Answer:
The smallest radius at the junction between the cross section that can be used to transmit the torque is 0.167 inches.
Explanation:
Torsional shear stress is determined by the following expression:
Where:
- Torque, measured in 
.
- Radius of the cross section, measured in inches.
- Torsion module, measured in quartic inches.
- Torsional shear stress, measured in pounds per square inch.
The radius of the cross section and torsion module are, respectively:
Where 
is the diameter of the cross section, measured in inches.
Then, the shear stress formula is now expanded and simplified as a function of the cross section diameter:
In addition, diameter is cleared:
![D = 2\cdot \sqrt[3] {\frac {2\cdot T}{\pi\cdot \tau}}](https://tex.z-dn.net/?f=D%20%3D%202%5Ccdot%20%5Csqrt%5B3%5D%20%7B%5Cfrac%20%7B2%5Ccdot%20T%7D%7B%5Cpi%5Ccdot%20%5Ctau%7D%7D)
If 
and 
, then:
![D = \sqrt[3]{\frac{2\cdot (710\,lbf\cdot in)}{\pi \cdot (12000\,psi)} }](https://tex.z-dn.net/?f=D%20%3D%20%5Csqrt%5B3%5D%7B%5Cfrac%7B2%5Ccdot%20%28710%5C%2Clbf%5Ccdot%20in%29%7D%7B%5Cpi%20%5Ccdot%20%2812000%5C%2Cpsi%29%7D%20%7D)
The smallest radius at the junction between the cross section that can be used to transmit the torque is 0.167 inches.
The answer is D concept. For example, I grasp the concept of the idea.
Answer:
Explanation:
The formula for critical stress is
K is the plane strain fracture toughness
Y is dimensionless parameters
We are to Determine the Critical stress
Now replacing the critical stress with 54.8
a with 0.2mm = 0.2 x 10⁻³
Y with 1
The fracture will not occur because this material can handle a stress of 2186.20Mpa before fracture. it is obvious that is greater than 2023Mpa
Therefore, the specimen does not failure for surface crack of 0.2mm
