MECHANICS OF MATERIALS - II

1. MECHANICS OF MATERIALS - II

2. THIN-WALLED TUBES

3. PREVIOUSLY WE HAVE DISCUSSED THE TORSIONAL BEHAVIOR OF CIRCULAR CROSS SECTIONS EITHER SOLID OR HOLLOW. SUCH SHAPES ARE COMMONLY USED IN MACHINERY. HOWEVER, IN LIGHT WEIGHT STRUCTURES SUCH AS AIRCRAFT AND SPACECRAFT, THIN-WALLED TUBULAR MEMBERS OF NON-CIRCULAR SHAPES ARE OFTEN USED TO RESIST TORSION. LET US CONSIDER A THIN-WALLED TUBE OF ARBITRARY CROSS-SECTIONAL SHAPE. THE TUBE IS CYLINDRICAL AND IS SUBJECTED TO PURE TORSION BY TORQUES “T” ACTING AT THE ENDS. THE THICKNESS “t” OF THE WALL OF THE TUBE MAY VARY AROUND THE X-SECTION BUT IT IS ASSUMED TO BE VERY SMALL COMPARING WITH THE TOTAL WIDTH OF THE TUBE. THIN-WALLED TUBES

6. AN ELEMENT OF THE TUBE IS CUT OUT BETWEEN TWO CROSS SECTIONS AT A DISTANCE “dx” APART. SHEAR STRESSES ACT ON THE CROSS SECTIONS AND ARE DIRECTED PARALLEL TO THE EDGES OF THESE CROSS SECTIONS AND THEY FLOW AROUND THE TUBE. THE INTENSITY OF THE SHEAR STRESSES VARIES SO SLIGHTLY ACROSS THE THICKNESS OF THE TUBE THAT FOR MANY PURPOSES SHEAR STRESSES MAY BE ASSUMED CONSTANT ACROSS THE THICKNESS. IN ORDER TO DETERMINE THE MAGNITUDE OF THE SHEAR STRESSES, CONSIDER A RECTANGULAR ELEMENT OBTAINED BY MAKING TWO LONGITUDINAL CUTS “ab” AND “cd”.

7. THIS ELEMENT IS ISOLATED AS A FREE BODY AND STRESSES ACTING ON X-SECTIONAL FACE “bc” ARE THE SHEAR STRESSES “τ”. IT IS FURTHER ASSUMED THAT INTENSITY OF SHEAR STRESSES MAY VARY AS WE MOVE AONG THE X-SECTION FROM “b” TO “c”. THUS SHEAR STRESSES AT THESE TWO POINTS ARE DENOTED AS “τb” AND “τc”. AS PER EQUILIBRIUM SHEAR STRESSES OF THE SAME MAGNITUDE ACT IN THE OPPOSITE DIRECTION ON THE OTHER CROSS SECTIONAL FACE “ad”. SHEAR STRESSES OF THE SAME MAGNITUDE AS THOSE ON THE X-SECTIONS ACT ON THE LONGITUDINAL FACES “ab” AND “cd” INASMUCH AS SHEAR STRESSES ARE EQUAL IN MAGNITUDE ON THE PERPENDICULAR PLANES.

8. HENCE THE CONSTANT SHEAR STRESSES THE ON FACES “ab” AND “cd” ARE EQUAL TO “τb” AND “τc” RESPECTIVELY. NOW SHEAR STRESSES ACTING ON THE LONGITUDINAL FACES PRODUCE FORCES “Fb” AND “Fc” AND THESE CAN BE OBTAINED BY MULTIPLYING THE STRESS BY THE AREAS. HENCE Fb = τbtbdx AND Fc = τctcdx NOW FOR EQUILIBRIUM CONDITION IN “X” DIRECTION, Fb = Fc τbtb = τctc

9. AS THE LOCATIONS OF LONGITUDINAL CUTS “ab” AND “cd” WERE ARBITRARILY SELECTED, THE PRODUCT OF THE SHEAR STRESS AND THE THICKNESS OF THE TUBE WOULD BE SAME AT EACH POINT IN THE CROSS SECTION OF THE TUBE. THIS PRODUCT IS KNOWN AS “SHEAR FLOW” AND IS DONOTED BY “f”, THEREFORE f = τt = CONSTANT IT MEANS THAT LARGERST SHEAR STRESSES OCCUR WHERE THE THICKNESS OF THE TUBE IS SMALLEST. OF COURSE REGIONS WHERE THE THICKNESS IS CONSTANT, SHEAR STRESSES ARE ALSO CONSTANT.

10. NOW IN ORDER TO RELATE SHEAR FLOW “f” WITH THE APPLIED TORQUE “T”, CONSIDER AN ELEMENT OF AREA OF LENGTH “ds” IN THE CROSS SECTION. THE DISTANCE “s” IS MEASURED ALONG THE MEDIAN LINE OF THE X-SECTION. THE TOTAL SHEAR FORCE ACTING ON THE ELEMENT OF AREA IS “f ds” AND THE MOMENT OF THIS FORCE ABOUT ANY POINT “O” WOULD BE dT = r f ds HERE “r” IS THE PERPENDICULAR DISTANCE FROM “O” TO THE LINE OF ACTION OF THE FORCE. NOW THE TOTAL TORQUE PRODUCED BY SHEAR STRESSES CAN BE OBTAINED BY INTEGRATING ALONG THE FULL LENGTH “Lm” OF THE MEDIAN LINE OF THE X-SECTION.

11. HENCE T = f ∫ r ds HERE THE QUANTITY “rds” REPRESENTS TWICE THE AREA OF THE SHADED TRIANGLE AS SHOWN IN FIGURE. NOTE THAT THE TRIANGLE HAS A BASE LENGTH “ds” AND A HEIGHT EQUAL TO “r”. THEREFORE, THE INTEGRAL REPRESENTS DOUBLE THE AREA ENCLOSED BY THE MEDIAN LINE OF THE CROSS SECTION. HENCE T = 2 f Am

12. HENCE FROM THIS EQUATION WE GET f = τt = T/2Am OR τ = T/2tAm FROM THESE TWO EQUATIONS THE SHEAR FLOW AND SHEAR STRESSES CAN BE CALCULATED FOR ANY THIN-WALLED TUBE. THE ANGLE OF TWIST “Φ” CAN BE CALCULATED BY CONSIDERING THE STRAIN ENERGY OF THE TUBE. AS WE KNOW STRAIN ENERGY DENSITY IN CASE OF PURE SHEAR IS GIVEN AS “τ² / 2G. NOW THE STRAIN ENERGY OF A SMALL ELEMENT OF THE TUBE WITH X-SECTIONAL AREA “t ds” AND LENGTH “dx” IS GIVEN AS

13. dU = τ²(t ds dx) / 2G = τ²t² (ds dx / t) / 2G dU = f² (ds dx / t) / 2G THEREFORE, THE TOTAL STRAIN ENERGY OF THE TUBE CAN BE CALCULATED AS FOLLOWS: U = ∫ dU = f² / 2G ∫ [ ∫ dx] ds / t NOW THE INNER INTEGRAL IS EQUAL TO LENGTH OF THE TUBE “L”, HENCE THE EQUATION BECOMES U = f²L / 2G ∫ ds / t NOW SUBSTITUTING THE VALUE OF SHEAR FLOW, f = T/2Am U = T²L / 8 G Am² ∫ ds / t

14. THIS IS THE EQUATION FOR CALCULATING STRAIN ENERGY FOR THE TUBE IN TERMS OF TORQUE “T”. NOW BY INTRODUCING A NEW CONSTANT “J”, KNOWN AS TORSION CONSTANT, ABOVE EQUATION MAY BE WRITTEN IN A MORE SIMPLER FORM. FOR A THIN-WALLED TUBE, THE TORSION CONSTANT (J) IS GIVEN AS J = 4 Am² / ∫ ds / t HENCE BY SUBSTITUTING THIS CONSTANT THE EQUATION FOR STRAIN ENERGY BECOMES U = T²L / 2GJ

15. IT IS TO BE NOTED THIS EQUATION HAS THE SAME FORM AS THE ONE FOR STRAIN ENERGY IN CIRCULAR BAR ( T²L / 2GIp). IN THE SPECIAL CASE OF A CROSS SECTION HAVING CONSTANT THICKNESS , “t”, THE EXPRESSION FOR “J” WOULD BECOME J = 4 TAm² / Lm VALUES OF J FOR DIFFERENT SHAPES CAN BE CALCULATED. FOR EXAMPLE CONSIDER A THIN-WALLED CIRCULAR TUBE WITH THICKNESS “t” AND RADIUS “r” TO THE MEDIAN. THE LENGTH OF THE MEDIAN LINE AND THE AREA ENCLOSED BY IT ARE Lm = 2∏ r AND Am = ∏ r²

16. HENCE THE TORSION CONSTANT AS OBTAINED FROM THE PREVIOUS EQUATION WOULD BE J = 2∏ r³ t NOW CONSIDER A THIN-WALLED RECTANGULAR TUBE WITH THICKNES “t1” ON THE SIDES AND THICKNESS “t2” ON THE TOP AND BOTTOM. THE HEIGHT AND WIDTH TO THE MEDIAN LINE OF THE CROSS SECTION ARE “h” AND “b” RESPECTIVELY. FOR THIS CROSS SECTIO, WE HAVE Lm = 2(b + h) AND Am = bh AND ∫ ds / t = 2 ∫ ds/t1 + 2 ∫ ds / t2 = 2(h/t1 + b/t2) J = 2b²h²t1t2 / bt1 + ht2

17. THE ANGLE OF TWIST “Φ” FOR A THIN-WALLED TUBE MAY BE DETERMINED BY EQUATING THE WORK DONE BY THE APPLIED TORQUES “T” TO THE STRAIN ENERGY OF THE BAR TΦ/2 = T²L/GJ FROM WHICH Φ = TL/GJ AGAIN IT CAN BE OBSERVED THAT THE EQUATION IS OF THE SAME FORM AS THAT FOR A CIRCULAR BAR. IF THE ANGLE OF TWIST PER UNIT LENGTH (θ) IS REQUIRED THEN θ = T/GJ THE QUANTITY GJ IS KNOWN AS TORSIONAL RIGIDITY OF BAR.

18. IT IS QUITE CLEAR THAT IN CASE OF CIRCULAR SHAFT, THE TORSION CONSTANT IS THE POLAR MOMENT OF INERTIA. FOR OTHER SHAPES OF X-SECTIONS, DIFFERENT FORMULAE ARE REQUIRED. FOR EXAMPLE FORMULAE FOR THIN-WALLED OPEN SECTIONS AND SOLID NON-CIRCULAR SECTIONS CAN BE OBTAINED BY MORE ADVANCED METHODS OF ANALYSIS.

19. LET US AGAIN CONSIDER THE CIRCULAR TUBE. THE SHEAR FLOW AND SHEAR STRESSES FOR THIS TUBE CAN BE CONSIDER BY FOLLOWING RELATIONSHIPS: f = T/2Am AND τ = T/2tAm f = T/2 ∏ r² AND τ = T/2 ∏ r² t WHERE Am = ∏ r² SIMILARLY THE ANGLE OF TWIST, Φ, CAN ALSO BE CALCULATED AS FOLLOWS: Φ = TL/GJ = TL/ 2∏Gr³t WHERE J = 2∏ r³ t VERIFICATION

20. IF WE RECALL THESE RESULTS AGREE WITH THOSE OBTAINED FROM THE EQUATIONS DERIVED FOR HOLLOW CIRCULAR BARS. IT CAN BE OBSERVED THAT IF THE HOLLOW BAR IS THIN-WALLED, THE POLAR MOMENT OF INERTIA IS APPROXIMATELY GIVEN BY Ip = 2∏ r³ t NOW USING THIS EXPRESSION FOR “Ip” IN THE TORSION FORMULA FOR HOLLOW BAR / SHAFT LEADS TO EXPRESSION TO CALCULATE SHEAR STRESSES FOR THIN-WALLED TUBE. IN A CASE A TUBE SUBJECTED TO TORSION HAS VERY THIN WALLS, THE POSSIBILITY OF BUCKLING OF WALLS MUST BE TAKEN INTO ACCOUNT.

21. FOR EXAMPLE, A LONG CIRCULAR TUBE CONSTRUCTED OF MILD STEEL WILL BUCKLE AT NORMAL WORKING STRESSES WHEN THE RATIO RADIUS TO THICKNESS (r/t) IS ABOUT 60. THEREFORE, IN ORDER TO AVOID THE FAILURE, BUCKLING OF THIN-WALLED TUBE, WALL THICKNESS MUST BE REASONABLY GREAT ENOUGH AS COMPARED WITH RADIUS. THE SAME WE DISCUSSED IN CASE OF HOWLLOW SHAFT WHICH CAN BUCKLE AND WRINKLE IN CASE WALL THICKNESS IS NOT LARGE ENOUGH.

22. DESIGN PROBLEM – I A 5 kN-m TORQUE IS APPLIED TO A HOLLOW STRUCTURE HAVING THE X-SECTION AS SHOWN IN FIG. NEGLECTING THE EFFECTS OF CONCENTRATIONS, DETERMINE THE SHEARING STRESS AT POINTS “a” AND “b”.

23. DESIGN PROBLEM – II A 750 N-m TORQUE IS APPLIED TO A HOLLOW SHAFT HAVING THE CROSS SECTION AS SHOWN IN FIG. AND A UNIFORM 6-mm WALT THICKNESS. AGAIN NEGLECTING THE EFFECT OF STRESS CONCENTRATIONS, DETERMINE THE SHEAR STRESS AT POINTS “a” AND “b”.

24. DESIGN PROBLEM – III A COOLING TUBE HAVING THE X-SECTION SHOWN IS FORMED FROM A SHEET OF STAINLESS STEEL OF 3 mm THICKNESS. THE RADII r1 = 150 mm AND r2 = 100 mm ARE MEASURED TO THE CENTRE LINE OF THE SHEET METAL. A TORQUE OF 3 kN-M IS APPLIED TO THE TUBE. DETERMINE (a) THE MAXIMUM SHEARING STRESS IN THE TUBE, (b) THE MAGNITUDE OF THE TORQUE CARRIED BY THE OUTER CIRCUAL SHEET.

25. DESIGN PROBLEM – IV THE ALUMINUM ROD BC WITH G = 26 GPa IS BONDED TO THE BRASS ROD AB WITH G = 40 GPa. EACH ROD IS SOLID AND HAS A DIAMETER OF 15 mm. DETERMINE THE ANGLE OF TWIST AT POINTS B AND C.

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