Bruhn - Analysis And Design Of Flight Vehicles Structures Episode 1 Part 6 doc

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AT 8 q= A=Ao(1-,*) q =409/in, 2Ô As=4in.3 constant L = 40in E =10* 108 psi +P, Fig A7.12 Soluticn:

After additicn of the Tictitious end load R the axial load from statics was found to 5e S(x) = R+q (L - x} Hence, since lceadings other than tenstie are of a secondary nature _ 1 fh stax u=5 | AE ° L + ~x rã] [? tạ xe AaE 1 - aE °

Before evaluating the integrals it was ob- served that the steps to follow, in which U was to be differentiated with respect to PR and the subsequent setting of R = o would drop out both the first and last terms Hence only the second term was evaluated - 2L° Raq Uex+ Se (1 - in 2) +x Then = ae 5 = oF 2L°q Agi (1 - In 2) DIFFERENTIATION UNDER THE INTEGRAL SIGN

An important labor savings may be had in the calculation of deflections by Castigliano’s thecrem

In the strain energy integrals arising in this class of problems, the load Pos with respect to which the deflection is to be found, acts as an independent parameter in the integral Pro- vided certain requirements for continuity of the functions are met - and they invariably are in these problems - the differentiation with respect to P may be carried out before the inte-

gration is made The resulting integrals gen- DEFLECTIONS OF STRUCTURES erally are easier to evaluate Example Problem 11 id the deflection at point B of the beam of rig A7.13 El constant Solution:

A fictitious load Pi was added at point B

and the bending moment diagram was drawn in two parts ` 2PL Mp > M==s= 3 O<X<L/3 QPL PY P.L PY , bx Ley TT trậT 0<1<L/3 `“ v.PL P3 2P.L 2P,2 ye Mp, MaRE - FB, BP ab 2Ps2 oc acs/s

Fig A7, 13a

‘Then neglecting the energy of shear as bein sma11* Ug 2 =k š U*“ Le + Pi) | on Lựa * 'tan| ° | -#Ð xÃ (e0) sy 1/2 a

-ar| (Es) w -se)] ae ;

Differentiation under the integral sign with respect to P, gave [Fea -3y) + Ba - sv) Ly le + pee ax ° ( : 3y) dy _eu _ 1 Sạ “3P, Tấn L/a | 1 *ø ~ 3g dz | 2(L - 3z) cee

* For beams of usual high length - to - depth ratios the shear

strain energy is small compared to the energy of flexure

Neglecting the shear energy is equivalent to neglecting the

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ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES

ÀT.9

The fictitious load P,, havingyserved its pur- The total loadings

pose, was set equal to zero before completing were the work, M=Mi-Ma=(To-PR)sin 0 Plan View L/a 2P a oe Ff xT T=f?¿+Ta= Tocos @ + PR M 7 ° (1 - cos 9) Ae , Ly Za N 2 1P ý r Thus +51), (2L - 3y)(L + 3y) cy - Fig AT 14b L a i 7⁄2 a a ,i2 ‘3p sg)^ ae UF oR ; (To = PR)* sin? @ Rd 9 =I 81), Tựz 7 PLS + whe ("Pt cos @ + PR(L - cos 0)]7 Rd 0 ae eGJd , B 486 EI °

, (Note use of "Rd@" for the length of a đi7- Example Problem 12 Tig A7.14 shows a ferential beam element instead of “dx"), Dif-

cantilever round rod of diameter D formed in a quarter circle and acted upon by a torque To Find the vertical movement of the free end

Fig AT, 14 Solution:

Fig A7.14a shows

the vector resolution of 7 Plan View

the applied torque T, on

beam elements T; (9) = | M

Tạcos 9 and the moment ZÀ\T:

M,(@) = Tạ sin'9, Ap- a

plication of a fictitious ‡

vertical load“P (down) at To

the point of desired de-

flection gave the loadings Fig AT 14a

shown in Fig A7.14b ferentiating under the au (T, - PR)R? P BÍ a aR as integral sign TỰ { 2 oina $ d8 ° TỰ2

{ [T¿eos @ + FR(L - cos 9) ](1 - cos 9) de

tting P, the fictItious load equal to zero and integrating gave 4 (t-#) Since J = 2l and G = "/2.6 the deflection was negative (UP)

AT.T Calculation of Structural Deflections by the Method of Dummy-Unit Loads (Method of Virtual Loads) The strict application of the calculus to Cast

number of cumbersome techniques ill-suited to the so

flexible approach, readily adapted to improved "book keeping” techniques

Unit Loads developed independently 5y J 7 Maxwell (1864) and 0 2 Monr (1874)

below are tains the

aopeal to the concepts of $

mechanics Based as they are

course, yield the same result

equations by 4 reinterpretation of ain energy

Derivation

That the equations for the Method of Dummy-Unt attested to by the great variety of names applied to

two derivations of the equations stemning from different viewpoints

the symools of Castigliano’s theorem - essentially an derivation uses the principles of rigid oody upon a common set of consistent assumptions, all the methods must, of The othe of Equations for the Method of Dummy-Un this (Virtual Loads ) I - From Castigliane’s Theorem

Beginning with the general expressicn for strain energy, eq (43)

————————-

® variously called the Maxweil-Mohr Method,

Loads, Dummy Loads, Method of Work, etc

Il - From the Pri

“when a system

it L

tgliano?s theorem as 1n Art A7.6, leads to a lution of large complex structures A more

ts the Method of Dumny-

t Loads may be derived in a number of ways is

method in the literature ® Presented One derivation ob-

oads

e of Virtual Work

orees whose resultant is

zero (a system in equilibrium) 1s displaced a

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AT 10 DEFLECTIONS OF STRUCTURES 1 Cont'd, a Cont'd — 2, a

U= Z Ị* + 3 [Sẽ +5 jae + = - = ete.| small amount without disturbing the equi

* balance, the work done is zero - obviously,

differentiate under the integral sign with since zero resultant force moving through a respect to P, to obtain distance does zero work

3S OM Consider now the set of equilibrium forces

6, 2 | Spree yp KR, LLL Lee etc.| applied to the truss of Fig A7.15(b) The set

+ + — may be divided into two parts: the "external

AE SỈ system", consisting of the unit load applisd at

Consider the symbols the point whose deflection is desired and the

3s om oT FL LL Lt ete two Teactions fixing the line of reference, and Py , oP, 1 aMHR “““*) the "internal system” consisting of the axial

Each of these is the "rate of change of so -and-so with respect to Py” or “how much so- and-so changes when P; changes a unit amount" OR EQUALLY, "the so-and-so loading for a unit load

Py"

Thus, to compute these partial derivative terms one need only compute the internal loadings due to a unit load (the virtual load) applied at the point of desired deflection For example, the term aM yp ù could be computed in either of the two ways shown in Fig A7.15a

RATE METHOD UNIT METHOD x 7 x —_ te 4 dụ Fig A7.18a M=P¿x m = dummy loading aM ap, ~ * =x (23 5% (- àP

Likewise, *54p., where P, 1s a load (real or fictitious) applied at joint ¢ of Fig A7.15b,]

is given by the loadings for the unit load ap-

plied as shown

In practice the use of the unit load ts

most convenient Using the notation as =aM = dt Spo MP5 t Pas oP, SP, aP, 2 5 289 SOP, 72 * 3p for the unit loadings, the deflection equation becomes 5 = ƒ 898% „ ( Mmdx „ (Ttdx 1 J AE EI Sử VWvdx addy AC 7 J St mm 8) -L0 — -LỢ & VN Ss b Gs lũ s 0 > a £ 2.50 | 80 Ci 50 | 50 4

5=Ra Fig AT7.15b tệ=Ra

"4" loads due to a unit (virtual) load

loads acting on the truss members to produce equilibrium These latter ars denoted by the symbol "u", Tunis set of ferces 1s considered small enough so as not to.affect the actual be-

havior of the structure during subsequent ap-

plication of a real set of major loads This unit load set is present solely for mathematical reasons and is called a “virtual loading” or

"dummy loading",

Assume now that the structure undergoes 4 deformation due to application of a set of real loads, the virtual leads "going along for the ride" Hach member of the structure suffers a deformation denoted by A 9 ,„ The virtual load~ ing system, being in equilibgium (zero resultant) by hypothesis, does work ("virtual work") equal ta zero Or, considering the subdivision of the virtual loading system, the work done dy the external virtual load must equal that absorbed by the internal virtual loads The work dene by the external virtual forces is equal to one pound times the deflection at joint C, the reactions Ri and Ra not moving That is

External Virtual Work = v x 84 The internal virtual work is the sum over the structure of the products of the member virtual loads u by the member distortions 4 That is,

Internal Virtual Work = 2 ued

Then equating these works,

1x 6, sZuea

If the deformations 4 are the result of elastic strains due to real member loads S then a= Sun for each member and one has

=y 5h $ Bau nr

The argument given above may be extended quickly to include the internal virtual work of virtual bending moments (m), torsion loads (t}, shear loads (v), and shear flows (3) doing work during deformations due to real moments (mM), torques (T), shear loads (V), and shear flows

(q) The general expression becomes - (8d „ rMmdx „ ¢ Tex 5%“ [TS + [mi J Sở Vvdx q dxd: ao {Va - (18)

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ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES AT.I1

In applying eq (19) the labor of a deflec-

tion calculation divides conveniently into onl 0 0 9 9 Q

several steps: >

7 ol

1 Calculation of the real (actual) load 2 ` % S © Fs

distribution (S,M,T, ete.) VY a 0 0 0

{1 Calculation of the unit (virtual) load =75 | -.78 3 4.0] 2.0 1.0

distribution (u,m,t, etc.) due to a unit tự 1#

(virtual) load applied at the point of desired "na! loads "ue" loads

deflection and reacted at the reference point(s) Fig, AT 16a Fig AT 18b

iii Calculation of flexibilities,

1 1 TABLE AT.3

Ts,’ El? etc AEx 107° tu SueL

Đạo Be MEM | Lin ie Her, 8 wa fue = SMsL r 106

iv Summation and/or Integration

AB} 30 4.783 6.270 10,300| 1.5 |0 98.75, 0

Here again note the general nature of the pe luol 307s ome | nan

terms "load" and "deflection" (See p A7.6) - - can 9 Ị9 9 ° The following examples apply the method of cD {30 | 3.07 9,739 2,250; 0 |0 a 0 dummy-unit loads to the determination of ab~ FF |30| 5.365 5.591 | ~6,250/-0 751 22.01 | -29,36 Noi san: top tàu deflections, both rotation Fo (30 | 5.388 5.501 |-8,280[-0 75/1 | 33.01 | ~20,38

GH |39| 3.48 9,851 9 |9 |1 9 9

Example Problem 13 BE | 50 | 10.15 4.9286 |-8,T50|-1.75|0 | %3.8 9 The pin-jointed truss of Fig 47.16 is pa [sol 3.48 14.368 3,000] 1.3510 | 80.80 ọ acted upon by the external loading system shown

The member loads are given on the figure Mem~ ĐG |40| 5.346 | 9.340 |-379g] g j0 9 1

ber properties are given in Table A7.3 Find SF | 40 | 3.074 13,014 2000| 0 |0 9 9 the vertical movement of joint G and the hort- ca [40] 3.07 | 13.012 |cu000) 0 |0 3 3 zontal movement of joint H DH | 40 3,014 13,014 44000] 0 9 a g 1000%# 1000# 1000# 2 = 286.45 2-88.72 $ 10,500}B 2280 tc 2250 Ìp & al os Answers: 5, = 0,286" S383 SA Soe VER x/ ä | Bf St 40" 5950 | -3250 9 “I | 6 = -,0587" (the negative sign means E H Gy a yor

30" peso 30"— the Joint moves to the LEFT since the unit load 20004 20008 was drawn to the RIGHT in Fig A7.16b)

1.1 Example Problem 14

Fig AT 16 For the truss of Fig a7.16 find the fol-

Solution: lowing relative displacements of joints:

Only the energy of axial loadings in the members was considered Unit (virtual) loads wera applied successively at joints G and H as

shown in Figs A7.16a and A7.16b, A11 § and u loadings were sntered in Table A7.3 and the sudx AE /AE terms for the members = 5 Sul Š5 “7n calculation for ð = f was carried out 5y Sul

forming the sum of

¢} the movement of joint ¢ in the direction

or a diagonal line Joining ¢ and F,

ad) the movement of joint G relative to a line joining points F and H

Relative deflections are determined by applying unit (virtual) loads at the points where the deflections are desired and by support-

ing such unit load systems at the reference

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AT 12 DEFLECTIONS OF STRUCTURES part (ad) the system of unit loads of Fig A7.16d

was used Table A7.4 completes the solution, the real loads and member ?lexibilities

(œ) being the same as for example problem 1%, 0 -.6 C0 0 > 375 = 375 Rotations, both absolutes and relative are (

determined by applying unit virtual) 2

to the member or portion of structure whose retation is

for the rotation

desired The unit couple i

ed by reactions placed on the line of r

Thus Figs a7.16e 4 show the unit (virtual) lcadings is resist- eference 47,3 for parts (e 2

and (f) respectively Table A7.5 completes the 1# le 2 `, 0 4, calculation, the real loads and member ?lexibil-

Sa Vl S| Sol VC | YM, ities °

<< (4z) being the same as fcr example

GFL -.6 9 o |Fo @ oH problem 15

x ' 3

= 3 1# ,

R=1# “uạ" loads cố 4" load R=1/40# 025.025 ; „025 `»D 1/504 9 D /.095C Y/50H 21/404 9

Fig AT 16¢ Fig A7.16d og s 2, 2

2 0 © 5 9 2 9 5 Ô 0

0 0 9 9 |}

R=i/T0W 7.025 =.025 G1000 1/50# YG” Rel/ 40+

TABLE A1.4 "ue" loads "ue" loads

es 10¢ AE Sul SuạL fig A7 16e Fig AT 16f a in/ib x 105 [Sask x 102 MEM see Tale | b |? we ™E tea | AE inch ‘AT 3) AB 6.270 |10,500) 0 -0 0 9 BC 9.159 | 2,250 -.358 | -13.1T -8.43 cp 9.199 | 2250| 0 -.38 9 -8.28 TABLE AT.5 = EF 5.581 |-5,250| 0 9 9 0 FG 5.501 |-5, 250) -.8 0 11.61 9 xg t 10" Sug Su,b a , - MEM.| coe qi] sty |uaaens| suiaee+| Uế x t0°| THẺ x1p* GH 8.621 9 a 9 0 0 AT.3) BE 4.926 |-8,750| 0 0 0 9 AE 6.270 |10,500| 025 9 1,68 9 BG 14.368 | 5,000! 1.0 625 | T1.84 “4.9 BC 9.759 | 2,280] 025 | -.025 35 +0 53 DG 3.320 |-3,750} 0 - 625 9 721.8 cp 9.758 | 2.250| 025 a 38 9 BE | 14.012 | 2,000| -.8 +50 | ~20,83 713.0 EF 5.591 |-§,250 | -.025 9 +.T4 9 c6 13.014 |-1,000| -.8 a +1041 0 FG §.591 |-5,250 | -.025 9 73 9 DH | 13.012 | 2,000/ 9 -.50 0 -18.0 GH 8.621 0 0 0 9 D & = 65.87 7 ==19,36 BE 4.926 | -8, 750 9 9 9 0 BG | 14.368 | 5,000} 0 015 ọ 1.08 DG 9.320 |-3,T50 | -.015 0 33 0 BF | 13.012 | 2,000] o@ 9 L 9 cG | 13.012 |-1,000] 0 9 a 9

Therefore the movement of joint c towards DH 13.012 1 m0 0 ọ 5 0 joint F was 6 = 06587 inches and the motion of - ~

joint G relative to a line between F and H was 224.73 220.53

6 = -.0194 inches, the negative sign indicating an upward movement

Example Problem 15

For the truss of Fig A7.16 determine e) the absolute rotation of member DG

†) the rotation of member BG relative to

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ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES Example Problem 16

Find the vertical deflection of point Ở for

the cantilever beam of : Av.17 carrying 2

concentrated load P at its end Also finde slope or elastic surve at ở Solution: With origin at B M=- Px (Fig 47.17) For virtual loading (Fig A7.17a) m= 0, for x<b m=-l (x - b), for x2b Hence Mmdx = - Px(-x + D)dx = (PX” - PbX)dX L i Whence 5, = c3 = { (x? - bx)dx = Al a -£) # =z 7 a ớ 3 ` ————— —_—— ° :.::.:: E—`'— —¬3 A c ao P poi g anit A Ss nit B Fig A7 175 : souple Mmdx 8= For virtual loading see Fig A7.17> m=0, x<bd, m=- 1, x>b dence Mmdx = ¢Px} (+1) ax = F If d= 0, ay = Example Problem 17

Y the uniformly loaded cantilever beam A?.18, find the deflection of point D

of Fig

relative to the line joining points © anc = on the elastic curve of the beam This is repre- sentative of a practical probiem in 2 ronautics,

in that 48 mizht represent 2 rear wing beam and H a 5 Ũ H " AT.13 points c, D, a, the attachment points of an aileron or a Slap The wing beam deflection bends the aileron or flap structure by applying a load at D thru aileron supporting oracket To know this force the deflection of the wing beam at D relative to line CE must be known Solution: Origin at B: <- M=30x-35 ã 18 x ˆ vi 30 lbm/tn io TTT Ww 30" - xe 20 ty 8 k—— áet — ? Fiøg AT.18 ae ret 50" C 8 sự Fig A7 18a Reed Ra=«2 Oo, x < 20 = -.5 (x-20) =-.5x + 10, when x = 20 to SO ~.5 (x- 20) +1 (x -50) =.5x-40, when x = 50 to 80 5 D (or) - ( Max El 20 Ề 11.72 + 6.25 + 300 - 400 + 76.8 ~ 102.4 ~ 11.72 + 25.9] 17,9 x 10% BH

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AT 14

Find the horizontal deflection of point C for the frame and loading of Fig A7.19 Also angular deflection of C with respect to line CD Solution:

Fig b shows the static moment curve for the given loading and Fig c the moment diagram for the virtual loading of a unit horizontal load applied at C and resisted at D, Mudx , m= WLX _ wx? og 4 CCH) [Fa = = Whe _ wt 2 2 =} " hence wl AE" * us) ne oc = ————ˆ LỆ - hax = | (WERE wee) a ~ T8.JBỊ RL a EI a = 2 walt TÊ “sr

To find angular deflection at C apply a unit

imaginary couple at C with reactions at C and D Fig A7.20 shows the virtual m diagram @ = “vmx 1 EI EL ° WLx2 wx2 _ 1 fx? wxt* a - ap) & Be" ep] F ° 1 wie B42 EI

Linear Deflection of Beams Due to Shear by Virtual Work

Generally speaking, shear deflections in beams are small compared to those due to bending except for comparatively short beams and therefore are usually neglected in deflection calculations A close approximation is sometimes made by using 2 modulus of elasticity slightly less than that for bending and using the bending deflection equations

The expression for shear deflection of a beam is derived fram the same reasoning as in previous derivations The virtual work equation for the hypothetical unit load system for a shear detrusion dy (Fig A7.21) considering only dx elastic is 1 x 6=vdy where v is Shear on section due to unit hypothetical load at point 0, and dy is the shear detrusion of the element dx due to any given load system or any other cause _>1 rờ uy Fig A721 1# *Sometimes "G", instead, See Pp AT.4 [CES Esa’ , 82 a7 mi DEFLECTIONS OF STRUCTURES V dy = ni =——

dy Sax and 3 BES?

area of beam at section and Z, =

where A = cross sectional

medulus of rigidity; and assuming that the shearing stress

q

i is uniform over the cross-section

Therefore 1 x 6 = _ » Then the total deflec- as, tion for the shear slips of all elements of the beam equals L 8 total = I Wer - (2) Es 9

where V is the shear at any section due to given loads v = Shear at any section due to unit hypothetical load at the point where the deflection is wanted and acting in the desired direction of the deflection The reactions to the nypo- thetical unit load fix the line of reference for the deflection,

A ts the cross-sectional area and Z, the modulus of rigidity Equation (a) is slizhtly

in error as the shearing stress is not uniform over the cross-section, ¢.g being varabolic for 4 rectangular section However, the average shearing stress gives close results

For a uniform load of w per unit length, center deflection on a simply supported beam is: - the the L L Wvdx _ 2 (WL _ wx L Scenter = 2 2 any ° (ah bas o AE " SAES

For bending deflection for a stmply supported beam uniformly loaded the center deflection is 5 wut ` S84 11 Hence wL®

aE, mm = 2a (F)", using Bg = 4E, mà _

2a, r = radius of gyration

For I beams and channels r is approximately 5 d and for rectangular sections r = » (4 = depth)

In aircraft structures a ratio of d is seldom greater than S$

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ANALYSIS AND DESIGN Example Problem 19 50° 100% Unit vét be 10" ~+| Load 1 A A Fig AT 22

Pind the vertical deflection of free end A due to shear deformation for beam of Fig A7.22 assuming shearing stress uniform over cross- section, and AES constant 6g = = VY = 100# for x =O to 10 V = 150 for x = 10 to 20 v=1 for x = 0 to 20

Method of Virtual Work Applied to Torsion of Cylindrical Bars

The angle of twist of a circular shaft due to a torsional moment may be found by similar reasoning as used in previous articles for find-

ing deflection due to bending or shear forces The resulting expressions are: -

In equation (A), for translation deflections, T = twisting moment at any section due to

applied twisting forces

t = torsional moment at any section due to a virtual unit 1 1b forces applied at the point where deflection is wanted and applied in the direction of the desired displacement (in lbs/1b) shearing modulus of elasticity for the material (also °G")

poler moment of inertia of the circular cross-section

In equation (B), for rotational deflections,

9 = angle in twist at any section due to

the applied twisting moments in planes

perpendicular to the shaft axis Angle in radians =H = 5 J= OF FLIGHT VERICLE STRUCTURES AT 15 t = torsional moment at any saction due to

@ unit virtual couple acting at section where angle of twist is desired and acting in the plane of the desired de-

flection (inch lLbs/inch lb) Exampie Problem

Example Problem 20

Fig A7.23 shows a cantilever landing gear strut-axle unit ABC lying in XY plane A load of 1000# is applied to axle at point A normal to XY plane Find the deflection of point A normal to XY plane Assume strut and axle are tubular and of constant section

Solution:

The loading shown causes doth bending and twisting of the strut axle unit First find pending and torsional moments on axle and strut due to 1000# load / 1# BUA ig AT 24 Member AB M = 1000 x, (for x = 0 to 3) T=0 Member BC Mpg = 3000 sin 20° + 1000 x, (for x = 0 to 36) Tạc = 3000 cos 20° constant between B and C ˆ Now ApPly ä un1b 1# force at A normal to xy

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AT.16 Subt 1000 x + 1026) Bị 3 1000 x + xớy + Xd aI) 36 + (36 (x + 1.028) ] dx + sơ], (2820) (2.82) ax 3 - 2 ( [see x 3] « [eas x 3 + 1008 + 2 + 1080 9 J"): 36 [7282] = 1 Ed 1 at wale (16,92500) + (286200)

Note: A practical landing gear strut would in- volve a tapered or reinforced section involving a variable I and J and the integration would nave to be done graphically or numerically Example Problem 21

For the thin-web aluminum beam of Fig A7.25 determine the deflection at point G under the loading shown Stringer section areas are given on the figure, Ac As 15 BACB ans 5 I te.032 || ts.032 |] t=.032 | " lậm 2 P Asoa E A=0sij | |{a=.o8 F G HỆ p.is00# E20» fas RE 20" 1 Flg A7.25 Solution:

It was assumed that the webs did not buckle and carried shear only Fig A7.26 is an ex-

ploded view of the beam showing the internal real loads carried as determined by statics? 3.0P Ị Đụ “ 2 04 3.0P— Fig, A7 26

* The equations of statics for tapered beam webs are

derived in Art, Al5.18, Ch A-15,

DEFLECTIONS OF STRUCTURES

The shear flows shown on che (nearly) noriz edges of the wed panels are averags values A7.27 is an exploded view of the beam showing

the unit (virtual) loads onte PL + 46 —— dl iS: Si >} i _=— 1#

Virtual loading, Fig A7.27

Since both axial loads and shear flows were considered, the form of deflection

equation used was

Sudx

on = ——— +

sr| 3|

Integrations in the flanges were made assuming linear load variations Such an integration carried out over a uniform flange of lengtn L whose real load varies from S; to Sj and whose virtual load varies from 4, So uy ylelds Gq dxdy Gt + 344 3 6 AZ [rae L (š tr AE Q + Shes Ss) ch tệ 4

The integrations in the trapezoidal sheet panels were made using the shear flows on the

(nearly) horizontal sides as average values, assumed constant over the panel, With this

day Gay oxdy = = §

simplification JJ = ay day

Gt where S is the panel area

The calculation was completed in Table A7.6 [ TABLA ATS # er er) 1 aa spa Sal by 40031469 c Giá 0: d9 [SE t0 Be pT wg iw sence!

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= (273-8 + 13.4) Px 107° =

.147 (1800) 107° in = 265 in

AT.8& Deflections Due to Thermal Strains

AS noted in the "virtual work derivation” of the dummy-unit load deflection equations, the real internal strains of the structure may be due to any cause including thermal effects Hence, provided the temperature distribution and thermal properties of a structure are known, the dumny~unit load method provides a ready means for computing thermal deflections

Example Problem 22

Find the axial movement at the free end of a uniform bar due to heat application to the

fixed end, resulting in the steady state temperature distribution shown in Fig A7.23, As~ sume material properties are not functions of temperature T = temperature above KX ambient tempera- To- T(X)= To (t-tanh =) ture y 7 K = an empirical con-

Input [— x stant depending

—— L upon thermal prop-

erties and rate of Fig, A7.28 heat addition

Solution:

The thermal coefficient of expansion of the rod material was @ Hence a rod element of

length dx experiemced a thermal deformation A2a-T + dx, Application of a unit load at the bar end gave u=i1 Therefore

Example Problem 23

Tr ized two-flange tilever Deam of Fig A7.2Sa undergoes rapid seating of the upper flange to a temperature T, uniform span- wise, above that of the lower flange Deter- mine the resulting displacement of the free end ATAT T r-z1 ° Fig A7.28 Solution:

The axial deformation of a differential element of the upper flange (subscript U) was assumed given by ÂU = aT dx where a was the material thermal coefficient of expansion The lower flange, having received no heating underwent no expansion

Inasmuch as a thermal expansion is uniform in all

directions no shear strain can occur on a material element

Hence no shear strain occurs in the web The apparent anomaly here - that web elements appear to undergo shear

deformations ¥= am (Fig AT 29b) - is explained as

follows: The temperature varies linearly over the beam depth The various horizontal beam "fibers" thus undergo axial deformations which vary linearly also in the manner

of Fig AT 29b giving the apparent shear deformation No virtual work is done during this web deformation since no axial virtual stresses are carried in the web

with the addition of a unit (virtual) load to the free end, the virtual loadings obtained in the flanges were: a Lx* U h u — ° „ “ of ayy s | at, 0-H Example Problem 24

™e “irst step in computing the thermal stresses in a closed ring (3 times indeterminate) involves cutting the ring to make 1t statically determinate and finding the relative movement of the two cut faces

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AT.18 DEFLECTIONS to vary linearly over the depth of the cross

Section, Find the relative movement of the cut surfaces shown in Fig A7.20b we ue 4 LX ah tho Raf | FSR dộ | JR=Td@i a b ° Fig A7,30 Solution:

An element of the beam of length Rdd 1s

shown in Fig A7.30c Due to the linear temperature variation an angular change dG = Ret dp occurred in the element The change in

Length of the midline (centroid) of the section Rat ab

was 4 = =

plied at: the cut surface as shown in Fig A7.30b giving the following unit loadings around the ring

From unit redundant couple (X)

(m positive if it tends to open =1 "x the ring) uy (= axial load) Unit redundant loads were ap— o (positive of tensile) From unit redundant axial (horizontal) load (¥) m, = -R (1 ~ cos $} uy = cos ý $ frơm unit redundant shear (vertical) load (2) m, = -R sin Ệ u, = - sin b The deflection equation by the dummy~unit load method is furarfa-ae 2n [os |F Ral ag = ° an oy = |:= 8a ab + a ) R(l-cos o) Rot 6 Then % = ad _ at R®gT (negative indicating movement to h the right) OF STRUCTURES 2n (-sta 9) et a + al ° #n (-R sin a Remarks:

In the three elementary examples given above no stresses were developed inasmuch as the idealizations yielded statically determinate structures which, with no loads applied, can have no stresses Indeterminate structures are treated in Chapter A.d

) Rel ag =o

AT.9 Matrix Methods in Deflection Calculations

Introduction There {Ss much to recommend the use of matrix methods for the handling of the quantity

of data arising in the solutions of stress and deflection calculations of complex structures: The data is presented in a form suitable for use

in the routine calculatory procedures of hign speed digital computers; a flexibility of operation 1s present which permits the solution of ad- ditional related problems by 4 simple expansion of the program; The notation itsel? suggests new and improved metheds both of theoretical approach and work division

The methods and notations employed here and later are essentially those presented by Wehle

and Lansing® in adapting the Method of Dumy-

Unit Loads to matrix notation Other appropriate references are listed in the bibliography

BASIS OF METHOD

Assume the structure to be analized has been idealized into a truss-like assembly of rods, bars, tubes and panels (sheets) upon which are acting the external loads applied as concentrated loads Py or Py, each with a different numerical Pin Pai FP, - my +, bes) + ‡ fa) (b)

Fig AT.31 Idealization into an assembly of bars and panels

For the reader not familiar with the elementary arithmetic rules of matrix operations employed here, a short appen-

dix has been included

9L B Weble Jr and Warner Lansing, A Method for Re-

ducing the Analysis of Complex Redundant Structures to a Routine Procedure, Journ of Aero Sciences, 19, October

T882

Trang 12

ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES subscript Thus the system of Fig A7.3la is

idealized into that of Fig A7.31>

With the above idealization an improved scheme may be employed to systematize the computation of deflection calculations The ?o1~ lowing steps summarize the procedure which is discussed in detail in succeeding sections

I A set of internal generalized forces, denoted by ay or q3 i, j are different numerical subscripts), 18 used to describe the internal stress distribution The q’s may represent axial loads, moments, shears, etc In con-

junction with a set of member flexibility coef ficients, a4, the q’s are employed fo exprass the strain energy U at gives the displacement or point i Per unit force at point d.*

II, f8quilibrium conditions are used to relate the internal generalized forces đi, qj to the external applied loads, Py or Py With this relationship the strain energy expression obtained in I, above is then transformed to give U as a function of the P’s,

T1I, Castigliano’s Theorem is used to

compute deflections

CHOICE OF GENERALIZED FORCES

Consider for example the problem of writing the strain energy (of flexure) of the stepped cantilever beam of Fig A? ,32a, assuming external loads are to be applied as transverse point loads at A and B The set of internal generalized forces of Pig A7.32b will com- pletely determine the bending moment distribu~ tion in the beam elements and hence the strain energy Set (b) then is a satisfactory choice of generalized forces,

It should be pointed out that set (b) is

not a unique set Other satisfactory choices (not an exhaustive display) are shown in Figs A7.d2c, d and e, The final selection may be “made for convenience or personal taste

Note that only as many generalized forces are used per element as are required to deter-—

mine the significant loadings in that element Ela A El, 3 _ 4 habs ha) qs (——) =——— Ga 8y (e) qa dig đa q Poe Yee as {b) (a) M=aqx O<x<L, ds đz qa 4—) ——'

M= qs + qay O<y<Le (e) Fig A7.32, Some possible choices of generalized forces

("Relative displacements in the individual member")

AT.19 THE STRAIN ENERGY

Observe that each of the integral terms in the above expression is a property of the structural element (variation of EI) and of the nature of the associated generalized force (exponent on variable), Introducing the notation = [xtex = (“yay fa) ° ar Gea =| 2 OD Eayay - [hs das = tT, Isa = =I, 5 3

Equation (19) is an expression for U which could have been written immediately from physi~ cal considerations Hach coefficient ayy is the displacement at point 1 per unit change in force at point J This identity 1s easily seen by applying Castigliano’s Theorem to eq (19), With this interpretation the first term in eq

(19), representing the strain energy in the outer team portion, is written by analogy to

1 8 L

eq (2) of Art, A7,3 ( 73 3£)"

Trang 13

AT 20

The general form of strain energy express- jon is (expanding eq (19) by induction) 2U = Q:Q: G11 + Gide Gia t+- - - +a Gy Oy i ' +QaQx Gai * đaQz dạa tan nan n 1 1 1 +dady 51 + Qađa 0á + - - - ~~ ee He ' tha Git + + - ee ee ee ee ' \ 1 1 ! 1 + N4 ÔN Tnhh TT TT + Qn Gy Sy In matrix notation this equation is written (see appendix)

au = [984 coo ay | Già Qia- di qa đay Gaa-————— S3 (e1) 1 1 ' ‘ SO Se Nah or, more concisely, 2U = Lai} [mua] 13st the matri: a iz

In the x [ tạ] many (if not most) of

the elements are zero In the specific example, aq (21) would be written

2u =q:qzqs | |a, co 0 qa

9 daz Gas Qa

9 đạa Iss Qs

The problem of computing and tabulating various a,,’s is considered in detail later _

RELATING THE INTERNAL GENERALIZED FORCES TO THE EXTERNAL APPLIED LOADS For the statically determinate structures considered in this chapter the internal forces are related to the external loads by use of the equations of static equilibrium A set of linear equations results Thus, in the specific example considered, if P, and P, are the external loads applied as in Fig A7.cef, then by

statics (refer to Fig A7.32b) DEFLECTIONS OF STRUCTURES qs = Pi qa = Pi + Pa Qa = Pa La ` ì P, Py 4 ‡ ị _—————— Am Flg AT.32f In matrix notation dd 19 Py đa) =#|1 1 Pa qs li 0 Symbolically these relationships are wrivven \ E || \ ig =| Gim| (Pm its voy

The matrix [2] is called the "unit load

distribution” inasmuch as any one column of [ein] ; say the mth column, gives the values of

the generalized forces (the q’s) for 2 unit

value of load Py, all other external loads zero

THE STRAIN ENERGY IN TERMS OF APPLIED LOATS ~ If eq (22) and its transpose are used to sub- stitute for the q,’s in eq (21) one gets {

=8J[x]Eelbzlll

In the notation here 1 and j are used tnter- changeably as are mand n Also [Gmi7 1s the

transpose of [#im] , 1.e interchange of subs scripts denotes transposition (see appendix)

If the matrix triple product in eq (23) 1s

formed and defined as

(oe J fie] ar] [Ba] === => so

Trang 14

[Pe] [ae] }Pa|

Eq (25) expresses the strain entrgy as a function of the external applied loads

In the specific example being used

di o 0 1 9

zl|]l1L

[aan | = 31° © Gas Gas 1 1

Ø Qsa dss

DEFLECTIONS BY CASTIGLIANG’S THEOREM Application of Castigliano’s theorem to eq (25), au = | Fa ì {ha - (25) yields i288 Lig le ip to ieee | OP, 4 on | [3a | thị (26)

The steps in passing to eq (26) may de demonstrated readily by writiag out eq (25) for, say, 2 set of three loads (m, n = 1,2,3),

differentiating successively with respect to BR, Pe and Ps and then re-collecting in matrix

form

The matrix [ann | gives the deflection at the external points "m" for unit values of the loads Po and is therefore, by definition, the matrix of influence coefficients

COMPARISON WITH DUMMY~UNIT LOADS EQUATIONS It is instructive to write eq (26) out as

teat ac] Bead [exe] fret

and compare the expression with a typical term from the dummy-unit load method equations, say

528 aS + In the mat#x equation (26a) the is the unit (virtual) matrix correspond~

ing to the symbol "u" in the simple sum The

`

are the member flexibilities correspond-—

G {pt

Tne matrix product Fin] ¡BỊ

gives the member load distributions due to the real applied loads, hence these are the "S" leads Finally the operation of matrix multi-

ing to stot 2

AT 21 plication itself yields a summation to complete the calculation

from this last discussion it {s clear that eq (26a) also may be derived by formulating the Dummy-Unit Load equations (Art A7.7) in matrix

notation

To illustrate the application of the matrix metheds presented thus far, 4 brief and element-

ary example is worked with those tools already developed

Example Problem 25

Determine the influence coefficient matrix for’ transverse forces to be applied to the uniform cantilever of Fig A7.33 at the tnree points indicated ® ,® 9® —————— L Rete — as 24, 44 T3 a Fig A7,33 Solution:

The choice and numbering of generalized forces are shown on the figure These forces were placed so that praviously derived express-

ions for the a’s could be used The following member flexibility coefficients were computed Note that the only non-zero coefficients of mixed subscripts (i not equal to J) are those

for loads common to an element Lyg

x?dx _ LS

fia =| ° “Er “ BIẾT

This expression was adopted from that developed for a transverse shear force on a cantilever beam segment in the preceding illustrative example L ⁄3 L SEI œ bey das = Ten= tal Fal

This expression is for a couple on the end cof a cantilever segment (of length L/3),

Ls 3

das = Gee = yay _ _L?

aa ee “EI ~ isst

°

Trang 15

AT, 22 o 0a L = ° 1 [aja kẻ: i ° 1 °

The unit load matrix [Ea j was computed by successively applying unit leads at points 1, 2 and 3 and computing the values of the q’s by statics 1 0 0 2 1 0 [#m]= | 1⁄4 9 0 11 2 2L/z bự 9

Note that the first colum of [2.2] gives the values of the q’s obtained for a unit load at point "1" with no other loads applied The second colum gives the q’s for 4 unit load at point "2" only, and so forth Finally, i 0 0 9 0ll1 a 9 111⁄4121⁄2|o to 0ll1 10 L L Z7 6L [Aon gar io oO 1 Be otf 0 6|f⁄2o o 0001 0 rail, yy 9 9 0 mất 2 Len yy 00 0 & Es 30 Multiplying (see appendix), 54 28 8 1® = [Aum |= sager 28 16 5 8 5

Should the deflections be desired at the three points one forms a 54 28 8 Py Lê 6a >= aber [28 16 5 Pa 5s 8 5 2 Ps DEFLECTIONS OF STRUCTURES as per eq (26)

Tne matrix [Bon | is seen to be symmetric about the main diagonal as it must be:

Maxwell?s reciprocal theorem A, 5 mn = A eq (20)) ?rem mm (see AT.10 Member Flexibility Coefficients: Compilation of a Library

Several member flexibility coeffic derived below for various members and 1:

BARS

The energy in a uniform bar under varying axial force (Fig 47.34) is q,-8 a m= —=—=ˆ b HmL ¬ Then referring to eq (20), Fig AT 34 vy a (eH) „9U ÄJj |dJ+r) dx {1 = 3q7 ¬ + ° ° eet fag SAE “ja/* and, 3? : _ UỦ _L |x _* "14° jqaq, AE LỆ T)% ve" ° =.L < ~ Sie (= 23)

Trang 16

es a

ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES AT 23

In the case of tapered members the

coefficients are determined by evaluating in- : U = s& || 31 = ra +q 3* K =, qj

tegrals of the form 2k1 L

c1 Le (x )ax and from which

“I°F) aR L Fig AT 34e

° S11 Far

“a 5

NOTE: Th -

Such a quadrature can always be made in these L ieients for nu

problems For the linearly tapered bar the O44 2 ay, = beams are directly results may be obtained as functions of the end 251) analogous (compare

area ratios Thus, Wehle and Lansing give several cases) so

_ L that for tapered

L Jd * gay beams one uses the

aii =+=-e buy results for tapered

oA 5 bars with EI in place

4q er SHEAR PANELS {of AE ._L Ay ‘aL N %1“ sa g- bY Fig AT 34b L #131 “ST T' $jj Fig AT 34¢ Area varies linearly, BEAMS The energy in the uniform beam o? A7.34d ts given oy Fle Then Fig AT 34d -2U by Git = 5q77 aay (= a) 2 U ò =k (2 041)

An alternate choice of generalized forces for the beam of Fig A7,.34d is shown in Fig

A7,24e,

For the rectangular shear panel with a uniform shear flow q, on all edges (Fig A7.34f) ai? errs Ue 26t Ss i sha † 8 = surface area { > 3 == OL Foe Fig AT, 34f

The trapezoidal shear

panel (Fig A?.34g) is treated approximately by using the average shear flow on the non-parallel

sides as though it were constant throughout the sheet Thus at? U = a8 S = surface area 3 aii =a Since by statics 4, = Fig AT 34g hạ

Sun; one could use q, a8

an alternate choice of generalized force and 933 7 Ge : TORSION BAR A uniform shaft under torque 4, has strain energy Qi®L

Pay Then aie,

AT.11 Application of Matrix Methods to Various Structures

Example Problem 26

The tubular steel truss of Fig A7.25 is to be analyzed for vertical deflections at points = and F under several load conditions in which vertical loads are to de applied to all joints excepting A and D The cross sectional areas of tube members are given on the figure

Trang 17

AT, 24 DEFLECTIONS OF STRUCTURES Constant axial † Loads Sun, 1s" > 4 Then mee yt aig « L/AE % ah

Fig AT.35 Fig A7.35a deflections at points & and 7

Solution:

The member flexibility coefficient fora uniform bar under constant axial load is U/ag Fig A7.36a gives the numbering scheme applied to the members and the q’s (these being one and the same, since q is constant in a given member) Fig A7.36b shows the numbering scheme adopted for the external loading points P, P 1 2 + 4 43 +P 5 tp,

Fig AT 36a Fig A7.36b

Member flexibility coefficients were collected in matrix form as ì 1 | 2 3 4 5 § ? 8 9 1192.2 8 0 Qo 6 Q ö 9 a 2) 0 92.2 9 QO 9 a 6 9

aio | o |izslol o |o|sl slo L: : 3 z‡|l4|o | 0 | o juss} o | ojo) o | o 5| 9 ° 9 0 55,7 9 8 9 9 6| 9 a @ 9 6 183 j 0 6 9 7}a | 9 | o} 0! 0 | o fis} o lo a; 0 ọ Qo 9 ọ 9 Ø6 $5.7 9 3, 0 0 9 8 oO 9 6 9 229

In the case of a pin jointed truss, where only a single generalized force is required to describe the strain energy per member, the matrix of member flexibility coeffictents is a

diagonal matrix as above

Unit lead distributions were obtained by placing unit loads successively at external loading points 1, 2, 3 and 4 (Fig A7.36b) The results were collected in matrix form as ÑJ 1 2 3 2 | 1] 0 1.0 9 1.0 210 bì 0 ọ 3| 0 -1.0 e 9 E =| 4) 0 1.25 9 1.25 5] 0 “1.03 9 ~1.08 8|~1.9 |- 75 ọ - 75 7| 1,13 848 | 1.13 848 8| -,885 | -.185 | -.825 | 71.65 9} 20 0.40 20 0,40

Then the matrix triple product

fin] si] bs] fe]

was formed giving, per eq (26), lá, 440 589 287-389 | | Ps da}, | 389 927 252 789 || Pa 4, = 257 252 257 252 |] Ps &, 389 78 252 789 || P„ inches

The results here give the deflections of all four points Since only the deflections of points 3 and 4 were desired the first two rows of [| may be dropped out The same result

could have been achieved by leaving out the

first two rows of [és :| ( transpose of se)

The matrix form of equation above is usefu in organizing the computation of deflections for a number of different loading conditions Thus, should there be several different sets of external loads P,, corresponding to various load~ ing conditions, each set 1s placed in column form giving the loads as the rectangular matrix

>

[?» ; k different numerical subscripts for the

Trang 18

vas the deflections at each point (m} for various load conditions (x)

2 Problem 27

Derlections at pcints 1, 2, 3 and 4 of the truss of Fig A7.25 are cesired for the follow~ ing loading conditions: ZandiEion No, Py Pa Ps P, (see Fig 1 2500 | 2000 ‘CO 450 A7.36b) 2 ~1200 | -300 | -2100 | +1750 3 1800 | 1470 | -1200 | -1100 Solution: The matrix product formed per eq (26b) was Set up as 440 589 257 489 | |2800 -1Z00 1800 = 1 389 927 252 789 2000 - 300 1470 1 Ễ Š | 8857 252 287 258 800 -2100 -1200 389 789 252 789 450 -1750 -1100 2xample Problem 28

For the landing gear unit of Sxample Prob- lem 20, Fig A7.23 find the matrix of influence coefficients relating deflections due to lift and drag loads acting at point A and torque about the axle 4-~B

Solution:

The structure was divided into

and the set of internal generalized forces applied as shown in Sig A7.37a (Torques and moments are shown vectorially by R.H rules) Axial stresses were neglected in C-B elements Bis B TA, el aes 3, # Nar Lx cà — tờ Oe Ps BA qa AaB 3a

Fig AT 37a Fig A7.37b

ollowing member flexibility coefficients were determined AT, 25 LS AB ; SEI L* L BC ; Ge, = BC SEI Gử L? L BC ; đau = Qạa = BC 2E1 EI (rot ing that dir = Gass a * + u a ø a " As = Vea Collected in matrix form az =#5) 1.35 1 3 4 5] 6 ] 7 a 1pe0l o} ol 9 of oefe ọ 2| o |3.s|o| 9 a}ola Q 310/10 |s0} © oj} alo ọ [es] =a |J+|9 |9 | 0 | 15,600! 648 | o | 0 6 sjololo | &ø |z.o| o | o ọ 6 9 a 9 9 0 36.0 9 s48 zlalojo} o 9 | o |4.8| 9 alojojo] o 9 | saa | o {25,600

The unit load distributions [era were obtained by applying unit external applied loads, numbered and directed as in Fig A7.37b À + | 2 | 3 1.0 9 9 2 1.0 9 đi 0 9 1,0 4| 342 9 9 6 9 -.937 1.026 7 9 342 2.311 3 9 9 1.0 cm At this point problem as solved, fo ig a routine cperation: FE JEIES example Problem 29

Tre beam of example problem is to be solved by the matrix methods oresented herein

nginesr may consider the @ ramaining computation

él re-

Trang 19

AT 26 DEFLECTIONS, OF STRUCTURES Influence coefficients for points F, G and H are to be found Solution: Fig A7.38 shows the choice and numbering of generalized forces (coetf's a coll'td fr two members) 1 „ Qe 93g ° 933 * 2 33 oS ee + We la Ld b ——— j ——— qe =————— "4 Fig A7.38

No forces were shown applied to the lower flange elements as these were known to be equal to those of the upper flange due to symmetry Entries were made for ty in matrix form as be~ low, Entries for aa, and d,, were quadrupled as these occur in two identical members each on top: and bottom Entries for dea, Geo and Goo were

doubled (See Art A7,10 for coefficient formulae } 10 2065 17.8 4,46 Bạn“ 1.8 ` a 4.48 9.89

Unit load values were obtained as in Pig A7.27, considered to be external loading num-

ber "two", Similar diagrams were drawn for unit

loads at points "one" (H) and "three" (Fl ior Fig A 7-25) Ẳ +1 2 3 1/ 1.0 9 9 2 «0667 9 9 3 | 1.201 9 9 [z~| -]+| 0 |i.o ọ 5 „0540 „0600 Q 6 | 3.184 |1.08 9 7 9 9 1,0 3 0447 „0496 20545 9 ] 5.00 2.00 11,00 The matrix triple preduct ñ)'E=IzIE4 completes the calculation Example Problem 30

Deflections of statically indeterminate structures often may be computed successfully 5y the methods of this chapter provided that some auxiliary means is employed to obtain an approximation to the true internal force distribution The exact internal force distribution 1s not necessarily required in making deflection calculations inasmuch as such a calculation amounts to an integration over the structure - an operation which tends to average out any errors, Thus one may use the engineering theory of bending (&.1,.8.), experimental data, previous expertence, etc to obtain reasonaple estimates of the internal force distribution for unit loadings

In the following problem the matrix of in~ fluence coefficients 1s determined for 4a singles cell, three-bay box beam (3 times indeterminate) by using the 5.T.3B

Fig A7,.39a shows an idealized doubly symmetric single cell cantilever box deam naving three bays Determine the matrix of influence coefficients for the six pcint net indicated 301 —— 20" Stringer Areas Constant = 33 in # Fig A7 39a Solution:

Fig A7.39b is an exploded view of the beam showing the placement and numbering of the internal generalized forces Note that only th upper side of the beam was numbered, the lower side being identical by symmetry

Trang 20

Fig A7.39b

Was accounted for Note also that entries for

G44; đam; Gog ANd Gig.i19 Were Te-doubled as

Note: VOID SPACES INDICATE ZEROS

Unit load distributions were obtained for successive applications of unit loads to points one through six (Pig A7.39a) The internal forces predicted by the £.T.B, for a load

through the shear center (center of beam, due to symmetry) were superposed on the uniform shear sot 2 7 3 4 5s § 1 |0.3 g.1 2 |0.1 |J¬9.1 „|5 |9.1 0.3 Ea] “ia 2 s2 2 § | 0878 | ,0125|0.15 | 9.05 7 | 0578 |~.075|0.05 | ¬0.08 8 | 0128| 0875|0.06 | 0.15 9 [2.67 |2.67 [1.93 | 1.33 10 | 2.67 2,87 1,5% 1.33 11 „9425 „0023 0713 0179 10.100 „=5 12 „020 =,020 „0267 | -.0267 0333 |- 0855 13 „0065 „0425 20179 „3715 0333} 100 1⁄4 3 2 1 1 15 3 2 1 J1 AT 2T

flow ( = 4) due to the torque developed in transferring the load to one side

The matrix triple product

iE [3] [&]

completes the solution

AT.12 Deflections and Angular Change of the Elastic Curve

of Simple Beams by the "Method of Elastic Weights"

{Mohr's Method)

In the calculation of structural deflections there occur many steps involving simple integral properties of elementary functions The Method of Elastic Weights (and the Area Moment Method to follow in Art A7.14) owes tts popularity in large measure to the fact that it enables the analyst to write down many of these integral properties almost by inspection, rely~ ing as it does upon the analyst’s familiarity with the properties of simple geometric figures

For finding the deflection of a point on a simply supported beam relative to a line joining the supports, the Method of Elastic Weights states:

distributed beam load ˆ

Spelled out in steps:

i - The # diagram is drawn just as it occurs due to the applied beam load

1i ~ This diagram is visualized as being ‘he loading on a second beam (the conjugate beam) supported at the points of reference for

the deflection desired

iii - The bending moment in this conjugate beam is found at the station where the deflection of the original beam was desired This bending moment is equal to the desired deflection

To prove the theorem, consider the dumy- unit load (virtual work) equation

7 ;:ơ»n‹-d6«n: Mdx

Trang 21

AT, 28

eq 12) pression, usin:

the bending

simple beam ioaded by ceflection at the point (

show that the deflecti above equation, is tne moment expression fer a an "elastic weight" == bề

In Fig A7-40, the loading of (a) produces the real moments of (D) Consider the deflections of points 8 and C due to the angular 2 as change Bes in a beam element at a* (F‡8 À7-4C2), Py Pa = A B be L L Lio ree + 4 # {a) ii sagan ¬ kdx (b) Mdx EL 7 {c} i

(d) mp, moment diagram for unit load acting downward at point B 3 tel (e) mg, unit load at point C Max (Er 3 Max ứ 1 Mdx 4 EI 4 ET Fig a7 40

For a unit load at point b, Fig d shows ¢ diagram The value of m at the midpoint of dx

(point a) = L/8 Hence Mdx L _ MLcx

S = ar ° 8 * ear

For deflection of point c, draw m diagram for a unit load at C (see Fig e) Value of m on element dx = L 16 Hence | _ Max L _ Mick ° * SI 16 HH

“For simplicity the points A, B and C were placed at the one- quarter span points The reader may satisfy himself with the general character of the proof by substituting x4, xp and X¢ for the point iocations and then following through the

argument once again DEFLECTIONS OF STRUCTURES Fig f consider Mcx as a load on a simply EI

supported beam and deter

at points 5 and c cue ta “te = L , Made tp “2 SEI = be Mo ấm cac

These values of che moments 40 points > and c are identical to the deflections at b and c by the virtual work equations The moment diagram m for a unit load at b and c

(Figs d and e) ts numerically precisely the same

as the influence line for moment at points b

and c

Therefore deflections of 4 simple beam can be determined by considering the M curve as an

KT

imaginary beam loading The bending moment at any point due to this M loading equals the de-

aI the team under

flection of the given loads

Likewise it is easily proved that the engu- lar_chanze at any section of a simply supported

beam is equal to the shear 2t that section due to the M diagram acting as a deam icad

Bị

A7.13 Example Problens

SXample Problem Fine ac

deflection and slope of points a ‘and b for beam and loading shown in Fig A?.41 The lower Fig shows the moment diagram for load P acting at center of 4 simple beam P 1L + h2 ¬ F5 = 1 EI Constant a b L P > 2 3 cE aL Fig, AT 31 + L k a L L pL s12 T8 3 ie PL? PL? PL? 64 16 18

Trang 22

=6 (Slope ts horizontal or no change from original direction of 5eam axis.)

2 Datarmine the deflection

loaded uniformly as shown in

bending moment expression for = wlx - wx® or parabolic as

2 2

.42a The deflection 2t mid- bending moment due to M dia- fa simple team : & A7.42 The UW shown in Fig A7 point equals the gram as a load F“=—L = su 2 wo 2 Fig A7 42 M Diagram 1 1 sz WL LẺ = 54 w 3 w Area 1 LS center Za 5 wLt 384 SĨ” L 1 1 đeenter {4 WLS ~ 54 a8) 8i” 9 3 3 1] wU LO at st = th eact so Slope supports @ reaction = 55 7 Exemole Probiem 2s

Fig A’.44 shows the plan view of one-nalf of a cantilever wing The aileron is supported on brackets at points D, 5 and F with sel?-al!zn- ing bearings The brackets are attached to the Wing rear beam at points A, B,and C When the wing bends under the air load the aileron must likewise bend since it 1s connected to wing at

three points In the design of the aileron bean

and similarly for cases of wihg flaps this de- lection produces critical bending moments As- suming that the running load distribited to the

rear beam as the wing bends 2s a unit 1s as shown| !

in the Fig., find the deflection of point 3 with respect to straight line joining points A and ¢, wnicn will be the deflection of © with respect to line joining D and F if bracket deflection is neglected The moment of inertia of the rear beam between A and C varies as indicated in

the Table A7.6 AT 29 © Airplane [FƑ# re |Ailsron > ge Rear Beam A Cc ) yo Front Bean Plan View 1/2 Wing Air Load 29#/in ey (1:1111111111111111111111T1111TT 111m

& Load on Rear Beam 13#/m 5

Fig A7 1â

Solution:- Due to tne beam variable moment of inertia the deam length between A and C will be divided into 10 equal strips of 10 inches each The bending moment M at the midpoint of each will be calculated The elastic weight for each strip will equal Mds, where ds = 10" and I the moment

+

of inertia at midpoint of the strip These elastic loads are then considered at loads on an imaginary beam of length AC and simply supported at A and C The bending moment on this imaginary bzam at 2oinE B will equal the deflection of B with respect to Line joining AC

The bending moment at c = 15 x gO x 15 + 10 x 15 x 10 = 8250”£

The shear load at C = (15 +25) x 30 = 600# 2

Bending moment expression between points C

and A equals, M = 6250 + 600x + 12.5x@, where

x =O to 100

Trang 23

AT 30 DEFLECTIONS OF STRUCTURES

Bending moment at point B due to above elastic loading = 7,100,000 * deflection at B relative to line aC = 7,100,000 = 71 inch

£=10,000, COO Example Problem 34

Fig A7.43a shows a section of a cantilever wing sea plane The wing beams are attached to the hull at points A and B Due to wing loads the wing will deflect vertically relative to attachment points AB Thus installations such as piping, controls, etc., must be so lo- cated as not to interfere with the wing deflections between A and B For {llustrative pur- poses 2 simplified loading has been assumed as shown in the figure EI has been assumed as constant whereas the practical case would in- volve variable I For the given loading deter~ mine the deflection of point C with respect to

the support points A and 8 Also determine tha vertical deflection of the tip points D and E So eS 2 5 S s eo ° 400" Sa 300" " s Zh 5 §r Trở | Fig AT 43a D A B ElConstant E Moment Diagram s in # t Fig AT 43b i „ { ' gi SI abe | ote | | 3 3, # giŠ|I|lSø gi Pi£ Aidie sen See ae | ~ rs a 4 Sale ¡| § § Sp 876" 4 “Fig AT 43Â Đ 720" 3

Solution:- Fig A7.43b shows the bending moment diagram for the given wing loading To find the deflection of C normal to line joining AB we treat the moment diagram as a load on a imaginary beam of length AB and simply supported at A and B (See Fig A7.43c.) The deflection of c ts equal numerically to the bending moment on this ficti-~ cious beam

Hence EI6, = 25920 x 40 = 25920 x 20 518000

EI

Ta find the tip deflection, we place the elastic loads (area of moment diagram) on an imaginary beam simply supported at the tip D and = (See Fig A7.43d) The bending moment on this imaginary beam at points A or B will equal numerically the deflection of these pointa with respect to the tip points D and E and since points A and B actually do not move this deflection will be the movement of the tip points with respect to the beam support points

Bending moment at A = 192420 x 700 - 40000 x! or Og =

AT 14 Deflections of Beams by Moment Area Method*

For certain types of beam problems the method of moment areas has advantages and this method is frequently used in routine analysis Angular Change Principle Fig A7.44 shows 4 cantilever beam Let it be required to determine the angular change of the elastic line between the polntsAand B due to any given loading From the equation of virtual work, we have Deflected 433 - 127500 x 124 = 102200000 " Stip = 102,200,000 EI Fig AT 44

section, distant x from B due to unit hypothetical couple applied at B But m = unity

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Referring to Fig A7.44, this expression repre-

sents the area of the M diagram between points BỊ

B and A Thus the first principle:- "The ghange in slope of the elastic line of a beam between any -wo points A and B is numerically equal to tne area of the M diagram between

EL these two points.”

Deflection Principle

In Fig A7.44 determine the deflection of point 83 normal to tangent of elastic curve at A In Flg A7.44 this deflection would be vertical since tangent to elastic line at A is horizontal

From virtual work expression 6g = a & m, where m 1s the moment at any section A distance x from B due to a unit hypothetical vertical

load acting at B Hence m= 1.x = x for any point between B and A

Hence A Mdx

53 = B TỈ + Ta1s

ession represents the lst moment of the diagram about a vertical thru B Thus the

Is

Ly T

deflection principle of the moment area method can de stated as follows:- "The deflection of a point A on the elastic line of 4 beam in dending normal to the tangent of the elastic line ata Doint 5 is equal numerically to the statical mor tient of the M aréa detween points "AT and Ta" B1 about point A" Illustrative Problems

Example Problem 25, Determine the slope and vertical deflection at the free end B of the cantilever beam shown in Fig A7.45 51 is constant Pp A ip a 1 L—— 2 77M Diagram 2 i Pu ; 2,

Solution:- The moment diagram for given load is triangular as shown in Fig A7.45 Since the beam 1s fixed at A, the elastic line at A is horizontal or slope is zero Therefore true slope at B equals angular change between A and B which equals area of moment diagram between A and B divided by EI Hence Fig AT.45 PL2 _#EI

he vertical deflec9lon at B 1s equal to the lst moment of the moment diagram about point B di-

vided by BI, since tangent to elastic curve at A tg horizontal due to 2ixed support

ap = (-PL L/2) # =

AT 31

fence B (PL 2,)2 2-28 2 °'3 7) 8 SEI

Example Problem 36

Fig A746 illustrates the same simplified wing and loading as used in example problem 34 Find the deflection of point © normal to line joining the support points A and B Also find the deflection of the tip points D and & relative to support points A and B 1000 2 B Fig AT 46 Moment Solution:-

Due to symmetry of loading, the tangent to the deflected elastic line at the center line of airplane is horizontal Therefore, we will find the deflection of points A or B away from the horizontal tangent of the deflected beam at point CG which is equivalent to vertical deflection of c with respect to line AB

Thus to find vertical deflection of A with respect to horizontal tangent at C take moments of the M diagram as a load between points A and BI CG about point A Whence (area) (arm) Og _ 1 (650+ 646 ) =

Toco (tangent atc) = 2 (S20 688 40x20

# (518400) = deflection of © normal to AB To find the vertical deflection of the tip point D with respect to line AB, first find deflection of D with respect to norizontal tangent at C and subtract deflection of a with respect to tangent atc

oD Mườd (respect to tangent at C) (re % 2) ei #ĩ (40000 x 267 + 127500 x 576 + 25920 x 720) = = {102700 , 000)

(See Fig A7.46 for areas and arms of M/SI diagram) Subtracting the deflection of A with respect to C as found above we obtain

OR (r ei -

B08 (respect to line AB} # (102,700,000

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AT 32

A7,15 Beam Fixed End Moments by Method of Area Moments

From the two principles of area moments 2s given in art A7.14, it is evident that the deflection and slope of the elastic curve depend on the amount of bending moment area and its lo- cation or its center of gravity

Fig A7.47 shows a beam fixed at the ends and carrying a single load P as shown The bend4 ing Moment shown in (c} can be considered as made up of two parts, namely that for a load P acting on a simply supported beam which gives the triangular diagram with value Pa (L-a)/L for the moment at the load point, and secondly a trapezoidal moment diagram of negative sign with values of Ma and Mp and of such magnitude as to make the slope of the beam elastic curve zero or horizontal at the support points A and B, since the beam is considered fixed at A and By

The end moments Ma and Mp are statically indeterminate, however, with the use of the two moment area principles they are eastly determin-

ed In Fig b the slope of elastic curve ata and B 1s zero or horizontal, thus the change in slope between A and B is zero By the lst ỊP jw a aA ¬1 El Constant of 2 (a) a L ——Ừ a ⁄ A lastic Curve Fig, A7 47

principle of area moments, this means that the algebraic sum of the moment areas between A and B equal zero Hence in Fig c

(+a -Mp) L, (Pa (L-a) LL

2 L "2

In Fig > the deflection of B away from a tangent to elastic curve at A is zero, and also deflection of A away from tangent to elastic curve at Bis zero

Thus by moment area principle, the moment the moment diagrams of Fig C about points A, B is zero

of or

Taking moments about point A:-

SA = Pa “(L-3).2a+Pa (La ) x(4 + Lò8)*HAL XU „ aL 3 2L 3 2 3 Mplx2 L=0 - (a) 2 ó Solving equations A and B for Mg and Mp DEFLECTIONS OF STRUCTURES M Mạ = - Pab° and Mp = - Pba°/L? xhere d= (L-a} > L

To find the fixed end moments Zor a team with variable moment of inertia use the M/I di in place of the moment diagrams,

g A?.48 shows a fix-ended beam carrying two concentrated loads Find the fixed-end moments Ma and Mp on 100# ˆ É 4 : ứ Mã Za : Re Mp (a) {b) Fig A7.48 Mal 15M + TT” —— Mp —— 20" —> _ Fig b shows the static moment diecran Solution:- agr

assuming the beam simply supported at A and 3 For simplicity in finding areas and taking moments oz the moment areas the moment diagram has been divided into the 4 simple shapes 2s shown The centroid of each portion is shown together with the area which is shown as a concentrated load at the centroids

Fig C shows the moment diagrams due to un- Known moments Ma 4nd Mg ‘he area of these tri- angles is shown as a concentrated load at the centroids

Since the change in slope of the elastic curve between A and B is zero, ‘he area of these moment diagrams must equal zero, hence

5265 + 14040 + 2160 + 6885 + 15M, + 15Mp = 0

or

1SMg + 15Mg + 28350 =0

The deflection of point A away from tang: elastic curve at B is zero, therefore the 2 moment of the moment diagrams about point A

equals zero Hence, 9265 x 6 + 15 X 14040 + 17 x 2160 + 24 x 6868 + 15CMA + irst 300 Mp =O or 150 Mạ + 500 Mp + 424600: 9a - - -(2) Solving equations (1) and (2), we obtain Ma = - 816 in, lb Mp = - 1074 in lbs

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