unit6 mit plane stress and strain

download unit6 mit plane stress and strain

of 19

Transcript of unit6 mit plane stress and strain

  • 8/7/2019 unit6 mit plane stress and strain

    1/19

    MIT - 16.20 Fall, 2002

    Unit 6Plane Stress and Plane Strain

    Readings:

    T & G 8, 9, 10, 11, 12, 14, 15, 16

    Paul A. Lagace, Ph.D.Professor of Aeronautics & Astronautics

    and Engineering Systems

    Paul A. Lagace 2001

  • 8/7/2019 unit6 mit plane stress and strain

    2/19

    MIT - 16.20 Fall, 2002

    There are many structural configurations where we do not

    have to deal with the full 3-D case. First lets consider the models

    Lets then see under what conditions we canapply them

    A. Plane Stress

    This deals with stretching and shearing of thin slabs.

    Figure 6.1 Representation of Generic Thin Slab

    Paul A. Lagace 2001 Unit 6 - p. 2

  • 8/7/2019 unit6 mit plane stress and strain

    3/19

    MIT - 16.20 Fall, 2002

    The body has dimensions such thath

  • 8/7/2019 unit6 mit plane stress and strain

    4/19

  • 8/7/2019 unit6 mit plane stress and strain

    5/19

    MIT - 16.20 Fall, 2002

    Stress-Strain (fully anisotropic)

    Primary (in-plane) strains1

    1 =E1

    [1 122 166] (3)1

    2 = E2 [ 21 1 + 2 266] (4)1

    6 =G6

    [61 1 622 + 6] (5)Invert to get:

    * = E Secondary (out-of-plane) strains (they exist, but they are not a primary part of the problem)

    13 =

    E3[ 311 322 366 ]

    Paul A. Lagace 2001 Unit 6 - p. 5

  • 8/7/2019 unit6 mit plane stress and strain

    6/19

    MIT - 16.20 Fall, 2002

    14 =

    G4

    [ 41 1 42 2 466 ]1

    5 =G5

    [51 1 522 566 ]Note: can reduce these for orthotropic, isotropic

    (etc.) as before.

    Strain - Displacement

    Primary

    11 = u1 (6)y122 = u2 (7)

    y2

    12 = 1 u1 + u2 (8)2 y2 y1

    Paul A. Lagace 2001 Unit 6 - p. 6

  • 8/7/2019 unit6 mit plane stress and strain

    7/19

    MIT - 16.20 Fall, 2002

    Secondary

    13 = 1 u1 + u3 2 y3 y1

    23 = 1 u2 + u3 2 y3 y2

    33 = u3y3Note: that for an orthotropic material

    (23 ) (13)4 = 5 = 0 (due to stress-strain relations)

    Paul A. Lagace 2001 Unit 6 - p. 7

  • 8/7/2019 unit6 mit plane stress and strain

    8/19

    MIT - 16.20 Fall, 2002

    This further implies from above

    (since = 0 )y3No in-plane variation

    u3= 0

    ybut this is not exactly true

    INCONSISTENCYWhy? This is an idealized model and thus an approximation. There

    are, in actuality, triaxial (zz, etc.) stresses that we ignore here asbeing small relative to the in-plane stresses!

    (we will return to try to define small)

    Final note: for an orthotropic material, write the tensorial

    stress-strain equation as:2-D plane stress

    = ( , , , , = 1 2)E

    Paul A. Lagace 2001 Unit 6 - p. 8

  • 8/7/2019 unit6 mit plane stress and strain

    9/19

  • 8/7/2019 unit6 mit plane stress and strain

    10/19

    MIT - 16.20 Fall, 2002

    Figure 6.3 Representation of Long Prismatic Body

    Dimension in z - direction is much, much larger than inthe x and y directions

    L >> x, yPaul A. Lagace 2001 Unit 6 - p. 10

  • 8/7/2019 unit6 mit plane stress and strain

    11/19

    MIT - 16.20 Fall, 2002

    (Key again: where are limits to >>??? wellconsider later)

    Since the body is basically infinite along z, the important loads are in thex - y plane (none in z) and do not change with z:

    = = 0

    y3 z

    This implies there is no gradient in displacement along z, so (excludingrigid body movement):

    u3 = w = 0

    Equations of elasticity become:

    Equilibrium:

    Primary11

    + 21 + f1 = 0 (1)y1 y2

    12+

    22+ f2 = 0 (2)y1 y2

    Paul A. Lagace 2001 Unit 6 - p. 11

  • 8/7/2019 unit6 mit plane stress and strain

    12/19

    MIT - 16.20 Fall, 2002

    Secondary13

    +23

    + f3 = 0y1 y213 and 23 exist but do not enter into primary

    consideration

    Strain - Displacement

    11 = u1 (3)y1

    22= u2

    (4)y2

    12 = 1 u1 + u2 (5)2 y2 y1

    Assumptions = 0, w = 0 give: y3

    13 = 23 = 33 = 0(Plane strain)

    Paul A. Lagace 2001 Unit 6 - p. 12

  • 8/7/2019 unit6 mit plane stress and strain

    13/19

    MIT - 16.20 Fall, 2002

    Stress - Strain

    (Do a similar procedure as in plane stress)

    3 Primary

    11 = ... (6)22 = ... (7)12 = ... (8)

    Secondary

    13 = 023 = 0

    orthotropic ( 0 for anisotropic)

    33 0INCONSISTENCY: No load along z,

    yet 33 (zz) is non zero.

    Why? Once again, this is an idealization. Triaxial strains (33)actually arise.

    You eliminate 33 from the equation set by expressing it in terms of

    via (33) stress-strain equation.Paul A. Lagace 2001 Unit 6 - p. 13

  • 8/7/2019 unit6 mit plane stress and strain

    14/19

    MIT - 16.20 Fall, 2002

    SUMMARY Plane Stress Plane Strain

    Eliminate 33 from eq.Set by using 33 - eq. and expressing 33in terms of

    Eliminate 33 from eq. setby using 33 = 0 - eq.and expressing 33 interms of

    Note:

    3333, u3SecondaryVariable(s):

    , , u, , uPrimaryVariables:

    i3 = 0i3 = 0ResultingAssumptions:

    only/y3 = 0

    33 > in-planedimensions (y1, y2)

    thickness (y3)

  • 8/7/2019 unit6 mit plane stress and strain

    15/19

    MIT - 16.20 Fall, 2002

    Examples

    Plane Stress:

    Figure 6.4 Pressure vessel (fuselage, space habitat) Skin

    in order of 70 MPa(10 ksi)

    po 70 kPa (~ 10 psi for living environment) zz

  • 8/7/2019 unit6 mit plane stress and strain

    16/19

    MIT - 16.20 Fall, 2002

    Plane Strain:

    Figure 6.5

    Dams

    waterpressure

    Figure 6.6 Solid Propellant Rockets

    high internal pressurePaul A. Lagace 2001 Unit 6 - p. 16

  • 8/7/2019 unit6 mit plane stress and strain

    17/19

    MIT - 16.20 Fall, 2002

    butwhen do these apply???

    Depends on

    loading

    geometry

    material and its response

    issues of scale how good do I need the answer

    what are we looking for (deflection, failure, etc.)

    Weve talked about the first two, lets look a little at each of the last three:

    --> Material and its response

    Elastic response and coupling changes importance /magnitude of primary / secondary factors

    (Key: are primary dominating the response?)

    --> Issues of scale What am I using the answer for? at what level?

    Example: standing on table

    --overall deflection or reactions in legs are not

    dependent on way I stand (tip toe or flat foot)Paul A. Lagace 2001 Unit 6 - p. 17

  • 8/7/2019 unit6 mit plane stress and strain

    18/19

    MIT - 16.20 Fall, 2002

    model of top of table as plate inbending is sufficient

    --stresses under my foot very sensitive tospecifics

    (if table top is foam, the way I standwill determine whether or not I

    crush the foam)

    --> How good do I need the answer?

    In preliminary design, need ballpark estimate; in finaldesign, need exact numbers

    Example: as thickness increases when is a plate nolonger in plane stress

    Paul A. Lagace 2001 Unit 6 - p. 18

  • 8/7/2019 unit6 mit plane stress and strain

    19/19

    MIT - 16.20 Fall, 2002

    Figure 6.7 Representation of the continuum from plane stress toplane strain

    very thick

    a continuumvery thin

    (plane stress)

    (plane strain)

    No line(s) of demarkation. Numbers

    approach idealizations but never getto it.

    Must use engineering judgment

    AND

    Clearly identify key assumptions in model and resulting limitations

    Paul A. Lagace 2001 Unit 6 - p. 19