Stress & Strain

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Stress and Strain Explanation

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  • Chapter Objectives

    To show how to use the method of sections for determining the internal loadings in a member.

    To introduce the concepts of normal and shear stress.

    To analyse and design members subject to axial load and direct shear.

    To define normal and shear strain, and show how they can be determined for various types of problems.

    Copyright 2011 Pearson Education South Asia Pte Ltd

  • Reading Quiz

    Applications

    Normal Stress

    Shear Stress

    Normal Strain

    Shear Strain

    Cartesian Strain Components

    Concept Quiz

    In-Class Activities

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  • READING QUIZ

    1. What is the normal stress in the bar if P=10

    kN and 500 mm?

    a) 0.02 kPa

    b) 20 Pa

    c) 20 kPa

    d) 200 N/mm

    e) 20 MPa

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  • READING QUIZ (cont)

    2. What is the average shear stress in the

    internal vertical surface AB (or CD), if F=20

    kN, and AAB

    =ACD

    =1000 mm?

    a) 20 N/mm

    b) 10 N/mm

    c) 10 kPa

    d) 200 kN/m

    e) 20 MPa

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  • READING QUIZ (cont)

    3. The center portion of the rubber balloon has

    a diameter of d = 100 mm. If the air pressure

    within it causes the balloons diameter to become d = 125 mm, determine the average

    normal strain in the rubber.

    a) 0.2

    b) 0.25

    c) 0.25

    d) 1.25

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  • READING QUIZ (cont)

    4. What is the unit of strain?

    a) mm

    b) mm/m

    c) Micron

    d) no unit

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  • APPLICATIONS

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    Where does the average normal stress occur?

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  • APPLICATIONS (cont)

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    Will the total shear force over the anchor length be equal to

    the total tensile force tensile A in the bar?

  • APPLICATIONS

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  • AVERAGE NORMAL STRESS

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    Will the total shear force over the anchor length be equal to the total tensile force tensile A in the bar?

    A

    P

  • NORMAL SHEAR STRESS

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    A

    FzA

    z

    0lim

    A

    F

    A

    F

    y

    Azy

    x

    Azx

    0

    0

    lim

    lim

  • EXAMPLE 1

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    The bar in Fig. 7-14a has a constant width of 35 mm and a thickness of 10 mm. Determine the maximum average normal stress in the bar when it is subjected to the loading shown.

  • EXAMPLE 1 (cont)

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    Graphically, the normal force diagram is as shown.

    By inspection, different sections have different internal forces.

    Solution

  • EXAMPLE 1 (cont)

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    By inspection, the largest loading is in region BC,

    Since the cross-sectional area of the bar is constant, the largest average normal stress is

    Solution

    kN 30BCP

    (Ans) MPa 7.8501.0035.0

    1030 3

    A

    PBCBC

  • DESIGN OF SIMPLE CONNECTION

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    For normal force requirement

    For shear force requirement

    allow

    PA

    allow

    VA

  • EXAMPLE 2

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    The rigid bar AB shown in Fig. 129a is supported by a steel rod AC having a diameter of 20 mm and an aluminum block having a cross sectional area of 1800 mm2. The 18-mm-diameter pins at A and C are subjected to single shear. If the failure stress for the steel and aluminum is and , respectively, and the failure shear stress for each pin is , determine the largest load P that can be applied to the bar. Apply a factor of safety of FS=2.

    MPa 680failst

    MPa 70failal

    MPa 900fail

  • EXAMPLE 2 (cont)

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    The allowable stresses are

    There are three unknowns and we apply the equations of equilibrium,

    Solution

    MPa 4502

    900

    ..

    MPa 352

    70

    ..

    MPa 3402

    680

    ..

    SF

    SF

    SF

    fail

    allow

    failal

    allowal

    failst

    allowst

    (2) 075.02 ;0

    (1) 0225.1 ;0

    PFM

    FPM

    BA

    ACB

  • EXAMPLE 2 (cont)

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    We will now determine each value of P that creates the allowable stress in the rod, block, and pins, respectively.

    For rod AC,

    Using Eq. 1,

    For block B,

    Using Eq. 2,

    Solution

    kN 8.10601.010340 26 ACallowstAC AF

    kN 171

    25.1

    28.106P

    kN 0.631018001035 66 BallowalB AF

    kN 168

    75.0

    20.63P

  • EXAMPLE 2 (cont)

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    For pin A or C,

    Using Eq. 1,

    When P reaches its smallest value (168 kN), it develops the allowable normal stress in the aluminium block. Hence,

    Solution

    kN 5.114009.010450 26 AFV allowAC

    kN 183

    25.1

    25.114P

    (Ans) kN 168P

  • NORMAL STRAIN

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    s

    ssavg

    '

    ss 1'

    s

    ss

    nAB

    'lim

    along

  • SHEAR STRAIN

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    'lim2

    along along

    tACnAB

    nt

  • CARTESIAN STRAIN COMPONENTS

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    The approximate lengths of the sides of the parallelepiped are

    The approximate angles between sides, again originally defined by the sides x, y and z are

    Notice that the normal strains cause a change in volume of rectangular element, whereas the shear strain cause a

    change in shape

    zyx zyx 1 1 1

    xzyzxy

    2

    2

    2

  • CARTESIAN STRAIN COMPONENTS (cont)

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  • EXAMPLE 3

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    The slender rod creates a normal strain in the rod of

    where z is in meters. Determine (a) displacement of end B

    due to the temperature increase, and (b) the average normal

    strain in the rod.

    2/131040 zz

  • EXAMPLE 3 (cont)

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    Part (a)

    Since the normal strain is reported at each point along the rod, it has a deformed length of

    The sum along the axis yields the deformed length of the rod is

    The displacement of the end of the rod is therefore

    Solution

    dzzdz 2/1310401'

    m 20239.010401'2.0

    0

    2/13 dzzz

    (Ans) mm39.2m00239.02.020239.0 B

  • EXAMPLE 3 (cont)

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    Part (b)

    Assumes the rod has an original length of 200 mm and a change in length of 2.39 mm. Hence,

    Solution

    (Ans) mm/mm 0119.0200

    39.2'

    s

    ssavg

  • EXAMPLE 4

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    Due to a loading, the plate is deformed into the dashed shape

    shown in Fig. 26a. Determine (a) the average normal strain along the side AB, and (b) the average shear strain in the plate

    at A relative to the x and y axes.

  • EXAMPLE 4 (cont)

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    Part (a)

    Line AB, coincident with the y axis, becomes line AB after deformation, thus the length of this line is

    The average normal strain for AB is therefore

    The negative sign indicates the strain causes a contraction of AB.

    Solution

    mm 018.24832250' 22 AB

    (Ans) mm/mm 1093.7240

    250018.248' 3

    AB

    ABABavgAB

  • EXAMPLE 4 (cont)

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    Part (b)

    As noted, the once 90 angle BAC between the sides of the plate, referenced from the x, y axes, changes to due to the displacement of B to B.

    Since then is the angle shown in the figure.

    Thus,

    Solution

    '2

    xy xy

    (Ans) rad 121.02250

    3tan 1

    xy

  • CONCEPT QUIZ

    1) The thrust bearing is subjected to the loads

    as shown. Determine the order of average

    normal stress developed on cross section

    through B, C and D.

    a) C > B > D

    b) C > D > B

    c) B > C > D

    d) D > B > C

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  • CONCEPT QUIZ (cont)

    2) The rectangular membrane has an

    unstretched length L1 and width L2. If the

    sides are increased by small amounts L1 and L2, determine the normal strain along the diagonal AB.

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    2221

    2

    2

    2

    1

    21

    21

    2

    2

    2

    1

    2

    2

    2

    1

    2

    2

    1

    1

    D) C)

    B) A)

    LL

    LL

    LL

    LL

    LL

    LL

    L

    L

    L

    L

  • CONCEPT QUIZ (cont)

    3) The rectangular plate is subjected to the

    deformation shown by the dashed line.

    Determine the average shear strain xy

    of the

    plate.

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    22 2001503

    200

    150

    200

    3

    150

    3

    D) C)

    B) A)