Fourth Edition MECHANICS OF MATERIALS OF Fourth Edition MECHANICS OF MATERIALS CHAPTER Ferdinand P. Beer E. Russell Johnston, Jr. John T. DeWolf Torsion Lecture Notes: J. Walt Oler

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  • MECHANICS OF Fourth Edition

    MECHANICS OF MATERIALS

    CHAPTER

    MATERIALS Ferdinand P. Beer E. Russell Johnston, Jr. John T. DeWolf

    Torsion

    Lecture Notes: J. Walt Oler Texas Tech Universityy

    © 2006 The McGraw-Hill Companies, Inc. All rights reserved.

  • MECHANICS OF MATERIALS

    Fourth Edition

    Beer • Johnston • DeWolf

    Contents

    Introduction

    Torsional Loads on Circular Shafts

    Statically Indeterminate Shafts

    Sample Problem 3.4

    Net Torque Due to Internal Stresses

    Axial Shear Components

    Design of Transmission Shafts

    Stress Concentrations

    Shaft Deformations

    Shearing Strain

    Stresses in Elastic Range

    Plastic Deformations

    Elastoplastic Materials

    Residual StressesStresses in Elastic Range

    Normal Stresses

    Torsional Failure Modes

    Residual Stresses

    Examples 3.08, 3.09

    Torsion of Noncircular Members

    Sample Problem 3.1

    Angle of Twist in Elastic Range

    Thin-Walled Hollow Shafts

    Example 3.10

    © 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 2

  • MECHANICS OF MATERIALS

    Fourth Edition

    Beer • Johnston • DeWolf

    Torsional Loads on Circular Shafts

    • Interested in stresses and strains of circular shafts subjected to twisting

    lcouples or torques.

    • Turbine exerts torque T on the shaft.

    • Generator creates an equal and

    • Shaft transmits the torque to the generator.

    • Generator creates an equal and opposite torque T’.

    © 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 3

  • MECHANICS OF MATERIALS

    Fourth Edition

    Beer • Johnston • DeWolf

    Net Torque Due to Internal Stresses

    • Net of the internal shearing stresses is an internal torque, equal and opposite to the

       dAdFT 

    applied torque,

    • 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 l d t b d if

    © 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 4

    loads cannot be assumed uniform.

  • MECHANICS OF MATERIALS

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    Beer • Johnston • DeWolf

    Axial Shear Components

    • Torque applied to shaft produces shearing stresses on the faces perpendicular to the

    iaxis.

    • Conditions of equilibrium require the existence of equal stresses on the faces of the

    • The existence of the axial shear components

    existence of equal stresses on the faces of the two planes containing the axis of the shaft.

    The existence of the axial shear components is demonstrated by considering a shaft made up of axial slats.

    • The slats slide with respect to each other when equal and opposite torques are applied to the ends of the shaft.

    © 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 5

  • MECHANICS OF MATERIALS

    Fourth Edition

    Beer • Johnston • DeWolf

    Shaft Deformations

    • From observation, the angle of twist,  of the shaft is proportional to the applied torque and to the shaft lengthto the shaft length.

    L

    T

    • 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.

    © 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 6

  • MECHANICS OF MATERIALS

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    Beer • Johnston • DeWolf

    Shearing Strain

    • Consider an interior section of the shaft. As a torsional load is applied, an element on the i t i li d d f i t h b

    tan in radianxy xy 

    interior cylinder deforms into a rhombus.

    • Since the ends of the element remain planar, the shear strain is equal to angle of twist

    • For small value of , It follows that

    the shear strain is equal to angle of twist.

    AA L 

    AA   or AA L

    L       (and are in radians)

    AA  

    AA 

    • Shear strain is proportional to twist and radius

    maxmax and  

    L c

    

    L

    (Shown next page)

    © 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 7

    cL tan AA

    L  

      

  • MECHANICS OF MATERIALS

    Fourth Edition

    Beer • Johnston • DeWolf

    Shearing Strain

    c

    At shaft surface, Max shearing strain

    max c L  

    Shearing strain at a distance  from the shaft axisShearing strain at a distance  from the shaft axis

    Shear strain inside the shaft, L L   

       

    max maxShear strain at the shaft's surface,

    Lc L c

    LL

      

     

      

    max max

    LL c c

        

        

    © 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 8

    maxc  

  • MECHANICS OF MATERIALS

    Fourth Edition

    Beer • Johnston • DeWolf

    Stresses in Elastic Range • Multiplying previous eq. by the shear modulus,

    max  G c

    G  c

    From Hooke’s Law,  G , so

    The shearing stress varies linearly4 2 1 cJ  max

     c

     The shearing stress varies linearly with the radial position in the section.

    F h ll li d

    max max 2

    From or c c      

    • For hollow cylinder

     414221 ccJ  

    2

    1 1 min max

    2

    Substitute cc c

        

    © 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 9

    J = Polar moment of inertia min 1

    max 2

    c c

     

  • MECHANICS OF MATERIALS

    Fourth Edition

    Beer • Johnston • DeWolf

    Stresses in Elastic Range

    • Recall that the sum of the moments from the internal stress distribution is equal to the torque on the shaft at the section

    J c

    dA c

    dAT max2max    

    the torque on the shaft at the section,

    Th lt k th l ti t i

    dAJ  2max c

    • The results are known as the elastic torsion formulas,

    andmax TTc  

       dAdFT 

    and max JJ 

    © 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 10

  • MECHANICS OF MATERIALS

    Fourth Edition

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    Normal Stresses • Elements with faces parallel and perpendicular

    to the shaft axis are subjected to shear stresses only (Element a). Normal stresses, shearing stresses or a combination of both may be found for other orientations (Element b).

    • Consider an element at 45o to the shaft axis,  

    max 0max

    45

    0max0max

    2 2

    245cos2

    o  

    

    

    

    A A

    A F

    AAF

    0 2AA

    • Element a is in pure shear. • Element c is subjected to a tensile stress on

    • Note that all stresses for elements a and c have

    j two faces and compressive stress on the other two.

    © 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 11

    the same magnitude.

  • MECHANICS OF MATERIALS

    Fourth Edition

    Beer • Johnston • DeWolf

    Torsional Failure Modes

    • Ductile materials generally fail in shear. Brittle materials are weaker inshear. 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 planesspecimen breaks along planes perpendicular to the direction in which tension is a maximum, i.e., along surfaces at 45o to the shaft

    © 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 12

    g axis.

  • MECHANICS OF MATERIALS

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  • MECHANICS OF MATERIALS

    Fourth Edition

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  • MECHANICS OF MATERIALS

    Fourth Edition

    Beer • Johnston • DeWolf

    Sample Problem 3.1

    SOLUTION:

    • Cut sections through shafts AB d BC d f iand BC and perform static

    equilibrium analyses to find torque loadings.

    Shaft BC is hollow with inner and outer

    • Apply elastic torsion formulas to find minimum and maximum stress on shaft BC.

    Shaft BC is hollow with inner and outer diameters of 90 mm and 120 mm, respectively. Shafts AB and CD are solid of diameter d. For the loading shown,

    • Given allowable shearing stress and applied torque, invert the elastic torsion formula to find the g ,

    determine (a) the minimum and maximum shearing stress in shaft BC, (b) the required diameter d of shafts AB and CD

    required diameter.

    © 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 15

    if the allowable shearing stress in these shafts is 65 MPa.

  • MECHANICS OF MATERIALS

    Fourth Edition

    Beer • Johnston • DeWolf

    SOLUTION Sample Problem 3.1 SOLUTION: • Cut sections through shafts AB and BC

    and perform static equilibrium analysis to find torque loadingsto find torque loadings.

     

    CDAB

    ABx

    TT

    TM

    

     mkN6

    mkN60     mkN20

    mkN14mkN60

    

    

    BC

    BCx

    T

    TM

    © 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 16

  • MECHANICS OF MATERIALS

    Fourth Edition

    Beer • Johnston • DeWolf

    A l l ti t i f l t Gi ll bl h i t d Sample Problem 3.1 • 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.

          444142 045.0060.022   ccJ mkN665 34max

      MPaTc

    J Tc

     

    46 m1092.13

    22 

       109213

    m060.0mkN20 46

    2 2max

      J

    cTBC m109.38 3

    3 2

    4 2

    c

    ccJ 

    mm8772  cd

    MPa2.86 m1092.13 46

     J

    mm45min1min   c

    mm8.772  cd

    dAB = dBC

    © 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 17

    MPa7.64

    mm60MPa2.86

    min

    2max

    

     c

    MPa7.64

    MPa2.86

    min

    max

  • MECHANICS OF MATERIALS

    Fourth Edition

    Beer • Johnston • DeWolf

    290 N-m

    Fixed end

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  • MECHANICS OF MATERIALS

    Fourth Edition

    Beer • Johnston • DeWolf

    2.8 kN-m

    Free end (T = 0)(T = 0)

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  • MECHANICS OF MATERIALS

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    Beer • Johnston • DeWolf

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  • MECHANICS OF MATERIALS

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  • MECHANICS OF MATERIALS

    Fourth Edition

    Beer • Johnston • DeWolf

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  • MECHANICS OF MATERIALS

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    Beer • Johnston • DeWolf

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  • MECHANICS OF MATERIALS

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    © 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 24

    Why ? Show on next page !!!CCD AB B

    rT T r

     

  • MECHANICS OF MATERIALS

    Fourth Edition

    Beer • Johnston • DeWolf

    Gear CGear C Gear BGear B

    rc=240 mm F

    FTCD rB=80 mm

    BC T=TAB= 900 N-m

    FCD

    TCD = (F)(rc) TAB = 900 = (F)(rB)

    CD ABT TF   C B

    C CD AB

    r r rT T r

    © 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 25

    Br

  • MECHANICS OF MATERIALS

    Fourth Edition

    Beer • Johnston • DeWolf

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  • MECHANICS OF MATERIALS

    Fourth Edition

    Beer • Johnston • DeWolf

    Angle of Twist in Elastic Range • Recall that the angle of twist and maximum

    shearing strain are related,

    L c max and max J

    T J

    Tc   L

    max

    • In the elastic range, the shearing strain and shear are related by Hooke’s Law,

    T 

    max JJ

    max max

    Tc c G JG L     

    • Equating the expressions for shearing strain and solving for the angle of twistsolving for the angle of twist,

    JG TL

    

    The relation shows that, within the elastic range, the angle of twist  is proportional to the torque T applied to the shaft. This is accordance with the experimental evidence. The

    © 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 27

    previous eqs also provides us with a convenient method for determining the modulus of rigidity of a given materials.

  • MECHANICS OF MATERIALS

    Fourth Edition

    Beer • Johnston • DeWolf

    Angle of Twist in Elastic Range

    • If the torsional loading or shaft cross-section changes along the length, the angle of rotation is found as the sum of segment rotationsfound as the sum of segment rotations

     i ii

    ii GJ LT

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  • MECHANICS OF MATERIALS

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    Beer • Johnston • DeWolf

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  • MECHANICS OF MATERIALS

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    Beer • Johnston • DeWolf

    © 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 30

  • MECHANICS OF MATERIALS

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    Beer • Johnston • DeWolf

    (A to B)

    (A to C)

    © 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 31

    See from A to C

  • MECHANICS OF MATERIALS

    Fourth Edition

    Beer • Johnston • DeWolf

    A/B B/C

    A to B B to C

    B/C

    A/B

    A to C

    © 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 32

    A to C

  • MECHANICS OF MATERIALS

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    Beer • Johnston • DeWolf

    © 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 33

  • MECHANICS OF MATERIALS

    Fourth Edition

    Beer • Johnston • DeWolf

    © 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 34

  • MECHANICS OF MATERIALS

    Fourth Edition

    Beer • Johnston • DeWolf

    Sample Problem 3.4

    SOLUTION:

    • Apply a static equilibrium analysis on th t h ft t fi d l ti hithe two shafts to find a relationship between TCD and T0 .

    • Apply a kinematic analysis to relate

    Two solid steel shafts are connected • Find the maximum allowable torque

    on each shaft – choose the smallest.

    the angular rotations of the gears.

    Two solid steel shafts are connected by gears. Knowing that for each shaft G = 77 GPa and that the allowable shearing stress is 55 MPa, determine

    • Find the corresponding angle of twist for each shaft and the net angular rotation of end A

    o eac s a t c oose t e s a est.

    g (a) the largest torque T0 that may be applied to the end of shaft AB, (b) the corresponding angle through which

    rotation of end A.

    © 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 35

    end A of shaft AB rotates.

  • MECHANICS OF MATERIALS

    Fourth Edition

    Beer • Johnston • DeWolf

    Sample Problem 3.4 SOLUTION:

    • Apply a static equilibrium analysis on the two shafts to find a relationship

    • Apply a kinematic analysis to relate the angular rotations of the gears.the two shafts to find a relationship

    between TCD and T0 . the angular rotations of the gears.

    60

    C CD AB

    B

    rT T r

    T T

    © 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 36

    22 2.73

    CD O

    CD O

    T T

    T T

  • MECHANICS OF MATERIALS

    Fourth Edition

    Beer • Johnston • DeWolf

    Fi d th T f th i Fi d th di l f t i t f h

    Sample Problem 3.4 • Find the T0 for the maximum

    allowable torque on each shaft – choose the smallest.

    • Find the corresponding angle of twist for each shaft and the net angular rotation of end A.

    D = 0 (Fix)

    8.05°

    10.2°

    61.8

    © 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 37

    inlb5610 T

  • MECHANICS OF MATERIALS

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    D = 0 (Fix)

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  • MECHANICS OF MATERIALS

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    Beer • Johnston • DeWolf

    © 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 39

  • MECHANICS OF MATERIALS

    Fourth Edition

    Beer • Johnston • DeWolf

    Design of Transmission Shafts

    • Determine torque applied to shaft at specified power and speed,

    fTTP 2

    f PPT

    fTTP

    

    

    2

    2

    

    

    • Designer must select shaft • Find shaft cross-section which will not

    exceed the maximum allowable shearing stress,

    • Designer must select shaft material and cross-section to meet performance specifications without exceeding allowable g

     h ftlid3

    max

    TJ J

    Tc 

    without exceeding allowable shearing stress.

    Angular velocity 2 f 

    1 Hz(rpm) Hz 60 rpm

    f  

     

       shafts hollow 2

    shaftssolid 2

    max

    4 1

    4 2

    22

    max

    3

     

    Tcc cc

    J

    c c

    

    Angular velocity, 2 f 

    © 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 40

    p 2 max22 cc

  • MECHANICS OF MATERIALS

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    Watts = N-m/s

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  • MECHANICS OF MATERIALS

    Fourth Edition

    Beer • Johnston • DeWolf

    Statically Indeterminate Shafts

    • Given the shaft dimensions and the applied torque, we would like to find the torque reactions at A and B.at A and B.

    • From a free-body analysis of the shaft,

    which is not sufficient to find the end torques. The problem is statically indeterminate.

    • Divide the shaft into two components which• Divide the shaft into two components which must have compatible deformations,

    • Substitute into the original equilibrium equation,

    © 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 42

  • MECHANICS OF MATERIALS

    Fourth Edition

    Beer • Johnston • DeWolf

    Stress Concentrations

    • The derivation of the torsion formula,

    J Tc

    max

    assumed a circular shaft with uniform cross-section loaded through rigid end plates.

    • The use of flange couplings, gears, and pulleys attached to shafts by keys in keyways and cross-section discontinuities

    • Experimental or numerically determined t ti f t li d

    keyways, and cross section discontinuities can cause stress concentrations.

    J TcKmax

    concentration factors are applied as

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  • MECHANICS OF MATERIALS

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    Beer • Johnston • DeWolf

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  • MECHANICS OF MATERIALS

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    © 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 45

  • MECHANICS OF MATERIALS

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    Plastic Deformations • With the assumption of a linearly elastic material,

    J Tc

    max

    • If the yield strength is exceeded or the material has a nonlinear shearing-stress-strain curve, this expression does not hold.

    • Shearing strain varies linearly regardless of material properties. Application of shearing-stress-strain curve allows determination of stress distribution

    • The integral of the moments from the internal stress distribution is equal to the torque on the shaft at the

    ti

    curve allows determination of stress distribution.

       cc

    ddT 0

    2

    0 22 

    section,

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  • MECHANICS OF MATERIALS

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    Elastoplastic Materials • At the maximum elastic torque,

    YYY cc JT  32

    1 c

    L Y Y

     

    • As the torque is increased, a plastic region ( ) develops around an elastic core ( )Y  Y

    Y 

      

     YL

      

       

     

     

       

      3

    3

    4 1

    3 4

    3

    3

    4 13

    3 2 11

    c T

    c cT YYYY

    

      YLY 

     cc

      

       

      3

    3

    4 1

    3 4 1

     Y

    YTT

    • As , the torque approaches a limiting value,0Y torque plastic TT YP  3

    4

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  • MECHANICS OF MATERIALS

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    Beer • Johnston • DeWolf

    Residual Stresses

    • Plastic region develops in a shaft when subjected to a large enough torque.

    Wh th t i d th d ti f t• When the torque is removed, the reduction of stress and strain at each point takes place along a straight line to a generally non-zero residual stress.

    • On a T- curve, the shaft unloads along a straight line to an angle greater than zero.

    • Residual stress found from principle of superposition.p p p p

    © 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 48

      0 dA J

    Tc m 

  • MECHANICS OF MATERIALS

    Fourth Edition

    Beer • Johnston • DeWolf

    Example 3.08/3.09

    SOLUTION:

    • Solve Eq. (3.32) for Y/c and evaluate the elastic core radius.

    • Solve Eq. (3.36) for the angle of twist.

    A solid circular shaft is subjected to a torque at each end. Assuming that the shaft is made of an l l i i l i h

    mkN6.4 T • Evaluate Eq. (3.16) for the angle

    which the shaft untwists when the torque is removed. The permanent t i t i th diff b t thelastoplastic material with

    and determine (a) the radius of the elastic core, (b) the angle of twist of the shaft When the

    MPa150Y GPa77G

    • Find the residual stress distribution by

    twist is the difference between the angles of twist and untwist.

    angle of twist of the shaft. When the torque is removed, determine (c) the permanent twist, (d) the distribution of residual stresses

    a superposition of the stress due to twisting and untwisting the shaft.

    © 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 49

    of residual stresses.

  • MECHANICS OF MATERIALS

    Fourth Edition

    Beer • Johnston • DeWolf

    SOLUTION Example 3.08/3.09 SOLUTION: • Solve Eq. (3.32) for Y/c and

    evaluate the elastic core radius 13 

    • Solve Eq. (3.36) for the angle of twist

      YY31

    341 3 3

    4 1

    3 4

     

      

     

     

       

     

    Y

    YY Y T

    T cc

    TT 

     m1025 321421  cJ    

      49- 3

    Pa1077m10614 m2.1mN1068.3

     

    

    

     

    Y Y

    YY

    JG LT

    cc

      m10614

    m1025

    49

    22

    

     

    Y Y

    Y Y

    JTcT

    cJ

    

       

    o3 3

    3

    8.50rad103.148rad104.93

    rad104.93

    Pa1077m10614

     

    

     

    Y

       m1025

    m10614Pa10150 3

    496

     

    

    Y

    YY

    T

    c T

    J  630.0

    o50.8

    mkN68.3 

    630.0 683 6.434

    3 1

      

       Y

    © 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 50

    68.3 c

    mm8.15Y

  • MECHANICS OF MATERIALS

    Fourth Edition

    Beer • Johnston • DeWolf

    Example 3.08/3.09

    • Evaluate Eq. (3.16) for the angle which the shaft untwists when the torque is removed The

    • Find the residual stress distribution by a superposition of the stress due to twisting and untwisting the shaftthe torque is removed. The

    permanent twist is the difference between the angles of twist and untwist.

    twisting and untwisting the shaft

       m10614

    m1025mN106.4 49-

    33 max

     

    J Tc

      3 21N1064 

    JG TL

    MPa3.187

         

    3

    949

    3

    6.69rad108.116

    Pa1077m1014.6 m2.1mN106.4

    

     

    o1 81

    69.650.8

    

     pφ

    © 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 51

    1.81 o81.1p

  • MECHANICS OF MATERIALS

    Fourth Edition

    Beer • Johnston • DeWolf

    Torsion of Noncircular Members • Previous torsion formulas are valid for

    axisymmetric or circular shafts.

    Pl ti f i l• Planar cross-sections of noncircular shafts do not remain planar and stress and strain distribution do not vary linearly

    TLT  

    • For uniform rectangular cross-sections,

    linearly.

    • At large values of a/b, the maximum

    Gabcabc 32 2

    1 max  

    shear stress and angle of twist for other open sections are the same as a rectangular bar.

    © 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 52

  • MECHANICS OF MATERIALS

    Fourth Edition

    Beer • Johnston • DeWolf

    Thin-Walled Hollow Shafts • Summing forces in the x-direction on AB,

        flowshear

    0

    

     qttt

    xtxtF

    BBAA

    BBAAx

    

    

    shear stress varies inversely with thickness

    qBBAA

        dAqpdsqdstpdFpdM 20  

    • Compute the shaft torque from the integral of the moments due to shear stress

    tA T

    qAdAqdMT

    2

    220

     

     tA2

     t ds

    GA TL

    24 

    • Angle of twist (from Chapter 11)

    © 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 53

    tGA4

  • MECHANICS OF MATERIALS

    Fourth Edition

    Beer • Johnston • DeWolf

    Example 3.10 Extruded aluminum tubing with a rectangular cross-section has a torque loading of 2.7 kN . m. Determine the shearing stress in each of the four walls with (a) uniform wall thickness of 4 mm and wall thicknesses of (b) 3 mm on AB and CD and 5 mm on CD and BD.

    SOLUTION:

    D t i th h fl th h th• Determine the shear flow through the tubing walls.

    • Find the corresponding shearing stress p g g with each wall thickness .

    © 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 54

  • MECHANICS OF MATERIALS

    Fourth Edition

    Beer • Johnston • DeWolf

    Example 3.10 SOLUTION:

    • Determine the shear flow through the tubing walls

    • Find the corresponding shearing stress with each wall thickness.

    tubing walls.

    With a uniform wall thickness,

    With a variable wall thicknessWith a variable wall thickness

    © 2006 The McGraw-Hill Companies, Inc. All rights reserved. 3 - 55

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