Answer:
![K_f = 1.422](https://tex.z-dn.net/?f=K_f%20%3D%201.422)
Explanation:
We know that relationship between fatigue and stress concentration factor
![K_f = 1+q(K_t-1)](https://tex.z-dn.net/?f=K_f%20%3D%201%2Bq%28K_t-1%29)
Where
=fatigue concentration factor
=stress concentration factor
q=Notch sensitivity
Here D/d=2 and r/d=0.125
Now from standard design data book ![K_t=1.79](https://tex.z-dn.net/?f=K_t%3D1.79)
For Sut=440 MPa and r= 4 mm
q=0.8
Now by putting the values
![K_f = 1+q(K_t-1)](https://tex.z-dn.net/?f=K_f%20%3D%201%2Bq%28K_t-1%29)
![K_f = 1.8(1.79-1)](https://tex.z-dn.net/?f=K_f%20%3D%201.8%281.79-1%29)
![K_f = 1.422](https://tex.z-dn.net/?f=K_f%20%3D%201.422)
Answer:
1.25 cm/day
Explanation:
An air thickness , (l) = 0.15 cm
Air Temperature =
![(T_a)=20^0C = (20+273)K\\(T_a)=293K](https://tex.z-dn.net/?f=%28T_a%29%3D20%5E0C%20%3D%20%2820%2B273%29K%5C%5C%28T_a%29%3D293K)
Mass Diffusion coefficient (D) = ![0.25cm^2/sec](https://tex.z-dn.net/?f=0.25cm%5E2%2Fsec)
If the air pressure ![(P_a) = 0.5 P_{sat}](https://tex.z-dn.net/?f=%28P_a%29%20%3D%200.5%20P_%7Bsat%7D)
We are to determine how fast will the water
level drop in a day.
From the property of air at T = 20° C
from saturated water properties.
The mass flow of
can be calculated as:
![H_2O = \frac{D}{\phi} \delta C](https://tex.z-dn.net/?f=H_2O%20%3D%20%5Cfrac%7BD%7D%7B%5Cphi%7D%20%5Cdelta%20C)
where:
![\delta C = \frac{P_{sat}*P_a}{RT }](https://tex.z-dn.net/?f=%5Cdelta%20C%20%3D%20%5Cfrac%7BP_%7Bsat%7D%2AP_a%7D%7BRT%20%7D)
R(constant) = 8.314 kJ/mol.K
![\delta C = \frac{2.34*0.5}{8.314*293 }](https://tex.z-dn.net/?f=%5Cdelta%20C%20%3D%20%5Cfrac%7B2.34%2A0.5%7D%7B8.314%2A293%20%7D)
![\delta C = 4.803*10^{-4}\\ \delta C =0.48*10^{-3} mol/m^3\\ \delta C = 0.48*10^{-6} mol/cm^3](https://tex.z-dn.net/?f=%5Cdelta%20C%20%3D%204.803%2A10%5E%7B-4%7D%5C%5C%20%5Cdelta%20C%20%3D0.48%2A10%5E%7B-3%7D%20mol%2Fm%5E3%5C%5C%20%5Cdelta%20C%20%3D%200.48%2A10%5E%7B-6%7D%20mol%2Fcm%5E3)
Since 1 mole = 18 cm ³ of water
will be: ![(0.48*10^{-6}mol/cm^3 *18)cm^3/cm^3](https://tex.z-dn.net/?f=%280.48%2A10%5E%7B-6%7Dmol%2Fcm%5E3%20%2A18%29cm%5E3%2Fcm%5E3)
![\delta C = 8.64 * 10^{-6}](https://tex.z-dn.net/?f=%5Cdelta%20C%20%3D%208.64%20%2A%2010%5E%7B-6%7D)
Again:
![H_2O = \frac{D}{\phi} \delta C](https://tex.z-dn.net/?f=H_2O%20%3D%20%5Cfrac%7BD%7D%7B%5Cphi%7D%20%5Cdelta%20C)
![= \frac{0.25}{0.15}*8.69*10^{-6} \frac{cm^2/sec}{cm}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B0.25%7D%7B0.15%7D%2A8.69%2A10%5E%7B-6%7D%20%5Cfrac%7Bcm%5E2%2Fsec%7D%7Bcm%7D)
![=1.4481*10^{-5} \frac{cm^2/sec}{cm}](https://tex.z-dn.net/?f=%3D1.4481%2A10%5E%7B-5%7D%20%5Cfrac%7Bcm%5E2%2Fsec%7D%7Bcm%7D)
Converting the above value to cm/day: we have:
![1.448*10^{-5}*3600*24\frac{cm}{s}*\frac{s}{yr}*\frac{yr}{day}](https://tex.z-dn.net/?f=1.448%2A10%5E%7B-5%7D%2A3600%2A24%5Cfrac%7Bcm%7D%7Bs%7D%2A%5Cfrac%7Bs%7D%7Byr%7D%2A%5Cfrac%7Byr%7D%7Bday%7D)
= 1.25 cm/day
∴ the rate at which the water level drop in a day = 1.25 cm/day
The answer is osha ensure that employers have right to succeed in what they do best receive information
Answer:
![\frac{dP}{P} = 6.25](https://tex.z-dn.net/?f=%5Cfrac%7BdP%7D%7BP%7D%20%3D%206.25)
Explanation:
Given data:
Sa = 2.1
![R = \frac{pl}{A}](https://tex.z-dn.net/?f=R%20%3D%20%5Cfrac%7Bpl%7D%7BA%7D)
![\frac{dR}{R} =\frac{dP}{P} +\frac{dL}{L} (1_2V)](https://tex.z-dn.net/?f=%5Cfrac%7BdR%7D%7BR%7D%20%3D%5Cfrac%7BdP%7D%7BP%7D%20%2B%5Cfrac%7BdL%7D%7BL%7D%20%281_2V%29)
![\frac{dR}{R} =\frac{dP}{P} +\epsilon (1_2V)](https://tex.z-dn.net/?f=%5Cfrac%7BdR%7D%7BR%7D%20%3D%5Cfrac%7BdP%7D%7BP%7D%20%2B%5Cepsilon%20%281_2V%29)
![Sa = \frac{\frac{dR}{R}}{\epsilon} =\frac{\frac{dP}{P}}{\epsilon} +\frac{\epsilon (1_2V)}{\epsilon}](https://tex.z-dn.net/?f=Sa%20%3D%20%5Cfrac%7B%5Cfrac%7BdR%7D%7BR%7D%7D%7B%5Cepsilon%7D%20%3D%5Cfrac%7B%5Cfrac%7BdP%7D%7BP%7D%7D%7B%5Cepsilon%7D%20%2B%5Cfrac%7B%5Cepsilon%20%281_2V%29%7D%7B%5Cepsilon%7D)
![Sa = (1+2v) + \frac{\frac{dP}{P}}{\epsilon}](https://tex.z-dn.net/?f=Sa%20%3D%20%281%2B2v%29%20%2B%20%5Cfrac%7B%5Cfrac%7BdP%7D%7BP%7D%7D%7B%5Cepsilon%7D)
change in specific resistance is given as ![\frac{dP}{P}](https://tex.z-dn.net/?f=%5Cfrac%7BdP%7D%7BP%7D)
........2
where v is elastic range = 0.30
![\epsilon = 0.08](https://tex.z-dn.net/?f=%5Cepsilon%20%3D%200.08)
![\frac{dP}{P} = \frac{2.1 -(1-2\times 0.30)}{0.08}](https://tex.z-dn.net/?f=%5Cfrac%7BdP%7D%7BP%7D%20%3D%20%5Cfrac%7B2.1%20-%281-2%5Ctimes%200.30%29%7D%7B0.08%7D)
![\frac{dP}{P} = 6.25](https://tex.z-dn.net/?f=%5Cfrac%7BdP%7D%7BP%7D%20%3D%206.25)
Answer:
a)True
Explanation:
In cutting action,three type of zone are presents
1.Primary zone :
The most of part of energy is converted in to heat.
2.Secondary zone:
Heat generation due to rubbing between tool and chip.This also called deformation zone.
3.Tertiary zone:
Heat generated due to flank of tool and already machined surface.
So the highest temperature is located close to the shear zone.