MAT LAB

HW9_1: AXIAL AND TRANSVERSE STRAINS - THE POISSON'S RATIO

When no loads are applied, the isotropic elastic block depicted below is a cube of side L0=200 mm.

When the distributed load (surface traction) q=200 MPa is applied as shown below, the side length along the direction of the applied load extends to measure La=210 mm, and the side lengths perpendicular to the direction of the applied load contract to measure Lt=196 mm.


HW9_1_1 : 40.0 POINTS

Obtain the numerical values for the Young's modulus (in GPa) and Poisson's ratio (dimensionless) of the material composing the block:

E=
GPa unanswered
ν=
unanswered
HW9_2: HOOKE'S LAW FOR NORMAL STRESSES AND STRAINS

When no loads are applied, an isotropic elastic block is a cube of side L0=100 mm with sides aligned with the Cartesian x, y, z axes. The modulus of the material is 30 GPa, and its Poisson's ratio is 0.4.


HW9_2_1 : 40.0 POINTS

If normal stresses σxx=−150MPa, σyy=−30MPa, and σzz=90MPa are applied to the faces of the block, obtain the numerical values (in mm) of the deformed side lengths Lx, Ly, and Lz of the block.

Note: these measurements will be very precise! You will need to provide your answer to the second decimal digit.

Lx=
mm unanswered
Ly=
mm unanswered
Lz=
mm unanswered

HW9_3: BLOCK BETWEEN RIGID WALLS

A cube L0×L0×L0 of isotropic linear elastic material (properties E, ν) is placed between two rigid frictionless walls at fixed distance L0 and compressed by a lateral pressure p along the x-axis, as indicated in the figure. No loads or constraints are acting along the z-axis. Under the effect of the applied lateral pressure, the cube elongates in the direction normal to the plane of the figure (z-direction) and the strain in this direction is measured to be ϵzz=ϵ0.


HW9_3_1 : 40.0 POINTS

Obtain symbolic expressions for the applied pressure p and for the constraining normal stress along the y direction σyy exerted by the walls on the block.

Write your expressions in terms of the measured strain along the z-axis, ϵ0 (enter as epsilon_0) and of the elastic properties E, ν (enter ν as nu).

p=
unanswered
σyy=
unanswered
No one has answered this question yet.
No one has answered this question yet.

  1. 0
  2. 0
  3. 5
asked by AMK
  1. 9_1_1

    Recall that for an isotropic material, when σx is applied on a material of modulus of elasticity E, then the strains are:
    εx=σx/E
    from which e can be calculated.
    Since
    εy=-νσx/E, and similarly
    εz=-νσx/E,
    we are able to calculate ν using one of the two equations.

    For the given case,
    εx=(210-200)/200=1/20
    E=σx / εx
    =200 MPa /(1/20)
    =4 GPa

    Using
    εy=-νσx/E =>
    ν=-εy*E/σx
    =-(196-200)/200 * 4000 Mpa / 200 Mpa
    = (4/200)*20
    =0.4

    1. 0
    2. 0
    posted by MathMate
  2. Note that 1/20 is a relatively large strain and the above equations are approximate to first order and may not be accurate for large strains.

    Use the relations:

    εx=[σx-ν(σy+σz)]/E ....(A)

    εy=[σy-ν(&sigmaz;+σx)]/E ....(B)

    εz=[σz-ν(σx+σy)]/E ....(C)

    1. 0
    2. 0
    posted by MathMate
  3. Note:
    Equations (A), (B) and (C) may be used to solve problem 9_2_1, namely
    Lx=L(1+εx),...

    Watch the signs of the stresses.

    1. 0
    2. 0
    posted by MathMate
  4. 9_1_1 4 &0.4
    9_2_1 99.42,99.98,100.54
    9_3_1 don't know...

    1. 0
    2. 0
    posted by NICK
  5. For 9_3, the question is not clear because we do not see the accompanying figure.

    However, on examining the data, it seems probable that the two "walls" (meaning no strain/movement along that direction) are in the y-direction, which means that εy=0.

    From the given information, we have
    σxx=p
    σyy=to be found
    σzz=0 (no constraints)
    εx=unknown
    εy=0 (between rigid walls)
    εz=ε0

    Assuming E and ν are known, we substitute the above values into equations (A), (B) and (C):


    εx=[σx-ν(σy+σz)]/E ....(A)
    εy=[σy-ν(&sigmaz;+σx)]/E ....(B)
    εz=[σz-ν(σx+σy)]/E ....(C)

    ex=(p-ν*σyy)/E ....(A1)
    0=(σyy-ν*(0+p))/E ...(B1)
    e0=(0-ν&(p+σyy))/E ...(C1)

    Solve for above system for
    ex,p and σyy in terms of e0,E and ν:
    p=-(ε0 E)/(ν²+ν)
    σyy=-(ε0 E)/(1+ν)
    εx=-ε0*(1-ν)/ν

    Check my work to make sure there are no typos or arithmetical errors.

    1. 0
    2. 0
    posted by MathMate
  6. p=(ε0 E)/(ν²+ν)
    σyy=-(ε0 E)/(1+ν)

    1. 0
    2. 0
    posted by Destroyer

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