What is the Optimal Thickness for 4x4 Steel Tubing in a Giant Monkey Bar Set?

[TK9320]

I'm building a giant set of monkey bars for a gym and I'm curious about the strength of material I plan to use.I'm going to use two 4x4 3/16 thick steel tubing that will run 20 feet long and sit on top of the same material at each end 9 feet high.I will brace the top corners at each end.
I'm curious if 3/16 will work or should I use 1/4 inch thick?
It won't have any large loads on it.It's just for chin up bars and maybe a punching bag hanging from it. Thanks for your help.

 

[nvn]

TK9320: I think 12 to 15 people could easily fit in this space doing chin-ups simultaneously. Please let us know the maximum number of people that can fit in this space.

 

[TK9320]

TK9320: I think 12 to 15 people could easily fit in this space doing chin-ups simultaneously. Please let us know the maximum number of people that can fit in this space.

There will 10 chin up bars on each end.They will only be 7 ft from the end,so they won't be in the center.There may be an 80lbs punching bag in the center.
I don't think they will ever have 20 people on it at the same time though. Thanks

 

[nvn]

TK9320: What is the distance between your two horizontal beams?

 

[TK9320]

TK9320: What is the distance between your two horizontal beams?

they will be 52 inches apart with 1.25 OD round tubing for the bars,drilled through the 4 inch

 

[nvn]

OK, I misinterpreted something earlier. My fault. Let me make sure I understand. You have two horizontal beams, 52 inches apart. And these two beams have one chin-up bar on each end, located 7 feet from each end. Therefore, the distance between the two chin-up bars is 6 feet. Plus, there is a punching bag bar in the center. And we can call this one "frame." It sounds like you have five of these frames, side by side, in the room, correct? Therefore, does each adjacent frame share a horizontal beam? In other words, you have six horizontal, 20-foot-long beams in the room, right? And each of the six 20-foot beams sits on a vertical 4 x 4 inch tube at each end, correct?

 

[TK9320]

OK, I misinterpreted something earlier. My fault. Let me make sure I understand. You have two horizontal beams, 52 inches apart. And these two beams have one chin-up bar on each end, located 7 feet from each end. Therefore, the distance between the two chin-up bars is 6 feet. Plus, there is a punching bag bar in the center. And we can call this one "frame." It sounds like you have five of these frames, side by side, in the room, correct? Therefore, does each adjacent frame share a horizontal beam? In other words, you have six horizontal, 20-foot-long beams in the room, right? And each of the six 20-foot beams sits on a vertical 4 x 4 inch tube at each end, correct?

There are only two 20 ft 4x4s with two 4x4s at each end 9ft high.There are two bars at each end one 3ft from the end and one 7ft from the end.There will two other bars that run horizontal on each side.So at each end there are two bars between the beams and two on the outside.So 6 total at each end.

 

[nvn]

I cannot imagine, yet, how there are ten chin-up bars on each end. By the way, did you know you can https://www.physicsforums.com/showthread.php?t=433015#post2906520" a diagram when you post? You could scan (or photograph) a freehand sketch, or draw a rough diagram using Start > Programs > Accessories > Paint, if you wish.

 

[TK9320]

I cannot imagine, yet, how there are ten chin-up bars on each end. By the way, did you know you can https://www.physicsforums.com/showthread.php?t=433015#post2906520" a diagram when you post? You could scan (or photograph) a freehand sketch, or draw a rough diagram using Start > Programs > Accessories > Paint, if you wish.

There are six at each end.I took some pics of my sketch but the file is too big to upload.I can email them you want.

 

[nvn]

Try going to the link I posted, then click the imageshack link, then upload your pictures. I think you can use the "image resize" pull-down menu in imageshack, if you wish. But it is probably not required. Then post direct links to the pictures.

 

[TK9320]

http://img693.imageshack.us/img693/6169/sketch2h.jpg
http://img219.imageshack.us/img219/694/sketch1.jpg

 

[nvn]

TK9320: Nice job uploading the diagrams. Assuming up to 20 people are doing chin-ups simultaneously, with an average mass per person of 91.8 kg, and assuming one punching bag in the center, I think you can use a dynamic amplification factor of daf = 2.0. I think the steel square tube minimum tensile yield strength is probably Sty = 275 MPa. You could use a yield factor of safety of FSy = 1.50. Therefore, for your 101.6 x 101.6 x 4.7625 mm square tubes, we have the following.

L = 6100 mm, M = 6.1803e6 N*mm, I1 = 2.8902e6 mm^4.
sigma = daf*M*c/I1 = 217.26 MPa.
Ry = FSy*sigma/Sty = 1.50(217.26 MPa)/(275 MPa) = 118.5 %.

Ry > 100 % indicates the two beams are currently overstressed using 101.6 x 101.6 x 4.7625 mm steel tubes.​

For your 101.6 x 101.6 x 6.35 mm square tubes, we have the following.

L = 6100 mm, M = 6.1803e6 N*mm, I2 = 3.6745e6 mm^4.
sigma = daf*M*c/I2 = 170.89 MPa.
Ry = FSy*sigma/Sty = 1.50(170.89 MPa)/(275 MPa) = 93.2 %.

Ry ≤ 100 % indicates your two beams are currently not overstressed using 101.6 x 101.6 x 6.35 mm steel tubes.​

Therefore, if you want to use 101.6 x 101.6 mm square tubes, then it appears they would need to have a wall thickness of 6.35 mm.

By the way, it appears your structure needs to be stabilized in the lateral direction. No end view is shown, so we do not know if you provided a corner brace on the ends.

The above results assume no one is hanging (bouncing) on the structure near midspan, which might be an unwise assumption.

 

[TK9320]

TK9320: Nice job uploading the diagrams. Assuming up to 20 people are doing chin-ups simultaneously, with an average mass per person of 91.8 kg, and assuming one punching bag in the center, I think you can use a dynamic amplification factor of daf = 2.0. I think the steel square tube minimum tensile yield strength is probably Sty = 275 MPa. You could use a yield factor of safety of FSy = 1.50. Therefore, for your 101.6 x 101.6 x 4.7625 mm square tubes, we have the following.

L = 6100 mm, M = 6.1803e6 N*mm, I1 = 2.8902e6 mm^4.
sigma = daf*M*c/I1 = 217.26 MPa.
Ry = FSy*sigma/Sty = 1.50(217.26 MPa)/(275 MPa) = 118.5 %.

Ry > 100 % indicates the two beams are currently overstressed using 101.6 x 101.6 x 4.7625 mm steel tubes.​

For your 101.6 x 101.6 x 6.35 mm square tubes, we have the following.

L = 6100 mm, M = 6.1803e6 N*mm, I2 = 3.6745e6 mm^4.
sigma = daf*M*c/I2 = 170.89 MPa.
Ry = FSy*sigma/Sty = 1.50(170.89 MPa)/(275 MPa) = 93.2 %.

Ry ≤ 100 % indicates your two beams are currently not overstressed using 101.6 x 101.6 x 6.35 mm steel tubes.​

Therefore, if you want to use 101.6 x 101.6 mm square tubes, then it appears they would need to have a wall thickness of 6.35 mm.

By the way, it appears your structure needs to be stabilized in the lateral direction. No end view is shown, so we do not know if you provided a corner brace on the ends.

The above results assume no one is hanging (bouncing) on the structure near midspan, which might be an unwise assumption.

Thanks.How does it work out if only 12 people max would be on it at one time?

 

[nvn]

It depends on where they are hanging. More people near midspan creates higher stresses.

 

[nvn]

I can make the structure fail using 14 people, 101.6 x 101.6 x 6.35 mm steel tubes, and daf = 2.0. I put two people on the bar holding the punching bag, one person on each beam at x = +/-450 mm from midspan, two people on each chin-up bar at x = +/-914 mm from midspan, and one person on each outside chin-up bar at x = +/-1524 mm from midspan. This gives the following stress level (and I haven't even subtracted for the holes in the beam).

M = 7.2828e6 N*mm, sigma = daf*M*c/I2 = 201.37 MPa.
Ry = FSy*sigma/Sty = 1.50(201.37 MPa)/(275 MPa) = 109.8 % > 100 %.​

However, I think using daf = 2.0 for all people simultaneously is conservative, and can be reduced to daf = 1.75, which gives the following.

M = 7.2828e6 N*mm, sigma = daf*M*c/I2 = 176.20 MPa.
Ry = FSy*sigma/Sty = 1.50(176.20 MPa)/(275 MPa) = 96.1 % < 100 %.​

Therefore, if you want to use 101.6 x 101.6 mm square tubes, this still indicates they would need to have a wall thickness of 6.35 mm.

Or, you might want to try 152.4 x 101.6 x 4.7625 mm rectangular tubes (which I have not tried). Post the material specification for your tubes, if you have it.

 

[nvn]

TK9320: However, the results in post 15 do not include self weight. When we include self weight, the second Ry in post 15 becomes Ry = 106.5 %, which indicates the two square beams are overstressed. It seems you need to use 152 x 102 x 4.763 mm rectangular tubes for your two horizontal beams. These would have the same weight as the square tube beams in post 15, but would instead give the following results for the post 15 load case using daf = 1.75, except including self weight.

L = 6100 mm, M = 8.3974e6 N*mm, I3 = 7.5903e6 mm^4.
sigma = daf*M*c/I3 = 147.53 MPa.
Ry = FSy*sigma/Sty = 1.50(147.53 MPa)/(275 MPa) = 80.5 % < 100 %.​