Answer:https://global.oup.com/us/companion.websites/9780199385423/student/ch6/mcq/ just go here
Explanation:
Answer:
4 number answer is correct.
Answer:
The condition does not hold for a compression test
Explanation:
For a compression test the engineering stress - strain curve is higher than the actual stress-strain curve and this is because the force needed in compression is higher than the force needed during Tension. The higher the force in compression leads to increase in the area therefore for the same scale of stress the there is more stress on the Engineering curve making it higher than the actual curve.
<em>Hence the condition of : on the same scale for stress, the tensile true stress-true strain curve is higher than the engineering stress-engineering strain curve.</em><em> </em>does not hold for compression test
Answer:
a. ε₁=-0.000317
ε₂=0.000017
θ₁= -13.28° and θ₂=76.72°
b. maximum in-plane shear strain =3.335 *10^-4
Associated average normal strain ε(avg) =150 *10^-6
θ = 31.71 or -58.29
Explanation:
![\epsilon _{1,2} =\frac{\epsilon_x + \epsilon_y}{2} \pm \sqrt{(\frac{\epsilon_x + \epsilon_y}{2} )^2 + (\frac{\gamma_xy}{2})^2} \\\\\epsilon _{1,2} =\frac{-300 \times 10^{-6} + 0}{2} \pm \sqrt{(\frac{-300 \times 10^{-6}+ 0}{2}) ^2 + (\frac{150 \times 10^-6}{2})^2}\\\\\epsilon _{1,2} = -150 \times 10^{-6} \pm 1.67 \times 10^{-4}](https://tex.z-dn.net/?f=%5Cepsilon%20_%7B1%2C2%7D%20%3D%5Cfrac%7B%5Cepsilon_x%20%2B%20%5Cepsilon_y%7D%7B2%7D%20%20%5Cpm%20%5Csqrt%7B%28%5Cfrac%7B%5Cepsilon_x%20%2B%20%5Cepsilon_y%7D%7B2%7D%20%29%5E2%20%2B%20%28%5Cfrac%7B%5Cgamma_xy%7D%7B2%7D%29%5E2%7D%20%5C%5C%5C%5C%5Cepsilon%20_%7B1%2C2%7D%20%3D%5Cfrac%7B-300%20%5Ctimes%2010%5E%7B-6%7D%20%2B%200%7D%7B2%7D%20%20%5Cpm%20%5Csqrt%7B%28%5Cfrac%7B-300%20%5Ctimes%2010%5E%7B-6%7D%2B%200%7D%7B2%7D%29%20%5E2%20%2B%20%28%5Cfrac%7B150%20%5Ctimes%2010%5E-6%7D%7B2%7D%29%5E2%7D%5C%5C%5C%5C%5Cepsilon%20_%7B1%2C2%7D%20%3D%20-150%20%5Ctimes%2010%5E%7B-6%7D%20%20%5Cpm%201.67%20%5Ctimes%2010%5E%7B-4%7D)
ε₁=-0.000317
ε₂=0.000017
To determine the orientation of ε₁ and ε₂
![tan 2 \theta_p = \frac{\gamma_xy}{\epsilon_x - \epsilon_y} \\\\tan 2 \theta_p = \frac{150 \times 10^{-6}}{-300 \times 10^{-6}-\ 0}\\\\tan 2 \theta_p = -0.5](https://tex.z-dn.net/?f=tan%202%20%5Ctheta_p%20%3D%20%5Cfrac%7B%5Cgamma_xy%7D%7B%5Cepsilon_x%20-%20%5Cepsilon_y%7D%20%5C%5C%5C%5Ctan%202%20%5Ctheta_p%20%3D%20%5Cfrac%7B150%20%5Ctimes%2010%5E%7B-6%7D%7D%7B-300%20%5Ctimes%2010%5E%7B-6%7D-%5C%200%7D%5C%5C%5C%5Ctan%202%20%5Ctheta_p%20%3D%20-0.5)
θ= -13.28° and 76.72°
To determine the direction of ε₁ and ε₂
![\epsilon _{x' }=\frac{\epsilon_x + \epsilon_y}{2} + \frac{\epsilon_x -\epsilon_y}{2} cos2\theta + \frac{\gamma_xy}{2}sin2\theta \\\\\epsilon _{x'} =\frac{-300 \times 10^{-6}+ \ 0}{2} + \frac{-300 \times 10^{-6} -\ 0}{2} cos(-26.56) + \frac{150 \times 10^{-6}}{2}sin(-26.56)\\\\](https://tex.z-dn.net/?f=%5Cepsilon%20_%7Bx%27%20%7D%3D%5Cfrac%7B%5Cepsilon_x%20%2B%20%5Cepsilon_y%7D%7B2%7D%20%20%2B%20%5Cfrac%7B%5Cepsilon_x%20-%5Cepsilon_y%7D%7B2%7D%20cos2%5Ctheta%20%20%2B%20%5Cfrac%7B%5Cgamma_xy%7D%7B2%7Dsin2%5Ctheta%20%5C%5C%5C%5C%5Cepsilon%20_%7Bx%27%7D%20%3D%5Cfrac%7B-300%20%5Ctimes%2010%5E%7B-6%7D%2B%20%5C%200%7D%7B2%7D%20%20%2B%20%5Cfrac%7B-300%20%5Ctimes%2010%5E%7B-6%7D%20-%5C%200%7D%7B2%7D%20cos%28-26.56%29%20%20%2B%20%5Cfrac%7B150%20%5Ctimes%2010%5E%7B-6%7D%7D%7B2%7Dsin%28-26.56%29%5C%5C%5C%5C)
=-0.000284 -0.0000335 = -0.000317 =ε₁
Therefore θ₁= -13.28° and θ₂=76.72°
b. maximum in-plane shear strain
![\gamma_{max \ in \ plane} =2\sqrt{(\frac{\epsilon_x + \epsilon_y}{2} )^2 + (\frac{\gamma_xy}{2})^2} \\\\\gamma_{max \ in \ plane} = 2\sqrt{(\frac{-300 *10^{-6} + 0}{2} )^2 + (\frac{150 *10^{-6}}{2})^2}](https://tex.z-dn.net/?f=%5Cgamma_%7Bmax%20%5C%20in%20%5C%20plane%7D%20%3D2%5Csqrt%7B%28%5Cfrac%7B%5Cepsilon_x%20%2B%20%5Cepsilon_y%7D%7B2%7D%20%29%5E2%20%2B%20%28%5Cfrac%7B%5Cgamma_xy%7D%7B2%7D%29%5E2%7D%20%5C%5C%5C%5C%5Cgamma_%7Bmax%20%5C%20in%20%5C%20plane%7D%20%3D%202%5Csqrt%7B%28%5Cfrac%7B-300%20%2A10%5E%7B-6%7D%20%2B%200%7D%7B2%7D%20%29%5E2%20%2B%20%28%5Cfrac%7B150%20%2A10%5E%7B-6%7D%7D%7B2%7D%29%5E2%7D)
=3.335 *10^-4
![\epsilon_{avg} =(\frac{\epsilon_x + \epsilon_y}{2} )](https://tex.z-dn.net/?f=%5Cepsilon_%7Bavg%7D%20%3D%28%5Cfrac%7B%5Cepsilon_x%20%2B%20%5Cepsilon_y%7D%7B2%7D%20%29)
ε(avg) =150 *10^-6
orientation of γmax
![tan 2 \theta_s = \frac{-(\epsilon_x - \epsilon_y)}{\gamma_xy} \\\\tan 2 \theta_s = \frac{-(-300*10^{-6} - 0)}{150*10^{-6}}](https://tex.z-dn.net/?f=tan%202%20%5Ctheta_s%20%3D%20%5Cfrac%7B-%28%5Cepsilon_x%20-%20%5Cepsilon_y%29%7D%7B%5Cgamma_xy%7D%20%5C%5C%5C%5Ctan%202%20%5Ctheta_s%20%3D%20%5Cfrac%7B-%28-300%2A10%5E%7B-6%7D%20-%200%29%7D%7B150%2A10%5E%7B-6%7D%7D)
θ = 31.71 or -58.29
To determine the direction of γmax
![\gamma _{x'y' }= - \frac{\epsilon_x -\epsilon_y}{2} sin2\theta + \frac{\gamma_xy}{2}cos2\theta \\\\\gamma _{x'y' }= - \frac{-300*10^{-6} - \ 0}{2} sin(63.42) + \frac{150*10^{-6}}{2}cos(63.42)](https://tex.z-dn.net/?f=%5Cgamma%20_%7Bx%27y%27%20%7D%3D%20%20-%20%5Cfrac%7B%5Cepsilon_x%20-%5Cepsilon_y%7D%7B2%7D%20sin2%5Ctheta%20%20%2B%20%5Cfrac%7B%5Cgamma_xy%7D%7B2%7Dcos2%5Ctheta%20%5C%5C%5C%5C%5Cgamma%20_%7Bx%27y%27%20%7D%3D%20%20-%20%5Cfrac%7B-300%2A10%5E%7B-6%7D%20-%20%5C%200%7D%7B2%7D%20sin%2863.42%29%20%20%2B%20%5Cfrac%7B150%2A10%5E%7B-6%7D%7D%7B2%7Dcos%2863.42%29)
= 1.67 *10^-4
Answer:
0.5m^2/Vs and 0.14m^2/Vs
Explanation:
To calculate the mobility of electron and mobility of hole for gallium antimonide we have,
(S)
Where
e= charge of electron
n= number of electrons
p= number of holes
mobility of electron
mobility of holes
electrical conductivity
Making the substitution in (S)
Mobility of electron
![8.9*10^4=(8.7*10^{23}*(-1.602*10^{-19})*\mu_e)+(8.7*10^{23}*(-1.602*10^{-19})*\mu_h)](https://tex.z-dn.net/?f=8.9%2A10%5E4%3D%288.7%2A10%5E%7B23%7D%2A%28-1.602%2A10%5E%7B-19%7D%29%2A%5Cmu_e%29%2B%288.7%2A10%5E%7B23%7D%2A%28-1.602%2A10%5E%7B-19%7D%29%2A%5Cmu_h%29)
![0.639=\mu_e+\mu_h](https://tex.z-dn.net/?f=0.639%3D%5Cmu_e%2B%5Cmu_h)
Mobility of hole in (S)
![2.3*10^5 = (7.6*10^{22}*(-1.602*10^{-19})*\mu_e)+(1*10^{25}*(-1.602*10^{-19}*\mu_h))](https://tex.z-dn.net/?f=2.3%2A10%5E5%20%3D%20%287.6%2A10%5E%7B22%7D%2A%28-1.602%2A10%5E%7B-19%7D%29%2A%5Cmu_e%29%2B%281%2A10%5E%7B25%7D%2A%28-1.602%2A10%5E%7B-19%7D%2A%5Cmu_h%29%29)
![0.1436 = 7.6*10^{-3}\mu_e+\mu_h](https://tex.z-dn.net/?f=0.1436%20%3D%207.6%2A10%5E%7B-3%7D%5Cmu_e%2B%5Cmu_h)
Then, solving the equation:
(1)
(2)
We have,
Mobility of electron ![\mu_e = 0.5m^2/V.s](https://tex.z-dn.net/?f=%5Cmu_e%20%3D%200.5m%5E2%2FV.s)
Mobility of hole is ![\mu_h = 0.14m^2/V.s](https://tex.z-dn.net/?f=%5Cmu_h%20%3D%200.14m%5E2%2FV.s)