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

C2.18 shows a pin ended column in 4 deflected neutral equilibrium position when carrying ths ultimate or critical load P, Assume that the shape of the deflected column follows a sine cu

a < ~ wt [>

C2.8

particular properties of the matertal Bo.7/F

Inserting values of F equation (3),

Gee) z )

= n7Et/(L'/p}* in

The use of tne curves in Fig C2.17 will

be illustrated later tn the example problem

ma4e with a non-uniform cross-section To

find the ultimate strength of such columns,

it is usually necessary to use a trial and

error method The general method of solution

involving a consideration of column deflection

will be illustrated for a case of a long

column with uniform cross-section

Fig C2.18 shows a pin ended column in 4 deflected neutral equilibrium position when

carrying ths ultimate or critical load P,

Assume that the shape of the deflected column

follows a sine curve relationship with the

deflection at midpoint equal to unity (see

mx y,sin “r”

By the well known "moment area” principle (see Chapter A7; Art A7.14), the deflection

of a point (A) on the elastic curve away from

a tangent to elastic curve (B) equals the first

moment of the M/EI diagram between (A) and (B)

about (A)

STRENGTH OF COLUMNS WITH STABLE CROSS-~SECTIONS

Thus in Fig C2.18, the deflection of

point (0) away from tangent at midpoint e

equals unity in our assumed conditions and it also equals the first moment of the area of

the M/EI diagram between (0) and (C) about (0)

(Fig C2.19)

The value of the ordinate for M/EI diagram

at any point x from O is si ¬° +

hence area = a and hal? arsa = =

The center of gravity of the nalf area 1S

which is the Ruler equation, and thus the

assumed sine curve was ths proper one for the

deflected elastic curve of the column

Suppose that the elastic curve of de- flected column had been assumed as a parabola with unit deflection at midpoint Fig c2.20 shows the M/SI diagram

The area of one-half the diagram =

which compares with P =

9.9E1/L® of the Euler equation or an error of

We can now apply the same procedure to

a column with non-uniform cross-section The

steps in this procedure for 4 column

symmetrical about the center point are as

follows:~

(1) Assume a sine curve for the deflected

column with unit deflection a center

point

(2) Plot a moment of inertia (I) curve for

column cross-section

(3) Find the bending moment curve due to end

load P times the lateral deflection

(4) Divide these moment values by the EI

values to obtain M/EI curve The modulus

of elasticity E ts considered constant

(5) Find the deflected column curve due to

this M/EI loading

(6) Compare the shape of the derived column

deflection curve with that originally

assumed 4S 4 sine curve This can be

done by multiplying the computed de-

flections by a factor that makes the

center deflection equal to unity Since

the assumed sine curve is not the true

column deflection curve, the computed

deflection will differ somewhat from the

sine curve

(7) With the computed deflection curve,

modified to give unity at center point,

repeat steos 3, 4, 5 and 6 The results

this time will show derived deflection

curve still closer to the assumed

deflection curve

(8) To obtain the desired accuracy, the pro-

cedure in step (7) will usually have to

(9) Salve for load P by writing and expression for the deflection at the center point which equals unity This is done by using the moment area principle as was done in the previous example problem in- volving a column with uniform section

In the above outlined procedure, E has been assumed constant or, in other words, the column failure is elastic or fatling stresses are below the proportional limit stress of the Material The practical problem usually involves a slenderness ratio where failure is due to inelastic bending and thus § 1s not constant For this case, a trial and error method of solution in necessary using the tangent modulus of elasticity which varies with stress in the inelastic stress range

C2.8 Design Column Curves for Columns with Non- Uniform Cross-Section

Figs C2.21 and C2.22 give curves for rapid solution of two types of stepped colums

Figs C2.23 and C2.24 gives curves for the rapid solution of two forms of tapered columns

Use of these curves will be illustrated later

in this chapter

C2.T Column Fixity Coefficients ¢ for Use with Columns with Elastic Side Restraints and Known End Bending Restraint

Figs C2.25 and C2.26 give curves for finding fixity coefficient ¢ for columns with one and two elastic lateral restraints and Fig C2.27 gives curves for finding ¢ when restraining moments at column ends are known

Use of these various curves will be illustrated later

C2.8 Selection of Materials for Elevated Temperature Conditions

Light weight is an important requirement

in aerospace structural design Por columns that fail in the inelastic range of stresses,

a comparison of the Fy,/w ratio of materials gives a fairly good picture of the effictency

of compression members when subjected ta elevated temperature conditions In this ratio Foy is the yield stress at the particular temperature and w is the weight per cu inch

of the material Fig C2.28 shows a plot of Fey/w for temperature ranges up to 600° F

with 1/2 nour time exposure for several im- portant aerospace materials

C2.9 Example Problems

Fie C2.29 shows a forged (I} section member 30 inches long, which is to be used as

CRITICAL LOADY-NON-UNIFORM COLUMNS Single Stepped - Pin Ended

CRITICAL LOADS -NON-UNIFORM COLUMNS

Duuble Stepped - Pio Ended

2.12 STRENGTH OF COLUMNS WITH

Fig C2.28

AISI Steel, Fry = 180,000

17-7 PH Stainless Steel, Fy,.= 210,000 7075-76 Alum Alloy

AZ31B Magnesium Alloy 6AL-4V Titanium Alloy

(2) (3) (4) (5)

100 200 300 400 500 600

TEMP oF

a compression member Find the ultimate

strength of the member 1f made from the

following materials and subjected to the given

temperature and time conditions

Case 1 Material 7079-T6é Alum Alloy hand

forging and room temperature

Case 2 Same as Case 1, but subjected 1/2

hour to a temperature of 300°Fr

Case 3 Same as Case 2, but for S600°F,

Case 4 Material 17-4 PH stainless steel,

hand forging at room temperature

of the radius of gyration of the cross-section, the first step in the solution will be the cal- culation of Iy and ly, from whichox and py can

be found

Calculating Iy: In Fig C2.30 the section will

be first considered a solid rectangle 2.5 x 2.75 and then the properties of portions (1) and (2) will be subtracted

=x 2.5 «2.755 = Te 4.32

Iy (rectangle) Portions (1) and (2)

Column strength is considerably influenced

by the end restraint conditions For failure

by bending about the x-x axis, the end restraint against rotation is zero as the single fitting bolt has an axis parallel to the x-x axis and thus c the fixity coefficient is l For failure by bending about the y-y axis we have end restraint which will depend on the rigidity

of the bolt and the adjacent fitting and structure For this example problem, this restraint will be such as to make the end fixity coefficient c = 1.5

For failure about x-x axis,

L! = LAV = ZO// 71 = 30, Li/py = 30/.88

= 36 For failure about y~y axis,

L' = 30// 1.5 = 24.6, L'/Py 24,6/.60

=41 Therefore failure 1s critical for bending

about y-y axis, with L'/p = 41

Case 1 The material is 7079-T6 Alum Alloy

hand forging Fig C2.14 gives the failing

stress Fg for this material plotted against

the L'/o ratio Thus using L'/p = 41 and the

room temperature curve, “we read Fo = 50500 pst

Thus the failing load if P = Fed = 50500 x

4.375 = 220,000 lbs

Case 2 Using the 300°F curve in Fig C2.14

for the same L'/o value, we read Fe = 40,400,

and thus P = 40,400 x 4.375 = 177,000

Case 3 Using the 600°F curve, F, reads 6100

and thus P = 6100 x 4.375 = 26700 lbs Thus

subjecting this member to a temperature of

600°F for 1/2 hour reduces its strength from

220,000 to 26,700 lbs., which means that

Alum Alloy 1s a poor material for carrying

loads under such temperatures since the

reduction in strength is quite large

case 4 Material 17-4 PH stainless steel

forging Fig C2.8 gives the column curves

for this material For L'/o = 41 and using

the room temperature curve we read Fy =

135,200 and thus P = 135,200 x 4.375 =

591,000 lbs

C2.10 Solution Without Using Column Curves

When primary bending failure occurs at

stresses above the proportional limit stress,

the failing stress is given by equation (5)

which 18,

Fo = n*85/(L' /o)*

Since Zp is the tangent modulus of

elasticity, 1t varies with Fy, and thus the

relation of Ey to Fg must be known before the

equation can be solved To plot column curves

for all materials in their many manufactured

forms plus the various temperature conditions

ould require several hundred individual

column charts The use of such curves can be

avotded if we know several values or parameters

regarding the material as presented by

Ramsburg and Osgood and expanded by Cozzone

and Melcon (see Arts C2.4 and C2.5) for use

Thus we make use of the curves in Fig

c2.17

Case 1 Material 7079-T6 Alum Alloy forging

Table B1.1 of Chapter Bl summarizes certain material properties The properties needed

to use Fig C2.17 are the shape factor n, the moduls E, and the stress F,,, Referring to Table Bl.l, we find that n = 26, Eo =

B= 535,800,000 (42) = 1-02

Using Fig C2,17 with 1.01 on bottom scale and projecting vertically upward to

n = 26 curve and then horizontal to scale at

left side of chart we read Fo/F,., = 842

Then Fy = 59,500 x 842 = 50,100, as compared to 50,500 in the previous solution

using Pig C2.14

Case 2, From Table Bl.1 for this material Subjected to a temperature of 600°F for 1/2 hour, we find n = 29, Fy = 9,400,000 and

Fo,7 = 46,500,

1 46 ,500

Then B = 5 ¥ 37400,000 (41) = 917 From Fig C2.17 for B = 917 and n = 2g,

we read F,/F,,, = 88, thus FP, = 46,500 x 88

= 40,900 as compared to 40,400 in the previous solution

EXAMPLE PROBLEM 2

Pig C2.31 shows an extruded (1) section

A member composed of this section is 32 inches long The member is

braced laterally in the

x direction, thus x failure will occur by

bending about x-x axis

The member 1s pin ended tỷ

and thus c # 1 The material is 7075-T6

®xtrusion The problem

is to find the failing stress Fg under room temperature conditions

This (1) section corresponds to Section

15 in Table A3.15 in Chapter A3, Reference

Wd Oo { > `&

temperature we read F, = 38,500 psi

Solution by using Fig C2.17, From Table B1.1 for this material we find

n = 16.6, Eẹ = 10,500,000 and F„,„ = 72,000

_1 / 72,000 - Then, B= 5 V rea cog (81.7) = 1.36

From Fig C2.17 for B = 1.36 and n = 16.6, we

read Fo/F,_, = 537, hence Fy = 537 x 72,000

29 ,000

Bod f— 292000 n ¥ 7,800,000 (51.7) = 1.00

From Fig C2.17 we find ?e/F¿,„ = 74

Then Fy = 74 x 29,000 = 21,450 pat

A very common aluminum alloy in aircraft

construction 1s 2014-Té extrusions Let it

be required to determine the allowable stress

Fg for our member when made of this material

Since we have not presented column curves for this material, we will use Fig C2.17

From Table Bl.1, for our material, we

find n 2 18.5, Ey = 10,700,000 and F, , =

53,000

52,000

„1 Then B= 5 ¥ 10,700,000 (51.7) = 1.16

From Fig Cl.17 for B = 1.16 and n = 18.5,

we read Fo/F,,, = 71, hence Fy = 71 x

53,000 = 37,600

The result shows that the 2014-T6 material gave a failing stress of 37,600 as compared to

38,900 for the 7075-T6 material which has a

Fey of 70,000 as compared to Foy = 63,000 for

the 2014-T6 material The reason for the

7075 material not showing much higher column

failing stress Fo over that for the 2014 alloy

STRENGTH OF COLUMNS WITH STABLE CROSS-SECTIONS

is due to the fact that the stress existing under a L'/p value of 51.7 is near the pro~

portional limit stress or Ey is not much different than E,, the elastic modulus

To illustrate a situation where the 7075 material becomes more efficient in comparison

to the 2014 alloy, let us assume that our member has a rigid connection at its end which will develop an end restraint equivalent to a fixity coefficient ¢ = 2,

From Fig C2.17 for B = 823 and n =

18.5, we read Fo/F,., = 87, when Fy = 87 x

53,000 = 46,100 as compared to 58,300 for the

7075 material, thus 7075 material would permit lighter weight of required structural material The student should realize that if the stress range is such as to make Ey = E,, then the bending failure 1s elastic instead of inelastic and equation (5), using Young’s modulus of elasticity Eg: can be solved directly without resort to column curves or

a consideration of E,, since E, is equal to

Eo:

The student should realize that equation (8) is for strength under orimary column failure due to bending as a whole and not due

to local buckling or crippling of the member

or by twisting failure The subject of column design when local failure is involved

is covered in a later chapter

In example problem 2, we have assumed that local crippling is not critical, which calculation will show is true as explained and covered in a later chapter

C2.11 Strength of Stepped Column

The use of curves in Pig C2.22 will be tllustrated by the solution for the strength

of two stepped columns in order to tllustrate both elastic and inelastic failure of such columns

Case 1, Elastic failure

Fig C2.32 shows a double stepped pin ended column The member is machined from a

1 inch diameter extruded rod made from

7075-T6 materfal The problem is to find the

maximum compressive load this member will

Fig C2 32 PORTION 1 PORTION 2

the Euler equation for failure under elastic

bending If the ratio a/L equals 1 or 2

uniform section, B becomes n* or 10 as shown

in Fig C2.22 The curves in Pig C2.22

apply only to elastic failure Since the

member in Fig C2.32 is rather slender we

will assume the fatlure is elastic and then

check this assumption

BI, 10,500,000 x 0491

51a * 10,500,000 x 0165 = 5:17, a/b 30/60

0.5 From Fig C2.22 for a/L = 0.5 and S1,/EIe

proportional limit stress of the matertal so

Eq is constant and our solution is correct

Case 2 Inelastic Failure

The column has been shortened to the dimensions as shown in Fig C2.33 The

diameters and material remain the same as in

p for portion 1 is 0.25 inches

Ð for portion 2 is 0.1875 Average p = (6 X 25 + 6 x 0.1875)/12

= 0.22 Then L/p = 12/0.22 = 54.5, use 55

Fig C2.11 is a column curve for 7075-T6 Alum Alloy extruded material With L/P = S5, we read allowable stress Fo = 33,500 psi

Therefore P= A = 33,500X0.7854 = 26,300 1b

f, = 33,500 and fa = 26,300/.4418

59,500, The stress f in portion 2 is above the proportional limit stress so a plasticity correction must be made in using the curves

in Pig (2.22

Referring ta Table Bl.1 in Chapter Bl,

we find the following values for 7075-T6 extrustons:- n= 16.6, Fo, = 72,000,

Referring to Pig C2.16 and using 0.465

and n = 16.6, we read E/E = 1.0, thus By =

EB and thus no plasticity correction for Portion 1

For Portion 2, f2/Fo.7 = 59,500/72,000

= 826 From Fig C2.16, we obtain Z¢/E = 675 whence, St = 675 x 10,500,000 = 7,090,000

C2, 16

Our guessed strength was 25,300 lb Our guessed strength and calculated strength must

be the same so we must try again

Trial 2 Assume a critical load P =

23500 1b

f, = 23500/.7854 = 29500

f2 = 23500/,4418 = 53100

Portion 1 f /Fo., * 29900/72000 = 415 From Fig €2.36 for n= 16.6, we read E,/E 21.0

Portion 2 f2/Fo., = 53100/72000 = 738 From Fig C2.16, Et/E = 90, whence

and if further accuracy is desired another

trial should be carried through

The other types of columns with non- uniform cross-sections as shown in Figs

C2.21, C2.23 and C2.24 are solved in a similar

manner These charts are to be used only with

pin ended columns The end fixity coefficient

ce for tapered columns is not the same as for

uniform section columns

C2.12 Column Strength With Known End Restraining

Moment

-Fig C2.27 shows curves for finding the end fixity coefficient ¢ for two conditions

of known end bending restraint

To tilustrate the use of these curves,

a simple preblem will be solved

Fig C2.34 shows a 3-bay welded steel tubular truss The problem is to determine

the allowable compressive stress for member

AB This strength is {tnfluenced by the fixity

existing at ends A and B The diameter and

wall thickness of each tube in the truss is

Shown on the figure The material is AISI

Steel, Fry = 90,000, Fey 2 70,000, B=

conservative, we will assume the far ends of

members coming into joints {A} and (B} as pinned, Thus yw = SEI/L The sum of p = SEI/L will be computed for the 3 members which form the support of member AB at end

= 3(,001289 + 00071 + 000962) 29,000,000

u = 258,000

In Fig C2.27 we need term L/Z1 The L/EI refers to member 4B Thus u L/EI = (258,000 x 30)/29,000,000 x ,0S67 = 7.28

We use the upper curve in Fig (2.27

since restraint at both ends of member AB

is the same Thus for u L/EI = 7.28, we read end fixity coefficient c = 2.58

allowable failing stress to be F,

psi = 55,200

If the far ends of the connecting members were assumed fixed instead of pinned, then

w = 4EI/L, or we can multiply previous value

of 7.28 by 4/3, which gives 9.7 which, used in

Fig C2.27, gives c = 2.80

L'/p = 50/V 2.B x 422 = 42.5 Then from Fig 02.3, Fg = 56,600 psi Since the far

ends are less than fixed, the assumption that

far ends are pinned gives fairly accurate

results

In a truss structure all members are carrying axial loads and axial loads effect

the ability of members to resist rotation of

their ends Art All.12 of Chapter All

explains how to take account of the effect

of axtal load upon the stiffness of a member

as required in calculating the end restraint

coefficient p

C2.13 Columns With Elastic Lateral Supports

Pigs C2.25 and C2.26 provide curves for finding the end fixity coefficient ¢ to take

care of elastic lateral supports at points

midway between the column ends

To illustrate the use of these charts, a round bar 0.5 inches in diameter and 24 inches

long is braced laterally as shown in Fig

c2.35 The bar is made of

AISI Steel, heat treated to

Fey = 125,000 The spring

constant for the lateral

and error approach must be used as illustrated

in the problem dealing with a tapered column

C2.14 Problems

(1)

(a) (>)

§061-T6 Aluminum Alloy sheet, heat-treated and aged has the following properties:

Under room temperature:- Po = 35,000 psi, E, = 10,100,000 pst, and n=31

For 1/2 exposure at ZO0°P:- Fy.7 =

29,000, Ey = 9,500,000 and n = 26,

For the above two cases (a) and (b),

determine Et (tangent modulus values) from Fig

C2.16 and then calculate and plot column curves for these 2 material conditions

(2) Fig

the

of a compression member

the pressive load under the

for bending about x-x

1,5 about axis y-y

Same as Problem (2) but member {s sub- jected to a temperature of 50°F for hour

extruded channel sections identical

to Section No 50 in Table A3.1il in Chapter A3, are riveted back to back used aS a column member If member

is 26 inches long and end fixity is

1 and material is 7075-Té extrusion, what is the failing compressive load

If member is fastened rigidly to adjacent structure which provides a fixity c = 2, what will be the failing load

Consider the column in problem (4) 13 made from 2014-T6 Aluminum Alloy extrusion Find falling load

axis and (3)

1/2 {4) Two

1" Sq Bar 3/4" Square Bar

a=81” ——e— “4

L = 30"

k—

Fig C2.37

C2, 18

(7) Same as Problem (5) but member is exposed

1/2 hour to a temperature of S009F,

(8) Same as Problem (5) but change dimension

(a) to 10 inches, and L to 14.28 inches

(9) Find the failing compressive load for

the doubly stepped column in Fig (2.28

if member is made from 7079-T6 hand

forging

1+3/8 Dia Rod { 1-1/8 Dia Rod h=12 —>+—a=12 —*©— b=12 ¬1

STRENGTH OF COLUMNS WITH STABLE CROSS-SECTIONS

(11) The cylindrical tapered member in Fig

02.39 is used as a compression member

If member is made from AISI Steel 4140, Foy = 125,000, what is the fatling load

1/2" Dia Rod

a +

(1) NACA Technical Note 902

(2) Non-dimensional Buckling Curves, by Cozzone Z Melcon, Jr of Aero Sciences, October, 1946

{3) Chart from Lockheed Aircraft Structures Manual

quite common in flight vehicle structures

The limit design loads on a structural member

must be carried without permanent distortion

and the ultimate design loads must be carried

without rupture or failure The well known

bending stress equation f, = Mc/I, assumes a

linear variation of stress with strain or, in

other words, the equation holds for stresses

below the proportional limit stress or, in

general, the elastic range Failure of a

member in bending, unless there ts local

weakness, does not occur at stresses in the

elastic range but occurs at stresses in the

inelastic range Since the ultimate strength

of a member ts nseded to compare against the

ultimate design load to be carried, a theory

or procedure is necessary which will accurately

determine the ultimate and yield strength of a

jected to stresses which fall in the inelastic

range of stresses This stress can be taken

as the ultimate tensile or compressive stress

of the material or limited to some stress or

deformation in the inelastic range To obtain

the true internal resisting moment, we must

Know how the normal tension and compressive

stress varies over the cross-section The

stress-strain curve for the material provides

the source for obtaining the true stress

picture If 4 material has a different shape

in the tensile and compressive inelastic

zones, the neutral axes does not coincide with

the centroidal axis, thus adding some difficult

to an analysis msthod The analysis procedure

for determining the true internal resisting

moment is dsst explained by an example

solution

C3.3 Bending Strength of a Solid Round Bar

Fig C3.la shows the cross-section of a

round solid bar made of aluminum alloy The

stress-strain curve up to a unit strain of

010 in per inch is given in Fig C3.2 Note

that the shape of the curve in the inelastic

compression In this example solution, we will find the internal resisting moment when we limit the unit strain at the extreme edge on the compressive side of the beam section to 9.010, Now plane sections remain plans after bending in both elastic and inslastic stress conditions when member 1s in pure bending

We will guess the neutral axis as located 0.0375 inches above the centroidal axis as shown in Fig C3.1b Having assumed the maximum unit strain as 010, we can draw the strain diagram of Fig C3.1b We now divide the cross-section in Fig C3.la tnto 20 narrow horizontal strips Having the strain curve in Fig C3.1b, we can find the unit