Answer:
to optimize the reliability of equipment and infrastructure;
- to ensure that equipment and infrastructure are always in good condition;
- to carry out prompt emergency repair of equipment and infrastructure so as to secure the best possible availability for production;
- to enhance, through modifications, extensions, or new low-cost items, the productivity of existing equipment or production capacity;
- to ensure the operation of equipment for production and for the distribution of energy and fluids;
- to improve operational safety;
- to train personnel in specific maintenance skills;
- to advise on the acquisition, installation and operation of machinery;
- to contribute to finished product quality;
- to ensure environmental protection.
Explanation:
pick whichever you want
Answer:
Here are some cool ideas that you could do
-Zero fuel aircraft
-Advanced Space Propulsion Technologies
-Smart Automation and Blockchain
These are some things I've been working on for a few years lol, maybe you will have more luck
Answer:
Explanation:
Given that:
The direction of the applied tensile stress =[001]
direction of the slip plane = [
01]
normal to the slip plane = [111]
Now, the first thing to do is to calculate the angle between the tensile stress and the slip by using the formula:
![cos \lambda = \Big [\dfrac{d_1d_2+e_1e_2+f_1f_2}{\sqrt{(d_1^2+e_1^2+f_1^2)+(d_2^2+e_2^2+f_2^2) }} \Big]](https://tex.z-dn.net/?f=cos%20%5Clambda%20%3D%20%5CBig%20%5B%5Cdfrac%7Bd_1d_2%2Be_1e_2%2Bf_1f_2%7D%7B%5Csqrt%7B%28d_1%5E2%2Be_1%5E2%2Bf_1%5E2%29%2B%28d_2%5E2%2Be_2%5E2%2Bf_2%5E2%29%20%7D%7D%20%5CBig%5D)
where;
![[d_1\ e_1 \ f_1]](https://tex.z-dn.net/?f=%5Bd_1%5C%20e_1%20%5C%20f_1%5D)
= directional indices for tensile stress
![[d_2 \ e_2 \ f_2]](https://tex.z-dn.net/?f=%5Bd_2%20%5C%20e_2%20%5C%20f_2%5D)
= slip direction
replacing their values;
i.e 
= 0 ,
= 0 
= 1 & 
= -1 , 
= 0 , 
= 1
![cos \lambda = \Big [\dfrac{(0\times -1)+(0\times 0) + (1\times 1) }{\sqrt{(0^2+0^2+1^2)+((-1)^2+0^2+1^2) }} \Big]](https://tex.z-dn.net/?f=cos%20%5Clambda%20%3D%20%5CBig%20%5B%5Cdfrac%7B%280%5Ctimes%20-1%29%2B%280%5Ctimes%200%29%20%2B%20%281%5Ctimes%201%29%20%7D%7B%5Csqrt%7B%280%5E2%2B0%5E2%2B1%5E2%29%2B%28%28-1%29%5E2%2B0%5E2%2B1%5E2%29%20%7D%7D%20%5CBig%5D)
Also, to find the angle 
between the stress [001] & normal slip plane [111]
Then;
![cos \ \phi = \Big [\dfrac{d_1d_3+e_1e_3+f_1f_3}{\sqrt{(d_1^2+e_1^2+f_1^2)+(d_3^2+e_3^2+f_3^2) }} \Big]](https://tex.z-dn.net/?f=cos%20%5C%20%20%5Cphi%20%3D%20%5CBig%20%5B%5Cdfrac%7Bd_1d_3%2Be_1e_3%2Bf_1f_3%7D%7B%5Csqrt%7B%28d_1%5E2%2Be_1%5E2%2Bf_1%5E2%29%2B%28d_3%5E2%2Be_3%5E2%2Bf_3%5E2%29%20%7D%7D%20%5CBig%5D)
replacing their values;
i.e = 0 , = 0 = 1 & 
= 1 , 
= 1 , 
= 1
![cos \ \phi= \Big [ \dfrac{ (0 \times 1)+(0 \times 1)+(1 \times 1)} {\sqrt {(0^2+0^2+1^2)+(1^2+1^2 +1^2)} } \Big]](https://tex.z-dn.net/?f=cos%20%5C%20%20%5Cphi%3D%20%5CBig%20%5B%20%5Cdfrac%7B%20%280%20%5Ctimes%201%29%2B%280%20%5Ctimes%201%29%2B%281%20%5Ctimes%201%29%7D%20%7B%5Csqrt%20%7B%280%5E2%2B0%5E2%2B1%5E2%29%2B%281%5E2%2B1%5E2%20%2B1%5E2%29%7D%20%7D%20%5CBig%5D)
However, the critical resolved SS(shear stress) 
can be computed using the formula:
where;
applied tensile stress 
13.9 MPa
∴
JT2 oaoaosnforeneomdocdmsnsksmsmsodnfnfj
