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

Question

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.

Answer 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

Answer 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)

Answer 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.

Answer 4

9_1_1 4 &0.4
9_2_1 99.42,99.98,100.54
9_3_1 don't know...

Answer 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.

Answer 6

p=(ε0 E)/(ν²+ν)
σyy=-(ε0 E)/(1+ν)