Max torque allowed on cypress wooden beam

Question

I am trying to calculate the maximum torque that a piece of cypress wooden beam can handle before it snaps or becomes structurally unsound. I know this will depend on the thickness and width of the wooden beam.

Does anyone know where I can lookup the property value (coefficient or some other value) that will allow me to calculate the max tolerable torque given the thickness and width? I am looking for the value for cypress wood, but if there is a database for other types of woods, that'd be even better.

For reference, I am trying to design a bathtub bench for sitting that will hold up to 300 pounds. The current plank is 1 inch thick x 3.5 inches wide x 30 inches long.

Thanks,

Answer 1

We really want to avoid putting wood members into torsional stress states. The previous answers are valid and applicable mostly for isotropic materials. Wood is an-isotropic and its material properties are the result of the complex structures naturally formed in wood.

Trying to rely on planks in torsion is particularly problematic because the wood is very weak when tension is applied perpendicular to the grain. Wood material has very little strength between the grain lamina.

Significant levels of torsion on a wood plank would almost certainly split the plys and have a non-ductile failure. This is a failure state the you would probably not capture by simply taking the allowable shear stress into the 'stress due to torsion' equation.

My recommendation is to add more framing to eliminate the reliance on torsion in wood.

Answer 2

You can always try engineering toolbox online, it may take some digging but they would have something to give you an idea. If you are a student you can use Fusion 360 to test various materials, using the simulation environment. It is not exact, but it should have a material you can use for wood. I've used this method to calculate steel pipe strengths. Hope this helps!

Answer 3

The angle of twist of a wood beam in Radians is $$ θ = \frac{TL}{ GK}\space and\space torsion\space stress\space is f= T/(\beta\times h\times b^2) $$ T applied torque, L member length, G shear modulus which could be approximated by E/16, K is cross-section shape factor for a rectangular beam and beta is approximately 3 for your member dimensions, 1 and 3.5 inches.

so the stress is

$$F_{shear} = \frac {T}{\beta\times h\times b^2} $$

h is the long side and b is the short side of your member section.