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:
a) Two points that differ in phase by π/8 rad are 0.0461 m apart.
b) The phase difference between two displacements at a certain point at times 1.6 ms apart is 4π/3.
Explanation:
f = 420 Hz, v = 310 m/s, λ = wavelength = ?
v = fλ
λ = v/f = 310/420 = 0.738 m
T = periodic time of the wave = 1/420 = 0.00238 s = 0.0024 s = 2.4 ms
a) Two points that differ in phase by π/8 rad
In terms of the wavelength of the wave, this is equivalent to [(π/8)/2π] fraction of a wavelength,
[(π/8)/2π] = 1/16 of a wavelength = (1/16) × 0.738 = 0.0461 m
b) two displacements at times 1.6 ms apart.
In terms of periodic time, 1.6ms is (1.6/2.4) fraction of the periodic time.
1.6/2.4 = 2/3.
This means those two points are 2/3 fraction of a periodic time away from each other.
1 complete wave = 2π rad
Points 2/3 fraction of a wave from each other will have a phase difference of 2/3 × 2π = 4π/3.
Answer:
The specific weight of unknown liquid is found to be 15 KN/m³
Explanation:
The total pressure in tank is measured to be 65 KPa in the tank. But, the total pressure will be equal to the sum of pressures due to both oil and unknown liquid.
Total Pressure = Pressure of oil + Pressure of unknown liquid
65 KPa = (Specific Weight of oil)(depth of oil) + (Specific Weight of unknown liquid)(depth of unknown liquid)
65 KN/m² = (8.5 KN/m³)(5 m) + (Specific Weight of Unknown Liquid)(1.5 m)
(Specific Weight of Unknown Liquid)(1.5 m) = 65 KN/m² - 42.5 KN/m²
(Specific Weight of Unknown Liquid) = (22.5 KN/m²)/1.5 m
<u>Specific Weight of Unknown Liquid = 15 KN/m³</u>
Answer:
the percent increase in the velocity of air is 25.65%
Explanation:
Hello!
The first thing we must consider to solve this problem is the continuity equation that states that the amount of mass flow that enters a system is the same as what should come out.
m1=m2
Now remember that mass flow is given by the product of density, cross-sectional area and velocity
(α1)(V1)(A1)=(α2)(V2)(A2)
where
α=density
V=velocity
A=area
Now we can assume that the input and output areas are equal
(α1)(V1)=(α2)(V2)

Now we can use the equation that defines the percentage of increase, in this case for speed

Now we use the equation obtained in the previous step, and replace values

the percent increase in the velocity of air is 25.65%