Bruhn - Analysis And Design Of Flight Vehicles Structures Episode 3 part 6 ppt

As the load is increased the sheet buckles between the stiffeners and does not carry a greater stress than the buckling stress, However as the stiffeners are approached, the skin being

ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES CT,7

C7.10 Ilustrative Problems ¡n Calculating Crippling

For this method, we use Fig

C7.3 The parameter for bottom scale of Fig

C7.3 is (a + b)/2t, where (a) and (b) are leg

lengths measured to centerline of adjacent leg

of angle For our case 4 = b = 1 ~ 025 =

0.978 Thus (a + b)2t = 1.95/0.1 = 19.5,

From Fig C7.3 using 19.5 on lower scale,

and the curve for two edges free, we read on

the left hand scale that Foc/VFoy8 = 0330

Since we have only one angle, the crippling

stress Feg = Foo = 0330 x y40,000 x 10,700,000

= 21600 psi

Solution by Method 3 (Gerard Method}

For angle sections we use equation C7.4

A plot of this equation is given in Fig C7.7

The parameter for lower scale is,

Po wan, where A = section area and g

equals the number of flanges plus cuts, or

g = 2 for an angle section Substituting,

Sy 08*) Tes700,000 > «2-188

Using this value in Fig C7.7, we read

Feg/Foey = 0.50 Therefore Fog = 0.50 x 40,000

= 30 , B60 psi

Problem 2 Same as Problem 1, but change

material to aluminum alloy 7075-T6

Problem 4

Find the crippling stress for the channel

section shown in Fig b if the material is

aluminum alloy 2024-T3, Fey = 40,000,

the channel is composed 16 || -75

of 2 equal angle units Area =.137 +

(1) and (2) Since they Fig.b —=ll*05

are the same size, we — b—n

need only calculate the failing stress for one angle

Feyvx/s _„ ,;157.,_ 40,000

Barre =( s327v, 420,000 (x73 „ =

“gã” ) 557706, 000" 8.50

A + From Fig C7.9 Fog/Fey = 65, whence, Fog 2 65 x 40,000 = 36,005 pst

Problem 5 Same as Problem 4 but

to aluminum alloy 7075-T6 Foy =

Fe = 10,500,000

change material 87,000,

Solution by Method 1

= 045 x V67,000 x 10,500,000 = 38200 pst Solution by Method 2

Fes

1137, 67,000 s/s „ Cas") To, 500, 000! 10.17

of 4 equal angles with no

bi/t = (a+b)/2t = 1.95/0.1 = 19.5 From Fig C7.3, using upper curve, we obtain

Pee/VfcyE = 0392 whence,

Fog = Fog = 0392 x V40,000 x 10,700,000 =

25,600 psi Solution by Method 2

Area A = 373 g = number of cuts plus flanges or 4+ 8 = 12,

For rectangular tubes we use equation C?7.4 or Fig C7.7

material to magnesium HK31A-0 Sheet, subjected

to a temperature of 300°F for 1/2 hour

CRIPPLING STRENGTH OF COMPOSITE SHAPES AND SHEET-STIFFENER PANELS IN COMPRESSION

-8 x 11,100 = 8900 psi, Thus unless tests substantiate higher values, the crippling stress

Should be taken as 8900 psi

Problem 8, Same as Problem 6, but change material to stainless steel 17~7PH(TH1O50), Fey = 162,000, E_ = 29,000,000

Method 1

Fog = 0892V162,000x 29,000,000 = 85,000 pst Method 2,

373 162,000 1/2 _ Sex 087) SS 00,000) = 83 From Fig 07.7, Fog/Fey = 1595 Pog = 162,000 x 595 = 96,200 psi

Ee = 10,700,000

Solution by Method 1 (Needham)

The section is divided into 6 angle units

by the dashed lines in Fig e They are

numbered (1) to (3) since we have symmetry

The procedure will be to find the failing load for each angle and add up the total ror the

6 angle units The crippling stress will then equal this total load divided by the

section area

Angle unit (1) (One edge free)

0'/E = (a+b)/Et = [(.375 - 02) + (.5 - 02)] /.08

= 10.44 > From Fig C7.3, Fee/Vfcy = 06

ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES Area of angle (1) = o3O9 =A

From Table C7.1, 1t {3 recommended for

multi-corner sections that Fy, maximum be

limited to 8 Fey unless tests can prove higher

values

Fes 7 „8 x 40,000 = 32,000 psi Since

this is less than the above calculated values,

it should be used

Problem 10 Find the

extruded bulb angle shown T >

2014-TS extruded aluminum (a) I bự

This particular bulb BI ng

section is taken from

Table A3.16 of Chapter A3 Fig f

c?.9

as Section No 2

Q.113 sq in The area from that table is

Solution by Method 2 (Gerard) For this material Foy = 53,000, Ey = 10,700,000

The first question that arises is the bulb size sufficient to give an end stiffness to the (a) leg so that the bulb may be equivalent to the normal corner

In Fig f, bp = 0.78, hence be/; = 15 Referring to Fig C7.12, we observe that

for a br/t value of 15 we need a D/t ratio of

at least 3.8 The D/t value for our bulb

angle is (7/32)/.05 = 4.4, thus bulb has

sufficient stiffness to develop a corner

next question that arises should the bulb angle still be classed as an angle section for which equation C7.4 applies or be classed

as a channel or 2 corner section with the bulb acting as a short thick leg of the channel

For this case, equation C7.6 would apply

The

The crippling stress will be calculated

by both equations

By equation C7.4 or Fig C7.7:-

If bulb is considered as a full corner

then g = 4 flanges plus 1 cut = 5

In Table C7.1, the so-called cut-off stress for angles is 7 Fey and channels 9 Foy If we use the average value or 8 Foy:

it gives Fog = 8 x 53,000 = 42,400 as maximum permissible because of limited test

results on bulb angles

For the case where the bulb or lip does N not develop the stiffness necessary to

assume a full corner, then the bulb is only

CT 10

considered as an additional flange and the ¢

count would be four instead of 5, thus reducing

the crippling stress

EFFECTIVE SHEET WIDTHS

CT 11 Introduction

The previous discussion in this chapter has dealt with the crippling stress of formed

or extruded sections when acting alone, that is

not fastened to any other structure along its

length However, the major structure of

aerospace vehicles, such as the wing or body,

involves a sheet covering which is strengthened

by attached or integral fabricated stiffeners

such as angles, zees, tees, etc Since the

sheet and stiffeners must deform together, the

sheet will therefore carry compressive load and

to neglect this load carrying capacity of the

sheet would be too conservative in aerospace

structural design where weight saving is very

important

C7.12 Sheet Effective Widths

Fig C7.13a illustrates a continuous flat

thin plate fastened to stiffener and the entire

unit is subjected to a uniform compressive load

Up to the buckling strength of the sheet the

compressive stress distribution 1s uniform over

both stiffeners and sheet as in (Fig b) assum-

ing same material for sheet and stiffeners As

the load is increased the sheet buckles between

the stiffeners and does not carry a greater

stress than the buckling stress, However as

the stiffeners are approached, the skin being

stabilized by the stiffeners to which it is

attached can take 2 higher stress and

immediately over the stiffeners the sheet can

take the same stress as the ultimate strength

of the stiffener, assuming that the sheet has

a@ continuous connection to the stiffener Fig

¢ Shows the general stress distribution after

the sheet has buckled This distribution

depends om the degree of restraint provided by

the stiffeners and the panel dimension

Various theoretical studies (Ref 6) have

been made to determine this stress distribution

after buckling In general they lead to long

and complicated equations To provide a simple

basis for design purposes, an attempt has been

made to find an effective width of sheet w

which would be considered as taking a uniform

stress (Fig d) which would give the same total

sheet strength as the sheet under the true

non-uniform stress distribution of Fig c

The question of sheet effective sheet has been considered by many individuals The

names of VonKarman, Sechler, Timoshenko,

Newell, Frankland, Margurre, Fischel, Gerard,

and many more are closely associated with the

present knowledge on effective sheet width

CRIPPLING STRENGTH OF COMPOSITE SHAPES AND SHEET-STIFFENER PANELS IN COMPRESSION

Fig 0 Sheet stress distribution before buckling

CEPT EPP Pp yp yyy

Sheet stress distribution after buckling

or “2 (1 -¥%*) ‘d

If we assume that the stiffener to which

the sheet is attached provides a boundary

restraint equal to a simple support, then k,

= 4.0, and if Poisson’s ratio V, 1s taken as

0.3, then equation C7.13 reduces to, For = 3.60E(t/b)*

The Von-Karman-Sechler method as first proposed consisted of solving equation 07.14

for a width (w) in place of (b), when Foy was

equal to the yield stress of the material since experiments had shown that the ultimate strength of a sheet simply supported at the edges was independent of the width of the sheet

Thus equation C7.14 changes to,

Foy = 3.60E(t/w)* , whence,

ws l.90t V8/Fey > - (07.15)

Since the crippling or local failing

stress of a stiffener can exceed the yield

} strength of the material, equation C7.15 was later changed by replacing Fey by the stress

in the stringer Ffop, thus giving,

ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES

Some early experiments by Newell indicated

the constant 1.90 was too high and for light

stringers a value of 1.7 was more realistic,

thus 1.7 has been widely used in industry

If we assume the stiffness of the stiffener

and its attachment to the sheet as developing

a fixed or clamped edge condition for the

Sheet, then

For = 6.35E(t/b)* or w= 2.52t VE/Fgp

For general design purposes, it is felt

that 1.9 or equation C7.15 is appropriate for

determining the effective width w If

stiffener is relatively light, use 1.7

C7.14 illustrates the effective width for

sheet-stiffener units which are fastened

together by a single attachment line for each

flange of the stiffener

Fig

2 = Rivet Lines

The crippling stress is determined for

the stiffener alone This stress is then used

in equation C7.15 to determine the effective

widths w The total area then equals the

stiffener area plus the area of the effective

sheet width w The radius of gyration should

include the effect of the effective skin area

Fig C7.15 illustrates the case where

stiffeners are fastened to sheet by two rows

of rivets on each stiffener flange In this

ease, the rivet lines are so close together

that the effective width w for each rivet line

would overlap considerably A common practice

in industry for such cases its to use the

effective width for one rivet line attachment

as per equation C7.16 to represent sheet width

to go with each stiffener flange However,

in calculating the crippling stress of the

stiffener alone, the stiffener flange which 1s

attached to sheet is considered as having 2

thickness equal to 3/4 the sum of the flange

thickness plus the sheet thickness

crippling stress The subject of inter-rivet

sheet duckling is discussed later in this

chapter

Pig C7.16 1llustrates a procedure to follow for determining the effective width w when sheet and stiffener are integral in

manufacture

tạ 3 tị « 2ts te 2 tg

Tee Section

tấn kế

Fig C7.18 For Case 1, find the crippling stress for the tee section alone, assuming the vertical stem of the tee has both ends simply supported

For value of t in equation C7.15, use

(tg + tr)/2 The effective stiffener area -

equals the area of the tee plus the area of the Sheet of width w

For Case 2, determine the crippling stress

for the I section acting alone Calculate w/2

from equation C7.15 to include as effective Sheet area The column properties should

include I section plus effective sheet

CT 12 Effective Width W, for Sheet with One Edge Free

In normal sheet-stiffener construction, the sheet usually ends on a stiffener and thus

we have a free edge condition for the sheet as

illustrated in Fig C7.16a The sheet ends at

Fig C7 16a

fed!

0

a distance b’ from the rivet line For a sheet

free on one edge, the buckling coefficient in equation C7.13 ts 0.43, thus equation C7.13

reduces to,

For obtain, er © 387E(t/b')7, and replacing b' by w , we

W, 2 62t vE/Fsn

C7.12

Then the total effective sheet width for

this end stiffener would thus equal w, + w/2

Cĩ.13 Effective Width When Sheet and Stiffener Have

Different Material Properties

In practical sheet-stiffener construction

it is common to use extruded stiffeners which

have different material strength properties in

the inelastic stress range as compared to the

Sheet to which the stiffener is attached For

example, in Pig C7.17 the stiffener material

could have the stress-strain curve represented

by curve (1) and the sheet to which it is

attached by the curve (2) Now when the

stiffener is stressed to point (B), the sheet

directly adjacent to the stiffener attachment

line must undergo the same strain as the

stiffener and thus the stress in the sheet

will be that given at point (A) in Fig C7.17

This difference in stress will influence the

effective width w Correction for this

condition can be made in equation C7.17 by

multiplying it by Fsy/Foep, which gives

w= 1.900(Fsw/Fer) (ge)

Where fst is the stiffener stress and fgy is the

sheet stress existing at the same strain as

existing for the stiffener and obtained from a

stress-strain curve of the sheet material

of the various effective widths theories as

compared, see article by Gerard (Ref 7)

Equation C7,.15 is in general conservative for

higher b/t ratios,

Cï.14 Inter-Rivet Buckling Stress

The effective sheet area is considered to act monolithically with the stiffener How-

ever, if the rivets or spot welds that fasten

the shest to the stiffener are spaced too far

apart, the sheet will buckle between the rivets

CRIPPLING STRENGTH OF COMPOSITE SHAPES AND SHEET-STIFFENER PANELS IN COMPRESSION

before the crippling stress of the stiffener

is reached, which means the sheet is less

subjected compressive load Thus,

to save structural weight, str ra Select rivet snacings that will »revent inte

rivet buckling of the shset In zeneral, the

Tivet spacing along the stiffeners in the upper surface of the wing will be closer to-

gether than on the bottom surface of ‘he wing since the design comsressive loads on the

top surface are considerably larger than those

on the bottom surface

The following method is widely used oy

engineers concerned with aerospace structures

relative to calculating inter-rivet buckling

stresses It is assumed that the sheet

between adjacent rivets acts as a column with

to 4 for a fixed end support

Tae effective column length L' = L/vC,

thus equation C7.19 can be written

Let p the rivet spacing be considered the column length L Assume a umit of sheet 1 inch wide and t its thickness Then moment of

inertia of cross section = 1 x t°/l2, and area

A=1xt=t Then radius of gyration 9 = 0.29t Then substituting in equation 27.20 to obtain the inter-rivet buckling stress Fir,

a Pip? eee TT - TT TT ~ (c7.21)

To vlot this equation, the tangent modulus

#y for the material must be Known, However, we

can use the various column curves fn Chapter ce

which show a plot of F, versus L'/p and in

equation C7.22 the term p/0.58t corresponds to

L'/o

The fixity coeffictent C = 4 can be used for flat head rivets For spot welds it should

be decreased to 3.5 For the Bragier rivet type

use C = 3 and for counter-sunk or dimpled rivets

use C=l1 au

ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES Figs C7.18 and C7,.19 show a plot of

equation C7.22 for aluminum alloy materials

If the inter-rivet buckling stress cal-

culates to be more than the crippling stress

of the stiffener, then the effective sheet area

can be added to the stiffener area to obtain the

total effective area This total effective

area times the stiffener crippling stress will

give the crippling load for the total sheet-

stiffener unit

When the sheet between rivets buckles

before the crippling stress of the stiffener is

reached, the sheet in the buckled state has the

ability to approximately nold this stress as

the stiffener continues to take load until it

reaches the stiffener crippling stress This

buckling sheet strength can be taken advantage

or by reducing the effective sheet area Thus

effective sheet width equals,

Noorrected * W(Fir/Fes) - - - - > = - (07.25)

The area of the corrected effective sheet

is then added to the area of the stiffener

The crippling load then equals the crippling

stress of the stiffener times the total area

The use of sheet effective widths in

finding the moment of inertia of a wing or

fuselage cross-section is a widely used pro-

cedure in the analysis for bending stresses in

conventional wing and fuselage construction

Reference should be made to article A19.13 of

Chapter Al9 and article AZ0.3 of Chapter A20

for practical illustrations in the use of

effective widths

C7.15 Olustrative Problem Involving

Effective Sheet, Conventional airplane wing construction

is illustrated in Fig C?7.20 The wing is

covered with sheet, generally referred to as

skin, and this skin is stiffened by attaching

formed or extruded shapes referred to as skin

stiffeners or skin stringers A typical wing

section involves one or several interior

straight webs and to tie these webs to the

skin, a stringer often referred to as the web

flangs member, is required to facilitate this

connection Fig C7.2l is a detail of the

flange member and the connection at point (1)

in Fig C7.20

The stiffener or flange member is an

extrusion of 7075-T6 aluminum alloy The skin

and web sheets are 7075-T6 aluminum alloy The

skin is fastened to stiffener by two rows of

1/8 inch diameter rivets of the Brazier head

type, spaced 7/8 inch apart The web is

attached to stiffener by one row of 3/16

diameter rivets spaced 1 inch apart The

problem is to determine the crippling stress of

the stiffener, the effective skin area and the total compressive load that the unit can carry

at the failure point Since the stiffener 1s

braced laterally by the web and the skin, column

bending action is prevented and thus the

crippling strength is the true resulting strength

of this corner member under longitudinal compression (Additional stresses are produced

on these corner members if web buckles under

shear stresses and web diagonal tension forces are acting This subject is treated in the chapter on semi-tension field beams.)

s +/a „Q.24 70,000 sa3/2 _

se Š ) ŠxO.72” (10,500,068) 0.83

Using this value, we read from Fig C7.7 that Fes/Foy = 0.82 Thus Fog = 82 x 70,000 = 57,400 psi The g¢ value of 6 was determined as

shown in Fig (a)

Effective Sheet Widths: fof

Equation C?7.16 will be

used to determine the sheet

f

effective widths Flanges (f) = 5

For the skin t = 05 No of cuts 5

Materfal {s 7075-Té6é aluminum

alloy Ec = 10,500,000, Fig a

= 1.9t /Fog 2 1.9 % 0.5 ¥10,500,000/57 ,400

= 1.28 in

Thus a piece of sheet 1.28/2 = 64 wide acts to

each Side of the rivet centerline Observation

2024-T36 2024-T6 2024-T81 2024-88 ~*, 250 T0T5-T6 016-.039 1075-T6 040-,240 DOr

2024-T36 2024-T6 2024-T81 2024-T86 ~< 064 1075-T6 .016-.039 1015-T6 = 040- 249 Wer

ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES

of the dimensions in Fig C7.21 shows the

effective width with each skin rivet line would

overlap slightly thus we use only the width

between rivet line (see Fig b)

For the web t = 064 Since the web has

a free edge, the effective width calculation

will be in two steps

compressive load that entire unit ce carry

before failure is then equal to AF,., = 435 x

57,400 = 25,000 lbs

This result assumes that no later-rivet

buckling occurs under the stress of 57,400 pst

in the sheet between rivets

The skin rivets are Brazier head type

spaced 7/8 inch apart or p= 7/8

As discussed under inter-rivet buckling,

the end coefficient c for this type of rivet

should be less than 4 or assumed as 3

Fig C7.18 gives the inter-rivet buckling

stress versus the p/t ratio This chart is

based on a clamped end condition or C = 4,

Since C = 3 will be used for the Brazier type

rivet, we correct the p/t ratio by the ratio

v4 / VS = 1.16

The corrected p/t = 1.16 x 875/.05 2 20.3

From Fig C7.18 using curve (8) which Is

our material, we read Fir = 50,000 psi, thus

skin will not buckle between rivets as

crippling stress is 57,400 psi

The web rivets are of the flat head type

and C = 4 can be used Spacing is 1 inch

Hence p/t = 1/,064 = 15.6 and from Fig C7.18,

curve 8, we read Fyp = 54,600, which is

considerably more than the unit crippling

stress

In wing construction, the skin rivets are

C?, 15 usually of the flush surface type, either countersunk or dimpled If we make the skin rivets of the countersunk type, the end fixity coeffictant must be reduced to 1 to be safe

Then the corrected p/t ratio to use with Fig C7,.18 would be (V4 / VI )p/t = 2 x 875/.05

= 35 From Fig C7.18 and curve 8, Fir = 29,000 which is far below the calculated crippling stress, thus the rivet spacing would have to be reduced Use 9/16 inch spacing corrected p/t = 2x 5625/.05 = 22 From Fig C7.18, Fir * 57,400 psi, which happens to be the crippling stress and therefore satisfactory

C?.16 Failing Strength of Short Sheet-Stiffener Panels

in Compression

Gerard (Refs 2, 3) from a comprehensive

study of test results on short sheet-stiffener panels in compression, has shown that his equation C7.4, or Fig C7.7, can be used to give the local monolithic crippling stress for

Sheet panels stiffened by Z, Y and H at shaped

stiffeners The method of calculating the value of the g factor is illustrated in Fig

C7.22, Fig C7.23 1s a photograph showing the erippling type of failure for a short panel involving the Z shape stiffener

C7.17 Failure by Inter-Rivet Buckling

Howland (Ref 8) assumed that the sheet

acts as a wide column which ts clamped at its ends and whose length 1s equal to the rivet spacing The inter-rivet buckling stress

= enn te

The end fixity coefficient C is taken as

4 for flat head rivets and reduced for other types as previously explained for equation

c?.21

Ris the plasticity correction factor

i is the clad correction factor

Vg is Poisson’s ratio (use 0.30)

ts = sheat thickness, inches

p = rivet spacing or pitch in inches

For non-clad materials the curves of Fig

c?’.24 can be used This figure is the same as Fig CS.8 of Chapter CS For the clad correction see Table C5.1 of Chapter C5

C7.18 Failure of Short Panels by Sheet Wrinkling

In a riveted sheet-stiffener panel, if the rivet spacing is relatively large, the sheet will buckle between rivets, such as illustrated in the Photograph of Fig C7.25 This inter-rivet buckling stress was discussed in the previous

ANALYSIS AND DESIGN OF FLIGHT VEHICLE STRUCTURES article This sheet buckling does not deform

the flange of the stiffener to which the sheet

is attached However, if the rivet or spot

weld spacing 1s such as to prevent inter-rivet

buckling of the sheet, then failure often occurs

by a larger wrinkling of the sheet as

tllustrated in Fiz C7.26 The larger wrinkle

Snape subjects to flange of the stiffener to

which the sheet is attached to lateral forces

and thus the stiffener flange often deforms

With the sheet wrinkle shape This deforming

of the stiffener flange produces stresses on

the stiffener wed, thus wrinkling failure is a

combination of sheet and stiffener failure

The action of the wrinkling sheet to deform

the stiffener flange places tension loads on

the rivets, thus rivet design enters into the

failing strength of sheet-stiffener panels

Several persons have studied this wrinkling

or forced crippling of riveted panels (Refs 9

and 10) A rather recent study was carried out

by Semonian and Peterson (Ref 11), which is

reviewed and simplified somewhat by Gerard in

(Ref 2) The results as given in (Refs 11

and 2} used to calculate the wrinkling stress

C1.19 Equation for Wrinkling Failing Stress Fy

From Ref 2 we obtain,

_ ky TT l8

W_ 12 (1-1)

Py Gee sect tee “sya

ky, the wrinkling coefficient is obtained

from Pig C7.27 This coefficient is a function

of the effective rivet offset 7 which is

obtained from Fig C7.23 Having determined

Ky from Fig C7.27, equation C7.25 can be solved

by use of Fig C7.24,

CT.20 Rivet Criterion for Wrinkling Failure

A criterion for the rivet pitch found from

test data which results in a wrinkling mode

failure 15,

8/Đg ~ 127/4 V122 =~ -

The lateral force required to make the

stringer attachment flange conform to the

wrinkled shset, loads the rivet in tension An

7,17 approximate criterion for rivet strength from Re?, 2 18,

O7 Dgp ,„ „* >a iw

re Sse 4

The tensile strengta of the rivet Sy is

defined in terms of the shank area and it may

be associated with either shank failure or

pulling of the countersunk head of the rivet

through the sheet

For aluminum alloy 2117-T4 rivets whose tensile strength is s = 57 ksi, the criteria

are:-

$= 57 ksi., dạ/ta„y = 1.87

(27.28) -_ 199 160

se“ Te/tay ˆ tay)! de/tgy > 1.67

where tay, is the average of sheat and stiffener

thickness in inches The effective diameter dạ

is the diameter for a rivet made from 2117-T4 material

The effective diameter of a rivet of

another material is,

dg/d = (S,/3}*/4 ~ - (c7.29)

where Sr is the tensile strength of a rivet defined as maximum tensile load divided by

shank area in ksi units

CT.21 Problem 1 Mlustrating Calculation of

Short Panel Failing Strength

Fig C7.29 snows a sheet-stiffener panel composed of formed Z stiffeners The material

ts aluminum alloy 2024-TS Fey = 40,000

F,,, 2 39,000, n = 14.5, Ey ® Ÿo,700,000 The problem is to determine the compressive failing

strength of a short length of this panel unit

Curves for finding Fir

Fig C?,27 Experimentally determined coefficients for failure in

wrinkling mode (Ref 2)