Change in internal volume change in internal volume=v volumetric strain=3hoop stra ε v = ε 1 =3 ) ( 1 2 1 E v= ε v v v = ) 1 ( 2 3 tE pr v 6 metry of the sphere the stresses set u ally perpendicular hoop or circumfere volumetric strain×original volume ain = ) 1 ( 3 1 E = ) 1 ( 2 3 tE pr up owing to internal ential stress of equal

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Cylindrical Vessels with H r=d/2 a) For the cylindrical portion c t Pr 1 hoop st c t 2 Pr 2 longitud ) ( 1 2 1 1 E = ε 1 = ) 2 ( 2 E t pr c b) For the spherical ends s t 2 Pr 1 ) ( 1 2 1 1 E = ε 1 = ) 1 ( 2 E t pr s Thus equating the two junction. ) 1 ( 2 E t pr s = 2 ( 2 E t pr c 2 1 c s t t 7 Hemispherical Ends: n tress dinal stress ) 2 Pr Pr ( 1 c c t t E hoop strain hoop stress ) 1 ( 1 E hoop strain o strains in order that there shall be ) no distortion of the
8 Example 31: A thin cylinder 75 mm internal diameter, 250 mm long with walls 2.5 mm thick is subjected to an internal pressure of 7 MN/m 2 . Determine the change in internal diameter and the change in length. If in addition to the internal pressure, the cylinder is subjected to a torque of 200 N.m find the magnitude and nature of the stresses set up in the cylinder. E=200 GN/m 2 , υ=0.3 . d= ) 2 ( 2 tE pr d d= 3 9 3 3 6 10 75 ] 3 . 0 2 [ 10 200 10 5 . 2 2 10 2 75 10 7 d=33.468×10 -6 m=33.468 μm L= ) 2 1 ( 2 tE pr L L= 3 9 3 3 6 10 250 ] 3 . 0 2 1 [ 10 200 10 5 . 2 2 10 2 75 10 7 L=26.25×10 -6 m=26.25 μm t pr 1 = 3 3 6 10 5 . 2 10 2 75 10 7 1 =105×10 6 N/m 2 =105 MN/m 2 t pr 2 2 = 3 3 6 10 5 . 2 2 10 2 75 10 7 2 =52.5×10 6 N/m 2 =52.5 MN/m 2 J Tr = ] [ 2 4 4 i o r r Tr = ] ) 10 5 . 37 ( ) 10 40 [( 2 10 40 200 4 3 4 3 3 =8.743862 MN/m 2

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9 Example 32: A cylinder has an internal diameter of 230 mm , has walls 5 mm thick and is 1 m long. It is found to change in internal volume by
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• Winter '15
• MAhmoudali

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