w(z) distance down wing [ft]

Transcription

1 Loading [lb/ft] SAE Aero Design Regular Class 2017 Analysis performed by Aida Hrickson Bing Moment on the Wing & Strut Location One of the primary advantages of using an external strut to support the wing is to reduce the bing moment on the wing. Therefore, the bing moment along the wing was calculated for varying strut locations along the wingspan to determine the strut location that would yield the smallest bing moment in the wing. The goal of this analysis was to find the location of the strut at which the bing moment between the fuselage and the strut was the same as the bing moment at the strut. The lift distribution was determined from Schrenk s Lift Distribution. A third order polynomial was used to approximate the distribution: w(z) y = x x x distance down wing [ft] Figure 1 Lift Distribution along the wing The wing lift distribution (equation 1) was simplified into two point-loads, one inboard of the strut and one outboard, acting through the centroid (equation 3) of the lift distribution on either side of strut location.

2 w(z) = z z z (1) b A = w(z)dz a z = 1 A b zw(z)dz, location of centroid of lift distribution a (2) (3) The attachment of the wing to fuselage, strut to wing, and strut to fuselage were all modeled as pin joints. The resulting free body diagram of the wing (Figure 2) was a simply supported beam with reaction forces F ry, F rx, F sy, and F sx and two point-loads F 1 and F 2. Figure 2 Free Body Diagrams Using the Method of Sections and iterating for strut locations starting one-quarter down the wingspan to three-quarters of the wingspan, the bing moment was calculated using MATLAB. Below is the MATLAB code used to calculate the bing moments.

3 %%Aida Hrickson November 2017 clc; clear all ; close all z=[0:0.125:5.5]; %increments along wingspan %importing data from Excel Sheet centroid(:,1)= xlsread('strut_wing_bingmoment.xlsx','data','a3:a24'); %strut location centroid(:,2)=xlsread('strut_wing_bingmoment.xlsx','data','c3:c24'); %zbar1: centroid inboard of strut centroid(:,3)=xlsread('strut_wing_bingmoment.xlsx','data','f3:f24'); %zbar2: centroid outboard of strut Force(:,1)=xlsread('Strut_Wing_BingMoment','data','A3:A24'); %strut location Force(:,2)=xlsread('Strut_Wing_BingMoment','data','H3:H24'); %F1 (lift distribution inboard of strut) Force(:,3)=xlsread('Strut_Wing_BingMoment','data','I3:I24'); %F2 (lift distribution outboard of strut) Force(:,4)=xlsread('Strut_Wing_BingMoment','data','J3:J24'); %Fsy (force at strut) Force(:,5)=xlsread('Strut_Wing_BingMoment','data','K3:K24'); %Fry (rxn force at wing-fuselage connection) figure(1); hold on M = zeros(length(z),length(centroid)); for i=1:length(centroid) for j=1:length(z) if (0<=z(j) && z(j)<centroid(i,2)) M(j,i)=-Force(i,5)*z(j); %1st section before F1 elseif (centroid(i,2)<=z(j) && z(j)<centroid(i,1)) M(j,i)=Force(i,2)*(z(j)-centroid(i,2))-Force(i,5)*z(j); %2nd section b/w F1 and Fs elseif (centroid(i,1)<=z(j) && z(j)<centroid(i,3)) M(j,i)=Force(i,2)*(z(j)-centroid(i,2))-Force(i,4)*(z(j)- centroid(i,1))-force(i,5)*z(j); %3rd section b/w Fs and F2 else M(j,i)=Force(i,3)*(z(j)-centroid(i,3))+Force(i,2)*(z(j)-centroid(i,2))- Force(i,4)*(z(j)-centroid(i,1))-Force(i,5)*z(j); %4th section after F2 if (z(j) == centroid(i,1)) strut_moment (i,1) = M(j,i); plot(z, M(:,i), centroid(i,1), strut_moment(i,1),'o');%bing moments at all locations % figure(1) % plot(z, M(:,13), centroid(13,1), strut_moment(13,1),'o')

4 % figure(2) % plot(z, M(:,14), centroid(14,1), strut_moment(14,1),'o') figure(3) %best strut location ft from fuselage plot(z, M(:,15), centroid(15,1), strut_moment(15,1),'o') Figure 3 Bing Moment for different strut locations Figure 4 Bing Moment for Strut Location at 3.125ft highlighted in blue

5 Spar Sizing Because the spars in the wing withstand most of the loads, it is essential that the spars are properly sized. Using the modulus of rupture of Sitka spruce with a factor of safety of 2.5 and the maximum bing moment with a load factor of 5, the minimum moment of inertia of the spars (top and bottom) was calculated as I min = 0.18 in 4. Considering common cross-sectional areas of spruce that would be available for purchase, a cross sectional area of both top and bottom spars is 5 16 W 1 4 T in2, which yields a moment of inertia of 0.2in 4. To confirm that the spars would not buckle from the compressive loads from the strut, the compressive strength of the strut along with the component of the strut force in the span wise direction were used to find a minimum area to resist yielding. The minimum area to resist the compressive load, 0.12in 2, was smaller than the spar area of 0.16in 2. Calculations are outlined below. σ all = 10,200 psi (modulus of rupture) σ all = 4080 psi with FOS = 2.5 M max = lb in (load factor n = 5) y t = in (at wing tip) σ = My I x I minimum = My σ = in4 I x = B 12 (H3 h 3 ) = in in 4 F sx = lb lb, (load factor n = 5) σ comp = 5610 psi σ = F sx A A = F sx σ = 0.12in2