Strength of materials - Chapter 9: Torsion - Nguyễn Sỹ Lâm

9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS
• Interested in stresses and strains of circular shafts subjected to twisting couples or torques 

Turbine exerts torque T on the shaft 

Shaft transmits the torque to the generato 

Generator creates an equal and opposite torque T 

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  1. CHAPTER 9: TORSION 9.0 Introduction 9.1 Torsional loads on circular shafts 9.2 Torsion of noncircular members 9.3 Helical spring under axial load
  2. 9.1 INTRODUCTION
  3. 9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS NET TORQUE DUE TO INTERNAL STRESSES • Net of the internal shearing stresses is an internal torque, equal and opposite to the applied torque, T dF  dA • Although the net torque due to the shearing stresses is known, the distribution of the stresses is not • Distribution of shearing stresses is statically indeterminate – must consider shaft deformations • Unlike the normal stress due to axial loads, the distribution of shearing stresses due to torsional loads can not be assumed uniform.
  4. 9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS SHAFT DEFORMATIONS • From observation, the angle of twist of the shaft is proportional to the applied torque and to the shaft length.   T   L • When subjected to torsion, every cross-section of a circular shaft remains plane and undistorted. • Cross-sections for hollow and solid circular shafts remain plain and undistorted because a circular shaft is axisymmetric. • Cross-sections of noncircular (non- axisymmetric) shafts are distorted when subjected to torsion.
  5. 9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS STRESSES IN ELASTIC RANGE • Multiplying the previous equation by the shear modulus, G G c max From Hooke’s Law,  G , so   c max The shearing stress varies linearly with the J 1 c4 2 radial position in the section. • Recall that the sum of the moments from the internal stress distribution is equal to the torque on the shaft at the section,  2  T  dA max dA max J c c • The results are known as the elastic torsion formulas, J 1 c4 c4 2 2 1 Tc T  and  max J J
  6. 9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS TORSIONAL FAILURE MODES • Ductile materials generally fail in shear. Brittle materials are weaker in tension than shear. • When subjected to torsion, a ductile specimen breaks along a plane of maximum shear, i.e., a plane perpendicular to the shaft axis. • When subjected to torsion, a brittle specimen breaks along planes perpendicular to the direction in which tension is a maximum, i.e., along surfaces at 45o to the shaft axis.
  7. 9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS EXAMPLES 9.01 SOLUTION: • Cut sections through shafts AB and BC and perform static equilibrium analysis to find torque loadings M x 0 6kNm TAB M x 0 6kNm 14kNm TBC TAB 6kNm TCD TBC 20kNm
  8. 9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS ANGLE OF TWIST IN ELASTIC RANGE • Recall that the angle of twist and maximum shearing strain are related, c  max L • In the elastic range, the shearing strain and shear are related by Hooke’s Law,  Tc  max max G JG • Equating the expressions for shearing strain and solving for the angle of twist, TL  JG • If the torsional loading or shaft cross-section changes along the length, the angle of rotation is found as the sum of segment rotations T L   i i i JiGi
  9. 9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS STATICALLY INDETERMINATE SHAFTS • Given the shaft dimensions and the applied torque, we would like to find the torque reactions at A and B. • From a free-body analysis of the shaft, TA TB 90lb ft which is not sufficient to find the end torques. The problem is statically indeterminate. • Divide the shaft into two components which must have compatible deformations, TAL1 TBL2 L1J2  1 2 0 TB TA J1G J2G L2J1 • Substitute into the original equilibrium equation, L1J2 TA TA 90lb ft L2J1
  10. 9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS EXAMPLE 9.02 SOLUTION: • Apply a static equilibrium analysis on • Apply a kinematic analysis to relate the two shafts to find a relationship the angular rotations of the gears between TCD and T0 rBB rCC  M B 0 F 0.875in. T0 rC 2.45in. B C C  MC 0 F 2.45in. TCD rB 0.875in. TCD 2.8T0 B 2.8C
  11. 9.1 TORSIONAL LOADS ON CIRCULAR SHAFTS DESIGN OF TRANSMISSION SHAFTS • Principal transmission shaft • Determine torque applied to shaft at performance specifications are: specified power and speed, - power P T 2 fT - speed P P T  2 f • Designer must select shaft material and cross-section to • Find shaft cross-section which will not meet performance specifications exceed the maximum allowable without exceeding allowable shearing stress, shearing stress. Tc  max J J T c3 solid shafts c 2 max J 4 4 T c2 c1 hollow shafts c2 2c2 max
  12. 9.2 TORSION OF NONCIRCULAR MEMEBERS TORSION OF NONCIRCULAR MEMBERS • Previous torsion formulas are valid for axisymmetric or circular shafts • Planar cross-sections of noncircular shafts do not remain planar and stress and strain distribution do not vary linearly • For uniform rectangular cross-sections, T TL   max 2 3 c1ab c2ab G • At large values of a/b, the maximum shear stress and angle of twist for other open sections are the same as a rectangular bar.
  13. 9.2 TORSION OF NONCIRCULAR MEMEBERS THIN-WALLED HOLLOW SHAFTS • Summing forces in the x-direction on AB,  Fx 0  A t A x  B t B x  t  t t q shear flow A A B B shear stress varies inversely with thickness • Compute the shaft torque from the integral of the moments due to shear stress dM0 p dF p t ds q pds 2q dA T dM0 2q dA 2qA T  2tA • Angle of twist (accepted, from energy method) TL ds  4A2G t
  14. 9.2 TORSION OF NONCIRCULAR MEMEBERS EXAMPLE 9.03 SOLUTION: • Find the corresponding shearing stress with each wall thickness • Determine the shear flow through the tubing walls with a uniform wall thickness, q 1.335kip in.  t 0.160in.  8.34ksi with a variable wall thickness 1.335kip in.   A 3.84in. 2.34in. 8.986in.2 AB AC 0.120in. T 24kip -in. kip q 1.335  AB BC 11.13ksi 2A 2 in. 2 8.986in. 1.335kip in.   BD CD 0.200in. BC CD 6.68ksi
  15. 9.3 HELICAL SPRING UNDER AXIAL LOAD INTRODUCTION
  16. 9.3 HELICAL SPRING UNDER AXIAL LOAD INTRODUCTION
  17. 9.3 HELICAL SPRING UNDER AXIAL LOAD BASIC FORMULAS FOR SPRING d : Wire diameter Gd 4 D : Mean coil diameter k 8nD3 n : Active coils D L : Free length 0.25 8PD d 흀 : Deflection  K K d 3 D P : Load 1 d k : Spring constant (stiffness) G : Shear modulus 8PDn  4 τ : Torsional corrected stress Gd K : Application correction factor OR