# 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. – Solar Mike Oct 29 '20 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 Oct 29 '20 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... – Solar Mike Oct 29 '20 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. – Wasabi Oct 29 '20 at 15:15

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 $$850 psi$$.