# MECHANICS OF MATERIALS - i

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MECHANICS OF MATERIALS - i

TORSION- II

TRANSMISSION OF POWER BY CIRCULAR SHAFTS
THE PRIME PURPOSE OF CIRCULAR SHAFT IS TO TRANSMIT POWER FROM A DEVICE AND MACHINE TO ANOTHER DEVICE AND MACHINE. EXAMPLES MAY BE SUCH AS THE DRIVE SHAFT OF AN AUTOMOBILE THE PROPELLER SHAFT OF A SHIP THE AXLE OF A BICYCLE, AND MANY MORE …… THE EFFICIENCY OF SUCH POWER TRANSMISSION MECHANISM DEPENDS UPON THE PERFORMANCE OF CIRCULAR SHAFT TRANSFERRING THE POWER A VERY COMMON DESIGN PROBLEM IN SUCH TYPE OF STRUCTURES IS THE DETERMINATION OF THE REQUIRED DIMENSIONS OF A SHAFT. a

SO THAT SHAFTS CAN TRANSMIT SPECIFIED AMOUNT OF POWER AT SPECIFIED SPEED OF REVOLUTION WITHOUT EXCEEDING THE ALLOWABLE STRESSES FOR THE MATERIAL. IN ORDER TO CALCULATE THE POWER A SHAFT CAN EFFICIENTLY TRANSMIT, LET US SUPPOSE THAT A MOTOR-DRIVEN SHAFT IS ROTATING AT AN ANGULAR SPEED “ω”. THE SHAFT IS TRANSMITTING A TORQUE “T”. IN GENERAL THE WORK DONE BY ANY TORQUE OF CONSTANT MAGNITUDE IS EQUAL TO THE PRODUCT OF THE TORQUE AND THE ANGLE THROUGH WHICH IT ROTATES AND IS GIVEN BY THE FOLLOWING RELATIONSHIP: W = T Φ a

WE KNOWN THAT POWER P CAN BE DEFINED AS THE TIME RATE AT WHICH THE WORK IS DONE AND, HENCE IS GIVEN AS FOLLOWS: P = dW/dt = T dΦ/dt THE RATE OF CHANGE dΦ/dt OF THE ANGULAR DISPLACEMENT Φ IS ALSO CALLED ANGULAR SPEED “ω” IN RADIANS / SEC, THEREFORE, P = Tω IT IS ALSO KNOWN THAT ANGULAR SPEED IS ALSO EXPRESSED AS THE FREQUENCY “f” OF REVOLUTION , OR NUMBER OF REVOLUTIONS PER UNIT TIME. THE UNIT OF FREQUENCY HERTZ (Hz) IS DEFINED AS ONE REVOLUTION PER SECOND. ω = 2∏f a

HENCE WE CAN WRITE THAT ANGULAR SPEED FOR ONE REVOLUTION, 2∏ RADIANS,
THEREFORE, THE EXPRESSION FOR POWER CAN ALSO BE DEFINED AS P = 2∏f T FREQUENCY IS MORE COMMONLY EXPRESSED AS THE NUMBER OF REVOLUTIONS PER MINUTE (rpm), GIVEN AS f = n/60 P = 2∏nT/60 POWER “P” IS TAKEN IN WATTS, TORQUE “T” IS TAKEN IN NEWTON-METER AND “n” IS IN rpm. a

ONE HORSE POWER (HP) IS EQUAL TO ABOUT 746 WATTS.
OFTEN POWER IS MORE FREQUENTLY EXPRESSED IN TERMS OF HORSE POWER (HP) WHICH IS EQUAL TO 550 FT-POUND PER SECOND. THEREFORE, HP = 2∏nT/33000 ONE HORSE POWER (HP) IS EQUAL TO ABOUT 746 WATTS. THE EQUATION DERIVED ABOVE RELATE TRANSMITTED POWER TO THE TORQUE “T” IN THE SHAFT. TORQUE, OF COURSE, IS RELATED TO THE SHEAR STRESSES, SHEAR STRAINS AND ANGLES OF TWIST WHICH CAN BE CALCULATED FROM THE EQUATIONS DERIVED EARLIER. a

DESIGN PROBLEM: POWER OF 65 KW HAS TO BE TRANSMITTED THROUGH THE SHAFT WITH 750 rpm. THE ALLOWABLE SHEAR STRESS IS ABOUT 45 MPA. WHAT SHOULD BE THE DIAMETER OF SHAFT THAT CAN TRANSMIT REQUIRED POWER WITHOUT EXCEEDING THE VALUE OF ALLOWABLE SHEAR STRESS? WHAT WOULD BE DIAMETER IF THE SHAFT ROTATES AT 1500 RPM? HINTS: P = 2∏nT/60, τmax = 16T/ ∏ d³ a

DESIGN PROBLEM: ONE OF THE HOLLOW DRIVE SHAFTS OF A CRUISE SHIP IS 40 m LONG AND ITS OUTER AND INNER DIAMETRS ARE 400 mm AND 200 mm RESPECTIVELY. THE SHAFT IS MADE OF A STEEL FOR WHICH ALLOWABLE SHEAR STRESS IS 60 MPa AND G = 77.2 GPa. KNOWING MAXIMUM SPEED OF ROTATION OF THE SHAFT IS 160 rpm, DETERMINE (a) THE MAXIMUM POWER THAT CAN BE TRANSMITTED BY ONE SHAFT TO ITS PROPELLER, (b) THE CORRESPONDING ANGLE OF TWIST OF THE SHAFT. HINTS: τmax = Tr/ Ip, f = n/60, P = 2∏f T, Ф = TL/G Ip a

a

STATICALLY INDETERMINATE TORSIONAL MEMBER
IN PREVIOUS EXAMPLES THE TORQUES ACTING AT ALL CROSS SECTIONS OF THE MEMBERS CAN BE DETERMINED BY STATIC EQUILIBRIUM. HENCE THESE EXAMPLES ARE CALLED STATICALLY DETERMINATE MEMBERS. IN MORE SIMPLE WORDS IF THE NUMBER OF UNKNOWN TORQUES EXCEEDS THE NUMBER OF APPLICABLE EQUILIBRIUM EQUATIONS, THE SUBJECT TORSIONAL MEMBER IS CALLED THE STATICALLY INDETERMINATE. SIMILARLY, IF THESE MEMBERS ARE CONSTRAINED BY MORE SUPPORTS THAN ARE REQUIRED TO HOLD THEM IN STATIC EQUILIBRIUM, THESE MEMBERS ARE CALLED STATICALLY INDETERMINATE MEMBERS. a

SUCH MEMBERS IN TORSION CAN ONLY BE SOLVED BY SUPPLEMENTING THE EQUILIBRIUM EQUATIONS WITH EQUATIONS PERTAINING TO THE DISPLACEMENT, THAT IS, BY COMPATIBILITY EQUATIONS. AS DISCUSSED EARLIER THAT IN ORDER TO SOLVE SUCH PROBLEMS, TWO METHODS OF STIFFNESS AND FLEXIBILITY WERE USED IN CASE OF AXIALLY LOADED MEMBERS. THESE METHODS CAN ALSO BE USED FOR TORSIONAL MEMBERS. FOR THIS TYPE OF TORSIONAL PROBLEMS ORDINARILY ENCOUNTERED, HOWEVER, ONLY THE FLEXIBILITY METHOD IS USED. a

LET US CONSIDER THE TORSION MEMBER AB WHICH IS FIXED AT TWO ENDS
LET US CONSIDER THE TORSION MEMBER AB WHICH IS FIXED AT TWO ENDS. THIS IS, THEREFORE, A STATICALLY INDETERMINATE MEMBER. IT IS ASSUMED THAT THE MATERIAL IS SAME THROUGHOUT THE BAR. THE BAR HAS DIFFERENT DIAMETERS da AND db IN RESPECTIVE TWO PARTS OF THE BAR, AC AND CB. IT IS LOADED BY A TORQUE “To” AT POINT C. BY THE ANALYSIS OF BAR, THE REACTIVE TORQUES AT THE TWO ENDS, Ta AND Tb, THE MAXIMUM SHEAR STRESSES, AND THE ANGLE OF ROTATION Φc AT THE SECTION WHERE To IS APPLIED, ARE TO BE DETERMINED.

FROM STATIC EQUILIBRIUM, FOLLOWING EQUATION RELATING THE TORQUES AT THREE POINTS ARE OBTAINED:
To = Ta + Tb …………… (1) IN ORDER TO OBTAIN A SECOND EQUATION RELATING TORQUES AT POINT A AND POINT B, A REDUNDANT TORQUE (Tb) IS SELECTED BY REMOVING THE SUPPORT B AND THE CORRESPONDING RELEASED STRUCTURE IS THEN ANALYZED. NOW THE TWO TORQUES To AND Tb ACT AS LOADS ON THIS RELEASED STRUCTURE AND PRODUCE AN ANGLE OF TWIST Φb AT END B.

THIS ANGLE OF TWIST IS EQUAL TO THE ALGEBRAIC SUM OF THE ANGLES OF TWIST OF THE TWO PARTS AC AND CB. THEREFORE Фb = Toa/G Ipa - Tba/G Ipa - Tbb/G Ipb HERE Ipa AND Ipb ARE THE POLAR MOMENTS OF INERTIA OF THE LEFT AND RIGHT SIDE PARTS OF THE BAR. IN FACT THE ANGLE OF ROTATION AT END B IN THE ORIGINAL BAR IS EQUAL TO ZERO, AND HENCE AS PER EQUATION OF COMPATIBILITY Фb = 0. THEREFORE, Toa/G Ipa - Tba/G Ipa - Tbb/G Ipb = 0

Ta = To {bIpa / (aIpb + bIpa)}
BY SIMPLY SOLVING THE EQUATION FOR THE REDUNDANT TORQUE Tb AND SUBSTITUTING IT IN EQUATION (1), WE GET FOLLOWING EQUATIONS: Tb = To {aIpb / (aIpb + bIpa)}, AND Ta = To {bIpa / (aIpb + bIpa)} IF THE BAR HAS A CONSTANT CROSS SECTION, THEN Ipa = Ipb = Ip, & a + b = L THEN ABOVE EQUATIONS SIMPLIFy AS FOLLOWS Ta = To b / L and Tb = To a / L THESE EQUATIONS ARE ANALOGOUS TO THOSE DETERMINED FOR INDETERMINATE AXIALLY LOADED STRUCTURES.

THE MAXIMUM SHEAR STRESSES IN EACH PART, AC & CB, OF THE BAR ARE OBTAINED DIRECTLY FROM THE TORSION FORMULA AS FOLLOW: τac= Ta da / 2Ipa τcb= Tb db / 2Ipb VALUES OF “Ta” AND “Tb” CAN BE SUBSTITUTED FROM PREVIOUS EQUATIONS AND MAXIMUM SHEAR STRESSES CAN BE CALCULATED. IN A SIMILAR WAY THE ANGLE OF ROTATION “Φ” AT POINT C WHERE THE TORQUE IS APPLIED CAN ALSO BE EASILY FOUND.

AS CAN BE ANTICIPATED FROM THE SYMMETRY OF THE BAR AND LOADING.
THIS ANGLE WOULD BE EQUAL TO THE ANGLE OF ROTATION OF EITHER PART AS BOTH ENDS WOULD HAVE THE SAME ANGLE OF ROTATION AS PER COMPATIBILITY EQUATION. THEREFORE, Φc = Ta a/Gipa = Tb b/GIpb = abTo/G(aIpb+bIpa) NOW IF Ipa = Ipb = Ip AND IF a = b = L/2, THE ANGLE OF ROTATION CAN BE CALCULATED AS FOLLOWS: Φc = To L / 4GIp AS CAN BE ANTICIPATED FROM THE SYMMETRY OF THE BAR AND LOADING. THE ABOVE-MENTIONED METHOD IS QUITE GENERAL AND MUST BE FOLLOWED WHEN ANALYZING A STATICALLY INDETERMINATE TORSIONAL SYSTEM BY THE FLEXIBILITY METHOD.

CALCULATION OF ANGLE OF TWIST AND SHEAR STRESS IN COMPOSITE BARS
IF THE TWO PARTS OF THE BAR ARE CONSTRUCTED OF THE SAME MATERIAL, SUCH BAR BEHAVES IN THE SAME MANNER AS IF IT WERE MADE OF ONE PIECE, AND ALL OF THE RELATIONSHIPS DERIVED EARLIER CAN BE USED DIRECTLLY. A COMPOSITE BAR IS MADE OF TWO CONCENTRIC CIRCULAR TORSIONAL BARS OF DIFFERENT MATERIALS WHICH ARE FIRMLY BONDED TOGETHER TO ACT AS A SINGLE MEMBER. IF SUCH BAR IS UNDER STATICALLY INDETERMINATE CONDITION, THEN A MORE DETAILED ANALYSIS IS REQUIRED. FOR A SUCH CASE, FOLLOWING ASSUMPTIONS HAVE TO BE MADE: a

Ga, Gb = SHEAR MODULUS OF ELASTICITY FOR INNER AND OUTER SHAFTS
da, db = DIAMETERS OF INNER AND OUTER SHAFTS Ipa, Ipb = polar moments of inertia for inner and outer shafts IT WOULD ALSO BE ASSUMED THAT TOTAL TORQUE “T” HAS THE FOLLOWING RELATIONSHIP WITH Ta AND Tb OF BOTH SHAFTS T = Ta + Tb IT IS FURTHER ASSUMED THAT ANGLE OF TWIST WOULD BE SAME FOR BOTH PARTS OF THE BAR Φ = TaL/GaIpa = TbL/GbIpb WHERE “L” IS THE LENGTH OF THE BAR. a

DESIGN PROBLEM: A SOLD STEEL SHAFT ABCD HAVING DIAMETER d = 75mm TURNS FREELY IN A BEARING AT D AND IS LOADED AT B AND C BY TORQUES T1 = 2.50 kN. m and T2 = 1.75 kN. m. THE SHAFT IS CONNECTED IN THE GEAR BOX AT A TO GEARS THAT ARE TEMPORARILY LOCKED IN POSITION. DETERMINE THE MAXIMUM SHEAR STRESS IN EARCH PART OF THE SHAFT AND THE ANAGLE OF TWIST AT END D. SUPPOSE L1 =0 .75 m, L2 = 1.25 m AND L3 = 1.50 m. a

ASSIGNMENT N0. 01 A STEEL SHAFT ABC 50 MM DIAMETER IS DRIVEN AT A BY A MOTOR THAT TRANSMITS 50 TO THE SHAFT AT 50 Hz. THE GEARS AT B AND C REMOVE 30 kW AND 20 kW RESPECTIVELY. DISTANCES BETWEEN AB IS 1.0 M AND BETWEEN BC 1.5 M. VALUE OF G MAY BE TAKEN AS 80 GPA. COMPUTE THE MAXIMUM SHEAR STRESS IN THE SHAFT AND THE ANGLE OF TWIST BETWEEN A AND C. a

ASSIGNMENT N0. 02 REFERRING THE PREVIOUS PROBLEM SUPPOSE THE SHAFTS HAVE THREE DIFFERENT DIAMETERS AND ARE MADE OF DIFFERENT MATERIALS. NOW SOLVE THE SAME PROBLEM WITH THIS SUPPOSITION. CALCULATE THE RELATIONSHIPS FOR FOLLOWINGS: TORQUE AT BAR A, TA TORQUE AT BAR B, TB ANGLE OF TWIST AT BAR A ANGLE OF TWIST AT BAR B SHEAR STRESSES AT BAR A SHEAR STRESSES AT BAR B, AND RATIO OF STRESSES BETWEEN OUTER AND INNER BAR a

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

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