The Ultimate strength of a round steel rod is 630.000MPa. If a factor of safety of 5.000 is required, what is the maximum permissible load for the rod if it has a diameter of 3.000cm?
ANS = kN (Round to 3 decimal places)
Safety = 5.000
Rod diameter = 3.000cm = 0.03m
Ultimate Strength
= 630.000MPa = 630000kPa
Allowable strength
= Ultimate strength/ Factor of safety
= 630000kPa/ 0.03m
= 21000000kPa
Load = Allowable strength * Area
Area = pi * d^2/ 4
= 21000000kPa * 3.1416 * 3^2/ 4
= 21000000kPa * 3.1416 * 9/ 4
= 593762400kPa/ 4
ANS = 148440.600kPa
Please check. Thank you.
To find the maximum permissible load for the round steel rod, we can follow these steps:
1. Convert the diameter from centimeters to meters. Given that the diameter is 3.000 cm, we divide it by 100 to get 0.03 m.
2. Calculate the allowable strength. This is derived by dividing the ultimate strength by the factor of safety. In this case, the ultimate strength is 630,000 MPa and the factor of safety is 5.000. So, the allowable strength is 630,000 MPa / 5.000 = 126,000 MPa.
3. Calculate the area of the rod. The formula for the area of a circle is A = π * r^2, where r is the radius of the rod. In this case, the diameter is given, so we need to divide it by 2 to get the radius. With a diameter of 0.03 m, the radius is 0.03 m / 2 = 0.015 m. Using this value, the area can be calculated as A = π * (0.015 m)^2 = 0.000706858 M^2 (rounded to nine decimal places).
4. Finally, calculate the maximum permissible load by multiplying the allowable strength by the area: Load = Allowable strength * Area = 126,000 MPa * 0.000706858 M^2 = 88.952548 kN.
Therefore, the maximum permissible load for the round steel rod with a diameter of 3.000 cm is approximately 88.953 kN (rounded to three decimal places).