Answer:
Coat new O-rings (D) with silicone oil or polyalkyleneglycol (PAG) oil, and pull them on the injectors.
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:
The ARM processor normally contains at least the Z, N, C, and V flags, which are updated by execution of data processing instructions.
Explanation: