Answer:
See explaination
Explanation:
2. 0-1 km shear value: taking winds at 1000mb and 850 mb
15 kts south easterly and 50 kts southerly
Vector difference 135/15 and 180/50 will be 170/61 or southerly 61 kts
3. 0-6 km shear value: taking winds at 1000 mb and 500 mb
15 kts south easterly and 40 kts westerly
Vector difference 135/15 and 270/40 will be 281/51 kts
please see attachment
Question in order:
A silicon carbide plate fractures in bending when a blunt load was applied to the plate center. The distance between the fracture origin and the mirror/mist boundary on the fracture surface was 0.796 mm. To determine the stress used to break the plate, three samples of the same material were tested and produced the following. What is the estimate of the stress present at the time of fracture for the original plate?
Mirror Radius (mm) Bending Failure Stress (MPa)
0.603 225
0.203 368
0.162 442
Answer:
191 MPa
Explanation:
Failure stress of bending is Inversely proportional to the mirror radius
At mirror radius 1 = 0.603 mm Bending stress 1 = 225 Mpa..............(1)
At mirror radius 2 = 0.203 mm Bending stress 2 = 368 Mpa...............(2)
At mirror radius 3 = 0.162 mm Bending stress 3 = 442 Mpa...............(3)
comparing case 1 and 2 using the above equation
Taking the natural logarithm of both side
ln(0.6114) = n ln(0.3366)
n₁ = ln(0.6114)/ln(0.3366)
n₁ = 0.452
comparing case 2 and 3 using the above equation
Taking the natural logarithm of both side
ln(0.8326) = n ln(0.7980)
n₂ = ln(0.8326)/ln(0.7980)
n₂ = 0.821
comparing case 1 and 3 using the above equation
Taking the natural logarithm of both side
ln(0.5090) = n ln(0.2687)
n₃ = ln(0.5090)/ln(0.2687)
n₃ = 0.514
average for n
n = 0.596
Hence to get bending stress x at mirror radius 0.796
stress x = 191 MPa
Answer:
L = 475.718
T = 240.89 ft
M = 23.0195
LC = 472.728
R = 1225 ft
Explanation:
See the attached file for the calculation.
