MECHANICS OF MATERIALS

1. MECHANICS OF MATERIALS

2. STRESSES IN BEAMS

3. AS ALREADY EXPLAINED THAT TBEAM IS STRUCTURAL MEMBER SUBJECTED TO LOADS ACTING TRANSVERSELY TO THE LONGITUDINAL AXIS. THESE LOADS PRODUCE INTERNAL ACTIONS OR STRESS RESULTANTS , IN THE FORM OF SHEAR FORCES AND BENDING MOMENTS AS DISCUSSED EARLIER. IN THIS LECTURE WE WOULD DISCUSS AND EXAMINE ISSUES RELATED WITH PRACTICAL APPLICATION AND IMPORTANCE OF DESIGN OF BEAMS BY EXAMINING THE STRESSES AND STRAINS ASSOCIATED WITH THE SHEAR FORCES AND BENDING MOMENTS. STRESSES OM BEAMS SUBJECTED TO DIFFERENT LOADS

4. IN DISCUSSING THE STRESSES AND STRAINS IN BEAMS WE CONSIDER A REFERENCE OF AXIS SUCH THAT +VE X-AXIS IS ON THE RIGHT SIDE AND +VE Y-AXIS IS DOWNWARD AND +VE Z-AXAIS IS DIRECTED INWARD. NOW CONSIDER A LATERAL LOAD “P” ACT ON THE FREE END OF A CANTILEVER BEAM “AB” WHICH CAUSE IT TO BEND OR FLEX AND THUS CAUSING THE BEAM INTO A CURVED LINE. OF COURSE BEFORE APPLING THE LOAD THE LONGITUDINAL AXIS IS A STRAIGHT LINE. AFTER LOADING THIS AXIS IS BENT INTO A CURVE AND THAT IS KNOWN AS THE DEFLECTION CURVE OF THE BEAM.

5. HERE IT IS ASSUMED THAT BEAM IS SYMMETRIC ABOUT THE “XY” PLANE . SIMILARLY ALL TYPES OF LOADS AFE ALSO ASSUMED TO ACT IN THE “XY” PLANE AND THUS THE BENDING DEFLECTIONS OCCUR IN THE SAME PLANE KNOWN AS PLANE OF BENDING. NOW AGAIN CONSIDER A CANTILEVER BEAM “AB” WITH LATERAL LOAD “P” WHICH CAUSES THE DEFLECTION OF BEAM. NOW CONSIDER TWO POINTS “m1” AND “m2” ON THE DEFLECTION CURVE.

8. POINT “m1” IS LOCATED AT DISTANCE “x” FROM Y AXIS AND POINT “m2” IS LOCATED AT A SMALL DISTANCE “ds” FURTHER AWAY ALONG THE CURVE. AT EACH POINT LINES NORMAL TO THE TANGENT TO THE LINE OF DEFELCTION ARE DRAWN. THESE TWO NORMAL LINES ULTIMATELY INTERSECT A POINT “O” THAT IS THE CALLED THE CENTRE OF CURVATURE FOR THE SUBJECT DEFLECTION CURVE AT A DISTANCE “x” AWAY FROM THE SUPPORT. THE LENGTH OF NORMAL LINE THAT IS THE DISTANCE FROM THE CENTRE OF CURVATURE TO THE CURVE IS KNOWN AS THE RADIUS OF CURVATURE DENOTED BY THE SYMBAL “ρ”. DEFLECTIONS OF BEAMS

9. THE CURVATURE, RECIPROCAL OF RADIUS OF CURVATURE, IS DENOTED BY “к”. THEREFORE к = 1/ρ NOW AS PER THE SIMPLE GEOMETRY FOLLOWING RELATIONSHIP CAN BE OBTAINED BETWEEN RADIUS OF CURVATURE, ANGLE OF CURVATURE AND THE DEFLECTION CURVE ALONG THE BEAM: ρ dθ = ds ---------------- (2) IT IS TO BE NOTED HERE THAT THE DEFELCTIONS OF BEAMS ARE VERY SMALL IN MOST OF THE COMMON SITUATIONS AND THEREFORE THE DEFLECTION CURVE IS ALMOST FLAT AND THE DISTANCE MAY BE SET EQUAL TO ITS HORIZONTAL PROJECTION.

10. NOW THE RELATIONSHIP BETWEEN THE CURVATURE “к” AND DEFLECTION OF CURVE CAN BE OBTAINED AS FOLLOWS: к = 1/ρ = dθ/dx ………….. (3) IT MEANS THAT CURVATURE IS A FUNCTION OF ”x” . EQUATION (3) WOULD BE USED TO CALCULATE STRAINS IN A BENT BEAM IN COMING DISCUSSION. BEFORE FURTHER MOVING WE MUST HAVE THE DIFFERENCE BETWEEN PURE BENDING AND NON-UNIFORM BENDING. PURE BENDING MEANS THAT THE BEAM IS UNDER CONSTANT BENDING MOMENT (M) AND THERE IS NO SHEAR FORCE (V) THAT IS dM/dx = V = 0.

11. ON THE OTHER END NON-UNIFORM BENDING, BENDING MOMENT ALWAYS CHANGES ALONG THE AXIS OF BEAM AND SHEAR FORCE IS PRESENT. IF ANY BEAM IS LOADED WITH COUPLES ONLY THEN IT MEANS THERE IS NO SHEAR FORCE AND ONLY BENDING MOMENT EXISTS. THIS IS AN EXAMPLE OF PURE BENDING. IN CONTRAST IF A BEAM IS SUBJECTED TO ANY LOAD, IT MEANS SHEAR FORCE IS PRESENT AND HENCE IT IS NON-UNIFORM BENDING. IN THE NEXT DISCUSSION WE WOULD BE FOCUSSING PURE BENDING IN ORDER TO DETERMINE STRESS & STRAIN IN THIS PARTICULAR CONDITION.

12. IN ORDER TO CALCULATE STRAINS IN PURE BENDING WE NEDD TO CONSIDER THE CURVATURE OF BEAM AND ASSOCIATED DEFORMATION IN IT. CONSIER THE PORTION OF A BEAM “ab” IN PURE BEDNING PRODUCED BY APPLIED COUPLES Mo APPLIED IN SUCH A WAY THAT THESE COUPLES PRODUCE POSITIVE CURVATURE OF THE BEAM. THE BEAM INITIALLY HAS A STRAIGHT LONGITUDINAL AXIS AND THE CROSS SECTIONS ARE SYMMETRIC ABOUT THE Y AXIS. AS THE COUPL Mo IS APPLIED DEFLECTS IN THE XY PLANE AND ITS AXIS IS BENT INTO A CIRCULAR CURVE. THE SYMMETRY OF THE BEAM AND ITS LOADING REQUIRES THAT ALL ELEMENTS OF THE BEAM DEFORM IN AN IDENTICAL MANNER. STRAINS IN PURE BENDING

15. HOWEVER, THIS IS POSSIBLE ONLY IF THE DEFLECTION CURVE IS CIRCULAR AND IF CROSS SECTIONS REMAIN PLANE DURING LOADING. THESE FACTS CAN BE PROVED THEORITICALLY EXPERIMENTALLY BY MAKING PRECISE MEASUREMENTS OF STRAINS. THESE CONCLUSTIONS ARE VALID FOR A BEAM OF ANY MATERIAL WHETHER ELASTIC OR INELASTIC. NOW AS A RESULT OF BENDING DEFORMATIONS CROSS SECTIONS “mn” AND “pq” ROTATE WITH RESPECT TO EACH OTHER ABOUT AXIS PERPENDICULAR TO THE XY PLANE. THE LONGITUDINAL SEGMENTS ON THE UPPER SIDES STRETCH WHERE THOSE SEGMENTS ON THE LOWER SIDES CONTRACT.

16. THEREFORE, THE UPPER SEGMENTS ARE IN TENSION AND LOWER SIDE SEGMENTS ARE IN COMPRESSION. OF COURSE BETWEEN THESE TWO EXTREMES THERE ARE SEGMENTS BETWEEN TOP AND BOTTOM OF THE BEAM WHICH DO NOT CHANGE IN LENGTH. THE SURFACE CONSISTED OF SUCH SEGMENTS IS KNOWN AS THE NEUTRAL SURACE OF THE BEAM AND ITS INTERSECTION WITH ANY X-SECTIONAL PLANE IS CALLED THE NEUTRAL AXIS OF THE X-SECTION. Z AXIS IS THE NEUTAL AXIS IN THIS CASE. THE PLANES OF X-SECTIONS mn AND pq OF THE DEFORMED BEAM INTERSECT IN A LINE THROUGH THE CENTRE OF CURVATURE O’. ANGLE BETWEEN THESE PLANES IS “dθ”.

17. DISTANCE FROM POINT “O” TO THE NEUTRAL SURFACE IS CALLED THE RADIUS OF CURVATURE “ρ”. HOWEVER, THE INITIAL DISTANCE “dx” BETWEEN THESE TWO PLANES AND THE NEUTRAL SURFACE REMAINS UNCHANGED, AND HENCE THE RELATIONSHIP “ρ dθ = dx” IS VALID. ALL OTHER LONGITUDINAL SEGMENTS OR FIBRES AT UPPER AND LOWER SIDES LENGTHEN OR SHORTEN AND THUS PRODUCES LONGITUDINAL STRAINS “ε”.

18. AS WE KNOW THAT LONGITUDINAL STRAIN IN BEAM SUBJECTED TO PURE BENDING CAN BE CALCULATED BY THE FOLLOWING RELATIONSHIP: ε = -y/ρ = -кy …………. (1) THIS EQUATION SHOWS THAT LONGITUDINAL OR AXIAL STRAIN IN THE BEAM IS PROPORTIONAL TO THE CURVATURE AND THAT IT VARIES LINEARLY WITH THE DISTANCE “y” FROM THE NEUTRAL SURFACE. SIMILARLY IT WAS ESTABLISHED THAT POSITIVE TRANSVERSE STRAIN IN THE BEAM WOULD BE CALCULATED BY THE FOLLOWING RELATIONSHIP: ε = -νε = νкy ………. (2) NORMAL STRESSES IN BEAMS

19. IN TERMS OF GEOMETRY THE TRANSVERSE STRAIN CAUSES THE WIDTH OF THE X-SECTION TO INCREASE BELOW THE Z AXIS AND DECREASES ABOVE IT. CONSEQUENTLY ALL STRAIGHT LINES IN THE X-SECTION BECOME SLIGHTLY CURVED. STRESSES ACTING NORMAL TO THE X-SECTION OF THE BEAM (NORMAL STRESSES) CAN BE COMPUTED FROM NORMAL STRAIN AS CAN BE CALCULATED FROM ABOVE-MENTIONED RELATIONSHIP. IN FACT EACH LONGITUDINAL FIBER OF THE BEAM IS SUBJECTED ONLY TO TENSION AND COMPRESSION, AND HENCE THE STRESS-STRAIN DIAGRAM WILL PROVIDE THE RELATIONSHIP STRESS AND STRAIN. BY HOOKE’S LAW σ = Eε = -Eкy ……….. (3)

21. THIS RELATIONSHIP SHOWS THAT NORMAL STRESSES ACTING ON THE X-SECTION VARY LINEARLY WITH DISTANCE y FROM THE NEUTRAL SURFACE. IN CASE OF POSITIVE CURVATURE, THE STRESSES ARE NEGATIVE BELOW THE NEUTRAL SURFACE AND POSITIVE ABOVE THE NEUTRAL SURFACE. IT IS TO BE NOTED THAT POSITIVE CURVATURE IS PRODUCED WHEN THE BENDING MOMENT IS NEGATIVE, THAT IS M = -Mo. NOW WE CONSIDER THE RESULTANT OF NORMAL STRESSES ACTING OVER THE X-SECTION. THIS RESULTANT IN GENERAL WOULD BE CONSISTED OF HORIZONTAL FORCE IN THE X-DIRECTION AND A COUPLE ACTING ABOUT THE Z-AXIS.

22. AS WE KNOW NO AXIAL FORCE ACTS ON THE X-SECTION, HENCE THE ONLY RESULTANT IS THE COUPLE, Mo. IN THIS SITUATION WE WOULD HAVE TWO EQUATIONS OF STATICS. FIRST EQUATION STATES THAT THE RESULTANT FORCE IN THE X DIRECTION WOULD BE EQUAL TO ZERO, AND THE SECOND EQUATION STATES THAT THE RESULTANT MOMENT IS EQUAL TO Mo. IN ORDER TO EVALUATE THESE STATED RESULTANTS, LET US SUPPOSE THAT AN ELEMENT OF AREA “dA” IN THE X-SECTION AT A DISTANCE “y” FROM THE NEUTRAL AXIS. THE FORCE ACTING ON THIS ELEMENT IS NORMAL TO THE X-SECTION AND HAS A MAGNITUDE “σdA”.

24. AS THERE IS NO RESULTANT NORMAL FORCE THAT ACTS ON THE X-SECTION, THE INTEGRAL OF THIS FORCE OVER THE ENTIRE X-SECTIONAL AREA MUST VANISH, THEREFORE, ∫σdA = - ∫EкydA = 0 ……….. (4) SINCE THE CURVATURE AND MODULUS OF ELASTICITY ARE CONSTANT AT THE X-SECTION, IT CAN BE CONCLUDED THAT FOR A BEAM IN PURE BENDING ∫ydA = 0 ………………. (5) THIS EQUATION SHOWS THAT THE FIRST MOMENT OF THE AREA OF THE X-SECTION WITH RESPECT TO THE Z-AXIS IS ZERO; HENCE THE Z-AXIS MUST PASS THROUGH THE CENTROID OF THE X-SECTION.

25. AS Z-AXIS IS ALSO THE NEUTRAL AXIS, THEREFORE, THE NEUTRAL AXIS PASSES THROUGH THE CENTOROID OF THE X-SECTION WHEN THE MATERIAL OF THE BEAM FOLLOWS HOOKE’S LAW. WE ALSO ASSUMED THAT Y-AXIS IS AN AXIS OF SYMMETRY, HENCE Y-AXIS MUST ALSO PASS THROUGH THE CENTROID. IT MEANS THAT THE ORIGIN OF COORDINATES “O” IS LOCATED AT THE CENTROID OF THE X-SECTION. BEING SYMMETRY OF ABOUT THE Y-AXIS ALSO MEANS THAT Y-AXIS IS A PRINCIPAL AXIS. Z-AXIS IS ALSO A PRINCIPAL AXIS AS IT IS PERPENDICULAR TO THE Y-AXIS. IT MAY BE CONCLUDED THAT Y AND Z AXIS ARE SAID TO BE PRINCIPAL CENTROIDAL AXES IN CASE BEAM OF LINEAR ELASTIC MATERIAL IS SUBJECTED TO PURE BENDING.

26. AS Z-AXIS IS ALSO THE NEUTRAL AXIS, THEREFORE, THE NEUTRAL AXIS PASSES THROUGH THE CENTOROID OF THE X-SECTION WHEN THE MATERIAL OF THE BEAM FOLLOWS HOOKE’S LAW. WE ALSO ASSUMED THAT Y-AXIS IS AN AXIS OF SYMMETRY, HENCE Y-AXIS MUST ALSO PASS THROUGH THE CENTROID. IT MEANS THAT THE ORIGIN OF COORDINATES “O” IS LOCATED AT THE CENTROID OF THE X-SECTION. BEING SYMMETRY OF ABOUT THE Y-AXIS ALSO MANS THAT Y-AXIS IS A PRINCIPAL AXIS. Z-AXIS IS ALSO A PRINCIPAL AXIS AS IT IS PERPENDICULAR TO THE Y-AXIS. IT MAY BE CONCLUDED THAT Y AND Z AXIS ARE SAID TO BE PRINCIPAL CENTROIDAL AXES IN CASE BEAM OF LINEAR ELASTIC MATERIAL IS SUBJECTED TO PURE BENDING.

27. NOW LET US COMPUTE THE MOMENT OF RESULTANT OF THE STRESSES ACTING OVER THE X-SECTION. THE ELEMENT OF FORCE “σdA” ON THE ELEMENT “dA” ACTS IN THE POSITIVE DIRECTION OF X-AXIS WHEN STRESS IS POSITIVE AND IN NEGATIVE DIRECTION WHEN STRESS IS NEGATIVE. THEREFORE, THE MOMENT OF AREA ABOUT Z-AXIS IS GIVEN BY THE FOLLOWING RELATIONSHIP” dMo = -σydA ………….. (6) THE INTEGRAL OF ALL SUCH ELEMENTAL MOMENTS OVER THE ENTIRE X-SECTIONAL AREA MUST RESULT IN THE TOTAL MOMENT. THERE FORE, Mo = -∫σ ydA …..……. (7)

28. AS WE ALSO ASSUMED THAT RESULTANT MOMENT “M” IS EQUAL TO “-Mo” M = ∫σ ydA = -κE ∫y²dA ……. (8) ABOVE EQUATION CAN ALSO WRITTEN IN A MORE SIMPLE FORM AS M = -кE I ……………. (9) WHERE I = ∫y²dA ………….… (10) “I” IS THE MOMENT OF INERTIA OF X-SECTIONAL AREA WITH RESPECT TO THE Z-AXIS.

29. EQUATION (8) CAN ALSO BE ARRANGED AS к = 1/ρ = -M/EI ……… (11) THIS EQUATION CLEARLY SHOWS THAT THE CURVATURE OF THE LONGITUDINAL AXIS OF A BEAM IS PROPORTIONAL TO THE BENDING MOMENT “M”, AND INVERSELY PROPORTIONAL TO THE PRODUCT “EI” WHICH IS ALSO KNOW AS THE FLEXURAL RIGIDITY OF THE BEAM. THE MINUS SIGN IS BECAUSE OF THE SIGN SCHEME WE ADOPTED. WE ASSUMED THAT POSITIVE BENDING PRODUCES NEGATIVE CURVATURE AND A NEGATIVE BENDING MOMENT PRODUCES POSITIVE CURVATURE.

30. NOW THE NORMAL STRESSES IN THE BEAM CAN EASILY BE RELATED TO THE BENDING MOMENT BY SUBSITUTING THE VALUE OF CURVATURE FROM EQUATION (3) AS FOLLOWS: σ = M y/I ……… (12) THIS EQUATION, ALSO KNOWN AS FLEXURE FORMULA, SHOWS THAT THE STRESSES: ARE PROPORTIONAL TO THE BENDING MOMENT “M”, INVERSELY PROPORTIONAL TO THE MOMENT OF INERTIA “I” OF THE X-SECTION, AND THESE VARY LINEARLY WITH THE DISTANCE “y” FROM THE NEUTRAL AXIS.

31. AGAINST APPLICATION OF POSITVE BENDING,THE RESULTING STRESSES WOULD BE POSITIVE OVER THE PART OF THE X-SECTION WHERE “y” IS POSITIVE. IN CASE OF NEGATIVE BENDING MOMENT, NEGATIVE STRESSES WOULD BE PRODUCED WHERE “y” IS POSITIVE. NOW THE MAXIMUM TENSILE AND COMPRESSIVE STRESSES IN THE BEAM UNDER PURE BENDING WOULD OCCUR AT POINTS LOCATED FARTHEST FROM THE NEUTRAL AXIS.

32. NOW THE MAXIMUM TENSILE AND COMPRESSIVE STRESSES ARE GIVE BY THE FOLLOWING RELATIONSHIP: σ1 = MC1/I = M/S1, S1 = I/C1 σ2 = -MC2/I = M/ S2, S2 = I/C2 “S1” ABD “S2” ARE KNOWN AS THE SECTION MODULI OF THE X-SECTIONAL AREAS AND HAVE THE DIMENSION OF LENGTH TO THE THIRD POWER. IF THE X-SECTION IS ALSO SYMMETRIC TO Z-AXIS, IN THAT CASE C1 = C2 = C AND THE RESULTING STRESSES WOULD BE AS FOLLOWS:

33. σ1 = -σ2 = MC/I = M/S AND S = I/C WHERE “S” IS THE SECTION MODULUS. NOW FOR BEAMS OF RECTANGULAR X-SECTION WITH WIDTH “b” AND HEIGHT “h”, THE MOMENT OF INERTIA AND SECTION MODULUS ARE I = bh³/12 AND S = bh²/6 FOR CIRCULAR X-SECTION I = ∏d⁴/64 AND S = ∏d³/32

34. COMPOSITE BEAMS ARE THOSE BEAMS WHICH ARE MADE OF MORE THAN ONE MATERIAL. EXAMPLES OF SUCH BEAMS INCLUDE BIMETALLIC BEAMS, SANDWICH BEAMS AND REINFORCED CONCRETE BEAMS. COMPOSITE BEAMS CAN ALSO BE ANALYZED BY THE SAME BENDING THEORY AS HAS BEEN DISCUSSED EARLIER FOR ORDINARY BEAMS IN CASE OF PURE BENDING. THIS IS BASED ON THE ASSUMPTION THAT CROSS SECTIONS THAT ARE PLANE BEFORE BENDING REMAINS PLANE AFTER BENDING IN CASE OF PURE BENDING. STRESSES & STRAINS IN COMPOSITE BEAMS

35. FROM THIS ASSUMPTIONS IT FOLLOWS THAT THE LONGITUDINAL STRESS & STRAIN VARIES LINEARLY FROM TOP TO BOTTOM OF THE BEAM. IN CASE COMPOSITE BEAMS THE POSITION OF NEUTRAL AXIS IS NOT AT THE CENTROID OF THE X-SECTIONAL AREA AS HAS BEEN EXPLAINED EARLIER. THE NORMAL STRESSES “σ” ACTING ON THE X-SECTION IN A COMPOSITE BEAM CAN OBTAINED FROM THE STRAINS “ε” BY USING THE STRESS-STRAIN RELATIONSHIPS FOR THE MATERIALS. IF IT IS ASSUMED THAT THE MATERIALS BEHAVE IN A LINEAR ELASTIC MANNER, THEN THE STRESSES IN EACH OF THE MATERIALS MAY BE OBTAINED BY MULTIPYING THE STRAINS BY THE APPROPRIATE MODULUS OF ELASTICITY.

36. IN THIS WAY THE NORMAL STRESSES AT ANY DISTANCE “y” FROM THE NEUTRAL AXIS CAN BE COMPUTED BY THE FOLLOWING EQUATIONS: σ1 = -E1кy AND σ2 = -E2кy THE POSITION OF NEUTRAL AXIS CAN BE FOUND BY USING THE CONDITION THAT RESULTANT AXIAL FORCE ACTING ON THE CROSS SECTION IS EQUAL TO ZERO. THEREFORE, ∫σ1dA + ∫σ2dA = 0 AND REPLACING σ1 AND σ2 WITH ABOVE-MENTIONED EQUATION, WE GET E1∫ydA + E2∫ydA = 0

37. A STEEL WIRE OF DIAMETER “d” IS BENT OVER A DRUM OF RADIUS “r”. CALCULATE THE MAXIMUM BENDING STRESS “σmax” AND BENDING MOMENT “M” IN THE WIRE. VALUES OF “E”, “d”, AND “r” ARE E = 200 GPa, d = 4 mm, r = 0.5 m RESPECTIVELY. HINTS ρ = r + d/2, σmax = Eкy σmax = Mmax/S, S = I/r DESIGN PROBLEM - I

38. DESIGN PROBLEM - II

39. DESIGN PROBLEM - III

40. σ = Eε = Eкy ……….. (3) к = 1/ρ = M/EI ……… (11) DESIGN PROBLEM - III

41. DERIVE THE FOLLOWING FLEXURE EQUATION FOR A COMPOSITE BEAM σ1 = MyE1/E1I1 +E2I2 AND σ2 = MyE2/E1I1 + E2I2 AND IN WHAT CASE THESE EQUATIONS CAN BE REDUCED TO σ = My/I ASSIGNMENT # 1 FLEXURE EQUATION FOR COMPOSITE BEAMS

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