Wing spar sizing


Well-Known Member
Dec 31, 2009
Barcelona, Spain.

After studying a bit of theory, it is now time to play with the formulas. The idea is to size a main spar for a given load, calculated from the specifications of a primary glider.

Although the equations are not very complicated, There is something wrong with the values I get for the box spar, in the last example.

The specs below, are from a primary glider of rectangular wing;

MTOW = 225kg,
Wingspan = 10m
Wing area = 10m2
AR = 10, C = 1m
Load factor = 3,5
Safety factor 1,5

First I get the total wing loading = (225*9,81*3,5*1,5)/(10)=1158,81N/m2

Then I divide that number by 2, to get the wing loading on the half wing.
I assume the fuselage S is negligible; the fuselage it is a 2D tubular weld structure. The resulting W/L for the half wing is 579,403 N/m2.

I use a basic flat lift distribution, that is equal to the half wing loading. The goal of this exercise is to learn how to use the basic formulas. Later on we can estimate a more accurate lift distribution using Schrenk equation.

My lift distribution is y = 579,4013 N/m2

I integrate the lift distribution to get the shear stress. I locate the origin on the tip of the wing, so that I do not have to calculate the attaching points loads, to get shear and moments over the half wingspan. This is to save time and get some quick numbers

My Shear is equation is 579,403*X N/m2
The maximum shear would be at the root (X=5 m); 579,403*5 = 2897,02 N/m2

I integrate Shear to get Moments over the wingspan;
M = 579,4013*X^2/2
The maximum flexion would be again on the root; 14,485 KN/m2
Although this calculations are intended for a cantilever wing aircraft. I pick the value of maximum flexion at half wing, as if there was a strut placed in there, intended to save weight on the main spar. If the strut was placed on the half wing, the maximum moment would be there.
The M at X = 2,5 m is 3,621 KN

If the maximum M, and the maximum bending strength of the material of the spar is known, we can calculate the required surface using Navier's Law Equation;
Max strength = Max M *y / I
In where I is the moment of inertia, and y is the distance between the neutral axis (center of the beam), and the upper or lower part of the beam

Frati in his book The Glider, provides some data for "Fir" at maximum compression and tension.
According to the given data, the maximum tension admitted by Fir is 1,26 times higher, than its maximum compression value. I therefore pick the minimum value as my limit; the maximum strength in my equation would be that of compression.

According to Frati the maximum allowable compression in fir is of 380Kg/Cm2 -> 37265270 N/m2 -> 37,26 MPa

I pick the maximum possible thick airfoil for my wing, to allow for the highest possible spar and thus save weight; the maximum compression and tension loads are located at the extreme parts of the spar. The NACA 4418 has roughly a maximum thickness of C/6.

If the chord of my wing is of 1 m, the maximum height would be of 0,166 m. I consider 15 cm/0,15 m as the value for my calculations.

Full main spar:

The moment of Inertia of a rectangle is (b*h^3)/12
The value h is given by the maximum thickness of the wing = 0,15. My only unknow is then only b.
I can find the value straight by using Navier's equation;

37,26*10^6 = (3,621*10^3*0,15)/(b*0,15^3/12)

b = 0,0259m, -> 2,6cm

I have searched on the internet an average value for fir density; 0,53*10^3(kg/m3).
The weight of the main spar would be the surface of the main spar * the spar lentgh * density;
W = 0,026*0,15*10*0,53*10^3 = 20,67Kg

It is crazy, the weight of the main spar is almost 10% of MTOW. I believe Billski pointed a value similar to that, for a cantilever wing.

Given that the maximum flexion happens on the outboard of the spar, the optimal solution is to use a hollow spar to reduce weight, and locate the maximum spar surface in the extremes; next to the upper and lower part of the wing.

Box spar.

Althought Frati uses a method on his book, it seems a bit confusing to me. Since we can use the same equation that we have used for a full spar, I calculate the moment of inertia of a hollow spar using Steiner's theorem.

I = I+A*d^2
I = [(b*h^3)/12+b*h(y/2-h/2)]
I calculate the moment of inertia of one cap, because the spar is symmetrical and both caps are equal.

Now the calculations are a bit more difficult, since we have two unknowns (b and h, the respective base and height of the cap), and only one equation.

Given that it is far more easy to isolate b than to isolate h. I isolate b, and I pick a random value for h;

As my first iteration, I pick h = 2,6 cm, the thickness of the full spar.
b/2(0,026^3/6+0,026*0,15-0,026^2) = b/2*0,0033227

I = My/(fir Max Compression)
I = (3,621*10^3*0,15/2)/(37,26*10^6) = 7,288*10^-6
then b = 2* (7,288*10^-6/3,227*10^-3) = 0,004518 m = 4,518mm

The value of b seems too low, there is probably an error on the calculations?
I check everything but I am not able find the mistake.

I pick h = 1,3 cm, half of the thickness of the full spar, for the second iteration.
I get b = 0,008184 m, 8,1 mm.

Now I pick h = 1 cm,
The resulting b is 0,010412 m, 1 cm.
That is a nice result, because it means a squared cap, which is more convenient for ease construction.

I calculate the resulting weight of the last iteration;
0,001*0,001*10*0,53*10^3= 1,06kg,

The resulting weight of the box spar, is 20 times lower than that of the full spar. Whereas in Frati example the difference is only 2 times higher. This means that, probably there is an error of one order of magnitude, somewhere in the calculations.

Thanks in advance for your comments and help!

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