Swing safety: maximum load of beam (supported at one end, free with load on the other)

Question

My husband and his dad built a fort for our two boys. Coming off of one side of the fort is a 4x6 wooden beam with a tire swing attached at the end. It is supported on one side by the fort, but not supported at all on the side where the tire swing sits.

The beam comes out about 5 feet from the fort. My boys weigh about 45 and 35 pounds, respectively, and like to be on the swing together at the same time. I'd like to trust that they'll be fine, but I'd be much more comfortable if the math proved this to be true.

Can anyone help me figure out if the wooden beam can support their weight--especially when they are riding wild?

And if not, what can be done to make the swing more structurally sound?

Thanks in advance!

Answer 1

I would recommend you have your boys wear a helmet while playing. Also cover the landing area with 8 inch deep bed of mulch or some soft material like foam sheets covered by a mat.

With all due respect to the enviably beautiful job your husband has done, 5 feet cantilever swing is the ideal unbalanced pendulum, overtime working its way to loosen up the nails and screws your husband and his dad have used to secure it to the tower.

A mischievous boy weighing 45 pound can jump up and down to entice the swing to go fastener creating an impulse of 200 pounds.

One chip or knot on the beam will turn it to an accident waiting to happen.

Lumber is a very good material if used in a project redundantly and repeatedly. For cases like this steel pipe or light gauge steel sections is better.

Edit

I drew a crude sketch trying to animate an exaggerated motion of the beam and the fort while kids are swinging to answer questions about the potential loosening of the fasteners.

Answer 2

Disclaimer: As a parent myself, I understand your anxiety to keep your boys safe. I have completed this calculation to the best of my knowledge, however I need to point out that I don't feel confident working with US units, nor do I have any experience in wooden structures.

One thing to note is the loads on this swing. I'll just point out, outright that I'll make the assumption that the bolts which attach the beam to the fort are rigid (because there is no way of determining from this photo).

Also, I will make the following assumptions that :

• the length of the swing (with the tire) is about L= 12 feet.
• your boys will take the swing at least h =3 ft up (knowing boys that might be an underestimation).

While they are swinging, and they pass through their lowest point their velocity will be:

$$u = \sqrt{2 g h}\approx 13.894 [ft/s]$$

At that point the centrifugal acceleration will be: $a_c=\frac{v^2}{L}= \frac{2gh}{L}\approx 16.09 ft/s^2$.

That means that while passing through the lowest point the load on the beam will be $$F_{total} = Weight + F_{c}= m\;g + m\;a_c= mg(1 + \frac{2 h}{L})$$

You need to note that the force will be higher, the higher they go. Also the mass $m$ should be the added mass of the boys plus the tire (since I expected them to try something like that).

So the maximum load while they are swinging with a height of 3 ft should be about $F_c = 135[lbf]$

The bending moment of that force on 5 ft cantilever beam: $$ M_b = F_{total}\cdot 5[ft] = 675[ft\cdot lbf]$$

The maximum stress on the wooden beam will be given by the following equation:

$$\sigma_{b,max} = \frac{M_b}{I}\frac{y}{2}$$

where:

• $M_b$ the bending moment
• $I$ the moment of inertia. From eyeballing the photo the 6 inches are vertical and 4 are horizontal, so $$I= \frac{4 6^3}{12}= 72 [in^4]$$
• $y$ is the vertical dimension i.e. $6[in]$

Provided I did not mess up the calculation, for the numbers state above the maximum stress should be: $$\sigma_b = 337.5 [psi]$$

Just to have an feeling, if the beam was positioned the wrong way around (4 vertical, 6 horizontal), the stress would be approximately 50% more (500[psi]).

As I said at the disclaimer, I don't usually work with wood or US units. So this is as far as I think I should go.

There are several links that point to strength of wood, from wood handbook, and others. The problem is that there is conflicting information. Also there are a lot of factors, that might affect it like moisture, rot, grain direction etc.

Suggestion: As a parent what I would do is try it myself. I'd try to swing as high as the boys might go. The good thing about wood, is that it usually gives you a fair warning- usually through a loud creak. So if something is wrong you should be able to hear it before something bad happens

Suggestion Update: kamran's improvements actually are much more useful than my calculations. I've upvoted them and I'll repeat them here:

• use steel pipe or light gauge steel sections (its behaviour is more predicatble than wood)
• cover the landing area with 8 inch deep bed of mulch or some soft material like foam sheets covered by a mat
• use more fasteners and if possible secure them (glue or other method).
• Have the boys wear protective gear.

Answer 3

Your kids will soon discover that you they can lift the tire up to the floor level of the fort and then jump off while holding it.

As such, a reasonable safety factor on their weight would be at least 10 times and probably more. You might need to consider the strength of the rope, as well as the supporting beams.

I would suggest increasing $F_c$ in NMech's answer from 135 to something like 800 or 1600.

And don't forget that kids grow bigger!

Answer 4

I’m not sure the size of the bolts (maybe 10 inches or longer), but they are lag bolts and they used two, each with a locking washer, an oversized washer and a nut.