(We are still calling for reduced speed [ed V_{max }] at high gross weight.
The analysis was written with varying assumptions to cover PHA and PLA conditions. NHA and NLA conditions are not critical since the wing is symmetrical structure up and down, except for the rear spar.
Also, the load factors are at +7.5 and –5.5 or 5.5/7.5=.733 or +CL_{max} = 1.5 and – CL_{max} =1 or 1/1.5=.666 and the profile angles are higher at – angles less chord load. ??
Gross Weight 740 lb.
Wing Weight 215 lb.
Corrected wing weight for the nonelliptical span wise distribution of wing weight 195 lb.
(740195)/2=272.5 lb. Per side.
Ultimate Load Factor = 7.5G
Air Load not resisted Locally by wing inertia is 272.5 * 7.5 = 2044 lbs. Per wing.
The semi span is (6 X 47" + 10" +.25" +.8" = 293" tip to CL (Aircraft centerline).
The wing has no aerodynamic twist and the chord distribution is near elliptical so the air load will be assumed elliptical:
.42 X b/2 * air load = M _{CL}
The wing reference plane is the mean line of the aft 2/3
of the airfoil...
... so it conservative to assume the
wing reference plane to be the 0 lift line µ
_{L0}.
The spars are normal to the wing reference plane.
The slope of the lift curve is C_{Lµ }= 2P / 1+2/A^{2} = 2P / 1 + 2/23.26 = 5.7857 per radian
Or 5.7857/57.3 = .100972 per degree.
Factor for use .1 C_{L} per degree or C_{Lµ ° }= .1
CL_{MAX} = 1.5 and 1.5/.1 = 15° = X_{W} at CL_{MAX}.
C_{DI} = (1.5)^{2} / P * 23.26 = .03079 and .03079/1.5 * 57.3 = 1.176°
The C_{D0} of the wing at CLMAX is about .012 so .012/1.5 * 57.3 = .458°
1 + .176 + .458 =1.63 . .458 is a small angle and it is conservative to neglect it.
[Ed Zero Lift Drag Coefficient… per Raymer .02 for a clean propeller driven aircraft, .015 for a jet. The high aspect ratio of sailplane wing probably justifies the slightly lower assumption (& low drag of a composite?) What is happening here is looking at the lift slope at cruise and stall angles of attack, since, when the airfoil is moving forward, not all of the load is bending the wing up and down (though this is the limiting condition to be tested), some of the force is pushing forward creating tension loads on the rear fitting. Since this is determined by the lift angle (15° ), it actually acts as a net force trying to rotate the wing about its CL; but the application is still tension on the rear fitting..]
So the normal component is
cos 15° = .966
and the forward component is
sin 15° = .259
Angles relative to the reference plane.
So the normal to the airstream ultimate moment at the CL is .966 * 2044 * 293 * .42 = 242,982 inch lb.
The Chordwise Ultimate moment at the CL is .259 * 2044 * 293 * .42 = 65,147 inch lb.
To account for the Rolling pull out at higher G’s the moment distribution is conservatively assumed to be that for rectangular distribution reference the plan view of the wing.
The fitting normal moment is 242,982 * (293 –5 / 293)^{2 }= 234,749 in lb. At the CL fitting
The Normal moment at the outboard side of the inboard rib is 242,982 * (293 10.25 / 293)^{2 }= 226279 in lb.
The normal moment at A [first foam 47" joint]:
is 242,982 * (293 10.25 / 29347)^{2 }= 157,305 in lb.
The normal moment at B [second foam 47" joint]:
is 242,982 * (293 10.25 / 293(47 * 2))^{2 }= 100,835 in lb.
The normal moment at C [third foam 47" joint]:
is 242,982 * (293 10.25 / 293(47 * 3))^{2 }= 56,870 in lb.
The normal moment at D [fourth foam 47" joint]:
is 242,982 * (293 10.25 / 293(47 * 4))^{2 }= 25,409 in lb.
The normal moment at E [fifth foam 47" joint]:
is 242,982 * (293 10.25 / 293(47 * 5))^{2 }=
6,453 in lb.
The ratio of the moments is:
Forward chord load 65147 / Normal chord Load 242982 = .268
So we can multiply the wing up bending * .268 to get chord loading.
Root rib chord loading:
226279 * .268 = 60,643 in. lb. ultimate chord loading.
Distance from CL front spar to rear wing fitting is 15".
60,643/15 = 4043 lb. tension on rear fitting. Note the shear from the root rib in the fore and aft direction is taken by the front spar. The rear spar has essentially 0 stiffness in this direction, relative to the front spar, inboard of the root rib.
Chord Bending Moments
:
[this is the part of the load trying to bend the wing, ie carrying the "weight" of the ariframe plus G forces]
M at the root rib = 60643 in.lb.
M at A 157,305 * .268 = 42,158 in. lb.
M at B 100,835 * .268 = 27,024 in. lb.
M at C 56,870 * .268 = 15,241 in lb.
M at D 25,409 * .268 = 6,810 in. lb.
M at E 6,453 * .268 = 1,729 in. lb.
The Dimensions from the CL of the front spar to
the CL of the rear spar are (x)
(x)  M/(x) tension  Cap Area Rear Spar  Rear Spar Stress tension psi  
Root Rib  14.65"  4139 lb.  .375  11037 psi 
A  12.9"  3268 lb.  .105  31124 psi 
B  11.15"  2424 lb.  .082  29561 psi 
C  9.33"  1634 lb.  .071  23014 psi 
D  7.52"  906 lb.  .053  17094 psi 
E  7.6"  228 lb.  .024  950 psi 
Spar assumption ½ Upper and ½ Lower cap area.
[As the distance between the surface of the cap & the skin becomes greater, a beam flange carries less of the surface load and become more a component of the shear web. The assumption is that the bottom half of each beam, acting with the dowels (we’ve neglected the foam, remember?) is the shear web of the composite beam, with the top halves being the flange. For the same effect, imagine a solid beam with lightening holes drilled in it. In fact, of course, the more the composite beam deflects, the greater force this additional material can assume…it may have no function under normal conditions but prevent absolute fracture under overstress conditions.]
Main Beam cap load
Thickness
H" 
A (sq. in.)  M ultimate  Mean Stress
F = M/HA 
Rear Spar loads are also compression in front spar.  Compressive Stress  
Root Rib  5.7"  2  226,279  19,849  1035  E = 20,884 
A  4.9"  .72  157,305  42,804  2269  E = 45,073 
B  4.1"  .55  100,835  44,726  2203  E = 46,919 
C  3.3"  .39  56,870  44,188  2095  E = 46,223 
D  2.5"  .24  25,409  42,348  1881  E = 44,229 
E  1.7"  .17  6,453  22,328  671  E = 22,999 
Peak fiber stress
is h/H where h is the overall height of spar.
h  H  h/H  f  F_{OF}= h/H X f
(outer fiber stress) 
Stress Chord bending psi  E outer fiber stress psi compression  
Root Rib  7.75"  5.7"  1.36  19,849  26,944  1035  28,029 
A  5.75"  4.9"  1.173  42,804  50,209  2269  52,478 
B  4.84"  4.1"  1.18  44,716  52,787  2203  54,990 
C  3.93"  3.3"  1.19  44,188  55,583  2095  54,678 
D  3.01"  2.5"  1.20  42,343  50,817  1281  52,698 
E  2.10"  1.7"  1.15  22,328  25,677  671  26,348 
[ For Carbon fibers in vacuum bagged layup, Holloman reports 130,000 tensile ultimate strength, 72,000 compressive ultimate strength. For Hand Laid UP EFiberglass Unidirectional ( used in spar caps of the lower aspect ratio wing & as reported to the SHA & tested without posts) he reports 70,000 tensile ultimate strength and 43,500 compressive ultimate strength Carbon is "twice" as good. These are relatively thin material sections, reported by physical test. The difference between the calculated stress & the rela stress of the materials is a significant part of our safety factor.]
Compression:
Cap Stability:
The compression cap is stabilized by the skin in the fore and aft direction. The Z direction is stabilized by posts through the foam between the upper and lower caps, and by the foam between the supports. Some years ago we built a test wing with no posts, just foam. This was the low aspect ratio wing; it carried ultimate with no posts…
(Paper describing this test is referenced)
Compression load:
PC = P ^{2}E_{C}I / L^{2} + E_{F}WL^{2 }/ hP
So we must evaluate the second post of station D. E_{F }= 10,000psi.
The effect of the force is: 10,000 X .489 X 34.5 / 2.05" X P ^{2} = 8,338 lbs.
P_{C} = 8,239 + 8,338 = 16,577.
The axial load …is 10,615 at D so 16,577/10,615 –1 = .56 margin.
This is conservative since we used the width of the cap for W instead of analyzing to find effective W.
[ Because the cap is directly bonded to a relatively thick skin on the top surface, the effective W of the top & bottom of the beam flange in carrying loads is somewhat greater than the width of the carbon itself. Imagine the wing carrying some sandbag weights with NONE actually applied to the area over the cap]
Compression load in posts (note the tension CAP is the same as the compression cap).
E_{C} = 6,000,000. The crushing load per running inch is f^{2} X 2 A_{caps}/Ech
The distance between posts is 5.875"
So f^{2} * 2 A_{C} * 5.875 / E_{c}H
= compressive load in posts.
f_{MEAN}  A_{CAPS}  H  F^{2} X A_{C} /HE_{c}  Ult Crush load on posts A*8*2* 5.875  
Inboard Rib  19,849  2  5.7"  23.0  270.25 
A  42,804  .72  4.9"  44.9  527.6 
B  44,716  .55  4.1"  44.7  525.2 
C  44,188  .39  3.3"  38.5  452.4 
D  42,348  .29  2.5"  28.7  334.9 
E  22,328  .17  1.7"  14.1  165.7 
We are using very conservative shear distribution span wise. Assuming that the shear is proportional to the span:
b/2 =293" and the distance from CL to the outboard side of the inner rib is 10.25". So shear at CL = 2044 lbs. ultimate. Shear at inner rib is 2044 * (29310.25)/293 = 1973 lb.
We will conservatively use the distance between the CL
of the caps h as the effective height. The external shear is:
shear  h  S/h  
Inner Rib  1,973 lb.  5.7"  346 
A  1,644 lb.  4.9"  336 
B  1,315 lb.  4.1"  321 
C  987 lb.  3.3"  299 
D  658 lb.  2.5"  263 
E  329 lb.  1.7"  188 
This slope is constant over the spar so the shear carried by the slope of the spar cap is
µ _{SC} = M/h
= D Shear internal
M  h  M/h * .01702  D Shear = shear/h  
Inner Rib  226,279  5.7"  676  119 
A  157,305  4.9"  546  111 
B  100,835  4.1"  419  102 
C  56,870  3.3"  293  89 
D  25,409  2.9"  173  69 
E  6,453  1.7"  65  38 
External shear Flow  Internal Shear Flow  E Vertical shear flow  Skin shear from vertical shear  
Inner Rib  346  119  227  113.5 
A  336  111  225  112.5 
B  321  102  219  109.5 
C  299  89  210  105 
D  263  69  194  97 
E  188  38  150  75 
Torque shear flow. The enclosed areas of the wings are:
Area of wing  2 * area  
Inner Rib  149 in sq.  298 
A  105 in sq.  210 
B  79 in sq.  158 
C  58 in sq.  116 
D  44 in sq.  88 
E  23 in sq.  46 
To be conservative, use shear center at 40% over the entire spar, and CP at .25C. .4.25 = .15C.
Note PLA condition is not critical since the C_{M} = .003 so the CP is not at 40% when pulling G’s at high speed.
The Chord At the stations =C
C  a=C * .15  D T=329 * a  S D T in Lb.  2A  S T/2A  
Root Rib  34.6"  5.19  1,708  7,858  298  26 
A  31.8"  4.77  1,569  6,150  210  29 
B  29"  4.35  1,431  4,586  158  29 
C  26"  3.9  1,283  3,155  116  27 
D  21.9"  3.29  1,082  1,872  88  21 
E  16"  2.4  790  790  46  17 
so the LE skin shears are
Vertical shear flow  Torque shear flow  S LE shear flow lbs. in.  
Root Rib  113.5  26  139.5 
A  112.5  29  141.5 
B  109.5  29  138.5 
C  105  27  132 
D  97  21  118 
E  75  17  92 
[the point here is that if you design
to prevent hanger rash & kiddie handprints then you are heavier than
one needs to be for air loads.]
We will not reduce the shear due to torque aft of the beam caps to account for of aileron deflection during High G (rolling Pull out).
Skin shear due to forward chord loads.
We will assume 60% in lower skin & 40% in upper skin,
or invert if critical. From prior table the ratio or shears is .259/.966=.268
a  a * .268  x front to rear spars  shear flow 100%  shear flow 60%  shear (table)  S shear flow aft of spar  
Root Rib  1,973  529  14.65"  36  22  113.5  135.5 
A  1,644  441  12.9"  34  20  112.5  132.5 
B  1,315  352  11.15"  32  19  109.5  128.5 
C  987  265  9.33"  28  17  105  122 
D  658  176  7.52"  23  14  97  111 
E  329  88  7.6"  12  7  75  82 
Fasteners:
The Net Spar Cap Area at the two outer bolts is:
1" X 2" –[(.375 * 1)*2] = 1.25 sq. in.
Tension side member stress is
39,530/1.25 = 31,624 psi ultimate.
The 2 outer bolts of the front spar fitting are .375 Dia or .1875 radius
I = P R^{4}/4 = P * (.1875)^{4}/4 = .00097
Allowable ultimate bending is 145,000 * bending form factor of 2 = 290,000psi
The Cap is 1"wide with .125 plywood on the sides, allowing for bearing on the fitting .05" the moment is P_{U}/2 * (1/4" + .125 +.05)=M or P_{U} * .2125 =M
So MR/I = P_{U} * .2125 * .1875/.00097 = 290000
or P_{U} = 290,000 * .00097 / .2125 * .1875 = 7,060 lbs. per Bolt.
CL Bolts for Disassembly
h = 6" Vertical distance between fastener bolt
Dia = .75" R=.375"
Area = .441 sq in.
I = .01553
the compression load from chord bending will be divided 50% upper & 50% lower, since the upper bolt is displaced up. The lower bolt is also slightly afore the mean line of the lower cap.
Upper bolt load 234,759/6=39,127+4,139 * .5 = 41197 lb.
Lower Bolt Load 234,759/6 = 39127 – 4139 * .5 = 37058 lb.
Tension on legs of lower fitting
.
two fittings fore and aft so 37,058 lb/2 = 18,529 lb.
tension load.
[drawing]
area is 2 * .7 * .25 = .35 sq. in.
18529/.35 = 52,940 psi at ultimate.
Normalized 4130 = 90000 ultimate 90000/52940 – 1 = .7 margin.
at limit load 52940/1.5 = 35293psi. with a stress concentration factor of 2.5 the bending part is 1.5 the form factor for rectangular sections in bending is 1.5 so
1.5/1.5 * 35,293 = 35, 293 so 35,293 + 35,293 = 70,586psi
Main fitting bolts at CL
.75 Dia R=.375
Upper Bolt Load = 41,197
2 shear planes so 41,197/2 = 20,599 lbs.
area is P * (.375)^{2 }=.44 sq in.
20,599/.44 = 46,816 psi OK
or the bolt is good for 33,000 lb ultimate
33,000/20,599 – 1 = .60 margin
Rear Beam Fitting
ultimate chord bending moment 60,642 in lb
15" CL front spar to rear spar
60,642/15" = 4043 lb tension
torque at root rib 7858 lbs
7858/15" =524 lb shear
[drawing]
horizontal forces only are treated.
4043 * .25 / 2" = 505 lbs tension in upper cap and
4043 –505=3,538 lbs tension in lower cap
vertical force moment resisted by horizontal force
524 * 3.3"/2" =865 lbs.
865 lbs. compression in upper cap
865 lbs tension in lower cap
S upper cap load 505 –865 = 360 lbs compression
S lower cap load 3538 + 865 = 4403 lbs tension
tear out in lower cap
4 X 8" X .5 = 3.2 sq in shear
at load 4403 lbs
so 4403 /3.2 = 1,376 psi shear in carbon epoxy cap (OK)
if fitting on aft side only
then 1,376 X 2 = 2752psi (OK)
tear out shear allowable = 4,000 psi.
Upper cap tear out OK
Lower cap bolts
4043/4 bolts = 1101 lbs per bolt.
in shear for ¼" bolts shear allowable = 3600 (OK).
3/8 pin allowable shear [(4043)^{2} + (425)^{2}]^{1/2} = 4077lb
4077/2 = 2039 lb
allowable shear 5400 lbs OK
note: vertical shear force can be taken by the center bolts into the plywood.
tension stress in carbon epoxy lower cap. The cap is .5" wide and .25" deep.
,85*.5 = .425 sq in.
the first bolt removes .25 * .5 = .125 sq in.
E = ,425.125 = .3sq in
so 4403/.3 = 14677 psi ) ok
the vertical load was transferred to the center of the fitting in the former analysis.
now we must transfer it 2"more to the root rib.
524 * 2" / 2" = 524 lb D compressions in the upper cap and tension in the lower cap.
So the cap loads at the root rib are:
upper cap load = 360 + 524 = 884 lbs compression
lower cap load = 4403 + 524 = 4927 lb tension.
The upper cap load must be taken in the skin of the wing in 47" or less because the upper cap fades out in the first foam block.
884/47 = 19lb per inch, use 25 for the possibility of the notch not being full outboard in the tapered out region. The shear flow 135.5 from prior plus 24 = 161 shear in skin assuming the shear adds and goes one direction only. The allowable shear is 400 lbs in.
Tension in rear beam attach lug
fitting is .25 in thick
area is .4 X .25 X 2 = .2 sq. in.
P/A = 4077/.2 = 20,385 psi OK
Rear Beam fitting on fuselage
M = 4077 X 1.2" = 4,892 in lb.
I = .187 * (1.75)^{2} / 12 = .0835 and .0835 X2 = .167 = I_{2}
MY/I = 4,892 * .875 / .167 = 25632 psi OK
Aadditional stuff:
Analysis of bending in wings – deflection predictions for proof loading tests.
Method 1 (Mohr’s Method): Method of elastic weights.
Method 2 (Greene): Moment area method.
Assumptions:
Fuselage Loads Moved to root ribs.
Moment loads averaged for wing sections, ie. section A – root rib to Station A (47") is Root load Plus ½ Station Load. Section B = ½ Station A load Plus ½ Station B load. This will reflect ability to place physical loads (sandbags or equal).
Calculated deflections are actually deflection of root rib "attach points" for imaginary beam supported at midpoint; (method per Bruhn).
Gross Weight 740 lb.
Wing Weight 215 lb.
Corrected wing weight for the nonelliptical span wise distribution of wing weight 195 lb.
(740195)/2=272.5 lb. Per side.
Ultimate Load Factor
=7.5G
Air Load not resisted Locally by wing inertia is 272.5 * 7.5 = 2044 lbs. Per wing.
4088 lb. total.
STATUTORY Aerobatic loading = 6G
Air Load not resisted Locally by wing inertia is 272.5 * 6.0 = 1635 lbs. Per wing
3270 lb. Total.
"HIGH PERFORMANCE SAILPLANES" (Per Thurston, Design for Flying) 5.33G
Air Load not resisted Locally by wing inertia is 272.5 * 5.33 = 2452 lbs. Per wing
2904 lb. Total.
UTILITY Category 4.4 G
Air Load not resisted Locally by wing inertia is 272.5 * 4.4 = 1199 lbs. Per wing
2398 lb. Total.
OVERWEIGHT – UTILITY CATEGORY FIXED GEAR MOTORGLIDER
(Option for utility category motorglider using fixed gear & engine adding 100 pounds additional empty weight. No other structural loads, ie. Engine mount, etc. have been checked. Roughly Based on simple spring steel gear & Subaru EA81 Engine, Direct drive)
Adding 100 Pounds, 342.5 * 4.4 = 1507 lb. Per wing, 3014 total
NORMAL Category 3.8G
Air Load not resisted Locally by wing inertia is 272.5 * 3.8 = 1036 lbs. Per wing
2072 lb. Total.
Test Loading Table: All loads carried
by wing as beam in bending.
Left Tip  Left A  Left B  Left
C 
Left
D 
Left
E 
Left
RR 
Rt.
RR 
Rt.
A 
Rt.
B 
Rt.
C 
Rt.
D 
Rt.
E 
Rt.
Tip 

Frac  0  .5%  2%  5%  9%  14%  20%  20%  14%  9%  5%  2%  .5%  0 
3.8  0  10  41  103  186  290  414  414  290  186  103  41  10  0 
4.4  0  12  48  120  216  335  479  479  335  216  120  48  12  0 
5.33  0  14  58  145  261  406  580  580  406  261  145  58  14  0 
6.0  0  16  65  163  294  458  654  654  458  294  163  65  16  0 
7.5  0  20  81  204  368  572  817  817  572  368  204  81  20  0 
OW  0  15  60  150  271  422  602  602  422  271  150  60  15  0 
Modulus of Elasticity E. Estimated for the Skin At 2.0 * 10^{6}. This is similar to tested BID.
( in tension, Neubert in Sport av. 1978).
Values in the 30 to 36 range have been reported for carbon. However, fabric density may be very low in spar. Tensile strength is likely to be much higher than compressive. Pending comparative test to aluminum and/or coupon test, estimate corresponding to 1/3 ultimate stress used, ie 12*10^{6}.
NOTE: Here’s my take on estimating the level of conservatism: We build & test a spar of carbon fiber, with wood posts as shear web. Suppose we built a wing with no spar & no carbon? If we calculate a simple beam of 15 meters(48’), area 100sf,with a nominal average depth of 4", a (blue foam) shear of 12psi, and a (skin only) E of 2*10^{6}
Per foot skin area A
A=2*12"*.3"=7.2 in^{2}
Distance of panel CL to NA
Y=.3"/2 + 4"/2=2.15 in.
I=7.2*2.15*21.5=33.282 in^{4}
S=I/C where C = Distance from beam neutral axis to skin
C= 4"/2+.3 = 2.3 in.
S = 33.282/2.3 = 14.47 in^{3}
We’ve all seen glass wings deflect
on sailplanes: what load would that imply?
D=12" = 1278*WL^{4}/195EI +WL^{2}/4(h+c)G Where G = Compressive stress = 43500psi
12" = ( 1278 * W * 5308416/195*2000000*33.232 + W * 2304 / 4 * 6.3 * 43500 )
12" =.523 W psf
ie 1ft. deflection = .523 psf load or about 50 lb .
At 48" deflection, about 204 Lb would be carried still within the material stress limits.
Estimating from the moment diagram that the weight over the supports & first ¼ panel of such as beam
as a wing equals half the load of a simple beam (the moment area method as calculated for the actual wing)
we can estimate about 400 pounds of load that could be carried by this simple foam and skin beam. In the past I’ve built eight foot sections that confirm this level of strength.
As a very rough estimate, then we can see that the safety factor of ignoring the skin & foam for bending
Is an allowance of about 10% of the calculated ultimate strength, at the assumed material numbers.
While the other simplifications
& conservative estimates have not been estimated, I’d believe we have
a safety factor of between 1.2 and 2.0 over the calculations, with the
biggest variable, of course, being the actual realized strength of fibers
in the carbon & Kyntex; these are subject to & will be part of
the test.
In any case, we’ll use the loads
tabled above to create a wing loading moment diagram, and take a pass at
forecasting actual wing deflections we’ll see under test, based on both
Mohr’s & Green’s methods.