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

9_1_1 4 &0.4

9_2_1 99.42,99.98,100.54
9_3_1 don't know...

9.1

a) 4
b) 0.4

To obtain the numerical values for the Young's modulus and Poisson's ratio in HW9_1_1, we can use the formulas for axial strain (ε_a) and transverse strain (ε_t).

Axial strain (ε_a) is calculated using the formula:

ε_a = (La - L0) / L0

Substituting the given values, we have:

ε_a = (210 mm - 200 mm) / 200 mm = 0.05

Similarly, transverse strain (ε_t) is calculated using the formula:

ε_t = (L0 - Lt) / L0

Substituting the given values, we have:

ε_t = (200 mm - 196 mm) / 200 mm = 0.02

Now, we can use the formula for Poisson's ratio (ν) to solve for it:

ν = ε_t / ε_a

Substituting the calculated values, we have:

ν = 0.02 / 0.05 = 0.4

Next, to obtain the numerical values of the deformed side lengths Lx, Ly, and Lz of the block in HW9_2_1, we can use Hooke's Law for normal strains and stresses.

Hooke's Law states:

σ = E * ε

Where σ is the stress, E is the Young's modulus, and ε is the strain.

From the given values in HW9_2_1, we have:

σxx = -150 MPa
εxx = (Lx - L0) / L0

σyy = -30 MPa
εyy = (Ly - L0) / L0

σzz = 90 MPa
εzz = (Lz - L0) / L0

Using the formula for stress, we can rearrange it to solve for the strain:

ε = σ / E

Substituting the given values, we have:

εxx = -150 MPa / 30 GPa = -0.005
εyy = -30 MPa / 30 GPa = -0.001
εzz = 90 MPa / 30 GPa = 0.003

Now, we can solve for the deformed side lengths Lx, Ly, and Lz using the strain formulas:

Lx = L0 * (1 + εxx)
Ly = L0 * (1 + εyy)
Lz = L0 * (1 + εzz)

Substituting the calculated values, we have:

Lx = 100 mm * (1 + (-0.005)) = 99.50 mm
Ly = 100 mm * (1 + (-0.001)) = 99.90 mm
Lz = 100 mm * (1 + 0.003) = 100.30 mm

Finally, in HW9_3_1, we need to 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.

Since the cube elongates only in the z-direction, we can use Hooke's Law for normal strains and stresses along the z-direction:

εzz = σzz / E = ϵ0

From this equation, we can solve for the applied pressure:

p = σzz = E * ϵ0

Similarly, since no loads or constraints are acting along the z-axis, the normal stress along the y-direction is zero:

σyy = 0

Therefore, the symbolic expressions for the applied pressure p and the constraining normal stress along the y direction σyy are:

p = E * ϵ0
σyy = 0