Hello,
Are six 3/8-inch steel rods as strong as four 1/2-inch steel rods? How is this calculated?
Thank you!
Hello,
Are six 3/8-inch steel rods as strong as four 1/2-inch steel rods? How is this calculated?
Thank you!
The question is too vague for an answer. "Strong" is a very nebulous term. More specifics are required. It all has to do with the geometry, the loading, the support, etc. Are the bars loaded in tension? Bending? Compression? Are they tied together, or are they loose? Or are they spaced apart at some distance? See what I mean? A diagram would help a lot.
Probably not as there is less cross sectional area with the 3/8 steel rods.. However not knowing the loading configuration or application your question cannot actually be analyzed.
On a wooden schooner, these rods go widthwise through a 5x5" transverse cypress beam, a three-foot-wide by six-foot long cypress mast partner laminate and then another 5x5" transverse cypress beam on the other end. Previously, four 1/2-inch diameter x four-foot-long rods ran through the six-foot-long mast partner, two about six inches apart from each other about 1.5 feet from the one end and two about six inches apart from each other about 1.5 feet from the other end. It proved impossible to drill through the entire approx. four-foot unit with the half-inch bit. The 3/8" bit worked. So, at the suggestion of the shipwright, we added a third rod in the middle *.* like that and went with 6 3/8" rods. Now, the Coast Guard wants us to show that this is as strong or stronger than the original configuration. Is it?
[QUOTE=Kelly_Bramble;18195]Probably not as there is less cross sectional area with the 3/8 steel rods.. However not knowing the loading configuration or application your question cannot actually be analyzed.[/QUOTE]
But there would be six rods instead of four, more than compensating for the cross-sectional area, no?
Do the math... There's actually less area with the 3/8 rods (3.14157 * r^2) = area of circle
[URL]https://www.engineersedge.com/calculators/section_square_case_11.htm[/URL]
Kelly's words are very true. Math is math, and you can't deny it. But I doubt the coast guard would be very impressed with the words of some guys you never met on an online forum. We can give you advice, but that's all it is. We could tell you anything but you have no way of verifying that we know what we're talking about. You should ask the coast guard what sort of evidence they would require. Written engineering calculations stamped by a registered professional engineer? Real world test results? What have other shipbuilders done in similar situations?
[QUOTE=Kelly_Bramble;18201]Do the math... There's actually less area with the 3/8 rods (3.14157 * r^2) = area of circle
[URL]https://www.engineersedge.com/calculators/section_square_case_11.htm[/URL][/QUOTE]
Thank you!
So with the area as the key factor, if we put in another rod, we should be good to go, no? 0.3927 vs. 0.4418
[QUOTE=jboggs;18203]Kelly's words are very true. Math is math, and you can't deny it. But I doubt the coast guard would be very impressed with the words of some guys you never met on an online forum. We can give you advice, but that's all it is. We could tell you anything but you have no way of verifying that we know what we're talking about. You should ask the coast guard what sort of evidence they would require. Written engineering calculations stamped by a registered professional engineer? Real world test results? What have other shipbuilders done in similar situations?[/QUOTE]
Coast Guard wants calculations. I'm just looking for a direction and then will consult with the Coast Guard and probably then certified engineers.
Thank you!