ASEN 2001Wing Design Lab

Design LabGroup 11 - December 13, 2017

Benjamin Vidaurre ∗ , Justin Wang † , Jarrod Puseman ‡ , Caleb Sytner §

University of Colorado - Boulder

This report intends to detail the design of a composite beam with the intent of minimiz-ing mass while maximizing strength under typical loading conditions. This is accomplishedwith an analysis of existing experimental data to create a model for determining the failurestresses of the material. The analysis yielded a loading constant (p0) of 985 Pa, a failureproducing shear stress (τfail) of 71.9 kPa, and a failure producing bending moment stress(σfail) of 10.3 MPa. This model is then used to determine the minimum necessary widthof the beam to support the load given its most likely mode of failure. The effectiveness ofthe design is then assessed heuristically after testing the beam using a wiffle tree systemto approximate an aerodynamic load. From testing, it is concluded that the wing failedunder shear stress at a force of 84.5 Newtons.

Nomenclature

σ = Stress as Result of Bending Moment in Wing [Pa]τ = Shear Stress in the Wing [Pa]Af = Cross-sectional Area of the Foam [m2]c = Half the Cross-Sectional Height: 0.01031875 mEb = Young’s Modulus of the Balsa Wood: 0.035483 GPaEf = Young’s Modulus of the Foam: 3.2953 GPaIb = Moment of Inertia of Balsa Wood [m4]If = Moment of Inertia of Foam [m4]L = Total Length of the Wing: .9144 mMfail = Bending Moment at Point of Failure [N*m]p(x) = Load Distribution on the Wing as Function of x [Pa]p0 = Loading Constant of the Wing [Pa]q(x) = Load Distribution on Wing as Function of x [N/m]Vfail = Shear Force at Failure [N]w(x) = Width of Wing as Function of x [m]x = Distance Along the Wing [m]

∗104286050†105751647‡104003252§106105221

ASEN 2001 Section 014 1 of 14 Fall 2017

Contents

I Introduction 3

II Strength Analysis 3II.A Beam Construction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3II.B Beam Testing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3II.C Analyzing Strength Test Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

III Wing Design 4III.A Finding P0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5III.B Finding the Widths . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5III.C Wiffle Tree Design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

IV Discussion 8

V Conclusion 8

2

I. Introduction

The purpose of this lab is to practice designing and analyzing composite beams. Beams have multipleapplications, and in aerospace engineering, beams are important components of wings. A composite beamis a beam that is made of multiple materials. In this lab, a composite beam that was made using GorillaGlue to connect extruded polystyrene foam and balsa wood was utilized. With both materials, the beambecomes stronger and lighter than if only one material was used.

Using failure data from previous lab groups that used the same composite beam, properties of the materialwere able to be established. After this was accomplished, an ideal wing design was developed given the beam’sabilities to withstand forces, and the composite beam was cut in order to model a wing. After this, a wiffletree design was utilized to test the wing to better understand the wing’s properties.

II. Strength Analysis

To analyze the strength of the composite material, data from the failure of several wings of the samematerial from a four point bending test were analyzed.

II.A. Beam Construction

The rectangular beams were constructed as two sheets of balsa wood glued to a foam block sandwichedbetween them. Construction of both the strength test specimens and the wing design specimens was han-dled by Bobby Hodgkinson and the course assistants before the lab. The beams for the strength test haddimensions of 36 inches by 2 inches.

II.B. Beam Testing

The testing of the strength specimens was done with a four-point loading test. See Fig. 1 for how thistest is set up. Two straps symmetrical about the center of the beam are used to apply downward pointforce on the beam. The other two forces are the upward support forces at the ends. The data of the forcelocations and breaking locations for each beam are then recorded for later analysis.

Figure 1: Four-Point Strength Test This is the four-point strength test used to gather strength data for the beamspecimens to be used later in the wing design.

II.C. Analyzing Strength Test Data

The first step in the strength test analysis was determining the method of failure of the beam. Based onthe shear and moment diagrams of a four-point strength test, the maximum shear forces occurred from theends in to the loaded straps, and the maximum internal moment occurs in the region between the straps(shear is also 0 here). If the the beam was recorded to break in shear (break location outside straps), theshear force was calculated and carried through to find the shear stress. If the beam broke on the internal ofthe straps, this is a bending moment failure and the bending moment stress is used as the failure.

ASEN 2001 Section 014 3 of 14 Fall 2017

Based on each group’s method of failure, values of the bending moments of failure σfail and shear stressesat failure τfail are found by using equation (1)

σ = − Mfailc

Ib +(

Ef

Eb

)If

(1)

if it broke in bending or (2)

τ =3

2

VfailAf

(2)

if it broke in shear. These values are then compared for each test, and outliers are removed if they liemore than 1.5 interquartile ranges below the first quartile or above the third quartile. This is plotted in Fig.2.

(a) Outliers for Bending Moment Failures This plotis all the bending moment failure stresses calculated fromthe provided experimental data. The statistical bound-aries for outliers are plotted as well.

(b) Outliers for Shear Failures This plot is all theshear failure stresses calculated from the provided exper-imental data. The statistical boundaries for outliers areplotted as well.

Figure 2: Removing Outliers from the Strength Test Data

From the data, a σfail is found by taking the average of the bending moments at failure. Likewise, aτfail is found by taking the average of the shear stresses at failure.

The safety factor of τ and σ are found using a desired failure probability and backwards calculating a zscore from a normal distribution. With a failure percentage of 1%, the safety factor required for the bendingstress would be 1.3 and that for the shear stress would be 1.75. Out of the two, the safety factor of τ isgreater giving a safety factor of 1.75.

III. Wing Design

Once the strength data of the composite have been obtained, it is possible to optimize the shape of awing given a specific load distribution. For this lab, the load on the wing is given by

p(x) = p0

√1−

(2x

L

)2

(3)

where p0, x, and L are as defined in the nomenclature. This load is intended to model the lift generatedby a wing. In general, force on a location is dependent on the width of the wing which yields an equationfor the load as a function of x.

q(x) = p0

√1−

(2x

L

)2

· w(x) (4)

The goal, then, is to design the wing to hold the load but be as light as possible (have the least material).

ASEN 2001 Section 014 4 of 14 Fall 2017

III.A. Finding P0

To design the wing, a value for p0 had to be obtained. Since the width of the wing is capped at 4 inches,we can calculate what value of p0 will cause the shear and bending moment stresses to match the allowabledesign stresses. After creating shear and moment diagrams for a 36 in wing (L = 36 in = .9144 m), theshear was predicted to be greatest at the end of the wing and bending moment greatest in the center. UsingMATLAB to symbolically integrate the load formula (Eq. (4)) and method of sections, an equation for shearforce in the wing as a function of p0 can be obtained (recall that x is set to a constant L/2 with the origin inthe center of the wing). Similarly, the shear equation can be integrated to get an equation for the maximummoment in the beam as a function of p0. Then, using the flexure and shear formulas (Eqs. (2), (1)), eachequation can be backwards solved for p0. Then, the smaller p0 is chosen as this the only load the wing willhave been designed to support. Based on the strength-test data and chosen safety factor, the value of p0used in the remainder of calculations is 984.94 Pa.

III.B. Finding the Widths

Figure 3: Shear and Moment Diagrams These are the shear andmoment diagrams for the wing once p0 is calculated.

With this value of p0 found, the widthof the wing can now be calculated. Todo this, the shear and moment diagramsfor the wing assuming constant width arecreated as a function of x (see Fig. 3).Then, for a number of points along thewing (we chose 1000 points), the momentand shear force in the four-inch widebeam are retrieved. These are taken tobe Mfail and Vfail respectively for com-putation of the width. Then, for each xposition along the wing, MATLAB com-putes the width by plugging Mfail andVfail into equations (1) and (2).

At this point, the minimum width foreach type of failure is known for somesample points along the beam. See Fig.4 The wing must then use the greater ofthe two required widths. This creates thewing profile given in Fig. 5.

III.C. Wiffle Tree Design

To test the wing, it is necessary to have a static test procedure that replicates the predicted lift loadingp(x). To do this, the wing was divided into 8 equal subsections in x. Due to the symmetry of the wing, onlyfour of these sections need to be considered and can be mirrored to the other half of the wing.

For each of these subsections, we can create an equivalent force acting on the wing. Simply, the magnitudeof the force is the area under the force distribution curve q(x) (4). The x-value of the centroid of this shapewill be the location this force is applied. It is not simple, however, to get this area. Recall that the load onthe wing is dependent on width of the wing. To solve this, MATLAB was used to get 5th order polynomialsto fit the width data. This polynomial was then multiplied by w(x) at the end of equation (4). With the loaddistribution as an integrable function of only x, the area for each subsection and location of its centroid arecomputable. These locations and magnitudes are reported in Table 1. Note that the locations are measuredfrom the origin of x at the center of the wing.

Table 1: Equivalent Loading Forces and Locations

Subsection 1 2 3 4

Location [m] -.3905 -.2803 -.1681 -.0562

Magnitude [N] 3.26 5.41 8.91 11.06

ASEN 2001 Section 014 5 of 14 Fall 2017

Figure 4: Required Widths These are the minimum overall required widths by both the shear and moment stressesallowed in the wing.

The wiffle tree needs to be designed to duplicate these forces. The wiffle tree essentially takes a group oftwo overhead forces and “combines” them into one force. Since the locations and magnitudes of the overheadforces are known, the location of where the downward force must be applied can be calculated with the sumof moments concept in static equilibrium. MATLAB was used to compute these locations. See Table 2. Notethat the distances here are reported as distance from the outer end of the beam. Also, a standard locationof the first upward force was taken to be at 2 cm in from the left end. Also note that the very bottom pieceof the wiffle tree will be completely symmetrical, so it’s location data is not required to be calculated. Thewiffle tree setup is shown in Fig. 6.

Table 2: Wiffle Tree Design Distances

Distance from left [cm] for Bar 1 for Bar 2

on Row 1 7.54 6.30

on Row 2 NA 16.38

ASEN 2001 Section 014 6 of 14 Fall 2017

Figure 5: Wing Profile This is the profile of the wing. Note that the wing is larger on the ends and middle as thisis where the maximum shear and bending moments occur, respectively.

Figure 6: Wiffle Tree Design This is the wiffle tree setup to model the given aerodynamic loading on the wing.

ASEN 2001 Section 014 7 of 14 Fall 2017

IV. Discussion

In testing, our wing held a total weight of 84.5 Newtons, which is 20.7% more weight than the designweight of 70 Newtons. After analyzing the wing at the break point, we concluded that the wing broke inshear. This is because the break point was very jagged, whereas we observed that wings that broke due inhad a more smooth break. In our predictions, we theorized that the wing would break in moment at thepoint where the wing broke in testing. This contradiction between the theoretical failure mode and actualfailure mode likely stems from how the wiffle tree reacted to the load. When the load was applied, the wingstarted bending, which caused the straps on the wiffle tree to rotate and slide toward the middle of the wing.This shifted the distribution of the weight and is likely the cause of the difference between the theoreticalfailure mode and the actual failure mode.

For a more realistic wing model, additional considerations must be taken into account. For example,a real wing will have mounts for engines and landing gear. These would add point loads to the wing inaddition to the distributed load from the aerodynamic loading. In addition, the cross-sectional area of thewing would change as opposed to the constant square cross-section we used in this lab. Both of theseadditional considerations would change the analysis required to predict the failure of the wing.

V. Conclusion

While the development of analytical models aids in estimating the failure conditions for the wing, thereare a myriad of reasons for the wing to have failed under unexpected circumstances. A possible contributingfactor to the breaking of the wing before the force on the wing reached Mfail or Vfail is any minor nicksor scoring in the balsa wood which contributes to the system breaking earlier than expected. We confirmedthis anecdotally after the initial test using the unbroken wing sections.

Another factor which may have contributed to the wing breaking earlier than predicted is imperfection inthe cutting of the wing. Differences in the shape of the wing from the intended shape would contribute to thewing reacting to the distributed loading differently than intended. This is a similar case to the shifting of thestraps used to approximate the loading conditions during testing. When the straps shifted as the beam bent,the loading conditions would have changed accordingly, leading to different failure conditions than wouldhave been predicted. Ultimately, the wing failed in between the expected minimum and maximum failureproducing values meaning that the developmental method for determining the expected failure conditionswas correct.

ASEN 2001 Section 014 8 of 14 Fall 2017

References

1Neogi, Sanghamitra. ASEN 2001 Lab 3: Composite Beam Bending. CU, 10 Oct 2017. PDF.2Hibbeler, R. C., Statics and Mechanics of Materials, 5th ed., Pearson Education, London EN, 2016, Chap 6, 11 and 12.3Neogi, Sanghamitra. Slides Lab3 November6 . CU, 06 Nov 2017. PDF.4Neogi, Sanghamitra. Slides Lab3 November13 . CU, 13 Nov 2017. PDF.5Neogi, Sanghamitra. Slides Lab3 November27 . CU, 27 Nov 2017. PDF.

Acknowledgments

Thanks to Bobby Hodgkinson and the CA’s and TA’s for their help and fabrication of the wings. Thanksto Professor Neogi for instruction throughout the duration of the lab.

Appendix A: Code

1 %This s c r i p t i s the product o f development in ASEN 2001 f o r Lab 3 : Wing2 % Design3 %lab 3 .4 %Created 11/7/175 %Last Modif ied : 12/2/176 %P r o f e s s o r Neogi78 %F i r s t be p o l i t e9 c l e a r ; c l o s e a l l ;

1011 %F i r s t th ing s second , get the data .12 data=x l s r e a d ( ' TestData . x l sx ' ) ;13 F=data ( : , 2 ) ;14 a=data ( : , 3 ) ;15 w=data ( : , 4 ) ;16 d=data ( : , 5 ) ;1718 %Revis ion # − nece s sa ry f o r f i l e s f o r p r o f i l e . Can save a ton o f time19 %Increment MANUALLY any time something changes in the code ( t h i s s c r i p t OR20 %the getWidths func t i on )21 % Var iab l e s that a f f e c t r e v i s i o n : L , c , Eb , Ef , I b , I f , tauFai l , s igmaFai l ,22 % data , p r o f i l e , T design , M design , k , kTau , kSigma , V, M, o u t l i e r s ,23 % P0 , e t c .24 %Var iab l e s that do NOT a f f e c t r e v i s i o n : samples25 Rev i s ion = 1 ;2627 %Use fu l l a t e r − These are our cons tant s28 L=.9144; %length o f the wing [m]29 Eb = 3 . 2 9 5 3 ; %[GPa] − okay s i n c e only used in a r a t i o30 Ef= 0 .035483 ; %[GPa] − okay s i n c e only used in a r a t i o31 c =.01031875; %h a l f c r o s s s e c t i o n a l he ight [m]32 WeightS = . 5 8 ; %[N]33 WeightB6 = 1 . 7 7 ; %[N]34 l en6 = . 1 5 2 4 ; %[m]35 WeightB12 = 2 . 7 4 ; %[N]36 len12 = . 3 0 4 8 ; %[m]37 WeightB18 = 3 . 9 2 ; %[N]38 I f = @(w) (1/12) ∗w∗ ( ( . 0 1 9 0 5 ) ˆ3) ; %Moment o f the foam [mˆ4 ]39 I b = @(w) 2∗ ( ( ( 1/1 2 ) ∗w∗ (7 .9375 e−4)ˆ3)+(w∗7.9375 e−4) ∗ (0 .009921875) ˆ2) ; %

moment o f ba l sa

ASEN 2001 Section 014 9 of 14 Fall 2017

404142 %% Shear and Bending Failure Stresses43 %Now we c a l c u l a t e the bending and shear f a i l u r e s t r e s s e s44 t a u F a i l I n d i v i d u a l = [ ] ; %S t a r t s as an empty vec to r s i n c e we don ' t know how many

we w i l l need to add45 s i gmaFa i l I nd iv idua l = [ ] ; %S t a r t s as an empty vec to r s i n c e we don ' t know how

many we w i l l need to add4647 %Loop through every data entry48 f o r i =1: l ength ( data )49 i f d ( i )>a ( i ) %Moment Fa i l u r e − break occured between po in t s o f l oad ing5051 %Based on moment diagram − l o ok s l i k e a t rapezo id and the top i s52 %t h i s he ight53 MFail=F( i ) /2∗a ( i ) ; %[N∗m]54 I f = I f (w( i ) ) ; %(1/12) ∗w( i ) ∗ ( ( . 0 1 9 0 5 ) ˆ3) ; %[mˆ4 ]55 Ib = I b (w( i ) ) ; % 2∗ ( ( ( 1/1 2 ) ∗w( i ) ∗ (7 .9375 e−4)ˆ3)+(w( i ) ∗7.9375 e−4)

∗ ( .009921875) ˆ2) ; %[mˆ4 ]565758 s i gmaFa i l I nd iv idua l =[ s i gmaFa i l I nd iv idua l (−MFail∗c ) /( Ib+(Ef/Eb) ∗ I f ) ] ;

%Concatenate the new c a l c u l a t i o n to the end o f the vec to r59 %t a u F a i l I n d i v i d u a l =[ t a u F a i l I n d i v i d u a l (3/2∗F( i ) /2/(w( i ) ∗ . 01905) ) ] ;6061 e l s e %Shear Fa i l u r e62 %Based on shear diagram of four po int t e s t − max shear i s ends ,63 %here constant F( i ) /264 t a u F a i l I n d i v i d u a l =[ t a u F a i l I n d i v i d u a l (3/2) ∗(F( i ) /2) /(w( i ) ∗ . 01905) ] ; %

Concatenate the new c a l c u l a t i o n to the end o f the vec to r656667 end68 end6970 %% Outliers7172 %Star t with the bending moments o f f a i l u r e73 IQRBend=i q r ( s i gmaFa i l I nd iv idua l ) ; %Inner−q u a r t i l e range74 Q1Bend=q u a n t i l e ( s i gmaFa i l Ind iv idua l , . 2 5 ) ; %F i r s t q u a r t i l e75 Q3Bend=q u a n t i l e ( s i gmaFa i l Ind iv idua l , . 7 5 ) ; %Third q u a r t i l e76 minBend=Q1Bend−1.5∗IQRBend ; %Data below t h i s are o u t l i e r s77 maxBend=Q3Bend+1.5∗IQRBend ; %Data above t h i s are o u t l i e r s7879 %What i t l ooks l i k e80 p l o t ( s i gmaFa i l Ind iv idua l , 'bx ' )81 t i t l e ( ' Bending Moments o f Fa i l u r e ( Before Removal o f O u t l i e r s ) [N∗m] ' )82 x l a b e l ( ' Test Specimen ' )83 y l a b e l ( ' Bending Moment o f Fa i l u r e ' )84 hold on85 p l o t ( [ 0 l ength ( s i gmaFa i l I nd iv idua l ) ] , [ maxBend maxBend ] )%l i n e s showing the

ranges ou t s id e which data are o u t l i e r s86 p l o t ( [ 0 l ength ( s i gmaFa i l I nd iv idua l ) ] , [ minBend minBend ] )87 legend ( ' Data ' , 'Maximum Cutof f f o r O u t l i e r s ' , 'Minimum Cutof f f o r O u t l i e r s ' )88 s e t ( gcf , ' Pos i t i on ' , [ 1 00 , 100 , 1100 , 7 0 0 ] )

ASEN 2001 Section 014 10 of 14 Fall 2017

89 p r i n t ( ' bendStrength ' , '−dpng ' )9091 %Delete the o u t l i e r s92 o u t l i e r s =[ f i n d ( s i gmaFa i l Ind iv idua l>maxBend) f i n d ( s i gmaFa i l Ind iv idua l<minBend )

] ;93 s i gmaFa i l I nd iv idua l ( o u t l i e r s ) = [ ] ;94 f p r i n t f ( ' %1. f o u t l i e r ( s ) removed from the data s e t o f bending moment f a i l u r e s

.\n ' , l ength ( o u t l i e r s ) )959697 %Now to do the shear s t r e s s o f f a i l u r e98 IQRShear=i q r ( t a u F a i l I n d i v i d u a l ) ;99 Q1Shear=q u a n t i l e ( t auFa i l I nd iv idua l , . 2 5 ) ;

100 Q3Shear=q u a n t i l e ( t auFa i l I nd iv idua l , . 7 5 ) ;101 minShear=Q1Shear−1.5∗ IQRShear ;102 maxShear=Q3Shear+1.5∗ IQRShear ;103104 %See what i t l ook s l i k e105 f i g u r e106 p l o t ( t auFa i l I nd iv idua l , 'bx ' )107 t i t l e ( ' Shear S t r e s s o f Fa i l u r e ( Before Removal o f O u t l i e r s ) [N/mˆ2 ] ' )108 x l a b e l ( ' Test Specimen ' )109 y l a b e l ( ' Bending Moment o f Fa i l u r e ' )110 hold on111 p l o t ( [ 0 l ength ( t a u F a i l I n d i v i d u a l ) ] , [ maxShear maxShear ] )112 p l o t ( [ 0 l ength ( t a u F a i l I n d i v i d u a l ) ] , [ minShear minShear ] )113 legend ( ' Data ' , 'Maximum Cutof f f o r O u t l i e r s ' , 'Minimum Cutof f f o r O u t l i e r s ' )114 s e t ( gcf , ' Pos i t i on ' , [ 1 00 , 100 , 1100 , 7 0 0 ] )115 p r i n t ( ' shearStrength ' , '−dpng ' )116117 %Delete the o u t l i e r s118 o u t l i e r s =[ f i n d ( tauFa i l I nd iv idua l>maxShear ) f i n d ( tauFa i l I nd iv idua l<minShear ) ] ;119 t a u F a i l I n d i v i d u a l ( o u t l i e r s ) = [ ] ;120 f p r i n t f ( ' %1. f o u t l i e r ( s ) removed from the data s e t o f shear moment f a i l u r e s .\n

' , l ength ( o u t l i e r s ) )121122 %% Use the Data We have accrued and vetted123 s igmaFai l = mean( s i gmaFa i l I nd iv idua l ) ; %Bending moment f a i l u r e we w i l l use f o r

the lab124 de l ta s i gma = std ( s i gmaFa i l I nd iv idua l ) / s q r t ( l ength ( s i gmaFa i l I nd iv idua l ) ) ; %

Standard dev i a t i on o f the mean125 tauFa i l = mean( t a u F a i l I n d i v i d u a l ) ; %Shear s t r e s s f a i l u r e we w i l l use126 d e l t a t a u = std ( t a u F a i l I n d i v i d u a l ) / s q r t ( l ength ( t a u F a i l I n d i v i d u a l ) ) ; %Standard

dev i a t i on o f the mean127128 %% Safety Factor129130 %Use the method from the t r u s s lab131 P=.01; %Fa i l u r e p r o b a b i l i t y132133 %Backwards c a l c u l a t e the maximum a l l owab l e va lue s134 T des ign=i c d f ( ' normal ' ,P, tauFai l , d e l t a t a u ) ;135 kTau=tauFa i l / T des ign ;136137 M design=i c d f ( ' normal ' ,P, abs ( s igmaFai l ) , de l ta s i gma ) ;

ASEN 2001 Section 014 11 of 14 Fall 2017

138 kSigma = abs ( s igmaFai l / M design ) ;139140 %Overa l l Sa f e ty Factor can be the g r e a t e r o f the se two c a l c u l a t i o n s141 k=max(kTau , kSigma ) ;142143 %New max a l l o w a b l e s with l a r g e r s a f e t y f a c t o r144 T des ign=abs ( tauFa i l /k ) ; %[ Pa}145 M design=abs ( s igmaFai l /k ) ; %[ Pa ]146147 %% Find p0148 % Loading func t i on149 syms x150 syms p0151152 %Max shear w i l l happen at the endpoints as found in the shear diagram153 V = − i n t ( . 1016∗ p0∗ s q r t (1−((2∗x ) /L) ˆ2) ,x , 0 , x ) ;% [N] %No IC adjustment because V

(0) == 0 N154 M = i n t (V, x , 0 , x ) − i n t (V, x , 0 ,L/2) ; %[N∗m]155 %Make sure we s t i l l l ook good to go156157158 VinTermsofp0=subs (V, x,−L/2) ; %Max shear happens at the endpoints we know from

diagram159 MinTermsofp0=subs (M, x , 0 ) ; %Max bending moment s t r e s s occurs in the cente r we

know from diagram160 %Get the max P0 s i n c e shear and moment w i l l a l low d i f f e r e n t P0 ' s161 P0 shear = s o l v e (3∗ VinTermsofp0 /(2∗ . 01905∗ . 1016 ) == T design , p0 ) ; %Reverse

shear formula162 P0 moment = s o l v e ( MinTermsofp0∗c /( I b ( . 1 0 1 6 ) +(Ef/Eb) ∗ I f ( . 1 0 1 6 ) )==M design , p0 )

; %Reverse f l e x u r e formula163 P0 = min ( double ( P0 shear ) , double (P0 moment ) ) ; %[N/mˆ2 ] or [ Pa ]164165166 %% Get widths167 %Now we have P0168 V=−i n t ( . 1016∗P0∗ s q r t (1−(2∗x/L) ˆ2) , x ) ; %[N]169 M = i n t (V, x , 0 , x )− i n t (V, x ,0 ,−L/2) ; %Adjustment r equ i r ed dependent on P0170171 %Make sure we look okay172 plotDiagrams (V,M,−L/2 ,L/2 ,P0)173174 %Do the math i t s e l f175 %%%%%%%%%%%%%%%%%%%% YOU MAY CHANGE ME %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%176 samples = 1000 ;177 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%178 p r o f i l e=l i n s p a c e (−L/2 ,L/2 , samples ) ;179 Widths = getWidths (V,M, samples , p r o f i l e , c , Eb , Ef , I f , I b , T design , M design ,

Rev i s ion ) ;180181182 %% Wiffle Tree Config183 f ina lWidths = max( Widths ) ;184 [ ˜ , idx ] = min ( f ina lWidths ) ;185 x C r i t i c a l = p r o f i l e ( idx ) ; %Distance in m186 n = 4 ;

ASEN 2001 Section 014 12 of 14 Fall 2017

187 shearwidths = p o l y f i t ( p r o f i l e ( 1 : idx ) , f ina lWidths ( 1 : idx ) ,n ) ;188 w s = @(x ) shearwidths (1 ) ∗xˆ4 + shearwidths (2 ) ∗xˆ3 + shearwidths (3 ) ∗xˆ2 +

shearwidths (4 ) ∗x + shearwidths (5 ) ;189 momentwidths = p o l y f i t ( p r o f i l e ( idx : samples /2) , f ina lWidths (1 , idx : samples /2) ,n ) ;190 w m = @(x ) momentwidths (1 ) ∗xˆ4 + momentwidths (2 ) ∗xˆ3 + momentwidths (3 ) ∗xˆ2 +

momentwidths (4 ) ∗x + momentwidths (5 ) ;191192 %area o f the wing193 area = 2∗ r e a l ( double ( i n t ( w s , x,−L/2 , x C r i t i c a l ) + i n t (w m, x , x C r i t i c a l , 0 ) ) ) ;194195 %Load d i s t r i b u t i o n s as a func t i on o f x => two because two curves make up196 %the width o f the beam197 q s = P0∗ s q r t (1−(2∗x/L) ˆ2) ∗w s ( x ) ;198 q m = P0∗ s q r t (1−(2∗x/L) ˆ2) ∗w m( x ) ;199200 % Now we need the f o r c e and c e n t r o i d s f o r the s u b s e c t i o n s201 s u b s e c t i o n s = l i n s p a c e (−L/2 ,0 ,5 ) ;202 f o r c e s = ze ro s (1 , 4 ) ;203 l o c a t i o n s 1 = ze ro s (1 , 4 ) ;204 f o r i = 1 :4 %Four s u b s e c t i o n s205 l im1 = s u b s e c t i o n s ( i ) ;206 l im2 = s u b s e c t i o n s ( i +1) ;207 i f l im1 < x C r i t i c a l && lim2 > x C r i t i c a l %Need to s p l i t208 f o r c e s ( i ) = r e a l ( i n t ( q s , x , lim1 , x C r i t i c a l ) + i n t (q m , x , x C r i t i c a l , l im2 )

) ;209 l o c a t i o n s 1 ( i ) = r e a l ( double ( ( i n t ( q s ∗x , x , lim1 , x C r i t i c a l ) + i n t (q m∗x , x

, x C r i t i c a l , l im2 ) ) / f o r c e s ( i ) ) ) ;210211 e l s e i f x C r i t i c a l>l im2 %Shear regime212 f o r c e s ( i ) = r e a l ( i n t ( q s , x , lim1 , l im2 ) ) ;213 l o c a t i o n s 1 ( i ) = r e a l ( double ( i n t ( q s ∗x , x , lim1 , l im2 ) / f o r c e s ( i ) ) ) ;214215 e l s e %Moment regime216 f o r c e s ( i ) = r e a l ( i n t (q m , x , lim1 , l im2 ) ) ;217 l o c a t i o n s 1 ( i ) = r e a l ( double ( i n t (q m∗x , x , lim1 , l im2 ) / f o r c e s ( i ) ) ) ;218 end219 end220221 %Now we need to do s t a t i c s problems down the t r e e222 l o c a l D i s t s = ze ro s (1 , 2 ) ; %Distance o f the downward hook as measured from the

l e f t end o f the bar ( l e f t h a l f o f wing223 f o r c e s 2 = ze ro s (1 , 2 ) ;224 f o r i = 1 :2225 F1 = f o r c e s (2∗ i −1)−WeightS ;226 F2 = f o r c e s (2∗ i ) − WeightS ;227 f o r c e s 2 ( i ) = F1+F2−WeightB6 ;228 l o c a l D i s t s ( i ) = (F2∗( abs ( l o c a t i o n s 1 (2∗ i −1)− l o c a t i o n s 1 (2∗ i ) ) ) − WeightB6∗(

l en6 /2 − . 0 2 ) ) / f o r c e s 2 ( i ) ;229 end230231 %Second row down232 F1 = f o r c e s 2 (1 )−WeightS ;233 F2 = f o r c e s 2 (2 )−WeightS ;234 FDown = F1+F2−WeightB12 ;

ASEN 2001 Section 014 13 of 14 Fall 2017

235 l o c a l D i s t 2 = (F2∗( abs ( ( l o c a t i o n s 1 (1 )+l o c a l D i s t s (1 ) )−( l o c a t i o n s 1 (3 )+l o c a l D i s t s(2 ) ) ) )−WeightB12 ∗( l en12 /2− .02) ) /(FDown) ;

236237 %Third and l a s t row needs to be exac t l y symmetrical and then we ' re done238 %with t h i s w i f f l e t r e e bus in e s s239 o v e r a l l B u s i n e s s = [ l o c a t i o n s 1 ; l o c a l D i s t s (1 ) 0 l o c a l D i s t s (2 ) 0 ; 0 0 l o c a l D i s t 2

0 ]∗1 0 0 / 2 . 5 4 ;240241 %% Expected Weight in the Bucket242 % Total weight the wing should support :243 t o t a l = P0∗ area ;244245 %Need to subract a l l the other s t u f f here246 add1 = t o t a l − 4∗WeightB6 − 2∗WeightB12 − WeightB18 − 15∗WeightS ;247 add2 = 2∗FDown − 3∗WeightS − WeightB18 ;248 %% Plot the Width Information249 f i g u r e250 p l o t ( p r o f i l e , Widths ( 1 , : ) )251 hold on252 p l o t ( p r o f i l e , Widths ( 2 , : ) )253 t i t l e ( ' Calcu lated Required Widths ' )254 x l a b e l ( ' Distance [m] ' )255 y l a b e l ( 'Width [m] ' )256 legend ( 'Minimum Required Shear Width ' , 'Minimum Required Bending Moment Width ' )257 s e t ( gcf , ' Pos i t i on ' , [ 1 00 , 100 , 1100 , 7 0 0 ] )258 p r i n t ( ' reqwidths ' , '−dpng ' )259260261 %% Plot the Profile of our bad boy i t s e l f262263 f i n a l w i d t h s=max( Widths ) /2/2 .54∗100 ;264 p r o f i l e = p r o f i l e ∗100/2 .54 ; %[ in ]265 f i g u r e266 p l o t ( p r o f i l e , f i na lw id th s , 'b ' )267 hold on268 p l o t ( p r o f i l e ,− f i na lw id th s , 'b ' )269 p l o t ( p r o f i l e , z e r o s (1 , l ength ( p r o f i l e ) ) , 'k−− ' )270 t i t l e ( ' Fina l Wing P r o f i l e ' )271 x l a b e l ( ' Distance [ in ] ' )272 y l a b e l ( ' Deviat ion from Center Line [ in ] ' )273 s e t ( gcf , ' Pos i t i on ' , [ 1 00 , 400 , 1200 , 4 0 0 ] )274 p r i n t ( ' p r o f i l e ' , '−dpng ' )

ASEN 2001 Section 014 14 of 14 Fall 2017