# Wood support calculation [closed]

Does anyone know how to calculate the width of the support (x) required for 6x10 lumber spanning 5’ supporting 2700 lb of uniform load and a load perpendicular to grain of 2000lb/ft2?

• What are the specs for the lumber? Once you have that there are many sites that show the calculations. Commented Oct 29, 2020 at 5:34
• Can you please point me to some of the sites? I haven’t found anything relatable. Perhaps I am using a wrong search words...
– SSan
Commented Oct 29, 2020 at 6:52
• "Beam loading" in Google gives as one of my first results: engineeringtoolbox.com/beam-stress-deflection-d_1312.html but beam loading seems an obvious search term... Commented Oct 29, 2020 at 7:05
• Welcome to Engineering! This seems to fall into our homework question category (even if it isn't homework!). In order for such questions to be answered in this site, we need you to add details describing the precise problem you're having. What have you tried to solve this yourself? Please edit your question to include this information. Commented Oct 29, 2020 at 15:15

## 1 Answer

I wouldn't be that much worried about the support bracket. There are standard brackets that would more or less be able to handle the load.

What i'd be more worried is that on the strong axis of beam there are 2700lb of load (vertical), while on the weak axis the applied load by my calculation is 11250lb.

Assuming x is the horizontal axis (6 in), y is the vertical (10 in) and z is the length of the beam (5 ft), then the ratio of the moments of inertia is $$\frac{I_{xx}}{I_{yy}}=2.778$$close to 3, while the corresponding moment ratio is $$\frac{M_{xx}}{M_{yy}}=0.24$$.

That results in $$\sigma_{yy}= \frac{M_{xx}}{I_{xx}}\cdot \frac{10 in}{2}= 202 psi$$, and $$\sigma_{xx} = \frac{M_{yy}}{I_{yy}}\cdot \frac{6 in}{2}= 1406.25 psi$$. The total stress in the corners would then be in the order of $$1600 psi$$.

In order to optimise the use of the material (although the stresses are not that high), I would recommend rotating the beam by 90 degrees. In that case, the stresses would be significantly lower ie. $$\sigma_{yy}=337.25$$, $$\sigma_{xx}=506.25$$, and the total stress in the corners would be in the order of $$850 psi$$.

• Thanks for your thought process. I was actually given a question to solve for’x’. And I haven’t figured out how. If any of you know, please share.
– SSan
Commented Oct 30, 2020 at 2:25
• Is the beam supported only on one end? Commented Oct 30, 2020 at 2:47