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

The set 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 syst

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

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 re-

spect to P may be carried out before the inte-

gration is made The resulting integrals gen-

A fictitious load Pi was added at point B

and the bending moment diagram was drawn in two

Fig A7, 13a

‘Then neglecting the energy of shear as bein sma11*

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

Differentiation under the integral sign with respect to P, gave

* 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

(see p AT 14) shear deflection contribution

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

quarter circle and acted

upon by a torque To

Find the vertical move-

ment 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

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

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

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

Method of Virtual Velocities, Method of Virtual Work, Method of Auxiliary

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

differentiate under the integral sign with since zero resultant force moving through a

respect to P, to obtain distance does zero work

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

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 load-

ings due to a unit load (the virtual load) ap-

plied 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

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

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 re- actions 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

=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)

@ Note that the deformations are not restricted to those due

to elastic strains only They may be the resuit of elastic

or inelastic strains, temperature strains or misalign- ment corrections,

AT.I1

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

tion calculation divides conveniently into onl 0 0 9 9 Q

7 ol

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

(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,

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

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

30" peso 30"— the Joint moves to the LEFT since the unit load

20004 20008 was drawn to the RIGHT in Fig A7.16b)

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

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

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

points of the motion Thus, for solution to part (c) a unit load system was applied as shown in fig A7.16c and for the solution of

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%,

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

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 °

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

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

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

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

= ,00053 radians

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

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

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

17,9 x 10%

BH Therefore deflection of point D relative to line Joining CB is down because result comes out nega— tive and therefore opposite to direction of vir- tual load,

Example Problem 13

zw 1h/£t

couple Fig AT 20

Ca)

for the virtual loading of a unit horizontal

load applied at C and resisted at D,

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

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 there-

fore are usually neglected in deflection calcu-

lations 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 equa-

tion 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

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 deflec- tion 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

wL®

aE, mm = 2a (F)", using Bg = 4E, mà _

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$

Thus the shearing deflection in percent of the bending deflection equals 4.1% for a $ ratio L

of Ẻ for I-beam sections and 1.4 percent for rectangular sections

ANALYSIS AND DESIGN Example Problem 19

due to shear deformation for beam of Fig A7.22

assuming shearing stress uniform over cross-

section, and AES constant

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

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

Member AB meil.x =x, (for x = 0 to 3)

to Member BC m=3 sin 20° +1 x (for x = 0 to 36)

t = 3 cos 20° constant between B and C

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,

ploded view of the beam showing the internal

real loads carried as determined by statics?

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 +

Virtual loading, Fig A7.27

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

equation used was

simplification JJ = ay day

Gt where S is the panel area

The calculation was completed in Table A7.6

sence!

Tap her | hime] bao col naw) bia) aie a vos eesti eat mee ma?

Meare ea nN 2 5 Cir Ther areal cette nF ms

LG TA HE HE ĐÔ Tan ae AE wor has te : 3

ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES

= (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 tem-

perature 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

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

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 ob- tained in the flanges were:

Fig A7.30a shows a uniform circular ring whose inside surface is neatgd to a temperature

T above the outside surface The temperature is constant around the circumference and ts assumed

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

An element of the beam of length Rdd 1s

shown in Fig A7.30c Due to the linear tem-

perature 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

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 opera- tion 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 ap- proach 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) +

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

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 com-

putation 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 numeri-

cal subscripts), 18 used to describe the inter-

nal 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 ob-

tained 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 ex-

ternal 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

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

It is next desired to write the strain energy as a function of the q’s Continuing the illustrative example, write

of the associated generalized force (exponent on variable), Introducing the notation

(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£)"

ing three terms, representing the energy stored

in the inner beam segment by qa and da, are like- wise readily written, with proper account taken for the cross influence of one force upon an~

other (the "das da dq” term)