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:
Judgement
Explanation:
Gilbert is required by the Judgement Principle to "disclose those conflicts of interest that cannot reasonably be avoided or escaped." Since Gilbert professionally believes that the software meets specifications, secures documents, and satisfies user requirements, it is not clear if he violated any principle. However, he could have informed his client of his interest in the software and also presented other software packages of different companies from which the client could make its independent choice.
Answer:
375 KPa
Explanation:
From the question given above, the following data were obtained:
Initial pressure (P₁) = 125 KPa
Initial temperature (T₁) = 300 K
Final temperature (T₂) = 900 K
Final pressure (P₂) =?
The new (i.e final) pressure of the gas can be obtained as follow:
P₁/T₁ = P₂/T₂
125 / 300 = P₂ / 900
Cross multiply
300 × P₂ = 125 × 900
300 × P₂ = 112500
Divide both side by 300
P₂ = 112500 / 300
P₂ = 375 KPa
Thus, the new pressure of the gas is 375 KPa
Answer:
f = c / λ = wave speed c (m/s) / wavelength λ (m). The formula for time is: T (period) = 1 / f (frequency). λ = c / f = wave speed c (m/s) / frequency f (Hz). The unit hertz (Hz) was once called cps = cycles per second.
Explanation: