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I have a "loop" made out of mild steel plate with a 1.125" I.D and a 2.125" O.D. - The material thickness is 5/16", giving me a cross section of about 0.156 square inches. I am going to be hanging a cable or chain from this loop and I am looking for a formula that will help me determine the load it will be able to support. Any help is most appreciated, Thank you.
Similar example calculation.
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[COLOR=#000000][FONT=sans-serif]Design load (P) = Safe Working Load (SWL) x Dynamic coefficient (Dynamic coefficient is taken as 1.3.)[/FONT][/COLOR]
[COLOR=#000000][FONT=sans-serif]Allowable stress[/FONT][/COLOR]
[COLOR=#000000][FONT=sans-serif]Tensile stress, σa = Yield limit / (1.5 k) = 235 ƒ1 / (1.5 k)[/FONT][/COLOR]
[COLOR=#000000][FONT=sans-serif]Shear stress, τa = σa / √3 = 235 ƒ1 / (1.5 k x √3 )[/FONT][/COLOR]
[COLOR=#000000][FONT=sans-serif]where k = 0.85 (mild steel) and 1.0 for high tensile steel, and ƒ1 = 1.0 (mild steel), 1.08 (AH27), 1.28 (AH32), 1.35 (AH34), 1.39 (AH36), 1.43 (AH40)[/FONT][/COLOR]
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[COLOR=#000000][FONT=sans-serif]Required section area for shear stress, Ars[/FONT][/COLOR]
[COLOR=#000000][FONT=sans-serif]For shear stress the area shown by the hatch on the righ section is applicable.[/FONT][/COLOR]
[COLOR=#000000][FONT=sans-serif]Ars = P / τa = 1.3 x SWL / (235 ƒ1 / (1.5 k x √3 )) = (3.38 x k x SWL) / 235 ƒ1[/FONT][/COLOR]
[COLOR=#000000][FONT=sans-serif]Actual section area for shear stress, Aas[/FONT][/COLOR]
[COLOR=#000000][FONT=sans-serif]Aas = b x t (as shown in the shaded area on the right section)[/FONT][/COLOR]
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[/FONT][/COLOR]
[COLOR=#000000][FONT=sans-serif]Required section for tensile stress, Art[/FONT][/COLOR]
[COLOR=#000000][FONT=sans-serif]For tensile stress, the area is shown by two shaded hatch on the lower part is applicable.[/FONT][/COLOR]
[COLOR=#000000][FONT=sans-serif]Art = P / σa = 1.3 x SWL / ( 235 ƒ1 / (1.5 k)) = (1.95 x k x SWL) / 235 ƒ1[/FONT][/COLOR]
[COLOR=#000000][FONT=sans-serif]Actual section area for tensile stress, Aat[/FONT][/COLOR]
[COLOR=#000000][FONT=sans-serif]Aat = 2 x b x t (as shown by two shaded hatch on the lower drawing)[/FONT][/COLOR]
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[COLOR=#000000][FONT=sans-serif]Evaluation[/FONT][/COLOR]
[COLOR=#000000][FONT=sans-serif]Aas ≥ Ars (satisfies shear stress requirement )and[/FONT][/COLOR]
[COLOR=#000000][FONT=sans-serif]Aat ≥ Art (satisfies tensile stress requirement)[/FONT][/COLOR]
Just a note of caution since you did not include a picture of your "loop". The sample configuration given in Kelly's example is an excellent design for a hanging fixture and will also apply to a horseshoe shape of loop as well (except that the example shown gives a much stronger top connection for welding); On the other hand, if your loop is actually a ring shape then not addressed in this example is the ovaling deflection of the ring under load that can result in severe bending stress and possible failure at either the ring top connection or the bottom chain or wire load point.
Thank you Kelly, that looks like it is exactly what I need. My problem now is that after running the equation for 'Required section area for shear stress', for a 10k load I come up with 122 Sq. Inches. That seems a bit high so I assume that I am doing something wrong.
Kelly, in looking at the hanger configuration and thinking about the failure distortion I am wondering if shear failure is really applicable in this case. It appears that the bottom stress is more like a curved beam with tearing from postivie tensile stress at the bottom of the loop being the failure mode and would be resisted by bending strength in the two vertical regions of the loop that would experience both tensile and bending stresses where the maximum tensile stress would be at the inside face of each vertical leg. What do you think?