The problem of transmitting a torque or rotary motion from one plane to another is frequently encountered in machine design.

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1 CHAPER ORSION

2 ORSION orsion refers to the twisting of a structural member when it is loaded by moments/torques that produce rotation about the longitudinal axis of the member he problem of transmitting a torque or rotary motion from one plane to another is frequently encountered in machine design. Normally circular bars are used for such transmissions chiefly because, in these bars, a plane section before twisting remains plane after twisting. 110

3 Assumption to determining the relationship of the shearing stress in circular shaft subjected to torsions: the material of the shaft is homogeneous the maximum shearing stress in the shaft is within the elastic limit the twist remains uniform along the whole length of the shaft the normal cross-section of the shaft which are plane and circular before the twist remain same after the twist the straight radial line of any cross section of the shaft remain straight. the distance between any two cross section of the shaft remain the same torques are applied on planes that are perpendicular to the axis of the shaft 111

4 orsional Deformation of Circular Bars Consider a bar of circular cross-section twisted by couples at the ends. Because the bar is subjected to torsion only, it is said to be in pure torsion. Assuming that the end B is fixed, then the torque will cause end A to rotate through a small angle Ф, known as the angle of twist. hus the longitudinal line AB on the surface of the bar will rotate through a small angle to position A'B 11

5 Since the ends of the element remain planar, the shear strain is equal to angle of twist,. It follows that BB' L r or According to Hooke s law, for linear elastic materials, shear stresses are proportional to shear strains and the constant of proportionality is the modulus of rigidity, G. Hence r L G G r G L Gr L r G L 113

6 orsion Formula: Relationship between and o determine the relationship between the applied torque and the stresses it produces, we consider equilibrium of the internal forces and the externally applied torque,. Considering an elemental area da within an elemental ring of thickness dr situated at radius r from the centre: df= x da df= x x.dx x x x r x r df x x dx r dm xdf x x x dx r 3 x dx r 3 M x dx r 11

7 Moments (M) from the internal stress distribution is equal to the torque (). L G r r dx x dx x r dx x r = Momen Luas Kedua Kutub/ the polar moment of inertia of the cross-sectional area

8 Polar moment of inertia for some structure i) Solid shaft ii) Hollow shaft d x 0 x d 3 3 dx d 0 max R D x d x 3 D 3 dx D d d max min R r Computation of Angle of wist L G L G 115

9 Example: a) Determine the torque which causes a maximum shearing stress of 70 MPa in the steel cylindrical shaft shown. b) Determine the maximum shearing stress caused by a torque of magnitude = 800 Nm.

10 61.67Nm r r 3 d r 7 6 max 7 max Solution (a) Solution (b) 87.3M Pa r 3 d r 7 max 7 max Solution:

11 Example: Knowing that the internal diameter of the hollow shaft shown is d = 3 mm, determine the maximum shearing stress caused by a torque of magnitude = 1.0 knm.

12 Solution: 7 i o o max i i o o m d d 3 r m 11.5mm 3 r 0.03m, 3mm d 0.0m 0mm 0 r 0.0m, 0mm d 89.3MPa m m Nm max

13 Example:

14 7.Nm 0.0m m Nm 10 5 r r m d d m 0.03m r 0.03m, 30mm d 0.0m 0.0m r 0.0m, 0mm d 7 6 o max o max 7 i o i i o o Solution:

15 Example:

16 6 i o o max i i o o m d d 3 r 0.0m 0.0m r 0.0m, 0mm d 0.03m 0.06m r 0.06m, 60mm d 70.6MPa m m 00Nm r 6 o max Solution: 0.08m r r Nm Nm r r 00Nm Nm r r r, r Solution (a) Solution (b)

17 Non Uniform orsion Non-uniform torsion arises when a bar made up of two or more segments of different diameters, materials or shape is subjected to torsion at several cross-sections. 117

18 EXAMPLE: he solid brass rod AB (G = 39 GPa) is bonded to the solid aluminium rod BC(G = 7 GPa). Determine the angle of twist (a) at B, (b) at A.

19 SOLUION: 180Nm C c C C 180Nm 0 180Nm A B AB AB 180Nm 180Nm 0 180Nm A AB BC 180Nm B BC BC 180Nm 0 180Nm A

20 SOLUION: G L L G rad Pa (0.03) Nm G L 9 AB AB rad Pa (0.036) Nm G L 9 BC BC BC AB A

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24 EXAMPLE: A 0mm diameter hole is drilled 3 m deep into the steel shaft. When the two torque are applied, determine (a) the maximum shear stress in the shaft, and (b) the angle of rotation of the free end of the shaft. Use G=83GPa for steel. ( AB =9.31MPa, BC =58.77MPa) A B C

25 SOLUION: ) 10 )(83 10 ( ) 10 )(83 10 ( G L G L 9.31M Pa r 58.77M Pa r m ) 0.0 ( di do m ) ( d 3 A A BC AB A BC AB A AB AB BC BC 6 BC 6 AB 000Nm BC 000Nm 6000Nm AB

26 EXAMPLE: Four pulleys are attached to the 50mm diameter aluminium shaft. If torques are applied to the pulleys as shown in the figure, determine the angle of rotation of pulley D relative to pulley A. Use G=8GPa for aluminium. Ans:6.3, clockwise when viewed from D toward A

27 SOLUION: AB AB 600Nm 0 600Nm BC BC 900Nm 600Nm 0 300Nm CD CD 1100Nm 900Nm 600Nm 0 800Nm BC AB CD D D D D AB BC CD L L G AB G BC L G CD (800 )

28 EXAMPLE: A solid steel shaft has a diameter for m of its length and a 50mm diameter for the remaining 1m of its length, as shown in the figure. A 100 mm long pointer CD is attached to the end of the shaft. he shaft is attached to a rigid support at the left end and is subjected to a 16 knm torque at the right end of the 100 mm section and a knm torque at the right end of the 50mm section. he modulus of rigidity G of the steel is 80GPa. Determine: (a)he maximum shearing stress in the 50 mm section of the shaft. (b)he maximum shearing stress in the 100mm section of the shaft.

29 SOLUION: AB BC 61.1M Pa 3 ) ( r 16.97M Pa 3 ) ( r 1000Nm 16000Nm 000Nm 0 000Nm 16000Nm 000Nm 0 000Nm BC BC BC BC AB AB BC BC

30 EXAMPLE: Shaft BC is hollow with inner and outer diameters of 90 mm and 10 mm, respectively. Shafts AB and CD are solid of diameter d. For the loading shown, determine (a) the minimum and maximum shearing stress in shaft BC, (b) the required diameter d of shafts AB and CD if the allowable shearing stress in these shafts is 65 MPa.

31 SOLUION: Cut sections through shafts AB and BC and perform static equilibrium analysis to find torque loadings Apply elastic torsion formulas to find minimum and maximum stress on shaft BC Given allowable shearing stress and applied torque, invert the elastic torsion formula to find the required diameter 118

32 SOLUION: Cut sections through shafts AB and BC and perform static equilibrium analysis to find torque loadings M AB x 0 6kN m 6kN m CD AB 0 6kN m 1kN m BC M x 0kN m BC 119

33 Apply elastic torsion formulas to find minimum and maximum stress on shaft BC max min max min m c c BCc 86. MPa c c 1 6.7MPa 0kNm0.060m min 86.MPa mm 60mm m max min 86. MPa 6.7 MPa 10

34 Given allowable shearing stress and applied torque, invert the elastic torsion formula to find the required diameter c max c c c 3 m 65MPa d 6kNm 3 c c 77.8mm d c 77.8mm 11

35 EXAMPLE: Under normal operating conditions, the electric motor exerts a torque of. knm on shaft AB. Knowing that each shaft is solid, determine the maximum shearing stress (a) in shaft AB (b) in shaft BC (c) in shaft CD 1

36 Power ransmission Power transmitted by a circular shaft, P - depends upon the magnitude of the torque and the speed of rotation,. Power is the rate at which work is done. P = where, P is power (watt) is torque (Nm) is angular speed (rad/s) he angular speed is often expressed as the frequency, f, of rotation, i.e. the number of revolutions per second. his means that = f where f is measured in Hetz (Hz) = s -1 ; 13

37 Hence, P = f Also, may be expressed in revolutions per minute (rpm), denoted by N. P f N 60 (where N is in rpm) Example 5.6 How much power P(kW) may be transmitted by a solid circular shaft of diameter 80 mm turning at 0.75 Hz if the shear stress is not to exceed 0 MPa? 1

38 EXAMPLE: A 1.6-m long tubular steel shaft (G = 77 GPa) of mm outer diameter d 1 and 30 mm inner diameter d is to transmit 10 kw between a turbine and a generator. Knowing that the allowable shearing stress is 65 MPa, determine the minimum frequency at which the shaft may rotate. 15

39 SOLUION:

40 EXAMPLE: One of two hollow drive shafts of an ocean liner is 38 m long, and its outer and inner diameters are 00 mm and 00 mm respectively. he shaft is made of a steel for which all = 60 MPa and G = 77 GPa. Knowing that the maximum speed of rotation of the shaft is.75 Hz, determine (a) the maximum power that can be transmitted by the one shaft to its propeller, (b) the corresponding angle of twist of the shaft.

41 SOLUION:

42 EXAMPLE:

43 SOLUION:

44 EXAMPLE: A tubular shaft having an inner diameter of 30mm and an outer diameter of mm is to be used to transmit 90 kw of power. Determine the frequency of rotation of the shaft so that the shear stress will not exceed 50 MPa.

45 Statically Indeterminate orsional Members When all internal and reaction torques cannot be determined using only equilibrium equations Solution steps for analyzing statically indeterminate torsional members: (i) generate equilibrium equations (F, M) (ii) generate compatibility equations (constrains) (iii) generate torque-displacement relations ( = L / G) (iv) simultaneously solve all equations 16

46 Statically Indeterminate orsional Members (i) (ii) generate equilibrium equations M=0 - A - B =0 generate compatibility equations (constrains) A/B =0 (iii) generate torque-displacement relations ( = L / G) A L G AC BL G (iv) simultaneously solve all equations Here, G is assumed to be constant L=L AC +L BC BC 0 A L L BC and B L L AC

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49 EXAMPLE: he design specifications of 1. m long solid circular transmission shaft require that the angle of twist of the shaft not exceed when a torque of 680 Nm is applied. Determine the required diameter of the shaft, knowing that the shaft is made of a steel with an allowable shearing stress of 83 MPa and a modulus of rigidity of 77 GPa.

50 SOLUION: