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.
9_1_1 4 &0.4
9_2_1 99.42,99.98,100.54
9_3_1 don't know...
9_3_1
(epsilon_0*E)/(nu^2+nu)
-nu*(epsilon_0*E)/(nu^2+nu)
MITx 2.091x
Can anyone help me on this question:
HW9_3: Design against buckling in a SD truss
We are considering again the truss in the figure, from homework problem HW4_2. The truss is loaded at joint C as indicated, and we had determined the axial forces in each of the bars using the method of joints.
From the solution to HW4_2, we had obtained:
N(AB)=2W
N(BC)=W
N(BD)=W
N(BE)=−2√W
N(CD)=−2√W
N(ED)=−W
Now you can take the magnitude of the applied load to be W=2 kN and the length L=4 m . Assume that the bars of the truss have a circular cross section and are made of steel, with a Young's modulus E=210 GPa and a failure strength (in tension and compression)
σ(f)=250 MPa
HW9_3_1
(24 points possible)
If you want a safety factor, SF=4 against both failure of the material, due to stresses that exceed its
failure strength, and collapse of the truss due to buckling of bars under compressive loading, obtain the numeric values, in mm, of the minimum radius of each of the bars of the truss.
r(AB) = not answered (in mm)
r(BC)= not answered (in mm)
r(BD) = not answered (in mm)
r(BE) = not answered (in mm)
r(CD) = not answered (in mm)
r(ED) = not answered (in mm)
To obtain the numerical values for the Young's modulus and Poisson's ratio in HW9_1, we can use the equations relating the strains and stresses in an isotropic elastic material under uniaxial load.
The Young's modulus (E) can be determined using the equation:
E = (σ / ε)
Where σ is the applied stress and ε is the corresponding strain.
In this case, we have the following information:
- The original length along the direction of the load (L0) is 200 mm.
- The final length along the direction of the load (La) is 210 mm.
- The final length perpendicular to the direction of the load (Lt) is 196 mm.
- The load applied (q) is 200 MPa.
To calculate the Young's modulus (E), we can use the formula:
E = σ / ε = q / (La - L0) = 200 MPa / (210 mm - 200 mm)
E = 200 MPa / 10 mm = 20 GPa
Therefore, the Young's modulus (E) is 20 GPa.
To calculate the Poisson's ratio (ν), we can use the formula:
ν = (Lt - L0) / L0 = (196 mm - 200 mm) / 200 mm = -0.02
Note: The negative sign is used to indicate contraction.
Therefore, the Poisson's ratio (ν) is -0.02.
For HW9_2, to obtain the numerical values of the deformed side lengths (Lx, Ly, and Lz) of the block, we can use Hooke's Law for normal stresses and strains.
Hooke's Law states that the strain (ε) is proportional to the applied stress (σ) with the proportionality constant being the modulus of elasticity (E) of the material.
In this case, we have the following information:
- The original side length (L0) is 100 mm.
- The modulus of elasticity (E) is 30 GPa.
- The Poisson's ratio (ν) is 0.4.
- The normal stresses applied (σxx, σyy, σzz) are -150 MPa, -30 MPa, and 90 MPa, respectively.
Using Hooke's Law, we can calculate the strains in each direction:
εxx = σxx / E = -150 MPa / 30 GPa = -0.005
εyy = σyy / E = -30 MPa / 30 GPa = -0.001
εzz = σzz / E = 90 MPa / 30 GPa = 0.003
To calculate the deformed side lengths, we use the equation:
Lx = L0 * (1 + εxx) = 100 mm * (1 - 0.005) = 99.5 mm
Ly = L0 * (1 + εyy) = 100 mm * (1 - 0.001) = 99.9 mm
Lz = L0 * (1 + εzz) = 100 mm * (1 + 0.003) = 100.3 mm
Therefore, the deformed side lengths are:
Lx = 99.5 mm
Ly = 99.9 mm
Lz = 100.3 mm
For HW9_3, we need to obtain symbolic expressions for the applied pressure (p) and the constraining normal stress along the y-direction (σyy).
The applied pressure (p) can be calculated using the formula:
p = E * ν * ϵ0
Where ϵ0 is the measured strain along the z-axis.
Therefore, the applied pressure (p) is:
p = E * ν * ϵ0 = E * ν * ϵ0
The constraining normal stress along the y-direction (σyy) can be determined using the formula:
σyy = -E * ν * ϵ0
Therefore, the constraining normal stress along the y-direction (σyy) is:
σyy = -E * ν * ϵ0 = -E * ν * ϵ0
Note: The negative sign is used to indicate compression.
Please note that the expressions provided above are symbolic and will depend on the values of the measured strain (ϵ0) and the elastic properties (E, ν) that are given in the problem statement.