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 [tex]=21.29mm[/tex]
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 [tex]\sigma_T[/tex].
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 [tex]\epsilon_T[/tex].
The mathematical relation between stress to strain on the plastic region of deformation is
[tex]\sigma _T =K\epsilon^n_T[/tex]
Where K is a constant
n is known as the strain hardening exponent
This constant K can be obtained as follows
[tex]K = \frac{\sigma_T}{(\epsilon_T)^n}[/tex]
No substituting [tex]345MPa \ for \ \sigma_T, \ 0.02 \ for \ \epsilon_T , \ and \ 0.22 \ for \ n[/tex] from the question we have
[tex]K = \frac{345}{(0.02)^{0.22}}[/tex]
[tex]= 815.82MPa[/tex]
Making [tex]\epsilon_T[/tex] the subject from the equation above
[tex]\epsilon_T = (\frac{\sigma_T}{K} )^{\frac{1}{n} }[/tex]
[tex]Substituting \ 411MPa \ for \ \sigma_T \ 815.82MPa \ for \ K \ and \ 0.22 \ for \ n[/tex]
[tex]\epsilon_T = (\frac{411MPa}{815.82MPa} )^{\frac{1}{0.22} }[/tex]
[tex]=0.0443[/tex]
From the definition we mentioned instantaneous length and this can be obtained mathematically as follows
[tex]l_i = l_o e^{\epsilon_T}[/tex]
Where
[tex]l_i[/tex] is the instantaneous length
[tex]l_o[/tex] is the original length
[tex]Substituting \ 470mm \ for \ l_o \ and \ 0.0443 \ for \ \epsilon_T[/tex]
[tex]l_i = 470 * e^{0.0443}[/tex]
[tex]=491.28mm[/tex]
We can also obtain the elongated length mathematically as follows
[tex]Elongated \ Length =l_i - l_o[/tex]
[tex]Substituting \ 470mm \ for l_o and \ 491.28 \ for \ l_i[/tex]
[tex]Elongated \ Length = 491.28 - 470[/tex]
[tex]=21.29mm[/tex]
