- 11 Jul '11 16:05I have an antique rocker, four legs about 25 X 30 mm in size and curved rocker pieces about the same size. So one of the rocker arms broke, joined together by 10 mm dowels and glued. The question is, is that the optimum size for the dowel? I wasn't sure if I could improve on the strength using wood.

Assuming the same compression and tensile strength and flexibility of the wood for both dowel and chair parts, what is the optimum size of the dowel? If you were doing a new piece where holes had not been drilled, would you gain anything by using lots of smaller dowels? Say 20 toothpick sized dowels requiring of course 20 small holes, say 1 mm in diameter.

So I bought 3/8 inch, 7/16 inch and 1/2 half inch dowels, about 9.5 mm, 11 mm and 13 mm respectively in metric.

Which one would give the strongest joint? Assuming good glue of course. I use Gorilla glue for such things myself. - 12 Jul '11 00:12 / 2 edits

Technically there will be less nominal bearing stress from the 20-1mm dia. pins, than the 1-13 mm dia pin...*Originally posted by sonhouse***I have an antique rocker, four legs about 25 X 30 mm in size and curved rocker pieces about the same size. So one of the rocker arms broke, joined together by 10 mm dowels and glued. The question is, is that the optimum size for the dowel? I wasn't sure if I could improve on the strength using wood.**

Assuming the same compression and tensile strength and ...[text shortened]... e the strongest joint? Assuming good glue of course. I use Gorilla glue for such things myself.

However, the stress concentration factor for your 20 -1mm dia pins will be greater than the 1-13mm dia pin, this means that their will be a slightly larger maximum shear stress in the first situation given any applied load,..so perhaps my strength analysis is inconclusive, but its obviously more work than its worth to drill 20 small holes in your rocking chair, given your rocking chair will probably never see the forces requred to shear either of the above connection scenerios off. So go with the one big one. - 12 Jul '11 01:42

Well SOMETHING sheered off the rocker arm! I get the feeling one of my kids had it on a tilt sideways and broke off the dowels. Nice way to treat a 150 year old rocker, eh. I was thinking the larger the dowel the less wood there is in the vertical support strut, so there seems to be a best size fit to such a joining. Clearly the stress was too much for the 3/8 inch dowels (about 9.6 mm). I guess I will try for the 13 mm version, 1/2 inch.*Originally posted by joe shmo***Technically there will be less nominal bearing stress from the 20-1mm dia. pins, than the 1-13 mm dia pin...**

However, the stress concentration factor for your 20 -1mm dia pins will be greater than the 1-13mm dia pin, this means that their will be a slightly larger maximum shear stress in the first situation given any applied load,..so perhaps my strength ...[text shortened]... rces requred to shear either of the above connection scenerios off. So go with the one big one. - 12 Jul '11 02:01

Well, your children probably loaded it to failure, but as you said it was 150 years old. Fatigue can be caued from cyclic loading which a rocking chair recieves. fractures in the wood can also appear from differnetials in temperature, humidity, ect...because it is an organic material.*Originally posted by sonhouse***Well SOMETHING sheered off the rocker arm! I get the feeling one of my kids had it on a tilt sideways and broke off the dowels. Nice way to treat a 150 year old rocker, eh. I was thinking the larger the dowel the less wood there is in the vertical support strut, so there seems to be a best size fit to such a joining. Clearly the stress was too much for the 3/8 inch dowels (about 9.6 mm). I guess I will try for the 13 mm version, 1/2 inch.** - 12 Jul '11 04:11

Any of them will do, or not, depending on the proper rearing of those children, which is the real problem.*Originally posted by sonhouse***I have an antique rocker, four legs about 25 X 30 mm in size and curved rocker pieces about the same size. So one of the rocker arms broke, joined together by 10 mm dowels and glued. The question is, is that the optimum size for the dowel? I wasn't sure if I could improve on the strength using wood.**

Assuming the same compression and tensile strength and ...[text shortened]... e the strongest joint? Assuming good glue of course. I use Gorilla glue for such things myself. - 12 Jul '11 18:45

Seriously tho, it seems to me that maximum surface area -- to glue the pieces together, and maximum thickness of wood to minimize chances of snapping off, imply the large diameter should be used.*Originally posted by sonhouse***Well that is one thing I don't have to worry too much about now, they are all but one gone.** - 12 Jul '11 21:47 / 1 editIf you glue the two bits together too, as well as dowelling them, that would help. Then the glue takes most of the shear, and the dowels just take the pulling apart force. Pulling apart force should be resisted by dowel surface area.

Surface area of a rod = 2*end caps + sides

= pi*r^2 + 2*pi*r*L

Say the dowel length is 2.5 cm, then 4 0.5 cm diameter dowels have a surface area of about 17.3 cm^2.

100 1mm diameter toothpicks would have the same drilled out area as that, and would have a surface area of 80cm^2

But you probably get a reasonable increase in surface area by using fairly large dowels with ridged sides. - 12 Jul '11 23:13

Typo!*Originally posted by iamatiger***If you glue the two bits together too, as well as dowelling them, that would help. Then the glue takes most of the shear, and the dowels just take the pulling apart force. Pulling apart force should be resisted by dowel surface area.**

Surface area of a rod = 2*end caps + sides

= pi*r^2 + 2*pi*r*L

Say the dowel length is 2.5 cm, then 4 0.5 cm diameter ...[text shortened]... obably get a reasonable increase in surface area by using fairly large dowels with ridged sides.

surface area of a rod = 2*pi*r^2 + 2*pi*r*L

= 2*pi(r^2 + r*L)

area of hole = pi*r^2

so, "goodness" of dowel, = surface_area / area of hole

= 2 + 2*L/r

so longer length or smaller radius => "better" dowel. - 12 Jul '11 23:41

Maybe in mathland, but in realityland you have to look at the material properties.*Originally posted by iamatiger***Typo!**

surface area of a rod = 2*pi*r^2 + 2*pi*r*L

= 2*pi(r^2 + r*L)

area of hole = pi*r^2

so, "goodness" of dowel, = surface_area / area of hole

= 2 + 2*L/r

so longer length or smaller radius => "better" dowel. - 13 Jul '11 00:13 / 1 edit

When you say the area of the hole, you are not including the length? You gave the area of just a circle.*Originally posted by iamatiger***Typo!**

surface area of a rod = 2*pi*r^2 + 2*pi*r*L

= 2*pi(r^2 + r*L)

area of hole = pi*r^2

so, "goodness" of dowel, = surface_area / area of hole

= 2 + 2*L/r

so longer length or smaller radius => "better" dowel.

To show the strength wouldn't you have to calculate the surface area of the entire dowel, which would be(2 PI *r^2)+ (2 PI*r*h). Since you would have glue on top and bottom of the dowel, that would add to the strength, I would assume. Then each piece of wood being joined would be one half that number in surface area, assuming equal lengths of holes in both pieces of wood, just deep enough to include the whole dowel with no gap at the ends.

So my question is what is the optimum size of dowel? Since the wood struts are 25 by 30 mm, clearly a 25 mm dowel would leave very little wood on the struts when you make that size hole.

So you start going down in size and there will be a best fit of hole size and strength I would think. Obviously if the dowel were to be 1 mm in diameter, it would clearly be a very low strength joint.

So what size would be optimum? I guess it could be generalized to a percentage of the smallest linear dimension of the strut. - 13 Jul '11 00:58 / 1 editI've always tried to match the (minimum) width of the "walls" around the drilled-out hole with the diameter of the hole.

I'm not sure if it's optimal, but you want to leave both the existing wooden structure with as much strength as possible (for torque on the joint), while using a thick enough dowel (for sheer strength).

Yes, the surface area of the glued surface matters as well, but i'd rather depend on the wood holding the weight; glue can crack.

This is assuming a single dowel for supporting the joint. I would say there woudl be a definite advantage to using multiple smaller (toothpick) dowels, but that's probably way more trouble than it's worth. - 13 Jul '11 02:29

You mean using a dowel that is 1/3 the size of the smallest dimension? So there is equal spacing from the dowel on either side? So you could theoretically get three dowels side by side, but only using one? That would make the 25 mm size good for about an 8 mm dowel or the 30 mm size a 10 mm dowel. Wonder if that is optimal for wood of the same strength, dowel and strut.*Originally posted by forkedknight***I've always tried to match the (minimum) width of the "walls" around the drilled-out hole with the diameter of the hole.**

I'm not sure if it's optimal, but you want to leave both the existing wooden structure with as much strength as possible (for torque on the joint), while using a thick enough dowel (for sheer strength).

Yes, the surface area o ...[text shortened]... ltiple smaller (toothpick) dowels, but that's probably way more trouble than it's worth. - 13 Jul '11 02:42 / 2 edits

Yes, that's exactly what I mean. I guess an easier way to say it would be to maximize your minimum dimension, assuming the materials are of equivalent strength.*Originally posted by sonhouse***You mean using a dowel that is 1/3 the size of the smallest dimension? So there is equal spacing from the dowel on either side? So you could theoretically get three dowels side by side, but only using one? That would make the 25 mm size good for about an 8 mm dowel or the 30 mm size a 10 mm dowel. Wonder if that is optimal for wood of the same strength, dowel and strut.**

A piece of furniture (or anything else) is no stronger than its weakest point. - 13 Jul '11 03:14

Out of the dials you have, the optimum size is the 13 mm dial...*Originally posted by sonhouse***When you say the area of the hole, you are not including the length? You gave the area of just a circle.**

To show the strength wouldn't you have to calculate the surface area of the entire dowel, which would be(2 PI *r^2)+ (2 PI*r*h). Since you would have glue on top and bottom of the dowel, that would add to the strength, I would assume. Then each piece ...[text shortened]... I guess it could be generalized to a percentage of the smallest linear dimension of the strut.

Three parameters can nail it down in my opinion.

Normal stress-Tension or compression from axial loading

Bearing stress -from transverse loading

Stress concentration factor - an experimentally obtained factor that is dependent on the geometry of the pieces to be mated by the pin.

The first.

The glues strength is porportional to the surface area to be mated by the pin and the member.

A quick internet search gives yeild strengths in excess of 3500 psi...this piece of data rules out slippage of the pin from the member in tensile loading in any circumstance this chair should ever meet in proper, and even impropper use.

Next, The bearing stress from transverse loading.

actual bearing stress are theoretically difficult to compute so an average model is used in practical applications.

Bearing Stress = Load/(2*Length_pin)*(1/radius_pin)

assuming load and length_pin remain constant between scenarios, then the bearing stress is just a function of the radius of the pin.

you can see from the above relation that as radius_pin increases, bearing stress decreases.

Finally, Stress Concentration Factor...

When a member has a discontinuity, such as a hole or fillet, that discontinuity obtains much higher than average stress near the discontinuity in the critical section of the member( ie the thinest section) So its also important to minimize this stress concentration factor-K.

As i mentioned above the K factor is obtained empirically, and in this particular case (centered hole) is dependent on the ratio

(radius_pin)/(material between hole and outside member)

and can be calculated by means of

(radius_pin)/(Width_member - 2*radius_pin)

K is a decreasing non-linear function, so as radius_pin increases, K decreases... no problem there.

Now we simply want to minimize the maximum stress

Max stress = K*Bearing stress

And using your dimensions the 13-mm dia pin we end up with a minimum for the size constraints to be

Bearing stress ~ .16* (load)/(length_pin)

man was that overkill, but good practice none the less...