CHAPTER 6 TORSION. its inner diameter d d / 2. = mm = = mm. π (122.16) = mm 2

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Opal Reed
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1 CHAPTER 6 TORSION Prolem. A olid circular haft i to tranmit 00 kw at 00 r.p.m. If the hear tre i not to exceed 80 N/mm, find the diameter of the haft. What percentage in aving would e otained if thi haft i replaced y a hollow one, whoe internal diameter i equal to 0.8 of the external diameter, the length, the material and the allowale maximum hear tre eing the ame? N 00 rpm, q 80 N/mm P 00 kw 00 0 watt N-mm/ec NT From the relation P, we get or 00 T T N-mm et d e the required diameter of olid haft. q T R q d d / 6 d i.e d 80 6 or d.6 mm. An. et d e the outer diameter of hollow haft and d it inner diameter d 0.8 d T { (0.8 ) q d d } d / i.e d 80 6 d 5.67 mm d 0.8 d mm. An. Cro ectional area of olid haft (.6) mm Cro ectional area of hollow haft ( d d ) ( ) 00.8 mm % Saving Weight of olid haft Weight of hollow haft 00 Wt. of olid haft 7

2 (.6) ρ 00 ( ) ρ i.e. % Saving 8.8. An. Prolem. A olid circular haft i to e deigned to tranmit.5 kw power at 00 r.p.m. If the maximum hear tre i not to exceed 80 N/mm and the angle of twit i not to exceed per metre length, determine the diameter of the haft. Take modulu of rigidity 80 kn/mm. P.5 kw N-mm/ec. N 00 rpm θ radian. 80 or 000 mm 80 N/mm G 80 0 N/mm P NT 00 T T N-mm et d e the diameter of the haft. d From the conideration of hear tre T d R 80 d d 0.89 mm From the conideration of angle of twit T d G θ d d 5.9 mm Minimum diameter of the haft to e ued i 5.9 mm. An. Prolem. A hollow circular haft m long i required to tranmit 00 kw power when running at a peed of 00 rpm. If the maximum hearing tre allowed in the haft i 80 N/mm and the ratio of inner diameter to the outer diameter i 0.75, find the dimenion of the haft and alo the angle of twit of one end of the haft relative to the other end. Modulu of rigidity of the material i 85 kn/mm. m 000 mm P 00 kw N-mm/ec 8

3 N 00 rpm 80 N/mm and G 85 0 N/mm et d e the outer diameter and d e the inner diameter d 0.75 d, given Now, P NT 00 T i.e T N-mm. From the torion formula, d (0.75 d) R T { } we get, { (0.75) } d 80 6 d d mm. d mm An. Again, from torion formula, T we get, ( ) G θ 85 0 θ 000 θ 0.89 radian Prolem. A olid haft of 00 mm diameter i propoed to e replaced y a hollow haft of external diameter D and internal diameter D/. If the ame power i to e tranmitted at the ame peed and at the ame level of hear tre, determine the diameter D. Diameter of olid haft d 00 mm et T e the torional reitance of olid haft and T h e the torional reitance of hollow haft. et N e revolution of haft per minute and e maximum hear tre permitted. Then T R d and T h 00 6 D D R D { 0.5 } D 6 9

4 Tranmiion of power y olid haft P NT N (00) 6 Tranmiion of power y the hollow haft Given condition i N (00) 6 P h T T h NT N ( 0.5 ) D 6 N ( 0.5 ) D 6 i.e. ( 0.5 ) D (00) D 0. mm An. Prolem 5. A hollow haft i m long and ha outer and inner diameter 00 mm and 50 mm o repectively. If the angle of twit mut not exceed in m and the maximum hearing tre i not to exceed 50 N/mm, find the maximum power that can e tranmitted at 00 rpm. Take modulu of rigidity of the material a 8 kn/mm. 000 mm d 00 mm d 50 mm o θ radian. 50 N/mm 80 G 8 0 N/mm N 00 rpm. From torion formula, T R { } 50 00/ N-mm From the conideration of angle of twit: θ G T { } N-mm NT P N-mm/ec 8.8 kw Thu, the maximum power tranmitted 8.7 kw. An. Prolem 6. A hollow haft ha to tranmit 0 kw power at 80 rpm. The maximum twiting moment may exceed the mean y 0%. Deign a uitale ection, if the permiile tre i 90 kn/mm. The diameter ratio i to e 0.8. What will e the angular twit meaured over a length of m if the modulu of rigidity 8 kn/mm. P 0 kw N-mm/ec N 80 rpm T max. T 90 N/mm G 8 0 N/mm 000 mm et d e outer diameter and d inner diameter of the hollow haft 50

5 or d 0.8 d NT From, P, we get 80 T T N-mm T max N-mm From the conideration of hear tre q T max {d (0.8d ) } R i.e From torion formula T max { (0.8) } d 90 6 d.6 mm d mm G θ 90 d / Tmax θ G { } rad..55 An. Prolem 7. The tepped haft hown in Fig. i ujected to torque at B and D of magnitude 00 N-m and 00 N-m. Find the rotation of free end A. Take modulu of rigidity 80 kn/mm. 6 E D C B A.5 m.0 m 0.5 m 80 mm 0 mm Section in CE Fig. T AB 0, T BC 00 N-m T CD 00 N-m, T DE 0 N-m Polar modulu of ection in the portion, Section in AC BC BC CD CD mm mm 80 5

6 DE DE 08.6 mm 80 Rotation of free end θ θ BC + θ CD + θ DE θ radian. [Note: rotation in portion AB 0] Prolem 8. A haft of length i fixed at end A and B and i ujected to torque T at C a hown in Fig. (a). Show that torque at A i T and at B it i equal to T. Fig. (a) et T A and T B e the reactive torque at A and B. Fig. () From the free ody diagram of AC and CB, T T A + T B... () The angle of twit at C with repect to A mut e the ame a that of C with repect to B. θ CA θ CB TA TB G G T A T... () Sutituting it in eqn. (), we get B + TB T TB + TB TB i.e. T B T Sutituting thi value in eqn. (), we get, T T T A. Prolem 9. A haft of 80 mm diameter i having a concentric ore of diameter 0 mm up to half the length a hown in Fig.. Find the total angle of twit for the loading hown. Take G 80 kn/mm. [Note: T AB kn-m, T BC kn-m, T CD 6 kn-m, θ θ AB + θ BC + θ CD ] 5

7 T C kn-m T kn-m B T kn-m A D C B A 0. m 0. m 0. m Fig. T AB kn-m T BC kn-m, T CD 6 kn-m AB (80 0 ) mm. BC mm and CD mm From torion formula, θ T G, θ θ AB + θ BC + θ CD radian Prolem 0. A teel rod of 5 mm diameter i tightly fitted to a ra tue of internal diameter 5 mm and external diameter 0 mm to form a compoite ar. If the permiile tree in ra and teel are 50 N/mm and 80 N/mm repectively, find the maximum torque the compoite ection can reit. Take G 0 kn/mm and G 80kN/mm. 0 mm 5 mm Fig. d 5 mm d 0 mm d 5 mm 50 N/mm q 80 N/mm G 80 0 N/mm G 0 0 N/mm mm 5

8 I (0 5 ) mm θ θ T T G G G T T G B ince T T If tre in teel govern the capacity T or T 598 N-mm R 5/ T N-mm T T + T N-mm... () If tre in ra govern the capacity, then T q T N-mm R 0 T N-mm T T + T N-mm () From () and () we can conclude that tre in teel govern the torque carrying capacity of the haft and it value i N-mm kn-m. An. Prolem. The rod AB taper from 00 mm to 50 mm diameter in a length of. m. Find the angular rotation of free end when a torque of kn-m i applied. Take G 80 kn/mm. What i the maximum tre induced and what i the percentage error in calculating angular rotation if the ar i treated a of uniform diameter with average value? 00 mm 50 mm. m Fig. 5 r 5 mm r 50 mm G 80 0 N/mm 00 mm T kn-m 0 6 N-mm Total angle of twit + + r r T r r r r G

9 0.09 rad. If the haft i treated a haft of average diameter, d mm T 0 00 θ rad. G % Error 00.. An The maximum tre will e at the maller end. T R or T R 6T 6 0 d N/mm Prolem. A cloe coiled helical pring i required to carry a maximum load of 800 N and to have tiffne of 5 N/mm. The mean diameter i to e 75 mm. The allowale hear tre 00 N/mm. Find the uitale diameter of the wire from which to make the pring and the approximate numer of turn required. W 800 N Stiffne of the pring W δ 5 N/mm Maximum hear tre i given y 75 R 7.5 mm 00 N/mm d?, n? 6WR d d d d.58 m. W Stiffne of pring δ W δ δ ut δ n mm 6WR n Gd δ Gd 80 0 (.58) 6WR (7.5) An. n 7. An. 55

10 Prolem. Find the torional contant of a rectangular ection of ize 00 mm 50 mm. (a) By taking one term in the exact equation () Taking two term in the exact equation (c) By approximate equation (d) What i over etimation of torional contant if it i taken a equal to the polar moment of inertia? (a) By taking one term in exact equation 9 50 d tan h 5 d mm () By taking two term in exact equation tan 50 h tan h tan h mm (c) Approximate equation d Since >.6 equation 6. from the chapter will e ued (d) Over etimation If warping i neglected, Over etimation i d 0.6 d mm. polar moment of inertia I h + I y time. 56

The example of shafts; a) Rotating Machinery; Propeller shaft, Drive shaft b) Structural Systems; Landing gear strut, Flap drive mechanism

The example of shafts; a) Rotating Machinery; Propeller shaft, Drive shaft b) Structural Systems; Landing gear strut, Flap drive mechanism TORSION OBJECTIVES: This chapter starts with torsion theory in the circular cross section followed by the behaviour of torsion member. The calculation of the stress stress and the angle of twist will be