Lecture Strength of Materials I: Chapter 6 - PhD. Tran Minh Tu

Lecture Strength of Materials I - Chapter 6: Torsion. The following will be discussed in this chapter: Introduction, torsional loads on circular shafts, strength condition and stiffness condition, statically indeterminate problem, strain energy, examples.

STRENGTH OF MATERIALS 1/10/2013 TRAN MINH TU - University of Civil Engineering, Giai Phong Str 55, Hai Ba Trung Dist Hanoi, Vietnam CHAPTER TORSION 1/10/2013 Contents 6.1 Introduction 6.2 Torsional Loads on Circular Shafts 6.3 Strength Condition and stiffness condition 6.4 Statically Indeterminate Problem 6.5 Strain Energy 6.6 Examples Home’s works 1/10/2013 6.1 Introduction 1/10/2013 6.1 Introduction 1/10/2013 6.1 Introduction Torsion members – the slender members subjected to torsional loading, that is loaded by couple that produce twisting of the member about its axis Examples – A torsional moment (torque) is applied to the lug-wrench shaft, the shaft transmits the torque to the generator, the drive shaft of an automobile • Torsional Loads on Circular Shafts: the torsional moment or couple A F B x C Q1 z y 1/10/2013 Q2 T t T t 6.1 Introduction  Internal torsional moment diagram • Using method of section • Sign convention of Mz - Positive: clockwise - Negative: counterclockwise M z 0 Mz > Mz = y y z z x 1/10/2013 x 6.2 Torsion of Circular Shafts 6.2.1 Simplifying assumptions 1/10/2013 6.2 Torsion of Circular Shafts => In the cross-section, only shear stress exists 6.2.2 Compatibility • Consider the portion of the shaft shown in the figure • CD – before deformation • CD’ – after deformation - From the geometry DD '   d   dz => The Shear strain:  d  dz - d – the angle of twist - Following Hooke’s law: 1/10/2013    G  G  d dz 6.2 Torsion of Circular Shafts 6.2.3 Equilibrium d d M z      dA  G  dA  G Ip  dz A dz A d M z    – the rate of twist dz GI p 6.2.3 Torsion formulas – Shearing stress Mz – internal torsional moment Mz    Ip 1/10/2013 Ip – polar moment of inertia  – radial position 10 6.3 Strength Condition – Stiffness condition  Strength condition:    0  max  Mz    Wp - Determine experimentally 0 n      - Third strength hypothesis      - Fourth strength hypothesis  Stiffness condition:  max 1/10/2013  Mz       GI   p max  rad / m  14 6.3 Strength Condition – Stiffness condition  Three main problems: For a circular shaft:  max Mz Mz           max 0.2D 0.1GD Investigating the strength condition, (stiffness condition)  max Mz     ??? 0.2D Determine the diameter of circular cross-section Mz D 0.2  Determine the maximum torque 1/10/2013 M z  0.2  D3 15 6.4 Statically Indeterminate Problem AD =  AD   AB   BD 1/10/2013 AB z AB p BD z BD p M a M 2a   GI GI 32 MA  M 33 MD M B a d A (2) MD  M a  M D 2a  AD   4 G  0,1  2d  G  0,1 d MD  M; 33 MA 2d • Assume that the reactions at the fixed ends MA, MD are shown in the figure • Equilibrium: MA + MD = M (1) • Compatibility condition: D 2a CD M M BD z AB z  MD  MD  M Mz MD D z M/33 0 Mz 32M/33 16 6.5 Strain Energy • For a shaft subjected to a torsional load, 2 U   xy 2G dV   T  2GJ dV • Setting dV = dA dx, T   U   dA dx    dA dx 2  2GJ 2GJ  A  0A L  xy  T J T 2 L L T2  dx 2GJ • In the case of a uniform shaft, T 2L U 2GJ 1/10/2013 17 6.6 Example Problem 1/10/2013 3M M B C 2a D A Circular shaft made from two segments, each having diameter of D and 2D The Shaft is subjected to the torques shown in the figure Draw the internal torsional moment diagram Determine the maximum shearing stress Determine the angle of twist of the end D with respect to B With M=5kNm; a=1m; D=10cm; G=8.103 kN/cm2 2D • D a 18 6.6 Example Problem M CD z B 2a M D a 3M  3M  15kNm MzCD Segment BC  z2  2a  BC z C D Internal 2D torsional moment diagram Segment CD  z1  a  3M M z1 M 3M  2M  10kNm MzBC z2 a 15 10 1/10/2013 Mz kNm 19 6.6 Example Problem 3M M   max CD max BC  M zBC 0,  D  B C 2a 10  102   0,625(kN / cm2 ) 0,  20 10   max  7,5(kN / cm2 ) D M zCD 15  102    7,5(kN / cm2 ) 3 0,2 D 0,2 10 2D Maximum shearing stress D a 15 Mz kNm Angle of twist of end D  D   BC  CD M zCD  a M zBC  2a D   CD GI p GI pBC 15  102 102 10 102  102 D    0,02(rad ) 4  10  0,110 10  0,1 20 1/10/2013 20 6.6 Example Problem 1/10/2013 21 6.6 Example Problem 1/10/2013 22 6.6 Example Problem 1/10/2013 23 Homework 1/10/2013 24 Homework 1/10/2013 25 Homeworks 1/10/2013 26 1/10/2013 27 THANK YOU FOR ATTENTION ! 1/10/2013 28 ... moment of inertia  – radial position 10 6. 2 Torsion of Circular Shafts - Maximum shearing stress  max Mz Mz  R Ip Wp - Wp - Section modulus of torsion – Angle of twist c  a b  O A B L From 6. 2.3:... y z z x 1/10/2013 x 6. 2 Torsion of Circular Shafts 6. 2.1 Simplifying assumptions 1/10/2013 6. 2 Torsion of Circular Shafts => In the cross-section, only shear stress exists 6. 2.2 Compatibility.. .CHAPTER TORSION 1/10/2013 Contents 6. 1 Introduction 6. 2 Torsional Loads on Circular Shafts 6. 3 Strength Condition and stiffness condition 6. 4 Statically Indeterminate Problem 6. 5 Strain