The ultimate strength of a steel rod is 600.000MPa. if the factor of safety 3.000 is required, what is the maximum permissible load for the rod if it has a diameter of 6.000cm?
ANS = kN (Round to 3 decimal place)
Ultimate strength = 600.000MPa = 600000000Pa = 600 * 10^6
Factor of Safety = 3.000
Rod diameter = 6.000cm = .06m
Allowable Strength
= Ultimate strength/ Factor of Safety
= 600* 10^6/ 3 = 200 * 10^6 (or)
= 600000000/3 = 200000000
Allowable Strength
= Load/ Area
Load
= Allowable Strength * Area
= Allowable Strength * (pi*d^2/4)
=600*10^6*pi*.06^2/4
=600000000*3.1416*.0036/4
=6785856/4
=1696464N
=1696.464kN
Please check. Thank you.
no. you need to divide the max load by 3, somehow you dropped that.
Load
= Allowable Strength * Area
= Allowable Strength * (pi*d^2/4)
=200*10^6*pi*.06^2/4
=200000000*3.1416*.0036/4
=2261952/4
=565488N
=565.488kN
Please check. Thank you.
Your calculation is almost correct, but there is a minor mistake in the decimal place. The correct calculation is as follows:
Allowable Strength = Ultimate strength / Factor of Safety
= 600 * 10^6 / 3
= 200 * 10^6
Now, let's calculate the maximum permissible load:
Area of the rod = pi * (diameter/2)^2
= 3.1416 * (0.06/2)^2
= 3.1416 * (0.03)^2
= 3.1416 * 0.0009
= 0.00282744 m²
Maximum permissible load = Allowable Strength * Area
= 200 * 10^6 * 0.00282744
= 565,488.8 N
= 565.4888 kN (rounded to 3 decimal places)
Therefore, the maximum permissible load for the steel rod with a diameter of 6.000 cm is approximately 565.489 kN.