was wondering if any one here can direct me to some pics showing the wing attachment for front and rear, it appears to me that the front area the front wing attachment is bolted to the top crossmember which is just butt jointed, is this right? I figure Im not seeing something because this doesn't seem very strong to me, as for the rear I don't have a clue on what Im seeing
The outboard wing attachments bolt through the spar and are pinned to a channel that runs across the fuselage below the instrument panel. There are no reports of wings detaching in flight.
was wondering if any one here can direct me to some pics showing the wing attachment for front and rear, it appears to me that the front area the front wing attachment is bolted to the top crossmember which is just butt jointed, is this right? I figure Im not seeing something because this doesn't seem very strong to me, as for the rear I don't have a clue on what Im seeing
My understanding is like this:
The load does not directly come on to the wood cross member. The two wings are held together by the aluminum channel to which the wings are attached. The wood cross member above the aluminum channel merely holds the channel in place.
In flight, the load on the joint is not lateral / sideways, but an up-down load, with fuselage pulling down and wing pulling up. There is also front to back load at the joint due to the drag on the wings caused by the airflow.
Also that wood member above the aluminum channel is not simply butt jointed. The top fuselage sheet acts as a massive gusset for that member. For that butt joint to give way, the top fuselage plywood sheet would have to rip at the joint.
The wing anchor points do carry lateral loads, some of the those are imparted by the diehedral of the wing and the other due to the loads which the struts place on the wing. Whether you have a mid wing or high wing aircraft, as long as the struts are place below the wing, the cross member which carries the wing attachments will be loaded in compression (as long as you are experiencing positive G's). I have not looked at the drawings in detail but I assume that the aluminum cross member is sized so that it can carry those lateral loads without buckling.
If anyone is interested, I could make a simple load diagram to illustrate the point.
There is no shear at the wing root because the lift struts attach to the wing under the center of the span-wise lift distribution which for all rectangular wings is at 46% of the span. Because the ailerons are full-span, the shape of the half-span lift distribution is unchanged when the aileron is deflected. The loads transferred from the main spar to the spar carry-through are compressive under positive g flight. Some of that compressive load is resolved anti-drag load at high angles of attack and resolved load from the rear strut pulling forward. The rear spar root attachment is relatively lightly loaded hence the carry-through is wooden. The structure does have to account for asymmetric wing loading. FAR Part 23 specifies that the structure has to take loads stemming from 100% of the design load on one wing while 70% of the load is imposed on the other for utility category airplanes, 60% for aerobatic airplanes. So for the maxes, the main spar carry-through must react about 660 lbs. The Minimax accomplishes this using 4 3/16" bolts bearing in the wooden cross member and carrying the stress into the plywood top sheet. The Himax reacts the load with the /\ cabane struts.
The rear spar carries less load then front spar in two spar wing structure but in minimax the rear strut carries more load than the front (1738 lbs vs 1072)according to the stress report.
The reply I gave above was for a cantilevered wing without struts, so it is not relevant to the original question.
Regardless, one issue which still puzzles me is why there is no shear loading on the spar to fuselage attachment point.
Let us take a simpler case of a Cessna 150, which only has a single strut for each wing. I consider a Cessna's fuselage weight lifted by the wings at 4 points - 2 per side, one of which is the root spar attachment point, and the other at strut-to-wing attachment point. So why is the root fitting not shear loaded ? There should still be some amount of shear loading at the root attachment, but much less than in a cantilevered / strut-less wing.
It is because the lift is linear across the span of each individual wing so all the lift is expressed as on the upper end of the strut because that is the mid point. Effectively, that is pulling the aircraft up on its own. Imagine the strut to wing joint was VERY strong and then cut the wing attachment straps to the fuselage. Even though it isn't actually attached, the same lift is applied to the strut so the aircraft rises just as it did before. It just seems like it shouldn't work, doesn't it?
It is because the lift is linear across the span of each individual wing so all the lift is expressed as on the upper end of the strut because that is the mid point. Effectively, that is pulling the aircraft up on its own. Imagine the strut to wing joint was VERY strong and then cut the wing attachment straps to the fuselage. Even though it isn't actually attached, the same lift is applied to the strut so the aircraft rises just as it did before. It just seems like it shouldn't work, doesn't it?
Actually, it is possible to lift the entire Max air-frame and keep it in flight by just the struts. Think of a high wing airplane, where the wings are placed a foot or two higher than the top of the fuselage and the wings do not directly attach to the fuselage, but the fuselage is slung below through a set of struts. We obviously need to have stronger struts to start with, that can take the load of the entire airplane minus the weight of the wings. (I am minus-ing the weight of the wings because the wings lift themselves in flight.)
However it is likely that is not how the Max is designed.
My Max 103's struts are not designed to lift the entire gross weight of my Max which is 460 lbs ( that is 525 lbs gross MINUS 65 lbs which is weight of wings ). If the struts cannot lift 460 lbs with some safety-margin built in, then it implies some load is lifted by the wing root attachment , and if that is so, then some shear load is carried by the wing root attachment.
The rear spar carries less load then front spar in two spar wing structure but in minimax the rear strut carries more load than the front (1738 lbs vs 1072)according to the stress report.
Any explaination for that.
KK
I believe the design case for the rear strut is at VD or dive speed whereas the design case for the main strut is VA. At the dive speed, the angle of attack is relatively low, only about 4°, while the airspeed is very high. The wing is lifting slightly more than at VA, however, the wing pitching moment(nose-down), acting at the quarter-chord point(behind the main spar), is very high, much higher than at VA. Because the wing pitching moment is centered behind the main spar, it relieves some of the tension on the main strut. Meanwhile, the rear strut has to react the very high pitching moment.
It is because the lift is linear across the span of each individual wing so all the lift is expressed as on the upper end of the strut because that is the mid point.
I have read that lift is linear across the wing's span, only from wing root till strut attachment point. From that point outward, the lift profile is elliptical. But I have no idea how this theory applies to the Max. I also remember reading that for cantilevered wings, the lift is elliptical across the entire span of the wing.
Take the wing off the aircraft and put it in a wind tunnel. It will rise straight up so the spanwise distribution must be equal side to side irrespective of the shape.. if you put the strut in the middle and it will then lift the aircraft from the axle, there can be no shear on the fuselage mountings.
I have read that lift is linear across the wing's span, only from wing root till strut attachment point. From that point outward, the lift profile is elliptical.
This is a simplified approximation.
The Schrenk method says the span-wise lift distribution of a wing (half-span) can be approximated by taking the average of the wing planform and a quarter ellipse. We want to put our strut at the center of the lift distribution so that the lift on either side is equal and there will be no vertical force on the root spar fitting. The center of a quarter-ellipse is given by 4/(3Pi) * half-span. The center of a rectangular wing is half-span/2. If we set the half-span equal to unity(1) we'll get a percentage of the half-span.
Then we have: (4/(3Pi) + 1/2)/2 = .46 So, the center of the span-wise lift distribution is 46% of the wing half-span out from the root. The Minimax wing is 139" and the strut is attached 64" out from the root. 64/139 = .46.
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