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.
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
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)
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.
9_1_1 4 &0.4
9_2_1 99.42,99.98,100.54
9_3_1 don't know...
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.
p=(ε0 E)/(ν²+ν)
σyy=-(ε0 E)/(1+ν)
To solve these problems, we need to apply the concepts of axial and transverse strains, Poisson's ratio, Hooke's Law, and the relationship between applied pressure and normal stress.
For HW9_1:
1. Given the initial side length (L0) and the deformed side lengths (La and Lt), we can calculate the axial strain and transverse strain using the formulas:
- Axial strain (εa) = (La - L0) / L0
- Transverse strain (εt) = (Lt - L0) / L0
2. Since we know the axial strain (εa) and transverse strain (εt), we can find Poisson's ratio (ν) using the formula:
- Poisson's ratio (ν) = -εt / εa
3. To find the Young's modulus (E), we use Hooke's Law which states that the stress and strain are directly proportional. Since we have the surface traction (q) applied to the block, we can calculate the stress using the formula:
- Stress (σ) = q / A
where A is the cross-sectional area of the block.
4. The Young's modulus (E) can be calculated using the formula:
- Young's modulus (E) = Stress (σ) / Axial strain (εa)
For HW9_2:
1. Given the initial side length (L0), normal stresses (σxx, σyy, σzz), Young's modulus (E), and Poisson's ratio (ν), we apply Hooke's Law to find the strains:
- Axial strain in x-direction (εxx) = σxx / E
- Axial strain in y-direction (εyy) = σyy / E
- Axial strain in z-direction (εzz) = σzz / E
2. The deformed side lengths (Lx, Ly, Lz) can be calculated using the formulas:
- Lx = L0 * (1 + εxx)
- Ly = L0 * (1 + εyy)
- Lz = L0 * (1 + εzz)
For HW9_3:
1. Given the initial side length (L0), strain in the z-direction (εzz), applied lateral pressure (p), and Poisson's ratio (ν), we can find the constraining normal stress along the y-direction (σyy) using the following relationship:
- σyy = E * (εzz - ν * p) / (1 - ν)
2. To express the applied pressure (p) symbolically in terms of the measured strain (ε0) and elastic properties (E, ν), we use the formula:
- p = E * ε0 / (1 - ν)
Please note that the calculations for each problem will require substituting the given values into the formulas to obtain the numerical values for the unknowns.