# Best plastic composite alternative to wood

I am planning to replace the material that is used for Cricket Stumps. Cricket is a game similar to baseball. One difference is the presence of 3 successive stumps between the hitter and catcher. The ball will hit the stumps if the hitter misses it. The combined weight of the stumps is 7 lbs. Each stump is a 28 inch long wooden cylindrical stump of 1.5 inch radius weight 2.3 lbs. The stump can be hollow inside. As of now the regular stumps are made of solid wood.

I am in search of a lightweight alternative for the wooden stump. The material should be able to withstand the force of the ball which typically travels at 80 mph.

Is there any cheap and lightweight plastic material that can drastically reduce the weight of the stump and still withstand the forces mentioned above?  • What do you mean by "drastically reduce"? Yes, plastic is lighter, so what level of "lightness" are you looking for. As it stands now, there could be a string of answers each giving a material that is slightly lighter than the previous one. – hazzey Jul 10 '15 at 1:14
• @rajGoogles4Code: For weekend cricket, 80 mph is OK, but for test cricket you need to cater for speeds around 100 mph, if Shoab Akhtar's record speed is to be accommodated. [sporteology.com/… – Fred Jul 10 '15 at 5:10
• You might consider looking into bamboo. Comparatively cheap; lightweight; fairly strong. – user16 Jul 10 '15 at 12:37
• @hazzey It can be as light as possible until it can withstand the force of the impact of the ball – raj'sCubicle Jul 10 '15 at 17:23
• @GlenH7 I am of the opinion that we can find a plastic material lighter and effective than bamboo. – raj'sCubicle Jul 10 '15 at 17:30

For impact:

$$\sigma_{max} = \frac{v}{V}E(1+\sqrt{\mu + \frac{2}{3}})$$

Where:

$V = \sqrt\frac{gE}{\delta}$

$g$ is Gravitational constant

$E$ is Young's modulus

$\delta$ is the weight per unit volume

$\mu$ is the ratio of ball mass to bar mass

$v$ is velocity of impacting ball

Per Roark's Formula's for stresses and strains 7th edition, this can be shown that as $\mu$ approaches $\infty$, then $\sigma_{max}$ will approach 2 times the case if the mass was placed on the bar and allowed to deflect under gravity. In that case, we'll run with 2 times the ball mass (163 gm) and treat it as a fixed beam. for simpler equations.

With round pegs we therefore have:

$$\delta = \frac{2mgL^3}{3EI}$$ $$\sigma = \frac{2mgLr}{I}$$

We need the deflection to stay the same, so the bails will move the same. We need to figure out how much that is.

With 7/3 pounds per peg, and a source saying they are willow wood, it is 15% air. That works to the inside being 0.581 inches radius hollow. So I'd calculate $\delta = 0.002 in$ for the wood.

Simplifying for a really thin material, $I = \pi r^3t$, so to match deflection we have $Et = 0.120 Msi*in$.

For a good fiberglass composite (the cheapest of all of them), the modulus would be 1.5 Msi, so the thickness should be a mere 0.08in - a drastic weight savings (1 pound per wicket). The stress would be also acceptable levels.

• Worth noting that fiberglass can break into many shards and slivers that can be quite sharp. That may present a safety issue given the context of how the poles are used. – user16 Jul 10 '15 at 12:39
• @GlenH7 - I did run a stress analysis. Normally for fiberglass with fatigue loading, you need to derate the material to 20% of the original strength. Here, we're well below that - I've got a factor of safety of over 50. Still, some of the rules depend a lot on the wicket's behavior - substitution may very well change a critical moment in the game. – Mark Jul 10 '15 at 17:24

You could use polycarbonate tubing, it is used for safety glasses and other unbreakable items and is lightweight. It is not stress crack resistant though so if you need to drill holes in in it there will be micro-cracks where you drill the hole. It is then advisable to anneal the plastic if you want the finished product to last in an outdoor environment.