|First Floor Plan of the Simple Cabin Design.|
When designing my cabin, if I do not choose correctly sized poles, the entire structure could collapse.
Less consequential, but more likely for the self-builder, is the use of larger structural members than required. This means more cost for little benefit. I don't want to waste money when a little calculation could help. The method of calculating the loads carried by the poles in the cabin is simplified from the engineering I learned in school, but since my cabin is so simple, the method is accurate enough. Furthermore a pole design booklet is available in personal document format here.
Since I have a sketch of the cabin, I know the spacing of the poles, location of the girts, and the layout of the walls, roof, and floors. The walls, roof, and floors are carried by the girts and poles, of course. The live load and dead load are added to give a total load for the first floor of fifty pounds per square foot; loft floor of thirty pounds per square foot; walls of ten pounds per square foot; and roof of fifty pounds per square foot. These building loads need to be multiplied by the square footage of each area, in order to determine total load. For instance the first floor has an estimated load of fifty pounds per square foot, so we need the total square feet of the first floor to determine how many pounds are being carried by the first floor.
Since the first floor is eight hundred square feet, then the total pounds are forty thousand, because fifty multiplied by eight hundred is forty thousand. The loft floor is four hundred square feet, with thirty pounds per square foot, to carry twelve thousand pounds. The walls are one thousand, three hundred, and ninety square feet, so the total wall load is thirteen thousand, nine hundred. The roof is one thousand, seven hundred, and thirty five square feet, so the total roof load is eighty six thousand, seven hundred and seventy five pounds.
So the totals are as follows:
- First floor = 40,000 pounds.
- Loft floor = 12,000 pounds.
- Walls = 13,900 pounds.
- Roof = 86,775 pounds.
Now we need to determine how many pounds each pole carries, by determining how many pounds each girt carries. For our calculations we will assume there are nine primary girts, that span forty feet each. (There is one exception that only spans twenty feet, but we will ignore it for the purposes of this estimation.) The total weight, 152,675, divided by 9, is 16,963. pounds per girt. Since each girt spans forty feet, then each foot of girt carries four hundred and twenty four pounds. Each post is spaced ten feet apart, so each post carries (424 x 10) = 4240 pounds, per girt.
Each pole carries three girts, so each pole carries twelve thousand, seven hundred and twenty pounds. The load on each pole divided by the soil bearing capacity will give the square footage required of the footings. (12720 / 4000) = 3.18 square feet. The four thousand pounds is an estimation that will have to be revised once the site determined. Therefore a square footing of 1.79 feet x 1.79 feet would meet this requirement. Alternatively a circular footing of 2.02 feet in diameter would suffice.
The safe limit for bearing capacity is given in American National Standard Specifications and Dimensions for Wood Poles. The safe load for the center row, which extends above the ground plane twenty five feet, is met by a class 2 pole with a maximum load of 16,500. A class 2 pole of 35 total feet, is 36.5 in circumference at the base; or in other words, 11.5 inches in diameter. The diameter is just under 8 inches at the top.
If I so wished, I could use a class 5 pole for the poles that do not extend to the gable apex, as they are only 13 feet above the ground plane. This means that the pole is 7.3 inches in diameter at the base, and 6.04 inches at the top. I don't think I will use thinner poles, but instead I will use the same class of pole throughout the Simple Cabin for uniformity. Perhaps this will change in the future if the cost between the pole classes is significant.
So now I have a better idea of the poles needed, and the footings needed. Hopefully I didn't make any mistakes in the calculations. Do you see any? Is there an easier way of engineering the cabin? Have any comments on unforeseen structural problems? Was my exercise helpful for you? Let me know in the comments.