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MECHANICS OF MATERIALS - II PowerPoint Presentation
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MECHANICS OF MATERIALS - II

MECHANICS OF MATERIALS - II

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MECHANICS OF MATERIALS - II

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  1. MECHANICS OF MATERIALS - II

  2. SPHERICAL AND CYLINDRICAL PRESSURE VESSELS (BIAXIAL STRESS)

  3. PRESSURE VESSELS ARE CLOSED STRUCTURES THAT MAY CONTAIN LIQUIDS OR GASES UNDER PRESSURE. EXAMPLES ARE WATER-STORAGE TANKS, CYLINDRICAL TANKS OR COMPRESSED AIR, PRESSURIZED PIPES ETC. THE CURVED WALLS OF SUCH STRUCTURES ARE VERY THIN WITH RESPECT TO RADIUS AND LENGTH AND AS SUCH THESE ARE KNOWN AS SHELLS. THIS MAY FURTHER INCLUDE THIN CURVED ROOFS, DOMES, FUSELAGES. AS A GENERAL RULE A RATIO OF 10 OR GREATER BETWEEN RADIUS (r) AND THINCKNESS (t) WOULD INCLUDE ALL STRUCTURES UNDER SUBJECT CATEGORY OF PRESSURE VESSELS OR SHELLS. SPHERICAL AND CYLINDRICAL PRESSURE VESSELS

  4. A TANK OF SPHERICAL SHAPE IS THE IDEAL CONTAINER FOR RESISTING THE INTERNAL PRESSURE. IN ORDER TO OBTAIN THE STRESSES AT THE WALLS, LET US CUT THROUGH THE SPHERE ON A VERTICAL DIAMETRAL PLANE. NOW ISOLATE HALF OF THE SHELL AND ITS CONTENTS AS A FREE BODY. ACTING ON THIS FREE BODY ARE THE STRESSES “σ” IN THE WALL AND THE INTERNAL PRESSURE “p”. THE PRESSURE ACTS HORIZONTALLY AGAINST THE PLANE CIRCULAR AREA FORMED BY THE CUT, AND THE RESULTANT FORCE SHOULD BE EQUAL TO p(Πr²). NOTE THAT THE PRESSURE IS THE NET INTERNAL PRESSURE OR GAGE PRESSURE. THE WEIGHT OF THE TANK AND ITS CONTENTS IS DISREGARDED IN THE ANALYSIS.

  5. THE TENSILE STRESS “σ” IN THE WALL OF THE SPHERE IS UNIFORM AROUND THE CIRCUMFERENCE OF THE TANK. ALSO BECAUSE OF VERY SMALL THICKNESS IT CAN ALSO ASSUMED THAT STRESS IS UNIFORM ACROSS THE THICKNESS. NOW THE RESULTANT FORCE OBTAINED FROM TENSILE STRESS IS EQUAL TO σ(2Πrmt). HERE “t” IS THE THICKNESS AND “rm” IS THE MEAN RADIUS. AS THICKNESS VERY SMALL HENCE rm = r. NOW THE EQUILIBRIUM OF FORCES IN THE HORIZONTAL DIRECTION YIELDS THE FOLLOWING: σ(2Πrt) - p(Πr²) = 0, AND σ = pr/2t

  6. BECAUSE OF THE SYMMETRY OF A SPHERICAL SHELL, THE SAME EQUATION WOULD BE OBTAINED IF SPHERE IS CUT THROUGH THE CENTRE OF THE SPHERE IN ANY DIRECTION. IT CAN BE CONCLUDED FROM THIS THAT A PRESSURIZED SPHERE IS SUBJECTED TO UNIFORM TENSILE STRESSES IN ALL DIRECTIONS. SUCH TYPE OF STRESSES WHICH ARE TANGENTIAL TO THE CURVED SURFACE RATHER THAN PERPENDICULAR ARE KNOWN AS MEMBRANCE STRESSES. IT CAN ALSO BE OBSERVED THAT AT THE OUTER SURFACE OF A SPHERICAL PRESSURE VESSEL, NO STRESSES ARE ACTING NORMAL TO THE SURFACE.

  7. THIS IS, THEREFORE, A SPECIAL CASE OF BIAXIAL STRESS IN WHICH THE VALUES OF σx AND σy ARE THE SAME. AS NO SHEAR STRESSES ACT ON THE ELEMENT, WE CAN GET THE SAME VALUES IF ROTATED TO AN ANGLE ABOUT THE Z-AXIS. THE PRINCIPAL STRESSES ARE, THEREFORE, GIVEN AS σ1 = σ2 = pr/2t HOWEVER, THE FACT SHOULD NOT BE OVERLOOKED THAT THE ELEMENT IS THREE-DIMENSIONAL AND THE THIRD PRINCIPAL STRESS IS ZERO.

  8. HENCE THE ABSOLUTE MAXIMUM SHEAR STRESS, OBTAINED BY ROTATING THE ELEMENT BY 45º ABOUT EITHER THE X OR Y AXIS, CAN BE OBTAINED AS FOLLOWS: τmax = σ/2 = pr/4t AGAINST THE OUTER SURFACE, AT INNER SURFACE OF THE WALL OF THE SPHERICAL VESSEL, STRESS ELEMENT HAS THE SAME MEMBRANCE STRESSES, BUT IN ADDITION A COMPRESSIVE STRESS EQUAL TO “p” ACTS IN THE Z DIRECTION. IN THIS CASE THE THREE NORMAL STRESSES ARE THE PRINCIPAL STRESSES AND ARE GIVEN AS FOLLOWS: σ1 = σ2 = pr/2t, σ3 = - p

  9. SIMILARLY THE MAXIMUM SHEAR STRESS, OBTAINED IN THE SAME WAY AS OBTAINED EARLIER, IS GIVEN AS τmax = (σ + p)/2 = pr/4t + p/2 IF THE RATIO r/t IS SUFFICIENTLY LARGE, THE LAST TERM OF THE EQUATION CAN BE DISREGARDED. CONSEQUENTLY THE EQUATION BECOMES THE SAME AS PREVIOUS EQUATION. BY THIS IT CAN BE ASSUMED THAT THE MAXIMUM SHEAR STRESS IS CONSTANT ACROSS THE THICKNESS OF THE SHELL. EVERY SPHERICAL TANK USED AS A PRESSURE VESSEL WILL HAVE AT LEAST ONE OPENING IN THE WALL AS WELL AS VARIOUS ATTACHMENTS AND SUPPORTS.

  10. SUCH FEATURES IN TURN RESULT IN NON-UNIFORMITIES IN THE STRESS DISTRIBUTION THAT CAN NOT BE ANALYZED BY ELEMENTARY METHODS. AS HIGH LOCALIZED STRESSES ARE PRODUCED NEAR THE CONTINUITIES IN THE SHELL, HENCE BECAUSE OF THIS REASON THESE DISCONTINUITIES MUST BE REINFORCED. THE EQUATIONS DERIVED FOR THE MEMBRANCE STRESSES ARE VALID EVERY WHERE IN THE WALL OF THE VESSEL EXCEPT CLOSE TO THE DISCONTINUITIES. OTHER POINTS TO BE CONSIDERED FROM DESIGN POINT OF VIEW MAY INCLUDE THE EFFECTS OF CORROSION, ACCIDENTAL IMPACTS, AND TEMPERATURE EFFECTS.

  11. NOW CONSIDER A THIN-WALLED CIRCULAR CYLINDRICAL TANK WITH CLOSED ENDS WHICH HAS INTERNAL PRESSURE “p”. NOW CONSIDER A STRESS ELEMENT WITH FACES PARALLEL AND PERPENDICULAR TO THE AXIS OF THE TANK. SIMILAR TO THE SPHERICAL SHELL, THE NORMAL STRESSES σ1 AND σ2 ACTING ON THE SIDE FACES OF THIS ELEMENT REPRESENT THE MEMBRANCE STRESSES IN THE WALL. BECAUSE OF THEIR DIRECTIONS NORMAL STRESS σ1 IS CALLED CIRCUMFERENTIAL STRESS OR THE HOOP STRESS AND σ2 IS CALLED THE LONGITUDINAL OR AXIAL STRESS. CYLINDRICAL PRESSURE VESSELS

  12. AS BECAUSE OF THE SYMMETRY OF THE VESSEL NO SHEAR STRESSES ACT ON THE FACES OF THE ELEMENT, THESE STRESSES, σ1 AND σ2, ARE THE PRINCIPAL STRESSES. THE STRESSES CAN BE DETERMINED FROM EQUILIBRIUM AND BY HAVING APPROPRIATE FREE-BODY DIAGRAMS. IN ORDER TO CALCULATE σ1, A FREE-BODY DIAGRAM IS OBTAINED BY MAKING TWO CUTS(mn & pq) AT A DISTANCE “b” APART AND PERPENDICULAR TO THE LONGITUDINAL AXIS. TWO TYPES OF HORIZONTAL FORCES ACT ON THE FREE-BODY DIAGRAM. ONE IS BECAUSE OF THE CIRCUMFERENTIAL STRESS AND THE OTHER IS BECAUSE OF THE INTERNAL PRESSURE. THESE TWO FORCES ACT IN THE OPPOSITE DIRECTIONS.

  13. AGAIN DISREGARDING THE WEIGHT OF THE VESSEL AND ITS CONTENTS, FOLLOWING EQUILIBRIUM EQUATION OF FORCES WOULD BE OBTAINED: σ1(2bt) - p(2br) = 0 σ1 = pr/t HERE “t” IS THE THICKNESS OF THE WALL AND “r” IS THE INSIDE RADIUS OF THE CYLINDER. THIS EQUATION GIVES US THE VALUE OF CIRCUMFERENTIAL STRESS WHICH IS UNIFORMLY DISTRIBUTED OVER THE THICKNESS OF THE ENTIRE WALL.

  14. THE LONGITUDINAL STRESS σ2 CAN BE OBTAINED BY THE FOLLOWING EQUATION OF EQUILIBRIUM σ2 (2Πrt) - p(Πr²) = 0 BY SOLVING THIS EQUATION, TAKING “r” AS THE INSIDE RADIUS OF THE SHELL, FOLLOWING RELATIONSHIP FOR σ2 IS OBTAINED: σ2 = pr/2t COMPARING EQUATIONS FOR σ1 AND σ2, IT CAN BE OBSERVED THAT LONGITUDINAL STRESS IN A CYLINDRICAL SHELL IS ONE-HALF THE CIRCUMFERENTIAL STRESS, THAT IS σ2 = σ1/2

  15. THE PRINCIPAL STRESSES ACTING IN THE X AND Y AXES ON THE OUTER SHELL OF THE VESSEL ARE σ1 AND σ2 WHILE THE THIRD PRINCIPAL STRESS IN THE Z DIRECTION IS ZERO. THE MAXIMUM SHEAR STRESS WOULD OCCUR ON THE ELEMENT WHEN IT IS ROTATED BY 45º ABOUT THE Z-AXIS. HENCE τmax = (σ1 - σ2)/2 = σ1/4 = pr/4t THE MAXIMUM SHEAR STRESSES OBTAINED BY ROTATING 45º ABOUT X AND Y AXES ARE τmax = σ1/2 = pr/2t AND τmax = σ2/2 = pr/4t

  16. OUT OF THESE TWO VALUES THE ABSOLUTE VALUE OF MAXIMUM SHEAR STRESS WOULD BE WHEN IT IS ROTATED AT 45˚ TO X AXIS. τmax = σ1/2 = pr/2t AT THE INNER SURFACE OF THE SHELL, THE PRINCIPAL NORMAL STRESSES WOULD BE AS FOLLOWS: σ1 = pr/t, σ2 = pr/2t AND σ3 = -p THE VALUES OF THREE MAXIMUM SHEAR STRESSES ABOUT X, Y AND Z AXES WOULD BE GIVEN AS FOLLOWS: τmax = (σ1+ p)/2 = pr/2t + p/2

  17. τmax = (σ2+ p)/2 = pr/4t + p/2 τmax = (σ1 - σ2)/2 = pr/4t IT IS TO BE NOTED THAT OUT OF THESE THREE VALUES FIRST VALUE IS THE LARGEST. HOWEVER, IN CASE OF SPHERICAL AND CYLINDRICAL SHELLS, THE ADDITIONAL TERM p/2 MAY BE DISREGARDED. THEREFORE, IT CAN BE ASSUMED THAT THE MAXIMUM VALUE OF SHEAR STRESS IS CONSTANT ACROSS THE THICKNESS AND IS GIVEN BY FOLLOWING EXPRESSION: τmax = pr/2t

  18. QUESTIONS AND QUERIES IF ANY! IF NOT THEN GOOD BYE SEE ALL OF YOU IN NEXT LECTURE ON-------------------------