Aerowerx
Well-Known Member
I have been amusing myself (OK, I'll confess---trying to stay awake at work) by reading about calculating the required spar cap size.
I have even worked out an example to see how it works.
My references are the comments by Billski in this old thread, and an old EAA article.
In particular, this paragraph from the article:
Some assumptions: gross weight 700 pounds at 4g, safety factor of 1.2, 36 foot span, sine(x) distribution, 5 foot root, 2 foot tip, 12% thick airfoils.
The spar caps will be laminated out of 1/8th inch Douglas Fir strips 1 inch wide, which has a compression limit of 6900 psi, and a modulus of rupture of 12500 psi.
I arbitrarily decide the minimum thickness would be 3/8th inch.
I then calculated how many plys would be needed at 1 foot intervals, using the methods mentioned by Billski and in the article.
For the top spar cap what I ended up with is 21 plys at the root. At 10 feet on the semispan, only 3 plys are needed and this is continued out to the tip. The bottom spar needs 12 plys at the root and reaches 3 plys at 8 feet.
Note that I rounded up to the next whole number to determine the number of 1/8th inch plys.
Does this sound reasonable?
One question I have is about the last line in the article clip above, where it says "dividing the area by the original cap height times two caps". Just what does that mean? Does the "two caps" refer to there being two strips on each side of the web, or to the top and bottom caps?
I have even worked out an example to see how it works.
My references are the comments by Billski in this old thread, and an old EAA article.
In particular, this paragraph from the article:
Some assumptions: gross weight 700 pounds at 4g, safety factor of 1.2, 36 foot span, sine(x) distribution, 5 foot root, 2 foot tip, 12% thick airfoils.
The spar caps will be laminated out of 1/8th inch Douglas Fir strips 1 inch wide, which has a compression limit of 6900 psi, and a modulus of rupture of 12500 psi.
I arbitrarily decide the minimum thickness would be 3/8th inch.
I then calculated how many plys would be needed at 1 foot intervals, using the methods mentioned by Billski and in the article.
For the top spar cap what I ended up with is 21 plys at the root. At 10 feet on the semispan, only 3 plys are needed and this is continued out to the tip. The bottom spar needs 12 plys at the root and reaches 3 plys at 8 feet.
Note that I rounded up to the next whole number to determine the number of 1/8th inch plys.
Does this sound reasonable?
One question I have is about the last line in the article clip above, where it says "dividing the area by the original cap height times two caps". Just what does that mean? Does the "two caps" refer to there being two strips on each side of the web, or to the top and bottom caps?