MECHANICS OF MATERIALS

MECHANICS OF MATERIALS - i

TORSION- I

TORSION / TORSIONAL LOADING IN PREVIOUS LECTURES WE DISCUSSED ABOUT SLENDER STRUCTURES SUBJECTED TO AXIAL LOADING, THAT IS, FORCES APPLIED ALONG THE LONGITUDINAL AXIS OF THE MEMBER. HERE SLENDER MEMBERS SUBJECTED TO TORSIONAL LOADING, THAT IS, LOADING BY COUPLES THAT PRODUCE TWISTING OF THE MEMBER ABOUT ITS AXIS WOULD BE DISCUSSED. A COUPLE IS A SYSTEM OF FORCES WITH A RESULTANT MOMENT BUT NO RESULTANT FORCE. ANOTHER TERM FOR A COUPLE IS A PURE MOMENT. ITS EFFECT IS TO CREATE ROTATION WITHOUT TRANSLATION, OR MORE GENERALLY WITHOUT ANY ACCELERATION OF THE CENTRE OF MASS.

THE SIMPLEST KIND OF COUPLE CONSISTS OF TWO EQUAL AND OPPOSITE FORCES WHOSE LINES OF ACTION DO NOT COINCIDE. THIS IS CALLED A "SIMPLE COUPLE”. THE FORCES HAVING A TURNING EFFECT OR MOMENT ARE ALSO CALLED A TORQUE ABOUT AN AXIS WHICH IS NORMAL TO THE PLANE OF THE FORCES. THE SI UNIT FOR THE TORQUE OF THE COUPLE IS NEWTON-METER. THE TERM “TORSION” REFERS TO THE TWISTING OF A STRUCTURAL MEMBER IN CASE IT IS LOADED BY COUPLES THAT PRODUCE ROTATION ABOUT ITS LONGITUDINAL AXIS.

LET CONSIDER A STRAIGHT BAR SUPPORTED AT ONE END AND LOADED BY TWO PAIRS OF FORCES ON THE OTHER END. EACH PAIR OF FORCES FORMS A COUPLE THAT TENDS TO TWIST THE BAR ABOUT ITS LONGITUDINAL AXIS. THE MOMENT OF A COUPLE IS EQUAL TO THE PRODUCT OF ONE OF THE FORCES AND THE DISTANCE BETWEEN THE LINES OF ACTION OF THE FORCES. THAT IS, IT IS THE PRODUCT OF FORCE “P” AND THE DISTANCE “d”. THE COUPLES THAT PRODUCE TWISTING OF A BAR ARE ALSO CALLED TORQUES, TWISTING COUPLES, OR TWISTING MOMENTS.

THE MOMENT OF A COUPLE CAN BE REPRESENTED BY MEANS OF A VECTOR IN THE FORM OF DOUBLE-HEADED ARROW WHICH IS PERPENDICULAR TO THE PLANE CONTAINING THE COUPLE. THE DIRECTION OF THE MOMENT IS USUALLY INDICATED BY A VERY SIMPLE RULE KNOWN AS RIGHT-HAND RULE FOR MOMENT VECTORS. ALTERNATELY THE MOMENT IS COUPLE IS INDICATED BY THE USE OF A CURVED ARROW ACTING IN THE DIRECTION OF TWIST.

MEMBERS IN TORSION MAY BE WITNESSED IN MANY ENGINEERING APPLICATIONS SUCH AS AXLES AND THE MOST COMMON APPLICATION OF TRANSMISSION SHAFTS. TRANSMISSION SHAFT ARE USED TO TRANSMIT POWER FROM ONE POINT TO ANOTHER, AND THESE SHAFTS MAY BE BOTH SOLID AND HOLLOW. IN ORDER TO CALCULATE TORSION OF A CIRCULAR BAR OR SHAFT OF CIRCULAR CROSS SECTION, LET US CONSIDER THAT IT IS TWISTED BY COUPLE “T” ACTING AT THE ENDS. IN THIS SPECIFIC CONDITION THE BAR OR SHAFT IS SAID TO BE IN PURE TORSION. WHY ?????????? TORSION OF A CIRCULAR SHAFT

CONSIDERING THE SYMMETRY IT CAN SIMPLY BE ASSUMED THAT CROSS SECTIONS OF THE CIRCULAR BAR ROTATE AS RIGID BODIES ABOUT THE LONGITUDINAL AXIS. IT IS ALSO ASSUMED THAT THE RADII REMAIN STRAIGHT AND CROSS SECTIONS REMAIN PLANE AND CIRCULAR. ALSO IF THE TOTAL ANGLE OF TWIST IS SMALL, NEITHER THE LENGTH OF THE BAR NOR ITS RADIUS WOULD CHANGE. AS THE BAR TWISTS, THERE WOULD BE ROTATION ABOUT THE LONGITUDINAL AXIS AT ONE END OF THE BAR WITH RESPECT TO THE OTHER END. IN THE SIMPLEST FORM IT MEANS THAT IF ONE END OF BAR IS FIXED, THE FREE END WOULD ROTATE THROUGH A SMALL ANGLE “Φ”.

THIS ANGLE IS KNOWN AS THE ANGLE OF TWIST OR ANGLE OF ROTATION, “Φ”. AT THE SAME TIME A LONGITUDINAL LINE ON THE SURFACE OF THE BAR WILL ROTATE THROUGH A SMALL ANGLE TO THE NEW POSITION. CONSEQUENTLY IT WILL BE DISTORTED INTO A RHOMBOID. BEFORE DISTORTION, THE ORIGINAL CONFIGURATION IS “abcd”. BECAUSE OF TORSION ONE SIDE OF CROSS SECTION ROTATES WITH RESPECT TO THE OTHER SIDE, AND POINTS “b” AND c MOVE TO b’ AND c’. DURING ROTATION THE LENGTHS OF THE SIDES OF THE ELEMENT DO NOT CHANGE BUT THE ANGLES AT THE CORNERS ARE NO LONGER AT 90 DEGREES AND THUS BEING THE ELEMENT IN PURE SHEAR STATE.

IN THIS CONDITION THE MAGNITUDE OF THE SHEAR STRAIN “ϒ” IS EQUAL TO THE DECREASE IN THE RIGHT ANGLE AT “a”, AND IS GIVEN BY THE FOLLOWING RELATIONSHIP: ϒ = bb’/ab HERE “bb’” IS THE LENGTH OF A SMALL ARC OF RADIUS “r” SUBTENDED BY THE ANGLE “dΦ” AND THUS bb’=rdΦ. SIMILARLY THE DISTANCE “ab” IS EQUAL TO “dx”. BY SUBSTITUTION THESE TWO VALUES, SHEAR STRAIN IS GIVEN BY ϒ = r dΦ / dx = rθ = rΦ/L SHEAR STRAIN

HERE “dΦ/ dx” IS THE RATE OF CHANGE OF THE ANGLE OF TWIST AND IS DENOTED BY “θ” AND IS REFERRED AS THE ANGLE OF TWIST PER UNIT LENGTH. “L” IS THE TOTAL LENGTH OF THE BAR. IT IS TO BE NOTED THAT THE ABOVE EQUATIONS ARE BASED ONLY UPON GEOMETRIC CONFIGURATIONS AND ARE EQUALLY VALID FOR A CIRCULAR BAR OF ANY MATERIAL WHETHER ELASTIC OR INELASTIC, LINEAR OR NONLINEAR. ACCORDING TO HOOKE’S LAW, FOR LINEAR ELASTIC MATERIALS SHEAR STRESS IS DIRECTLY PROPORTIONAL TO SHEAR STRAIN AND GIVEN BY THE FOLLOWING RELATIONSHIP: Τ = Gϒ = Grθ WHERE G IS SHEAR MODULUS OF ELASTICITY OR MODULUS OF RIGIDITY AS IS E FOR AXIAL LOADING.

STRAINS AND STRESSES WITHIN THE INTERIOR OF THE SHAFT CAN ALSO BE DETERMINED IN A SIMILAR MANNER AS ARE DETERMINED FOR OUTER OF THE SHAFT BECAUSE RADII IN THE CROSS SECTIONS OF THE BAR REMAIN STRAIGHT AND UNDISTORTED DURING TWISTING. THE INTERIOR OF THE CYLINDER IS ALSO IN A STATE OF PURE SHEAR, AND IF “ρ” IS THE RADIUS OF THE INTERIOR OF THE CYLINDER, THEN IN A SIMILAR WAY VALUES OF STRAIN AND STRESS CAN BE CALCULATED BY FOLLOWINGS: ϒ = ρθ AND τ = Gρθ

THESE EQUATIONS SHOW THAT THE SHEAR STRAIN AND STRESS IN A CIRCULAR BAR VARY LINEARLY WITH THE RADIAL DISTANCE FROM THE CENTRE, AND ARE MINIMUM AT THE CENTROID AND MAXIMUM AT THE OUTER ENDS. THE SHEAR STRESSES ACTING ON THE PLANE OF THE CROSS SECTION ARE ALSO ACCOMPANIED BY SHEAR STRESSES OF THE SAME MAGNITUDE ACTING ON THE LONGITUDINAL PLANES OF THE BAR. THIS IS TRUE FROM THE FACT THAT EQUAL SHEAR STRESSES ALWAYS EXIST ON MUTUALLY PERPENDICULAR PLANES OF A MEMBER.

IF A MATERIAL IS WEAKER IN SHEAR ON LONGITUDINAL PLANES THAN ON CROSS-SECTIONAL PLANES, THE FIRST CRACKS WILL APPEAR ON THE SURFACE IN THE LONGITUDINAL DIRECTION BECAUSE OF TWISTING. THE STATE OF PURE SHEAR STRESS AT THE SURFACE OF THE SHAFT IS EQUIVALENT TO EQUAL TENSILE AND COMPRESSIVE STRESSES ON AN ELEMENT ROTATED THROUGH AN ANGLE OF 45 DEGREES. IF A TWISTED BAR IS MADE OF A MATERIAL THAT IS WEAKER IN TENSION THAN IN SHEAR, FAILURE WILL OCCUR IN TENSION ALONG THE AXIS INCLINED AT 45 DEGREES.

THE RELATIONSHIP BETWEEN APPLIED TORQUE “T” AND THE CORRESPONDING ANGLE OF TWIST “Ф” CAN BE DETERMINED FROM THE CONDITION THAT THE RESULTANT COUPLE OF THE SHEAR STRESSES ACTING OVER THE CROSS SECTION MUST BE STATICALLY EQUIVALENT TO THE APPLIED TORQUE. THEREFORE THE SHEAR FORCE ACTING ON AN ELEMENT OF AREA “dA” IS EQUAL TO “τdA”, AND MOMENT DUE TO THIS FORCE WOULD BE EQUAL TO “τρdA”. NOW SUBSTITUTING THE VALUE OF “τ = Gρθ”, FOLLOWING EQUATION FOR TORQUE “T” IS OBTAINED: RELATIONSHIP B/W TORQUE AND ANGLE OF TWIST

T = ∫ Gθρ2 dA T = Gθ ∫ ρ2 dA = Gθ Ip WHERE Ip = ∫ ρ2 dA “Ip” IS THE POLAR MOMENT OF INERTIA OF THE CIRCULAR CROSS SECTION. FOR A CIRCLE WITH RADIUS “r” AND DIAMETER “d”, POLAR MOMENT OF INERTIA IS GIVEN BY Ip = ∏ r⁴/2 = ∏ d⁴/32 THEREFORE, T = Gθ Ip θ = T/G Ip

THE PRODUCT OF “G Ip” IS CALLED THE TORSIONAL RIGIDITY, AND ANGLE OF TWIST PER LENGTH IS INVERSELY PROPORTIONAL TO IT. TOTAL ANGLE OF TWIST “Ф” IS EQUAL TO “θL”. BY SUBSTITUTION FOLLOWING FINAL EQUATION IS OBTAINED: Ф = TL/G Ip (δ = PL/EA) WITH THE PREVIOUS EQUATION SHEAR MODULUS OF RIGIDITY “G” CAN BE DETERMINED FOR DIFFERENT MATERIALS BY CONDUCTING TORSION TESTS AND THUS DETERMINING ANGLE OF TWIST AGAINST THE APPLICATION OF TORQUE.

SIMILAR TO THE RELATIONSHIPS FOR AXIALLY LOADED MEMBERS, STIFFNESS IS DEFINED FOR TORSIONAL LOADING AS THE TORQUE REQUIRED TO PRODUCE A UNIT ANGLE OF ROTATION AT ONE END WITH RESPECT TO THE OTHER END. k = GIp/L (k = EA/L) SIMILARLY THE TORSIONAL FLEXIBILITY IS DEFINED AS THE RECIPROCAL OF STIFFNESS, AND IS EQUAL TO THE ROTATION PRODUCED BY A UNIT TORQUE. f = L/GIp (f = L/EA)

THE VALUE OF MAXIMUM SHEAR STRESS “ΤMAX” IN A CIRCULAR SHAFT SUBJECTED TO TORSION CAN BE COMPUTED AS FOLLOWS: Ф = TL/G Ip θ = T/G Ip τ = Gϒ = Grθ τmax = Tr/ Ip THIS EQUATION SHOWS THAT MAXIMUM SHEAR STRESS IS DIRECTLY PROPORTIONAL TO THE APPLIED TORQUE AND THE RADIUS OF THE SHAFT, AND IS INVERSELY PROPORTIONAL TO POLAR MOMENT OF INERTIA. THE ABOVE EQUATION IS KNOWN AS TORSION FORMULA. CALCULATION OF MAXIMUM SHEAR STRESS

BY SUBSTITUTING r = d/2 AND Ip = Π d⁴/32, FOLLOWING FINAL EQUATION IS OBTAINED TO CALCULATE THE VALUE OF MAXIMUM SHEAR STRESS “ΤMAX” IN A CIRCULAR SHAFT: τmax = 16T/ ∏ d³

DESIGN PROBLEM FIND THE MAXIMUM PERMISSIBLE VALUE OF TORQUE THAT CAN BE SAFELY APPLIED TO THE SHAFT IF THE RADIUS AND LENGTH OF THE SHAFT ARE 50MM AND 3 M, ALLOWABLE SHEAR STRESS IS EQUAL TO 60 MPA, ALLOWABLE ANGLE OF TWIST IS 3 DEGREES PER UNIT LENGTH, AND G= 80 GPA.

IN FACT AS COMPARED TO SOLID CIRCULAR SHAFTS, HOLLOW CIRCULAR SHAFTS ARE MUCH MORE EFFICIENT IN RESISTING TORSIONAL LOADS. FOR THE SAKE OF WEIGHT REDUCTION AND SAVINGS OF MATERIAL (BOTH ARE IMPORTANT FROM ENGINEERING ANALYSIS POINT OF VIEW!), HOLLOW SHAFTS ARE ADVISED TO BE USED AS MOST OF THE MATERIAL IN A SOLID SHAFT IS STRESSED SIGNIFICANTLY BELOW THE ALLOWABLE SHEAR STRESS. RELATIONSHIPS DERIVED FOR ANGLE OF TWIST AND MAXIMUM STRESS FOR SOLID SHAFTS ARE EQUALLY APPLICABLE TO THE HOLLOW SHAFTS AS ANALYSIS OF TORSION FOR HOLLOW AND SOLID SHAFTS ARE ALMOST IDENTICAL. TORSION OF HOLLOW CIRCULAR SHAFTS

Ip ~ 2∏r³t = ∏/4d³t HERE R AND D ARE THE AVERAGE RADIUS AND DIAMETER. FOLLOWING EQUATIONS DERIVED FOR SOLID SHAFTS TO CALCULATE θ, Φ AND τ MAY ALSO BE USED FOR HOLLOW SHAFTS: θ = T/G Ip, Ф = TL/G Ip, τmax = Tr/ Ip HOWEVER, IP HAS TO BE CALCULATED BY THE TWO EQUATIONS MENTIONED ABOVE. IT MUST BE KEPT IN MIND THAT THICKNESS OF SHAFT MUST BE LARGE ENOUGH IN ORDER TO AVOID THE WRINKLING OR BUCKLING OF THE WALL.

DESIGM PROBLEM: A TUBULAR SHAFT OF LENGTH 5 M LONG HAS AN OUTER RADIUS OF 135 MM AND AN INNER RADIUS OF HALF OF OUTER RADIUS AND IS MADE OF A MATERIAL WITH G = 70 GPa. IT IS SUBJECTED TO A TORQUE OF 1750 N.m. DETERMINE THE FOLLOWINGS: (A) MAXIMUM SHEAR STRESS IN THE SHAFT (B) DETERMINE THE ANGLE OF TWIST OF THE SHAFT

DESIGM PROBLEM: THE SHIP AT A HAS JUST STARTED TO DRILL FOR OIL ON THE OCCEAN FLOOR AT A DEPTH OF 1500m. THE TOP OF THE 200 mm DIAMETER STEEL DRILL PIPE WITH G = 77 GPa ROTATES TROUGH TWO COMPLETE REVOLUTIONS BEFORE THE DRIL BIT AT “B” STARTS TO OPERATE. DETERMINE THE MAXIMUM SHEAR STRESS CAUSED IN THE PIPE BY TORSION. HINTS Ф = TL/G Ip, τmax = Tr/ Ip τ = G Ф r/ L

SO FAR WE HAVE DISCUSSED PURE SHEAR WHICH MEANS TORSION OF A PRISMATIC BAR SUBJECTED TO TORQUES ACTING ONLY AT THE ENDS. NONUNIFORM TORSION IS DIFFERENT FROM THIS TYPE OF TORSION. NONUNIFORM TORSION MEANS THAT THE BAR IS NOT REQUIRED TO BE PRISMATIC BAR AND THE APPLIED LOADS MAY VARY ALONG THE LENGTH OF THE SHAFT. IN SUCH CASE MAGNITUDE AND DIRECTION OF THE INTERNAL TORQUE AT EACH POINT OF APPLICATION OF TORQUE IS INDIVIDUALLY DETERMINED. NON-UNIFORM TORSION

NOW FROM THE CALCULATED VALUES OF TORQUE, MAXIMUM SHEAR STRESS AND ANGLE OF TWIST FOR EACH POINT OR REGION CAN BE CALCULATED BY FOLLOWING EQUATIONS AS MENTIONED EARLIER: τmax = Tr/ Ip and Ф = TL/G Ip THE TOTAL ANGLE OF TWIST OF ONE END OF THE BAR WITH RESPECT TO THE OTHER END IS OBTAINED BY SUMMATION. HIGH LOCALIZED STRESSES MAY OCCUR AT THE SECTIONS WHERE THE DIAMETER CHANGES ABRUPTLY. HOWEVER, THESE STRESSES HAVE LITTLE EFFECT ON THE TOTAL ANGLE OF TWIST. THEREFORE, Ф = Σ TiLi/GiIpi

IN CASE IF THE TORQUE OR THE CROSS SECTION CHANGES CONTINUOUSLY ALONG THE AXIS OF THE BAR, THE SUMMATION FORMULA HAS TO BE CHANGED BY INTEGRAL OF THE WHOLE SHAFT. FOR THIS PARTICULAR CASE A TAPERED BAR SUBJECTED TO A TORQUE OF INTENSITY OF “q” PER UNIT DISTANCE ALONG THE AXIS OF THE BAR CAN BE VISUALIZED. THE TORQUE Tx AT A CROSS SECTION LOCATED AT DISTANCE “x” FROM THE END OF THE BAR MAY BE FOUND BY STATICS. NOW THE DIFFERENTIAL ANGLE OF ROTATION FOR AN ELEMENT OF LENGTH “dx” IS GIVEN BY dΦ = Tx dx / G Ipx

NOW THE TOTAL ANGLE OF TWIST BETWEEN THE TWO ENDS OF THE BAR CAN BE CALCULATED BY INTEGRATING THE DIFFERENTIAL ANGLE OF TWIST FOR THE WHOLE LENGTH OF THE BAR Ф = ∫ dФ = ∫ Txdx/G Ipx THIS EQUATION CAN EQUALLY BE USED FOR BOTH SOLID AND HOLLOW BARS OF CIRCULAR CROSS SECTION. THE INTEGRAL CAN BE EVALUATED IN ANALYTICAL FORM IN SOME CASES OTHERWISE IT MUST BE EVALUATED BY NUMERICAL METHODS.

DESIGM PROBLEM: THE ELECTRIC MOTOR EXERTS 500 N–m TORQUE ON THE ALUMINUM SHAFT ABCD WHEN IT IS ROTATING AT A CONSTANT SPEED. THE TORQUES EERTED ON PULLEYS B AND C ARE 200 AND 300 N-m RESPECTIVELY. DETERMINE THE ANGLE OF TWIST BETWEEN (a) B & C, (b) B & D. HINTS: Ф = Σ TiLi/GiIpi

DESIGM PROBLEM: THE DESIGN SPECIFICATONS OF A 1.2 m LONG SOLID TRANSMISSION SHAFT REQUIRES THAT THE ANGLE OF TWIST OF THE SHAFT NOT EXCEED 4º WHNE A TORQUE OF 750 N-m IS APPLIED. DETERMINE THE REQUIRED DIAMETER OF THE SHAFT WHICH IS MADE OF STEEL WITH AB ALLOWABLE SHEARING STRESS OF 90 MPa AND A MODULUS OF RIGIDITY OF 77 GPa. HINTS: Ф = TL/G Ip = 2TL/GΠr⁴, τmax = Tr / Ip = 2Tr / Πr⁴

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

MECHANICS OF MATERIALS - i