Complete Question
For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of this material elongate when a true stress of 411 MPa (59610 psi) is applied if the original length is 470 mm (18.50 in.)?Assume a value of 0.22 for the strain-hardening exponent, n.
Answer:
The elongation is 
Explanation:
In order to gain a good understanding of this solution let define some terms
True Stress
A true stress can be defined as the quotient obtained when instantaneous applied load is divided by instantaneous cross-sectional area of a material it can be denoted as
.
True Strain
A true strain can be defined as the value obtained when the natural logarithm quotient of instantaneous gauge length divided by original gauge length of a material is being bend out of shape by a uni-axial force. it can be denoted as
.
The mathematical relation between stress to strain on the plastic region of deformation is

Where K is a constant
n is known as the strain hardening exponent
This constant K can be obtained as follows

No substituting
from the question we have


Making
the subject from the equation above




From the definition we mentioned instantaneous length and this can be obtained mathematically as follows

Where
is the instantaneous length
is the original length



We can also obtain the elongated length mathematically as follows



Answer:
Actualmente estoy trabajando en una pregunta diferente en este momento.
Explanation:
Actualmente estoy trabajando en una pregunta diferente en este momento.
Answer:
The governing ratio for thin walled cylinders is 10 if you use the radius. So if you divide the cylinder´s radius by its thickness and your result is more than 10, then you can use the thin walled cylinder stress formulas, in other words:
- if
then you have a thin walled cylinder
or using the diameter:
- if
then you have a thin walled cylinder
Answer:
See explaination and attachment.
Explanation:
Navier-Stokes equation is to momentum what the continuity equation is to conservation of mass. It simply enforces F=ma in an Eulerian frame.
The starting point of the Navier-Stokes equations is the equilibrium equation.
The first key step is to partition the stress in the equations into hydrostatic (pressure) and deviatoric constituents.
The second step is to relate the deviatoric stress to viscosity in the fluid.
The final step is to impose any special cases of interest, usually incompressibility.
Please kindly check attachment for step by step solution.