Bruhn - Analysis And Design Of Flight Vehicles Structures Episode 3 Part 2 pdf

Although in practi- cal structures uniform stress distribution is not common, still sufficient accuracy for design practice is provided by using the stress relationships based on unifo

* Properties for sheet and platy are taken parallel {o the direction of rolling ‘Tranaverse propertles * Mochanical properties are based upon the guerantecd tensile properties from sepurately-cast

aro equal to or greater than the longitudluel properties

* Reference should be made ta the specific requirements of the procuring or certificating agency with regard (o the use of the above values in the design of castings

test bara, The mechanical properties of bara cut trom castings may be as low as 75 percent of the

‘Strangin af temoercture Exposure up 1o 000 nr

MECHANICAL AND PHYSICAL PROPERTIES OF METALLIC MATERIALS FOR FLIGHT VEHICLE STRUCTURES

» PLATZ & SAND CASTINGS) (Cont.)

Strength at tenperctune Expeaue up 10 1000 Ww

Fig B2.94 Effect of temperature on the ultimate tensile strength (Fry) of HK31A-T6 magnesium alloy (sand casting)

ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES B2 31

AZ61A, AZGZA, AZ8OA MAGNESIUM (EXTRUSIONS, FORGINGS, CASTINGS)

Table B2.22 Design Mechanical and Physical Properties of AZ61A7 Magnesium Alloy

(Extrusions and Forgings)

4 Properties or extruded bars, rods, shapes, tubes, and forgings are

AZ61A, AZ6ZA, AZSOA MAGNESIUM (EXTRUSIONS, FORGINGS, CASTINGS) (Cont.}

Table 52.23 Design Mechanical and Physical Properties of AZ80A2 Magnesium Alloy

(Extrusions and Forgings)

AZG1A, AZ6ZA, AZSOA MAGNESIUM (EXTRUSIONS, FORGINGS, CASTINGS) (Cont.)

Table B2.24 Material Specifications for AZ63A

Sand castings Permanent-mold castings

TABLE 4.2.3.0(b) Design Mechanical and Physical

Properties of AZ63A Magnesiums Alloy (Castings)

Sand and permanent-

° * a e a @

toán, QOOI infin

Tangent Modulus, 0" pa Fig B2.98 Typical stress-strain and tangent~modulus curves for AZ63A-T4 magnesium alloy (sand casting)

at room temperature

ArTansile stress-strain QsCompressive stress-strain

Ce Tensile tangent modulus OsComprensive tangent modulus

Swein, GOOI 1n Tangent Modulus, 10° pa

Fig B2.97 Typical stress-strain and tangent- modulus curves for AZ63A-F magnesium alloy (sand casting) at room temperature

Table B2.25 Design Mechanical and Physical Properties

of 8Mn Titanium Alloy

00-800

Tamperaiure, F Fig B2,98 Effect of temperature on the ulttmate tensile strength (Fty) of BMn annealed titanium

yield strength (Fty) of 8 Mn annealed titaniuin

alloy

Strength ot lemperctive Exposure up to 1000 he

‘Strength af temperature Exposure up 10 1000 he

6A1-4V TITANIUM ALLOY (BAR & SHEET)

Table BZ.26 Design Mechanical and Physical Properties

of 6Al-4V Titanium Alloy

Longuwvanat Strength at temperature

and bar}

Fig B2,112 Effect of temperature on

the tensile modulus (E) of Inconel X

Fig, B2.111 Effect of temperature on

the ultimate shear strength (Fgy) of

precipitation heat treated Inconel X nickel alloy,

Fig, B2.114 Effect of temperature on the bearing yield strength (Fpry) of precipitation heat treated Inconel X nickel alloy,

PART C

PRACTICAL STRENGTH ANALYSIS &

DESIGN OF STRUCTURAL COMPONENTS

CHAPTER C1 COMBINED STRESSES THEORY OF YIELD AND ULTIMATE FAILURE

1,1 Uniform Stress Condition

Aircraft structures are subjected to many

cause axial, bending and shearing stresses

designed satisfactorily, combined stress re-

lationships must be known Although in practi-

cal structures uniform stress distribution is

not common, still sufficient accuracy for

design practice is provided by using the stress

relationships based on uniform stress assump-

the Greek letter signa (a) will represent a

stress intensity normal to the surface and thus

a tensile or compressive stress and the Greek

letter tau (t) will represent a stress intensity

parallel to the surface and thus 4 shearing

Fig Cl.1 shows a circular solid shaft

subjected to a torsional moment The portion

{A) of the shaft exerts a shearing stress Ty 0n

section (1-1) and portion (B) exerts a resist-

ing shearing stress tz on section (2-2) Fig

Cl.2 illustrates a differential cube cut fram

shaft between sections (1-1) and (2-2) For

equilibrium a resisting couple must exist on

about lower left edge of cube:

Ty dxdy (dz) - Tz dzdy (dx) = 0

Thus if a shearing unit stress occurs on

one plane at a point in 4 bedy, a shearing unit

stress cf same intensity exists on planes at

right angles to the first plane

Cl.3 Simple Shear Produces Tensile and Compressive

Stresses,

Fig Cl.3 shows an elementary dlocx of

hence, ơ = 8 (+ 1 cos 4s ) cos 45° _ ~~~ (8)

Therefore when a point in a body is sub- jected to pure shear stresses of intensity 1+;

normal stresses of the same intensity as the

shear stresses are produced on a plane at 45 with the shearing planes

C1.4 Principal Stresses For a body subjected to any combination of stresses 3 mutually perpendicular planes can be

found on which the shear stresses are zero The

normal stresses on these planes of zero shear stress are referred to as principal stresses

C1.5 Shearing Stresses Resulting From Principal Stresses

In Fig C1.5 the differential block 1s subjected to tensile principal stresses Gy and

cut along a diagonal section giving the free

section have been resolved into stress compon- ents parallel and normal to the section as shown

For equilibrium the summation of the stresses along the axes (1-1) and (2-2) must equal zero

Gy dudy - oy dzdy cos © - og dydx sin @ = 0,

whence oy = 2X TT 9, % đực sin 8

hence, T= gg sin 6 cos @ - oy cos 9 sin @

(1/2) (og - oy) sin 2 0, where og is maximum principal stress and oy is minimum

principal stress

Since sin 2 9 1s maximum when @ = 45°,

Stated in words, the maximum value of the shear-

ing unit stress at a point in a stressed body

ig one-half the algebraic differences of the

maximum and minimum principal unit stresses

COMBINED STRESSES, THEORY OF YIELD AND ULTIMATE FAILURE

C1.6 Combined Stress Equations Fig Cl.7 shows a differential block sub-

jected to normal stresses on two planes at right

angles to each other and with shearing forces on

ing unit stresses will be determined

Pig Cl1.8 shows a free body dlagram of a portion cut by a diagonal plane at angle 6 as

(Gn<Ơx) cos @ + (tT - ty) sin @ =O - = + ~(6)

ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES (a + oz) sin 9 - (T- Tyg) cos 9=O - - -(7)

The maximum normal stress gy will be maximum

when @ equals such angle @' as to make t = zero

Thus 1f T= 0 amd @ = 9’ tn equations (6) and

(7), we obtain,

" oO

(Gn - Gx) cos @' - Tyz Sin ©@'

" a (on - Gz) sin @' - ty, cos @!

In equations (8) and (9) oy represents he

another to eliminate 9',

On* ~ (Ox - 0g) 1 + OZ BZ %_*, oF

Ox + Oy tA (E22) at 2) + tu = - (10)

In equation (10), tensile normal stress is

plus sign before radical and minus sign for

minimum oy

Ơn =

To find the plane of the principal stress-

es, the value of 6' may be solved for from

equations (8) and (9), which gives:

9' is measured from the plane of the

largest normal stress oy or gz The direction

of rotation of 9' from this plane is best de-

termined by inspection Thus if only the

shearing stresses +xz were acting, the maximum

principal stress would be one of the 45° planes,

the particular 45° plane being easily deter~

mined by inspection of the sense of the shear

stresses Furthermore if only the largest

normal stress were acting it would be the maxi~

mum principal stress and 9' would equal zero

Thus if both o and tT act, the plane of the

principal stress will be between the plane on

which o acts and the 45° plane As stated

before o refers to either oy or o, whichever

is the largest

The maximum value of t from equation (3)

equals,

“max Ð (Ơn(max,) ~ nian ))/?

Substituting the maximum and minimum values of

Gy from (10) tn (12), we obtain maximum shear-

ing stress as follows:

It 1s sometimes convenient to solve

graphically for the principal stresses and the maximum shear stress Mohr's circle furnishes

a graphical solution (Fig Cl.9a}) In the

Mohr method, two rectangular axes x and 2 are chosen to represent the normal and shearing

origin lay off to scale the normal stresses oy

gion, they are laid off to right of point O and

stress Tyg 1s laid off parallel to Og and with

the sense of the Shear stress on the face DC of

E the midpoint of AB as the center and with radius EC describe a circle cutting OB at F and

shear on face AB of Fib b It can be proven that OF and 0G are the principal stresses Cmax, and Ggin respectively and EC 1s the maximum

shear stress Tmgy, The principal stresses occur

on planes that are parallel to CF and ca (See

on two sections parallel to CH and CI where HEI

then O would coincide with a

to find the stress components on other planes

x and 2 represent the normal and shear stresses

off to scale on ox giving points D and § respec~

C1.4

normal stress on the plane defc of Fig Cl.10,

and CB represents the shear stress + on this

The maximum normal and shear stresses will be

determined for the block loaded as shown in

Fig Cl.12

The graphical solution making use of Mohr's circle is shown in Fig C1.13 From reference

axes x and 2 thru point 0, the given normal

stresses ¢x = 10000 is laid off to scale on ox

shear stress Tyz = 5000 1s laid off parallel to

OB as the center of the circle and with radius

EC a circle is drawn which cuts the Ox axis at

stresses are then equal to oF and oG which

Mum shear stress equals EC or 7070

Tmax ~Onmin,)) (Ref -Ba.22)

Example Problem 2

The maximum normal and shear stresses will

be determined for the block loaded as shown in Fig C1.14

ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES Flg Cl.15 shows the graphical solution

and dz = -20000 are laid off equal to OB and OA

respectively." Tyg equal to 12000 is laid off

midpoint of AB as the center of a circle of

radius EC a circle 1s drawn which cuts the ox

stresses are indicated on the figure

C1.10 Triaxial or Three Dimensionai Stresses

For bodies which are stressed in three

directions, the state of stress can be defined

completely by the six stress components as

procedure as was carried out for a two-dimen-

sional stress system, it can be shown that there

are three principal stresses o,, o, and os,

whose values are the three roots of o in the

y Fig C1 16

Fig C1.17 shows the principal stress system which replaces the system of Fig Cl.16

It can be shown that the maximum shear stress

max, 18 one of the following values

etl ymax = 75 {o1 ~ Øa)

+ or’ max ~

largest of the shear stresses in equations (15) depends on the magnitude and signs of the principal stresses, remembering that tension

is plus and compression 1S minus when making the substitution in equations (15)

The strains under combined stresses are usually expressed as strains in the direction

simple tension as illustrated in Fig Cl.19, The stress o, causes a lengthening unit strain é€ in the direction of the stress o,, anda

Shortening unit strain e«’ in a direction at

right angles to the StTreSS Ơi,

The ratio of e’ toe is called Poisson's

ratio and is usually given the symbol » Thus,

h=e'/e

295 eo xó?

it ! terms of o from equations (17) into equation

efi thet 1

og, = 0 and equation (20) becomes e'2pa/E - (18)

1 a = ẹ

C1.20 subjected to the three principal stresses

unit strain e, in the direction of stress o,

stretch the element in the direction of o,

whereas stresses go, and o, tend to shorten the

element in the direction of a,, hence,

For a two-dimensional stress system, that

is, stresses acting in one plane, Øạ = 0 and

the principal strains become,

es =H (% ~ po)

ts = Elo, +g)

Equations 17 and 16 give the strains when

For compressive principal stresses use a minus

sign when substituting the principal stresses

in the equations

Ci, 12 Elastic Strain Energy

The strain energy in the elastic range for the unit cube in Fig Cl.20 when subjected ta

combined stresses 1s equal to the work done by

the three gradually applied principal stresses

equal to e,, 6, and e, and thus the work done

per unit volume equais the strain energy Thus

if U equals the strain energy, we obtain,

U «Sm + Saxe + Sả» wwe eee (19)

commercial airliner is to carry passengers and

cargo from place to place at the lowest cost

To carry out this job a certain amount of flight and ground maneuvering is required and

the loads due to these maneuvers must be

carried safely and efficiently by the structure

A military fighter airplane must be maneuvered

in flight far more severely to accomplish its

desired Job as compared to the commercial air-

liner, thus the flight acceleration factors for the military fighter airplane will be considerably higher than that of the airliner

In other words, every type of flight vehicle will undergo a different loan environment,

which may be repeated frequently or infre-

quently during the life of the vehicle The

load environment may involve many factors such

as flight maneuvering loads, air gust loads, take off and landing loads, repeated loads, high and low temperature conditions, etc

maximum loads which may be subjected to the flight vehicle in carrying out the job it 1s designed to accomplish during its life time

of use The term limit was no doubt chosen

because every flight vehicle is limited relative to the extent of its operations A flight vehicle could easily be designed for loads greater than the limit loads, but such extra strength which is not necessary for safety would only increase the weight of the structure and decrease the commercial or military payload or in general be detrimental

to the design

Factor of safety can be

defined as the ratio considered in structural

design of the strength of the structure to the maximum calculated operational loads, that is, the limit loads,