Bruhn - Analysis And Design Of Flight Vehicles Structures Episode 1 Part 9 potx

CHAPTER A9 BENDING MOMENTS IN FRAMES AND RINGS BY ELASTIC CENTER METHOD A9,1 Introduction In observing the inside of an airplane fuselage or seaplane null one sees 4 large num- ber of

ANALYSIS AND DESIGN OF F influence coefficients

(12) Re-solve the doubly redundant beam

of Example Problem B, page A8.3 by matrix

methods The redundant reactions should be

given "q" symbols (See Example Problem 12a;

page A8.20)

(13) Re-solve Example Problem 5, page

AS.12 by matrix methods For simplicity, make

your choice of generalized forces ineluding

those designated as X and Y in the example so

that Figs A8.21 and A8.22 can be used to give

the Zir loadings

(14) By matrix methods re-solve Example

Problem 4, p A8.12 using 3 equal bay divisions

along the panel (3 times redundant) Use the

same structural dimensions as in Example Prob-

lem 20, p A8.36 Compare the results with

those obtained from the formulas developed in

Example Problem 4

(15) Show that the matrix equation

eq (21) is modified to cover the initial stress

problems of Art A8.8 by wr1ting

(rd (x = Ln (4 - [el 4}

where A; is the initial imperfection associated with force qị Refer to the argument leading

to eq (11) of Art as.8

(16) Using the equation of problem (15), above, re-solve Example Problem 6, p Á8.14

(17) Using the equation of problem (15)

above, re-solve Example Problem 7, p A8.14

(18) Using the matrix methods of Art

A8.13, re~solve Example Problem 9, p A8.15

(19) For the doubly symmetric four flange box beam of Example Problem 15, p A8.24, de-

termine the redundant stresses da, q7 and qia lf

one flange is heated to a temperature T, uniform spanwise, above the remainder of the structure

See references at the end of Chapter A-7

Douglas DC-8 airplane Photograph showing simulated aerodynamic load being applied to

main entrance door of fuselage test section

DOUGLAS DC-8 AIRPLANE, An outboard engine pylon mounted on a section of wing for static and flutter

tests The steel box represents the weight and moment of inertia of the engine

CHAPTER A9 BENDING MOMENTS IN FRAMES AND RINGS

BY ELASTIC CENTER METHOD

A9,1 Introduction

In observing the inside of an airplane

fuselage or seaplane null one sees 4 large num-

ber of structural rings or closed frames Some

appear quite light and are essentially used to

maintain the shape of the body metal shell and

to provide stabilizing supports for the longi-

tudinal shell stringers At points where large

load concentrations are transferred between body

and tail, wing power plant, landing gear, etc.,

relatively heavy frames will be observed In

null construction, the bottom structural fram-

ing transfers the water pressure in landing to

the bottom portion of the null frames woich in

turn transfers the load to tne hull shell

In general the frames are of such shape

ani the load distribution of such character

that these frames or rings undergo bending

forces in transferring the applied loads to the

other

=

a iB

- b

resisting portions of the airplane body

e cending forces produce frame stresses in

which are of major importance in the

sth proportioning of the frame, and thus

a reasonable close approximation of such benc-

ing forces is necessary

Such frames are statically indeterminate

relative to internal resisting stress and thus

consideration must be given to section and

physical properties to obtain a solution of the

distribution of the internal resisting forces

General Methods of Analysis:

There are many methods of applying the

principles of continuity to obtain the solution

gor the redundant forces in closed rings or

frames and bents The author prefers the one

is generally erred to as the "alastic

usea it for many years {nl

nm The method was orizi- The main difference

to most other methods

2 yuller-Breslau, H., Die Neveren Hethden der

Festigkeitslehre und der Statik der Baukon-

In computing distortions plane sections are assumed to remain plane after bending This is not strictly true because the curvature of tne frame chantes this linear distribution of 5end~

ing stresses on 2 frame cross~section (Correc- tions for curvature influence are given in

Chapter AlS

Furthermore it is assumed that stress is proportional to strain Since the airplane stress analyst must calculate the ultimate strength of a frame, this assumption obviously does not hold with heavy frames where the rup- turing stresses for the frame are above the pro~

portional limit of the frame material

This chapter will deal only with the the~

oretical analysis for bending moments in frames and rings by the elastic center methed Prac- tical questions of body frame design are covered

in a later chapter

The following photographs of a vcortion of the structural framing of the hull of a sea- plane illustrate both light and heavy frames

A9.2 BENDING MOMENTS

A9.2 Derivation of Equations, Unsymmetrical Frame

Fig AQ9.1 shows an unsymmetrical curved beam fixed at ends (4) and (B) and carrying

some external loading Pi, Pa, etc This

structure is statically indeterminate to the

third degree because the reactions at (A) and

(B) have three unknown elements, namely, magni-

tude, direction and line of action, making a

total of six unknowns with only three equations

of static equilibrium available

Fig AQ 2

In Pig A9.2 the reaction at (A) has been replaced by its 3 components, namely, the forces

Ky and Ys and the moment M, and the structure

is mow treated as a cantilever beam fixed at

end (B) and carrying the redundant loads at (A)

and the known external loading P,, Pz, etc

Because joint (A) is actually fixed it does not

suffer translation or rotation when structure

ts loaded, thus the movement of end (A) under

the loading system of Fig A9.2 must be zero

Therefore, three equations of fact can be

written stating that the horizontal, vertical

and angular deflection of soint (A) must equal

IN FRAMES AND _ RINGS

Consider a small element ds of the curved beam as shown in Fig A9.2 Let M, equal the bending moment on this small element due to the given external load system The total vending moment on the element ds thus equals,

M=Msg + My - Xay + Ygx

(Moments which cause tension on the inside fibers of the frame are regarded as posi- tive moments )

The following deflection equations for point (A) must equal zero:-

@ +0, (angular rotation of (A) = zero}

dy = 0, (movement of (A) in x direction = 0)

dy 20 {movement of (A) in y direction = 0} from Chapter À7, which dealt with deflection theory, we have the following equations for the movement of point (A):-

In equation (2) the term m is the Dending moment on a element ds due to a unit moment applied at point (A) (See Fig A9.3) The bend- ing moment is thus

equal one or unity A

of frame thit moment

Fig A9.3

Then substi- tuting in equation (2) and using value of M from

a = PSE + mynd NUIẾT + YUẾT =o (5)

In equation (3) the term m represents the bending moment on a element ds due to a unit load applied at point (A) and acting in the x direction, as illustrated in Fig Ag.4

The applied unit y

load has a positive 1 ds sign as it has been ‘A +

assumed acting toward wien

the right The Ị B

distance y to the ds ỹ element is a plus Fig A9.4

distance as it is

ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES measured upward from axis x-x through (A)

However the bending moment on the ds element

shown is negative (tension in top fibers), thus

the value of m= - (1) y = -y The minus sign

is necessary to give the correct bending moment

sign

Substituting in Equation (3) and using M

from Equation (1):-

2

Ác = 815 - tuyỆt ‹ xyr OS _ va35 5» os (6)

In equation (4) the term m represents the

bending moment on a element ds due to a unit

load at point (A) acting in Y direction as

illustrated in Fig A9.5 Hence, m = l(x) =x

Equations 5, 6, 7 can now be used to solve

for the redundant forces My, X4 and Yy With

these values known the true bending moment at

any point on structure follows from equation

(1)

REFERRING REDUNDANTS TO ELASTIC CENTER

For the purpose of simplifying equations

Š, 6, 7, let it be assumed that end A is

attached to a inelastic arm terminating at a

paint (0) as tllustrated in Fig A9.6 The

point (0) coincides with the centroid of the

ds/EI values for the structure Reference

axes x and y will now be taken with point (0)

as the origin The redundant reactions will

now be placed at point (o) the end of the

elastic Fig A9.6

inelastic bracket, as shown in Fig A9.6

Since point A suffers no movement in the actual

structure, then we can say that point (o) must

undergo no movement since (0) is connected to

point (A) by a rigid am

Thus equations 5, 6, and 7 can be re-

written using the redundants X,, Yj, and M, in

Ag, 3

place of Xa, Yy and Mg respectively

The axes x and y through the point (o) are centroidal axes for the values ds/EI of the structure This fact means that the summations-~

5 ar 7° and 5 Er 7°

The expressions 2 x"ds/EI, & y*ds/EI and

2 xyds/El also appear in equations 6 and 7

These terms will be referred to as elastic

moments of inertia and product of inertia of the frame about y and x axes through the elastic

center of the frame, and for simplicity will be

given the following symbols

xfds _ y 24s „ xyds _

Bar “ly: 3 gr Flee BAGS hy

Equations 5, 6 and 7 will now be rewritten using the redundant forces at point (0)

The term MgZ yds/EI is zero since % yds/EI

is zero, thus My drops out when substituting in Equation (6)

This angle change which actually is equal in value to the area of the M,/BI diagram on the element ds will be given the symbol @,, that is,

as = Msds/BI with this symbol substitution, equations 8, 9, 10 can now be rewritten as follows: -

Xốxy ~ 3Øsx (4) Table A9.1

Xo 2 NET Trẻ am (4) 7 i iy

Tx (a xX L) tien | XÊT from from we) wy Fwy trig twx2viy

A9.3 Equations for Structure with Symmetry About One Axis 1x? 79533" at:

through Elastic Center

30 wy222250 |wx2=1440

I7 the strueture 1s such that either the x | | CD |fÿ71012 j 12| 120| 1850| „ rap tự rÊn a0

or y axis through the elastic center is a axis

of symmetry than the product of inertia sum! 32 I 0 | 660 16800 3456

ixyds/EI = lyy = zero Thus making the term

lxy = 0 in equations ll, 14 and 15 we obtain,

S8 5,

Họ “sgặiy TT TT TT Tan (16)

Yo 228s 2 y TT ST (18) A3.4 Example Problem Solutions

One Axis of Symmetry Structures with at least Example Problem 1

termine the bend-

ing moment dia~ |

gram under this !

find the location of the elastic center of the

frame and the elastic moments of inertia 1x

and ly

Due to symmetry of the structure about the

Y axis the centroidal ¥ axis 1s located midway

between the sides of the frame, and thus the

elastic center (0) lies on this axis

Table A9.1 shows some of the necessary calculations to determine the location of the

elastic center and the elastic moments of

inertia The reference axes used are x*=x?

are the elastic moment

of inertia of each sort {on of the frame about its centroidal x and y exes Since I is con- Stant over each pertion the centroidal moment

of inertia of each portion is identical to that

of a rectangle about its centroidal axis

The terms 1, and

To explain for member AB:- Referr.ng to Fig a,

t Ÿ = one =i x 20x iis 09 (negligibie) “12 12 a:

The distance from the a two reference axes to the

Ix “ ly - 3⁄4(ÿ*) = 16800 ~ 32 x 20.6262= 2188

ly = Tự ~ Ä3w(X?®) = 3456 ¬ 32(o) = 3456 The problem now consists in solving equa- tions (16), (17) and (18) for the redundants at the elastic center, namely

uM #2 25 _ Area of static M/I diagram

0 548/1 Total elastic weight of structure

Moment of static M/I diagram about

Thus to solve these three equations we

must assume a static frame condition consistent

with the given frame and loading In general

there are a number of static conditions that can

be chosen For example in this problem we

might select one of the statically determinate

conditions illustrated in Fig A9.8 cases 1 to

To illustrate the use of different static

conditions, three solutions will be presented

with each using a different static condition

Solution No 1

In this solution we will use Case 3 as the

static frame concition The bending moment on

the frame for this static frame condition is

given in Fig A9.9 The equations

45% ig diagram 32.5 Ms/I diagram

for the redundants require J, the area of the

Ms/I diagrem, Fig 49.10 shows the Mj/I curve

xàxch is obtained by dividing the values in

Fig, A9.9 by the term 2 which is the moment of

inertia of member BC as given in the problem

Since the equations for Xg and Y, require the

moment of the Ms/I diagram as a load about

axes through the elastic center of frame, the

area of the M,/I diacram will be concentrated

at the centroid of the diagram and along the

centerline of the frame, or more accurately

Ag.5 along the neutral axis of the frame members

In Fig 49,10 the area of the M,/I diagram equals @g = 22.5 x 24/2 = 270 The centroid by simple calculations of this triangle would fall

10 inches from B Fig A9.11 now shows the frame with its Mg/I or its @; load @, is Øs= 270 Z“H Lan positive since Mg 1s

step is to solve the equations for the

x -¬¬gE—~-+ x redundant at the

20,625" elastic center The

Signs of the distances

A ' D + x and y from the axes

Fig AQ 11 tonal

These resulting values are plotted on Fig

AQ.12 to give the bending moment diagram due to che redundant forces at the elastic center

AS 6 BENDING MOMENTS

Adding this bending moment diagram to the static

bending diagram of Pig A9.9 we obtain the final

bending moment diagram of Fig A9.13

The final bending moments could also be obtained by substituting directly in equation

(1) using subscript (o) instead of (A) Thus,

MaMg + My = Xo¥ + ¥ox

For example, determine bending moment at point 8

For point B, x substituting in (19) = -12 and y = 9.375, Mg 20

=O + (-8.437) = 7939 x 9.375 + 1562 (-12) = -17,75 as previously

found

Mp

AT POINT D x = le, y = 20.625, Mg = 0

Mp = 0 + (-8.437) - 7939 (-20.625) + +1562 x 12 = 9.81 in.ib

Solution No 2

In this solution we will use Case 4 (See

Fig A9.8) as the assumed static condition,

that is two cantilever beams with half the

external load or 5 lb acting om each canti-

lever Fig A9.14 shows the static bending

moment diagram and Pig .A9.15 the M,/I diagram

the Mg/I diagram and its centroid location

Substituting in the equations for the re~

In this solution we will use Case 5 (Fig A9.8) as the assumed static condition, namely a frame with 3 hinges at points A, D and

5, as tllustrated in Fig A9.16

Before the bending moment dia- gram can be caleu~ lated the reacticns

at A and D are necessary

Then using OFy =o for entire frame, we obtain

Fy = -1 + Hg = 0, hence Hg = 1,

ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES The frame static bending moment ciazram

The moment diagram is labeled in 6 parts

1 to 6 as indicated by the values in the small

circles on each portion Most of the calcu-

lations from this point onward can be done

conveniently in table form as illustrated in

Mom z 1 „|x dist.| y = dist

Dia- | Area] ro ®s “| trom Y | from X

gram| of Beam a | axis to | axis to Osx | Oey

Por- | Mom Section} 1 C.g of | C.g of

In order to take moments of the Js values

in column (4) of the table, the centrotd of

each portion of the diagram must be determined

For example, the centroid of the two triangular

bending moment portions marked 1 and 6 is

«867 x 30 from the lower end or 20 inches as

shown in Fig A9.17 Thus the distances x

na y from this Zs location to the y and x

Consider point B:-

x =-12, y= 9.375, Ms = -30 Substituting -

A is continuous through the Joint but the reaction is applied through a sin at the center

of the joint The vroblem is to daternine the bending moment di

4

Ap 152 Fig A9 18

Msg diagram for load Py;

Spf 240

240P

Fig AQ 18

Fig A9.19 Fig Ag 20

Mg diagram for w icading Mg diagram for Pa load

Tne Tirst step is to find the location of the slastic center of the frame Due to

syunet TY of ?rame about the y axis, the elastic

enter will be on a y axis through the middle

or the frame The vertical distance ¥

measured from 2 axis through AD equais,

axes through the elastic center of the frame

Moment of inertia about x axis = Iy t- Members AB and DC,

1 =($ xậ x 14.88) ø = 653.9 x\5*5 - ~ ` ( xặx 9.879] 2 = 200.9

Member BC

= (30/4) (14,337) = 1540.0 (30/2) (3.67) = 1402.7

Ty = 3797.5 Moment of inertia about y axis = ly is Members AD and 3c,

Table A9.3

Portion | Area Og > = dist.ly = dist.|

Dlagremi Boruen Ha Axis Ans | 2s% | Gạy

7 ~108000/3/-36000 15 2.33 |~5400001- 83880

8 ~ 67500|21-33750) 5 ~ 9.67 |-188750| 326363

9 3240/3} 1080 15 - 3.67 16200|- 3964

10 5400/2) 2700 5 - 9.67 13500l~ 26109 sum |- 77700} ~T47225|- 4845

a0 ®

snce to P12, A9,21 are - 3

ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES

-84 4153 —15—

saat 14.3 I

Combining the bending moment diagrams of

Flgs A9, 18, 19, 20 with Fig A9.22 would

give the true or final oending moment diagram

Example Problem 3 Circular Ring

Fig A9.23 shows a circular ring of

constant cross-section subjected to a sym-

metrical loading as shown The problem is to

determine the bending moment diagram

+8

a Center

Solution Due to symmetry of the ring

structure the elastic center falls at the

center of the ring Since the ring has been

assumed with constant cross-section, a relative

value of one will be used for 1

Ix = ly = or® = 1 x 1e* = 18300

The next step in the solution is to assume

a static ring condition and determine the static (Mg) diagram In general it is good practice

to try and assume a static condition such that the M, diagram is symmetrical about one or if possible about both x and y axes through the elastic center, thus making one or both of the redundants Xo and Yo zero and thus reducing considerably the amount of numerical calcula- tion for the solution of the problem

In order to obtain symmetry of the Mg diagram dnd also the Mg/I diagram since I is constant, the static condition as shown in Fig A9.24 is assumed, namely, a pin at (e) and rollers at (f) The static bending moment at points (a), (b), (c) and (d) are the same

magnitude and equal,

is divided into similar portions labeled (1) and (2) Hence

zi.) = area of portion (1) = Pr’(q -sin a), where P = 50 lb and a = 45°

Substituting and multiplying by 4 since there are four portions labeled (1).,

The area of portion labeled (2) equals,

Pr*e(1 - cos a)

Since there are two areas (2) we obtain,

Bs (a)=2[50 x 18* x2 (1~0.707) ] = 16000

Hence, total Bs = 15000 + 5052 = 20052 Since the centroid of the Mg diagram due

to symmetry about both x and y axes coincides

Ag 10 BENDING MOMENTS

with the center or elastic center of the frame,

the terms 29,x and Z%sy will be zero

Substituting ta determine the value of tne redundants at the elastic center we obtain,

produced by these forces Adding the bending

moment diagram of Fig A$.25 which is a constant

value over entire frame of -177 to the static

moment diagram of Pig A9.24 gives the final

bending moment diagram as shown in Fig A9.26,

Example Problem 4 Huil Frame

Fig A9.27 shows a closed frame subjected

to the loads as shown The problem is to

determine the bending moment diagram

Yq Will De zero due to this symmet

In the last column of Table 49.4 the tern

iy 1s the moment of inertia of a particular member about its own centroidal x axis, Thus for member BDB;

= Aedes axeae=e22 i2*2

Let ¥ = distance from X'X' ref axis to

centroidal elastic axis X-x H3

Swy 21468 3w ` Z21.2 2 6.54 in

By parallel axis theorem,

Ix = Ig = 6.847 (Ew)

= 262140 - 6.84" (222.2) = 252400

The next step in the solution is to com- pute the static moment elastic weights 3 and their centroid locations In Fig A9.28, the static frame condition assumed is a pin at point

A and rollers at point A', which gives the