Determine the required nominal diameter of a threaded steel rod to carry an axial load of 16,000 pounds in tension if the tensile stress of 20,000 psi is permitted.
Find the area...
Area*20kpsi=16kpouds
area= 800 in^2
PI*(Diameter/2)^2=800
solve for diameter in inches.
Thanks bob mind helping me with one more? The questions all ready posted, same subject.
To determine the required nominal diameter of a threaded steel rod, we need to use the formula for tensile stress:
Tensile Stress (σ) = Force (F) / Area (A)
In this case, the tensile stress is given as 20,000 psi, and the axial load is 16,000 pounds.
First, we need to convert the load from pounds to force in pounds-force (lbf):
1 pound-force (lbf) = 1 pound (lb) * acceleration due to gravity (32.174 ft/s^2)
So, 16,000 pounds = 16,000 lbf * 32.174 ft/s^2 = 515,584 lbf.
Now, we can rearrange the formula to solve for the required area:
Area (A) = Force (F) / Tensile Stress (σ)
A = 515,584 lbf / 20,000 psi = 25.78 in^2 (square inches)
Since the area of a circle is given by the formula A = π * r^2, where r is the radius, we need to determine the radius of the rod.
π * r^2 = 25.78 in^2
Dividing both sides of the equation by π and then taking the square root of both sides, we get:
r = √(25.78 in^2 / π) = √(8.191) = 2.862 in (rounded to three decimal places)
Finally, the nominal diameter of the rod is twice the radius:
Nominal Diameter = 2 * r = 2 * 2.862 in = 5.724 in
Therefore, the required nominal diameter of the threaded steel rod is 5.724 inches to carry an axial load of 16,000 pounds in tension, with a maximum tensile stress of 20,000 psi.